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https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Reactivity_of_Alkanes/Cracking_Alkanes

What is cracking? Cracking is the name given to breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high pressures and temperatures without a catalyst, or lower temperatures and pressures in the presence of a catalyst. The source of the large hydrocarbon molecules is often the naphtha fraction or the gas oil fraction from the fractional distillation of crude oil (petroleum). These fractions are obtained from the distillation process as liquids, but are re-vaporized before cracking. There isn't any single unique reaction happening in the cracker. The hydrocarbon molecules are broken up in a fairly random way to produce mixtures of smaller hydrocarbons, some of which have carbon-carbon double bonds. One possible reaction involving the hydrocarbon \(C_{15}H_{32}\) might be: \[ C_{15}H_{32} \rightarrow \underset{\text{ethene}}{2C_2H_4} + \underset{\text{propene}}{C_3H_6} + \underset{\text{octane}}{C_8H_{18}}\] Or, showing more clearly what happens to the various atoms and bonds: This is only one way in which this particular molecule might break up. The ethene and propene are important materials for making plastics or producing other organic chemicals. The octane is one of the molecules found in petrol (gasoline). Modern cracking uses zeolites as the catalyst. These are complex aluminosilicates, and are large lattices of aluminium, silicon and oxygen atoms carrying a negative charge. They are, of course, associated with positive ions such as sodium ions. You may have come across a zeolite if you know about ion exchange resins used in water softeners. The alkane is brought into contact with the catalyst at a temperature of about 500 °C and moderately low pressures. The zeolites used in catalytic cracking are chosen to give high percentages of hydrocarbons with between 5 and 10 carbon atoms - particularly useful for petrol (gasoline). It also produces high proportions of branched alkanes and aromatic hydrocarbons like benzene. The zeolite catalyst has sites which can remove a hydrogen from an alkane together with the two electrons which bound it to the carbon. That leaves the carbon atom with a positive charge. Ions like this are called carbonium ions (or carbocations). Reorganisation of these leads to the various products of the reaction. In thermal cracking, high temperatures (typically in the range of 450 °C to 750 °C) and pressures (up to about 70 atmospheres) are used to break the large hydrocarbons into smaller ones. Thermal cracking gives mixtures of products containing high proportions of hydrocarbons with double bonds - alkenes. This is a gross oversimplification; tn fact, there are several versions of thermal cracking designed to produce different mixtures of products. These use completely different sets of conditions. Thermal cracking does not go via ionic intermediates like catalytic cracking. Instead, carbon-carbon bonds are broken so that each carbon atom ends up with a single electron. In other words, free radicals are formed. Reactions of the free radicals lead to the various products. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/09%3A_Chemical_Bonding_and_Molecular_Structure/9.05%3A_Molecular_Geometry

Make sure you thoroughly understand the following essential ideas: The Lewis electron-dot structures you have learned to draw have no geometrical significance other than depicting the order in which the various atoms are connected to one another. Nevertheless, a slight extension of the simple shared-electron pair concept is capable of rationalizing and predicting the geometry of the bonds around a given atom in a wide variety of situations. The that we describe here focuses on the bonding and nonbonding electron pairs present in the outermost (“valence”) shell of an atom that connects with two or more other atoms. Like all electrons, these occupy regions of space which we can visualize as — regions of negative electric charge, also known as — whose precise character can be left to more detailed theories . The covalent model of chemical bonding assumes that the electron pairs responsible for bonding are concentrated into the region of apace between the bonded atoms. The fundamental idea of VSEPR thoery is that these regions of negative electric charge will repel each other, causing them (and thus the chemical bonds that they form) to stay as far apart as possible. Thus the two electron clouds contained in a simple triatomic molecule AX will extend out in opposite directions; an angular separation of 180° places the two bonding orbitals as far away from each other they can get. We therefore expect the two chemical bonds to extend in opposite directions, producing a linear molecule. If the central atom also contains one or more pairs of electrons, these additional regions of negative charge will behave very much like those associated with the bonded atoms. The orbitals containing the various bonding and nonbonding pairs in the valence shell will extend out from the central atom in directions that minimize their mutual repulsions. If the central atom possesses partially occupied -orbitals, it may be able to accommodate five or six electron pairs, forming what is sometimes called an “ ”. As we stated above, a simple triatomic molecule of the type \(AX_2\) has its two bonding orbitals 180° apart, producing a molecule that we describe as having geometry. Examples of triatomic molecules for which VSEPR theory predicts a linear shape are BeCl (which, you will notice, doesn't possess enough electrons to conform to the octet rule) and CO . If you write out the electron dot formula for carbon dioxide, you will see that the C-O bonds are double bonds. This makes no difference to VSEPR theory; the central carbon atom is still joined to two other atoms, and the electron clouds that connect the two oxygen atoms are 180° apart. In an AX molecule such as BF , there are three regions of electron density extending out from the central atom. The repulsion between these will be at a minimum when the angle between any two is (360° ÷ 3) = 120°. This requires that all four atoms be in the same plane; the resulting shape is called , or simply . Methane, CH , contains a carbon atom bonded to four hydrogens. What bond angle would lead to the greatest possible separation between the electron clouds associated with these bonds? In analogy with the preceding two cases, where the bond angles were 360°/2=180° and 360°/3=120°, you might guess 360°/4=90°; if so, you would be wrong. The latter calculation would be correct if all the atoms were constrained to be in the same plane (we will see cases where this happens later), but here there is no such restriction. Consequently, the four equivalent bonds will point in four in three dimensions corresponding to the four corners of a centered on the carbon atom. The angle between any two bonds will be 109.5°. This is called . This is the most important coordination geometry in Chemistry: it is imperative that you be able to sketch at least a crude perspective view of a tetrahedral molecule. It is interesting to note that the tetrahedral coordination of carbon in most of its organic compounds was worked out in the nineteenth century on purely geometrical grounds and chemical evidence, long before direct methods of determining molecular shapes were developed. For example, it was noted that there is only one dichloromethane, CH Cl . If the coordination around the carbon were square, then there would have to be two isomers of CH Cl , as shown in the pair of structures here. The distances between the two chlorine atoms would be different, giving rise to differences in physical properties would allow the two isomers to be distinguished and separated. The existence of only one kind of CH Cl molecule means that all four positions surrounding the carbon atom are geometrically equivalent, which requires a tetrahedral coordination geometry. If you study the tetrahedral figure closely, you may be able to convince yourself that it represents the connectivity shown on both of the "square" structures at the top. A three-dimensional ball-and-stick mechanical model would illustrate this very clearly. Similar chains having the general formula H C–(CH ) –CH (or C H ) can be built up; a view of , C H , is shown below. Notice that these "straight chain hydrocarbons" (as they are often known) have a carbon "backbone" structure that is not really straight, as is illustrated by the zig-zag figure that is frequently used to denote hydrocarbon structures. refers to the number of electron pairs that surround a given atom; we often refer to this atom as the even if this atom is not really located at the geometrical center of the molecule. If all of the electron pairs surrounding the central atom are shared with neighboring atoms, then the is the same as the . The application of VSEPR theory then reduces to the simple problem of naming (and visualizing) the geometric shapes associated with various numbers of points surrounding a central point (the central atom) at the greatest possible angles. Both classes of geometry are named after the shapes of the imaginary geometric figures (mostly regular solid polygons) that would be centered on the central atom and would have an electron pair at each vertex. If one or more of the electron pairs surrounding the central atom is not shared with a neighboring atom (that is, if it is a lone pair), then the molecular geometry is simpler than the coordination geometry, and it can be worked out by inspecting a sketch of the coordination geometry figure. In the examples we have discussed so far, the shape of the molecule is defined by the coordination geometry; thus the carbon in methane is tetrahedrally coordinated, and there is a hydrogen at each corner of the tetrahedron, so the molecular shape is also tetrahedral. It is common practice to represent bonding patterns by "generic" formulas such as \(AX_4\), \(AX_2E_2\), etc., in which "X" stands for bonding pairs and "E" denotes lone pairs. This convention is known as the "AXE Method." The bonding geometry will not be tetrahedral when the valence shell of the central atom contains nonbonding electrons, however. The reason is that the are also in orbitals that occupy space and repel the other orbitals. This means that in figuring the coordination number around the central atom, we must count both the bonded atoms and the nonbonding pairs. In the water molecule, the central atom is O, and the Lewis electron dot formula predicts that there will be two pairs of nonbonding electrons. The oxygen atom will therefore be tetrahedrally coordinated, meaning that it sits at the center of the tetrahedron as shown below. Two of the coordination positions are occupied by the shared electron-pairs that constitute the O–H bonds, and the other two by the non-bonding pairs. Thus although the oxygen atom is tetrahedrally coordinated, the bonding geometry (shape) of the H O molecule is described as There is an important difference between bonding and non-bonding electron orbitals. Because a nonbonding orbital has no atomic nucleus at its far end to draw the electron cloud toward it, the charge in such an orbital will be concentrated closer to the central atom. As a consequence, nonbonding orbitals exert more repulsion on other orbitals than do bonding orbitals. Thus in H O, the two nonbonding orbitals push the bonding orbitals closer together, making the H–O–H angle 104.5° instead of the tetrahedral angle of 109.5°. The electron-dot structure of NH places one pair of nonbonding electrons in the valence shell of the nitrogen atom. This means that there are three bonded atoms and one lone pair, for a coordination number of four around the nitrogen, the same as occurs in H O. We can therefore predict that the three hydrogen atom will lie at the corners of a tetrahedron centered on the nitrogen atom. The lone pair orbital will point toward the fourth corner of the tetrahedron, but since that position will be vacant, the NH molecule itself cannot be tetrahedral. Instead, it assumes a shape. More precisely, the shape is that of a (i.e., a pyramid having a triangular base). The hydrogen atoms are all in the same plane, with the nitrogen above (or below, or to the side; molecules of course don’t know anything about “above” or “below”!) The fatter orbital containing the non-bonding electrons pushes the bonding orbitals together slightly, making the H–N–H bond angles about 107°. Compounds of the type AX are formed by some of the elements in Group 15 of the periodic table; PCl and AsF are examples. In what directions can five electron pairs arrange themselves in space so as to minimize their mutual repulsions? In the cases of coordination numbers 2, 3, 4, and 6, we could imagine that the electron pairs distributed themselves as far apart as possible on the surface of a sphere; for the two higher numbers, the resulting shapes correspond to the regular polyhedron having the same number of sides. The problem with coordination number 5 is that there is no such thing as a regular polyhedron with five vertices. In 1758, the great mathematician Euler proved that there are only five regular convex polyhedra, known as the platonic solids: tetrahedron (4 triangular faces), octahedron (6 triangular faces), icosahedron (20 triangular faces), cube (6 square faces), and dodecahedron (12 pentagonal faces). Chemical examples of all are known; the first icosahedral molecule, \(LaC_{60}\) (in which the La atom has 20 nearest C neighbors) was prepared in 1986. Besides the five regular solids, there can be 15 semi-regular isogonal solids in which the faces have different shapes, but the vertex angles are all the same. These geometrical principles are quite important in modern structural chemistry. The shape of PCl and similar molecules is a . This consists simply of two triangular-base pyramids joined base-to-base. Three of the chlorine atoms are in the plane of the central phosphorus atom ( positions), while the other two atoms are above and below this plane ( positions). Equatorial and axial atoms have different geometrical relationships to their neighbors, and thus differ slightly in their chemical behavior. In 5-coordinated molecules containing lone pairs, these non-bonding orbitals (which you will recall are closer to the central atom and thus more likely to be repelled by other orbitals) will preferentially reside in the equatorial plane. This will place them at 90° angles with respect to no more than two axially-oriented bonding orbitals. Using this reasoning, we can predict that an AX E molecule (that is, a molecule in which the central atom A is coordinated to four other atoms “X” and to one nonbonding electron pair) such as SF will have a “see-saw” shape; substitution of more nonbonding pairs for bonded atoms reduces the triangular bipyramid coordination to even simpler molecular shapes, as shown below. Just as four electron pairs experience the minimum repulsion when they are directed toward the corners of a tetrahedron, six electron pairs will try to point toward the corners of an . An octahedron is not as complex a shape as its name might imply; it is simply two square-based pyramids joined base to base. You should be able to sketch this shape as well as that of the tetrahedron. The shaded plane shown in this octahedrally-coordinated molecule is only one of three equivalent planes defined by a four-fold symmetry axis. All the ligands are geometrically equivalent; there are no separate axial and equatorial positions in an AX molecule. At first, you might think that a coordination number of six is highly unusual; it certainly violates the octet rule, and there are only a few molecules (SF is one) where the central atom is hexavalent. It turns out, however, that this is one of the most commonly encountered coordination numbers in inorganic chemistry. There are two main reasons for this: There are well known examples of 6-coordinate central atoms with 1, 2, and 3 lone pairs. Thus all three of the molecules whose shapes are depicted below possess octahedral coordination around the central atom. Note also that the orientation of the shaded planes shown in the two rightmost images are arbitrary; since all six vertices of an octahedron are identical, the planes could just as well be drawn in any of the three possible vertical orientations. The VSEPR model is an extraordinarily powerful one, considering its great simplicity. Its application to predicting molecular structures can be summarized as follows: While VSEPR theory is quite good at predicting the general shapes of most molecules, it cannot yield exact details. For example, it does not explain why the bond angle in H O is 104.5°, but that in H S is about 90°. This is not surprising, considering that the emphasis is on electronic repulsions, without regard to the detailed nature of the orbitals containing the electrons, and thus of the bonds themselves. The Valence Shell Electron Repulsion theory was developed in the 1960s by Ronald Gillespie of McMaster University (Hamilton, Ontario, Canada) and Ronald Nyholm (University College, London). It is remarkable that what seems to be a logical extension of the 1916 Lewis shared-electron pair model of bonding took so long to be formulated; it was first presented in the authors' classic article published in the 1957 (Vol 11, pg 339). Although it post-dates the more complete quantum mechanical models, it is easy to grasp and within a decade had become a staple of every first-year college chemistry course.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/21%3A_Nuclear_Chemistry/21.1%3A_Nuclear_Structure_and_Stability

is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of \(\ce{^1_1H}\), neutrons. Recall that the number of protons in the nucleus is called the atomic number (\(Z\)) of the element, and the sum of the number of protons and the number of neutrons is the mass number (\(A\)). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term and identify it by the notation: \[\ce{^{A}_{Z}X} \label{Eq1} \] where Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, \(\ce{^{14}_6C}\) is called “carbon-14.” Protons and neutrons, collectively called , are packed together tightly in a nucleus. With a radius of about 10 meters, a nucleus is quite small compared to the radius of the entire atom, which is about 10 meters. Nuclei are extremely dense compared to bulk matter, averaging \(1.8 \times 10^{14}\) grams per cubic centimeter. For example, water has a density of 1 gram per cubic centimeter, and iridium, one of the densest elements known, has a density of 22.6 g/cm . If the earth’s density were equal to the average nuclear density, the earth’s radius would be only about 200 meters (earth’s actual radius is approximately \(6.4 \times 10^6\) meters, 30,000 times larger). Example \(\Page {1}\) demonstrates just how great nuclear densities can be in the natural world. Neutron stars form when the core of a very massive star undergoes gravitational collapse, causing the star’s outer layers to explode in a supernova. Composed almost completely of neutrons, they are the densest-known stars in the universe, with densities comparable to the average density of an atomic nucleus. A neutron star in a faraway galaxy has a mass equal to 2.4 solar masses (1 solar mass = \(M_☉\) = mass of the sun = \(\mathrm{1.99 \times 10^{30}\; kg}\)) and a diameter of 26 km. We can treat both the neutron star and the U-235 nucleus as spheres. Then the density for both is given by: \[\rho = \dfrac{m}{V} \nonumber \] with \[V = \dfrac{4}{3} \pi r^3 \nonumber \] (a) The radius of the neutron star is \(\mathrm{\dfrac{1}{2}\times 26\; km = \dfrac{1}{2} \times 2.6 \times 10^4\; m = 1.3 \times 10^4\; m}\) so the density of the neutron star is: \[ \begin{align*} \rho &= \dfrac{m}{V} \\[4pt] &=\dfrac{m}{\frac{4}{3}\pi r^3} \\[4pt] &= \dfrac{2.4(1.99 \times 10^{30}\;kg)}{\frac{4}{3} \pi (1.3 \times 10^4m)^3} \\[4pt] &=5.2 \times 10^{17}\;kg/m^3 \end{align*} \nonumber \] (b) The radius of the U-235 nucleus is \(\mathrm{\dfrac{1}{2} \times 15 \times 10^{−15}\;m=7.5 \times 10^{−15}\;m}\), so the density of the U-235 nucleus is: \[ \begin{align*} \rho &=\dfrac{m}{V} \\[4pt] &=\dfrac{m}{\frac{4}{3}\pi r^3} \\[4pt] &= \dfrac{235\;amu \left(\frac{1.66 \times 10^{-27}\;kg}{1\;amu}\right)}{ \frac{4}{3} \pi (7.5 \times 10^{-15}m)^3} \\[4pt] &=2.2 \times 10^{17} \; kg/m^3 \end{align*} \nonumber \] These values are fairly similar (same order of magnitude), but the nucleus is more than twice as dense as the neutron star. Find the density of a neutron star with a mass of 1.97 solar masses and a diameter of 13 km, and compare it to the density of a hydrogen nucleus, which has a diameter of 1.75 fm (\(\mathrm{1\; fm = 1 \times 10^{–15}\; m}\)). The density of the neutron star is \(\mathrm{3.4 \times 10^{18}\; kg/m^3}\). The density of a hydrogen nucleus is \(\mathrm{6.0 \times 10^{17}\; kg/m^3}\). The neutron star is 5.7 times denser than the hydrogen nucleus. To hold positively charged protons together in the very small volume of a nucleus requires very strong attractive forces because the positively charged protons repel one another strongly at such short distances. The force of attraction that holds the nucleus together is the . (The strong force is one of the four fundamental forces that are known to exist. The others are the electromagnetic force, the gravitational force, and the nuclear weak force.) This force acts between protons, between neutrons, and between protons and neutrons. It is very different from the electrostatic force that holds negatively charged electrons around a positively charged nucleus (the attraction between opposite charges). Over distances less than 10 meters and within the nucleus, the strong nuclear force is much stronger than electrostatic repulsions between protons; over larger distances and outside the nucleus, it is essentially nonexistent. As a simple example of the energy associated with the strong nuclear force, consider the helium atom composed of two protons, two neutrons, and two electrons. The total mass of these six subatomic particles may be calculated as: \[ \underset{\Large\text{protons}}{(2 \times 1.0073\; \text{amu})} + \underset{\Large\text{neutrons}}{(2 \times 1.0087\; \text{amu})} + \underset{\Large\text{electrons}}{(2 \times 0.00055\; \text{amu})}= 4.0331\; \text{amu }\label{Eq2} \] However, mass spectrometric measurements reveal that the mass of an \(\ce{_2^4 He}\) atom is 4.0026 amu, less than the combined masses of its six constituent subatomic particles. This difference between the calculated and experimentally measured masses is known as the of the atom. In the case of helium, the mass defect indicates a “loss” in mass of 4.0331 amu – 4.0026 amu = 0.0305 amu. The loss in mass accompanying the formation of an atom from protons, neutrons, and electrons is due to the conversion of that mass into energy that is evolved as the atom forms. The is the energy produced when the atoms’ nucleons are bound together; this is also the energy needed to break a nucleus into its constituent protons and neutrons. In comparison to chemical bond energies, nuclear binding energies are greater, as we will learn in this section. Consequently, the energy changes associated with nuclear reactions are vastly greater than are those for chemical reactions. The conversion between mass and energy is most identifiably represented by the as stated by Albert Einstein: \[E=mc^2 \label{Eq3} \] where is energy, is mass of the matter being converted, and is the speed of light in a vacuum. This equation can be used to find the amount of energy that results when matter is converted into energy. Using this mass-energy equivalence equation, the nuclear binding energy of a nucleus may be calculated from its , as demonstrated in Example \(\Page {2}\). A variety of units are commonly used for nuclear binding energies, including , with 1 eV equaling the amount of energy necessary to the move the charge of an electron across an electric potential difference of 1 volt, making \(\mathrm{1\; eV = 1.602 \times 10^{-19}\; J}\). Determine the binding energy for the nuclide \(\ce{^4_2 He}\) in: The mass defect for a \(\ce{^4_2He}\) nucleus is 0.0305 amu, as shown previously. Determine the binding energy in joules per nuclide using the mass-energy equivalence equation. To accommodate the requested energy units, the mass defect must be expressed in kilograms (recall that 1 J = 1 kg m /s ). (a) First, express the mass defect in g/mol. This is easily done considering the of atomic mass (amu) and molar mass (g/mol) that results from the definitions of the amu and mole units (refer to the previous discussion in the chapter on atoms, molecules, and ions if needed). The mass defect is therefore 0.0305 g/mol. To accommodate the units of the other terms in the mass-energy equation, the mass must be expressed in kg, since 1 J = 1 kg m /s . Converting grams into kilograms yields a mass defect of \(\mathrm{3.05 \times 10^{–5}\; kg/mol}\). Substituting this quantity into the mass-energy equivalence equation yields: \[\begin{align*} E &=mc^2 \\[4pt] &= \dfrac{3.05 \times 10^{-5}\;kg}{mol} \times \left(\dfrac{2.998 \times 10^8\;m}{s}\right)^2 \\[4pt] &= 2.74×10^{12}\:kg\:m^2s^{-2}mol^{-1} \\[4pt] &=2.74 \times 10^{12}\;J/mol=2.74\: /mol \end{align*} \nonumber \] (b) The binding energy for a single nucleus is computed from the molar binding energy using Avogadro’s number: \[\begin{align*} E &= 2.74×10^{12}\:J\:mol^{-1}×\dfrac{1\: mol}{6.022×10^{23}\:nuclei} \\[4pt] &=4.55×10^{-12} \: J =4.55\: pJ \end{align*} \nonumber \] (c) Recall that \(\mathrm{1\; eV = 1.602 \times 10^{-19}\; J}\). Using the binding energy computed in part (b): \[\begin{align*} E &= 4.55×10^{-12} \: J× \dfrac{1\: eV}{1.602×10^{-19}\:J} \\[4pt] &=2.84×10^7\:eV=28.4\: MeV \end{align*} \nonumber \] What is the binding energy for the nuclide\(\ce{^{19}_9F}\) (atomic mass: 18.9984 amu) in MeV per nucleus? 148.4 MeV Because the energy changes for breaking and forming bonds are so small compared to the energy changes for breaking or forming nuclei, the changes in mass during all ordinary chemical reactions are virtually undetectable. As described in the chapter on thermochemistry, the most energetic chemical reactions exhibit enthalpies on the order of of kJ/mol, which is equivalent to mass differences in the nanogram range (10 g). On the other hand, nuclear binding energies are typically on the order of of kJ/mol, corresponding to mass differences in the milligram range (10 g). A nucleus is stable if it cannot be transformed into another configuration without adding energy from the outside. Of the thousands of nuclides that exist, about 250 are stable. A plot of the number of neutrons versus the number of protons for stable nuclei reveals that the stable isotopes fall into a narrow band. This region is known as the (also called the belt, zone, or valley of stability). The straight line in Figure \(\Page {1}\) represents nuclei that have a 1:1 ratio of protons to neutrons (n:p ratio). Note that the lighter stable nuclei, in general, have equal numbers of protons and neutrons. For example, nitrogen-14 has seven protons and seven neutrons. Heavier stable nuclei, however, have increasingly more neutrons than protons. For example: iron-56 has 30 neutrons and 26 protons, an n:p ratio of 1.15, whereas the stable nuclide lead-207 has 125 neutrons and 82 protons, an n:p ratio equal to 1.52. This is because larger nuclei have more proton-proton repulsions, and require larger numbers of neutrons to provide compensating strong forces to overcome these electrostatic repulsions and hold the nucleus together. The nuclei that are to the left or to the right of the band of stability are unstable and exhibit . They change spontaneously (decay) into other nuclei that are either in, or closer to, the band of stability. These nuclear decay reactions convert one unstable isotope (or ) into another, more stable, isotope. We will discuss the nature and products of this radioactive decay in subsequent sections of this chapter. Several observations may be made regarding the relationship between the stability of a nucleus and its structure. Nuclei with even numbers of protons, neutrons, or both are more likely to be stable (Table \(\Page {1}\)). Nuclei with certain numbers of nucleons, known as , are stable against nuclear decay. These numbers of protons or neutrons (2, 8, 20, 28, 50, 82, and 126) make complete shells in the nucleus. These are similar in concept to the stable electron shells observed for the noble gases. Nuclei that have magic numbers of both protons and neutrons, such as \(\ce{^4_2He}\), \(\ce{^{16}_8O}\), \(\ce{^{40}_{20}Ca}\), and \(\ce{^{208}_{82}Pb}\) and are particularly stable. These trends in nuclear stability may be rationalized by considering a quantum mechanical model of nuclear energy states analogous to that used to describe electronic states earlier in this textbook. The details of this model are beyond the scope of this chapter. The relative stability of a nucleus is correlated with its , the total binding energy for the nucleus divided by the number or nucleons in the nucleus. For instance, the binding energy for a \(\ce{^4_2He}\) nucleus is therefore: \[\mathrm{\dfrac{28.4\; MeV}{4\; nucleons}=7.10\; MeV/nucleon} \label{Eq3a} \] The binding energy per nucleon of a nuclide on the curve shown in Figure \(\Page {2}\) The iron nuclide \(\ce{^{56}_{26}Fe}\) lies near the top of the binding energy curve (Figure \(\Page {2}\)) and is one of the most stable nuclides. What is the binding energy per nucleon (in MeV) for the nuclide \(\ce{^{56}_{26}Fe}\) (atomic mass of 55.9349 amu)? As in Example, we first determine the mass defect of the nuclide, which is the difference between the mass of 26 protons, 30 neutrons, and 26 electrons, and the observed mass of an \(\ce{^{56}_{26}Fe}\) atom: \[\begin{align*} \mathrm{Mass\: defect}&=\mathrm{[(26×1.0073\: amu)+(30×1.0087\: amu)+(26×0.00055\: amu)]−55.9349\: amu}\\ &=\mathrm{56.4651\: amu−55.9349\: amu}\\ &=\mathrm{0.5302\: amu} \end{align*} \nonumber \] We next calculate the binding energy for one nucleus from the mass defect using the mass-energy equivalence equation: \[\begin{align*} E&=mc^2=\mathrm{0.5302\: amu×\dfrac{1.6605×10^{-27}\:kg}{1\: amu}×(2.998×10^8\:m/s)^2}\\ &=\mathrm{7.913×10^{−11}\:\textrm{kg⋅m}/s^2}\\ &=\mathrm{7.913×10^{−11}\:J} \end{align*} \nonumber \] We then convert the binding energy in joules per nucleus into units of MeV per nuclide: \[\mathrm{7.913×10^{−11}\:J×\dfrac{1\: MeV}{1.602×10^{−13}\:J}=493.9\: MeV} \nonumber \] Finally, we determine the binding energy per nucleon by dividing the total nuclear binding energy by the number of nucleons in the atom: \[\textrm{Binding energy per nucleon}=\mathrm{\dfrac{493.9\: MeV}{56}=8.820\: MeV/nucleon} \nonumber \] Note that this is almost 25% larger than the binding energy per nucleon for \(\ce{^4_2He}\).(Note also that this is the same process as in Example \(\Page {2}\, but with the additional step of dividing the total nuclear binding energy by the number of nucleons.) What is the binding energy per nucleon in \(\ce{^{19}_9F}\) (atomic mass, 18.9984 amu)? 7.810 MeV/nucleon An atomic nucleus consists of protons and neutrons, collectively called nucleons. Although protons repel each other, the nucleus is held tightly together by a short-range, but very strong, force called the strong nuclear force. A nucleus has less mass than the total mass of its constituent nucleons. This “missing” mass is the mass defect, which has been converted into the binding energy that holds the nucleus together according to Einstein’s mass-energy equivalence equation, = . Of the many nuclides that exist, only a small number are stable. Nuclides with even numbers of protons or neutrons, or those with magic numbers of nucleons, are especially likely to be stable. These stable nuclides occupy a narrow band of stability on a graph of number of protons versus number of neutrons. The binding energy per nucleon is largest for the elements with mass numbers near 56; these are the most stable nuclei.   ).

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Basics_of_Catalysts

A catalyst is another substance than reactants products added to a reaction system to alter the speed of a chemical reaction approaching a chemical equilibrium. It interacts with the reactants in a cyclic manner promoting perhaps many reactions at the atomic or molecular level, but it is not consumed. Another reason for using a catalyst is that it promote the production of a selected product. A catalyst changes the activation energy, , of a reaction by providing an alternate pathway for the reaction. The and rate constant of a reaction are related to in the following ways: = * function of concentration = exp ( / ) where is a constant related to collision rates. Thus, a change in changes the rate of a reaction. A catalyst in the same phase (usually liquid or gas solution) as the reactants and products is called . A catalyst that is in a separate phase from the reactants is said to be a , or . Contact catalysts are materials with the capability of adsorbing molecules of gases or liquids onto their surfaces. An example of heterogeneous catalysis is the use of finely divided platinum to catalyze the reaction of carbon monoxide with oxygen to form carbon dioxide. This reaction is used in catalytic converters mounted in automobiles to eliminate carbon monoxide from the exhaust gases. are not catalysts by themselves but increase the effectiveness of a catalyst. For example, alumina Al O , is added to finely divided iron to increase the ability of the iron to catalyze the formation of ammonia from a mixture of nitrogen and hydrogen. A reduces the effectiveness of a catalyst. For example, lead compounds poison the ability of platinum as a catalyst. Thus, leaded gasoline shall not be used for automobils equiped with catalytic converters. Because heterogeneous catalysts often are used in high temperatures reactions, they are usually high melting (refractory) materials, or else they can be supported by refractory materials such as alumina. Today, catalysts design is a challenge for chemists and engineers for effective productions, pollution prevention, and waste treatments. As mentioned in solid defects, solid surfaces are two-dimensional defects. They offer a potential for attraction to molecules of gases and liquid. Adsorption takes place as molecules are attracted to the surface, and when molecules penetrate through the bulk material, the term absorption is used. Absorption with no chemical bonds formed or broken is called physical absorption or physisorption, whereas chemisorptions refer to processes when new bonds are formed or broken. by Swaddle (page 117) gives an excellent example to illustrate the chemisorption of hydrogen by a nickel catalyst. The bond energy of H is 435 kJ/mol. Thus, in a hydrogenation reaction, energy has to be made available for the reactions: H -> 2 H, = 435 kJ | | >C=C< + 2 H -> H-C--C-H | | In the above reaction, the activation energy, is close to 435 kJ. However, when hydrogen is absorbed by nickel, the breakage of the H-H bond is facilitated by a series of steps. 2 Ni + H --->2 Ni...H ----> 2 Ni-H solid gas physisorption chemisorption The activation energy is thus lowered due to the formation of Ni-H bonds. A change in activation energy changes the rate of reaction. In the activation of O by a metal M, the O=O bond is weakend or broken via these steps: O=O O--O O O O- O- | | || || | | -M--M- ==> -M---M- ==> -M M- ==> -M M- In these steps, the oxygen is activated at various stages. With sophisticated experimental techniques, we can study the chemisorbed species in details. For example, the chemisorbed ethylene is believed to be an The chemisorbed ethylidune radical. If the absorbed species are very stable, and much energies are release in the chemisorption process, the absorbed species are not reactive. Their absorbtions prevent further absorption of other species, making the catalyst inactive. These phenomena are known as catalyst poisoning. A reduces the effectiveness of a catalyst. Tetraethyl lead has allways been additive to the gasoline. For environmental protection, catalytic converters have been installed in automobiles to oxidize carbon monoxide and hydrocarbons. However, lead compounds poison the ability of platinum as a catalyst. Thus, leaded gasoline should not be used for automobiles equipped with catalytic converters. There are many types of catalyst in the market place, for example MIRATECH oxidation catalyst can also reduce carbon monoxide and hydrocarbon emissions. The most common catalytic converter uses Pt metal. Recently, there is a concern over the reduction of sulfur in gasoline and other engine fuels for the purpose of reducing sulfur oxides emission. Technically, sulphur compounds are not catalyst poisons (i.e. they do not cause an irreversible reduction in catalyst efficiency). However, they will occupy part of the precious metal surface, thereby reducing the active conversion of exhaust emissions until the sulphur gets de-sorbed from the precious metal sites again (short-term effect). The first period of transition metals are represented by these metals. Sc Ti V Cr Mn Fe Co Ni Cu and Zn Typical common features among them are the presences of electrons, and in many of them, and their unfilled orbitals. As a result, transition metals form compounds of variable oxidation states. Thus, these metals are that lend out electrons at appropriate time, and store them for chemical species at other times. Tranisition metals are used in mentioned earlier. These reaction are represented by | | >C=C< + 2 H -> H-C--C-H | | For example, the hydrogenation of unsaturated oil in the manufacture of margarine is such an application. Special catalysts such as ICT-3-25-P is made of palladium supported on the special wide-porous carbon carrier Sibunit. Other processes catalyzed by transition metals are oxidation-reduction reactions: NH + 5/4 O = NO + H O 2 CO + O = 2 CO The oxidation of CO takes place in catalytic converters, platinum is often, but not always used as a catalyst in them. The picture shown here is a dual catalytic converter showing its gas flow path. For most transition metals except gold, the chemisorption strength follows a general sequence for gaseous reagents: O > alkynes > alkenes > CO > H > CO > N The chemisorption strength also varies with the metals. In general, the chemisorption is the strongest for metals on the left, and it decreases for transition metals in a period as the atomic number increases: Sc Ti V Cr Mn Fe Co Ni Cu Zn Y Zr Nb Mo Tc Ru Rh Pd Ag Cd La Hf Ta W Re Os Ir Pt Au Hg The chemisorptions are too strong for Sc, Ti, V, Cr, and Mn groups and these metals are not effective catalysts. These relative chemisorption strengths enable us to make some simple predictions regarding their sutability as catalysts for specific reactions. For example, a catalyst for the Haber process to produce ammonia must chemisorb nitrogen. Iron, ruthenium, or osmium may be considered. For hydrogenation reactions, the catalyst must chemisorb H . Metals Co, Rh, Ir, Ni, Pd, and Pt are suitable. Availability and costs are additional factors for the consideration. Nickel is actually a good choice, all considered. These guidelines are very crude, and each case must be carefully studied. Fortunately, many catalysts are commercially available. The research and development of catalyst are left for many companies. Syngas is a general term used to mean synthetic gases suitable as fuel or for the production of liquid fuel. Often, it is a mixtuure of H and CO, and this mixture can be converted into methanol, CH OH. The well known catalysts are Pt and Rh, but other technology such as memberanes are also used for syngas productions. Selection of a catalyst is important in industrial productions. For example, using rhodium or platinum as catalysts have shown to give very different distribution of products when methane or ethane were used. CH (65%) + O (35%) ---Rh--> H (60%) + CO (30%) + CO (2%) + H ) (5%) When platinum is used, more of the undsirable products H O and CO were obtained. Swaddle has described the difference between using these two metals as catalysts ( , page 120), but much more details is required when it comes to application. The data provided evidence to show that a slight difference in chemisorption led to very different results. The surface area per unit weight is an important consideration when solids are used as catalysts. There are many studies related to the study of surface area of particulate metals. Various methods are developed to measure the surface areas of solid materials. One such method is the surfact area determination from gas adsorption. Clusters are the limiting sizes of metal particles, each of which are made up only a few atoms. There is no need to rigorously define the number of atoms in a particulate to be called clusters, but a general view is that when the number of atoms at the surface of the particle is more than the nuber of atoms in the , the particle is a . Thus, a cluster can have as few as 3 atoms, and as large as a few tens of atoms. By the way, the term cluster have been used in other areas of study. For example, in organometallic chemistry, compounds with a few metals bonded together by metal-metal bonds are also called metal clusters. Many carbonyl compounds belong to this category. For example, Co (u-CO) (CO) , (u-CO meaning CO bridged between two metal atoms) Mn (CO) Fe (CO) Co (CO) Rh (CO) CFe (CO) Rh (CO) Os (CO) Metal carbonyls have been studies as homogeneous catalysts. They are mentioned here so that you will be able to appreciate their usage in other literatures. All catalytic activities occur at the surface, because the surface atoms have tendencies for chemisorption of gas molecules. Thus, clusters will naturally be excellent potential catalysts. Thus, the study of heterogeneous catalysts may involve the study of metal cluster ion chemistry, and encapsulated silver clusters as oxidation catalysts. Clusters can be made from vapour deposition. The title of this link sounds very interesting: Metal Atom Vapor Chemistry: A Field Awaits Its Breakthrough. Due to their ability to have various oxidation state, transition metals form non-stoichiometric oxides, and they have excellent potentials for oxidation and reduction (redox) reactions, because they can both give and accept electrons. M => M + e M + e => M Furthermore, they resemble metals, and they catalyze hydrogenation and isomerization reactions. A metal oxide have excess positive charges in the solid, and they can adsorb oxygen to form anions such as O , O , O , and O on their surfaces. Nickel oxide is such an oxide. It turns out that the adsorbed O species is the most active, O (g) + 2 Ni => 2 O (ads) + 2 Ni 2 O (ads) + 2 CO (ads) => 2 CO + 2 e 2 Ni + 2 e => 2 Ni When an oxide gives up oxygen, electrons were left behind and the negative charge in it makes it a oxide. Zinc oxide is such an oxide, and the reaction mechanism may be represented as follows: CO (g) + 2 O (lattice) => CO (lattice) + 2 e 0.5 O + 2 e => O CO (lattice) = CO + O (lattice) The overall reaction is --- CO + 0.5 O => CO In these primary steps, the oxygen is consumed via adsorption on the solid. A sulfide, such as MoS , can loose sulfur atoms to become a solid, Mo S or gain a sulfur atom to become a solid Mo S , depending on the vapour pressure of S gas surrounding the solid. By doping MoS with oxide can also make it a solid for a catalyst. One of the useful applications of MoS as a catalyst is to reduce sulfur in gasoline. For example, the cyclic thiophene C H S can be converted to a hydrocarbon by using a MoS , C H S + 4 H == MoS == > C H + H S This is accomplished by a typical commercial hydrodesulfurization catalyst, which may contain 14% MoO , and 3% CoO on alumina support. It has been well known that metal oxides dissolve in water to form basic solutions whereas non-metalic oxides dissolve in water to give acidic solutions. Some metal oxides such as Al O , Fe O , Cr O etc dissolve in strong acid and bases. Thus, we can divide oxides into acidic and basic oxides for catalytic activities. Acidic oxides such as Al O and SiO catalyze dehydration reactions such as R-CH CH OH (g) == (Al O , 600 K) ==> R-CH=CH If we consider the oxide a Lewis acid, it adsorbs the OH group, facilitating the reaction in the following steps. R-CH CH OH (g) => R-CH -CH + OH (adsorbed) R-CH -CH => R-CH -CH R-CH -CH + OH (adsorbed) => R-CH=CH + H O Zeolites, which are alumniosilicates, function as acidic catalysts. They also catalyze isomerization, cracking, alkylation and other organic reactions. Basic oxides such as MgO and ZrO favor reactions involving . When a proton, H , is adsorbed onto the surface close to an O ion in the metal oxide, an OH group is formed, leaving the organic molecule a negative charge. CH -CH -CN + MO (solid) => -CH -CH CN + M-OH (solid) => CH =CH-CN + MOH (solid) = + oxygen => CH =CH-CN + MO (solid) + H O (product) The over all reaction is a selective oxidation CH -CH -CN + MO (solid) + 0.5 O => CH =CH-CN + MO (solid) + 0.5 H O The oxidation eliminated two hydrogen atoms per molecule in the process, and the proposed mechanism suggests a two step elimination process. Mixtures of basic oxides have been used as catalysts in the . In some cases, special reactors and catalysts are designed for this type of application. TAP Reactor is one such an application. In this case, zeolites were used. Reactions caused by photons, bundles of radiation energy, are called photolysis. Photocatalyic reactions imply photolysis in the presence of a catalyst. In most cases, however, the catalysts are semiconductors and the reactions are semiconductor assisted photolysis reactions. In this aspect, the photocatalyst has a slightly different function than those in thermal chemical process. The simulation below shows that when the yellow beam strikes the semiconductor TiO , an electrons are excited from the valance band into the conduction band. This band gap is 3.2 V. The excited electron then promote the production of H . The holes take electrons from OH groups converting them to active OH radicals. The radicals break up forming O or react with CHCl converting it into the CO , H , and Cl . This simulation of photocatalysts is prepared by a Japanese group, and it illustrates the concept rather well. In reality, the process is rather complicated. In the photodecomposition of water, the excited electrons react with hydrogen ions (protons) 2 H + 2 e = H 2 OH + 2 e (hole) = H O + 0.5 O Thus, the products, H and O , are potential fuels for the supply of energy, especially for fuel cells. As another example, fluoroboric acid is used in electroplating and metal finishing. To treat wastewater from these industries requires the removal of fluoroboric acid. Existing methods of adsorption, coagulation, precipitation methods do not work. Thus, photocatalytic decomposition of fluoroboric acid has been studied, and it showed that TiO being rather effective when it doped with Cr and Fe oxides. The above link showed that doping of Cr or Fe drastically enhanced the activity. Moreover, 0.5 wt% Cr/TiO2 and 1.0 wt% Fe/TiO2 showed maximum activities of 61 % and 41 %, respectively. Recently, a news article has an attractive claim on Indoor Air Cleaner. Judge it yourself to see if it is something worth investigating. The energy gap of TiO is 3.2 V. What is the frequency of the photons that has just eneough energy to excite the electrons from the covalent band of TiO into the conduction band? The energy to excite an electron up 3.2 V is 3.2 eV. 1.6022e-19 J 1 3.2 eV ------------- ------------- = 7.74e14 Hz 1 eV 6.626e-34 J s The wavelength of these photons are 3e8 m/s -------- = 387e9 m (or 387 nm) 7.74 /s These photons are at the just within visible limit of 350 - 700 nm. A mole of photons is called an einstein. How much energy in J is an einstein of the photons described in Example 1. The energy is 1.6022e-19 J 6.022e23 3.2 eV -------------- ------------- = 308000 J = 308 kJ 1 eV 1 photon When we discuss Gibbs energy, we have learned that the enthalpy of formation for H O is - 285.83 kJ. This means that we need a minimum of 286 kJ to decompose water. Thus, 1 einstein of photons has more energy to decompose a mole of water than the minimum. However, an overpotential is required to decompose water. The titanium oxide is mixed with platinum metal and ruthenium oxide to facilitate the formation of bubbles in these process. (See by Swaddle(page 125). Describe purposes. Differentiate chemical process. Describe the requirements for an effective catalyst. Give a list of chemicals in decreasing order of their adsorption strength for transition metals. Know the trend of adsorption strength of metals from their position in the periodic table of chemical elements. Explain why transition metals form non-stoichiometric oxides.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.07%3A_Exact_Calculations_and_Approximations

The methods for dealing with acid-base equilibria that we developed in the earlier units of this series are widely used in ordinary practice. Although many of these involve approximations of various kinds, the results are usually good enough for most purposes. Sometimes, however — for example, in problems involving very dilute solutions, the approximations break down, often because they ignore the small quantities of H and OH ions always present in pure water. In this unit, we look at exact, or "comprehensive" treatment of some of the more common kinds of acid-base equilibria problems. The usual definition of a “strong” acid or base is one that is completely dissociated in aqueous solution. Hydrochloric acid is a common example of a strong acid. When HCl gas is dissolved in water, the resulting solution contains the ions H O , OH , and Cl , but except in very concentrated solutions, the concentration of HCl is negligible; for all practical purposes, molecules of “hydrochloric acid”, HCl, do not exist in dilute aqueous solutions. To specify the concentrations of the three species present in an aqueous solution of HCl, we need three independent relations between them. These relations are obtained by observing that certain conditions must hold for aqueous solutions: The next step is to combine these three limiting conditions into a single expression that relates the hydronium ion concentration to \(C_a\). This is best done by starting with an equation that relates several quantities and substituting the terms that we want to eliminate. Thus we can get rid of the \([Cl^–]\) term by substituting Equation \(\ref{1-3}\) into Equation \(\ref{1-4}\) : \[[H^+] = [OH^–] + C_a \label{1-5}\] The \([OH^–]\) term can be eliminated by the use of Equation \(\ref{1-1}\): \[[H^+] = C_a + \dfrac{K_w}{[H^+]} \label{1-6}\] This equation tells us that the hydronium ion concentration will be the same as the nominal concentration of a strong acid as long as the solution is not very dilute. As the acid concentration falls below about 10 M, however, the second term predominates; \([H^+]\) approaches \(\sqrt{K_w}\) or \(10^{–7} M\) at 25 °C. The hydronium ion concentration can of course never fall below this value; no amount of dilution can make the solution alkaline! No amount of dilution can make the solution of a strong acid alkaline! Notice that Equation \(\ref{1-6}\) is a quadratic equation; in regular polynomial form it would be rewritten as \[[H^+]^2 – C_a[H^+] – K_w = 0 \label{1-7}\] Most practical problems involving strong acids are concerned with more concentrated solutions in which the second term of Equation \(\ref{1-7}\) can be dropped, yielding the simple relation \[[H^+] \approx [A^–]\] In more concentrated solutions, interactions between ions cause their “effective” concentrations, known as their , to deviate from their “analytical” concentrations. Thus in a solution prepared by adding 0.5 mole of the very strong acid HClO to sufficient water to make the volume 1 liter, freezing-point depression measurements indicate that the concentrations of hydronium and perchlorate ions are only about 0.4 . This does not mean that the acid is only 80% dissociated; there is no evidence of HClO molecules in the solution. What has happened is that about 20% of the H O and ClO ions have formed ion-pair complexes in which the oppositely-charged species are loosely bound by electrostatic forces. Similarly, in a 0.10 solution of hydrochloric acid, the activity of H is 0.81, or only 81% of its concentration. (See the green box below for more on this.) Activities are important because only these work properly in equilibrium calculations. The relation between the concentration of a species and its activity is expressed by the \(\gamma\): \[a = \gamma C \label{1-8}\] As a solution becomes more dilute, \(\gamma\) approaches unity. At ionic concentrations below about 0.001 , concentrations can generally be used in place of activities with negligible error. Recall that pH is defined as the negative logarithm of the hydrogen ion , not its concentration. Activities of single ions cannot be determined, so activity coefficients in ionic solutions are always the average, or , of those for all ionic species present. This quantity is denoted as \(\gamma_{\pm}\). At very high concentrations, activities can depart wildly from concentrations. This is a practical consideration when dealing with strong mineral acids which are available at concentrations of 10 M or greater. In a 12 solution of hydrochloric acid, for example, the mean ionic activity coefficient* is 207. This means that under these conditions with [H ] = 12, the activity {H } = 2500, corresponding to a pH of about –3.4, instead of –1.1 as might be predicted if concentrations were being used. These very high activity coefficients also explain another phenomenon: why you can detect the odor of HCl over a concentrated hydrochloric acid solution even though this acid is supposedly "100% dissociated". At these high concentrations, a pair of "dissociated" ions \(H^+\) and \(Cl^–\) will occasionally find themselves so close together that they may momentarily act as an HCl unit; some of these may escape as \(HC before thermal motions break them up again. Under these conditions, “dissociation” begins to lose its meaning so that in effect, dissociation is no longer complete. Although the concentration of \(HCl will always be very small, its own activity coefficient can be as great as 2000, which means that its escaping tendency from the solution is extremely high, so that the presence of even a tiny amount is very noticeable. Most acids are weak; there are hundreds of thousands of them, whereas there are no more than a few dozen strong acids. We can treat weak acid solutions in exactly the same general way as we did for strong acids. The only difference is that we must now include the equilibrium expression for the acid. We will start with the simple case of the pure acid in water, and then go from there to the more general one in which strong cations are present. In this exposition, we will refer to “hydrogen ions” and \([H^+]\) for brevity, and will assume that the acid \(HA\) dissociates into \(H^+\) and its conjugate base \(A^-\). In addition to the species H , OH , and A− which we had in the strong-acid case, we now have the undissociated acid HA; four variables, requiring four equations. \[ [H^+,OH^–] = K_w \label{2-1}\] \[ K_a = \dfrac{[H^+,A^–]}{[HA]} \label{2-2}\] \[ C_a = [HA] + [A^–] \label{2-3}\] \[[H^+] = [OH^–] + [HA^–] \label{2-4}\] To eliminate [HA] from Equation \(\ref{2-2}\), we solve Equation \(\ref{2-4}\) for this term, and substitute the resulting expression into the numerator: \[ K_a =\dfrac{[H^+]([H^+] - [OH^-])}{C_a-([H^+] - [OH^-]) } \label{2-5}\] The latter equation is simplified by multiplying out and replacing [H ,OH ] with . We then get rid of the [OH ] term by replacing it with /[H ] \[[H^+] C_b + [H^+]^2 – [H^+,OH^–] = K_a C_a – K_a [H^+] + K_a [OH^–]\] \[[H^+]^2 C_b + [H^+]^3 – [H^+] K_w = K_a C_a – K_a [H^+] + \dfrac{K_a K_w}{[H^+]}\] Rearranged into standard polynomial form, this becomes \[[H^+]^3 + K_a[H^+]^2 – (K_w + C_aK_a) [H^+] – K_a K_w = 0 \label{2-5a}\] For most practical applications, we can make approximations that eliminate the need to solve a cubic equation. Unless the acid is extremely weak or the solution is very dilute, the concentration of OH can be neglected in comparison to that of [H ]. If we assume that [OH ] ≪ [H ], then Equation \(\ref{2-5a}\) can be simplified to \[K_a \approx \dfrac{[H^+]^2}{C_a-[H^+]} \label{2-6}\] which is a quadratic equation: \[[H^+]^2 +K_a[H^+]– K_aC_a \approx 0 \label{2-7}\] and thus, from the quadratic formula, \[ [H^+] \approx \dfrac{K_a + \sqrt{K_a + 4K_aC_a}}{2} \label{2-8}\] Calculate the pH of a 0.0010 M solution of acetic acid, \(K_a = 1.74 \times 10^{–5}\). First approximation: \[[H^+] = \sqrt{(1.0 \times 10^{–3}) × (1.74 \times 10^{–5}} = \sqrt{1.74 \times 10^{–8}} = 1.3 \times 10^{–4}\; M. \nonumber \] Applying the "5% test", \[\dfrac{1.3 \times 10^{–4}}{1.0 \times 10^{–3}} = 0.13\nonumber \] This exceeds 0.05, so we must explicitly solve the quadratic Equation \(\ref{2-7}\) to obtain two roots: \(+1.2 \times 10^{–4}\) and \(–1.4 \times 10^{-4}\). Taking the positive root, we have \[pH = –\log (1.2 \times 10^{–4}) = 3.9 \nonumber \] If the acid is fairly concentrated (usually more than 10 M), a further simplification can frequently be achieved by making the assumption that \([H^+] \ll C_a\). This is justified when most of the acid remains in its protonated form [HA], so that relatively little H is produced. In this event, Equation \(\ref{2-6}\) reduces to \[ K_a \approx \dfrac{[H^+]^2}{C_a} \label{2-9}\] or \[[H^+] \approx \sqrt{K_aC_a} \label{2-10}\] Calculate the pH and percent ionization of 0.10 M acetic acid "HAc" (CH COOH), \(K_a = 1.74 \times 10^{–5}\). It is usually best to start by using Equation \(\ref{2-9}\) as a first approximation: \[[H^+] = \sqrt{(0.10)(1.74 \times 10^{–5})} = \sqrt{1.74 \times 10^{–6}} = 1.3 \times 10^{–3}\; M\nonumber \] This approximation is generally considered valid if [H ] is less than 5% of ; in this case, [H ]/ = 0.013, which is smaller than 0.05 and thus within the limit. This same quantity also corresponds to the ionization fraction, so the percent ionization is 1.3%. The pH of the solution is \[pH = –\log 1.2 \times 10^{-3} = 2.9\nonumber \] If the acid is very weak or its concentration is very low, the \(H^+\) produced by its dissociation may be little greater than that due to the ionization of water. However, if the solution is still acidic, it may still be possible to avoid solving the cubic equation \(\ref{2-5a}\) by assuming that the term \(([H^+] - [OH^–]) \ll C_a\) in Equation \(\ref{2-5}\): \[ K_a = \dfrac{[H^+]^2}{C_a - [H^+]} \label{2-11}\] This can be rearranged into standard quadratic form \[[H^+]^2 + K_a [H^+] – K_a C_a = 0 \label{2-12}\] For dilute solutions of weak acids, an exact treatment may be required. With the aid of a computer or graphic calculator, solving a cubic polynomial is now far less formidable than it used to be. However, round-off errors can cause these computerized cubic solvers to blow up; it is generally safer to use a quadratic approximation. Boric acid, B(OH) ("H BO ") is a weak acid found in the ocean and in some natural waters. As with many boron compounds, there is some question about its true nature, but for most practical purposes it can be considered to be monoprotic with \(K_a = 7.3 \times 10^{–10}\): \[Bi(OH)_3 + 2 H_2O \rightleftharpoons Bi(OH)_4^– + H_3O^+\nonumber \] Find the [H ] and pH of a 0.00050 M solution of boric acid in pure water. Because this acid is quite weak and its concentration low, we will use the quadratic form Equation \(\ref{2-7}\), which yields the positive root \(6.12 \times 10^{–7}\), corresponding to pH = 6.21. Notice that this is only six times the concentration of \(H^+\) present in pure water! It is instructive to compare this result with what the quadratic approximation would yield, which yield \([H^+] = 6.04 \times 10^{–7}\) so \(pH = 6.22\). The weak bases most commonly encountered are: \[A^– + H_2O \rightleftharpoons HA + OH^–\] \[CO_3^{2–} + H_2O \rightleftharpoons HCO_3^– + OH^–\] \[NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^–\] \[CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^++ H_2O\] Note that, in order to maintain electroneutrality, anions must be accompanied by sufficient cations to balance their charges. Thus for a solution of the salt NaA in water, we have the following conditions: Na , A , HA, H O, H , OH \[[H^+,OH^–] = K_w \label{2-13}\] \[ K_b =\dfrac{[HA,OH^-]}{[A^-]} \label{2-14}\] \[C_b = [Na^+] = [HA] + [A^–] \label{2-15}\] \[[Na^+] + [H^+] = [OH^–] + [A^–] \label{2-16}\] Replacing the [Na ] term in Equation \(\ref{2-15}\) by \(C_b\) and combining with \(K_w\) and the mass balance, a relation is obtained that is analogous to that of Equation \(\ref{2-5}\) for weak acids: \[K_b =\dfrac{[OH^-] ([OH^-] - [H^+])}{C_b - ([OH^-] - [H^+])} \label{2-17}\] The approximations \[ K_b \approx \dfrac{[OH^-]^2}{C_b - [OH^-]} \label{2-18}\] and \[[OH^–] \approx \sqrt{K_b C_b} \label{2-19}\] can be derived in a similar manner. Calculate the pH and the concentrations of all species in a 0.01 M solution of methylamine, CH NH (\(K_b = 4.2 \times 10^{–4}\)). We begin by using the simplest approximation Equation \(\ref{2-14}\): \[[OH^–] = \sqrt{(K_b C_b}- = \sqrt{(4.2 \times 10^{-4})(10^{–2})} = 2.1 \times 10^{–3}\nonumber \] To see if this approximation is justified, we apply a criterion similar to what we used for a weak acid: [OH ] must not exceed 5% of . In this case, \[ \dfrac{[OH^–]}{ C_b} = \dfrac{(2.1 \times 10^{-3}} { 10^{–2}} = 0.21\nonumber \] so we must use the quadratic form Equation \(\ref{2-12}\) that yields the positive root \(1.9 \times 10^{–3}\) which corresponds to \([OH^–]\) \[[H^+] = \dfrac{K_w}{[OH^–} = \dfrac{1 \times 10^{-14}}{1.9 \times 10^{–3}} = 5.3 \times 10^{-12}\nonumber \] and \[pH = –\log 5.3 \times 10^{–12} = 11.3.\nonumber \] From the charge balance equation, solve for \[[CH_3NH_2] = [OH^–] – [H^+] \approx [OH^–] = 5.3 \times 10^{–12}\; M. \nonumber \] For the concentration of the acid form (methylaminium ion CH NH ), use the mass balance equation: \[[CH_3NH_3^+] = C_b – [CH_3NH_2] = 0.01 – 0.0019 =0.0081\; M.\nonumber \] Many practical problems relating to environmental and physiological chemistry involve solutions containing more than one acid. In this section, we will restrict ourselves to a much simpler case of two acids, with a view toward showing the general method of approaching such problems by starting with charge- and mass-balance equations and making simplifying assumptions when justified. In general, the hydrogen ions produced by the stronger acid will tend to suppress dissociation of the weaker one, and both will tend to suppress the dissociation of water, thus reducing the sources of H that must be dealt with. Consider a mixture of two weak acids HX and HY; their respective nominal concentrations and equilibrium constants are denoted by , , and , Starting with the charge balance expression \[ [H^+] = [X^–] + [Y^–] + [OH^–] \label{3-1}\] We use the equilibrium constants to replace the conjugate base concentrations with expressions of the form \[ [X^-] = K_x \dfrac{[HX]}{[H^+]} \label{3-2}\] to yield \[ [H^+] = \dfrac{[HX]}{K_x} + \dfrac{[HY]}{K_y} + K_w \label{3-3}\] If neither acid is very strong or very dilute, we can replace equilibrium concentrations with nominal concentrations: \[ [H^+] \approx \sqrt{C_cK_x + C_yK_y K_w} \label{3-4}\] Estimate the pH of a solution that is 0.10M in acetic acid (\(K_a = 1.8 \times 10^{–5}\)) and 0.01M in formic acid (\(K_a = 1.7 \times 10^{–4}\)). Because is negligible compared to the products, we can simplify \Equation \(ref{3-4}\): \[[H^+] = \sqrt{1.8 \times 10^{–6} + 1.7 \times 10^{-6}} = 0.0019\nonumber \] Which corresponds to a pH of \(–\log 0.0019 = 2.7\) Note that the pH of each acid separately at its specified concentration would be around 2.8. However, if 0.001 M chloroacetic acid ( = 0.0014) is used in place of formic acid, the above expression becomes \[ [H^+] \approx \sqrt{ 1.4 \times 10^{-6} + 1.75 \times 10^{-14}} = 0.00188 \label{3-5}\] which exceeds the concentration of the stronger acid; because the acetic acid makes a negligible contribution to [H ] here, the simple approximation given above \Equation \(\ref{3-3}\) is clearly invalid. We now use the mass balance expression for the stronger acid \[[HX] + [X^–] = C_x \label{3-6}\] to solve for [X ] which is combined with the equilibrium constant to yield \[[X^-] = C_x - \dfrac{[H^+,X^]}{K_x} \label{3-7}\] Solving this for [X ] gives \[ [X^-] = \dfrac{C_xK_x}{K_x + [H^+]} \label{3-8}\] The approximation for the weaker acetic acid (HY) is still valid, so we retain it in the substituted electronegativity expression: \[ [H^+] \dfrac{C_xK_x}{K_x+[H^+]} + \dfrac{C_yK_y}{[H^+]} \label{3-9}\] which is a cubic equation that can be solved by approximation. Several methods have been published for calculating the hydrogen ion concentration in solutions containing an arbitrary number of acids and bases. These generally involve iterative calculations carried out by a computer. See, for example, 67(6) 501-503 (1990) and 67(12) 1036-1037 (1990). Owing to the large number of species involved, exact solutions of problems involving polyprotic acids can become very complicated. Thus for phosphoric acid H PO , the three "dissociation" steps yield three conjugate bases: Fortunately, it is usually possible to make simplifying assumptions in most practical applications. In the section that follows, we will show how this is done for the less-complicated case of a diprotic acid. A diprotic acid HA can donate its protons in two steps, yielding first a monoprotonated species HA and then the completely deprotonated form A . Since there are five unknowns (the concentrations of the acid, of the two conjugate bases and of H and OH ), we need five equations to define the relations between these quantities. These are \[[H^+,OH^–] = K_w \label{4-1}\] \[ K_1 = \dfrac{[H^+,HA^-]}{[H_2A]} \label{4-2}\] \[ K_1 = \dfrac{[H^+,HA^{2-}]}{[HA^-]} \label{4-3}\] \[C_a = [H_2A] + [HA^–] + [A^{2–}] \label{4-4}\] \[[H^+] = [OH^–] + [HA^–] + 2 [A^{2–}] \label{4-5}\] (It takes 2 moles of \(H^+\) to balance the charge of 1 mole of \(A^{2–}\)) Solving these five equations simultaneously for \(K_1\) yields the rather intimidating expression \[ K_1 = \dfrac{[H^+] \left( [H^+] - [OH^-] \dfrac{2K_2[H^+] - [OH^-]}{[H^+ + 2K_2} \right)}{C_a - \left( [H^+] - [OH^-] \dfrac{K_2 [H^+] -[OH^-]}{[H^+] + 2K_2} \right)} \label{4-6}\] which is of little practical use except insofar as it provides the starting point for various simplifying approximations. If the solution is even slightly acidic, then ([H ] – [OH ]) ≈ [H ] and \[ K_1 = \dfrac{[H^+] \left( [H^+] \dfrac{2K_2[H^+]}{[H^+ + 2K_2} \right)}{C_a - \left( [H^+] \dfrac{K_2 [H^+]}{[H^+] + 2K_2} \right)} \label{4-7}\] For any of the common diprotic acids, \(K_2\) is much smaller than \(K_1\). If the solution is sufficiently acidic that \(K_2 \ll [H^+]\), then a further simplification can be made that removes \(K_2\) from Equation \(\ref{4-7}\); this is the starting point for most practical calculations. \[ K_1 \approx \dfrac{[H^+]^2}{C_a-[H^+]} \label{4-8}\] Finally, if the solution is sufficiently concentrated and \(K_1\) sufficiently small so that \([H^+] \ll C_a\), then Equation \(\ref{4-8}\) reduces to: \[ [H^+] \approx \sqrt{K_a C_a}\] Solutions containing a weak acid together with a salt of the acid are collectively known as . When they are employed to control the pH of a solution (such as in a microbial growth medium), a sodium or potassium salt is commonly used and the concentrations are usually high enough for the Henderson-Hasselbalch equation to yield adequate results. In this section, we will develop an exact analytical treatment of weak acid-salt solutions, and show how the H H equation arises as an approximation. A typical buffer system is formed by adding a quantity of strong base such as sodium hydroxide to a solution of a weak acid HA. Alternatively, the same system can be made by combining appropriate amounts of a weak acid and its salt NaA. A system of this kind can be treated in much the same way as a weak acid, but now with the parameter in addition to . Na , A , HA, H O, H , OH \[[H^+,OH^–] = K_w \label{5-1}\] \[K_a = \dfrac{[H^+,A^-]}{[HA]} \label{5-2}\] \[C_a + C-b = [HA] + [A^–] \label{5-3}\] \[C_b = [Na^+] \label{5-4}\] \[[Na^+] + [H^+] = [OH^–] + [A^–] \label{5-5}\] Substituting Equation \(\ref{5-4}\) into Equation \(\ref{5-5}\) yields an expression for [A ]: \[[A^–] = C_b + [H^+] – [OH^–] \label{5-6}\] Inserting this into Equation \(\ref{5-3}\) and solving for [HA] yields \[[HA] = C_b + [H^+] – [OH^–] \label{5-7)}\] Finally, we substitute these last two expressions into the equilibrium constant (Equation \(\ref{5-2}\)): \[ [H^+] = K_a \dfrac{C_a - [H^+] + [OH^-]}{C_b + [H^+] - [OH^-]} \label{5-8}\] which becomes cubic in [H ] when [OH ] is replaced by ( / [H ]). \[[H^+]^3 +(C_b +K_a)[H^+]^2 – (K_w + C_aK_a) [H^+] – K_aK_w = 0 \label{5-8a}\] In almost all practical cases it is possible to make simplifying assumptions. Thus if the solution is known to be acidic or alkaline, then the [OH ] or [H ] terms in Equation \(\ref{5-8}\) can be neglected. In acidic solutions, for example, Equation \(\ref{5-8}\) becomes \[ [H^+] = K_a \dfrac{C_a - [H^+]}{C_b + [H^+]} \label{5-9}\] which can be rearranged into a quadratic in standard polynomial form: \[ [H^+]^2 + (C_b + C_a)  [H^+] – K_a C_a = 0 \label{5-10}\] If the concentrations and are sufficiently large, it may be possible to neglect the [H ] terms entirely, leading to the commonly-seen . \[ \color{red} [H^+] \approx K_a \dfrac{C_a}{C_b} \label{5-11}\] It's important to bear in mind that the Henderson-Hasselbalch Approximation is an "approximation of an approximation" that is generally valid only for combinations of and concentrations that fall within the colored portion of this plot. Most buffer solutions tend to be fairly concentrated, with and typically around 0.01 - 0.1 . For more dilute buffers and larger 's that bring you near the boundary of the colored area, it is safer to start with Equation \(\ref{5-9}\). Chlorous acid HClO has a of 1.94. Calculate the pH of a solution made by adding 0.01 M/L of sodium hydroxide to a -.02 M/L solution of chloric acid. In the resulting solution, = = 0.01 . On the plots shown above, the intersection of the log = –2 line with the plot for = 2 falls near the left boundary of the colored area, so we will use the quadratic form \(\ref{5-10}\). Substitution in Equation \(\ref{5-10}\) yields \[H^+ + 0.02 H^+ – (10^{–1.9} x 10^{–2}) = 0 \nonumber\] which yields a positive root 0.0047 = [H ] that corresponds to pH = . Note: Using the Henderson-Hassalbach Approximateion (Equation \(\ref{5-11}\)) would give pH = = 1.9.

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Changes in reaction conditions can have a tremendous effect on the course of a redox reaction. For example, under standard conditions, the reaction of \(\ce{Co(s)}\) with \(\ce{Ni^{2+}(aq)}\) to form \(\ce{Ni(s)}\) and \(\ce{Co^{2+}(aq)}\) occurs spontaneously, but if we reduce the concentration of \(\ce{Ni^{2+}}\) by a factor of 100, so that \(\ce{[Ni^{2+}]}\) is 0.01 M, then the reverse reaction occurs spontaneously instead. The relationship between voltage and concentration is one of the factors that must be understood to predict whether a reaction will be spontaneous. Electrochemical cells convert chemical energy to electrical energy and vice versa. The total amount of energy produced by an electrochemical cell, and thus the amount of energy available to do electrical work, depends on both the cell potential and the total number of electrons that are transferred from the reductant to the oxidant during the course of a reaction. The resulting electric current is measured in , an unit that measures the number of electrons passing a given point in 1 s. A coulomb relates energy (in joules) to electrical potential (in volts). Electric current is measured in ; 1 A is defined as the flow of 1 C/s past a given point (1 C = 1 A·s): \[\dfrac{\textrm{1 J}}{\textrm{1 V}}=\textrm{1 C}=\mathrm{A\cdot s} \label{20.5.1} \] In chemical reactions, however, we need to relate the coulomb to the charge on a mole of electrons. Multiplying the charge on the electron by Avogadro’s number gives us the charge on 1 mol of electrons, which is called the , named after the English physicist and chemist Michael Faraday (1791–1867): \[\begin{align}F &=(1.60218\times10^{-19}\textrm{ C})\left(\dfrac{6.02214 \times 10^{23}\, J}{\textrm{1 mol e}^-}\right) \\[4pt] &=9.64833212\times10^4\textrm{ C/mol e}^- \\[4pt] &\simeq 96,485\, J/(\mathrm{V\cdot mol\;e^-})\end{align} \label{20.5.2} \] The total charge transferred from the reductant to the oxidant is therefore \(nF\), where \(n\) is the number of moles of electrons. Faraday was a British physicist and chemist who was arguably one of the greatest experimental scientists in history. The son of a blacksmith, Faraday was self-educated and became an apprentice bookbinder at age 14 before turning to science. His experiments in electricity and magnetism made electricity a routine tool in science and led to both the electric motor and the electric generator. He discovered the phenomenon of electrolysis and laid the foundations of electrochemistry. In fact, most of the specialized terms introduced in this chapter (electrode, anode, cathode, and so forth) are due to Faraday. In addition, he discovered benzene and invented the system of oxidation state numbers that we use today. Faraday is probably best known for “The Chemical History of a Candle,” a series of public lectures on the chemistry and physics of flames. The maximum amount of work that can be produced by an electrochemical cell (\(w_{max}\)) is equal to the product of the cell potential (\(E^°_{cell}\)) and the total charge transferred during the reaction (\(nF\)): \[ w_{max} = nFE_{cell} \label{20.5.3} \] Work is expressed as a negative number because work is being done by a system (an electrochemical cell with a positive potential) on its surroundings. The change in free energy (\(\Delta{G}\)) is also a measure of the maximum amount of work that can be performed during a chemical process (\(ΔG = w_{max}\)). Consequently, there must be a relationship between the potential of an electrochemical cell and \(\Delta{G}\); this relationship is as follows: \[\Delta{G} = −nFE_{cell} \label{20.5.4} \] A spontaneous redox reaction is therefore characterized by a negative value of \(\Delta{G}\) and a positive value of \(E^°_{cell}\), consistent with our earlier discussions. When both reactants and products are in their standard states, the relationship between ΔG° and \(E^°_{cell}\) is as follows: \[\Delta{G^°} = −nFE^°_{cell} \label{20.5.5} \] A spontaneous redox reaction is characterized by a negative value of ΔG°, which corresponds to a positive value of E° . Suppose you want to prepare elemental bromine from bromide using the dichromate ion as an oxidant. Using the data in , calculate the free-energy change (ΔG°) for this redox reaction under standard conditions. Is the reaction spontaneous? redox reaction \(ΔG^o\) for the reaction and spontaneity As always, the first step is to write the relevant half-reactions and use them to obtain the overall reaction and the magnitude of \(E^o\). From , we can find the reduction and oxidation half-reactions and corresponding \(E^o\) values: \[\begin{align*} & \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \\ & \textrm{anode:} &\quad & \mathrm{2Br^{-}(aq)} \rightarrow \mathrm{Br_2(aq)} +\mathrm{2e^-} &\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \end{align*} \nonumber \] To obtain the overall balanced chemical equation, we must multiply both sides of the oxidation half-reaction by 3 to obtain the same number of electrons as in the reduction half-reaction, remembering that the magnitude of \(E^o\) is not affected: \[\begin{align*} & \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \\[4pt] & \textrm{anode:} &\quad & \mathrm{6Br^{-}(aq)} \rightarrow \mathrm{3Br_2(aq)} +\mathrm{6e^-} &\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \\[4pt] \hline & \textrm{overall:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{6Br^{-}(aq)} + \mathrm{14H^+(aq)} \rightarrow \mathrm{2Cr^{3+}(aq)} + \mathrm{3Br_2(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cell}} =\textrm{0.27 V} \end{align*} \nonumber \] We can now calculate ΔG° using Equation \(\ref{20.5.5}\). Because six electrons are transferred in the overall reaction, the value of \(n\) is 6: \[\begin{align*}\Delta G^\circ &=-(n)(F)(E^\circ_{\textrm{cell}}) \\[4pt] & =-(\textrm{6 mole})[96,485\;\mathrm{J/(V\cdot mol})(\textrm{0.27 V})] \\& =-15.6 \times 10^4\textrm{ J} \\ & =-156\;\mathrm{kJ/mol\;Cr_2O_7^{2-}} \end{align*} \nonumber \] Thus \(ΔG^o\) is −168 kJ/mol for the reaction as written, and the reaction is spontaneous. Use the data in to calculate \(ΔG^o\) for the reduction of ferric ion by iodide: \[\ce{2Fe^{3+}(aq) + 2I^{−}(aq) → 2Fe^{2+}(aq) + I2(s)}\nonumber \] Is the reaction spontaneous? −44 kJ/mol I ; yes     Although list several half-reactions, many more are known. When the standard potential for a half-reaction is not available, we can use relationships between standard potentials and free energy to obtain the potential of any other half-reaction that can be written as the sum of two or more half-reactions whose standard potentials are available. For example, the potential for the reduction of \(\ce{Fe^{3+}(aq)}\) to \(\ce{Fe(s)}\) is not listed in the table, but two related reductions are given: \[\ce{Fe^{3+}(aq) + e^{−} -> Fe^{2+}(aq)} \;\;\;E^° = +0.77 V \label{20.5.6} \] \[\ce{Fe^{2+}(aq) + 2e^{−} -> Fe(s)} \;\;\;E^° = −0.45 V \label{20.5.7} \] Although the sum of these two half-reactions gives the desired half-reaction, we cannot simply add the potentials of two reductive half-reactions to obtain the potential of a third reductive half-reaction because \(E^o\) is not a state function. However, because \(ΔG^o\) is a state function, the sum of the \(ΔG^o\) values for the individual reactions gives us \(ΔG^o\) for the overall reaction, which is proportional to both the potential and the number of electrons (\(n\)) transferred. To obtain the value of \(E^o\) for the overall half-reaction, we first must add the values of \(ΔG^o (= −nFE^o)\) for each individual half-reaction to obtain \(ΔG^o\) for the overall half-reaction: \[\begin{align*}\ce{Fe^{3+}(aq)} + \ce{e^-} &\rightarrow \ce\mathrm{Fe^{2+}(aq)} &\quad \Delta G^\circ &=-(1)(F)(\textrm{0.77 V})\\[4pt] \ce{Fe^{2+}(aq)}+\ce{2e^-} &\rightarrow\ce{Fe(s)} &\quad\Delta G^\circ &=-(2)(F)(-\textrm{0.45 V})\\[4pt] \ce{Fe^{3+}(aq)}+\ce{3e^-} &\rightarrow \ce{Fe(s)} &\quad\Delta G^\circ & =[-(1)(F)(\textrm{0.77 V})]+[-(2)(F)(-\textrm{0.45 V})] \end{align*} \nonumber \] Solving the last expression for ΔG° for the overall half-reaction, \[\Delta{G^°} = F[(−0.77 V) + (−2)(−0.45 V)] = F(0.13 V) \label{20.5.9} \] Three electrons (\(n = 3\)) are transferred in the overall reaction, so substituting into Equation \(\ref{20.5.5}\) and solving for \(E^o\) gives the following: \[\begin{align*}\Delta G^\circ & =-nFE^\circ_{\textrm{cell}} \\[4pt] F(\textrm{0.13 V}) & =-(3)(F)(E^\circ_{\textrm{cell}}) \\[4pt] E^\circ & =-\dfrac{0.13\textrm{ V}}{3}=-0.043\textrm{ V}\end{align*} \nonumber \] This value of \(E^o\) is very different from the value that is obtained by simply adding the potentials for the two half-reactions (0.32 V) and even has the opposite sign. Values of \(E^o\) for half-reactions cannot be added to give \(E^o\) for the sum of the half-reactions; only values of \(ΔG^o = −nFE^°_{cell}\) for half-reactions can be added. We can use the relationship between \(\Delta{G^°}\) and the equilibrium constant \(K\), to obtain a relationship between \(E^°_{cell}\) and \(K\). Recall that for a general reaction of the type \(aA + bB \rightarrow cC + dD\), the standard free-energy change and the equilibrium constant are related by the following equation: \[\Delta{G°} = −RT \ln K \label{20.5.10} \] Given the relationship between the standard free-energy change and the standard cell potential (Equation \(\ref{20.5.5}\)), we can write \[−nFE^°_{cell} = −RT \ln K \label{20.5.12} \] Rearranging this equation, \[E^\circ_{\textrm{cell}}= \left( \dfrac{RT}{nF} \right) \ln K \label{20.5.12B} \] For \(T = 298\, K\), Equation \(\ref{20.5.12B}\) can be simplified as follows: \[ \begin{align} E^\circ_{\textrm{cell}} &=\left(\dfrac{RT}{nF}\right)\ln K \\[4pt] &=\left[ \dfrac{[8.314\;\mathrm{J/(mol\cdot K})(\textrm{298 K})]}{n[96,485\;\mathrm{J/(V\cdot mol)}]}\right]2.303 \log K \\[4pt] &=\left(\dfrac{\textrm{0.0592 V}}{n}\right)\log K \label{20.5.13} \end{align} \] Thus \(E^°_{cell}\) is directly proportional to the logarithm of the equilibrium constant. This means that large equilibrium constants correspond to large positive values of \(E^°_{cell}\) and vice versa. Use the data in to calculate the equilibrium constant for the reaction of metallic lead with PbO in the presence of sulfate ions to give PbSO under standard conditions. (This reaction occurs when a car battery is discharged.) Report your answer to two significant figures. redox reaction \(K\) The relevant half-reactions and potentials from are as follows: \[\begin{align*} & \textrm {cathode:} & & \mathrm{PbO_2(s)}+\mathrm{SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}+\mathrm{2e^-}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2H_2O(l)} & & E^\circ_\textrm{cathode}=\textrm{1.69 V} \\[4pt] & \textrm{anode:} & & \mathrm{Pb(s)}+\mathrm{SO_4^{2-}(aq)}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2e^-} & & E^\circ_\textrm{anode}=-\textrm{0.36 V} \\[4pt] \hline & \textrm {overall:} & & \mathrm{Pb(s)}+\mathrm{PbO_2(s)}+\mathrm{2SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}\rightarrow\mathrm{2PbSO_4(s)}+\mathrm{2H_2O(l)} & & E^\circ_\textrm{cell}=\textrm{2.05 V} \end{align*} \nonumber \] Two electrons are transferred in the overall reaction, so \(n = 2\). Solving Equation \(\ref{20.5.13}\) for log K and inserting the values of \(n\) and \(E^o\), \[\begin{align*}\log K & =\dfrac{nE^\circ}{\textrm{0.0591 V}}=\dfrac{2(\textrm{2.05 V})}{\textrm{0.0591 V}}=69.37 \\[4pt] K & =2.3\times10^{69}\end{align*} \nonumber \] Thus the equilibrium lies far to the right, favoring a discharged battery (as anyone who has ever tried unsuccessfully to start a car after letting it sit for a long time will know). Use the data in to calculate the equilibrium constant for the reaction of \(\ce{Sn^{2+}(aq)}\) with oxygen to produce \(\ce{Sn^{4+}(aq)}\) and water under standard conditions. Report your answer to two significant figures. The reaction is as follows: \[\ce{2Sn^{2+}(aq) + O2(g) + 4H^{+}(aq) <=> 2Sn^{4+}(aq) + 2H2O(l)} \nonumber \] \(5.7 \times 10^{72}\) Figure \(\Page {1}\) summarizes the relationships that we have developed based on properties of the system—that is, based on the equilibrium constant, standard free-energy change, and standard cell potential—and the criteria for spontaneity (ΔG° < 0). Unfortunately, these criteria apply only to systems in which all reactants and products are present in their standard states, a situation that is seldom encountered in the real world. A more generally useful relationship between cell potential and reactant and product concentrations, as we are about to see, uses the relationship between \(\Delta{G}\) and the reaction quotient \(Q\).     A coulomb (C) relates electrical potential, expressed in volts, and energy, expressed in joules. The current generated from a redox reaction is measured in amperes (A), where 1 A is defined as the flow of 1 C/s past a given point. The faraday (F) is Avogadro’s number multiplied by the charge on an electron and corresponds to the charge on 1 mol of electrons. The product of the cell potential and the total charge is the maximum amount of energy available to do work, which is related to the change in free energy that occurs during the chemical process. Adding together the ΔG values for the half-reactions gives ΔG for the overall reaction, which is proportional to both the potential and the number of electrons (n) transferred. Spontaneous redox reactions have a negative ΔG and therefore a positive E . Because the equilibrium constant K is related to ΔG, E° and K are also related. Large equilibrium constants correspond to large positive values of E°.

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This page looks at the manufacture of arenes such as benzene and methylbenzene (toluene) by the catalytic reforming of fractions from petroleum (crude oil). Reforming takes straight chain hydrocarbons in the C to C range from the gasoline or naphtha fractions and rearranges them into compounds containing benzene rings. Hydrogen is produced as a by-product of the reactions. For example, hexane, C H , loses hydrogen and turns into benzene. As long as you draw the hexane bent into a circle, it is easy to see what is happening. Similarly, methylbenzene (toluene) is made from heptane: Methylbenzene is much less commercially valuable than benzene. The methyl group can be removed from the ring by a process known as "dealkylation". The methylbenzene is mixed with hydrogen at a temperature of between 550 and 650°C, and a pressure of between 30 and 50 atmospheres, with a mixture of silicon dioxide and aluminium oxide as catalyst. Jim Clark ( )

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A flat tire is not very useful. It does not cushion the rim of the wheel and creates a very uncomfortable ride. When air is added to the tire, the pressure increases as more molecules of gas are forced into the rigid tire. The amount of air that should be put into a tire depends on the pressure rating for that tire. Too little pressure and the tire will not hold its shape. Too much pressure and the tire could burst. You have learned about Avogadro's hypothesis: equal volumes of any gas at the same temperature and pressure contain the same number of molecules. It follows that the volume of a gas is directly proportional to the number of moles of gas present in the sample. states that the volume of a gas is directly proportional to the number of moles of gas, when the temperature and pressure are held constant. The mathematical expression of Avogadro's Law is: \[V = k \times n \: \: \: \text{and} \: \: \: \frac{V_1}{n_1} = \frac{V_2}{n_2}\nonumber \] (Where \(n\) is the number of moles of gas and \(k\) is a constant). Avogadro's Law is in evidence whenever you blow up a balloon. The volume of the balloon increases as you add moles of gas to the balloon by blowing it up. If the container holding the gas is rigid rather than flexible, pressure can be substituted for volume in Avogadro's Law. Adding gas to a rigid container makes the pressure increase. A balloon has been filled to a volume of \(1.90 \: \text{L}\) with \(0.0920 \: \text{mol}\) of helium gas. If \(0.0210 \: \text{mol}\) of additional helium is added to the balloon while the temperature and pressure are held constant, what is the new volume of the balloon? . Note that the final number of moles has to be calculated by adding the original number of moles to the moles of added helium. Use Avogadro's Law to solve for the final volume. First, rearrange the equation algebraically to solve for \(V_2\). \[V_2 = \frac{V_1 \times n_2}{n_1}\nonumber \] Now substitute the known quantities into the equation and solve. \[V_2 = \frac{1.90 \: \text{L} \times 0.1130 \: \text{mol}}{0.0920 \: \text{mol}} = 2.33 \: \text{L}\nonumber \] Since a relatively small amount of additional helium was added to the balloon, its volume increases slightly.

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\( \newcommand{\tx}[1]{\text{#1}}      % text in math mode\)  \( \newcommand{\subs}[1]{_{\text{#1}}} % subscript text\)  \( \newcommand{\sups}[1]{^{\text{#1}}} % superscript text\)  \( \newcommand{\st}{^\circ}            % standard state symbol\)  \( \newcommand{\id}{^{\text{id}}}      % ideal\)  \( \newcommand{\rf}{^{\text{ref}}}     % reference state\)  \( \newcommand{\units}[1]{\mbox{$\thinspace$#1}}\)  \( \newcommand{\K}{\units{K}}  % kelvins\)  \( \newcommand{\degC}{^\circ\text{C}} % degrees Celsius\)  \( \newcommand{\br}{\units{bar}}  % bar (\bar is already defined)\)  \( \newcommand{\Pa}{\units{Pa}}\)  \( \newcommand{\mol}{\units{mol}}  % mole\)  \( \newcommand{\V}{\units{V}}  % volts\)  \( \newcommand{\timesten}[1]{\mbox{$\,\times\,10^{#1}$}}\)  \( \newcommand{\per}{^{-1}}  % minus one power\)  \( \newcommand{\m}{_{\text{m}}}  % subscript m for molar quantity\)  \( \newcommand{\CVm}{C_{V,\text{m}}} % molar heat capacity at const.V\)  \( \newcommand{\Cpm}{C_{p,\text{m}}} % molar heat capacity at const.p\)  \( \newcommand{\kT}{\kappa_T} % isothermal compressibility\)  \( \newcommand{\A}{_{\text{A}}}  % subscript A for solvent or state A\)  \( \newcommand{\B}{_{\text{B}}}  % subscript B for solute or state B\)  \( \newcommand{\bd}{_{\text{b}}}  % subscript b for boundary or boiling point\)  \( \newcommand{\C}{_{\text{C}}}  % subscript C\)  \( \newcommand{\f}{_{\text{f}}}  % subscript f for freezing point\)  \( \newcommand{\mA}{_{\text{m},\text{A}}} % subscript m,A (m=molar)\)  \( \newcommand{\mB}{_{\text{m},\text{B}}} % subscript m,B (m=molar)\)  \( \newcommand{\mi}{_{\text{m},i}}        % subscript m,i (m=molar)\)  \( \newcommand{\fA}{_{\text{f},\text{A}}} % subscript f,A (for fr. pt.)\)  \( \newcommand{\fB}{_{\text{f},\text{B}}} % subscript f,B (for fr. pt.)\)  \( \newcommand{\xbB}{_{x,\text{B}}}       % x basis, B\)  \( \newcommand{\xbC}{_{x,\text{C}}}       % x basis, C\)  \( \newcommand{\cbB}{_{c,\text{B}}}       % c basis, B\)  \( \newcommand{\mbB}{_{m,\text{B}}}       % m basis, B\)  \( \newcommand{\kHi}{k_{\text{H},i}}      % Henry's law constant, x basis, i\)  \( \newcommand{\kHB}{k_{\text{H,B}}}      % Henry's law constant, x basis, B\)  \( \newcommand{\arrow}{\,\rightarrow\,} % right arrow with extra spaces\)  \( \newcommand{\arrows}{\,\rightleftharpoons\,} % double arrows with extra spaces\)  \( \newcommand{\ra}{\rightarrow} % right arrow (can be used in text mode)\)  \( \newcommand{\eq}{\subs{eq}} % equilibrium state\)  \( \newcommand{\onehalf}{\textstyle\frac{1}{2}\D} % small 1/2 for display equation\)  \( \newcommand{\sys}{\subs{sys}} % system property\)  \( \newcommand{\sur}{\sups{sur}} % surroundings\)  \( \renewcommand{\in}{\sups{int}} % internal\)  \( \newcommand{\lab}{\subs{lab}} % lab frame\)  \( \newcommand{\cm}{\subs{cm}} % center of mass\)  \( \newcommand{\rev}{\subs{rev}} % reversible\)  \( \newcommand{\irr}{\subs{irr}} % irreversible\)  \( \newcommand{\fric}{\subs{fric}} % friction\)  \( \newcommand{\diss}{\subs{diss}} % dissipation\)  \( \newcommand{\el}{\subs{el}} % electrical\)  \( \newcommand{\cell}{\subs{cell}} % cell\)  \( \newcommand{\As}{A\subs{s}} % surface area\)  \( \newcommand{\E}{^\mathsf{E}} % excess quantity (superscript)\)  \( \newcommand{\allni}{\{n_i \}} % set of all n_i\)  \( \newcommand{\sol}{\hspace{-.1em}\tx{(sol)}}\)  \( \newcommand{\solmB}{\tx{(sol,$\,$$m\B$)}}\)  \( \newcommand{\dil}{\tx{(dil)}}\)  \( \newcommand{\sln}{\tx{(sln)}}\)  \( \newcommand{\mix}{\tx{(mix)}}\)  \( \newcommand{\rxn}{\tx{(rxn)}}\)  \( \newcommand{\expt}{\tx{(expt)}}\)  \( \newcommand{\solid}{\tx{(s)}}\)  \( \newcommand{\liquid}{\tx{(l)}}\)  \( \newcommand{\gas}{\tx{(g)}}\)  \( \newcommand{\pha}{\alpha}        % phase alpha\)  \( \newcommand{\phb}{\beta}         % phase beta\)  \( \newcommand{\phg}{\gamma}        % phase gamma\)  \( \newcommand{\aph}{^{\alpha}}     % alpha phase superscript\)  \( \newcommand{\bph}{^{\beta}}      % beta phase superscript\)  \( \newcommand{\gph}{^{\gamma}}     % gamma phase superscript\)  \( \newcommand{\aphp}{^{\alpha'}}   % alpha prime phase superscript\)  \( \newcommand{\bphp}{^{\beta'}}    % beta prime phase superscript\)  \( \newcommand{\gphp}{^{\gamma'}}   % gamma prime phase superscript\)  \( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\)  \( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\)  \( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\)  \( \newcommand{\dif}{\mathop{}\!\mathrm{d}}   % roman d in math mode, preceded by space\)  \( \newcommand{\Dif}{\mathop{}\!\mathrm{D}}   % roman D in math mode, preceded by space\)  \( \newcommand{\df}{\dif\hspace{0.05em} f} % df\)  \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)  \( \newcommand{\dq}{\dBar q} % heat differential\)  \( \newcommand{\dw}{\dBar w} % work differential\)  \( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)  \( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\)  \( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\)  \( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\)  \( \newcommand{\Del}{\Delta}\)  \( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\)  \( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\)  \( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\)  \( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\)  \( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\)  \( \newcommand{\dotprod}{\small\bullet}\)  \( \newcommand{\fug}{f} % fugacity\)  \( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)  \( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)  \( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)  \( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\)  \( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\)  \( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\) \( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\)  \( \newcommand{\D}{\displaystyle} % for a line in built-up\)  \( \newcommand{\s}{\smash[b]} % use in equations with conditions of validity\)  \( \newcommand{\cond}[1]{\\[-2.5pt]{}\tag*{#1}}\)  \( \newcommand{\nextcond}[1]{\\[-5pt]{}\tag*{#1}}\)  \( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}}     % gas constant value\)  \( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \) \( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \) \( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \)  An is a system in which passage of an electric current through an electrical circuit is linked to an internal cell reaction. A , or voltaic cell, is an electrochemical cell that, when isolated, has an electric potential difference between its terminals; the cell is said to be a . The cell reaction in a galvanic cell differs in a fundamental way from the same reaction (i.e., one with the same reaction equation) taking place in a reaction vessel that is not part of an electrical circuit. In the reaction vessel, the reactants and products are in the same phase or in phases in contact with one another, and the reaction advances in the spontaneous direction until reaction equilibrium is reached. This reaction is the . The galvanic cell, in contrast, is arranged with the reactants physically separated from one another so that the cell reaction can advance only when an electric current passes through the cell. If there is no current, the cell reaction is constrained from taking place. When the electrical circuit is open and the cell is isolated from its surroundings, a state of thermal, mechanical, and transfer equilibrium is rapidly reached. In this state of or , however, reaction equilibrium is not necessarily present—that is, if the reactants and products were moved to a reaction vessel at the same activities, there might be spontaneous advancement of the reaction. As will be shown, measurements of the cell potential of a galvanic cell are capable of yielding precise values of molar reaction quantities of the cell reaction and thermodynamic equilibrium constants, and of mean ionic activity coefficients in electrolyte solutions.

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/16%3A_Appendix/16.16%3A_Countercurrent_Separations

In 1949, Lyman Craig introduced an improved method for separating analytes with similar distribution ratios [Craig, L. C. , , 519–534]. The technique, which is known as a countercurrent liquid–liquid extraction, is outlined in Figure 16.16.1 and discussed in detail below. In contrast to a sequential liquid–liquid extraction, in which we repeatedly extract the sample containing the analyte, a countercurrent extraction uses a serial extraction of both the sample and the extracting phases. Although countercurrent separations are no longer common—chromatographic separations are far more efficient in terms of resolution, time, and ease of use—the theory behind a countercurrent extraction remains useful as an introduction to the theory of chromatographic separations. To track the progress of a countercurrent liquid-liquid extraction we need to adopt a labeling convention. As shown in Figure 16.16.1 , in each step of a countercurrent extraction we first complete the extraction and then transfer the upper phase to a new tube that contains a portion of the fresh lower phase. Steps are labeled sequentially beginning with zero. Extractions take place in a series of tubes that also are labeled sequentially, starting with zero. The upper and lower phases in each tube are identified by a letter and number, with the letters U and L representing, respectively, the upper phase and the lower phase, and the number indicating the step in the countercurrent extraction in which the phase was first introduced. For example, U is the upper phase introduced at step 0 (during the first extraction), and L is the lower phase introduced at step 2 (during the third extraction). Finally, the partitioning of analyte in any extraction tube results in a fraction remaining in the upper phase, and a fraction remaining in the lower phase. Values of are calculated using Equation \ref{16.1}, which is identical to in Chapter 7. \[(q_\text{aq})_1 = \frac {(\text{mol aq})_1} {(\text{mol aq})_0} = \frac {V_\text{aq}} {D V_\text{org} + V_\text{aq}} \label{16.1}\] The fraction , of course is equal to 1 – . Typically and are equal in a countercurrent extraction, although this is not a requirement. Let’s assume that the analyte we wish to isolate is present in an aqueous phase of 1 M HCl, and that the organic phase is benzene. Because benzene has the smaller density, it is the upper phase, and 1 M HCl is the lower phase. To begin the countercurrent extraction we place the aqueous sample that contains the analyte in tube 0 along with an equal volume of benzene. As shown in Figure \(\Page {1}\text{a}\), before the extraction all the analyte is present in phase L . When the extraction is complete, as shown in Figure \(\Page {1}\text{b}\), a fraction of the analyte is present in phase U , and a fraction is in phase L . This completes step 0 of the countercurrent extraction. If we stop here, there is no difference between a simple liquid–liquid extraction and a countercurrent extraction. After completing step 0, we remove phase U and add a fresh portion of benzene, U , to tube 0 (see Figure \(\Page {1}\text{c}\)). This, too, is identical to a simple liquid-liquid extraction. Here is where the power of the countercurrent extraction begins—instead of setting aside the phase U , we place it in tube 1 along with a portion of analyte-free aqueous 1 M HCl as phase L (see Figure \(\Page {1}\text{c}\)). Tube 0 now contains a fraction of the analyte, and tube 1 contains a fraction of the analyte. Completing the extraction in tube 0 results in a fraction of its contents remaining in the upper phase, and a fraction remaining in the lower phase. Thus, phases U and L now contain, respectively, fractions and of the original amount of analyte. Following the same logic, it is easy to show that the phases U and L in tube 1 contain, respectively, fractions and of analyte. This completes step 1 of the extraction (see Figure \(\Page {1}\text{d}\)). As shown in the remainder of Figure 16.16.1 , the countercurrent extraction continues with this cycle of phase transfers and extractions. In a countercurrent liquid–liquid extraction, the lower phase in each tube remains in place, and the upper phase moves from tube 0 to successively higher numbered tubes. We recognize this difference in the movement of the two phases by referring to the lower phase as a stationary phase and the upper phase as a mobile phase. With each transfer some of the analyte in tube moves to tube \(r + 1\), while a portion of the analyte in tube \(r - 1\) moves to tube . Analyte introduced at tube 0 moves with the mobile phase, but at a rate that is slower than the mobile phase because, at each step, a portion of the analyte transfers into the stationary phase. An analyte that preferentially extracts into the stationary phase spends proportionally less time in the mobile phase and moves at a slower rate. As the number of steps increases, analytes with different values of eventually separate into completely different sets of extraction tubes. We can judge the effectiveness of a countercurrent extraction using a histogram that shows the fraction of analyte present in each tube. To determine the total amount of analyte in an extraction tube we add together the fraction of analyte present in the tube’s upper and lower phases following each transfer. For example, at the beginning of step 3 (see Figure \(\Page {1}\text{g}\)) the upper and lower phases of tube 1 contain fractions and 2 of the analyte, respectively; thus, the total fraction of analyte in the tube is 3 . Table 16.16.1 summarizes this for the steps outlined in Figure 16.16.1 . A typical histogram, calculated assuming distribution ratios of 5.0 for analyte A and 0.5 for analyte B, is shown in Figure 16.16.2 . Although four steps is not enough to separate the analytes in this instance, it is clear that if we extend the countercurrent extraction to additional tubes, we will eventually separate the analytes. Figure 16.16.1 and Table 16.16.1 show how an analyte’s distribution changes during the first four steps of a countercurrent extraction. Now we consider how we can generalize these results to calculate the amount of analyte in any tube, at any step during the extraction. You may recognize the pattern of entries in Table 16.16.1 as following the binomial distribution \[f(r, n) = \frac {n!} {(n - r)! r!} p^{r} q^{n - r} \label{16.2}\] where ( , ) is the fraction of analyte present in tube at step of the countercurrent extraction, with the upper phase containing a fraction \(p \times f(r, n)\) of analyte and the lower phase containing a fraction \(q \times f(r, n)\) of the analyte. The countercurrent extraction shown in Figure 16.16.2 is carried out through step 30. Calculate the fraction of analytes A and B in tubes 5, 10, 15, 20, 25, and 30. To calculate the fraction, , for each analyte in the lower phase we use Equation \ref{16.1}. Because the volumes of the lower and upper phases are equal, we get \[q_\text{A} = \frac {1} {D_\text{A} + 1} = \frac {1} {5 + 1} = 0.167 \quad \quad q_\text{B} = \frac {1} {D_\text{B} + 1} = \frac {1} {0.5 + 1} = 0.667 \nonumber\] Because we know that \(p + q = 1 \), we also know that is 0.833 and that is 0.333. For analyte A, the fraction in tubes 5, 10, 15, 20, 25, and 30 after the 30th step are \[f(5,30) = \frac {30!} {(30 - 5)! 5!} (0.833)^{5} (0.167)^{30 - 5} = 2.1 \times 10^{-15} \approx 0 \nonumber\] \[f(10,30) = \frac {30!} {(30 - 10)! 10!} (0.833)^{10} (0.167)^{30 - 10} = 1.4 \times 10^{-9} \approx 0 \nonumber\] \[f(15,30) = \frac {30!} {(30 - 15)! 5!} (0.833)^{15} (0.167)^{30 - 15} = 2.2 \times 10^{-5} \approx 0 \nonumber\] \[f(20,30) = \frac {30!} {(30 - 20)! 20!} (0.833)^{20} (0.167)^{30 - 20} = 0.013 \nonumber\] \[f(25,30) = \frac {30!} {(30 - 25)! 25!} (0.833)^{25} (0.167)^{30 - 25} = 0.192 \nonumber\] \[f(30,30) = \frac {30!} {(30 - 30)! 30!} (0.833)^{30} (0.167)^{30 - 30} = 0.004 \nonumber\] The fraction of analyte B in tubes 5, 10, 15, 20, 25, and 30 is calculated in the same way, yielding respective values of 0.023, 0.153, 0.025, 0, 0, and 0. Figure 16.16.3 , which provides the complete histogram for the distribution of analytes A and B, shows that 30 steps is sufficient to separate the two analytes. Constructing a histogram using Equation \ref{16.2} is tedious, particularly when the number of steps is large. Because the fraction of analyte in most tubes is approximately zero, we can simplify the histogram’s construction by solving Equation \ref{16.2} only for those tubes containing an amount of analyte that exceeds a threshold value. For a binomial distribution, we can use the mean and standard deviation to determine which tubes contain a significant fraction of analyte. The properties of a binomial distribution were covered in Chapter 4, with the mean, \(\mu\), and the standard deviation, \(\sigma\), given as \[\mu = np \nonumber\] \[\sigma = \sqrt{np(1 - p)} = \sqrt{npq} \nonumber\] Furthermore, if both and are greater than 5, then a binomial distribution closely approximates a normal distribution and we can use the properties of a normal distribution to determine the location of the analyte and its recovery [see Mark, H.; Workman, J. , , 55–56]. Two analytes, A and B, with distribution ratios of 9 and 4, respectively, are separated using a countercurrent extraction in which the volumes of the upper and lower phases are equal. After 100 steps determine the 99% confidence interval for the location of each analyte. The fraction, , of each analyte that remains in the lower phase is calculated using Equation \ref{16.1}. Because the volumes of the lower and upper phases are equal, we find that \[q_\text{A} = \frac {1} {D_\text{A} + 1} = \frac {1} {9 + 1} = 0.10 \quad \quad q_\text{B} = \frac {1} {D_\text{B} + 1} = \frac {1} {4 + 1} = 0.20 \nonumber\] Because we know that \(p + q = 1 \), we also know that is 0.90 and is 0.80. After 100 steps, the mean and the standard deviation for the distribution of analytes A and B are \[\mu_\text{A} = np_\text{A} = (100)(0.90) = 90 \text{ and } \sigma_\text{A} = \sqrt{np_\text{A}q_\text{A}} = \sqrt{(100)(0.90)(0.10)} = 3 \nonumber\] \[\mu_\text{B} = np_\text{B} = (100)(0.80) = 80 \text{ and } \sigma_\text{A} = \sqrt{np_\text{A}q_\text{A}} = \sqrt{(100)(0.80)(0.20)} = 4 \nonumber\] Given that , , , and are all greater than 5, we can assume that the distribution of analytes follows a normal distribution and that the confidence interval for the tubes containing each analyte is \[r = \mu \pm z \sigma \nonumber\] where is the tube’s number and the value of is determined by the desired significance level. For a 99% confidence interval the value of is 2.58 (see Appendix 4); thus, \[r_\text{A} = 90 \pm (2.58)(3) = 90 \pm 8 \nonumber\] \[r_\text{B} = 80 \pm (2.58)(4) = 80 \pm 10 \nonumber\] Because the two confidence intervals overlap, a complete separation of the two analytes is not possible using a 100 step countercurrent extraction. The complete distribution of the analytes is shown in Figure 16.16.4 . For the countercurrent extraction in Example 16.16.2 , calculate the recovery and the separation factor for analyte A if the contents of tubes 85–99 are pooled together. From Example 16.16.2 we know that after 100 steps of the countercurrent extraction, analyte A is normally distributed about tube 90 with a standard deviation of 3. To determine the fraction of analyte A in tubes 85–99, we use the single-sided normal distribution in Appendix 3 to determine the fraction of analyte in tubes 0–84, and in tube 100. The fraction of analyte A in tube 100 is determined by calculating the deviation \[z = \frac {r - \mu} {\sigma} = \frac {99 - 90} {3} = 3 \nonumber\] and using the table in to determine the corresponding fraction. For 3 this corresponds to 0.135% of analyte A. To determine the fraction of analyte A in tubes 0–84 we again calculate the deviation \[z = \frac {r - \mu} {\sigma} = \frac {84 - 90} {3} = -1.67 \nonumber\] From we find that 4.75% of analyte A is present in tubes 0–84. Analyte A’s recovery, therefore, is \[100\% - 4.75\% - 0.135\% \approx 95\% \nonumber\] To calculate the separation factor we determine the recovery of analyte B in tubes 85–99 using the same general approach as for analyte A, finding that approximately 89.4% of analyte B remains in tubes 0–84 and that essentially no analyte B is in tube 100. The recovery for B, therefore, is \[100\% - 89.4\% - 0\% \approx 10.6\% \nonumber\] and the separation factor is \[S_\text{B/A} = \frac {R_\text{A}} {R_\text{B}} = \frac {10.6} {95} = 0.112 \nonumber\]

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis

After you have completed Chapter 9, you should be able to Addition reactions not only dominate the chemistry of alkenes, they are also the major class of reaction you will encounter. This chapter discusses an important difference between (terminal) alkynes and alkenes, that is, the acidity of the former; it also addresses the problem of devising organic syntheses. Once you have completed this chapter you will have increased the number of organic reactions in your repertoire, and should be able to design much more elaborate multistep syntheses. As you work through Chapter 9, you should notice the many similarities among the reactions described here and those in Chapters 7 and 8.

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/16%3A_Appendix/16.05%3A_Critical_Values_for_F-Test

The following tables provide values for \(F(0.05, \nu_\text{num}, \nu_\text{denom})\) for one-tailed and for two-tailed -tests. To use these tables, we first decide whether the situation calls for a one-tailed or a two-tailed analysis and calculate \[F_\text{exp} = \frac {s_A^2} {s_B^2} \nonumber\] where \(S_A^2\) is greater than \(s_B^2\). Next, we compare exp to \(F(0.05, \nu_\text{num}, \nu_\text{denom})\) and reject the null hypothesis if \(F_\text{exp} > F(0.05, \nu_\text{num}, \nu_\text{denom})\). You may replace with \(\sigma\) if you know the population’s standard deviation.

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/16%3A_Appendix/16.02%3A_Propagation_of_Uncertainty

In Chapter 4 we considered the basic mathematical details of a propagation of uncertainty, limiting our treatment to the propagation of measurement error. This treatment is incomplete because it omits other sources of uncertainty that contribute to the overall uncertainty in our results. Consider, for example, , in which we determined the uncertainty in a standard solution of Cu prepared by dissolving a known mass of Cu wire with HNO , diluting to volume in a 500-mL volumetric flask, and then diluting a 1-mL portion of this stock solution to volume in a 250-mL volumetric flask. To calculate the overall uncertainty we included the uncertainty in weighing the sample and the uncertainty in using the volumetric glassware. We did not consider other sources of uncertainty, including the purity of the Cu wire, the effect of temperature on the volumetric glassware, and the repeatability of our measurements. In this appendix we take a more detailed look at the propagation of uncertainty, using the standardization of NaOH as an example. Because solid NaOH is an impure material, we cannot directly prepare a stock solution by weighing a sample of NaOH and diluting to volume. Instead, we determine the solution’s concentration through a process called a standardization. A fairly typical procedure is to use the NaOH solution to titrate a carefully weighed sample of previously dried potassium hydrogen phthalate, C H O K, which we will write here, in shorthand notation, as KHP. For example, after preparing a nominally 0.1 M solution of NaOH, we place an accurately weighed 0.4-g sample of dried KHP in the reaction vessel of an automated titrator and dissolve it in approximately 50 mL of water (the exact amount of water is not important). The automated titrator adds the NaOH to the KHP solution and records the pH as a function of the volume of NaOH. The resulting titration curve provides us with the volume of NaOH needed to reach the titration's endpoint. The example below is adapted from Ellison, S. L. R.; Rosslein, M.; Williams, A. EURACHEM/CITAC Guide: Quantifying Uncertainty in Analytical Measurement, 3nd Edition, 2012. See for further details about standardizations and see for further details about titrations The end point of the titration is the volume of NaOH that corresponds to the stoichiometric reaction between NaOH and KHP. \[\ce{NaOH}(aq) + \ce{C8H5O4K}(aq) \ce{->} \ce{C8H4O4^{2-}}(aq) + \ce{K+}(aq) + \ce{Na+}(aq) + \ce{H2O}(l) \nonumber\] Knowing the mass of KHP and the volume of NaOH needed to reach the endpoint, we use the following equation to calculate the molarity of the NaOH solution. \[C_\ce{NaOH} = \frac {1000 \times m_\ce{KHP} \times P_\ce{KHP}} {FW_\ce{KHP} \times V_\ce{NaOH}} \nonumber\] where is the concentration of NaOH (in mol KHP/L), is the mass of KHP taken (in g), is the purity of the KHP (where = 1 means the KHP is pure and has no impurities), is the molar mass of KHP (in g KHP/mol KHP), and is the volume of NaOH (in mL). The factor of 1000 simply converts the volume in mL to L. Although it seems straightforward, identifying sources of uncertainty requires care as it easy to overlook important sources of uncertainty. One approach is to use a cause-and-effect diagram, also known as an Ishikawa diagram—named for its inventor, Kaoru Ishikawa—or a fish bone diagram. To construct a cause-and-effect diagram, we first draw an arrow that points to the desired result; this is the diagram's trunk. We then add five main branch lines to the trunk, one for each of the four parameters that determine the concentration of NaOH ( , , , and ) and one for the method's repeatability, . Next we add additional branches to the main branch for each of these five factors, continuing until we account for all potential sources of uncertainty. Figure 16.2.1 shows the complete cause-and-effect diagram for this analysis. Before we continue, let's take a closer look at Figure 16.2.1 to make sure that we understand each branch of the diagram. To determine the mass of KHP, , we make two measurements: taring the balance and weighing the gross sample. Each of these measurements is subject to a calibration uncertainty. When we calibrate a balance, we essentially are creating a calibration curve of the balance's signal as a function of mass. Any calibration curve is subject to an uncertainty in the -intercept (bias) and an uncertainty in the slope (linearity). We can ignore the calibration bias because it contributes equally to both ( ) and ( ) , and because we determine the mass of KHP by difference. \[m_\ce{KHP} = \left( m_\ce{KHP} \right)_\text{gross} - \left( m_\ce{KHP} \right)_\text{tare} \nonumber\] The volume of NaOH, , at the end point has three sources of uncertainty. First, an automated titrator uses a piston to deliver NaOH to the reaction vessel, which means the volume of NaOH is subject to an uncertainty in the piston's calibration. Second, because a solution’s volume varies with temperature, there is an additional source of uncertainty due to any fluctuation in the ambient temperature during the analysis. Finally, there is a bias in the titration’s end point if the NaOH reacts with any species other than the KHP. Repeatability, , is a measure of how consistently we can repeat the analysis. Each instrument we use—the balance and the automated titrator—contributes to this uncertainty. In addition, our ability to consistently detect the end point also contributes to repeatability. Finally, there are no secondary factors that affect the uncertainty of the KHP's purity, , or its molar mass, . To complete a propagation of uncertainty we must express each measurement’s uncertainty in the same way, usually as a standard deviation. Measuring the standard deviation for each measurement requires time and is not always practical. Fortunately, most manufacture provides a tolerance range for glassware and instruments. A 100-mL volumetric glassware, for example, has a tolerance of \(\pm 0.1 \text{ mL}\) at a temperature of 20 C. We can convert a tolerance range to a standard deviation using one of the following three approaches. Figure 16.2.2 a shows a uniform distribution between the limits of \(\pm x\), in which each result between the limits is equally likely. A uniform distribution is the choice when the manufacturer provides a tolerance range without specifying a level of confidence and when there is no reason to believe that results near the center of the range are more likely than results at the ends of the range. For a uniform distribution the estimated standard deviation, , is \[s = \frac {x} {\sqrt{3}} \nonumber\] This is the most conservative estimate of uncertainty as it gives the largest estimate for the standard deviation. Figure 16.2.2 b shows a triangular distribution between the limits of \(\pm x\), in which the most likely result is at the center of the distribution, decreasing linearly toward each limit. A triangular distribution is the choice when the manufacturer provides a tolerance range without specifying a level of confidence and when there is a good reason to believe that results near the center of the range are more likely than results at the ends of the range. For a triangular distribution the estimated standard deviation, , is \[s = \frac {x} {\sqrt{6}} \nonumber\] This is a less conservative estimate of uncertainty as, for any value of , the standard deviation is smaller than that for a uniform distribution. Figure 16.2.2 c shows a normal distribution that extends, as it must, beyond the limits of \(\pm x\), and which is centered at the mid-point between – and + . A normal distribution is the choice when we know the confidence interval for the range. For a normal distribution the estimated standard deviation, , is \[s = \frac {x} {z} \nonumber\] where is 1.96 for a 95% confidence interval and 3.00 for a 99.7% confidence interval. Now we are ready to return to our example and determine the uncertainty for the standardization of NaOH. First we establish the uncertainty for each of the five primary sources—the mass of KHP, the volume of NaOH at the end point, the purity of the KHP, the molar mass for KHP, and the titration’s repeatability. Having established these, we can combine them to arrive at the final uncertainty. After drying the KHP, we store it in a sealed container to prevent it from readsorbing moisture. To find the mass of KHP we first weigh the container, obtaining a value of 60.5450 g, and then weigh the container after removing a portion of KHP, obtaining a value of 60.1562 g. The mass of KHP, therefore, is 60.5450 – 60.1562 = 0.3888 g, or 388.8 mg. To find the uncertainty in this mass we examine the balance’s calibration certificate, which indicates that its tolerance for linearity is \(\pm 0.15 \text{ mg}\). We will assume a uniform distribution because there is no reason to believe that any result within this range is more likely than any other result. Our estimate of the uncertainty for any single measurement of mass, ( ), is \[u(m) = \frac {0.15 \text{ mg}} {\sqrt{3}} = \pm 0.087 \text{ mg} \nonumber\] Because we determine the mass of KHP by subtracting the container’s final mass from its initial mass, the uncertainty in the mass of KHP ( ), is given by the following propagation of uncertainty. \[u(m_{\text{KHP}}) = \sqrt{\left( 0.087 \text{ mg} \right)^2 + \left( 0.087 \text{ mg} \right)^2} =\pm 0.12 \text{ mg} \nonumber\] After we place the sample of KHP in the automated titrator’s reaction vessel and dissolve the KHP with water, we complete the titration and find that it takes 18.64 mL of NaOH to reach the end point. To find the uncertainty in this volume we need to consider, as shown in Figure 16.2.1 , three sources of uncertainty: the automated titrator’s calibration, the ambient temperature, and any bias in determining the end point. To find the uncertainty from the automated titrator’s calibration we examine the instrument’s certificate, which indicates a range of \(\pm 0.03 \text{ mL}\) for a 20-mL piston. Because we expect that an effective manufacturing process is more likely to produce a piston that operates near the center of this range than at the extremes, we will assume a triangular distribution. Our estimate of the uncertainty due to the calibration, ( ) is \[u(V_\text{cal}) = \frac {0.03 \text{ mL}} {\sqrt{6}} = \pm 0.012 \text{ mL} \nonumber\] To determine the uncertainty due to the lack of temperature control, we draw on our prior work in the lab, which has established a temperature variation of \(\pm 3 \text{°C}\) with a confidence level of 95%. To find the uncertainty, we convert the temperature range to a range of volumes using water’s coefficient of expansion \[(2.1 \times 10^{-4} \text{°C}) \times (\pm 3 \text{°C}) \times 18.64 \text{ mL} = \pm 0.012 \text{ mL} \nonumber\] and then estimate the uncertainty due to temperature, ( ) as \[u(V_\text{temp}) = \frac {\pm 0.012 \text{ mL}} {1.96} = \pm 0.006 \text{ mL} \nonumber\] Titrations using NaOH are subject to a bias due to the adsorption of CO , which can react with OH , as shown here. \[\ce{CO2}(aq) + \ce{2OH-}(aq) \ce{->} \ce{CO3^{2-}}(aq) + \ce{H2O}(l) \nonumber\] If CO is present, the volume of NaOH at the end point includes both the NaOH that reacts with the KHP and the NaOH that reacts with CO . Rather than trying to estimate this bias, it is easier to bathe the reaction vessel in a stream of argon, which excludes CO from the automated titrator’s reaction vessel. Adding together the uncertainties for the piston’s calibration and the lab’s temperature gives the uncertainty in the uncertainty in the volume of NaOH, ( ) as \[u(V_\ce{NaOH}) = \sqrt{(0.012 \text{ mL})^2 + (0.006 \text{ mL})^2} = \pm 0.013 \text{ mL} \nonumber\] According to the manufacturer, the purity of KHP is \(100 \% \pm 0.05 \%\), or \(1.0 \pm 0.0005\). Assuming a rectangular distribution, we report the uncertainty, ( ) as \[u(P_\ce{KHP}) = \frac {\pm 0.0005} {\sqrt{3}} = \pm 0.00029 \nonumber\] The molar mass of C H O K is 204.2212 g/mol, based on the following atomic weights: 12.0107 for carbon, 1.00794 for hydrogen, 15.9994 for oxygen, and 39.0983 for potassium. Each of these atomic weights has an quoted uncertainty that we can convert to a standard uncertainty assuming a rectangular distribution, as shown here (the details of the calculations are left to you). quoted uncertainty (per atom) standard uncertainty (per atom) Adding together these uncertainties gives the uncertainty in the molar mass, ( ), as \[u(FW_\ce{KHP}) = \sqrt{(0.00368)^2 + (0.00020)^2 + (0.00068)^2 + (0.0.000058)^2} = \pm 0.0037 \text{ g/mL} \nonumber\] To estimate the uncertainty due to repeatability we complete five titrations, obtaining the following results for the concentration of NaOH: 0.1021 M, 0.1022 M, 0.1022 M, 0.1021 M, and 0.1021 M. The relative standard deviation, , for these titrations is \[s_{rel} = \frac {s} {\overline{X}} = \frac {5.48 \times 10^{-5}} {0.1021} = \pm 0.0005 \nonumber\] If we treat the ideal repeatability as 1.0, then the uncertainty due to repeatability, ( ), is the relative standard deviation, or, in this case, 0.0005. The table below summarizes the five primary sources of uncertainty. As described earlier, we calculate the concentration of NaOH we use the following equation, which is slightly modified to include a term for the titration’s repeatability, which, as described above, has a value of 1.0. \[C_\ce{NaOH} = \frac {1000 \times m_\ce{KHP} \times P_\ce{KHP}} {FW_\ce{KHP} \times V_\ce{NaOH}} \times R \nonumber\] Using the values from our table, we find that the concentration of NaOH is \[C_\ce{NaOH} = \frac {1000 \times 0.3888 \times 1.0} {204.2212 \times 18.64} \times 1.0 = 0.1021 \text{ M} \nonumber\] Because the calculation of includes only multiplication and division, the uncertainty in the concentration, ( ) is given by the following propagation of uncertainty. \[\frac {u(C_\ce{NaOH})} {C_\ce{NaOH}} = \frac {u(C_\ce{NaOH})} {0.1021} = \sqrt{\frac {(0.00012)^2} {(0.3888)^2} + \frac {(0.00029)^2} {(1.0)^2} + \frac {(0.0037)^2} {(204.2212)^2} + \frac {(0.013)^2} {(18.64)^2} + \frac {(0.0005)^2} {(1.0)^2}} \nonumber\] Solving for ( ) gives its value as \(\pm 0.00010 \text{ M}\), which is the final uncertainty for the analysis. Figure 16.2.3 shows the relative uncertainty in the concentration of NaOH and the relative uncertainties for each of the five contributions to the total uncertainty. Of the contributions, the most important is the volume of NaOH, and it is here to which we should focus our attention if we wish to improve the overall uncertainty for the standardization.

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/09%3A_Titrimetric_Methods/9.04%3A_Redox_Titrations

Analytical titrations using oxidation–reduction reactions were introduced shortly after the development of acid–base titrimetry. The earliest took advantage of chlorine’s oxidizing power. In 1787, Claude Berthollet introduced a method for the quantitative analysis of chlorine water (a mixture of Cl , HCl, and HOCl) based on its ability to oxidize indigo, a dye that is colorless in its oxidized state. In 1814, Joseph Gay-Lussac developed a similar method to determine chlorine in bleaching powder. In both methods the end point is a change in color. Before the equivalence point the solution is colorless due to the oxidation of indigo. After the equivalence point, however, unreacted indigo imparts a permanent color to the solution. The number of redox titrimetric methods increased in the mid-1800s with the introduction of \(\text{MnO}_4^-\), \(\text{Cr}_2\text{O}_7^{2-}\), and I as oxidizing titrants, and of Fe and \(\text{S}_2\text{O}_3^{2-}\) as reducing titrants. Even with the availability of these new titrants, redox titrimetry was slow to develop due to the lack of suitable indicators. A titrant can serve as its own indicator if its oxidized and its reduced forms differ significantly in color. For example, the intensely purple \(\text{MnO}_4^-\) ion serves as its own indicator since its reduced form, Mn , is almost colorless. Other titrants require a separate indicator. The first such indicator, diphenylamine, was introduced in the 1920s. Other redox indicators soon followed, increasing the applicability of redox titrimetry. To evaluate a redox titration we need to know the shape of its titration curve. In an acid–base titration or a complexation titration, the titration curve shows how the concentration of H O (as pH) or M (as pM) changes as we add titrant. For a redox titration it is convenient to monitor the titration reaction’s potential instead of the concentration of one species. You may recall from that the Nernst equation relates a solution’s potential to the concentrations of reactants and products that participate in the redox reaction. Consider, for example, a titration in which a titrand in a reduced state, , reacts with a titrant in an oxidized state, . \[A_{red} + B_{ox} \rightleftharpoons B_{red} + A_{ox} \nonumber\] where is the titrand’s oxidized form, is the titrant’s reduced form, and the stoichiometry between the two is 1:1. The reaction’s potential, , is the difference between the reduction potentials for each half-reaction. \[E_{rxn} = E_{B_{ox}/B_{red}} - E_{A_{ox}/A_{red}} \nonumber\] After each addition of titrant the reaction between the titrand and the titrant reaches a state of equilibrium. Because the potential at equilibrium is zero, the titrand’s and the titrant’s reduction potentials are identical. \[E_{B_{ox}/B_{red}} = E_{A_{ox}/A_{red}} \nonumber\] This is an important observation as it allows us to use either half-reaction to monitor the titration’s progress. Before the equivalence point the titration mixture consists of appreciable quantities of the titrand’s oxidized and reduced forms. The concentration of unreacted titrant, however, is very small. The potential, therefore, is easier to calculate if we use the Nernst equation for the titrand’s half-reaction \[E_{rxn} = E_{A_{ox}/A_{red}}^{\circ} - \frac{RT}{nF}\ln{\frac{[A_{red}]}{[A_{ox}]}} \nonumber\] After the equivalence point it is easier to calculate the potential using the Nernst equation for the titrant’s half-reaction. \[E_{rxn} = E_{B_{ox}/B_{red}}^{\circ} - \frac{RT}{nF}\ln{\frac{[B_{red}]}{[B_{ox}]}} \nonumber\] Although the Nernst equation is written in terms of the half-reaction’s standard state potential, a matrix-dependent often is used in its place. See for the standard state potentials and formal potentials for selected half-reactions. Let’s calculate the titration curve for the titration of 50.0 mL of 0.100 M Fe with 0.100 M Ce in a matrix of 1 M HClO . The reaction in this case is \[\text{Fe}^{2+}(aq) + \text{Ce}^{4+}(aq) \rightleftharpoons \text{Ce}^{3+}(aq) + \text{Fe}^{3+}(aq) \label{9.1}\] Because the equilibrium constant for reaction \ref{9.1} is very large—it is approximately \(6 \times 10^{15}\)—we may assume that the analyte and titrant react completely. In 1 M HClO , the formal potential for the reduction of Fe to Fe is +0.767 V, and the formal potential for the reduction of Ce to Ce is +1.70 V. The first task is to calculate the volume of Ce needed to reach the titration’s equivalence point. From the reaction’s stoichiometry we know that \[\text{mol Fe}^{2+} = M_\text{Fe}V_\text{Fe} = M_\text{Ce}V_\text{Ce} = \text{mol Ce}^{4+} \nonumber\] Solving for the volume of Ce gives the equivalence point volume as \[V_{eq} = V_\text{Ce} = \frac{M_\text{Fe}V_\text{Fe}}{M_\text{Ce}} = \frac{(0.100 \text{ M})(50.0 \text{ mL})}{(0.100 \text{ M})} = 50.0 \text{ mL} \nonumber\] Before the equivalence point, the concentration of unreacted Fe and the concentration of Fe are easy to calculate. For this reason we find the potential using the Nernst equation for the Fe /Fe half-reaction. \[E = +0.767 \text{ V} - 0.05916 \log{\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}} \label{9.2}\] For example, the concentrations of Fe and Fe after adding 10.0 mL of titrant are \[[\text{Fe}^{2+}] = \frac{(\text{mol Fe}^{2+})_\text{initial} - (\text{mol Ce}^{4+})_\text{added}}{\text{total volume}} = \frac{M_\text{Fe}V_\text{Fe} - M_\text{Ce}V_\text{Ce}}{V_\text{Fe} + V_\text{Ce}} \nonumber\] \[[\text{Fe}^{2+}] = \frac{(0.100 \text{ M})(50.0 \text{ mL}) - (0.100 \text{ M})(10.0 \text{ mL})}{50.0 \text{ mL} + 10.0 \text{ mL}} = 6.67 \times 10^{-2} \text{ M} \nonumber\] \[[\text{Fe}^{3+}] = \frac{(\text{mol Ce}^{4+})_\text{added}}{\text{total volume}} = \frac{M_\text{Ce}V_\text{Ce}}{V_\text{Fe} + V_\text{Ce}} \nonumber\] \[[\text{Fe}^{3+}] = \frac{(0.100 \text{ M})(10.0 \text{ mL})}{50.0 \text{ mL} + 10.0 \text{ mL}} = 1.67 \times 10^{-2} \text{ M} \nonumber\] Substituting these concentrations into Equation \ref{9.2} gives the potential as \[E = +0.767 \text{ V} - 0.05916 \log{\frac{6.67 \times 10^{-2}}{1.67 \times 10^{-2}}} = +0.731 \text{ V} \nonumber\] After the equivalence point, the concentration of Ce and the concentration of excess Ce are easy to calculate. For this reason we find the potential using the Nernst equation for the Ce /Ce half-reaction in a manner similar to that used above to calculate potentials before the equivalence point. \[E = +1.70 \text{ V} - 0.05916 \log{\frac{[\text{Ce}^{3+}]}{[\text{Ce}^{4+}]}} \label{9.3}\] For example, after adding 60.0 mL of titrant, the concentrations of Ce and Ce are \[[\text{Ce}^{3+}] = \frac{(\text{mol Fe}^{2+})_\text{initial}}{\text{total volume}} = \frac{M_\text{Fe}V_\text{Fe}}{V_\text{Fe}+V_\text{Ce}} \nonumber\] \[[\text{Ce}^{3+}] = \frac{(0.100 \text{ M})(50.0 \text{ mL})}{50.0 \text{ mL} + 60.0 \text{ mL}} = 4.55 \times 10^{-2} \text{ M} \nonumber\] \[[\text{Ce}^{4+}] = \frac{(\text{mol Ce}^{4+})_\text{added}-(\text{mol Fe}^{2+})_\text{initial}}{\text{total volume}} = \frac{M_\text{Ce}V_\text{Ce}-M_\text{Fe}V_\text{Fe}}{V_\text{Fe}+V_\text{Ce}} \nonumber\] \[[\text{Ce}^{4+}] = \frac{(0.100 \text{ M})(60.0 \text{ mL})-(0.100 \text{ M})(50.0 \text{ mL})}{50.0 \text{ mL} + 60.0 \text{ mL}} = 9.09 \times 10^{-3} \text{ M} \nonumber\] Substituting these concentrations into Equation \ref{9.3} gives a potential of \[E = +1.70 \text{ V} - 0.05916 \log{\frac{4.55 \times 10^{-2} \text{ M}}{9.09 \times 10^{-3} \text{ M}}} = +1.66 \text{ V} \nonumber\] At the titration’s equivalence point, the potential, , in Equation \ref{9.2} and Equation \ref{9.3} are identical. Adding the equations together to gives \[2E_{eq} = E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ} - 0.05916 \log{\frac{[\text{Fe}^{2+},\text{Ce}^{3+}]}{[\text{Fe}^{3+},\text{Ce}^{4+}]}} \nonumber\] Because [Fe ] = [Ce ] and [Ce ] = [Fe ] at the equivalence point, the log term has a value of zero and the equivalence point’s potential is \[E_{eq} = \frac{E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ}}{2} = \frac{0.767 \text{ V} + 1.70 \text{ V}}{2} = +1.23 \text{ V} \nonumber\] Additional results for this titration curve are shown in Table 9.4.1 and Figure 9.4.1 . Calculate the titration curve for the titration of 50.0 mL of 0.0500 M Sn with 0.100 M Tl . Both the titrand and the titrant are 1.0 M in HCl. The titration reaction is \[\text{Sn}^{2+}(aq) + \text{Tl}^{3+} \rightleftharpoons \text{Tl}^+(aq) + \text{Sn}^{4+}(aq) \nonumber\] The volume of Tl needed to reach the equivalence point is \[V_{eq} = V_\text{Tl} = \frac{M_\text{Sn}V_\text{Sn}}{M_\text{Tl}} = \frac{(0.050 \text{ M})(50.0 \text{ mL})}{(0.100 \text{ M})} = 25.0 \text{ mL} \nonumber\] Before the equivalence point, the concentration of unreacted Sn and the concentration of Sn are easy to calculate. For this reason we find the potential using the Nernst equation for the Sn /Sn half-reaction. For example, the concentrations of Sn and Sn after adding 10.0 mL of titrant are \[[\text{Sn}^{2+}] = \frac{(0.050 \text{ M})(50.0 \text{ mL}) - (0.100 \text{ M})(10.0 \text{ mL})}{50.0 \text{ mL} + 10.0 \text{ mL}} = 0.0250 \text{ M} \nonumber\] \[[\text{Sn}^{4+}] = \frac{(0.100 \text{ M})(10.0 \text{ mL})}{50.0 \text{ mL} + 10.0 \text{ mL}} = 0.0167 \text{ M} \nonumber\] and the potential is \[E = +0.139 \text{ V} - \frac{0.05916}{2} \log{\frac{0.0250 \text{ M}}{0.0167 \text{ M}}} = +0.134 \text{ V} \nonumber\] After the equivalence point, the concentration of Tl and the concentration of excess Tl are easy to calculate. For this reason we find the potential using the Nernst equation for the Tl /Tl half-reaction. For example, after adding 40.0 mL of titrant, the concentrations of Tl and Tl are \[[\text{Tl}^{+}] = \frac{(0.050 \text{ M})(50.0 \text{ mL})}{50.0 \text{ mL} + 40.0 \text{ mL}} = 0.0278 \text{ M} \nonumber\] \[[\text{Tl}^{3+}] = \frac{(0.100 \text{ M})(40.0 \text{ mL}) - (0.050 \text{ M})(50.0 \text{ mL})}{50.0 \text{ mL} + 40.0 \text{ mL}} = 0.0167 \text{ M} \nonumber\] and the potential is \[E = +0.77 \text{ V} - \frac{0.05916}{2} \log{\frac{0.0278 \text{ M}}{0.0167 \text{ M}}} = +0.76 \text{ V} \nonumber\] At the titration’s equivalence point, the potential, , potential is \[E_{eq} = \frac{0.139 \text{ V} + 0.77 \text{ V}}{2} = +0.45 \text{ V} \nonumber\] Some additional results are shown here. To evaluate the relationship between a titration’s equivalence point and its end point we need to construct only a reasonable approximation of the exact titration curve. In this section we demonstrate a simple method for sketching a redox titration curve. Our goal is to sketch the titration curve quickly, using as few calculations as possible. Let’s use the titration of 50.0 mL of 0.100 M Fe with 0.100 M Ce in a matrix of 1 M HClO . This is the same example that we used in developing the calculations for a redox titration curve. You can review the results of that calculation in and . We begin by calculating the titration’s equivalence point volume, which, as we determined earlier, is 50.0 mL. Next, we draw our axes, placing the potential, , on the -axis and the titrant’s volume on the -axis. To indicate the equivalence point’s volume, we draw a vertical line that intersects the -axis at 50.0 mL of Ce . Figure 9.4.2 a shows the result of the first step in our sketch. Before the equivalence point, the potential is determined by a redox buffer of Fe and Fe . Although we can calculate the potential using the Nernst equation, we can avoid this calculation if we make a simple assumption. You may recall from that a redox buffer operates over a range of potentials that extends approximately ±(0.05916/ ) unit on either side of \(E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ}\). The potential at the buffer’s lower limit is \[E = E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} - 0.05916 \nonumber\] when the concentration of Fe is \(10 \times\) greater than that of Fe . The buffer reaches its upper potential of \[E = E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + 0.05916 \nonumber\] when the concentration of Fe is \(10 \times\) smaller than that of Fe . The redox buffer spans a range of volumes from approximately 10% of the equivalence point volume to approximately 90% of the equivalence point volume. Figure 9.4.2 b shows the second step in our sketch. First, we superimpose a ladder diagram for Fe on the -axis, using its \(E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ}\) value of 0.767 V and including the buffer’s range of potentials. Next, we add points for the potential at 10% of (a potential of 0.708 V at 5.0 mL) and for the potential at 90% of (a potential of 0.826 V at 45.0 mL). We used a similar approach when sketching the acid–base titration curve for the titration of acetic acid with NaOH; see for details. The third step in sketching our titration curve is to add two points after the equivalence point. Here the potential is controlled by a redox buffer of Ce and Ce . The redox buffer is at its lower limit of \[E = E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ} - 0.05916 \nonumber\] when the titrant reaches 110% of the equivalence point volume and the potential is \(E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ}\) when the volume of Ce is \(2 \times V_{eq}\). We used a similar approach when sketching the complexation titration curve for the titration of Mg with EDTA; see for details. Figure 9.4.2 c shows the third step in our sketch. First, we superimpose a ladder diagram for Ce on the -axis, using its \(E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ}\) value of 1.70 V and including the buffer’s range. Next, we add points representing the potential at 110% of (a value of 1.66 V at 55.0 mL) and at 200% of (a value of 1.70 V at 100.0 mL). Next, we draw a straight line through each pair of points, extending the line through the vertical line that indicates the equivalence point’s volume (Figure 9.4.2 d). Finally, we complete our sketch by drawing a smooth curve that connects the three straight-line segments (Figure 9.4.2 e). A comparison of our sketch to the exact titration curve (Figure 9.4.2 f) shows that they are in close agreement. Sketch the titration curve for the titration of 50.0 mL of 0.0500 M Sn with 0.100 M Tl . Both the titrand and the titrant are 1.0 M in HCl. The titration reaction is \[\text{Sn}^{2+}(aq) + \text{Tl}^{3+}(aq) \rightleftharpoons \text{Tl}^{+}(aq) + \text{Sn}^{4+}(aq) \nonumber\] Compare your sketch to your calculated titration curve from . The figure below shows a sketch of the titration curve. The two points before the equivalence point = 2.5 mL, = +0.109 V and = 22.5 mL, = +0.169 V are plotted using the redox buffer for Sn /Sn , which spans a potential range of +0.139 ± 0.5916/2. The two points after the equivalence point = 27.5 mL, = +0.74 V and = 50 mL, = +0.77 V are plotted using the redox buffer for Tl /Tl , which spans the potential range of +0.139 ± 0.5916/2. The black dots and curve are the approximate sketch of the titration curve. The points in are the calculations from . A redox titration’s equivalence point occurs when we react stoichiometrically equivalent amounts of titrand and titrant. As is the case for acid–base titrations and complexation titrations, we estimate the equivalence point of a redox titration using an experimental end point. A variety of methods are available for locating a redox titration’s end point, including indicators and sensors that respond to a change in the solution conditions. For an acid–base titration or a complexometric titration the equivalence point is almost identical to the inflection point on the steeply rising part of the titration curve. If you look back at and , you will see that the inflection point is in the middle of this steep rise in the titration curve, which makes it relatively easy to find the equivalence point when you sketch these titration curves. We call this a . If the stoichiometry of a redox titration is 1:1—that is, one mole of titrant reacts with each mole of titrand—then the equivalence point is symmetric. If the titration reaction’s stoichiometry is not 1:1, then the equivalence point is closer to the top or to the bottom of the titration curve’s sharp rise. In this case we have an . Derive a general equation for the equivalence point’s potential when titrating Fe with \(\text{MnO}_4^-\). \[5\text{Fe}^{2+}(aq) + \text{MnO}_4^-(aq) + 8\text{H}^+(aq) \rightarrow 5\text{Fe}^{3+}(aq) + \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) \nonumber\] The half-reactions for the oxidation of Fe and the reduction of \(\text{MnO}_4^-\) are \[\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^- \nonumber\] \[\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5 e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) \nonumber\] for which the Nernst equations are \[E = E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} - 0.5916 \log{\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}} \nonumber\] \[E = E_{\text{MnO}_4^{-}/\text{Mn}^{2+}}^{\circ} - \frac{0.5916}{5} \log{\frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^{-},\text{H}^+]^8}} \nonumber\] Before we add together these two equations we must multiply the second equation by 5 so that we can combine the log terms; thus \[6E_{eq} = E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + 5E_{\text{MnO}_4^{-}/\text{Mn}^{2+}}^{\circ} - 0.05916 \log{\frac{[\text{Fe}^{2+},\text{Mn}^{2+}]}{[\text{Fe}^{3+},\text{MnO}_4^{-},\text{H}^+]^8}} \nonumber\] At the equivalance point we know that \[[\text{Fe}^{2+}] = 5 \times [\text{MnO}_4^-] \text{ and } [\text{Fe}^{3+}] = 5 \times [\text{Mn}^{2+}] \nonumber\] Substituting these equalities into the previous equation and rearranging gives us a general equation for the potential at the equivalence point. \[6E_{eq} = E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + 5E_{\text{MnO}_4^{-}/\text{Mn}^{2+}}^{\circ} - 0.05916 \log{\frac{5[\text{MnO}_4^{-},\text{Mn}^{2+}]}{5[\text{Mn}^{2+},\text{MnO}_4^{-},\text{H}^+]^8}} \nonumber\] \[E_{eq} = \frac{E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + 5E_{\text{MnO}_4^{-}/\text{Mn}^{2+}}^{\circ}}{6} - \frac{0.05916}{6} \log{\frac{1}{[\text{H}^+]^8}} \nonumber\] \[E_{eq} = \frac{E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + 5E_{\text{MnO}_4^{-}/\text{Mn}^{2+}}^{\circ}}{6} + \frac{0.05916 \times 8}{6} \log{[\text{H}^+]} \nonumber\] \[E_{eq} = \frac{E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + 5E_{\text{MnO}_4^{-}/\text{Mn}^{2+}}^{\circ}}{6} - 0.07888 \text{pH} \nonumber\] Our equation for the equivalence point has two terms. The first term is a weighted average of the titrand’s and the titrant’s standard state potentials, in which the weighting factors are the number of electrons in their respective half-reactions. The second term shows that for this titration is pH-dependent. At a pH of 1 (in H SO ), for example, the equivalence point has a potential of \[E_{eq} = \frac{0.768 + 5 \times 1.51}{6} - 0.07888 \times 1 = 1.31 \text{ V} \nonumber\] Figure 9.4.3 shows a typical titration curve for titration of Fe with \(\text{MnO}_4^-\). Note that the titration’s equivalence point is asymmetrical. Derive a general equation for the equivalence point’s potential for the titration of U with Ce . The unbalanced reaction is \[\text{Ce}^{4+}(aq) + \text{U}^{4+}(aq) \rightarrow \text{UO}_2^{2+}(aq) + \text{Ce}^{3+}(aq) \nonumber\] What is the equivalence point’s potential if the pH is 1? The two half reactions are \[\text{Ce}^{4+}(aq) + e^- \rightarrow \text{Ce}^{3+}(aq) \nonumber\] \[\text{U}^{4+}(aq) +2\text{H}_2\text{O}(l) \rightarrow \text{UO}_2^{2+}(aq)) + 4\text{H}^+(aq) +2e^- \nonumber\] for which the Nernst equations are \[E = E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ} - 0.05916 \log{\frac{[\text{Ce}^{3+}]}{[\text{Ce}^{4+}]}} \nonumber\] \[E = E_{\text{UO}_2^{2+}/\text{U}^{4+}}^{\circ} - \frac{0.05916}{2} \log{\frac{[\text{U}^{4+}]}{[\text{UO}_2^{2+},\text{H}^+]^4}} \nonumber\] Before adding these two equations together we must multiply the second equation by 2 so that we can combine the log terms; thus \[3E = E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ} + 2E_{\text{UO}_2^{2+}/\text{U}^{4+}}^{\circ} - 0.05916 \log{\frac{[\text{Ce}^{3+},\text{U}^{4+}]}{[\text{Ce}^{4+},\text{UO}_2^{2+},\text{H}^+]^4}} \nonumber\] At the equivalence point we know that \[[\text{Ce}^{3+}] = 2 \times [\text{UO}_2^{2+}] \text{ and } [\text{Ce}^{4+}] = 2 \times [\text{U}^{4+}] \nonumber\] Substituting these equalities into the previous equation and rearranging gives us a general equation for the potential at the equivalence point. \[3E = E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ} + 2E_{\text{UO}_2^{2+}/\text{U}^{4+}}^{\circ} - 0.05916 \log{\frac{2[\text{UO}_2^{2+},\text{U}^{4+}]}{2[\text{U}^{4+},\text{UO}_2^{2+},\text{H}^+]^4}} \nonumber\] \[E = \frac{E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ} + 2E_{\text{UO}_2^{2+}/\text{U}^{4+}}^{\circ}}{3} - \frac{0.05916}{3} \log{\frac{1}{[\text{H}^+]^4}} \nonumber\] \[E = \frac{E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ} + 2E_{\text{UO}_2^{2+}/\text{U}^{4+}}^{\circ}}{3} + \frac{0.05916 \times 4}{3} \log{[\text{H}^+]^4} \nonumber\] \[E = \frac{E_{\text{Ce}^{4+}/\text{Ce}^{3+}}^{\circ} + 2E_{\text{UO}_2^{2+}/\text{U}^{4+}}^{\circ}}{3} - 0.07888\text{pH} \nonumber\] At a pH of 1 the equivalence point has a potential of \[E = \frac{1.72 + 2 \times 0.327}{3} - 0.07888 \times 1 = +0.712 \text{ V} \nonumber\] Three types of indicators are used to signal a redox titration’s end point. The oxidized and reduced forms of some titrants, such as \(\text{MnO}_4^-\), have different colors. A solution of \(\text{MnO}_4^-\) is intensely purple. In an acidic solution, however, permanganate’s reduced form, Mn , is nearly colorless. When using \(\text{MnO}_4^-\) as a titrant, the titrand’s solution remains colorless until the equivalence point. The first drop of excess \(\text{MnO}_4^-\) produces a permanent tinge of purple, signaling the end point. Some indicators form a colored compound with a specific oxidized or reduced form of the titrant or the titrand. Starch, for example, forms a dark purple complex with \(\text{I}_3^-\). We can use this distinct color to signal the presence of excess \(\text{I}_3^-\) as a titrant—a change in color from colorless to purple—or the completion of a reaction that consumes \(\text{I}_3^-\) as the titrand— a change in color from purple to colorless. Another example of a specific indicator is thiocyanate, SCN , which forms the soluble red-colored complex of Fe(SCN) in the presence of Fe . The most important class of indicators are substances that do not participate in the redox titration, but whose oxidized and reduced forms differ in color. When we add a to the titrand, the indicator imparts a color that depends on the solution’s potential. As the solution’s potential changes with the addition of titrant, the indicator eventually changes oxidation state and changes color, signaling the end point. To understand the relationship between potential and an indicator’s color, consider its reduction half-reaction \[\text{In}_\text{ox} + ne^- \rightleftharpoons \text{In}_\text{red} \nonumber\] where In and In are, respectively, the indicator’s oxidized and reduced forms. For simplicity, In and In are shown without specific charges. Because there is a change in oxidation state, In and In cannot both be neutral. The Nernst equation for this half-reaction is \[E = E_{\text{In}_\text{ox}/\text{In}_\text{red}}^{\circ} - \frac{0.05916}{n} \log{\frac{[\text{In}_\text{red}]}{[\text{In}_\text{ox}]}} \nonumber\] As shown in Figure 9.4.4 , if we assume the indicator’s color changes from that of In to that of In when the ratio [In ]/[In ] changes from 0.1 to 10, then the end point occurs when the solution’s potential is within the range \[E = E_{\text{In}_\text{ox}/\text{In}_\text{red}}^{\circ} \pm \frac{0.05916}{n} \nonumber\] This is the same approach we took in considering acid–base indicators and complexation indicators. A partial list of redox indicators is shown in Table 9.4.2 . Examples of an appropriate and an inappropriate indicator for the titration of Fe with Ce are shown in Figure 9.4.5 . Another method for locating a redox titration’s end point is a potentiometric titration in which we monitor the change in potential while we add the titrant to the titrand. The end point is found by examining visually the titration curve. The simplest experimental design for a potentiometric titration consists of a Pt indicator electrode whose potential is governed by the titrand’s or the titrant’s redox half-reaction, and a reference electrode that has a fixed potential. Other methods for locating the titration’s end point include thermometric titrations and spectrophotometric titrations. You will a further discussion of potentiometry in . The best way to appreciate the theoretical and the practical details discussed in this section is to carefully examine a typical redox titrimetric method. Although each method is unique, the following description of the determination of the total chlorine residual in water provides an instructive example of a typical procedure. The description here is based on Method 4500-Cl B as published in , 20th Ed., American Public Health Association: Washington, D. C., 1998. The chlorination of a public water supply produces several chlorine-containing species, the combined concentration of which is called the total chlorine residual. Chlorine is present in a variety of chemical states, including the free residual chlorine, which consists of Cl , HOCl and OCl , and the combined chlorine residual, which consists of NH Cl, NHCl , and NCl . The total chlorine residual is determined by using the oxidizing power of chlorine to convert I to \(\text{I}_3^-\). The amount of \(\text{I}_3^-\) formed is then determined by titrating with Na S O using starch as an indicator. Regardless of its form, the total chlorine residual is reported as if Cl is the only source of chlorine, and is reported as mg Cl/L. Select a volume of sample that requires less than 20 mL of Na S O to reach the end point. Using glacial acetic acid, acidify the sample to a pH between 3 and 4, and add about 1 gram of KI. Titrate with Na S O until the yellow color of \(\text{I}_3^-\) begins to disappear. Add 1 mL of a starch indicator solution and continue titrating until the blue color of the starch–\(\text{I}_3^-\) complex disappears (Figure 9.4.6 ). Use a blank titration to correct the volume of titrant needed to reach the end point for reagent impurities. 1. Is this an example of a direct or an indirect analysis? This is an indirect analysis because the chlorine-containing species do not react with the titrant. Instead, the total chlorine residual oxidizes I to \(\text{I}_3^-\), and the amount of \(\text{I}_3^-\) is determined by titrating with Na S O . 2. Why does the procedure rely on an indirect analysis instead of directly titrating the chlorine-containing species using KI as a titrant? Because the total chlorine residual consists of six different species, a titration with I does not have a single, well-defined equivalence point. By converting the chlorine residual to an equivalent amount of \(\text{I}_3^-\), the indirect titration with Na S O has a single, useful equivalence point. Even if the total chlorine residual is from a single species, such as HOCl, a direct titration with KI is impractical. Because the product of the titration, \(\text{I}_3^-\), imparts a yellow color, the titrand’s color would change with each addition of titrant, making it difficult to find a suitable indicator. 3. Both oxidizing and reducing agents can interfere with this analysis. Explain the effect of each type of interferent on the total chlorine residual. An interferent that is an oxidizing agent converts additional I to \(\text{I}_3^-\). Because this extra \(\text{I}_3^-\) requires an additional volume of Na S O to reach the end point, we overestimate the total chlorine residual. If the interferent is a reducing agent, it reduces back to I some of the \(\text{I}_3^-\) produced by the reaction between the total chlorine residual and iodide; as a result, we underestimate the total chlorine residual. Although many quantitative applications of redox titrimetry have been re- placed by other analytical methods, a few important applications continue to find relevance. In this section we review the general application of redox titrimetry with an emphasis on environmental, pharmaceutical, and industrial applications. We begin, however, with a brief discussion of selecting and characterizing redox titrants, and methods for controlling the titrand’s oxidation state. If a redox titration is to be used in a quantitative analysis, the titrand initially must be present in a single oxidation state. For example, iron is determined by a redox titration in which Ce oxidizes Fe to Fe . Depending on the sample and the method of sample preparation, iron initially may be present in both the +2 and +3 oxidation states. Before titrating, we must reduce any Fe to Fe if we want to determine the total concentration of iron in the sample. This type of pretreatment is accomplished using an auxiliary reducing agent or oxidizing agent. A metal that is easy to oxidize—such as Zn, Al, and Ag—can serve as an . The metal, as a coiled wire or powder, is added to the sample where it reduces the titrand. Because any unreacted auxiliary reducing agent will react with the titrant, it is removed before we begin the titration by removing the coiled wire or by filtering. An alternative method for using an auxiliary reducing agent is to immobilize it in a column. To prepare a reduction column an aqueous slurry of the finally divided metal is packed in a glass tube equipped with a porous plug at the bottom. The sample is placed at the top of the column and moves through the column under the influence of gravity or vacuum suction. The length of the reduction column and the flow rate are selected to ensure the analyte’s complete reduction. Two common reduction columns are used. In the the column is filled with amalgamated zinc, Zn(Hg), which is prepared by briefly placing Zn granules in a solution of HgCl . Oxidation of zinc \[\text{Zn(Hg)}(s) \rightarrow \text{Zn}^{2+}(aq) + \text{Hg}(l) + 2e^- \nonumber\] provides the electrons for reducing the titrand. In the the column is filled with granular Ag metal. The solution containing the titrand is acidified with HCl and passed through the column where the oxidation of silver \[\text{Ag}(s) + \text{Cl}^- (aq) \rightarrow \text{AgCl}(s) + e^- \nonumber\] provides the necessary electrons for reducing the titrand. Table 9.4.3 provides a summary of several applications of reduction columns. Several reagents are used as , including ammonium peroxydisulfate, (NH ) S O , and hydrogen peroxide, H O . Peroxydisulfate is a powerful oxidizing agent \[\text{S}_2\text{O}_8^{2-}(aq) + 2e^- \rightarrow 2\text{SO}_4^{2-}(aq) \nonumber\] that is capable of oxidizing Mn to \(\text{MnO}_4^-\), Cr to \(\text{Cr}_2\text{O}_7^{2-}\), and Ce to Ce . Excess peroxydisulfate is destroyed by briefly boiling the solution. The reduction of hydrogen peroxide in an acidic solution \[\text{H}_2\text{O}_2(aq) + 2\text{H}^+(aq) + 2e^- \rightarrow 2\text{H}_2\text{O}(l) \nonumber\] provides another method for oxidizing a titrand. Excess H O is destroyed by briefly boiling the solution. If it is to be used quantitatively, the titrant’s concentration must remain stable during the analysis. Because a titrant in a reduced state is susceptible to air oxidation, most redox titrations use an oxidizing agent as the titrant. There are several common oxidizing titrants, including \(\text{MnO}_4^-\), Ce , \(\text{Cr}_2\text{O}_7^{2-}\), and \(\text{I}_3^-\). Which titrant is used often depends on how easily it oxidizes the titrand. A titrand that is a weak reducing agent needs a strong oxidizing titrant if the titration reaction is to have a suitable end point. The two strongest oxidizing titrants are \(\text{MnO}_4^-\) and Ce , for which the reduction half-reactions are \[\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightleftharpoons \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) \nonumber\] \[\text{Ce}^{4+}(aq) + e^- \rightleftharpoons \text{Ce}^{3+}(aq) \nonumber\] A solution of Ce in 1 M H SO usually is prepared from the primary standard cerium ammonium nitrate, Ce(NO ) •2NH NO . When prepared using a reagent grade material, such as Ce(OH) , the solution is standardized against a primary standard reducing agent such as Na C O or Fe (prepared from iron wire) using ferroin as an indicator. Despite its availability as a primary standard and its ease of preparation, Ce is not used as frequently as \(\text{MnO}_4^-\) because it is more expensive. The standardization reactions are \[\text{Ce}^{4+}(aq) + \text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + \text{Ce}^{3+}(aq) \nonumber\] \[2\text{Ce}^{4+}(aq) + \text{H}_2\text{C}_2\text{O}_4(aq) \rightarrow 2\text{Ce}^{3+}(aq) + 2\text{CO}_2(g) + 2\text{H}^+(aq) \nonumber\] A solution of \(\text{MnO}_4^-\) is prepared from KMnO , which is not available as a primary standard. An aqueous solution of permanganate is thermodynamically unstable due to its ability to oxidize water. \[4\text{MnO}_4^-(aq) + 2\text{H}_2\text{O}(l) \rightleftharpoons 4\text{MnO}_2(s) + 3\text{O}_2 (g) + 4\text{OH}^-(aq) \nonumber\] This reaction is catalyzed by the presence of MnO , Mn , heat, light, and the presence of acids and bases. A moderately stable solution of permanganate is prepared by boiling it for an hour and filtering through a sintered glass filter to remove any solid MnO that precipitates. Standardization is accomplished against a primary standard reducing agent such as Na C O or Fe (prepared from iron wire), with the pink color of excess \(\text{MnO}_4^-\) signaling the end point. A solution of \(\text{MnO}_4^-\) prepared in this fashion is stable for 1–2 weeks, although you should recheck the standardization periodically. The standardization reactions are \[\text{MnO}_4^-(aq) + 5\text{Fe}^{2+}(aq) + 8\text{H}^+(aq) \rightarrow \text{Mn}^{2+}(aq) + 5\text{Fe}^{3+}(aq) + 4\text{H}_2\text{O}(l) \nonumber\] \[2\text{MnO}_4^-(aq) + 5\text{H}_2\text{C}_2\text{O}_4(aq) + 6\text{H}^+(aq) \rightarrow 2\text{Mn}^{2+}(aq) + 10\text{CO}_2(g) + 8\text{H}_2\text{O}(l) \nonumber\] Potassium dichromate is a relatively strong oxidizing agent whose principal advantages are its availability as a primary standard and its long term stability when in solution. It is not, however, as strong an oxidizing agent as \(\text{MnO}_4^-\) or Ce , which makes it less useful when the titrand is a weak reducing agent. Its reduction half-reaction is \[\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \rightleftharpoons 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) \nonumber\] Although a solution of \(\text{Cr}_2\text{O}_7^{2-}\) is orange and a solution of Cr is green, neither color is intense enough to serve as a useful indicator. Diphenylamine sulfonic acid, whose oxidized form is red-violet and reduced form is colorless, gives a very distinct end point signal with \(\text{Cr}_2\text{O}_7^{2-}\). Iodine is another important oxidizing titrant. Because it is a weaker oxidizing agent than \(\text{MnO}_4^-\), Ce , and \(\text{Cr}_2\text{O}_7^{2-}\), it is useful only when the titrand is a stronger reducing agent. This apparent limitation, however, makes I a more selective titrant for the analysis of a strong reducing agent in the presence of a weaker reducing agent. The reduction half-reaction for I is \[\text{I}_2(aq) + 2e^- \rightleftharpoons 2\text{I}^-(aq) \nonumber\] Because iodine is not very soluble in water, solutions are prepared by adding an excess of I . The complexation reaction \[\text{I}_2(aq) + \text{I}^-(aq) \rightleftharpoons \text{I}_3^-(aq) \nonumber\] increases the solubility of I by forming the more soluble triiodide ion, \(\text{I}_3^-\). Even though iodine is present as \(\text{I}_3^-\) instead of I , the number of electrons in the reduction half-reaction is unaffected. \[\text{I}_3^-(aq) + 2e^-(aq) \rightleftharpoons 3\text{I}^-(aq) \nonumber\] Solutions of \(\text{I}_3^-\) normally are standardized against Na S O using starch as a specific indicator for \(\text{I}_3^-\). The standardization reaction is \[\text{I}_3^-(aq) + 2\text{S}_2\text{O}_3^{2-}(aq) \rightarrow 3\text{I}^-(aq) + 2\text{S}_4\text{O}_6^{2-} (aq) \nonumber\] An oxidizing titrant such as \(\text{MnO}_4^-\), Ce , \(\text{Cr}_2\text{O}_7^{2-}\), and \(\text{I}_3^-\), is used when the titrand is in a reduced state. If the titrand is in an oxidized state, we can first reduce it with an auxiliary reducing agent and then complete the titration using an oxidizing titrant. Alternatively, we can titrate it using a reducing titrant. Iodide is a relatively strong reducing agent that could serve as a reducing titrant except that its solutions are susceptible to the air-oxidation of I to \(\text{I}_3^-\). \[3\text{I}^-(aq) \rightleftharpoons \text{I}_3^- (aq) + 2e^- \nonumber\] A freshly prepared solution of KI is clear, but after a few days it may show a faint yellow coloring due to the presence of \(\text{I}_3^-\). Instead, adding an excess of KI reduces the titrand and releases a stoichiometric amount of \(\text{I}_3^-\). The amount of \(\text{I}_3^-\) produced is then determined by a back titration using thiosulfate, \(\text{S}_2\text{O}_3^{2-}\), as a reducing titrant. \[2\text{S}_2\text{O}_3^{2-}(aq) \rightleftharpoons \text{S}_4\text{O}_6^{2-}(aq) + 2e^- \nonumber\] Solutions of \(\text{S}_2\text{O}_3^{2-}\) are prepared using Na S O •5H O and are standardized before use. Standardization is accomplished by dissolving a carefully weighed portion of the primary standard KIO in an acidic solution that contains an excess of KI. The reaction between \(\text{IO}_3^-\) and I \[\text{IO}_3^-(aq) + 8\text{I}^-(aq) + 6\text{H}^+(aq) \rightarrow 3\text{I}_3^-(aq) + 3\text{H}_2\text{O}(l) \nonumber\] liberates a stoichiometric amount of I-3 . By titrating this \(\text{I}_3^-\) with thiosulfate, using starch as a visual indicator, we can determine the concentration of \(\text{S}_2\text{O}_3^{2-}\) in the titrant. The standardization titration is \[\text{I}_3^-(aq) + 2\text{S}_2\text{O}_3^{2-}(aq) \rightarrow 3\text{I}^-(aq) + \text{S}_4\text{O}_6^{2-}(aq) \nonumber\] which is the same reaction used to standardize solutions of \(\text{I}_3^-\). This approach to standardizing solutions of \(\text{S}_2\text{O}_2^{3-}\) is similar to that used in the determination of the total chlorine residual outlined in . Although thiosulfate is one of the few reducing titrants that is not readily oxidized by contact with air, it is subject to a slow decomposition to bisulfite and elemental sulfur. If used over a period of several weeks, a solution of thiosulfate is restandardized periodically. Several forms of bacteria are able to metabolize thiosulfate, which leads to a change in its concentration. This problem is minimized by adding a preservative such as HgI to the solution. Another useful reducing titrant is ferrous ammonium sulfate, Fe(NH ) (SO ) •6H O, in which iron is present in the +2 oxidation state. A solution of Fe is susceptible to air-oxidation, but when prepared in 0.5 M H SO it remains stable for as long as a month. Periodic restandardization with K Cr O is advisable. Ferrous ammonium sulfate is used as the titrant in a direct analysis of the titrand, or, it is added to the titrand in excess and the amount of Fe produced determined by back titrating with a standard solution of Ce or \(\text{Cr}_2\text{O}_7^{2-}\). One of the most important applications of redox titrimetry is evaluating the chlorination of public water supplies. , for example, describes an approach for determining the total chlorine residual using the oxidizing power of chlorine to oxidize I to \(\text{I}_3^-\). The amount of \(\text{I}_3^-\) is determined by back titrating with \(\text{S}_2\text{O}_3^{2-}\). The efficiency of chlorination depends on the form of the chlorinating species. There are two contributions to the total chlorine residual—the free chlorine residual and the combined chlorine residual. The free chlorine residual includes forms of chlorine that are available for disinfecting the water supply. Examples of species that contribute to the free chlorine residual include Cl , HOCl and OCl . The combined chlorine residual includes those species in which chlorine is in its reduced form and, therefore, no longer capable of providing disinfection. Species that contribute to the combined chlorine residual are NH Cl, NHCl and NCl . When a sample of iodide-free chlorinated water is mixed with an excess of the indicator , -diethyl- -phenylenediamine (DPD), the free chlorine oxidizes a stoichiometric portion of DPD to its red-colored form. The oxidized DPD is then back-titrated to its colorless form using ferrous ammonium sulfate as the titrant. The volume of titrant is proportional to the free residual chlorine. Having determined the free chlorine residual in the water sample, a small amount of KI is added, which catalyzes the reduction of monochloramine, NH Cl, and oxidizes a portion of the DPD back to its red-colored form. Titrating the oxidized DPD with ferrous ammonium sulfate yields the amount of NH Cl in the sample. The amount of dichloramine and trichloramine are determined in a similar fashion. The methods described above for determining the total, free, or combined chlorine residual also are used to establish a water supply’s chlorine demand. Chlorine demand is defined as the quantity of chlorine needed to react completely with any substance that can be oxidized by chlorine, while also maintaining the desired chlorine residual. It is determined by adding progressively greater amounts of chlorine to a set of samples drawn from the water supply and determining the total, free, or combined chlorine residual. Another important example of redox titrimetry, which finds applications in both public health and environmental analysis, is the determination of dissolved oxygen. In natural waters, such as lakes and rivers, the level of dissolved O is important for two reasons: it is the most readily available oxidant for the biological oxidation of inorganic and organic pollutants; and it is necessary for the support of aquatic life. In a wastewater treatment plant dissolved O is essential for the aerobic oxidation of waste materials. If the concentration of dissolved O falls below a critical value, aerobic bacteria are replaced by anaerobic bacteria, and the oxidation of organic waste produces undesirable gases, such as CH and H S. One standard method for determining dissolved O in natural waters and wastewaters is the Winkler method. A sample of water is collected without exposing it to the atmosphere, which might change the concentration of dissolved O . The sample first is treated with a solution of MnSO and then with a solution of NaOH and KI. Under these alkaline conditions the dissolved oxygen oxidizes Mn to MnO . \[2\text{Mn}^{2+}(aq) + 4\text{OH}^-(aq) + \text{O}_2(g) \rightarrow 2\text{MnO}_2(s) + 2\text{H}_2\text{O}(l) \nonumber\] After the reaction is complete, the solution is acidified with H SO . Under the now acidic conditions, I is oxidized to \(\text{I}_3^-\) by MnO . \[\text{MnO}_2(s) + 3\text{I}^-(aq) + 4\text{H}^+(aq) \rightarrow \text{Mn}^{2+}(aq) + \text{I}_3^-(aq) + 2\text{H}_2\text{O}(l) \nonumber\] The amount of \(\text{I}_3^-\) that forms is determined by titrating with \(\text{S}_2\text{O}_3^{2-}\) using starch as an indicator. The Winkler method is subject to a variety of interferences and several modifications to the original procedure have been proposed. For example, \(\text{NO}_2^-\) interferes because it reduces \(\text{I}_3^-\) to I under acidic conditions. This interference is eliminated by adding sodium azide, NaN , which reduces \(\text{NO}_2^-\) to N . Other reducing agents, such as Fe , are eliminated by pretreating the sample with KMnO and destroying any excess permanganate with K C O . Another important example of redox titrimetry is the determination of water in nonaqueous solvents. The titrant for this analysis is known as the Karl Fischer reagent and consists of a mixture of iodine, sulfur dioxide, pyridine, and methanol. Because the concentration of pyridine is sufficiently large, I and SO react with pyridine (py) to form the complexes py•I and py•SO . When added to a sample that contains water, I is reduced to I and SO is oxidized to SO . \[\text{py}\cdot\text{I}_2 + \text{py}\cdot\text{SO}_2 + \text{H}_2\text{O} + 2\text{py} \rightarrow 2\text{py}\cdot\text{HI} + \text{py}\cdot\text{SO}_3 \nonumber\] Methanol is included to prevent the further reaction of py•SO with water. The titration’s end point is signaled when the solution changes from the product’s yellow color to the brown color of the Karl Fischer reagent. Redox titrimetry also is used for the analysis of organic analytes. One important example is the determination of the chemical oxygen demand (COD) of natural waters and wastewaters. The COD is a measure of the quantity of oxygen necessary to oxidize completely all the organic matter in a sample to CO and H O. Because no attempt is made to correct for organic matter that is decomposed biologically, or for slow decomposition kinetics, the COD always overestimates a sample’s true oxygen demand. The determination of COD is particularly important in the management of industrial wastewater treatment facilities where it is used to monitor the release of organic-rich wastes into municipal sewer systems or into the environment. A sample’s COD is determined by refluxing it in the presence of excess K Cr O , which serves as the oxidizing agent. The solution is acidified with H SO , using Ag SO to catalyze the oxidation of low molecular weight fatty acids. Mercuric sulfate, HgSO , is added to complex any chloride that is present, which prevents the precipitation of the Ag catalyst as AgCl. Under these conditions, the efficiency for oxidizing organic matter is 95–100%. After refluxing for two hours, the solution is cooled to room temperature and the excess \(\text{Cr}_2\text{O}_7^{2-}\) determined by a back titration using ferrous ammonium sulfate as the titrant and ferroin as the indicator. Because it is difficult to remove completely all traces of organic matter from the reagents, a blank titration is performed. The difference in the amount of ferrous ammonium sulfate needed to titrate the sample and the blank is proportional to the COD. Iodine has been used as an oxidizing titrant for a number of compounds of pharmaceutical interest. Earlier we noted that the reaction of \(\text{S}_2\text{O}_3^{2-}\) with \(\text{I}_3^-\) produces the tetrathionate ion, \(\text{S}_4\text{O}_6^{2-}\). The tetrathionate ion is actually a dimer that consists of two thiosulfate ions connected through a disulfide (–S–S–) linkage. In the same fashion, \(\text{I}_3^-\) is used to titrate mercaptans of the general formula RSH, forming the dimer RSSR as a product. The amino acid cysteine also can be titrated with \(\text{I}_3^-\). The product of this titration is cystine, which is a dimer of cysteine. Triiodide also is used for the analysis of ascorbic acid (vitamin C) by oxidizing the enediol functional group to an alpha diketone and for the analysis of reducing sugars, such as glucose, by oxidizing the aldehyde functional group to a carboxylate ion in a basic solution. An organic compound that contains a hydroxyl, a carbonyl, or an amine functional group adjacent to an hydoxyl or a carbonyl group can be oxidized using metaperiodate, \(\text{IO}_4^-\), as an oxidizing titrant. \[\text{IO}_4^-(aq) + \text{H}_2\text{O}(l) + 2e^- \rightleftharpoons \text{IO}_3^-(aq) + 2\text{OH}^-(aq) \nonumber\] A two-electron oxidation cleaves the C–C bond between the two functional groups with hydroxyl groups oxidized to aldehydes or ketones, carbonyl groups oxidized to carboxylic acids, and amines oxidized to an aldehyde and an amine (ammonia if a primary amine). The analysis is conducted by adding a known excess of \(\text{IO}_4^-\) to the solution that contains the analyte and allowing the oxidation to take place for approximately one hour at room temperature. When the oxidation is complete, an excess of KI is added, which converts any unreacted \(\text{IO}_4^-\) to \(\text{IO}_3^-\) and \(\text{I}_3^-\). \[\text{IO}_4^-(aq) + 3\text{I}^-(aq) + \text{H}_2\text{O}(l) \rightarrow \text{IO}_3^-(aq) + \text{I}_3^-(aq) + 2\text{OH}^-(aq) \nonumber\] The \(\text{I}_3^-\) is then determined by titrating with \(\text{S}_2\text{O}_3^{2-}\) using starch as an indicator. The quantitative relationship between the titrand and the titrant is determined by the stoichiometry of the titration reaction. If you are unsure of the balanced reaction, you can deduce its stoichiometry by remembering that the electrons in a redox reaction are conserved. The amount of Fe in a 0.4891-g sample of an ore is determined by titrating with K Cr O . After dissolving the sample in HCl, the iron is brought into a +2 oxidation state using a Jones reductor. Titration to the diphenylamine sulfonic acid end point requires 36.92 mL of 0.02153 M K Cr O . Report the ore’s iron content as %w/w Fe O . Because we are not provided with the titration reaction, we will use a conservation of electrons to deduce the stoichiometry. During the titration the analyte is oxidized from Fe to Fe , and the titrant is reduced from \(\text{Cr}_2\text{O}_7^{2-}\) to Cr . Oxidizing Fe to Fe requires a single electron. Reducing \(\text{Cr}_2\text{O}_7^{2-}\), in which each chromium is in the +6 oxidation state, to Cr requires three electrons per chromium, for a total of six electrons. A conservation of electrons for the titration, therefore, requires that each mole of K Cr O reacts with six moles of Fe . The moles of K Cr O used to reach the end point is \[(0.02153 \text{ M})(0.03692 \text{ L}) = 7.949 \times 10^{-4} \text{ mol K}_2\text{Cr}_2\text{O}_7 \nonumber\] which means the sample contains \[7.949 \times 10^{-4} \text{ mol K}_2\text{Cr}_2\text{O}_7 \times \frac{6 \text{ mol Fe}^{2+}}{\text{mol K}_2\text{Cr}_2\text{O}_7} = 4.769 \times 10^{-3} \text{ mol Fe}^{2+} \nonumber\] Thus, the %w/w Fe O in the sample of ore is \[4.769 \times 10^{-3} \text{ mol Fe}^{2+} \times \frac{1 \text{ mol Fe}_2\text{O}_3}{2 \text{ mol Fe}^{2+}} \times \frac{159.69 \text{g Fe}_2\text{O}_3}{\text{mol Fe}_2\text{O}_3} = 0.3808 \text{ g Fe}_2\text{O}_3 \nonumber\] \[\frac{0.3808 \text{ g Fe}_2\text{O}_3}{0.4891 \text{ g sample}} \times 100 = 77.86 \text{% w/w Fe}_2\text{O}_3 \nonumber\] Although we can deduce the stoichiometry between the titrant and the titrand in Example 9.4.2 without balancing the titration reaction, the balanced reaction \[\text{K}_2\text{Cr}_2\text{O}_7(aq) + 6\text{Fe}^{2+}(aq) + 14\text{H}^+(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 2\text{K}^+(aq) + 6\text{Fe}^{3+}(aq) + 7\text{H}_2\text{O}(l) \nonumber\] does provide useful information. For example, the presence of H reminds us that the reaction must take place in an acidic solution. The purity of a sample of sodium oxalate, Na C O , is determined by titrating with a standard solution of KMnO . If a 0.5116-g sample requires 35.62 mL of 0.0400 M KMnO to reach the titration’s end point, what is the %w/w Na C O in the sample. Because we are not provided with a balanced reaction, let’s use a conservation of electrons to deduce the stoichiometry. Oxidizing \(\text{C}_2\text{O}_4^{2-}\), in which each carbon has a +3 oxidation state, to CO , in which carbon has an oxidation state of +4, requires one electron per carbon or a total of two electrons for each mole of \(\text{C}_2\text{O}_4^{2-}\). Reducing \(\text{MnO}_4^-\), in which each manganese is in the +7 oxidation state, to Mn requires five electrons. A conservation of electrons for the titration, therefore, requires that two moles of KMnO (10 moles of ) react with five moles of Na C O (10 moles of ). The moles of KMnO used to reach the end point is \[(0.0400 \text{ M KMnO}_4)(0.03562 \text{ L})=1.42 \times 10^{-3} \text{ mol KMnO}_4 \nonumber\] which means the sample contains \[1 .42 \times 10^{-3} \text{ mol KMnO}_4 \times \frac{5 \text{ mol Na}_2\text{C}_2\text{O}_4}{2 \text{ mol KMnO}_4} = 3.55 \times 10^{-3} \text{ mol Na}_2\text{C}_2\text{O}_4 \nonumber\] Thus, the %w/w Na C O in the sample of ore is \[3.55 \times 10^{-3} \text{ mol Na}_2\text{C}_2\text{O}_4 \times \frac{134.00 \text{ g Na}_2\text{C}_2\text{O}_4}{\text{mol Na}_2\text{C}_2\text{O}_4} = 0.476 \text{ g Na}_2\text{C}_2\text{O}_4 \nonumber\] \[\frac{0.476 \text{ g Na}_2\text{C}_2\text{O}_4}{0.5116 \text{ g sample}} \times 100 = 93.0 \text{% w/w Na}_2\text{C}_2\text{O}_4 \nonumber\] As shown in the following two examples, we can easily extend this approach to an analysis that requires an indirect analysis or a back titration. A 25.00-mL sample of a liquid bleach is diluted to 1000 mL in a volumetric flask. A 25-mL portion of the diluted sample is transferred by pipet into an Erlenmeyer flask that contains an excess of KI, reducing the OCl to Cl and producing \(\text{I}_3^-\). The liberated \(\text{I}_3^-\) is determined by titrating with 0.09892 M Na S O , requiring 8.96 mL to reach the starch indicator end point. Report the %w/v NaOCl in the sample of bleach. To determine the stoichiometry between the analyte, NaOCl, and the titrant, Na S O , we need to consider both the reaction between OCl and I , and the titration of \(\text{I}_3^-\) with Na S O . First, in reducing OCl to Cl the oxidation state of chlorine changes from +1 to –1, requiring two electrons. The oxidation of three I to form \(\text{I}_3^-\) releases two electrons as the oxidation state of each iodine changes from –1 in I to –1⁄3 in \(\text{I}_3^-\). A conservation of electrons, therefore, requires that each mole of OCl produces one mole of \(\text{I}_3^-\). Second, in the titration reaction, \(\text{I}_3^-\) is reduced to I and \(\text{S}_2\text{O}_3^{2-}\) is oxidized to \(\text{S}_4\text{O}_6^{2-}\). Reducing \(\text{I}_3^-\) to 3I requires two elections as each iodine changes from an oxidation state of –1⁄3 to –1. In oxidizing \(\text{S}_2\text{O}_3^{2-}\) to \(\text{S}_4\text{O}_6^{2-}\), each sulfur changes its oxidation state from +2 to +2.5, releasing one electron for each \(\text{S}_2\text{O}_3^{2-}\). A conservation of electrons, therefore, requires that each mole of \(\text{I}_3^-\) reacts with two moles of \(\text{S}_2\text{O}_3^{2-}\). Finally, because each mole of OCl produces one mole of \(\text{I}_3^-\), and each mole of \(\text{I}_3^-\) reacts with two moles of \(\text{S}_2\text{O}_3^{2-}\), we know that every mole of NaOCl in the sample ultimately results in the consumption of two moles of Na S O . The moles of Na S O used to reach the titration’s end point is \[(0.09892 \text{ M})(0.00896 \text{ L}) = 8.86 \times 10^{-4} \text{ mol Na}_2\text{S}_2\text{O}_3 \nonumber\] which means the sample contains \[8.86 \times 10^{-4} \text{ mol Na}_2\text{S}_2\text{O}_3 \times \frac{1 \text{ mol NaOCl}}{\text{mol Na}_2\text{S}_2\text{O}_3} \times \frac{74.44 \text{ g NaOCl}}{\text{mol NaOCl}} = 0.03299 \text{ g NaOCl} \nonumber\] Thus, the %w/v NaOCl in the diluted sample is \[\frac{0.03299 \text{ g NaOCl}}{25.00 \text{ mL}} \times 100 = 0.132 \text{% w/v NaOCl} \nonumber\] Because the bleach was diluted by a factor of \(40 \times\) (25 mL to 1000 mL), the concentration of NaOCl in the bleach is 5.28% w/v. The balanced reactions for this analysis are: \[\text{OCl}^-(aq) + 3\text{I}^-(aq) + 2\text{H}^+(aq) \rightarrow \text{I}_3^-(aq) + \text{Cl}^-(aq) + \text{H}_2\text{O}(l) \nonumber\] \[\text{I}_3^-(aq) + 2\text{S}_2\text{O}_3^{2-}(aq) \rightarrow \text{S}_4\text{O}_6^{2-}(aq) + 3\text{I}^-(aq) \nonumber\] The amount of ascorbic acid, C H O , in orange juice is determined by oxidizing ascorbic acid to dehydroascorbic acid, C H O , with a known amount of \(\text{I}_3^-\), and back titrating the excess \(\text{I}_3^-\) with Na S O . A 5.00-mL sample of filtered orange juice is treated with 50.00 mL of 0.01023 M \(\text{I}_3^-\). After the oxidation is complete, 13.82 mL of 0.07203 M Na S O is needed to reach the starch indicator end point. Report the concentration ascorbic acid in mg/100 mL. For a back titration we need to determine the stoichiometry between \(\text{I}_3^-\) and the analyte, C H O , and between \(\text{I}_3^-\) and the titrant, Na S O . The later is easy because we know from Example 9.4.3 that each mole of \(\text{I}_3^-\) reacts with two moles of Na S O . In oxidizing ascorbic acid to dehydroascorbic acid, the oxidation state of carbon changes from +2⁄3 in C H O to +1 in C H O . Each carbon releases 1⁄3 of an electron, or a total of two electrons per ascorbic acid. As we learned in Example 9.4.3 , reducing \(\text{I}_3^-\) requires two electrons; thus, a conservation of electrons requires that each mole of ascorbic acid consumes one mole of \(\text{I}_3^-\). The total moles of \(\text{I}_3^-\) that react with C H O and with Na S O is \[(0.01023 \text{ M})(0.05000 \text{ L}) = 5.115 \times 10^{-4} \text{ mol I}_3^- \nonumber\] The back titration consumes \[0.01382 \text{ L Na}_2\text{S}_2\text{O}_3 \times \frac{0.07203 \text{ mol Na}_2\text{S}_2\text{O}_3}{\text{ L Na}_2\text{S}_2\text{O}_3} \times \frac{1 \text{ mol I}_3^-}{2 \text{ mol Na}_2\text{S}_2\text{O}_3} = 4.977 \times 10^{-4} \text{ mol I}_3^- \nonumber\] Subtracting the moles of \(\text{I}_3^-\) that react with Na S O from the total moles of \(\text{I}_3^-\) gives the moles reacting with ascorbic acid. \[5.115 \times 10^{-4} \text{ mol I}_3^- - 4.977 \times 10^{-4} \text{ mol I}_3^- = 1.38 \times 10^{-5} \text{ mol I}_3^- \nonumber\] The grams of ascorbic acid in the 5.00-mL sample of orange juice is \[1.38 \times 10^{-5} \text{ mol I}_3^- \times \frac{1 \text{ mol C}_6\text{H}_8\text{O}_6}{\text{mol I}_3^-} \times \frac{176.12 \text{ g C}_6\text{H}_8\text{O}_6}{\text{mol C}_6\text{H}_8\text{O}_6} = 2.43 \times 10^{-3} \text{ g C}_6\text{H}_8\text{O}_6 \nonumber\] There are 2.43 mg of ascorbic acid in the 5.00-mL sample, or 48.6 mg per 100 mL of orange juice. The balanced reactions for this analysis are: \[\text{C}_6\text{H}_8\text{O}_6(aq) + \text{I}_3^- (aq) \rightarrow 3\text{I}^-(aq) + \text{C}_6\text{H}_6\text{O}_6(aq) + 2\text{H}^+(aq) \nonumber\] \[\text{I}_3^-(aq) + 2\text{S}_2\text{O}_3^{2-}(aq) \rightarrow \text{S}_4\text{O}_6^{2-}(aq) + 3\text{I}^-(aq) \nonumber\] A quantitative analysis for ethanol, C H O, is accomplished by a redox back titration. Ethanol is oxidized to acetic acid, C H O , using excess dichromate, \(\text{Cr}_2\text{O}_7^{2-}\), which is reduced to Cr . The excess dichromate is titrated with Fe , giving Cr and Fe as products. In a typical analysis, a 5.00-mL sample of a brandy is diluted to 500 mL in a volumetric flask. A 10.00-mL sample is taken and the ethanol is removed by distillation and collected in 50.00 mL of an acidified solution of 0.0200 M K Cr O7. A back titration of the unreacted \(\text{Cr}_2\text{O}_7^{2-}\) requires 21.48 mL of 0.1014 M Fe . Calculate the %w/v ethanol in the brandy. For a back titration we need to determine the stoichiometry between \(\text{Cr}_2\text{O}_7^{2-}\) and the analyte, C H O, and between \(\text{Cr}_2\text{O}_7^{2-}\) and the titrant, Fe . In oxidizing ethanol to acetic acid, the oxidation state of carbon changes from –2 in C H O to 0 in C H O . Each carbon releases two electrons, or a total of four electrons per C H O. In reducing \(\text{Cr}_2\text{O}_7^{2-}\), in which each chromium has an oxidation state of +6, to Cr , each chromium loses three electrons, for a total of six electrons per \(\text{Cr}_2\text{O}_7^{2-}\). Oxidation of Fe to Fe requires one electron. A conservation of electrons requires that each mole of K Cr O (6 moles of ) reacts with six moles of Fe (6 moles of ), and that four moles of K Cr O (24 moles of ) react with six moles of C H O (24 moles of ). The total moles of K Cr O that react with C H O and with Fe is \[(0.0200 \text{ M K}_2\text{Cr}_2\text{O}_7)(0.05000 \text{ L})=1.00 \times 10^{-3} \text{ mol K}_2\text{Cr}_2\text{O}_7 \nonumber\] The back titration with Fe consumes \[(0.1014 \text{ M Fe}^{2+})(0.02148 \text{ L}) \times \frac{1 \text{ mol K}_2\text{Cr}_2\text{O}_7}{6 \text{ mol Fe}^{2+}} = 3.63 \times 10^{-4} \text{ mol K}_2\text{Cr}_2\text{O}_7 \nonumber\] Subtracting the moles of K Cr O that react with Fe from the total moles of K Cr O gives the moles that react with the analyte. \[(1.00 \times 10^{-3} \text{ mol K}_2\text{Cr}_2\text{O}_7) - (3.63 \times 10^{-4} \text{ mol K}_2\text{Cr}_2\text{O}_7) = 6.37 \times 10^{-4} \text{ mol K}_2\text{Cr}_2\text{O}_7 \nonumber\] The grams of ethanol in the 10.00-mL sample of diluted brandy is \[6.37 \times 10^{-4} \text{ mol K}_2\text{Cr}_2\text{O}_7 \times \frac{6 \text{ mol C}_2\text{H}_6\text{O}}{4 \text{ mol K}_2\text{Cr}_2\text{O}_7} \times \frac{46.07 \text{ g C}_2\text{H}_6\text{O}}{\text{mol C}_2\text{H}_6\text{O}} = 0.0440 \text{ g C}_2\text{H}_6\text{O} \nonumber\] The %w/v C H O in the brandy is \[\frac{0.0440 \text{ g C}_2\text{H}_6\text{O}}{10.0 \text{ mL diluted brandy}} \times \frac{500.0 \text{ mL diluted brandy}}{5.00 \text{ mL brandy}} \times 100 = 44.0 \text{% w/v C}_2\text{H}_6\text{O} \nonumber\] The scale of operations, accuracy, precision, sensitivity, time, and cost of a redox titration are similar to those described earlier in this chapter for an acid–base or a complexation titration. As with an acid–base titration, we can extend a redox titration to the analysis of a mixture of analytes if there is a significant difference in their oxidation or reduction potentials. Figure 9.4.7 shows an example of the titration curve for a mixture of Fe and Sn using Ce as the titrant. A titration of a mixture of analytes is possible if their standard state potentials or formal potentials differ by at least 200 mV.

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Make sure you thoroughly understand the following essential ideas: In the previous section you learned that an electron standing-wave pattern characterized by the quantum numbers ( ) is called an . According to the , no two electrons in the same atom can have the same set of quantum numbers ( ). This limits the number of electrons in a given orbital to two ( = ±1), and it requires that atom containing more then two electrons must place them in standing wave patterns corresponding to higher principal quantum numbers , which means that these electrons will be farther from the nucleus and less tightly bound by it. In this chapter, we will see how the Pauli restrictions on the allowable quantum numbers of electrons in an atom affect the electronic configuration of the different elements, and, by influencing their chemical behavior, governs the structure of the periodic table. Let us begin with atoms that contain only a single electron. Hydrogen is of course the only species of this kind, but by removing electrons from heavier elements we can obtain one-electron ions such as \(He^+\) and \(Li^{2+}\), etc. Each has a ground state configuration of 1 , meaning that its single electron exhibits a standing wave pattern governed by the quantum numbers =1, =0 and =0, with the spin quantum number undefined because there is no other electron to compare it with. All have simple emission spectra whose major features were adequately explained by Bohr's model. The most important feature of a single-electron atom is that the energy of the electron depends only on the principal quantum number . As the above diagram shows, the quantum numbers and have no effect on the energy; we say that all orbitals having a given value of are . Thus the emission spectrum produced by exciting the electron to the =2 level consists of a single line, not four lines. The wavelength of this emission line for the atoms H, He and Li will diminish with atomic number because the greater nuclear charge will lower the energies of the various levels. For the same reason, the energies required to remove an electron from these species increases rapidly as the nuclear charge increases, because the increasing attraction pulls the electron closer to the nucleus, thus producing an even greater attractive force. It takes 1312 kJ of energy to remove the electron from a mole of hydrogen atoms. What might we expect this value to be for helium? Helium contains two electrons, but its nucleus contains two protons; each electron "sees" both protons, so we might expect that the electrons of helium would be bound twice as strongly as the electron of hydrogen. The ionization energy of helium should therefore be twice 1312 kJ/mol, or 2612 kJ/mol. However, if one looks at the spectrum of helium, the continuum is seen to begin at a wavelength corresponding to an ionization energy of 2372 kJ/mol, or about 90% of the predicted value. Why are the electrons in helium bound less tightly than the +2 nuclear charge would lead us to expect? The answer is that there is another effect to consider: the between the two electrons; the resulting electron-electron repulsion subtracts from the force holding the electron to the nucleus, reducing the local binding of each. is a major factor in both the spectra and chemical behavior of the elements heavier than hydrogen. In particular, it acts to "break the degeneracy" (split the energies) of orbitals having the same value of but different . The diagram below shows how the energies of the - and -orbitals of different principal quantum numbers get split as the result of electron-electron repulsion. Notice the contrast with the similar diagram for one-electron atoms near the top of this page. The fact that electrons preferentially fill the lowest-energy empty orbitals is the basis of the rules for determining the electron configuration of the elements and of the structure of the periodic table. The German word means "building up", and this term has traditionally been used to describe the manner in which electrons are assigned to orbitals as we carry out the imaginary task of constructing the atoms of elements having successively larger atomic numbers. In doing so, we are effectively "building up" the periodic table of the elements, as we shall shortly see. The preceding diagram illustrates the main idea here. Each orbital is represented as a little box that can hold up to two electrons having opposite spins, which we designated by upward- or downward-pointing arrows. Electrons fill the lowest-energy boxes first, so that additional electrons are forced into wave-patterns corresponding to higher (less negative) energies. Thus in the above diagram, the "third" electron of lithium goes into the higher-energy 2s orbital, giving this element an electron configuration which we write 1s 2s . What is the electron configuration of the atom of phosphorus, atomic number 15? The number of electrons filling the lowest-energy orbitals are: 1 : 2 electrons, 2 : 2 electrons; 2 : 6 electrons, 3 : 2 electrons. This adds up to 12 electrons. The remaining three electrons go into the 3 orbital, so the complete electron configuration of P is 1 2 2 3 3 . You should be able to reproduce it from memory up to the 6 level, because it forms the fundamental basis of the periodic table of the elements. Inspection of a table of electron configurations of the elements reveals a few apparent non-uniformities in the filling of the orbitals, as is illustrated here for the elements of the so-called first transition series in which the 3 orbitals are being populated. These anomalies are a consequence of the very small energy differences between some of the orbitals, and of the reduced electron-electron repulsion when electrons remain unpaired (Hund's rule), as is evident in chromium, which contains six unpaired electrons. The other anomaly here is copper, which "should" have the outer-shell configuration 3 4 . The actual configuration of the Cu atom appears to be 3 4 . Although the 4 orbital is normally slightly below the 3 orbital energy, the two are so close that interactions between the two when one is empty and the other is not can lead to a reversal. Detailed calculations in which the shapes and densities of the charge distributions are considered predict that the relative energies of many orbitals can reverse in this way. It gets even worse when -orbitals begin to fill! Because these relative energies can vary even for the same atom in different chemical environments, most instructors will not expect you to memorize them. This diagram shows how the atomic orbitals corresponding to different principal quantum numbers become interspersed with one another at higher values of . The actual situation is more complicated than this; calculations show that the energies of and orbitals vary with the atomic number of the element. The relative orbital energies illustrated above and the Pauli exclusion principle constitute the fundamental basis of the periodic table of the elements which was of course worked out empirically late in the 19th century, long before electrons had been heard of. The periodic table of the elements is conventionally divided into sections called , each of which designates the type of "sub-orbital" ( ) which contains the highest-energy electrons in any particular element. Note especially that The above diagram illustrates the link between the electron configurations of the elements and the layout of the periodic table. Each row, also known as a , commences with two s-block elements and continues through the block. At the end of the rows corresponding to >1 is an element having a configuration, a so-called . At values of 2 and 3, - and -block element sequences are added. The table shown above is called the of the periodic table; for many purposes, we can use a "short form" table in which the -block is shown below the - and - block "representative elements" and the -block does not appear at all. Note that the "long form" would be even longer if the -block elements were shown where they actually belong, between La-Hf and Ac-Db.

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/16%3A_Appendix/16.04%3A_Critical_Values_for_t-Test

Assuming we have calculated , there are two approaches to interpreting a -test. In the first approach we choose a value of \(\alpha\) for rejecting the null hypothesis and read the value of \(t(\alpha,\nu)\) from the table below. If \(t_\text{exp} > t(\alpha,\nu)\), we reject the null hypothesis and accept the alternative hypothesis. In the second approach, we find the row in the table below that corresponds to the available degrees of freedom and move across the row to find (or estimate) the a that corresponds to \(t_\text{exp} = t(\alpha,\nu)\); this establishes largest value of \(\alpha\) for which we can retain the null hypothesis. Finding, for example, that \(\alpha\) is 0.10 means that we retain the null hypothesis at the 90% confidence level, but reject it at the 89% confidence level. The examples in this textbook use the first approach. The values in this table are for a two-tailed -test. For a one-tailed test, divide the \(\alpha\) values by 2. For example, the last column has an \(\alpha\) value of 0.005 and a confidence interval of 99.5% when conducting a one-tailed -test.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Carbohydrates/Monosaccharides/Galactose

is more commonly found in the disaccharide, lactose or milk sugar. It is found as the monosaccharide in peas. Galactose is classified as a monosaccharide, an aldose, a hexose, and is a reducing sugar. One baby out of every 18,000 is born with a genetic defect of not being able to utilize galactose. Since galactose is in milk as part of lactose, it will build up in the blood and urine. Undiagnosed it may lead to mental retardation, failure to grow, formation of cataracts, and in sever cases death by liver damage. The disorder is caused by a deficiency in one or more enzymes required to metabolize galactose. The treatment for the disorder is to use a formula based upon the sugar sucrose rather than milk with lactose. The galactose free diet is critical only in infancy, since with maturation another enzyme is developed that can metabolize galactose. The chair form of galactose follows the same pattern as that for glucose. The is the center of a functional group. A carbon that has both an ether oxygen and an alcohol group is a hemiacetal. he is defined as the -OH being on the same side of the ring as the C # 6. In the chair structure this results in a (Haworth - an upwards projection). The is defined as the -OH being on the opposite side of the ring as the C # 6. In the chair and this results in a . The position of the -OH group on the distinction between glucose and galactose. is defined as the -OH on C # 4 in a in the chair form, (down in the Haworth structure). is defined as the -OH on C # 4 in a in the chair form,(also upward in the Haworth structure). Both glucose and galactose may be either alpha or beta on the anomeric carbon, so this is not distinctive between them. Which carbon in the structure on the left is the anomeric carbon on galactose?

https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Biochemical_Cycles/Nitrogen_Cycle

The main component of the nitrogen cycle starts with the element nitrogen in the air. Two nitrogen oxides are found in the air as a result of interactions with oxygen. Nitrogen will only react with oxygen in the presence of high temperatures and pressures found near lightning bolts and in combustion reactions in power plants or internal combustion engines. Nitric oxide, NO, and nitrogen dioxide, NO , are formed under these conditions. Eventually nitrogen dioxide may react with water in rain to form nitric acid, HNO . The nitrates thus formed may be utilized by plants as a nutrient. Nitrogen in the air becomes a part of biological matter mostly through the actions of bacteria and algae in a process known as nitrogen fixation. Legume plants such as clover, alfalfa, and soybeans form nodules on the roots where nitrogen fixing bacteria take nitrogen from the air and convert it into ammonia, NH . The ammonia is further converted by other bacteria first into nitrite ions, NO , and then into nitrate ions, NO . Plants utilize the nitrate ions as a nutrient or fertilizer for growth. Nitrogen is incorporate in many amino acids which are further reacted to make proteins. Ammonia is also made through a synthetic process called the Haber Process. Nitrogen and hydrogen are reacted under great pressure and temperature in the presence of a catalyst to make ammonia. Ammonia may be directly applied to farm fields as fertilizer. Ammonia may be further processed with oxygen to make nitric acid. The reaction of ammonia and nitric acid produces ammonium nitrate which may then be used as a fertilizer. Animal wastes when decomposed also return to the earth as nitrates. To complete the cycle other bacteria in the soil carry out a process known as denitrification which converts nitrates back to nitrogen gas. A side product of this reaction is the production of a gas known as nitrous oxide, N O. Nitrous oxide, also known as "laughing gas" - mild anesthetic, is also a greenhouse gas which contributes to global warming  

https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Aquatic_Chemistry/Water_Chemistry

Water is an unusual compound with unique physical properties. As a result, its the compound of life. Yet, its the most abundant compound in the biosphere of Earth. These properties are related to its electronic structure, bonding, and chemistry. However, due to its affinity for a variety of substances, ordinary water contains other substances. Few of us has used, seen or tested pure water, based on which we discuss its chemistry. The chemistry of water deals with the fundamental chemical property and information about water. Water chemistry is discussed in the following subtitles. Water consists of only hydrogen and oxygen. Both elements have natural stable and radioactive isotopes. Due to these isotopes, water molecules of masses roughly 18 (H O) to 22 (D O) are expected to form. Isotopes and their abundances of H and O are given below. From these data, we can estimate the relative abundances of all isotopic water molecules. The predominant water molecules H O have a mass of 18 amu, but molecules with mass 19 and 20 occur significantly. Because the isotopic abundances are not always the same due to their astronomical origin, The isotopic distribution of water molecules depends on its source and age. Its study is linked to other sciences. ( Dojlido, J.R. & Best, G.A. (1993) , Ellis Harwood for isotopic distribution of water.) In particular, D O is called heavy water, and it is produced by enrichment from natural water. Properties of heavy water are particularly interesting due to its application in nuclear technology. Pure water, H O, has a unique molecular structure. The O-H bondlengths are 0.096 nm and the H-O-H angle = 104.5°. This strange geometry can be explained by various methods. From carbon to neon, the numbers of valence electrons increase from 4 to 8. These elements require 4, 3, 2, 1, and 0 H atoms to share electrons in order to complete the octet requirement. Their are shown on the right, and note the trend in bondlengths. There are six valance electrons on the oxygen, and one each from the hydrogen atom in the water molecule. The eight electrons form two H-O bonds, and left two lone pairs. The long pairs and bonds stay away from each other and they extend towards the corners of a tetrahedron. Such an ideal structure should give H-O-H bond angle of 109.5°, but the lone pairs repel each other more than they repel the O-H bonds. Thus, the O-H bonds are pushed closer, making the H-O-H angle less than 109°. After the introduction of quantum mechanics, the electronic configuration for the valence electron of oxygen are 2 2 . Since the energy levels of 2 and 2 are close, valence electrons have characters of both and . The mixture is called These hybridized orbitals are shown on the right. The structures of CH , NH , and H O can all explained by these hybrid orbitals of the central atoms. The above approach is the valence bond theory, and both the C-H bonds and lone electron pairs are counted as in the Valence-shell Electron-Pair Repulsion (VSEPR) model, according to which, the four groups point to the corners of a tetrahedron. For triatomic molecules such as water, molecular orbital (MO) approach can also be applied to discuss the bonding. The result however is similar to the valence bond approach, but the MO theory gives the energy levels of the electron for further exploration. Atoms in a molecule are never at rest, and for each type of molecule, there are some normal vibration modes. For the water molecule, the three normal modes of vibrations are symmetric stretching, bending and asymmetric stretching. quantized HDO The ideal transition bands are centered in the given wavenumbers. However, these wavenumbers are calculated based on isolated molecules with no interaction with any neighbor. When molecules interact with each other, the energy levels are modified, and the bands shift. Many more less intense absorption bands extend into the green part of the visible spectrum. The absorption spectrum of water may contribute to the blue color for lake, river and ocean waters. The water molecules are rather symmetric in that there are two mirror planes of symmetry, one containing all three atoms and one perpendicular to the plane passing through the bisector of the H-O-H angle. Furthermore, if the molecules are rotated 180° (360°/2) the shape of the molecule is unperturbed. This indicates that the molecules have a 2-fold rotation axis. The three symmetry elements are 2-fold rotation, and two mirror planes. Both mirror planes contain the rotation axis, and this type of symmetry belongs to the . A has a definite number of symmetry elements arranged in certain fashion. Molecules can be classified according to their point groups. Molecules of the same point group have similar spectroscopic characters. Other molecules of point group are CH =O, CH Cl , the bent O etc. . This statement is from Linus Pauling (1939) in his book . He gave the ion [F:H:F] as an example. At that time, the hydrogen bond was recognized as mainly ionic in nature. The energy associated with hydrogen bond is 8 to 40 kJ/mol. Normally, the melting point and boiling point of a substance increase with molecular mass. For example the melting points of inert gases are 0.95, 24.48, 83.8, and 116.6 K respectively for He, Ne, Ar, and Kr. In this table, the melting and boiling points for water are particular high for its small molecular mass. This is usually attributed to the formation of hydrogen bonds. The small electronegative atoms F, O and N are somewhat negatively charged when they are bonded to hydrogen atoms. The negative charges on F, O and N attract the slightly positive hydrogen atoms, forming a strong interaction called hydrogen bond. A graph showing the melting points and boiling points of group 16 provided by Prof. J. Boucher illustrates the same point. Based on the observed absorption at 3546 and 3691 cm , Van Thiel, Becker, and Pinmentel (1957, 386) suggested the formation of water dimer when trapped in a matrix of nitrogen. Due to hydrogen bonding, water molecules form dimers, trimers, polymers, and clusters. The hydrogen bonds are not necessarily liner. Ice occurs in many places, including the Antarctic. If all the ice melted, the water level of the oceans will rise about 70 m. The structure of ice and the caption are from this link. The density of ice is dramatically smaller than that of water, due to the regular arrangement of water molecule via hydrogen bonds. In an idealized structure of ice, every hydrogen atom is involved in hydrogen bond. Every oxygen atom is surrounded by four hydrogen bonds. This diagram from caltech.edu, shows the structure of hexagonal ice in (a) and cubic ice in (b). A rod here represents a hydrogen bond. Since the hydrogen bonds are not linear, the real structure is a little more complicated. The tetrahedral coordination opens up the space between molecules. On each hydrogen bond, shown by a rod joining the oxygen atoms, lies one proton in an asymmetric position (not shown). Bond lengths, 275 pm, are indicated. Ordinary ice is hexagonal. and the hexagonal axis is labelled 732 pm, and one of the hexagonal axes is labelled 450 pm. If water vapor condenses on very cold substrate at 143-193 K (-130 to -80ºC) a cubic phase is formed. In (b) the cubic unit cell is outlined with dashed lines; dimensions are in pm determined at 110 K. These diagrams can also be used to represent the two forms of diamond, and in this case, the rods joining the atoms represent C-C bonds. Each C-C bondlength is 154 pm. Silicon and germanium crystals have the same structure, but their bondlengths are longer. The two diamond types of structure are related to the packing of spheres. The hexagonal type has the ABABAB... sequence, whereas the cubic type has the ABCABC... sequence. In both cases, half of the tetrahedral sites are occupied by tetrahedrally bonded carbon atoms. Hexagonal diamonds have been observed in meteorites. The four hydrogen bonds around an oxygen atom form a tetrahedron in a fashion found in the two types of diamonds. Thus, ice, diamond, and close packing of spheres are somewhat topologically related. A phase diagram of water shows 9 different solid phases (ices). Ice Ih is the ordinary ice. In addition to ice Ic from vapor deposition, conditions for nine phases are shown. Aside from ice I, other phases are formed and observed under high pressure generated by machines built by scientists. So far, ten different forms of ice have been observed, and some ice forms exist at very high pressure. The pressure deep under the polar (Antarctic) ice cap is very high, but we are not able to make any direct observation or study. There is a report of the 11th ice, and the ice phase diagram and drawings of ice structures given here is extremely interesting. The Autoionization of Water in the formation of ions according to HOH(l) + HOH(l) = H3O+ + OH- This is an equilibrium process and is characterised by an equilibrium constant, : [H3O+] [OH-] K'w = ------------ [H2O] Since [H O] = 1000/18 = 55.56 M, and remains rather constant under any circumstance, we usually write Kw = [H3O+] [OH-] = 10-14 (or 1e-14) pKw = -log Kw (defined) = 14 (at 298 K) For neutral water, [H O ] = [OH ] = 1e-7 at this temperature. Furthermore, we define pH = -log[H3O+] pOH = -log[OH-] pH = pOH = 7 at 298 K; (in neutral solutions) It is important to realize that depends on temperature as shown in the Table here. The strength of strong acids and bases is dominated by the autoionization of water. In aqueous solutions, the strongest acid and base are the hydronium ion, H O , and the hydroxide ion OH respectively. Acids HCl, HBr, HI, HNO , HClO , HClO , and H SO completely ionize in water, making them as strong as H O due to the of water. Furthermore, strong acids, strong bases, and salts completely ionize in their aqueous solutions. For example, HCl is a stronger acid than H O, and the reaction takes place as HCl dissolves in water. HCl + H2O = Cl- + H3O+ A similar equation can be written for another strong acid. On the other hand, a stong base also react with water to give the stong base species, OH . H2O + B- = OH- + HB For example, O , CH O , and NH are strong bases. The leveling effect also apply to bases. Equilibria of acids and bases, are interesting chemistry. When an acid and a base differ by a proton, they are called a conjugate acid-base pair. A water molecule is a weak acid and base, due to its ability to accept or donate a proton. Such properties make water an . In fact, H O , H O and OH are amphiprotic, as are some other conjugate acid-base pairs of weak acids and bases. If several acids and bases are dissolved in water, all equilibria must be considered. To estimate the pH of these solutions requires the exact treatment of several equilibrium constants. For example, many species dissolve in rain water, and many equilibria must be considered. Detail consideration and examples are given in Acid-Base Reactions Carbon dioxide in the air dissolve in rain water, lakes and rivers. A solution of CO involves the following reaction: These complicated equilibria make natural water a buffer. Assume that the partial pressure of carbon dioxide causes a total concentration of carbonic species to be 8e-4 M. Estimate the pH of this solution. From the given data, we have the following five equations and five unknowns: Solving these equations for the 5 unknowns can be done using Maple, Mathcad, spread sheet, or approximation. In any case, we are interested in the pH, and we can make the following approximations or assumptions Combining (1) and (6) gives [H ] = x = 8.0e-4 * 5.0e-7 = 4.0e-10. Therefore, [H+] = 2.0e-5 pH = -log(2.0e5) = 4.7 Generally speaking, rain water has a pH about 5, rather acidic. It dissolves limestone and marble readily. Due to the dissolved carbon dioxide, rain water is a buffer solution. Increased carbon dioxide level forces an increase in dissolved carbon dioxide. Would this causes pH of rain water to decrease or increase? Justify your answer by giving the reasons. Since [H ] = 2.0e-5, [OH ] = 5e-9, the amount of H from ionization of water is also 5.0e-9, small with respect to 2.0e-5 from ionization of H CO . Similarly, the ionization from HCO3- « CO32- + H+ is also small. Most of the C-containing species is H CO H CO is a weak acid, its ionization is small indeed. Now, you may proceed to evaluate other concentrations: [OH ], [HCO ], and [CO ] Alkali metals react with water readily. Contact of cesium metal with water causes immediate explosion, and the reactions become slower for potassium, sodium and lithium. Reaction with barium, strontium, calcium are less well known, but they do react readily. Warm water may be needed to react with calcium metal, however. Many metals displace H ions in acidic solutions. This is often seen as a property of acids. The for liquid water, H O(l), is -285.830 and that of water vapour is -241.826 kJ/mol. The difference is the heat of vaporization at 298 K. Liquid water and vapor entropies ( ) are 69.95 and 188.835 kJ K mol respectively, ( Thermodynamic Data. These are entropies, not standard entropies of formation. The entropy of formation for water is obtained by, DSof water = Sowater - SoH2 - 0.5 SoO2 = 69.95 - 130.68 - 0.5*205.14 (data from Thermodynamic Data) = - 163.3 J K-1 mol-1 DGowater = DH - T DS (note H in kJ/mol and S in J/mol) DGowater = -285.83 - 298.15 * 163.3/1000 = -237.13 kJ The equilibrium constant and Gibb's energy are related, DGo = - R T ln K K = exp(- DGo / R T) = 3.5e41 atm-3/2 This is a very large value for the formation of water, H2 + 0.5 O2 = 0.5 H2O(l). In other words, the reaction is complete, and the possibility of water dissociated into hydrogen and oxygen is very small. A negative value for indicates an exothermic reaction. The Gibb's energy is the energy released other than pressure-volume work. This redox reaction to form water can be engineered to proceed in a Daniel cell. In this case, the energy is converted into electric energy according to this equation. DGowater = - n F E = -237.13 kJ where is the number of electrons (= 2) in the redox equation, is the Faraday constant (= 96485 C), and is the potential of the Daniel cell. Thus, - 237130 J E = - ------------- 2*96485 C = 1.23 V Ideally, a reverse voltage of 1.23 V is required for the electrolysis of water. But in reality, a little over voltage is required to carry out the electrolysis to decompose water. Furthermore, pure water does not conduct electricity, and acid, base or salt is often added for the electrolysis of water. This link has a demonstration. In order to carry out the electrolysis of water, 1.50 V is applied. Assume the energy not converted to chemical energy is converted to heat. How much heat is generated for the electrolysis of 1 mole water? Ideally, 1.23 V will be used for the electrolysis. Energy due to the over voltage of 1.50 - 1.23 = 0.27 V is converted to heat. Heat = 0.27 V * 2 * 96485 C = 52102 J = 52 kJ The excess energy can also be evaluated using Heat = n F *1.50 - 237130 This problem also illustrates the principle of conservation of energy. H2O(l) -> H2(g) + 0.5 OH2(g) Hint: 2.6e-64 Evaluate this value please!

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Book3A_Medicines_by_Design/04%3A_Molecules_to_Medicines/4.05%3A_Act_Like_a_Membrane

Researchers know that high concentrations of chemotherapy drugs will kill every single cancer cell growing in a lab dish, but getting enough of these powerful drugs to a tumor in the body without killing too many healthy cells along the way has been exceedingly difficult. These powerful drugs can do more harm than good by severely sickening a patient during treatment. Some researchers are using membrane-like particles called liposomes to package and deliver drugs to tumors. Liposomes are oily, microscopic capsules that can be filled with biological cargo, such as a drug. They are very, very small—only one one-thousand the width of a single human hair. Researchers have known about liposomes for many years, but getting them to the right place in the body hasn't been easy. Once in the bloodstream, these foreign particles are immediately shipped to the liver and spleen, where they are destroyed. Scientists who study anesthetic medicines have a daunting task—for the most part, they are "shooting in the dark" when it comes to identifying the molecular targets of these drugs. Researchers do know that anesthetics share one common ingredient: Nearly all of them somehow target membranes, the oily wrappings surrounding cells. However, despite the fact that anesthesia is a routine part of surgery, exactly how anesthetic medicines work in the body has remained a mystery for more than 150 years. It's an important problem, since anesthetics have multiple effects on key body functions, including critical processes such as breathing. Scientists define anesthesia as a state in which no movement occurs in response to what should be painful. The problem, is even though a patient loses a pain response, the anesthesiologist can't tell what is happening inside the person's organs and cells. Further complicating the issue, scientists know that many different types of drugs—with little physical resemblance to each other—can all produce anesthesia. This makes it difficult to track down causes and effects. Anesthesiologist Robert Veselis of the Memorial Sloan-Kettering Institute for Cancer Research in New York City clarified how certain types of these mysterious medicines work. Veselis and his coworkers measured electrical activity in the brains of healthy volunteers receiving anesthetics while they listened to different sounds. To determine how sedated the people were, the researchers measured reaction time to the sounds the people heard. To measure memory effects, they quizzed the volunteers at the end of the study about word lists they had heard before and during anesthesia. Veselis' experiments show that the anesthetics they studied affect separate brain areas to produce the two different effects of sedation and memory loss. The findings may help doctors give anesthetic medicines more effectively and safely and prevent reactions with other drugs a patient may be taking. Materials engineer David Needham of Duke University in Durham, North Carolina, is investigating the physics and chemistry of liposomes to better understand how the liposomes ad their cancer-fighting cargo can travel through the body. Needham worked for 10 years to create a special kind of liposome that melts at just a few degrees above body temperature. The end result is a tiny molecular "soccer ball" made from two different oils that wrap around a drug. At room temperature, the liposomes are solid and they stay solid at body temperature, so they can be injected into the bloodstream. The liposomes are designed to spill their drug cargo into a tumor when heat is applied to the cancerous tissue. Heat is know to perturb tumors, making the blood vessels surrounding cancer cells extra-leaky. As the liposomes approach the warmed tumor tissue, the "stitches" of the miniature soccer balls begin to dissolve, rapidly leaking the liposome's contents. Needham and Duke oncologist Mark Dewhirst teamed up to do animal studies with the heat-activated liposomes. Experiments in mice and dogs revealed that, when heated, the drug-laden capsules flooded tumors with a chemotherapy drug and killed the cancer cells inside. Researchers hope to soon begin the first stage of human studies testing the heat-triggered liposome treatment in patients with prostate and breast cancer. The results of these and later clinical trials will determine whether liposome therapy can be a useful weapon for treating breast and prostate cancer and other hard-to-treat solid tumors. David Needham designed liposomes resembling tiny molecular "soccer balls" made from two different oils that wrap around a drug. LAWRENCE MAYER, LUDGER ICKENSTEIN, KATRINA EDWARDS

https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Aquatic_Chemistry/Water_Treatment

Water treatment is` a process of making water suitable for its application or returning its natural state. Thus, water treatment required before and after its application. The required treatment depends on the application. For example, treatment of greywater (from bath, dish and wash water) differs from the black water (from flush toilets). Composting toilet is not allowed in urban dwelling. Yet, composting toilets are used in a 30,000-square-foot office complex at the Institute of Asian Research, University of British Columbia. Water treatment involves science, engineering, business, and art. The treatment may include mechanical, physical, biological, and chemical methods. As with any technology, science is the foundation, and engineering makes sure that the technology works as designed. The appearance and application of water is an art. In terms of business, RGF Environmental, Water Energy Technologies, Aquasana Store, Vitech, Recalyx Industrial SDN BHD and PACE Chemicals ltd are some of many companies that offer various processes for water treatment. Millipore, a Fisher Scientific partner, offers many lines of products to produce , using a combination of active charcoal membranes, and reverse osmosis filter. Internet sites of these companies offer useful information regarding water. An environmental scientist or consultant matches the service provider, modify if necessary, with the requirement. Water is a renewable resource. All water treatments involve the removal of solids, bacteria, algae, plants, inorganic compounds, and organic compounds. Removal of solids is usually done by filtration and sediment. Bacteria digestion is an important process to remove harmful . Converting used water into environmentally acceptable water or even drinking water is Water in the Great Lakes Region is an organization dealing with the water resources. Ontario Clean Water Agency (OCWA) is a provincial Crown corporation in business to provide environmentally responsible and cost-efficient water and wastewater services. It currently operates more than 400 facilities for 200 municipalities. This web site provides information on water and water treatment. In April 1993, 403,000 people in Milwaukee were ill as a result of contaimination of water due to spring run off. This outbreak caused the more stringent regulations to be implemented in the public dringking water system. The measures were aimed at removing cryptosporidium. In May 2000, due to torrential downpour surface water got into shallow wells in a small town Walkerton, Ontario, Canada. On May 17, some residents complained of fever, bloody diarrhea and vomiting. This was know as the Walkerton Outbreak. Nearly half of the population of the town fell ill, and several people died due to the E. Coli O157:H7 infection. A public inquiry recommended many measures to prevent similar outbreaks. These measures were aimed at eliminating E. Coli. As a general discussion, let us look at a typical process in sewage treatment. A flow diagram for a general sewage treatment plant from Water Education, Department of Computer Science, University of Exeter, U.K., is shown below: Sewage is SCREENED to remove large solid chunks, which are disposed in LANDFILL SITE. It flows over to the SETTLEMENT TANK to let the fine particles to settle. The settlement is called the activated SLUDGE. The supernatant is then PERCOLATING FILTERED and/or AERATED. The water can be filtered again, and then disinfected (chlorinated in most cases). When there is no other complication, the water is returned to nature back to the ecological cycle. The SLUDGE removed from the settlement is composed of living biological material. A portion of it may be returned to the AERATION TANK, but the raw SLUDGE is digested by both microorganism. Anaerobic (without oxygen) and aerobic (with air) bacteria digestions are used. At the digestion stage, carbon dioxide, ammonia, and methane gases are evolved. Volume of the digested sludge is reduced, and it is acceptable as a fertilizer supplement in farming. Although the sewage water may be discharged back to the ecological system after AERATED DIGESTION and PERCOLATING FILTRATION, but in some cases, further treatment is required. Some general consideration of water treatment is given below. A rather recent book, by S.D. Faust and O.M. Aly, 2nd Ed. (1998) [TD433 F38 1998], addresses the problem of quality natural and treated water. The first three chapters discuss the criteria and standards for drinking water quality, organic compounds in waters, taste and order of water. Understandably, the standards change over the years. So are the standards of treated waters. Guidelines are available from government agencies such as Environment Canada which is equivalent to U.S. Public Health Service and the Environment Protection Agency (EPA). We have talked about drinking water in Water Biology. Next seven chapters deal with the removal of the following: There is a chapter dealing with These items cover the chemistry, biology, and physics involved in the treatment of water. Some of these topics have been discussed in chemistry of water, physical properties of water, biology of water, and natural water. Introductions are going to be given to some selected topics below. Treatment by activated carbon is mostly due to adsorption or absorption. When a chemical species is adhered to the surface of a solid, it is an . When partial chemical bonds are formed between adsorbed species or when the absorbate got into the channels of the solids, we call it . However, these two terms are often used to mean the same, because to distinguish one from type from the other is very difficult. Application of activated charcoal for the removal of undesirable order and taste in drinking water has been recognized at the dawn of civilization. Using bone char and charred vegetation, gravel, and sand for the filtration of water for domestic application has been practised for thousands of years. Active research and production of activated charcoal was accelerated during the two world wars. The use of poison gas prompted the development of masks. They are still in use today. Charcoal absorbs many substances, ranging from colored organic particulates to inorganic metal ions. Charcoal has been used to remove the colour of raw sugar from various sources. Charcoal consists of microcrystallites of graphite. The particles are so small in charcoal that they were considered amorphous. The crystal structure of graphite consists of layers of hexagonal networks, stacked on top of each other. Today, making activated carbon is a new and widely varied industry. Other molecules attach themselves to the porous surface and dangling carbons in these microcrystallites. Carbon containing substances are charred at less than 900 K to produce carbon in the manufacture of . However, the carbon is activated at 1200 K using oxidizing agent to selectively oxidize portions of the char to produce pores in the material. Because of the special process to produce used, these materials with high surface to mass ratio, they are called rather than activated charcoal. Factors affecting the absorption are particle size, surface area, pore structure, acidity (pH), temperature, and the nature of the material to be absorbed. Usually, adsorption (absorption) equilibria and rate of adsorption must be considered for effective applications. Natural and wastewater containing small particulates. They are suspended in water forming a . These particles carry the same charges, and repulsion prevents them from combining into larger particulates to settle. Thus, some chemical and physical techniques are applied to help them settle. The phenomenon is known as . A well known method is the addition of electrolyte. Charged particulates combine with ions neutralizing the charges. The neutral particulates combine to form larger particles, and finally settle down. Another method is to use high-molecular-weight material to attract or trap the particulates and settle down together. Such a process is called . Starch and multiply charged ions are often used. Historically, dirty water is cleaned by treating with alum, Al (SO ) 12 H O, and lime, Ca(OH) . These electrolytes cause the pH of the water to change due to the following reactions: Al2(SO4)3.12 H2O, -> Al(aq)3+ + 3 SO4(aq)2- + 12 H2O SO4(aq)2- + H2O -> HSO4(aq)- + OH- (causing pH change) Ca(OH)2 -> Ca(aq)2+ + 2 OH- (causing pH change) The slightly basic water causes Al(OH) , Fe(OH) and Fe(OH) to precipitate, bringing the small particulates with them and the water becomes clear. Some records have been found that Egyptians and Romans used these techniques as early as 2000 BC. Suspension of iron oxide particulates and humic organic matter in water gives water the yellow muddy appearance. Both iron oxide particulates and organic matter can be removed from coagulation and flocculation. The description given here is oversimplified, and many more techniques have been applied in the treatment of water. Coagulation is a major application of lime in the treatment of wastewater. Other salts such as iron sulfates Fe (SO ) and FeSO , chromium sulfate Cr (SO ) , and some special polymers are also useful. Other ions such as sodium, chloride, calcium, magnesium, and potassium also affect the coagulation process. So do temperature, pH, and concentration. Disposal of coagulation sludge is a concern, however. let the water sit around to let the floculated or coagulated particles to settle out. It works best with relatively dense particles (e.g. silt and minerals), while flotation works better for lighter particles (e.g. algae, color). A settling tank should be big enough so that it takes a long time (ideally 4 hours +) to get through. Inlets and outlets are designed so the water moves slowly in the tank. Long and narrow channels are installed to let the water to snake its way through the tank. The settled particles, sludge, must occasionally be removed from the tanks. The water is next ready to be filtered. Sedimentation is used in pre-treatment and wastewater treatment. is the process of removing solids from a fluid by passing it through a porous medium. Coarse, medium, and fine porous media have been used depending on the requirement. The filter media are artificial membranes, nets, sand filter, and high technological filter systems. The choice of filters depends on the required filtering speed and the requirement. The flow required for filtration can be achieved using gravity or pressure. In pressure filtration, one side of the filter medium is at higher pressure than that of the other so that the filter plane has a pressure drop. Some portion of this filter type must be enclosed in a container. The process of removing the clogged portion of the filter bed by reversing the flow through the bed and washing out the solid is called . During this process, the solid must be removed out of the system, but otherwise the filters must be either replaced or taken out of service to be cleaned. manufacture water filters as shown here. This unit consists of four filters. Regarding the filtering system, its technical info gave the following statement. The units are designed for emergency and perhaps undeveloped countries. AquaSelect of Mississauga has a pitcher water filter system, and , its web site claims. Brita filters is very popular. Bringing air into intimate contact with water for the purpose of exchanging certain components between the two phases is called . Oxygenation is one of the purposes of aeration. Others are removal of volatile organic substances, hydrogen sulfide, ammonia, and volatile organic compounds. A gas or substance dissolved in water may further react with water. Such a reaction is called . Ionic substance dissolve due to hydration, for example: \[\ce{HCl (g) + x H2O <=> H(H2O)_{x}^{+} + Cl(aq)^{-}}\] \[\ce{H2S <=> H^{+}(aq) + HS^{-}(aq)}\] These reactions are reversible, and aeration may also causes dehydration resulting in releasing the gas from water. Henry's law is applicable to this type of equilibrium for consideration. Methods of aeration are In the following discussion, a dilute solution and a concentrated solution are considered. The dilute solution can be a clean water whereas the concentrated solution contains undesirable solute (electrolyte or others). When a compartment containing a dilute solution is connected to another compartment containing a concentrated solution by a semipermeable membrane, water molecules move from the dilute solution to concentrated solution. This phenomenon is called . Pig bladders are natural semipermeable membranes. As the water molecules migrate through the semipermeable membrane, water level in the solution will increase until the (osmotic) pressure prevents a net migration of water molecules in one direction. A pressure equivalent to the height difference is called the . The illustration given on the right is from the PurePro, one of the many companies that manufacture reverse osmosis water filter devices. Millipore also use this technique. By applying pressure in the higher concentration solution, water molecules migrate from a high concentration solution to a low concentration solution. This method is called . The concept of reverse osmosis is illustrated in the diagram here from . In this technique, the membrane must be able to tolerate the high pressure, and prevent solute molecules to pass through. Regarding membranes, PurePro made the following statement: This technology certainly works, and it has been used to convert salt (ocean or sea) water into fresh water. With this technique, the water with higher concentration is discharged. Thus, this technology is costly in regions where the water cost is high. Free Drinking Water also uses reverse osmosis filter system for domestic applications. The Environment Canada's (Atlantic Region) Waste Management and Remediation Section contains links to detailed information on the programs and activities relating to Petroleum and Allied Petroleum Storage Tank Systems, Ocean Disposal, Contaminated Sites, and Hazardous Waste Disposal Advice for the provinces of Newfoundland and Labrador, Nova Scotia, New Brunswick and Prince Edward Island. It also contains links to sites within and outside of Environment Canada with related information.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Averill_et_al./11.E%3A_Liquids_(Exercises)

   " by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact During cold periods, workers in the citrus industry often spray water on orange trees to prevent them from being damaged, even though ice forms on the fruit. ♦ Relative humidity is the ratio of the actual partial pressure of water in the air to the vapor pressure of water at that temperature (i.e., if the air was completely saturated with water vapor), multiplied by 100 to give a percentage. On a summer day in the Chesapeake, when the temperature was recorded as 35°C, the partial pressure of water was reported to be 33.9 mmHg. The following table gives the vapor pressure of water at various temperatures. Calculate the relative humidity. ♦ Liquids are frequently classified according to their physical properties, such as surface tension, vapor pressure, and boiling point. Such classifications are useful when substitutes are needed for a liquid that might not be available. ♦ In the process of freeze drying, which is used as a preservation method and to aid in the shipping or storage of fruit and biological samples, a sample is cooled and then placed in a compartment in which a very low pressure is maintained, ≈0.01 atm. ♦ Many industrial processes for preparing compounds use “continuous-flow reactors,” which are chemical reaction vessels in which the reactants are mixed and allowed to react as they flow along a tube. The products are removed at a certain distance from the starting point, when the reaction is nearly complete. The key operating parameters in a continuous-flow reactor are temperature, reactor volume, and reactant flow rate. As an industrial chemist, you think you have successfully modified a particular process to produce a higher product yield by substituting one reactant for another. The viscosity of the new reactant is, however, greater than that of the initial reactant. Describe the changes that take place when a liquid is heated above its critical temperature. How does this affect the physical properties? What is meant by the term ? What is the effect of increasing the pressure on a gas to above its critical pressure? Would it make any difference if the temperature of the gas was greater than its critical temperature? Do you expect the physical properties of a supercritical fluid to be more like those of the gas or the liquid phase? Explain. Can an ideal gas form a supercritical fluid? Why or why not? What are the limitations in using supercritical fluids to extract organic materials? What are the advantages? Describe the differences between a molten salt and an ionic liquid. Under what circumstances would an ionic liquid be preferred over a molten salt? Describe the common structural features of molecules that form liquid crystals. What kind of intermolecular interactions are most likely to result in a long-chain molecule that exhibits liquid crystalline behavior? Does an electrical field affect these interactions? What is the difference between an liquid and an liquid? Which is more anisotropic—a cholesteric liquid crystal or a nematic liquid crystal?

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Carbohydrates/Polysaccharides/Cellulose

Polysaccharides are carbohydrate polymers consisting of tens to hundreds to several thousand monosaccharide units. All of the common polysaccharides contain glucose as the monosaccharide unit. Polysaccharides are synthesized by plants, animals, and humans to be stored for food, structural support, or metabolized for energy. The major component in the rigid cell walls in plants is cellulose. Cellulose is a linear polysaccharide polymer with many glucose monosaccharide units. The acetal linkage is beta which makes it different from starch. This peculiar difference in acetal linkages results in a major difference in digestibility in humans. Humans are unable to digest cellulose because the appropriate enzymes to breakdown the beta acetal linkages are lacking. Indigestible cellulose is the fiber which aids in the smooth working of the intestinal tract. Animals such as cows, horses, sheep, goats, and termites have symbiotic bacteria in the intestinal tract. These symbiotic bacteria possess the necessary enzymes to digest cellulose in the GI tract. They have the required enzymes for the breakdown or hydrolysis of the cellulose; the animals do not, not even termites, have the correct enzymes. No vertebrate can digest cellulose directly. Even though we cannot digest cellulose, we find many uses for it including: Wood for building; paper products; cotton, linen, and rayon for clothes; nitrocellulose for explosives; cellulose acetate for films. The structure of cellulose consists of long polymer chains of glucose units connected by a beta acetal linkage. The graphic on the left shows a very small portion of a cellulose chain. All of the monomer units are beta-D-glucose, and all the beta acetal links connect C # 1 of one glucose to C # 4 of the next glucose. Carbon # 1 is called the anomeric carbon and is the center of an acetal functional group. A carbon that has two ether oxygens attached is an acetal. The Beta position is defined as the ether oxygen being on the same side of the ring as the C # 6. In the chair structure this results in a horizontal or up projection. This is the same definition as the -OH in a hemiacetal. Cellulose: Beta glucose is the monomer unit in cellulose. As a result of the bond angles in the beta acetal linkage, cellulose is mostly a linear chain. Starch: Alpha glucose is the monomer unit in starch. As a result of the bond angles in the alpha acetal linkage, starch-amylose actually forms a spiral much like a coiled spring. Dietary fiber is the component in food not broken down by digestive enzymes and secretions of the gastrointestinal tract. This fiber includes hemicelluloses, pectins, gums, mucilages, cellulose, (all carbohydrates) and lignin, the only non-carbohydrate component of dietary fiber. High fiber diets cause increased stool size and may help prevent or cure constipation. Cereal fiber, especially bran, is most effective at increasing stool size while pectin has little effect. Lignin can be constipating. Fiber may protect against the development of colon cancer, for populations consuming high fiber diets have a low incidence of this disease. The slow transit time (between eating and elimination) associated with a low fiber intake would allow more time for carcinogens present in the colon to initiate cancer. But constipated people do not have a higher incidence of colon cancer than fast eliminators, so fiber's role in colon cancer remains unclear. Dietary fiber may limit cholesterol absorption by binding bile acids. High fiber diets lower serum cholesterol and may prevent cardiovascular disease. Some fibers, such as pectin and rolled oats, are more effective than others, such as wheat, at lowering serum cholesterol. Dietary fiber is found only in plant foods such as fruits, vegetables, nuts, and grains. Whole wheat bread contains more fiber than white bread and apples contain more fiber than apple juice, which shows that processing food generally removes fiber.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Concept_Development_Studies_in_Chemistry_(Hutchinson)/11%3A__The_Ideal_Gas_Law

We assume as our starting point the atomic molecular theory. That is, we assume that all matter is composed of discrete particles. The elements consist of identical atoms, and compounds consist of identical molecules, which are particles containing small whole number ratios of atoms. We also assume that we have determined a complete set of relative atomic weights, allowing us to determine the molecular formula for any compound. The individual molecules of different compounds have characteristic properties, such as mass, structure, geometry, bond lengths, bond angles, polarity, diamagnetism, or paramagnetism. We have not yet considered the properties of mass quantities of matter, such as density, phase (solid, liquid, or gas) at room temperature, boiling and melting points, reactivity, and so forth. These are properties which are not exhibited by individual molecules. It makes no sense to ask what the boiling point of one molecule is, nor does an individual molecule exist as a gas, solid, or liquid. However, we do expect that these material or bulk properties are related to the properties of the individual molecules. Our ultimate goal is to relate the properties of the atoms and molecules to the properties of the materials which they comprise. Achieving this goal will require considerable analysis. In this Concept Development Study, we begin at a somewhat more fundamental level, with our goal to know more about the nature of gases, liquids, and solids. We need to study the relationships between the physical properties of materials, such as density and temperature. We begin our study by examining these properties in gases. It is an elementary observation that air has a "spring" to it: if you squeeze a balloon, the balloon rebounds to its original shape. As you pump air into a bicycle tire, the air pushes back against the piston of the pump. Furthermore, this resistance of the air against the piston clearly increases as the piston is pushed farther in. The "spring" of the air is measured as a pressure, where the pressure \(P\) is defined \[P = \dfrac{F}{A}\] \(F\) is the force exerted by the air on the surface of the piston head and \(A\) is the surface area of the piston head. For our purposes, a simple pressure gauge is sufficient. We trap a small quantity of air in a syringe (a piston inside a cylinder) connected to the pressure gauge, and measure both the volume of air trapped inside the syringe and the pressure reading on the gauge. In one such sample measurement, we might find that, at atmospheric pressure \(\left( 760 \: \text{torr} \right)\), the volume of gas trapped inside the syringe is \(29.0 \: \text{mL}\). We then compress the syringe slightly, so that the volume is now \(23.0 \: \text{mL}\). We feel the increased spring of the air, and this is registered on the gauge as an increase in pressure to \(960 \: \text{torr}\). It is simple to make many measurements in this manner. A sample set of data appears in Table 11.1. We note that, in agreement with our experience with gases, the pressure increases as the volume decreases. These data are plotted in Figure 11.1. An initial question is whether there is a quantitative relationship between the pressure measurements and the volume measurements. To explore this possibility, we try to plot the data in such a way that both quantities increase together. This can be accomplished by plotting the pressure versus the inverse of the volume, rather than versus the volume. The data are given in Table 11.2 and plotted in Figure 11.2. Notice also that, with elegant simplicity, the data points form a straight line. Furthermore, the straight line seems to connect to the origin \(\{ 0, 0 \}\). This means that the pressure must simply be a constant multiplied by \(\dfrac{1}{V}\): \[P = k \times \dfrac{1}{V}\] If we multiply both sides of this equation by \(V\), then we notice that \[PV = k\] In other words, if we go back and multiply the pressure and the volume together for each experiment, we should get the same number each time. These results are shown in the last column of Table 11.2, and we see that, within the error of our data, all of the data points give the same value of the product of pressure and volume. (The volume measurements are given to three decimal places and hence are accurate to a little better than \(1\%\). The values of Pressure \(\times\) Volume are all within \(1\%\) of each other, so the fluctuations are not meaningful.) We should wonder what significance, if any, can be assigned to the number \(22040 \: \text{torr} \cdot \text{mL}\) we have observed. It is easy to demonstrate that this "constant" is not so constant. We can easily trap any amount of air in the syringe at atmospheric pressure. This will give us any volume of air we wish at \(760 \: \text{torr}\) pressure. Hence, the value \(22040 \: \text{torr} \cdot \text{mL}\) is only observed for the particular amount of air we happened to choose in our sample measurement. Furthermore, if we heat the syringe with a fixed amount of air, we observe that the volume increases, thus changing the value of the \(22040 \: \text{torr} \cdot \text{mL}\). Thus, we should be careful to note that the . This observation is referred to as , dating to 1662. The data given in Table 11.1 assumed that we used air for the gas sample. (That, of course, was the only gas with which Boyle was familiar.) We now experiment with varying the composition of the gas sample. For example, we can put oxygen, hydrogen, nitrogen, helium, argon, carbon dioxide, water vapor, nitrogen dioxide, or methane into the cylinder. In each case we start with \(29.0 \: \text{mL}\) of gas at \(760 \: \text{torr}\) and \(25^\text{o} \text{C}\). We then vary the volumes as in Table 11.1 and measure the pressures. Remarkably, we find that the pressure of each gas is exactly the same as every other gas at each volume given. For example, if we press the syringe to a volume of \(16.2 \: \text{mL}\), we observe a pressure of \(1360 \: \text{torr}\), no matter which gas is in the cylinder. This result also applies equally well to mixtures of different gases, the most familiar being air, of course. We conclude that the pressure of a gas sample depends on the volume of the gas and the temperature, but not on the composition of the gas sample. We now add to this result a conclusion from a previous study. Specifically, we recall the Law of Combining Volumes, which states that, when gases combine during a chemical reaction at a fixed pressure and temperature, the ratios of their volumes are simple whole number ratios. We further recall that this result can be explained in the context of the atomic molecular theory by hypothesizing that equal volumes of gas contain equal numbers of gas particles, independent of the type of gas, a conclusion we call . Combining this result with Boyle's Law reveals that the of a gas depends on the of gas particles, the in which they are contained, and the of the sample. The pressure does depend on the type of gas particles in the sample or whether they are even all the same. We can express this result in terms of Boyle's Law by noting that, in the equation \(PV = k\), the "constant" \(k\) is actually a function which varies with both number of gas particles in the sample and the temperature of the sample. Thus, we can more accurately write \[PV = k \left( N, t \right)\] explicitly showing that the product of pressure and volume depends on \(N\), the number of particles in the gas sample, and \(t\), the temperature. It is interesting to note that, in 1738, Bernoulli showed that the inverse relationship between pressure and volume could be proven by assuming that a gas consists of individual particles colliding with the walls of the container. However, this early evidence for the existence of atoms was ignored for roughly 120 years, and the atomic molecular theory was not to be developed for another 70 years, based on mass measurements rather than pressure measurements. We have already noted the dependence of Boyle's Law on temperature. To observe a constant product of pressure and volume, the temperature must be held fixed. We next analyze what happens to the gas when the temperature is allowed to vary. An interesting first problem that might not have been expected is the question of how to measure temperature. In fact, for most purposes, we think of temperature only in the rather non-quantitative manner of "how hot or cold" something is, but then we measure temperature by examining the length of mercury in a tube, or by the electrical potential across a thermocouple in an electronic thermometer. We then briefly consider the complicated question of just what we are measuring when we measure the temperature. Imagine that you are given a cup of water and asked to describe it as "hot" or "cold". Even without a calibrated thermometer, the experiment is simple: you put your finger in it. Only a qualitative question was asked, so there is no need for a quantitative measurement of "how hot" or "how cold". The experiment is only slightly more involved if you are given two cups of water and asked which one is hotter or colder. A simple solution is to put one finger in each cup and to directly compare the sensation. You still don't need a calibrated thermometer or even a temperature scale at all. Finally, imagine that you are given a cup of water each day for a week at the same time and are asked to determine which day's cup contained the hottest or coldest water. Since you can no longer trust your sensory memory from day to day, you have no choice but to define a temperature scale. To do this, we make a physical measurement on the water by bringing it into contact with something else whose properties depend on the "hotness" of the water in some unspecified way. (For example, the volume of mercury in a glass tube expands when placed in hot water; certain strips of metal expand or contract when heated; some liquid crystals change color when heated; etc.) We assume that this property will have the same value when it is placed in contact with two objects which have the same "hotness" or temperature. Somewhat obliquely, this defines the temperature measurement. For simplicity, we illustrate with a mercury-filled glass tube thermometer. We observe quite easily that when the tube is inserted in water we consider "hot", the volume of mercury is larger than when we insert the tube in water that we consider "cold". Therefore, the volume of mercury is a measure of how hot something is. Furthermore, we observe that, when two very different objects appear to have the same "hotness", they also give the same volume of mercury in the glass tube. This allows us to make quantitative comparisons of "hotness" or temperature based on the volume of mercury in a tube. All that remains is to make up some numbers that define the scale for the temperature, and we can literally do this in any way that we please. This arbitrariness is what allows us to have two different, but perfectly acceptable, temperature scales, such as Fahrenheit and Centigrade. The latter scale simply assigns zero to be the temperature at which water freezes at atmospheric pressure. We then insert our mercury thermometer into freezing water, and mark the level of the mercury as "0". Another point on our scale assigns 100 to be the boiling point of water at atmospheric pressure. We insert our mercury thermometer into boiling water and mark the level of mercury as "100". Finally, we just mark off in increments of \(\dfrac{1}{100}\) of the distance between the "0" and the "100" marks, and we have a working thermometer. Given the arbitrariness of this way of measuring temperature, it would be remarkable to find a quantitative relationship between temperature and any other physical property. Yet that is what we now observe. We take the same syringe used in the previous section and trap it in a small sample of air at room temperature and atmospheric pressure. (From our observations above, it should be clear that the type of gas we use is irrelevant.) The experiment consists of measuring the volume of the gas sample in the syringe as we vary the temperature of the gas sample. In each measurement, the pressure of the gas is held fixed by allowing the piston in the syringe to move freely against atmospheric pressure. A sample set of data is shown in Table 11.3 and plotted in Figure 11.3. We find that there is a simple linear (straight line) relationship between the volume of a gas and its temperature as measured by a mercury thermometer. We can express this in the form of an equation for a line: \[V = \alpha t + \beta\] where \(V\) is the volume and \(t\) is the temperature in \(^\text{o} \text{C}\). \(\alpha\) and \(\beta\) are the slope and intercept of the line, and in this case, \(\alpha = 0.335\) and \(\beta = 91.7\). We can rewrite this equation in a slightly different form: \[V = \alpha \left( t + \dfrac{\beta}{\alpha} \right)\] This is the same equation, except that it reveals that the quantity \(\dfrac{\beta}{\alpha}\) must be a temperature, since we can add it to a temperature. This is a particularly important quantity: if we were to set the temperature of the gas equal to \(-\dfrac{\beta}{\alpha} = -273^\text{o} \text{C}\), we would find that the volume of the gas would be exactly 0! (This assumes that this equation can be extrapolated to that temperature. This is quite an optimistic extrapolation, since we haven't made any measurements near to -273^\text{o} \text{C}\). In fact, our gas sample would condense to a liquid or solid before we ever reached that low temperature.) Since the volume depends on the pressure and the amount of gas (Boyle's Law), then the values of \(\alpha\) and \(\beta\) also depend on the pressure and amount of gas and carry no particular significance. However, when we repeat our observations for many values of the amount of gas and the fixed pressure, we find that the \(-\dfrac{\beta}{\alpha} = -273^\text{o} \text{C}\) does not vary from one sample to the next. Although we do not know the physical significance of this temperature at this point, we can assert that it is a true constant, independent of any choice of the conditions of the experiment. We refer to this temperature as , since a temperature below this value would be predicted to produce a negative gas volume. Evidently, then, we cannot expect to lower the temperature of any gas below this temperature. This provides us an "absolute temperature scale" with a zero which is not arbitrarily defined. This we define by adding 273 (the value of \(\dfrac{\beta}{\alpha}\) to temperatures measured in \(^\text{o} \text{C}\), and we define this scale to be in units of degrees Kelvin \(\left( K \right)\). The data in Table 11.3 are now recalibrated to the absolute temperature scale in Table 11.4 and plotted in Figure 11.4. Note that in degrees Kelvin, \[V = kT\] provided that the pressure and amount of gas are held constant. This result is known as , dating to 1787. As with Boyle's Law, we must now note that the "constant" \(k\) is not really constant, since the volume also depends on the pressure and quantity of gas. Also as with Boyle's Law, we note that Charles' Law does not depend on the of gas on which we make the measurements, but rather depends only on the number of particles of gas. Therefore, we slightly rewrite Charles' Law to explicitly indicate the dependence of \(k\) on the pressure and number of particles of gas \[V = k \left( N, P \right) T\] We have been measuring four properties of gases: pressure, volume, temperature, and "amount", which we have assumed above to be the number of particles. The results of three observations relate these four properties pairwise. Boyle's Law relates the pressure and volume at constant temperature and amount of gas: \[P \times V = k_1 \left( N, T \right)\] Charles' Law relates the volume and temperature at constant pressure and amount of gas: \[V = k_2 \left( N, P \right) T\] The Law of Combining Volumes leads to Avogadro's Hypothesis that the volume of a gas is proportional to the number of particles \(\left( N \right)\) provided that the temperature and pressure are held constant. We can express this as \[V = k_3 \left( P, T \right) N\] We will demonstrate below that these three relationships can be combined into a single equation relating \(P\), \(V\), \(T\), and \(N\). Jumping to the conclusion, however, we can more easily show that these three relationships can be considered as special cases of the more general equation known as the : \[PV = nRT\] where \(R\) is a constant and \(n\) is the number of moles of gas, related to the number of particles \(N\) by Avogadro's number, \(N_A\) \[n = \dfrac{N}{N_A}\] In Boyle's Law, we examine the relationship of \(P\) and \(V\) when \(n\) (not \(N\)) and \(T\) are fixed. In the Ideal Gas Law, when \(n\) and \(T\) are constant, \(nRT\) is constant, so the product \(PV\) is also constant. Therefore, Boyle's Law is a special case of the Ideal Gas Law. If \(n\) and \(P\) are fixed in the Ideal Gas Law, then \(V = \dfrac{nR}{P} T\) and \(\dfrac{nR}{P}\) is a constant. Therefore, Charles' Law is also a special case of the Ideal Gas Law. Finally, if \(P\) and \(T\) are constant, then in the Ideal Gas Law, \(V = \dfrac{RT}{P} n\) and the volume is proportional to the number of moles or particles. Hence, Avogadro's Hypothesis is a special case of the Ideal Gas Law. We have now shown that each of our experimental observations is consistent with the Ideal Gas Law. We might ask, though, how did we get the Ideal Gas Law? We would like to derive the Ideal Gas Law from the three experimental observations. To do so, we need to learn about the functions \(k_1 \left( N, T \right)\), \(k_2 \left( N, P \right)\), and \(k_3 \left( P, T \right)\). We begin by examining Boyle's Law in more detail: if we hold \(N\) and \(P\) fixed in Boyle's Law and allow \(T\) to vary, the volume must increase with the temperature in agreement with Charles' Law. In other words, with \(N\) and \(P\) fixed, the volume must be proportional to \(T\). Therefore, \(k_1\) in Boyle's Law must be proportional to \(T\): \[k_1 \left( N, T \right) = k_4 \left( N \right) \times T\] where \(k_4\) is a new function which depends only on \(N\). The Boyle's Law equation then becomes \[P \times V = k_4 \left( N \right) T\] Avogadro's Hypothesis tells us that, at constant pressure and temperature, the volume is proportional to the number of particles. Therefore \(k_4\) must also increase proportionally with the number of particles: \[k_4 \left( N \right) = k \times N\] where \(k\) is yet another new constant. In this case, however, there is no variables left, and \(k\) is truly a constant. Combining equations gives \[P \times V = k \times N \times T\] This is very close to the Ideal Gas Law, except that we have the number of particles, \(N\), instead of the number of moles, \(n\). We put this result in the more familiar form by expressing the number of particles in terms of the number of moles, \(n\), by dividing the number of particles by Avogadro's number \(N_A\). Then, the equation becomes \[P \times V = k \times N_A \times n \times T\] The two constants, \(k\) and \(N_A\), can be combined into a single constant, which is commonly called \(R\), the gas constant. This produces the familiar conclusion of \(PV = nRT\). We referred briefly above to the pressure of mixtures of gases, noting in our measurements leading to Boyle's Law that the total pressure of the mixture depends only on the number of moles of gas, regardless of the types and amounts of gases in the mixture. The Ideal Gas Law reveals that the pressure exerted by a mole of molecules does not depend on what those molecules are, and our earlier observation about gas mixtures is consistent with that conclusion. We now examine the actual process of mixing two gases together and measuring the total pressure. Consider a container of fixed volume \(25.0 \: \text{L}\). We inject into that container \(0.78 \: \text{mol}\) of \(\ce{N_2}\) gas at \(298 \: \text{K}\). From the Ideal Gas Law, we can easily calculate the measured pressure of the nitrogen gas to be \(0.763 \: \text{atm}\). We now take an identical container of fixed volume \(25.0 \: \text{L}\), and we inject into that container \(0.22 \: \text{mol}\) of \(\ce{O_2}\) gas at \(298 \: \text{K}\). The measured pressure of the oxygen gas is \(0.215 \: \text{atm}\). As a third measurement, we inject \(0.22 \: \text{mol}\) of \(\ce{O_2}\) gas at \(298 \: \text{K}\) into the container which already has \(0.78 \: \text{mol}\) of \(\ce{N_2}\). (Note that the mixture of gases we have prepared is very similar to that of air.) The measured pressure in this container is now found to be \(0.975 \: \text{atm}\). We note now that the pressure of the mixture of \(\ce{N_2}\) and \(\ce{O_2}\) in the container is equal to the sum of the pressures of the \(\ce{N_2}\) and \(\ce{O_2}\) samples taken separately. We now define the of each gas in the mixture to be the pressure of each gas as if it were the only gas present. Our measurements tell us that the partial pressure of \(\ce{N_2}\), \(P_{N_2}\), is \(0.763 \: \text{atm}\), and the partial pressure of \(\ce{O_2}\), \(P_{O_2}\), is \(0.215 \: \text{atm}\). With this definition, we can now summarize our observation by saying that the total pressure of the mixture of oxygen and nitrogen is equal to the sum of the partial pressures of the two gases. This is a general result: . The total pressure of a mixture of gases is the sum of the partial pressures of the component gases in the mixture. Sketch a graph with two curves showing Pressure vs. Volume for two different values of the number of moles of gas, with \(n_2 > n_1\), both at the same temperature. Explain the comparison of the two curves. Sketch a graph with two curves showing Pressure vs. 1/Volume for two different values of the number of moles of gas, with \(n_2 > n_1\), both at the same temperature. Explain the comparison of the two curves. Sketch a graph with two curves showing Volume vs. Temperature for two different values of the number of moles of gas, with \(n_2 > n_1\), both at the same pressure. Explain the comparison of the two curves. Sketch a graph with two curves showing Volume vs. Temperature for two different values of the pressure of the gas, with \(P_2 > P_1\), both for the same number of moles. Explain the comparison of the two curves. Explain the significance of the fact that, in the volume-temperature experiments, \(\dfrac{\beta}{\alpha}\) is observed to have the same value, independent of the quantity of gas studied and the type of gas studied. What is the significance of the quantity \(\dfrac{\beta}{\alpha}\)? Why is it more significant than either \(\beta\) or \(\alpha\)? says that the pressure of a gas is proportional to the absolute temperature for a fixed quantity of gas in a fixed volume. Thus, \(P = k \left( N, V \right) T\). Demonstrate that Amonton's Law can be derived by combining Boyle's Law and Charles' Law. Using Boyle's Law in your reasoning, demonstrate that the "constant" in Charles' Law, i.e. \(k_2 \left( N, P \right)\), is inversely proportional to \(P\). Explain how Boyle's Law and Charles' Law may be combined to the general result that, for constant quantity of gas, \(P \times V = kT\). Using Dalton's Law and the Ideal Gas Law, show that the partial pressure of a component of a gas mixture can be calculated from \[P_i = PX_i\] Where \(P\) is the total pressure of the gas mixture and \(X_i\) is the of component \(i\), defined by \[X_i = \dfrac{n_i}{n_\text{total}}\] Dry air is \(78.084\%\) nitrogen, \(20.946\%\) oxygen, \(0.934\%\) argon, and \(0.033\%\) carbon dioxide. Determine the mole fractions and partial pressures of the components of dry air at standard pressure. Assess the accuracy of the following statement: "Boyle's Law states that \(PV = k_1\), where \(k_1\) is a constant. Charles' Law states that \(V = k_2 T\), where \(k_2\) is a constant. Inserting \(V\) from Charles' Law into Boyle's Law results in \(P k_2 T = k_1\). We can rearrange this to read \(PT = \dfrac{k_1}{k_2} = \text{a constant}\). Therefore, the pressure of a gas is inversely proportional to the temperature of the gas." In your assessment, you must determine what information is correct or incorrect, provide the correct information where needed, explain whether the reasoning is logical or not, and provided logical reasoning where needed.   ; Chemistry)

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The earliest studies of electricity focused on electrostatics. can be produced when certain materials are rubbed together, like silk or hair on some metals or plastics (this is called the ). This leads to a separation of , with positive charge on the silk or hair and negative charge on the metal or plastic. You've probably noticed how this can lead to things sticking to each other if they have opposite charges, and you can also observe that the same charges will repel each other. Benjamin Franklin (one of the few early American scientists) proposed that the positive and negative charges resulted from having either too much or too little of the same "electrical fluid". The had been studied by many scientists, some of whom suggested that it followed a law similar to Newton's law for gravity, but Coulomb gets credit for this, because he did many experiments that improved understanding. In 1785, he published the law in its current form. The equation is: \[F=\frac{kQq}{r^{2}}\] where F is the force, k is a constant, Q and q are two charges, and r is the distance between Q and Q. Like charges (+/+ or -/-) repel and opposite charges (+/-) attract. , like the electricity used in any electrical device today, was discovered a little later. Galvani was studying physiology, and noticed that the legs of dead frogs twitched when in contact with two different metals. He attributed this to "animal electricity." Volta did many more experiments and discovered that animals were unnecessary for the effect. He put a piece of paper soaked in saltwater (the , because it conducts electricity) between disks of two different metals, then connected the metals with wire and noticed that electrical current flowed. When he stacked these on top of each other, alternating the metals, the effect increased. This device for generating current came to be called a "voltaic pile." Volta published his pile reports in 1800. That same year, two English scientists published their results of splitting water into hydrogen and oxygen using a Voltaic pile. This is an example of , which means using electricity to break chemical compounds. Humphry Davy was a scientist and popular lecturer who had discovered "laughing gas" or nitrous oxide, which is still used by doctors and dentists. He did many electrochemical studies after 1800. He noticed that during electrolysis of various compounds, hydrogen, metals, or bases appear around the negative pole, and oxygen or acid appears around the positive pole. Based on this, he guessed he could use electricity to break chemical bonds. The strong bases KOH and NaOH (potassium hydroxide and sodium hydroxide) had been known for a long time. (A neutralizes an .) Sodium and potassium were believed to be elements but had not been isolated. Davy was able to isolate them by melting and electrolyzing the solid sodium hydroxide and potassium hydroxide. The solid bases didn't conduct electricity, but once melted they did conduct, and then they separated into shiny metal and a gas. K and Na are . When Davy isolated them, he found that they were light, soft metals that react vigorously and spontaneously with water and air. So they have to be stored under oil. Davy also isolated magnesium, calcium, strontium, and barium (Mg, Ca, Sr, and Ba) using a slightly modified procedure involving mercury. These are the . They are also soft, light metals, but they are not quite as reactive as the alkali metals. Berzelius was a very influential chemist who worked very hard and very carefully to determine atomic weights, without using Gay-Lussac's law or Avogadro's hypothesis, and whom we mentioned earlier because he introduced the idea of isomerism. He also did many electrochemical studies and took Davy's hypothesis, that chemical and electrical attraction are the same force, much farther. His said: elements have positive or negative polarity, and chemical reactions partially neutralize that polarity. For example: (positive) + (negative) → (approximately neutral) He used the terms and , which are still used today with a somewhat different meaning. He thought oxygen was the most electronegative element, while metals were generally electropositive. He also considered the element polarity to be on a spectrum, so that sulfur, for instance, was positive with respect to oxygen and negative with respect to metals. Thus it could combine with both; this is very different from the modern understanding of Coulomb attraction because he did not have a clear sense of neutrality. are usually composed of a metal and a non-metal. For instance, salt, and many rocks, are ionic compounds. They are different from molecular compounds because they do not have distinct molecules that vaporize as a unit. They are held together by charges, not quite as Berzelius thought, but close enough that his theory worked all right for them. However, Berzelius' theory did not work well for molecular compounds! He said Avogadro must be wrong because two oxygen (O) atoms would have the same polarity and thus repel each other. Now we know that chemical bonding is more complicated than Berzelius realized. O atoms don't repel each other, and do form a diatomic molecule. Berzelius' theory delayed the progress of science for about 50 years because he didn't believe Avogadro's hypothesis, which was needed for the next big breakthrough in chemistry. Faraday was born to a poor family, and discovered chemistry when working in a book-making shop, where we worked on Jane Marcet's Conversations on Chemistry (one of the most popular textbooks at the time, it is surprising that it was written by a woman). He became Davy's assistant, and became a great experimentalist. He helped invent the words used for electrochemistry, such as ion, anion, cation, electrolyte and electrolysis. At that time, and meant the parts of the electrolyte (the salt used to make water conductive) that appeared at each electrode. Faraday studied the amount of current necessary to produce an amount of an electrolysis product. For instance, the current that produced 1g of hydrogen produced 8g of oxygen, 36g of chlorine, 125g of iodine, 104g of lead or 58g of tin. Faraday called the O, Cl and I anions, meaning that they were produced at the positive pole, and Sn and Pb cations, meaning that they were produced at the negative pole. These numbers could have solved the atomic weights problem, just like Avogadro's hypothesis, but Faraday wasn't interested in atomic weights or theories, and Berzelius, who was, didn't believe Faraday's electrical laws, just like he didn't believe Avogadro's hypothesis.

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Make sure you thoroughly understand the following essential concepts: Acid-base reactions pervade every aspect of industrial-, physiological-, and environmental chemistry. In this unit we touch on a few highlights that anyone who studies or practices chemical science should be aware of. Buffer solutions and the buffering effect they produce are extremely important in many practical applications of chemistry. The reason for this is that many chemical processes are quite sensitive to the pH; the extent of the reaction, its rate, and even the nature of the products can be altered if the pH is allowed to change. Such a change will tend to occur, for example, when the reaction in question, or an unrelated parallel reaction, consumes or releases hydrogen- or hydroxide ions. Many reactions that take place in living organisms fall into this category. Most biochemical processes are catalyzed by enzymes whose activities are highly dependent on the pH; if the local pH deviates too far from the optimum value, the enzyme may become permanently deactivated. Buffers are employed in a wide variety of laboratory procedures and industrial processes: Acid-base chemistry plays a crucial role in physiology, both at the level of the individual cell and of the total organism. The reasons for this are twofold: About two-thirds of the weight of an adult human consists of water. About two-thirds of this water is located within cells, while the remaining third consists of extracellular water, mostly interstial fluid that bathes the cells, and the blood plasma. The latter, amounting to about 5% of body weight (about 5 L in the adult), serves as a supporting fluid for the blood cells and acts as a means of transporting chemicals between cells and the external environment. It is basically a 0.15 M solution of NaCl containing smaller amounts of other electrolytes, the most important of which are HCO and protein anions. Respiration, the most important physiological activity of a cell, is an acid-producing process. Carbohydrate substances are broken down into carbon dioxide, and thus carbonic acid: \[C(H_2O)_{n} + O_2 → CO_2 + n H_2O\] It is remarkable that the pH of most cellular fluids can be kept within such a narrow range, given the large number of processes that tend to upset it. This is due to the exquisite balance between a large number of interlinked processes operating at many different levels. Acid-base balance in the body is maintained by two general mechanisms: selective excretion of acids or bases, and by the buffering action of weak acid-base systems in body fluids. There is a huge industry aimed at the notoriously science-challenged "alternative health" market that promotes worthless nostrums such as that are claimed to restore or preserve "the body's acid-base balance". They commonly point out that most foods are "acidic", (i.e., they are metabolized to H CO ), but they never explain that most of this acid is almot immediately exhaled in the form of CO . The implication is that our health is being ruined by the resulting "acidification" of the body; some further imply this can be a cause of cancer and other assorted ailments. People who fall for these expensive scams are in effect paying a tax on scientific ignorance. Those who, like you, have studied Chemistry, should consider that they have a social duty to counter this kind of predatory and deceptive pseudoscience. Deviations of the blood plasma pH from its normal value of 7.4 by more than about ±.1 can be very serious. These conditions are known medically as and . They can be caused by metabolic disturbances such as diabetes and by kidney failure in which excretion of H PO , for example, is inhibited. Numerous other processes lead to temporary unbalances. Thus hyperventilation, which can result from emotional upset, leads to above-normal loss of CO , and thus to alkalosis. Similarly, hypoventilation can act as a compensatory mechanism for acidosis. On the other hand, retention of CO caused by bronchopneumonia, for example, can give rise to acidosis. Acidosis can also result from diarrhea (loss of alkaline fluid from the intestine), while loss of gastric contents by vomiting promotes alkalosis. The natural, unpolluted atmosphere receives acidic, basic, and neutral substances from natural sources (volcanic emissions, salt spray, windblown dust and microbial metabolism) as well as from pollution (Figure 11.6.X). These react in a kind of gigantic acid-base titration to give a solution in which hydrogen ions must predominate to maintain charge balance (indicated by the identical widths of the bar graphs at the bottom labeled "anions" and "cations"). Carbon dioxide, however, is the major source of acidity in the unpolluted atmosphere. As will be explained farther on, CO makes up 0.032 vol-% of dry air, and dissolves in water to form carbonic acid: \[CO_{2(g) }+ H_2O(l) \rightleftharpoons H_2CO_{3(aq)}\] Thus rain is “acidic” in the sense that it contains dissolved CO which will reduce its pH to 5.7. The term is therefore taken to mean rain whose pH is controlled by pollutants which can lower the pH into the range of 3-4, resulting in severe damage to the environment. Acid rain originates from emissions of SO and various oxides of nitrogen (known collectively as "NO ") that are formed during combustion processes, especially those associated with the burning of coal. Incineration of wastes, industrial operations such as smelting, and forest fires are other combustion-related sources; these also release particulate material into the atmosphere which plays an important role in concentrating and distributing these acid-forming substances. Nitrogen oxides are formed naturally by soil bacteria acting on nitrate ions, an essential plant nutrient and itself a product of natural nitrogen fixation. Ammonia, a product of bacterial decomposition of organic matter, is eventually transformed into NO by bacteria into nitrogen oxides. The quantity of fixed nitrogen produced industrially to support intensive agriculture now exceeds the amount formed naturally, and has become the major source of anthropogenic nitrogen oxides. The gases noted above react with atmospheric oxygen, with each other, and with particulate materials to form sulfuric, nitric, and hydrochloric acids. There are two basic mechanisms by which acidic material is transported through the environment. The acid that dissolves in the water droplets that form clouds and precipitation, and is eventually incorporated into fog and snowflakes, undergoes ; this is ordinarily what is referred to as "acid rain". Because it is widely dispersed, often high in the atmosphere, it can travel very large distances from the original source. About 40% of the acidic substances introduced into the atmosphere becomes adsorbed onto particulate matter, which can be soot, smoke, windblown dust, and salt particles formed naturally from sea spray. This spreads less widely (typically around 30 km from the source), and falls onto surfaces by " ". Once the acid-laden particles land on surfaces, ordinary rain, fog, and dew release the acidic components, often in considerably more concentrated form than occurs through dry deposition alone. Because soils support the growth of plants and of the soil microorganisms that are essential agents in the recycling of dead plant materials, acid rain has a indirect but profound effect on soil health and plant growth. Soils containing alkaline components (most commonly limestone CaCO and other insoluble carbonates) can neutralize added acid and mitigate its effects. But soils in high mountain regions tend to be thin and unable to provide adequate buffering capacity. The same is true in almost half of Canada, in which granite rock of the Canadian Shield is very close to the surface; the eastern provinces of the country are strongly impacted by acid rain. The effects on soils noted above affect plants most strongly. However, direct impingment of acidic rain and fog on leaves has other effects which can be especially serious when air pollutants such as SO are present. Lakes are subject not only to wet and dry acidic deposition, but also by the water they receive from streams and surface runoff. Any toxic elements released by the action of acid rain on soils and sediments can thus be conveyed to, and concentrated within lakes and the streams that empty them. Lakes in poorly-buffered areas such as are found in alpine regions (western North America, Colorado and much of Switzerland) or on the Canadian Shield and in the Adirondacs and Appalachians) are highly sensitive to acid deposition. Severely acidified lakes (such as the one depicted at the above right) can be so devoid of life, including algae, that the water appears to be perfectly clear and bright blue in color. ↓ on image to enlarge pH tolerances of some aquatic organisms Aquatic organisms are generally adapted to "ordinary" pH conditions of 6-8, but vary greatly in their tolerance of low pH. As pH declines below 6, the diversity of aquatic animals, plants, and microorganisms diminishes. ↓ on image to enlarge Fish damaged by acid deposition "It has often been observed that the stones and bricks of buildings, especially under projecting parts, crumble more readily in large towns where coal is burnt... I was led to attribute this effect to the slow but constant action of acid rain." Robert Angus Smith, 1856 Acid deposition most strongly affects heritage buildings made of limestone and similar carbonate-containing stone materials. The principal chemical process involves the reaction of sulfuric acid with calcium carbonate: \[\ce{CO3(s) + H2SO4 → CaSO4(s) + CO2}.\] The acid first erodes and breaks up the surface of the stone. As the hydrated calcium sulfate (gypsum) forms, it picks up iron and other components of the stone and forms an unsightly black coating. Some of this gradually blisters off, exposing yet more stone suface. The gypsum crystals can sometimes grow into the stone, further undermining the surface. Very old structures such as the Taj Mahal, Notre Dame, the Colosseum and Westminster Abbey have all been affected. Statues and monuments, including those made from marble, are also susceptible to erosion and damage from acid deposition. Modern buildings are less affected, although acid rain can erode some painted surfaces. The carbonate system, consisting as it does of a soluble gas CO , soluble ions HCO and CO , and sparingly soluble carbonate salts, spans all four realms of nature: the atmosphere, hydrosphere (mainly the oceans), the lithosphere, and the biosphere. And owing to the acid-base equilibria that govern the transformations between these carbonate species, carbon is readily transported between these geochemical reservoirs. This chapter presents an overview of carbon geochemistry. Carbon is the fourth most abundant element in the universe. Within the Earth's crust it ranks 15th, mostly in the form of carbonates in limestones and dolomites. Kerogens, which are fossilized plant-derived carbon mostly in the form of oil shale, constitute another major repository of terrestrial carbon. The carbonate system encompasses virtually all of the environmental compartments– the atmosphere, hydrosphere, biosphere, and major parts of the lithosphere. The complementary processes of photosynthesis and respiration drive a global cycle in which carbon passes slowly between the atmosphere and the lithosphere, and more rapidly between the atmosphere and the oceans. Thus the "carbon cycle" can be divided into "fast" and "slow" parts, operating roughly on annual and geological time scales. Another excellent depiction of the carbon cycle, by the U.S. Dept of Energy. ( ): CO has probably always been present in the atmosphere in the relatively small absolute amounts now observed. Precambrian limestones, possibly formed by reactions with rock-forming silicates, e.g. \[\ce{CaSiO_3 + CO_2 → CaCO_3 + SiO_2} \label{4-1}\] have likely had a moderating influence on the CO abundance throughout geological time. The volume-percent of CO in dry air is 0.032%, leading to a partial pressure of 3 × 10 (10 ) atm. In a crowded and poorly-ventilated room, can be as high as 100 × 10 atm. About 54 × 10 moles per year of CO is taken from the atmosphere by photosynthesis, divided about equally between land and sea. Of this, all except 0.05% is returned by respiration (mostly by microorganisms); the remainder leaks into the slow, sedimentary part of the geochemical cycle where it can remain for thousands to millions of years. Trends in the growth of atmospheric CO for the past five years can be seen on . Since the advent of large-scale industrialization around 1860, the amount of CO in the atmosphere has been increasing. Most of this has been due to fossil-fuel combustion; in 1966, about 3.6 × 10 g of C was released to the atmosphere; this is about 12 times greater than the estimated natural removal of carbon into sediments. The large-scale destruction of tropical forests, which has accelerated greatly in recent years, is believed to exacerbate this effect by removing a temporary sink for CO . About 30-50% of the CO released into the atmosphere by combustion remains there; the remainder enters the hydrosphere and biosphere. The oceans have a large absorptive capacity for CO by virtue of its transformation into bicarbonate and carbonate in a slightly alkaline aqueous medium, and they contain about 60 times as much inorganic carbon as is in the atmosphere. However, effcient transfer takes place only into the topmost (100 m) wind-mixed layer, which contains only about one atmosphere equivalent of CO ; mixing time into the deeper parts of the ocean is of the order of 1000 years. For this reason, only about ten percent of added CO is taken up by the oceans. Most of the carbon in the oceans is in the form of bicarbonate, as would be expected from the pH which ranges between 7.8 and 8.2. In addition to atmospheric CO , there is a carbonate input to the ocean from streams. This is mostly in the form of HCO which derives from the weathering of rocks and terrestrial carbonate sediments, and the acid-base reaction \[H_2CO_3 + CO_3^{2–} → 2 HCO_3^– \label{4-2}\] which can be considered to be the source of bicarbonate in seawater. In this sense, the ocean can be considered the site of a gigantic acid-base titration in which atmospheric acids (mainly CO but also SO , HCl, and other acids of volcanic origin) react with bases that originate from rocks and are introduced through carbonate-bearing streams or in windblown dust. Carbon dioxide is slightly soluble in water: is followed up to a CO pressure of about 5 atm: \[[\ce{CO2 (aq)}] = 0.032 P_{\ce{CO_2}} \label{4-3}\] “Dissolved carbon dioxide” consists mostly of the hydrated oxide CO2(aq) together with a small proportion of carbonic acid: \[[\ce{CO2 (aq)}] = 650 [\ce{H2CO3}] \label{4-4}\] The acid dissociation constant \(K_{a1}\) that is commonly quoted for "H CO " is really a equilibrium constant that includes the above equilibrium. This means that "pure" H CO (which cannot be isolated) is a considerably stronger acid than is usually appreciated. Water exposed to the atmosphere with = 10 atm will take up carbon dioxide, which becomes distributed between the three carbonate species CO , HCO and CO in proportions that depend on and and on the pH. The "total dissolved carbon" is given by the mass balance \[C_t = [H_2CO_3] + [HCO_3^–] + [CO_3^{2–}] \label{4-5}\] The distribution of these species as a function of pH can best be seen by constructing a log -pH diagram for = 10 . This Sillén plot is drawn for two different concentrations of the carbonate system. The lower one, in heavier lines, is for a 10 solution, corresponding roughly to atmospheric CO in equilibrium with pure water. The upper plot, for a 10 solution, is representative of many natural waters such as lakes and streams where the water is in contact with sediments. It is important to note that this diagram applies only to a system in which is constant. In a solution that is open to the atmosphere, this will not be the case at high pH values where the concentration of CO is appreciable. Under these conditions this ion will react with H CO and the solution will absorb CO from the atmosphere, eventually resulting in the formation of a solid carbonate precipitate. The lower of the two above plots can be used to predict the pH of 10^{5} solutions of carbon dioxide, sodium bicarbonate, and sodium carbonate in pure water. The pH values are estimated by using the for each solution as noted below. The reasoning leading to these calculations is explained in the discussion of the in a previous chapter. Natural waters aquire carbon from sediments they are in contact with, and of course also from the atmosphere. The pH is an important factor here; CO and solid carbonates are more soluble at high pH, and pH also controls the distribution of carbon species, as is seen in the plot just above. At the slightly alkaline pH of most bodies of water (of which the oceans constitute 97% of the earth's surface waters), bicarbonate is the principal dissolved carbon species. The quantity of organic carbon is fairly small.

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The great distances between atoms and molecules in a gaseous phase, and the corresponding absence of any significant interactions between them, allows for simple descriptions of many physical properties that are the same for all gases, regardless of their chemical identities. As described in the final module of the chapter on gases, this situation changes at high pressures and low temperatures—conditions that permit the atoms and molecules to interact to a much greater extent. In the liquid and solid states, these interactions are of considerable strength and play an important role in determining a number of physical properties that depend on the chemical identity of the substance. In this chapter, the nature of these interactions and their effects on various physical properties of liquid and solid phases will be examined.  

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Most of the known chemical elements are metals, and many of these combine with each other to form a large number of . The special properties of metals— their bright, lustrous appearance, their high electrical and thermal conductivities, and their malleability— suggest that these substances are bound together in a very special way. The fact that the metallic elements are found on the left side of the periodic table offers an important clue to the nature of how they bond together to form solids. These points lead us to the simplest picture of metals, which regards them as a lattice of positive ions immersed in a “sea of electrons” which can freely migrate throughout the solid. In effect the electropositive nature of the metallic atoms allows their valence electrons to exist as a mobile fluid which can be displaced by an applied electric field, hence giving rise to their high . Because each ion is surrounded by the electron fluid in all directions, the bonding has no directional properties; this accounts for the high and of metals. This view is an oversimplification that fails to explain metals in a quantitative way, nor can it account for the differences in the properties of individual metals. A more detailed treatment, known as the bond theory of metals, applies the idea of resonance hybrids to metallic lattices. In the case of an alkali metal, for example, this would involve a large number of hybrid structures in which a given Na atom shares its electron with its various neighbors. Metallic solids possess special properties that set them apart from other classes of solids and make them easy to identify and familiar to everyone. All of these properties derive from the liberation of the valence electrons from the control of individual atoms, allowing them to behave as a highly mobile fluid that fills the entire crystal lattice. What were previously valence-shell orbitals of individual atoms become split into huge numbers of closely-spaced levels known as bands that extend throughout the crystal. The strength of a metal derives from the electrostatic attraction between the lattice of positive ions and the fluid of valence electrons in which they are immersed. The larger the nuclear charge (atomic number) of the atomic kernel and the smaller its size, the greater this attraction. As with many other periodic properties, these work in opposite ways, as is seen by comparing the melting points of some of the Group 1-3 metals (right). Other factors, particularly the lattice geometry are also important, so exceptions such as is seen in Mg are not surprising. In general, the transition metals with their valence-level electrons are stronger and have higher melting points: Fe, 1539°C; Re 3180, Os 2727; W 3380°C. These terms refer respectively to how readily a solid can be shaped by pressure (forging, hammering, rolling into a sheet) and by being drawn out into a wire. Metallic solids are known and valued for these qualities, which derive from the non-directional nature of the attractions between the kernel atoms and the electron fluid. The bonding within ionic or covalent solids may be stronger, but it is also directional, making these solids subject to fracture (brittle) when struck with a hammer, for example. A metal, by contrast, is more likely to be simply deformed or dented. In order for a substance to conduct electricity, it must contain charged particles ( ) that are sufficiently mobile to move in response to an applied electric field. In the case of ionic solutions and melts, the ions themselves serve this function. (Ionic solids contain the same charge carriers, but because they are fixed in place, these solids are insulators.) In metals the charge carriers are the electrons, and because they move freely through the lattice, metals are highly conductive. The very low mass and inertia of the electrons allows them to conduct high-frequency alternating currents, something that electrolytic solutions are incapable of. In terms of the band structure, application of an external field simply raises some of the electrons to previously unoccupied levels which possess greater momentum. The conductivity of an electrolytic solution as the temperature due to the decrease in "viscosity" which inhibits ionic mobility. The mobility of the electron fluid in metals is practically unaffected by temperature, but metals do suffer a slight conductivity decrease (opposite to ionic solutions) as the temperature rises; this happens because the more vigorous thermal motions of the kernel ions disrupts the uniform lattice structure that is required for free motion of the electrons within the crystal. Silver is the most conductive metal, followed by copper, gold, and aluminum. Metals conduct electricity readily because of the essentially infinite supply of higher-energy empty MOs that electrons can populate as they acquire higher kinetic energies. This diagram illustrates the overlapping band structure (explained farther on) in beryllium. The MO levels are so closely spaced that even thermal energies can provide excitation and cause heat to rapidly spread through the solid. Electrical conductivities of the metallic elements vary over a wide range. Notice that those of silver and copper (the highest of any metal) are in classes by themselves. Gold and aluminum follow close behind. Everyone knows that touching a metallic surface at room temperature produces a colder sensation than touching a piece of wood or plastic at the same temperature. The very high thermal conductivity of metals allows them to draw heat out of our bodies very efficiently if they are below body temperature. In the same way, a metallic surface that is above body temperature will feel much warmer than one made of some other material. The high thermal conductivity of metals is attributed to vibrational excitations of the fluid-like electrons; this excitation spreads through the crystal far more rapidly than it does in non-metallic solids which depend on vibrational motions of atoms which are much heavier and possess greater inertia. We usually recognize a metal by its “metallic luster”, which refers to its ability of reflect light. When light falls on a metal, its rapidly changing electromagnetic field induces similar motions in the more loosely-bound electrons near the surface (this could not happen if the electrons were confined to the atomic valence shells.) A vibrating charge is itself an emitter of electromagnetic radiation, so the effect is to cause the metal to re-emit, or , the incident light, producing the shiny appearance. What color is a metal? With the two exceptions of copper and gold, the closely-spaced levels in the bands allow metals to absorb all wavelengths equally well, so most metals are basically black, but this is ordinarily evident only when the metallic particles are so small that the band structure is not established. The distinctive color of gold is a consequence of Einstein's theory of special relativity acting on the extremely high momentum of the inner-shell electrons, increasing their mass and causing the orbitals to contract. The outer (5d) electrons are less affected, and this gives rise to increased blue-light absorption, resulting in enhanced reflection of yellow and red light. The electrons within the electron fluid have a distribution of velocities very much like that of molecules in a gas. When a metal is heated sufficiently, a fraction of these electrons will acquire sufficient kinetic energy to escape the metal altogether; some of the electrons are essentially “boiled out” of the metal. This , which was first observed by Thomas Edison, was utilized in vacuum tubes which served as the basis of electronics from its beginning around 1910 until semiconductors became dominant in the 1960’s. Most metals are made of atoms that have an outer configuration of , which we would expect to completely fill the band of MO’s we have described. With the band completely filled and no empty levels above, we would not expect elements such as beryllium to be metallic. What happens is that the empty orbitals also split into a band. Although the energy of the 2 orbital of an isolated Be atom is about 160 kJ greater than that of the 2 orbital, the bottom part of the 2 band overlaps the upper part of the 2 band, yielding a continuous that has plenty of unoccupied orbitals. It is only when these bands become filled with 2 electrons that the elements lose their metallic character. In most metals there will be bands derived from the outermost -, -, and atomic levels, leading to a system of bands, some of which will overlap as described above. Where overlap does not occur, the almost continuous energy levels of the bands are separated by a forbidden zone, or . Only the outermost atomic orbitals form bands; the inner orbitals remain localized on the individual atoms and are not involved in bonding. In its mathematical development, the band model relies strongly on the way that the free electrons within the metal interact with the ordered regularity of the crystal lattice. The alternative view shown here emphasizes this aspect by showing the inner orbitals as localized to the atomic cores, while the valence electrons are delocalized and belong to the metal as a whole, which in a sense constitutes a huge molecule in its own right.

https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.12%3A_Mole_Fraction

Sulfur dioxide is a by-product of many processes, both natural and human-made. Massive amounts of this gas are released during volcanic eruptions. Humans produce sulfur dioxide by burning coal. When in the atmosphere, the gas has a cooling effect by reflecting sunlight away from the earth. However, sulfur dioxide is also a component of smog and acid rain, both of which are harmful to the environment. Many efforts have been made to reduce \(\ce{SO_2}\) levels to lower acid rain production. However, \(\ce{SO_2}\) reduction efforts have an unforeseen complication: as we lower the concentration of this gas in the atmosphere, we lower its ability to cool, and consequently have global warming concerns. One way to express relative amounts of substances in a mixture is with the mole fraction. is the ratio of moles of one substance in a mixture to the total number of moles of all substances. For a mixture of two substances, \(\ce{A}\) and \(\ce{B}\), the mole fractions of each would be written as follows: \[X_A = \frac{\text{mol} \: \ce{A}}{\text{mol} \: \ce{A} + \text{mol} \: \ce{B}} \: \: \: \text{and} \: \: \: X_B = \frac{\text{mol} \: \ce{B}}{\text{mol} \: \ce{A} + \text{mol} \: \ce{B}}\nonumber \] If a mixture consists of \(0.50 \: \text{mol} \: \ce{A}\) and \(1.00 \: \text{mol} \: \ce{B}\), then the mole fraction of \(\ce{A}\) would be \(X_A = \frac{0.5}{1.5} + 0.33\). Similarly, the mole fraction of \(\ce{B}\) would be \(X_B = \frac{1.0}{1.5} = 0.67\). Mole fraction is a useful quantity for analyzing gas mixtures in conjunction with Dalton's law of partial pressures. Consider the following situation... A 20.0 liter vessel contains \(1.0 \: \text{mol}\) of hydrogen gas at a pressure of \(600 \: \text{mm} \: \ce{Hg}\). Another 20.0 liter vessels contains \(3.0 \: \text{mol}\) of helium at a pressure of \(1800 \: \text{mm} \: \ce{Hg}\). These two gases are mixed together in an identical 20.0 liter vessel. Because each will exert its own pressure according to Dalton's law, we can express the partial pressures as follows: \[P_{H_2} = X_{H_2} \times P_\text{Total} \: \: \: \text{and} \: \: \: P_{He} = X_{He} \times P_\text{Total}\nonumber \] The partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure. For our mixture of hydrogen and helium: \[X_{H_2} = \frac{1.0 \: \text{mol}}{1.0 \: \text{mol} + 3.0 \: \text{mol}} = 0.25 \: \: \: \text{and} \: \: \: X_{He} = \frac{3.0 \: \text{mol}}{1.0 \: \text{mol} + 3.0 \: \text{mol}} = 0.75\nonumber \] The total pressure according to Dalton's law is \(600 \: \text{mm} \: \ce{Hg} + 1800 \: \text{mm} \: \ce{Hg} = 2400 \: \text{mm} \: \ce{Hg}\). So, each partial pressure will be: \[P_{H_2} = 0.25 \times 2400 \: \text{mm} \: \ce{Hg} = 600 \: \text{mm} \: \ce{Hg}\nonumber \] \[P_{He} = 0.75 \times 2400 \: \text{mm} \: \ce{Hg} = 1800 \: \text{mm} \: \ce{Hg}\nonumber \] The partial pressures of each gas in the mixture do not change, since they were mixed into the same size vessel and the temperature was not changed. A flask contains a mixture of 1.24 moles of hydrogen gas and 2.91 moles of oxygen gas. If the total pressure is \(104 \: \text{kPa}\), what is the partial pressure of each gas? First, the mole fraction of each gas can be determined. Then, the partial pressure can be calculated by multiplying the mole fraction by the total pressure. \[\begin{array}{ll} X_{H_2} = \frac{1.24 \: \text{mol}}{1.24 \: \text{mol} + 2.91 \: \text{mol}} = 0.299 & X_{O_2} = \frac{2.91 \: \text{mol}}{1.24 \: \text{mol} + 2.91 \: \text{mol}} = 0.701 \\ P_{H_2} = 0.299 \times 104 \: \text{kPa} = 31.1 \: \text{kPa} & P_{O_2} = 0.701 \times 104 \: \text{kPa} = 72.9 \: \text{kPa} \end{array}\nonumber \] The hydrogen is slightly less than one third of the mixture, so it exerts slightly less than one third of the total pressure.

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We saw earlier that the inherent stiffness of a material can be assessed by its Young's modulus. The Young's modulus is the ratio of the stress-induced in a material under an applied strain. The strain is the amount of deformation in the material, such as the change in length in an extensional experiment, expressed as a fraction of the beginning length. The stress is the force exerted on the sample divided by the cross-sectional area of the sample. If the strain is limited to a very small deformation, then it varies linearly with stress. If we graph the relationship, then the slope of the line gives us Young's modulus, . That's the proportionality constant between stress and strain in Hooke's Law. = σ / ε Hooke's Law is sometimes used to describe the behavior of mechanical springs. The modulus can be thought of the resistance to stretching a spring; the more resistance the spring offers, the greater the force needed to stretch it. The same force is what snaps the spring back into place once you let it go. In the experiments we saw earlier, we didn't let go. We continued to stretch the material farther and farther, applying generally increasing stress until the material finally broke. Now we will look at a much more limited approach. Instead of stretching the material as far as we can, we will only stretch it a tiny bit, then release the stress so that it snaps back to its original length. We can then stress it again and release it again. We can keep repeating. Instead of a continuously increasing strain, this sample is subjected to an oscillatory strain, one that repeats in a cycle. This approach is called dynamic mechanical analysis. We can use dynamic mechanical analysis to measure the modulus of the material. Instead of continuously moving all the way through the linear elastic region, beyond which Hooke's law breaks down, we carefully keep the sample in the Hookean region for the entire experiment. Now, one experiment should be good enough to extract the modulus, but we are letting go and doing it over again. Why? The principle reason for running the experiment this way is to get some additional information. We can get this information because polymers don't quite follow Hooke's Law perfectly. In reality, even within the linear elastic region, the stress-strain curve is not quite linear. In the picture below, the curvature is exaggerated quite a bit, just for illustrative purposes. Even if the relationship is not quite linear, then as we release the strain, the stress in the material should simply follow the curve back down to zero. It does not. Instead, there is a phenomenon called hysteresis at work. Hysteresis just means that a property of the material depends on how the material came to be in its current situation. In this case, Hooke's Law seems to imply that a specific sample subjected to a specific strain would experience a specific stress (or vice versa). However, it depends whether we are stretching the sample or letting it relax again. As we let the sample relax back to its initial length, the strain is different from what we saw when we were stretching it. Typically, it's lower. Again, we can see this in the curve below, where the curvature has been exaggerated. The difference between the loading curve (when the stress was first applied) and the unloading curve (when the stress was removed) represents an energy loss. A force was applied to move a sample or a portion of a sample, some distance. When the sample snapped back the same distance, the force was unequal to the one that was initially applied. Some energy was therefore lost. The slope of the loading curve, analogous to Young's modulus in a tensile testing experiment, is called the storage modulus, '. The storage modulus is a measure of how much energy must be put into the sample in order to distort it. The difference between the loading and unloading curves is called the loss modulus, ". It measures energy lost during that cycling strain. Why would energy be lost in this experiment? In a polymer, it has to do chiefly with chain flow. The resistance to deformation in a polymer comes from entanglement, including both physical crosslinks and more general occlusions as chains encounter each other while undergoing conformational changes to accommodate the new shape of the material. Once the stress is removed, the material springs back to its equilibrium shape, but there is no reason chains would have to follow the exact same conformational pathway to return to their equilibrium conformations. Because they have moved out of their original positions, they are able to follow a lower-energy pathway back to their starting point, a pathway in which there is less resistance between neighboring chains. For that reason, stretching a polymer is not quite the same as stretching a mechanical spring. A "spring-and-dashpot" analogy is often invoked to describe soft materials. Whereas a spring simply bounces back to its original shape after being pulled, a dashpot does not. If you don't know what a dashpot is, picture the hydraulic arms that support the hatchback on a car when you open it upward. There is some resistance to opening the hatchback because a piston is being pulled through a hydraulic fluid as the arm stretches. When we stop lifting, the arms stay at that length, because the hydraulic fluid also resists the movement of the piston back to its original position. The dashpot has a tendency to stay put rather than spring back. Polymers display a little of both properties. They have an elastic element, rooted in entanglement, that makes them resist deformation and return to their original shapes. They also have a viscous element, rooted in chain flow. That viscous element means that, when we distort polymeric materials, they might not return to exactly the same form as when they started out. Taken together, these behaviors are described as viscoelastic properties. Many materials have viscoelastic properties, meaning they display some aspects of elastic solids and some aspects of viscous liquids. So far, we have concentrated on extensional deformations of materials: we have been looking at what happens when we stretch them. It's worth looking at another type of deformation because it is very commonly used in materials testing. This second approach uses shear instead of an extension to probe how the material will respond. A shear force is applied unevenly to a material so that it tilts or twists rather than stretching. One of the reasons this approach is used so often is because it is very easy to do. A sample is sandwiched between two plates. The bottom plate is held in place while the top plate is twisted, shearing the material held in between. If we take a closer look at a layer of the sample, maybe at the surface, along the edge of the sandwich, we can imagine breaking it down into individual layers. Under shear strain, those layers move different amounts. The top layer, right beneath that top plate moves the most. The bottom layer, sitting on the stationary lower plate, doesn't move at all. In between, each layer moves a little further than the one beneath it. This gradation of deformation across the sample is very much like what we saw in the analysis of the of liquids. The difference is that viscosity looks at the variation of strain with time. Nevertheless, modulus in solids is roughly analogous to viscosity in liquids. We can use this parallel plate geometry to obtain values for storage modulus and loss modulus, just like we can via an extensional geometry. The values we get are not quite the same. For this reason, modulus obtained from shear experiments is given a different symbol than modulus obtained from extensional experiments. In a shear experiment, = σ / ε That means storage modulus is given the symbol and loss modulus is given the symbol . Apart from providing a little more information about how the experiment was actually conducted, this distinction between shear modulus and extension modulus is important because the resulting values are quite different. In general, the value of the storage modulus obtained from an extensional experiment is about three times larger than the value of storage modulus obtained from a shear experiment. = 3 The reason for the difference is that extension actually involves deformation of the material in three directions. As the material is stretched in one direction (let's say it's the y-direction), in order to preserve the constant volume of the material (there is still the same amount of stuff before and after stretching), the material compresses in both the other two directions (x and z). Metric prefixes are frequently encountered when reading about modulus. Rank the following units of stress from smallest to largest, and in each case provide a conversion factor to Pa. GPa kPa MPa Pa

https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Basics/Discovering_Sub-atomic_Particles

At the end of the last section ( ), we mentioned the problem of the masses of iodine and tellurium, which were "out of order" in the periodic table. This was only one of several problems facing chemists at this point. Another was the (the lanthanoides and actinoides, at the bottom of the periodic table). The rare earth elements were extremely similar to each other, so it was very hard to separate them, determine the masses, and fit them into the periodic table. Another problem was . Ions were charged chemical entities that moved through solutions carrying electric current when electricity was applied. Chemists knew they existed, but could not explain how they formed. Cations were positively charged, while anions were negatively charged. Hydrogen was known to form positive ions with a particular charge. But where did the charges come from? Berzelius had thought that the atoms themselves were charged, but this didn't make sense after Avogadro's hypothesis was accepted, since in that case the two O atoms in O would repel each other, not form a molecule. Of course, this was mostly a problem to physical chemists, because physicists mostly still didn't believe in atoms. Experiments by Davy, Faraday and others showed that solids usually conduct electricity without changing themselves; liquids are decomposed by electrical current, and gases conduct badly and only at high voltage, but become better conductors at low pressure. Glass tubes of low pressure gas glow; examples include neon lights in store windows, fluorescent light bulbs, and yellow sodium lamps sometimes used at street lights. When glass tubes full of very low pressure gas were made by expert glassblower Geissler after 1850, a glowing spot would appear somewhere on the glass when the tube was connected to a high voltage. This spot could be moved by a magnet, suggesting that it was negatively charged. A beam or ray was coming out of the cathode (the electrode to which cations move), and this beam caused the glass to glow. (This is exactly how old TVs work, from before there were flat screens.) But what was the beam that came out of the cathode? Some thought it was negatively-charged particles, while others (especially physicists) thought it was a wave. Thomson convinced himself that the beam is a negatively charged particle, which can be moved using electric or magnetic fields. He estimated that the particle has a charge/mass ratio 1000x greater than the hydrogen ion, but with the opposite charge. The ray that comes out of the cathode is the same whatever the material, and he found that you can generate the same particle by heating metal filaments very hot or exposing a metal surface to UV light (the light that gives you sunburns). What are cathode rays? ! When Thomson suggested in 1897 that either they had a much higher charge or a much smaller mass than hydrogen ions, many people were skeptical. However, careful measurements by Millikan (starting 1908, continuing for almost 10 years!) of falling droplets of mist in an electric or magnetic field revealed that electrons have the same charge as hydrogen ions, but are almost ~1/2000 less massive, so in fact Thomson guessed electrons were twice as big as they really are. In summary, electrons have an equal but opposite charge to hydrogen ions, but hydrogen ions weigh 2000x more than electrons. Electrons are part of every kind of atom. Other than electrons, what are atoms made of? X-rays were discovered in 1895 when Roentgen noticed a ray that passed through the glass of his cathode ray tube and developed photographic film placed in its path, even if it was wrapped in dark paper. Inspired by this discovery, Becquerel studied the fluorescence (this means glowing after being exposed to light) of minerals and whether this could develop film through dark paper. He used sunlight to make the minerals fluoresce, but discovered that some minerals fluoresce even without sunlight. Minerals that contain the elements uranium and thorium have this property, which is called . Three types of radioactivity were discovered, α (positive charge), β (electrons, same as cathode rays) and γ (light, similar to x-rays). This radiation is a property of the element, and is not changed by chemical combinations. It was also discovered that radioactive elements decay into products that closely resemble known elements, except that the masses are different. These decay products have exactly the same chemical properties as the known elements, and once mixed together, they can't be separated by chemical means. But the masses were different. The evidence against Dalton's "identical and indivisible atoms of each element" was getting strong. Rutherford studied the α-radiation, and found that it had a positive charge, and an e/m (charge/mass) ratio 1/2 that of a hydrogen ion. Later he proved that it is a charged form of the element helium (He). Finally, using an early version of the mass spectrometer, which measures e/m by accelerating charged particles in an electric or magnetic field, Aston determined that neon (Ne) atoms have masses of 20 or 22, not 20.2, then the accepted value. Here's another case where knowing the precision of your experiment is important: if he couldn't tell the difference between 20.0 and 20.2, this wouldn't have been useful! The two versions of Ne are called . Isotopes are atoms of the same element with have the same chemical properties, but have different masses. What is the structure of an atom? At this point, it was clear that Dalton had not been entirely correct, because atoms of one element could exist in different forms, or isotopes. It was also becoming clear that atoms could be divided into smaller parts, like electrons. But how were the electrons and the positive parts (like the part of He that is α-radiation) arranged, and what was the positive particle? It seemed like there was a lot of empty space in atoms, because cathode rays could pass through aluminum windows. Many theories were considered, but ultimately two of Rutherford's students, Geiger and Marsden, found an answer. They observed that α-rays directed at a thin gold (Au) foil usually passed through the foil (though not usually quite straight), but sometimes the foil knocked them back in the direction they came from. These α-particles must have bumped into (or been repelled by a large positive charge in) something very heavy, to have so large a change in direction. In 1911, Rutherford proposed that atoms consist of a small dense particle, the , which contains most of the atom's mass and all of its positive charge, with electrons moving through a large empty space around the nucleus. But they were a little confused about the mass and charge of the nucleus. In 1913, a very young chemist named Henry Moseley measured the fluorescence wavelength produced when x-rays shine on pure elemental samples. He discovered that if you graph the square root of the fluorescence frequency vs the element's location in the periodic table (such as H, 1; C, 6; Hg, 80) you get a straight line. The location in the periodic table is called . This confirmed the ordering of Te/I in the periodic table and a few other similar problems. He was also able to determine where undiscovered elements were missing in the table, and clarify identification of rare earth elements, which was very difficult because they were so similar. He was 26 years old, and he died a year later in WWI. Moseley's work showed that the ordering of the periodic table is based on the atomic number, which is the number of positive charges in the nucleus and also the number of electrons. The unit of positive charge is a hydrogen nucleus, or . The mass of the nucleus continued to be confusing even after the amount of positive charge had been determined by Moseley's atomic numbers. The atomic weights were about double what they should be if the nucleus held only protons. A proposal to explain this was that there were some electrons in the nucleus, canceling out the charge of half the protons. Later radiochemistry experiments showed the existence of the , which has about the mass of a proton and no charge. This was understood to be the extra mass in nuclei. Now, though, we know that neutrons can decay into protons and electrons (and some other less familiar particles), so the earlier idea wasn't far off. Atoms are composed of protons, neutrons, and electrons. Protons and neutrons make most of the atom's mass and all of its positive charge in the . are subatomic particles that contain a positive charge. Essentially, they are hydrogen nuclei. are subatomic particles that have the same mass as a proton but not charge. are subatomic particles with a negative charge that are found outside of the atom's nucleus. An element's provides its location on the periodic table as well as describe how many protons are in an atoms of that particular element. are atoms that are either positively or negatively charged due to the number of protons or electrons in an atom not being equal to each other. are atoms that have a different number of neutrons in the nucleus than normal. There are three main types of : α , β, and γ. α (Alpha) radiation involves the emission of a helium nucleus, which is two protons and two neutrons. β (Beta) radiation involves the emission of an electron from the atom's nucleus. γ (Gamma) radiation involves the emission of high-energy electromagnetic radiation.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Medicinal_Chemistry/Antimicrobial_Drugs

The modern era of the chemotherapy of infection started with the clinical use of sulfanilamide in 1936. The "golden age" of antimicrobial therapy began with the production of penicillin in 1941, when this compound was mass-produced and first made available for limited clinical trial. More than 30% of all hospitalized patients now receive one or more courses of therapy with antibiotics, and millions of potentially fatal infections have been cured. However, at the same time, these pharmaceutical agents have become among the most misused of those available to the practicing physician. One result of widespread use of antimicrobial agents has been the emergence of antibiotic-resistant pathogens, which in turn has created an ever-increasing need for new drugs. Many of these agents have also contributed significantly to the rising costs of medical care. An antibiotic is any substance produced by a microorganism that is excreted to harm or kill another microorganism. Technically, antibiotics are microbial or fungal products. But these substances can be synthesized and mass produced in the laboratory to use against harmful microorganisms in the environment. Thus, the synthetic chemist has added greatly to our therapeutic armamentarium. Synthetic drugs such as isonaizid and theambutol represent important contributions for the treatment of tuberculosis. While many such antimicrobial agents are not properly termed antibiotics, since they are not produced by living organisms, little distinction should now be made between compounds of natural and synthetic origin. Antibiotics can further be grouped under the broader heading of chemotherapuetic agents, chemical agents used to treat disease. Good chemotherapuetic agents are able to kill or inhibit the target pathogen without too much damage to the host organism. The basis for this selective toxicity lies in the differences between prokaryotic cells of microorganisms and our own eukaryotic cells. The prokaryotic cells of microorganisms differ in a number of ways from eukaryotic cells, such as absence of cell walls, different size of ribosomes, and details of metabolism. Thus, the goal of antibiotic therapy is to choose or design drugs that target these differences in host and pathogen cells. Alexander Fleming loved to play, both in the laboratory and out. He always loved snooker and golf and had many whimsical variants on the rules. In the lab he made "germ paintings," in which he would draw with his culture loop using spores of highly pigmented bacteria, which were invisible when he made the painting, but when cultured developed into brightly colored scenes. He followed what Max Delbruck would later call the "principle of limited sloppiness." Fleming abhorred a tidy, meticulous lab; he left culture dishes lying around for weeks and would often discover interesting things in them. Though the story has been told in many sometimes conflicting ways, something like this resulted in the discovery of penicillin. He seems to have left a culture dish lying on the lab bench and then gone away on vacation. When he returned a few spores of an unusual mold had germinated on the plate. When he cultured the bacteria on the plate he found that they grew up to within a few centimeters of the mold, but there were killed. A crude extract of the mold was then shown to have antibacterial properties. Fleming made this discovery in 1928 and by 1929 had named it penicillin (he was told by a colleague that the mold was a type of Penicillium and "penicillozyme" must have seemed cumbersome). Fleming continued to use penicillin in his lab but not with any great enthusiasm and certainly not to the exclusion of many other projects. He never developed it into a clinically useful compound, though in 1929 he suggested that it might have important clinical applications. Because he was a bacteriologist and not a chemist, Fleming did not attempt to purify penicillin. He seems to have run into a dead end with penicillin and so during the 1930s, though he kept it in his lab, he did not do much with it. In the late 1930s Australian Howard Florey came to London to work with Charles Sherrington. He worked on lysozyme for a while and then became interested in penicillin. It was Florey, with Chain and other of his group that developed penicillin into a clinical antibiotic. They did this during 1940-41. Fleming, Florey, and Chain shared the 1945 Nobel Prize in Physiology for Medicine. Fleming became world-famous for penicillin, and was rightly acknowledged as the father of modern antibiotics, but Florey was just as rightly miffed at being denied much of the credit for creating the powerful medical tool we now know. Evidence does not suggest that Fleming deliberately denied Florey his due credit, but Fleming's peculiar, dry sense of humor seems to have caused him not to deny even the wildest attributions to him. Cephalosporins are the second major group of b-lactam antibiotics. They differ from penicillins by having the b-lactam ring fused to a dihydrothiazine ring rather than a thiazolidine. The other difference, which is more significant from a medicinal chemistry stand point, is the existence of a functional group (R) at position 3 of the fused ring system. This now allows for molecular variations to be introduced at the 7-NH2 group, as in the penicillins, as well as to effect changes in properties by diversifying the moieties at position 3. The first member of the newer series of b-lactams was isolated in 1956 from extracts of Cephalosporium acremonium, a sewer fungus. This species actually produced several antibiotics: cephalosporin C, cephalosporins P1 - P5 and penicillin N. This was also true with Fleming's Penicillium notatum. However, that was not discovered until research on the chemistry of penicillin was worked on. Cephalosporin has also been identified from other fungi such as Emericellopsis and Paecilomyces, two genera that are morphologically similar to Penicillium. Like penicillin, cephalosporins are valuable because of their low toxicity and their broad spectrum of action against various diseases. In this way, cephalosporin is very similar to penicillin. Cephalosporins are one of the most widely used antibiotics, and economically speaking, has about 29% of the antibiotic market. The cephalosporins are possibly the single most important group of antibiotics today and are equal in importance to penicillin. The structure and mode of action of the cephalosporins are similar to that of penicillin. They affect bacterial growth by inhibiting cell wall synthesis, in Gram-positive and -negative bacteria. Penicillin consists of a thiolidine ring fused to a -lactam ring, to which a variable R group is attached by a peptide bond. This structure can undergo a variety of rearrangements, which accounts for the instability first encountered by Flemming. In particular, the b-lactam ring is very labile. In 1957, it was shown that bacteria ordinarily susceptible to penicillin could be grown in its presence if a hypertonic medium were used. The organisms obtained this way are devoid of a cell wall and consequently lyse when transferred to a normal medium. Hence it was inferred that penicillin interferes with the synthesis of the bacterial cell wall. The cell walls of bacteria are essential for their normal growth and development. Peptidoglycan is a heteropolymeric component of the cell wall that provides rigid mechanical stability by virtue of its highly cross-linked lattice work structure, which prevents bacteria from bursting from their high internal osmotic pressure. The peptidoglycan is composed of glycan chains, which are linear strands of alternating pyranoside residues of two amino sugars (N-acetylgucosamine and N-acetylmuramic acid), that are cross-linked by peptide chains. In gram-positive microorganisms, the cell wall is 50 to 100 molecules thick, but it is only 1 or 2 molecules thick in gram-negative bacteria. In 1965, it was deduced that penicillin blocks the last step in cell wall synthesis, namely the cross-linking of different peptioglycan strands. This cross-linking is accomplished by a transpeptidation reaction that occurs outside the cell membrane. The transpeptidase itself, called glycopeptide transpeptidase, is membrane bound. In the formation of the cell wall of Staphylococcus aureus, the transpeptidase normally forms an acyl-enzyme intermediate with the penultimate D-analine residue of the D-Ala-D-Ala-peptide. This covalent acyl-enzyme intermediate then reacts with the amino group of the terminal glycine in another peptide to form the cross-link (see figure below). Therefore, the end result is that the amino group at one end of a pentaglycine chain attacks the peptide bond between two D-analine residues in another peptide unit, a peptide bond is formed between glycine and one of the D-alanine residues, and the other D-alanine residues is released. It should be noted that bacteria cell walls are unique in containing D amino acids, which form cross-links by a different mechanism from that used to synthesize proteins. Penicillin mimics the D-Ala-D-Ala moiety of the normal substrate and is welcomed into the active site of the transpeptidase. Bound penicillin then forms a covalent bond with a serine residue at the active site of the enzyme. This penicilloyl-transpeptidase does not further react and the enzyme is irreversibly inhibited (see figure below). The reason penicillin is so effective in inhibiting glycopeptide transferase is the four-membered -lactam ring is strained, which makes it highly reactive. Also, the conformation of this part of penicillin is probably very similar to that of the transition state of the normal substrate, a species that interacts strongly with the enzyme. Because of the high use of penicillin, some bacteria have developed resistance by producing molecules that can disable penicillin. is an enzyme produced by certain penicillin-resistant bacteria which reacts irreversibly with the b-lactam ring. Scientists have responded with other drugs that inturn react and disable penicillinase. One such drug is clavulanic acid. This compound irreversibly binds to penicillinase and prevent the enzyme from working. Therefore, sometimes clavulanic acid is given along with one of the semi-synthetic penicillins. Erythromycin is an orally effective antibiotic discovered in 1952 in the metabolic products of a strain of Streptocyces erythreus, originally obtained from a soil sample collected in the Philippine Archepelago. Erythromycin is one of the macrolide antibiotics, so named because they contain a many-membered lactone ring to which are attached one or more deoxy sugars. It is a white crystalline compound, soluble in water to the extent of 2 mg/ml. the structuras formula of erythromycine is a follows: Erythromycin may be either bacteriostatic or bactericidal, depending on the microorganism and the concentration of the drug. The bactericidal activity is greatest against a small number of rapidly dividing microorganisms and increases markedly as the pH of the medium is raised over the range of 5.5 to 8.5. the antibiotic is most effective in vitro against gram positive cocci such as Strep. pyogens and Strep. Pneumoniae. Resistant strains of these bacteria are rare and are usually isolated from populations of people who have been recently exposed to macrolide antibiotic. For example, in 1979, only 5% of group-A streptococcal strains isolated in Oklahoma were resistant to erythromycin, but 60% of such strains were found to be resistant in Japan. the difference likely reflects the wide use of erythromycin in Japan for respiratory infections. Erythromycin and other macrolide antibiotics inhibit protein synthesis by binding to 50 S ribosomal subunits of sensitive microorganisms. (Humans do not have 50 S ribosomal subunits, but have ribosomes composed of 40 S and 60 S subunits). Certain resistant microorganisms with mutational changes in components of this subunit of the ribosome fail to bind the drug. The association between erythromycin and the ribosome is reversible and takes place only when the 50 S subunit is free from tRNA molecules bearing nascent peptide chains. The production of small peptides goes on normally in the presence of the antibiotic, but that of highly polymerized homopeptides is suppressed. Gram-positive bacteria accumulate about 100 times more erythromycin than do gram-negative microorganisms. The nonionized from of the drug is considerably more permeable to cells, and this probably explains the increased antimicrobial activity that is observed in alkaline pH. The development of the tetracycline antibiotics was the result of a systematic screening of soil specimens collected from many parts of the world for antibiotic-producing microorganisms. The first of these compounds, chlortetracycline, was introduced in 1948. Soon after there initial development, the tetracyclines were found to be highly effective against rickettsiae, a number of gram-positive and gram-negative bacteria, and the agents responsible for conjunctivitis, and psittacosis, and hence became known as "broad spectrum" antibiotics. They are also effective to agents that exert their effects on the bacterial cell wall, such as rickettsiae, Chlamydia, and amebae. , tetracycline drugs are primarily bacteriostatic and only multiplying microorganisms are affected. These drugs are closely related bacteriostatic antibiotics, similar in antibacterial spectrum and toxicity. The site of action of tetracyclines is the 30 S subunit of the ribosome, but at least two processes appear to be required for these antibiotics to gain access to the ribosomes of gram-negative bacteria. The first is passive diffusion through hydrophobic pores in the outer cell membrane. The second process involves an active transport system that pumps all tetracyclines through the inner cytoplasmic membrane. Although permeation of these drugs into gram-positive bacteria is less will understood, it too requires an active transport system. Once the tetracyclines gain access to the bacteria cell, they inhibit protein synthesis and bind specifically to 30 S ribosomes. They appear to prevent access of aminoacyl tRNA to the acceptor site on the mRNA-ribosome complex. This prevents the addition of amino acids to the growing peptide chain. Only a small portion of the drug is irreversibly bound, and the inhibitory effects of the tetracyclines can be reversed by washing. Therefore, it is probable that the reversibly bound antibiotic is responsible for the antibacterial action. These compounds also impair protein synthesis in mammalian cells at high concentrations; however the host cells lack the active transport system found in bacteria. Tetracycline-resistant strains of pneumococci account for about 5% of isolates from pneumococcal pneumonia patients. Infections due to Group A beta-hemolytic streptococci should not be treated with a tetracycline, since as many as 25% of the organisms may be resistant when tested in vitro. Serious staphylococcal disease is also not a primary indication for tetracyclines. Bacterial resistance to one tetracycline is generally accompanied by cross-resistance to the others. The tetracyclines are variably absorbed after oral administration. Food interferes with absorption of tetracyclines, with the exception of doxycycline and minocycline. Absorption of tetracyclines is decreased in the presence of antacids containing aluminum, Ca, and Mg, and in preparations containing iron. The half-lives in plasma are about 8 h for oxytetracycline and tetracycline; 13 h for demeclocycline and methacycline; and 16 to 20 h for doxycycline and minocycline. Tetracyclines penetrate into most tissues and body fluids. However, CSF levels are not reliably therapeutic. Minocycline, because of its high lipid solubility, is the only tetracycline that penetrates into tears and saliva in levels high enough to eradicate the meningococcal carrier state. All tetracyclines, except doxycycline, are excreted primarily in the urine by glomerular filtration, and their blood levels increase in the presence of renal insufficiency. Doxycycline is excreted mainly in the feces. All tetracyclines are partially excreted in bile, resulting in high biliary levels. They are then partially reabsorbed. The sulfonamides are synthetic bacteriostatic antimicrobial agents with a wide spectrum encompassing most gram-positive and many gram-negative organisms. These drugs were the first effective chemotherapeutic agents to be employed systematically for the prevention and cure of bacterial infections in man. The considerable medical and public health importance of their discovery and their subsequent widespread use were quickly reflected in the sharp decline in morbidity and mortality figures for the treatable infectious decease. Before penicillin became generally available, the sulfonamides were the mainstay of antibacterial chemotherapy. While the advent of antibiotics has diminished the usefulness of the sulfonamides, they continue to occupy an important, although relatively small, place in the therapeutic armamentarium of the physician. Sulfonimides are structural analogs and competitive antagonists of para-aminobenzoic acid (PABA), and thus prevent normal bacterial utilization of PABA for the synthesis of the vitamin folic acid. (The role of folic acid in RNA synthesis was already discussed in ). More specifically, sulfonamides are competitive inhibitors of the bacterial enzyme sulfihydropteroate synthase, which is responsible for the conversion of PABA into dihydrofolic acid, the immediate precursor of folic acid. Sensitive microorganisms are those that must synthesis their own folic acid; bacteria that can utilize preformed folic acid are not affected. Bacteriostasis induced by sulfonamides is counteracted by PABA competitively. Sulfonamides do not affect mamalian cells by this mechanism, since they require preformed folic acid and cannot synthesis it. Most sulfonamides are readily absorbed orally, the small intestine being the major site of absorption. Parenteral administration is difficult, since the soluble sulfonamide salts are highly alkaline and irritating to the tissues. The sulfonamides are widely distributed throughout all tissues. High levels are achieved in pleural, peritoneal, synovial, and ocular fluids. CSF levels are effective in meningeal infections, but sulfonamides are rarely used for this indication. When given in pregnancy, high levels are achieved in the fetus. Sulfonamides are loosely and reversibly bound in varying degrees to serum albumin. Since the bound sulfonamide is inactive and nondiffusible, the degree of binding can affect antibacterial effectiveness, distribution, and excretion. The antibacterial action is inhibited by pus. The sulfonamides are metabolized mainly by the liver to acetylated forms and glucuronides, both therapeutically inactive. Excretion is primarily renal by glomerular filtration with minimal tubular secretion or reabsorption. The relative insolubility of most sulfonamides, especially their acetylated metabolites, may cause precipitation in the renal tubules. The more soluble analogs, such as sulfisoxazole and sulfamethoxazole, should be chosen for systemic therapy, and the patient must be well hydrated. To avoid crystalluria and renal damage, fluid intake should be sufficient to produce a urinary output of 1200 to 1500 mL/day. Sulfonamides should not be used in renal insufficiency. Chloroquine is one of the cheif agents for the chemotherapy of malaria. Although chlorouqine causes a number of effects that singly or in combination may relate to its promary mechanism of plasmodicidal action, this process is not yet shown. From early work, it has been found that chloroquine inhibits the incorporation of phosphate into RNA and DNA. Later it was shown that chloroquine combines strongly with double-stranded DNA. The drug is also reported to inhibit DNA polymerase activity markedly by combing with the DNA primer. A common misuse of these agents is in infections that have been proven to be untreatable. The vast majority of the diseases due to the true viruses will not respond to any of the presently available anti-infective compounds. Thus, the antimicrobial therapy of measles, chickenpox, mumps, and at least 90% of infections of the upper respiratory tract is totally ineffective and, therefore, worse than useless Fever of undetermined etiology may be of two types: one that is present for only a few days to a week and another that persists for an extended period. Both of these are frequently treated with antimicrobial agents. Most instances of pyrexia of short duration, in the absence of localizing signs, are probably associated with undefined viral infections, often of the upper respiratory tract, and do not respond to antibiotics. In the bulk of these cases, defervescence takes place spontaneously within a week or less. Studies of prolonged fever have shown that three common infectious causes are tuberculosis, often of the disseminated variety, hidden pyogenic intraabdominal abscess and infectious endocarditis. Also, the so-called collagen disorders and various neoplasms, especially lymphoma are frequently responsible for prolonged and significant degrees of fever. Various types of cancer, metabolic disorders, hepatitis, asymptomatic regional enteritis, atypical rheumatoid arthritis, and a number of other noninfectious disorders may present themselves as cases of fever of unknown etiology. The most rational approach to the problem of fever of unknown origin is not one that concentrates of the elevated temperature alone but one that involves a thorough search for its cause. The patient should not be unnecessarily exposed to chemotherapy in the hope that, if an agent is not effective, another one or combination of drugs will be helpful. Because of misuse of antibiotics many strains of bacteria have become resistant to the effects of these drugs. Here are some examples of resistance that has occurred in staphylococcus species:

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Make sure you thoroughly understand the following essential concepts: Acid-base reactions, in which protons are exchanged between donor molecules (acids) and acceptors (bases), form the basis of the most common kinds of equilibrium problems which you will encounter in almost any application of chemistry. In order to thoroughly understand the material in this unit, you are expected to be familiar with the following topics which were covered in the separate unit Introduction to Acid-Base Chemistry: You should also have some familiarity with the principles of chemical equilibrium, and know how to write and use equilibrium constants. As you should recall from your earlier introduction to acids and bases, the +1 electric charge of the tiny proton (a bare hydrogen nucleus) is contained in such a miniscule volume of space that the resulting is far too large to enable its independent existence in solution; it will always attach to, and essentially bury itself in, the non-bonding orbitals of a solvent. Thus in aqueous solution, what we commonly represent as the "hydrogen ion" H is more accurately described as the H O . is a in which protons are exchanged between two chemical species. A molecule or ion that loses or "donates" a proton is acting as an ; a species that receives or "accepts" that proton plays the role of a . Defining an acid as a substance that simply "dissociates" into its component parts \[HA \rightarrow H^+ + AB^–\] fails to capture the concept outlined in the above box. It is therefore important that you become comfortable with the Brønsted-Lowry model of acid-base behavior that represents the generalized acid-base reaction as a : \[\underbrace{HA}_{\text{acid}} + \underbrace{B}_{base} \rightarrow \underbrace{A^-}_{\text{conjugate base}} + \underbrace{HB^+}_{\text{conjugate acid}} \] The actual electric charges of the reaction products will of course depend on the particular nature of the species \(A\), but the base will always have one more negative charge than the acid HA. Although anyone familiar with chemicals tends to regard certain substances as "acids" and others as "bases", it is important to bear in mind that these labels have meaning only in terms of the particular process being considered. For example, in the absence of water, ammonia, which we usually think of as a base, can lose a proton (and thus act as an acid) to a "stronger" base \(B\) to form the amide ion: \[NH_3 + B → NH_2^– + HB^+\] Similarly, anhydrous sulfuric acid can accept a proton from a stronger acid AH: \[H_2SO_4 + AH \rightarrow H_3SO_4^+ + A^–\] However, this is rather exotic stuff that is beyond the scope of this introductory lesson. The acid-base processes we describe in this set of lessons take place in aqueous solution, so we will always assume that \(H_2O\), which can act either as a proton donor or proton acceptor, plays an active role: \[HA + H_2O → H_3O^+ + A^–\] \[B + H_2O → BH^+ + OH^–\] A small number of acids are themselves amphiprotic, meaning that one molecule can transfer a proton to another molecule of the same kind. The best-known example of this process, known as , is of course that of water: \[2 H_2O_{(l)} \rightleftharpoons OH^– + H_3O^+ \label{1-1}\] Pure liquid ammonia and sulfuric acid are other well-known examples: \[2 NH_{3(l)} \rightleftharpoons NH_2^– + NH_4^+\] \[2 H_2SO_{4(l)} \rightleftharpoons HSO_4^– + H_3SO_4^+\] Autoprotolysis reactions such as these typically proceed to only a very small extent. The equilibrium constant for Equation \(\ref{1-1}\) is commonly written as \[K_w = [H^+,OH^–] \label{1-2}\] and is usually known as the or of water. For most purposes in elementary courses, we can use the value 1.0 × 10 for . But you should understand that this value is correct only for pure water at 25° C and 1 atm pressure. varies considerably with temperature, pressure, and the ionic content of water: It should be obvious that anyone working in the fields of physiological or oceanographic chemistry routinely use \(1.0 \times 10^{–14}\) for ! Acids and bases vary greatly in their "strength" — that is, their tendencies to donate or accept protons. As with any chemical reaction, we can define an equilibrium constant \(K_c\) whose numerical value reflects the extent of the reaction — that is, the relative concentrations of the products and reactants when the reaction reaches equilibrium. The strength of an acid \(HA\) in water can be defined by the equilibrium \[HA + H_2O \rightleftharpoons H_3O^+ + A^–\] \[K_a =\dfrac{[H_3O^+,A^-]}{[HA]} \label{1-3a}\] Similarly, the strength of the base \(A^−\) in water is defined by \[A^− + H_2O \rightleftharpoons HA + OH^–\] \[K_b = \dfrac{'[HA,OH^-]}{[A^-]} \label{1-3b}\] How are and related? The answer can found by multiplying the above two expressions for and : (1-3) Alternatively, we can simply add the two corresponding reaction equations: and — as you should recall from your earlier study of equilibrium, the equilibrium constant of the sum of two reactions is the product of the two K's. So either way you figure it, the \(K_a\) for an acid and \(K_b\) for its conjugate base are related in a very simple way: \[K_a K_b = K_w \label{1-4}\] Clearly, as the strength of a series of acids increases, the strengths of their conjugate bases will decrease, this inverse relation between K and K is implicit in Equation \(\ref{1-4}\). You will recall that the pH scale serves as a convenient means of compressing a wide range of [H ]-values into a small range of numbers. Just as we defined the pH as the negative logarithm of the hydrogen ion concentration, we can define \[pK = −\log K \label{2-5}\] for any equilibrium constant. Acid and base strengths are very frequently expressed in terms of \(pK_a\) and \(pK_b\). From Equation \(\ref{1-4}\), it should be apparent that \[pK_a + pK_b = pK_w \;(14.0\; at\; 25^o C) \label{1-5}\] In order to convert into , simply take the negative antilog of . Thus, for acetic acid with \[pK_a = 4.7\] \[K_a = 10^{–4.7} = 2.0 \times 10^{–5} \] The following table gives the p values for a number of commonly-encountered acid-base systems which are listed in order of decreasing acid strength. It's worth taking some time to study this table in some detail. The concept of the (more properly, ), is not nearly as intimidating as it might sound; it is easy to grasp, requires no mathematics, and will enable you to achieve a far better understanding of acid-base chemistry. Most courses (and their textbooks) do a good job of explaining the concept of proton transfer, but few of them provide you with a clear basis for understanding and making useful predictions about the and this transfer will take. For example, when the gas HCl is added to water, you probably know that the reaction proceeds strongly to the right: \[HCl + H_2O \rightleftharpoons H_3O^+ + Cl^– \label{2-1}\] ...but what principle determines this, and what is to prevent the hydronium ion (also an acid) from pushing one of its own protons back onto a chloride ion, and thus reversing the reaction? The concept of the was introduced in R.W. Gurney's classic (1953). The best detailed exposition can be found in the first (1970) edition of by W. Stumm and J.L. Morgan. The answer is that a proton is never really "pushed"; it will spontaneously seek out the lowest potential energy state (that is, the strongest base) it can find. Think how a macroscopic object such as a book will tend to fall to a location where its [gravitational] potential energy is as low as possible. Of course, the potential energy of a proton has nothing to do with gravity, but rather with how tightly it can bind to a base. This kind of potential energy is known as (PFE) — but in the context of first-year chemistry, we can also just call it "proton energy". You are not sure what free energy is? Don’t worry about it for the time being; just think of it as you would any other form of potential energy: something that falls when chemical reactions take place. See for the lesson on free energy. For the present, all you need to know is that a proton will tend to "fall" to the lowest potential energy state it can find. In keeping with this idea, you can think of an acid as a , and a base as a . Using this terminology, we can depict the generalized process HA + B → AB + HB as in this plot, which depicts the proton on HA "falling" to the lower-PFE "sink" provided by the base B, leaving AB and HB as products. We will be making a lot of use of schematics like this farther on, so take a moment to familiarize yourself with information it depicts. This "source/sink" nomenclature recalls the tendency of water to flow down from a high elevation to a lower one; this tendency (which is related to the amount of energy that can be extracted in the form of electrical work if the water flows through a power station at the bottom of the dam) will be directly proportional to the difference in elevation (difference in ) between the source (top of the dam) and the sink (bottom of the dam). Thus you can regard proton free energy as entirely analogous to the gravitational potential energy of the water contained in a dam; in this case, we are dealing with potential energy, more commonly known as . Take particular note of the following: When you were first introduced to the Brønsted-Lowry model, you learned that the proper way to represent the "dissociation" of a strong acid like hydrochloric is \[HCl + H_2O \rightarrow H_3O^+ + Cl^- \label{2-1a}\] Thus water is revealed as an active participant in the behavior of an acid, rather than as an inert spectator or solvent. Although analogous acid-base reactions can take place in other solvents such as liquid ammonia, virtually all of the acid-base chemistry we encounter in daily life is based on water. Since water is acting as the proton acceptor here, it must be fulfilling the role of the base B in the schematic given above. We can therefore construct a similar plot specifically for HCl (or, in fact, for any strong acid). Recall that, for strong acids, this reaction is essentially complete. This means that a "solution of hydrochloric acid" is in reality a solution of hydronium and chloride ions; except in solutions much more concentrated than 1 M, the species HCl is only present in trace amounts. When acetic acid CH COOH is dissolved in water, the resulting solution displays only very mild acidic properties. For simplicity we will represent acetic acid and the acetate ion CH COO by HAc and Ac , respectively: \[HAc + H_2O \rightleftharpoons H_3O^+ + Ac^– \label{2-2}\] But although this equation has the same form as (2-1) above, it hides an important qualitative difference: HAc is a acid, meaning that the proton transfer takes place to only a tiny extent; the equilibrium strongly favors the left side: HAc + H O H O + Ac The corresponding proton free energy diagram now shows the H O -H O system at the top, with the proton-transfer arrow pointing upward, indicating that the protons get kicked up to a free-energy level in this process. Where does this additional energy come from? Simply from the random thermal energy in the solution. As you might expect, this is not a thermodynamically favorable process, so it only happens to a very small extent. In contrast to the HCl example we described previously in which 99.99+ percent of the HCl molecules end up as hydronium ions, almost all of the acetic acid molecules remain unchanged, and only tiny amounts of H O and Ac are produced. Strong and weak acids both donate protons to water, yielding hydrogen ions and thus rendering the solution acidic. What's different is that virtually of the protons of the strong acid "fall" to the H O level, whereas only a tiny fraction of the weak acid molecules acquire enough thermal energy to promote their protons up to the H O level. The fundamental difference between strong acids and weak acids is clearly evident when we combine the preceding two proton-free energy diagrams into one. Strong acids (strong "conjugate-pairs") are always above the PFE of the H O -H O system, while the weak conjugate pairs are below it. The other big difference with acetic acid is that water is acting as a here, rather than as an as it did in the HCl example. So it's not surprising that water, the basis of almost all acid-base chemistry, should appear in the middle of these diagrams. Substances that can both donate and accept protons are said to be . Many oxides and hydroxides behave in this way, and since H O falls into both of these classes, it is not surprising that water is amphiprotic: ("AH" above represents any acid; B stands for any base) These two opposing tendencies can be represented by In the case of water, the contest is a draw; not only does neither side win, but the fraction of H O molecules that break up in this way is miniscule: In pure water, only about one H O molecule out of 10 is “dissociated” at any instant: H O H + OH (2-4) The actual reaction, of course, is the proton transfer (2-5) for which the equilibrium constant = [H O ,OH ] (2-6) is known as the . The value of at room temperature is 1.008 x 10 . We can also express this in logarithmic terms: = 14.0. As with any equilibrium constant, the value of is affected by the Because most practical calculations involving refer to ionic solutions rather than to pure water, the common practice of using 10 as if it were a universal constant is unwise; under the conditions commonly encountered in the laboratory, can vary from about 11 to almost 15. In seawater, is 6.3 × 10 . Notice that under conditions when differs significantly from 1.0 × 10 , the pH of a neutral solution will not be 7.0. For example, at a pressure of 93 kbar and 527°C, = 10 , the pH of pure water would be 1.5. Similarly unusual conditions apply to deposits of water in geological formations and in undersea vents. At 60° C, the ion product of water is \(9.6 \times 10^{-14}\). What is the pH of a neutral solution at this temperature? Under these conditions, [H ,OH ] = 9.6E–14. If the solution is neutral, [H ] = [OH ] = (9.6E–14) = 3.1E–7, corresponding to a pH of log 3.1E-7 = . The strengths of strong acids and bases are "leveled" in water. This means that . To see this more clearly, notice that the free energies of protons in all of these acids (even nitric) are sufficiently above the H O /H O level that their "dissociation" is essentially complete. Thus The strongest acid that can exist in water is H O ; The strongest base that can exist in water is OH Similarly, the hydroxide ion is the strongest base that can exist in water. So what happens if you add a soluble oxide such as Na O to water? Since O is a proton sink to H O, it will react with the solvent, leaving OH as the strongest base present: Na O + H O → 2 OH + Na . Thus — simply because they are all converted to OH . In the preceding section, we have employed simple PFE (proton free energy) diagrams to help you better understand the concepts of strong and weak acids, water as an acid or a base, and the leveling effect. But we can gain a more generalized and comprehensive view of acid-base chemistry with the aid of a PFE diagram that shows a whole series of acid-base systems, arranged in order of decreasing acid strengths. As we explained near the beginning of this lesson, we ordinarily express the strength of an acid by quoting its "dissociation constant" , which is the equilibrium constant for transfer of a proton to the base H O. HA + H O → H O + A We also introduced the concept of PFE as an alternative way of expressing the strength of an acid. If both PFE and express the same thing, you may wonder how these two quantities are related. If you had an introduction to thermodynamics and specifically to free energy , you might wish to look at the of this lesson to review the relationship between the fall in free energy and the value of . A thorough answer to this question requires a bit of thermodynamics, which you may not have yet studied, but you don't really need any of the math here. It should suffice to show the results in the form of a plot, which yields a simple straight-line relationship when plotted on a semi-logarithmic scale. Note that On the PFE diagram shown below, each acid-base system is represented by a horizontal line connecting an acidic species with its conjugate base. Each line is placed at a height on the diagram that is proportional to its , as measured on the logarithmic scale (labeled "pH") at the right. What we have, then, is a list of acid-base systems arranged in order of decreasing acid strength. Think of this chart as an alternative view of the table of acid strengths displayed near the top of this page, in which some of the information that is "hidden" in the table is now revealed in an easily-grasped way. Take particular note of the following points: The virtue of a PFE chart is that you can see, at a glance, the relations between conjugate acid-base pairs that enables you to make quick, qualitative determinations of what is going on even in complex mixtures of acids and bases. The speed and convenience this affords enables you to build a much deeper conceptual understanding of acid-base chemistry. In order to do the same thing by consulting a simple list of acid 's or 's, you also need to be familiar with the principles of chemical equilibrium and thermodynamics — and then take the time to work through the arithmetic. In the preceding sections we have shown how the PFE-chart approach can clarify the distinction between strong and weak acids. We will now go on to describe a variety of other principles that are easily grasped by consulting a PFE chart. You will have noticed the scale labeled "pH" at the left side of the PFE plot. First, please understand what this scale does NOT do; This pH will depend on both the strength ( or ) of the acid, and on its concentration in the resulting solution; you will learn how to carry out pH calculations in later lessons. All but the very weakest acids can drive the pH down to near the bottom of the scale if their solutions are sufficiently concentrated. The only thing you can be sure of is that the pH of a solution of a solution of an acid in pure water will never be greater than 7. When pH was first introduced in 1909, it was defined in terms of the "hydrogen ion concentration" [H ]. This has since been amended to the hydrogen ion {H }. The latter term refers to the "effective" concentration of these ions — that is to the "availability" of protons (regardless of whether they physically exist as H O units or in other forms) to react with bases. But from the standpoint of thermodynamics, "availability" is just a loose term for "free energy". So we can also say that pH is a measure of the average PFE in an aqueous solution The more negative the pH, the higher the proton free energy in the solution. (The inverse relation is a consequence of the negative-logarithm definition of pH.) The pH has the dual advantages of being both dimensionless and readily observable. More importantly, we can easily alter the pH of a solution by adding some strong acid or base. And when we do this, we are in effect using the pH as a tool for controlling the average PFE in the solution. More on this in a moment; first, let us introduce an important relation that you may have seen in an earlier chemistry course, and which we will discuss more in a later lesson in this group. Recalling the equilibrium expression for a weak acid \[K_a = \dfrac{[H^+,A^-]}{[HA]} \label{3-1}\] We can solve this for \([H^+]\): \[H^+] = K_a \dfrac{[HA]}{[A^-]} \label{3-2}\] Re-writing this in terms of negative logarithms, this becomes (3-3) or, since = – log , we invert the ratio to preserve the positive sign: We will make extensive use of this equation in on acid-base buffering and titration. (3-4) This extremely important relation tells us that the pH of a solution containing a weak acid-base system controls the relative concentrations of the acid and base forms of that system. Of special interest is the case in which the pH of a solution of an acid-base system is set to the value of its . According to the above equation, when pH = , the log term becomes zero, so that the ratio [AB ] / [HA] = 10 = 1, meaning that [HA] = [AB ]. In other words, This pH adjustment can be easily made by adding appropriate amounts of a strong acid or strong base to the solution. When the pH of a solution is set to the value of the of an acid-base pair, the concentrations of the acid- and base forms will be identical. This condition can be represented schematically: This is the basis for the arrangement of the various acid-base systems depicted in the PFE diagram; each conjugate pair is placed at a location on the vertical pH scale that corresponds to the acid's . It is instructive to consider how sensitively the concentration ratio [A ] / [HA] depends on the pH. Hypochlorous acid, HOCl, has a of 8.0. Substituting in Eq 3-4, the log term becomes (8.2 – 8.0) = 0.2, so = 10 = 1.6. Note that it does not matter whether the initial solution consists of HOCl or NaOCl or some mixture thereof; adjusting the pH the system composition to the ratio calculated above.) In the resulting solution, the base form OCl is the predominant species, the mole ratio being 1.6/1. The mole fraction of OCl will therefore be 1.6/2.6 = 0.62, making that of HOCl (1.0 – .62) = 0.38. The actual concentrations will be 0.1 times these values, reflecting the 0.10 concentration of the solution. To what pH must we adjust a solution of acetic acid ( 4.7) in order to convert 20 percent of the acid into its base form? In the final solution, the mole fractions will be [AB ] = .20, [HA] = 0.80; the ratio [AB ]/[HA] = 0.20/0.80 = 0.25. Substituting into Eq 2-9, we have \[pH = 4.7 + \log\' 0.25 = 4.7 – 0.6 = 4.1\] We can explore the relation between the pH and the distribution of conjugate species between the acid and base forms by carrying out calculations similar to those in Problem Example 2 for an acid-base system at different pH values. The orange bars and blue numbers show the mole fractions of the acid and base forms at pH values of the and the ± 1. The green numbers give the percent dissociation of the acid at each pH. The apparent symmetry between the two extreme pH values suggests a simple relation between the log of the ratio [AB ] / [HA] and the pH. This is seen very clearly when this ratio is plotted as a function of pH. Solutions containing many different weak acid-base systems are very commonly encountered in nature, especially in biological fluids or natural waters, including the ocean. When the pH of such solutions is altered by the addition of strong acid or base (as can occur, for example, when wastes from industrial processes or mine drainage, or acidic rain, enter a lake or stream); the distribution of all conjugate acid-base systems will change. It is important to note that the [AB ]  / [HA] ratios (calculated here from Eq 3-4) yield only a qualitative picture and do not take into account the concentrations of the different acid-base systems or proton transfers between them. The latter processes would cause the lowest proton-vacant levels (that is, the strongest bases) to be completely filled from the bottom up, depending on the number of protons available. An exact calculation would require solving a set of simultaneous equations and is beyond the scope of this lesson. The Brønsted-Lowry model defines the of an acid HA as its tendency to donate protons to a base B. But bases vary in their abilities to protons, so the tendency for the complete transaction HA + B → BH + A (A-1) to occur depends on the processes HA → H + A (A-2) B + H → BH (A-3) 's are always expressed on a scale relative to the of the solvent Because it is impossible to study either of these steps independently, what we might call the "absolute" strength of an acid (Eq A-2) cannot be measured. For this reason, we define a "standard" base, usually the solvent, and most commonly, water. Thus we replace Eq A-3 with H O + H → H O (A-4) so that Eq A 1 now expresses the strength of HA : HA + H O → H O + A (A-5) which, for convenience, we write in its abbreviated form HA → H + A (A-5) whose equilibrium constant is (A-6) This value of is, of course, numerically identical to that for Eq A-5. This section may be of interest to readers who have studied elementary thermodynamics, and have some familiarity with . it is not required for understanding and using the concept of PFE. From elementary thermodynamics, the driving force of a chemical reaction is given by the standard free energy change: Δ ° = ln (B-1) Solving this equation for yields (B-2) For a weak acid we know that (B-3) Substituting Eq B-2 into this expression, we get (B-4) Next, we take the negative log of each term, recalling that ln = 2.3 log : (B-5) or (B-6) For the special case in which the concentrations of the conjugate pairs [H ] and [HA] are equal, pH = , and the rightmost term above disappears, leaving (B-7) Thus the pH is a direct measure of the average proton free energy in a solution. Construct a PFE diagram showing H O , acetic acid HAc ( = 4.76), and H O. Solving Eq A 13 for ΔG° yields ΔG° = (2.3RT) × , or, at 25°C, ΔG° = (2.3 × 8.314 J K mol ) × = 5698 J K mol × The proton in H O is already at the level of that in the standard base H O (Eq A 5), so for this transfer is [by definition] ΔG° = 5698 J K mol × 0 = The increase in the free energy required to transfer a proton from its level in HAc to that in H O is ΔG° = 5698 J K mol × 4.76 = To transfer a proton from H O to another H O (autoprotolysis), its free energy must increase by 5698 J K mol × 14.0 =

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Chemical change is one of the two central concepts of chemical science, the other being . The very origins of Chemistry itself are rooted in the observations of transformations such as the combustion of wood, the freezing of water, and the winning of metals from their ores that have always been a part of human experience. It was, after all, the quest for some kind of constancy underlying change that led the Greek thinkers of around 200 BCE to the idea of elements and later to that of the atom. It would take almost 2000 years for the scientific study of matter to pick up these concepts and incorporate them into what would emerge, in the latter part of the 19th century, as a modern view of chemical change. Chemical change occurs when the atoms that make up one or more substances rearrange themselves in such a way that new substances are formed. These substances are the components of the chemical reaction system; those components which decrease in quantity are called reactants, while those that increase are products. A given chemical reaction system is defined by a which is conventionally written as \[\text{reactants} \rightarrow \text{products}\] The first thing we need to know about a chemical reaction represented by a balanced equation is whether it can actually take place. If the reactants and products are all substances capable of an independent existence, then in principle, the answer is always "yes". This answer must be qualified, however, by the following considerations: A reaction that is thermodynamically possible but for which no reasonably rapid mechanism is available is said to be . Conversely, one that occurs rapidly but only to a small extent is . As you will see later, there are often ways of getting around both kinds of limitations, and their discovery and practical applications constitute an important area of industrial chemistry. Basically, the term refers to what we might call a "balance of forces". In the case of mechanical equilibrium, this is its literal definition. A book sitting on a table top remains at rest because the downward force exerted by the earth's gravity acting on the book's mass (this is what is meant by the "weight" of the book) is exactly balanced by the repulsive force between atoms that prevents two objects from simultaneously occupying the same space, acting in this case between the table surface and the book. If you pick up the book and raise it above the table top, the additional upward force exerted by your arm destroys the state of equilibrium as the book moves upward. If you wish to hold the book at rest above the table, you adjust the upward force to exactly balance the weight of the book, thus restoring equilibrium. An object is in a state of when it is either static (motionless) or in a state of unchanging motion. From the relation = , it is apparent that if the net force on the object is zero, its acceleration must also be zero, so if we can see that an object is not undergoing a change in its motion, we know that it is in mechanical equilibrium. Another kind of equilibrium we all experience is . When two objects are brought into contact, heat will flow from the warmer object to the cooler one until their temperatures become identical. Thermal equilibrium arises from the tendency of thermal energy to become as dispersed or "diluted" as possible. A metallic object at room temperature will feel cool to your hand when you first pick it up because the thermal sensors in your skin detect a flow of heat from your hand into the metal, but as the metal approaches the temperature of your hand, this sensation diminishes. The time it takes to achieve thermal equilibrium depends on how readily heat is conducted within and between the objects; thus a wooden object will feel warmer than a metallic object even if both are at room temperature because wood is a relatively poor thermal conductor and will therefore remove heat from your hand more slowly. Thermal equilibrium is something we often want to avoid, or at least postpone; this is why we insulate buildings, perspire in the summer and wear heavier clothing in the winter. When a chemical reaction takes place in a container which prevents the entry or escape of any of the substances involved in the reaction, the quantities of these components change as some are consumed and others are formed. Eventually this change will come to an end, after which the composition will remain unchanged as long as the system remains undisturbed. The system is then said to be in its , or more simply, "at equilibrium". What is the nature of the "balance of forces" that drives a reaction toward chemical equilibrium? It is essentially the balance struck between the tendency of energy to reside within the chemical bonds of stable molecules, and its tendency to become dispersed and diluted. Exothermic reactions are particularly effective in this, because the heat released gets dispersed in the infinitely wider world of the surroundings. In the reaction represented here, this balance point occurs when about 60% of the reactants have been converted to products. Once this equilibrium state has been reached, no further net change will occur. The only spontaneous changes that are allowed follow the arrows pointing toward maximum dispersal of energy. Chemical equilibrium is something you definitely want to avoid for yourself as long as possible. The myriad chemical reactions in living organisms are constantly moving equilibrium, but are prevented from getting there by input of reactants and removal of products. So rather than being in equilibrium, we try to maintain a "steady-state" condition which physiologists call — maintenance of a constant internal environment. For the time being, it's very important that you know this definition: The in which a chemical reaction is written (and thus which components are considered reactants and which are products) is arbitrary. Consider the following two reactions: \[\underset{\text{synthesis of hydrogen iodide}}{H_2 + I_2 \rightarrow 2 HI} \label{10.1}\] \[\underset{\text{dissociation of hydrogen iodide}}{2 HI \rightarrow H_2 + I_2} \label{10.2}\] Equations \(\ref{10.1}\) and \(\ref{10.2}\) represent the same chemical reaction system in which the roles of the components are reversed, This is central to the concept of chemical equilibrium. It makes no difference whether we start with two moles of HI or one mole each of H and I ; once the reaction has run to completion, the quantities of these two components will be the same. In general, then, we can say that the composition of a chemical reaction system will tend to change in a direction that brings it closer to its equilibrium composition Once this equilibrium composition has been attained, no further change in the quantities of the components will occur as long as the system remains undisturbed. The composition of a chemical reaction system will tend to change in a direction that brings it closer to its equilibrium composition. The two diagrams below show how the concentrations of the three components of this chemical reaction change with time. Examine the two sets of plots carefully, noting which substances have zero initial concentrations, and are thus "products" of the reaction equations shown. Satisfy yourself that these two sets represent the same , but with the reactions occurring in opposite directions. Most importantly, note how the final (equilibrium) concentrations of the components are the same in the two cases. The composition is independent of the direction from which it is approached (i.e., the initial conditions). A chemical equation of the form represents the transformation of A into B, but it does not imply that of the reactants will be converted into products, or that the reaction cannot also occur. In general, processes (forward and reverse) can be expected to occur, resulting in an containing finite amounts of of the components of the reaction system. (We use the word when we do not wish to distinguish between reactants and products.) If the equilibrium state is one in which significant quantities of both reactants and products are present (as in the hydrogen iodide example given above), then the reaction is said to or . The latter term is preferable because it avoids confusion with "complete" in its other sense of being completed or finished, implying that the reaction has run its course and is now at equilibrium. Note that there is no fundamental difference between the meanings of A → B and A B. Some older textbooks just use A = B. In principle, all chemical reactions are reversible, but this reversibility may not be observable if the fraction of products in the equilibrium mixture is very small, or if the reverse reaction is very slow (the chemist's term is " ") We can thank Napoleon for bringing the concept of reaction reversibility to Chemistry. \[Na_2CO_3 + CaCl_2 \rightarrow CaCO_3 + 2 NaCl \label{10.3}\] which was known to proceed to completion in the laboratory. He immediately realized that the Na CO must have been formed by the reverse of this process brought about by the very high concentration of salt in the slowly-evaporating waters. This led Berthollet to question the belief of the time that a reaction could only proceed in a single direction. His famous textbook (1803) presented his speculations on chemical affinity and his discovery that an excess of the product of a reaction could drive it in the reverse direction. Unfortunately, Berthollet got a bit carried away by the idea that a reaction could be influenced by the amounts of substances present, and maintained that the same should be true for the compositions of individual compounds. This brought him into conflict with the recently accepted Law of Definite Proportions (that a compound is made up of fixed numbers of its constituent atoms), so his ideas (the good along with the bad) were promptly discredited and remained largely forgotten for 50 years. (Ironically, it is now known that certain classes of compounds do in fact exhibit variable composition of the kind that Berthollet envisioned.) Berthollet's ideas about reversible reactions were finally vindicated by experiments carried out by others, most notably the Norwegian chemists (and brothers-in-law) Cato Guldberg and Peter Waage. During the period 1864-1879 they showed that an equilibrium can be approached from either direction (see the hydrogen iodide illustration above), implying that any reaction \[aA + bB \rightarrow cC + dD\] is really a competition between a "forward" and a "reverse" reaction. When a reaction is at equilibrium, the rates of these two reactions are identical, so no (macroscopic) change is observed, although individual components are actively being transformed at the microscopic level. Guldberg and Waage showed that for a reaction \[aA + bB \rightarrow cC + dD\] the rate (speed) of the reaction in either direction is proportional to what they called the "active masses" of the various components: in which the proportionality constants are called and the quantities in square brackets represent concentrations. If we combine the two reactants and , the forward reaction starts immediately; then, as the products and begin to build up, the reverse process gets underway. As the reaction proceeds, the rate of the forward reaction diminishes while that of the reverse reaction increases. Eventually the two processes are proceeding at the same rate, and the reaction is at equilibrium: rate of forward reaction = rate of reverse reaction \[k_f [A]^a [B]^b = k_r [C]^c [D]^d\] It is very important that you understand the significance of this relation. The equilibrium state is one in which there is no change in the quantities of reactants and products. But do not confuse this with a state of "no change"; at equilibrium, the forward and reverse reactions continue, but , essentially cancelling each other out. Be sure you understand the difference between the of a reaction and a . The latter, usually designated by , relates the reaction rate to the concentration of one or more of the reaction components — for example, = [A]. At equilibrium the rates of the forward and reverse processes are identical, but the rate are generally different. To see how this works, consider the simplified reaction A → B in the following three scenarios. The images shown below offer yet another way of looking at these three cases. The plots show how the relative concentrations of the reactant and product change during the course of the reaction. The plots differ in the assumptions we make about the ratio of to . The equilibrium composition of the system is illustrated by the proportions of A and B in the horizontal parts of each plot where the composition remains unchanged. In each case, the two rate constants are sufficiently close in magnitude that each reaction can be considered "incomplete". The Law of Mass Action is thus essentially the statement that the equilibrium composition of a reaction mixture can vary according to the quantities of components that are present. This of course is just what Berthollet observed in his Egyptian salt ponds, but we now understand it to be a consequence of the dynamic nature of chemical equilibrium. Clearly, if we observe some change taking place— a change in color, the release of gas bubbles, the appearance of a precipitate, or the release of heat, we know the reaction is not yet at equilibrium. However, . The equilibrium state is one in which not only no change in composition take place, but also one in which no energetic tendency for further change is present. Unfortunately, "tendency" is not a property that is directly observable! Consider, for example, the reaction representing the synthesis of water from its elements: \[2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(g)} \label{10.5}\] You can store the two gaseous reactants in the same container indefinitely without any observable change occurring. But if you create an electrical spark in the container or introduce a flame, ! After you pick yourself up off the floor and remove the shrapnel from what's left of your body, you will know very well that the system was not initially at equilibrium! It happens that this particular reaction has a tremendous tendency to take place, but for reasons that we will discuss in a later chapter, nothing can happen until we "set it off" in some way— in this case by exposing the mixture to a flame or spark, or (in a more gentle way) by introducing a platinum wire, which acts as a . The similar reaction of hydrogen and iodine \[H_{2(g)} + I_{2(g)} \rightarrow 2 HI_{(g)} \label{10.6}\] by contrast is only moderately favored thermodynamically (and is thus incomplete), but its kinetics are both unspectacular and reasonably facile. Make sure you thoroughly understand the following essential ideas which have been presented above.

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The storage modulus measures the resistance to deformation in an elastic solid. It's related to the proportionality constant between stress and strain in Hooke's Law, which states that extension increases with force. In the dynamic mechanical analysis, we look at the stress (σ), which is the force per cross-sectional unit area, needed to cause an extension in the sample, or the strain (ε). \[E = \dfrac{σ}{ ε}\] Alternatively, in a shear experiment: \[G = \dfrac{σ}{ε}\] The dynamic mechanical analysis differs from simple tensile testing by performing the experiment cyclically. The sample is stretched and released (or sheared and released). It can then be subjected to additional stress and released again. There is an element of time involved here. Specifically, because the experiment is cyclic, it can be carried out at different frequencies. When you do that, and you plot the resulting modulus against frequency, you can get additional information about the sample. The results would typically be presented in a graph like this one: What the graph tells us is that frequency clearly matters. When the experiment is run at higher frequencies, the storage modulus is higher. The material appears to be stiffer. In contrast, the loss modulus is lower at those high frequencies; the material behaves much less like a viscous liquid. In particular, the sharp drop in loss modulus is related to the relaxation time of the material. In this context, that's the time it takes the chains to flow into new conformations in response to the applied stress. If they don't have time to flow, then that viscous response of the material is lost. The material behaves much more like an elastic solid when subjected to high-frequency cyclic deformation. That's important to know, because a material might be subjected to vibrations or other stimuli during everyday use, and its properties might change accordingly. Another variation on this kind of experiment is called dynamic mechanical thermal analysis. Instead of changing the frequency of the stimulus throughout the experiment, the frequency is held constant and the temperature is changed instead. As a result, we can again see how the material responds under different conditions, which might tell us how it will behave in everyday applications. The result of the experiment might be a graph like the one below: At this point, you are already familiar with the . It shouldn't be surprising that the properties are dependent on temperature. At the glass transition temperature, the expanding volume of the material with increasing temperature becomes sufficient to allow chain flow. As a result, the material suddenly behaves much more like a viscous liquid. Loss modulus increases. The stiffness of the material drops as the entangled chains not longer resist deformation as strongly. Storage modulus decreases. The dynamic mechanical thermal analysis thus provides an alternative way to determine the glass transition temperature. Because it is actually measuring a different physical phenomenon than differential scanning calorimetry, the obtained from a DMTA experiment may not agree exactly with one obtained from a DSC experiment. Nevertheless, it is often useful to have different ways of assessing properties. In order to facilitate the analysis of the in this experiment, a different quantity is usually displayed. Tan delta is just the ratio of the loss modulus to the storage modulus. It peaks at the glass transition temperature. The term "tan delta" refers to a mathematical treatment of storage modulus; it's what happens in-phase with (or at the same time as) the application of stress, whereas loss modulus happens out-of-phase with the application of stress. Because it would take some time for the chains to move into new confirmation when they are subjected to stress, the strain actually lags behind the stress in these experiments. Delta refers to the phase lag, the amount of time between application of stress and the observation of maximum strain. You may remember that a sine curve and cosine curve are out of phase with each other. Storage modulus is described as being proportional to δ whereas loss modulus is proportional to . The ratio of to is just . Why does peak at the glass transition temperature? Clearly, as chains begin to move more freely, loss modulus increases. Consequently, the material also becomes less stiff and more rubbery. The storage modulus drops. If tan delta is the ratio of loss modulus to storage modulus, it should increase at that point -- and it does. Why does it drop again? That's because loss modulus refers to an energy loss, but because the material has gotten softer, less stress (and less energy) is put into the sample in the first place, so the energy loss also gets smaller. As a result, tan delta goes up at the glass transition but drops again shortly beyond that point. Estimate the storage and loss, modulus in the glassy phase and rubbery phase in each of the following cases. Estimate in each of the following cases.

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Up to this point we have discussed only the properties of pure solids and liquids. Of more importance to a chemist, though, are the properties of solutions. Very few chemical reactions involve only pure substances―almost all involve a solution of some sort. We defined a solution as a homogeneous mixture of two or more substances , that is, a mixture which appears to be uniform throughout. Under this definition we would refer to sugar or salt dissolved in water as solutions, but we would not apply the term to muddy water or to milk. A close inspection of muddy water reveals that it is not uniform in appearance but consists of small solid particles dispersed in water. We refer to such a mixture as a . Under the microscope, milk can also be seen to be nonuniform, It consists of small drops of milk fat dispersed throughout an aqueous phase. A more obvious example of a suspension is chia seeds in water, seen below. The chia seeds are randomly suspended in the cup of water, a perfect example of a suspension. Our definition of a solution in terms of the homogeneity of a mixture is somewhat unsatisfactory since it does not tell us where to draw the line. Field-emission microscopes and electron microscopes have now been developed which can just about ”see” a single atom. With such a microscope virtually matter looks nonuniform and hence not homogeneous. If our definition extends to such microscopes, then true solutions do not exist. In practice we draw the line somewhere around the 5 nanometer (nm) mark, even though some molecules are larger than this. Strictly speaking, the term applies to any homogeneous mixture, but we will concentrate our discussion on those solutions which involve liquids since these are the most common. It should be realized, though, that other types of solutions also exist. Air is a solution of a large number of gases (oxygen is the most concentrated) in another gas (nitrogen). A 5-cent coin is made from an alloy in which one solid (nickel) is dissolved in another (copper). Solutions of hydrogen gas in solid palladium and some other metals are also possible. As mentioned in the brief discussion of solutions , it is sometimes difficult to decide which component of a solution is the solute and which is the solvent. Usually the amount of solvent is much larger than that of the solute. If the pure components were initially in separate phases (a gas and a solid, for example), the phase corresponding to the state of the solution is taken to be the solvent. In the case of H ( ) and Pd(s) mentioned above, for example, Pd would be the solvent because the solution is a solid phase.

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It should be clear by now that elements follow a periodic law, but atomic weight is inadequate for fully explaining this phenomenon. If ordered by atomic weight, chemical and physical properties will repeat at regular intervals, but there are clear exceptions to this behavior, for instance, K has a smaller atomic weight than Ar, but its properties would place it after Ar in terms of periodic behavior. All of this implies that there is another aspect to atomic structure to explain this law. Today, we know that such properties arise from a nuclear model of organization for atomic structure. The atom is not indivisible as Dalton suggested, but rather made up of a small, dense nucleus containing neutrons and protons, as well as rapidly moving electrons which take up the greater volume of an atom.   What led to this model? Beginning in the late 19th century a number of major experiments were performed. The results of these experiments helped scientists to formulate the current view of atomic structure. From this model of atomic structure, the origin of the periodicity of the elements can be more deeply explained. The order of the elements in the periodic table is dictated mostly by atomic weight which is directly related to the number of protons in the nucleus and the number of electrons surrounding the nucleus. The model of the Nuclear Atom reveals how the arrangement of electrons dictates the Periodic Law.

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Elimination reactions of 4º-ammonium salts are termed . Since the counter anion in most 4º-ammonium salts is halide, this is often replaced by the more basic hydroxide ion through reaction with silver hydroxide (or silver oxide). The resulting hydroxide salt must then be heated (100 - 200 ºC) to effect the E2-like elimination of a 3º-amine. Example #1 below shows a typical Hofmann elimination. Obviously, for an elimination to occur one of the alkyl substituents on nitrogen must have one or more beta-hydrogens, as noted earlier in examining elimination reactions of alkyl halides. In example #2 above, two of the alkyl substituents on nitrogen have beta-hydrogens, all of which are on methyl groups (colored orange & magenta). The chief product from the elimination is the alkene having the more highly substituted double bond, reflecting not only the 3:1 numerical advantage of those beta-hydrogens, but also the greater stability of the double bond. Example #3 illustrates two important features of the Hofmann elimination: The tendency of Hofmann eliminations to give the less-substituted double bond isomer is commonly referred to as the , and contrasts strikingly with the Zaitsev Rule formulated for dehydrohalogenations and dehydrations. In cases where other activating groups, such as phenyl or carbonyl, are present, the Hofmann Rule may not apply. Thus, if 2-amino-1-phenylpropane is treated in the manner of example #3, the product consists largely of 1-phenylpropene (E & Z-isomers). To understand why the base-induced elimination of 4º-ammonium salts behaves differently from that of alkyl halides it is necessary to reexamine the nature of the E2 transition state, first described for dehydrohalogenation. The energy diagram shown earlier for a single-step bimolecular E2 mechanism is repeated on the right. The E2 transition state is less well defined than is that of S 2 reactions. More bonds are being broken and formed, with the possibility of a continuum of states in which the extent of C–H and C–X bond-breaking and C=C bond-making varies. For example, if the bond to the leaving group (X) is substantially broken relative to the other bond changes, the transition state approaches that for an E1 reaction (initial ionization followed by a fast second step). At the other extreme, if the acidity of the beta-hydrogens is enhanced, then substantial breaking of C–H may occur before the other bonds begin to be affected. For most simple alkyl halides it was proper to envision a balanced transition state, in which there was a synchronous change in all the bonds. Such a model was consistent with the Zaitsev Rule. When the leaving group X carries a positive charge, as do the 4º-ammonium compounds discussed here, the inductive influence of this charge will increase the acidity of both the alpha and the beta-hydrogens. Furthermore, the 4º-ammonium substituent is much larger than a halide or hydroxyl group and may perturb the conformations available to substituted beta-carbons. It seems that a combination of these factors acts to favor base attack at the least substituted (least hindered and most acidic) set of beta-hydrogens. The favored anti orientation of the leaving group and beta-hydrogen, noted for dehydrohalogenation, is found for many Hofmann eliminations; but syn-elimination is also common, possibly because the attraction of opposite charges orients the hydroxide base near the 4º-ammonium leaving group. Three additional examples of the Hofmann elimination are shown in the following diagram. Example #1 is interesting in two respects. First, it generates a 4º-ammonium halide salt in a manner different from exhaustive methylation. Second, this salt is not converted to its hydroxide analog prior to elimination. A concentrated aqueous solution of the halide salt is simply dropped into a refluxing sodium hydroxide solution, and the volatile hydrocarbon product is isolated by distillation. Example #2 illustrates an important aspect of the Hofmann elimination. If the nitrogen atom is part of a ring, then a single application of this elimination procedure does not remove the nitrogen as a separate 3º-amine product. In order to sever the nitrogen function from the molecule, a second Hofmann elimination must be carried out. Indeed, if the nitrogen atom was a member of two rings (fused or spiro), then three repetitions of the Hofmann elimination would be required to sever the nitrogen from the remaining molecular framework. Example #3 is noteworthy because the less stable trans-cyclooctene is the chief product, accompanied by the cis-isomer. An anti-E2-transition state would necessarily give the cis-cycloalkene, so the trans-isomer must be generated by a syn-elimination. The cis-cyclooctene produced in this reaction could also be formed by a syn-elimination. Cyclooctane is a conformationally complex structure. Several puckered conformations that avoid angle strain are possible, and one of the most stable of these is shown on the right. Some eclipsed bonds occur in all these conformers, and transannular hydrogen crowding is unavoidable. Since the trimethylammonium substituent is large (about the size of tert-butyl) it will probably assume an equatorial-like orientation to avoid steric crowding. An anti-E2 transition state is likely to require an axial-like orientation of this bulky group, making this an unfavorable path. ),

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Sufficiently accurate measurement of pressure, temperature, volume, and amount of any gas will reveal that is never obeyed exactly. This is why the molar volumes in were not all exactly 22.414 liters. A convenient way to detect deviations from the ideal gas law is to calculate under various conditions. According to , is two-thirds the total molecular kinetic energy and should remain constant at a given temperature for a given amount of gas. That it does not is evident from Figure \(\Page {1}\), where for 1 mol CH ( ) is plotted versus . At high pressures, PV is always larger than would be predicted by the ideal gas law. As the temperature decreases, deviations occur at lower pressures, and drops below the predicted horizontal line before rising again with pressure. At high pressures, increases above the ideal gas value because is no longer valid. As pressure increases, the molecules are squeezed close to one another, and the volume of the molecules themselves becomes a significant fraction of the volume of the container. This is shown in Figure \(\Page {2}\). The space which other molecules are prevented from occupying is called the . The measured volume of the container, , is the sum of the volume available to the gas molecules, , and this excluded volume. Since is larger than , the experimentally measured is too large. Intermolecular forces cause to drop below the ideal gas prediction at low temperatures and medium pressures. Consider a gas molecule which is about to hit the wall of the container (Figure \(\Page {3}\) ). Kinetic theory assumes that its neighbors exert no forces on such a molecule except during a collision (postulate 5), but we know that such forces exist. When a molecule is near the wall, the attractions between it and its neighbors are unbalanced, tending to pull it away from the wall. The molecule produces slightly less impact than it would if there were no intermolecular forces. All collisions with the walls are softer, and the pressure is less than would be predicted by the ideal gas law. This effect of intermolecular forces is more pronounced at lower temperatures because under those conditions the kinetic energies of the molecules are smaller. The potential energy of intermolecular attraction is comparable to that kinetic energy and can have a significant effect. For gases such as hydrogen, oxygen, nitrogen, helium, or neon, deviations from the ideal gas law are less than 0.1 percent at room temperature and atmospheric pressure. Other gases, such as carbon dioxide or ammonia, have stronger intermolecular forces and consequently greater deviation from ideality. Nonideal behavior is quite pronounced for any gas at very high pressures or at temperatures just above the boiling point. Under these conditions molecular volume or intermolecular attractions can have maximum effect.

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Freshly-baked bread is light and fluffy as a result of the action of yeast on sugar. The yeast converts the sugar to carbon dioxide, which at high temperatures causes the dough to expand. The end result is an enjoyable treat, especially when covered with melted butter. French physicist Jacques Charles (1746-1823) studied the effect of temperature on the volume of a gas at constant pressure. states that the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant. The absolute temperature is temperature measured with the Kelvin scale. The Kelvin scale must be used because zero on the Kelvin scale corresponds to a complete stoppage of molecular motion. Mathematically, the direct relationship of Charles's Law can be represented by the following equation: \[\frac{V}{T} = k\nonumber \] As with Boyle's Law, \(k\) is constant only for a given gas sample. The table below shows temperature and volume data for a set amount of gas at a constant pressure. The third column is the constant for this particular data set and is always equal to the volume divided by the Kelvin temperature. When this data is graphed, the result is a straight line, indicative of a direct relationship, shown in the figure below. Notice that the line goes exactly toward the origin, meaning that as the absolute temperature of the gas approaches zero, its volume approaches zero. However, when a gas is brought to extremely cold temperatures, its molecules would eventually condense into the liquid state before reaching absolute zero. The temperature at which this change into the liquid state occurs varies for different gases. Charles's Law can also be used to compare changing conditions for a gas. Now we use \(V_1\) and \(T_1\) to stand for the initial volume and temperature of a gas, while \(V_2\) and \(T_2\) stand for the final volume and temperature. The mathematical relationship of Charles's Law becomes: \[\frac{V_1}{T_1} = \frac{V_2}{T_2}\nonumber \] This equation can be used to calculate any one of the four quantities if the other three are known. The direct relationship will only hold if the temperatures are expressed in Kelvin. Temperatures in Celsius will not work. Recall the relationship that \(\text{K} = \: ^\text{o} \text{C} + 273\). A balloon is filled to a volume of \(2.20 \: \text{L}\) at a temperature of \(22^\text{o} \text{C}\). The balloon is then heated to a temperature of \(71^\text{o} \text{C}\). Find the new volume of the balloon. Use Charles's Law to solve for the unknown volume \(\left( V_2 \right)\). The temperatures have first been converted to Kelvin. First, rearrange the equation algebraically to solve for \(V_2\). \[V_2 = \frac{V_1 \times T_2}{T_1}\nonumber \] Now substitute the known quantities into the equation and solve. \[V_2 = \frac{2.20 \: \text{L} \times 344 \: \text{K}}{295 \: \text{K}} = 2.57 \: \text{L}\nonumber \] The volume increases as the temperature increases. The result has three significant figures.

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If you want to measure ΔH, you usually use calorimetry, which just means measuring heat. The usual way this is done is by measuring how much the temperature of a system increases when the process occurs. For instance, perhaps we have 2 solutions (like an acid and a base solution) and we mix them in a thermos. We measure the temperature of the solutions before mixing and also after the reaction. Because we run the reaction in a thermos, we expect that almost all the heat from the reaction will stay in the thermos. Also, we don't close the thermos all the way, so the pressure is always atmospheric pressure. The reaction enthalpy is related to the temperature change, but how, exactly? tells us how much heat is needed to increase the temperature of an object or substance by a certain amount. The unit is the energy needed to increase the temperature of 1 g of water by 1 degree (either Celsius or Kelvin). 1 calorie = 4.18 J. Generally, the heat capacity is \[C=\frac{q}{\Delta T}\] Depending on what is convenient, heat capacity can be defined in different ways. Sometimes it is defined using heat transferred at constant pressure, and other times heat transferred at constant volume. Specific heat is the heat capacity per unit of mass: \[C=\dfrac{q}{\Delta T\; \times m}\] Molar heat capacity is the heat capacity per mole of substance. If you know the heat capacity of the system, you can calculate ΔH using the temperature change data from a calorimetry experiment. In the example of the acid-base reaction in the thermos, you would want the heat capacity of the thermos and you would add the heat capacity of the solution, which you could calculate using the specific or molar heat of water. First, you need to be a little careful about whether the experiment was done at constant pressure or constant volume. This will determine whether you calculate enthalpy or internal energy of reaction. Second, make sure you figure out the heat capacity of the system correctly. You might have to add up heat capacities of different parts of the system using (mass x specific heat) or (moles x molar heat capacity) of each part. Once you've figured that out, you can usually think of it as a "unit conversion" and use to combine all the quantities you know to find the quantity you want.

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1. The following masses were recorded for 12 different U.S. quarters (all given in grams): Report the mean, median, range, standard deviation and variance for this data. 2. A determination of acetaminophen in 10 separate tablets of Excedrin Extra Strength Pain Reliever gives the following results (in mg) (a) Report the mean, median, range, standard deviation and variance for this data. (b) Assuming that \(\overline{X}\) and are good approximations for \(\mu\) and for \(\sigma^2\), and that the population is normally distributed, what percentage of tablets contain more than the standard amount of 250 mg acetaminophen per tablet? The data in this problem are from Simonian, M. H.; Dinh, S.; Fray, L. A. , , 37–47. 3. Salem and Galan developed a new method to determine the amount of morphine hydrochloride in tablets. An analysis of tablets with different nominal dosages gave the following results (in mg/tablet). (a) For each dosage, calculate the mean and the standard deviation for the mg of morphine hydrochloride per tablet. (b) For each dosage level, and assuming that \(\overline{X}\) and are good approximations for \(\mu\) and for \(\sigma^2\), and that the population is normally distributed, what percentage of tablets contain more than the nominal amount of morphine hydro- chloride per tablet? The data in this problem are from Salem, I. I.; Galan, A. C. Anal. Chim. Acta 1993, 283, 334–337. 4. Daskalakis and co-workers evaluated several procedures for digesting oyster and mussel tissue prior to analyzing them for silver. To evaluate the procedures they spiked samples with known amounts of silver and analyzed the samples to determine the amount of silver, reporting results as the percentage of added silver found in the analysis. A procedure was judged acceptable if its spike recoveries fell within the range 100±15%. The spike recoveries for one method are shown here. Assuming a normal distribution for the spike recoveries, what is the probability that any single spike recovery is within the accepted range? The data in this problem are from Daskalakis, K. D.; O’Connor, T. P.; Crecelius, E. A. , , 2303– 2306. See to learn more about using a spike recovery to evaluate an analytical method. 5. The formula weight ( ) of a gas can be determined using the following form of the ideal gas law \[FW = \frac {g \text{R} T} {P V} \nonumber\] where is the mass in grams, R is the gas constant, is the temperature in Kelvin, is the pressure in atmospheres, and is the volume in liters. In a typical analysis the following data are obtained (with estimated uncertainties in parentheses) = 0.118 g (± 0.002 g) R = 0.082056 L atm mol K (± 0.000001 L atm mol K ) = 298.2 K (± 0.1 K) = 0.724 atm (± 0.005 atm) = 0.250 L (± 0.005 L) (a) What is the compound’s formula weight and its estimated uncertainty? (b) To which variable(s) should you direct your attention if you wish to improve the uncertainty in the compound’s molecular weight? 6. To prepare a standard solution of Mn , a 0.250 g sample of Mn is dissolved in 10 mL of concentrated HNO (measured with a graduated cylinder). The resulting solution is quantitatively transferred to a 100-mL volumetric flask and diluted to volume with distilled water. A 10-mL aliquot of the solution is pipeted into a 500-mL volumetric flask and diluted to volume. (a) Express the concentration of Mn in mg/L, and estimate its uncertainty using a propagation of uncertainty. (b) Can you improve the concentration’s uncertainty by using a pipet to measure the HNO , instead of a graduated cylinder? 7. The mass of a hygroscopic compound is measured using the technique of weighing by difference. In this technique the compound is placed in a sealed container and weighed. A portion of the compound is removed and the container and the remaining material are reweighed. The difference between the two masses gives the sample’s mass. A solution of a hygroscopic compound with a gram formula weight of 121.34 g/mol (±0.01 g/mol) is prepared in the following manner. A sample of the compound and its container has a mass of 23.5811 g. A portion of the compound is transferred to a 100-mL volumetric flask and diluted to volume. The mass of the compound and container after the transfer is 22.1559 g. Calculate the compound’s molarity and estimate its uncertainty by a propagation of uncertainty. 8. Use a propagation of uncertainty to show that the standard error of the mean for determinations is \(\sigma / \sqrt{n}\). 9. Beginning with and , use a propagation of uncertainty to derive . 10. What is the smallest mass you can measure on an analytical balance that has a tolerance of ±0.1 mg, if the relative error must be less than 0.1%? 11. Which of the following is the best way to dispense 100.0 mL if we wish to minimize the uncertainty: (a) use a 50-mL pipet twice; (b) use a 25-mL pipet four times; or (c) use a 10-mL pipet ten times? 12. You can dilute a solution by a factor of 200 using readily available pipets (1-mL to 100-mL) and volumetric flasks (10-mL to 1000-mL) in either one step, two steps, or three steps. Limiting yourself to the glassware in , determine the proper combination of glassware to accomplish each dilution, and rank them in order of their most probable uncertainties. 13. Explain why changing all values in a data set by a constant amount will change \(\overline{X}\) but has no effect on the standard deviation, . 14. Obtain a sample of a metal, or other material, from your instructor and determine its density by one or both of the following methods: : Determine the sample’s mass with a balance. Calculate the sample’s volume using appropriate linear dimensions. : Determine the sample’s mass with a balance. Calculate the sample’s volume by measuring the amount of water it displaces by adding water to a graduated cylinder, reading the volume, adding the sample, and reading the new volume. The difference in volumes is equal to the sample’s volume. Determine the density at least five times. (a) Report the mean, the standard deviation, and the 95% confidence interval for your results. (b) Find the accepted value for the metal’s density and determine the absolute and relative error for your determination of the metal’s density. (c) Use a propagation of uncertainty to determine the uncertainty for your method of analysis. Is the result of this calculation consistent with your experimental results? If not, suggest some possible reasons for this disagreement. 15. How many carbon atoms must a molecule have if the mean number of C atoms per molecule is at least one? What percentage of such molecules will have no atoms of C? 16. In we determined the probability that a molecule of cholesterol, C H O, had no atoms of C. (a) Calculate the probability that a molecule of cholesterol, has 1 atom of C. (b) What is the probability that a molecule of cholesterol has two or more atoms of C? 17. Berglund and Wichardt investigated the quantitative determination of Cr in high-alloy steels using a potentiometric titration of Cr(VI). Before the titration, samples of the steel were dissolved in acid and the chromium oxidized to Cr(VI) using peroxydisulfate. Shown here are the results ( as %w/w Cr) for the analysis of a reference steel. Calculate the mean, the standard deviation, and the 95% confidence interval about the mean. What does this confidence interval mean? The data in this problem are from Berglund, B.; Wichardt, C. , , 399–410. 18. Ketkar and co-workers developed an analytical method to determine trace levels of atmospheric gases. An analysis of a sample that is 40.0 parts per thousand (ppt) 2-chloroethylsulfide gave the following results (a) Determine whether there is a significant difference between the experimental mean and the expected value at \(\alpha = 0.05\). (b) As part of this study, a reagent blank was analyzed 12 times giving a mean of 0.16 ppt and a standard deviation of 1.20 ppt. What are the IUPAC detection limit, the limit of identification, and limit of quantitation for this method assuming \(\alpha = 0.05\)? The data in this problem are from Ketkar, S. N.; Dulak, J. G.; Dheandhanou, S.; Fite, W. L. , , 267–270. 19. To test a spectrophotometer’s accuracy a solution of 60.06 ppm K Cr O in 5.0 mM H SO is prepared and analyzed. This solution has an expected absorbance of 0.640 at 350.0 nm in a 1.0-cm cell when using 5.0 mM H SO as a reagent blank. Several aliquots of the solution produce the following absorbance values. Determine whether there is a significant difference between the experimental mean and the expected value at \(\alpha = 0.01\). 20. Monna and co-workers used radioactive isotopes to date sediments from lakes and estuaries. To verify this method they analyzed a Po standard known to have an activity of 77.5 decays/min, obtaining the following results. Determine whether there is a significant difference between the mean and the expected value at \(\alpha = 0.05\). The data in this problem are from Monna, F.; Mathieu, D.; Marques, A. N.; Lancelot, J.; Bernat, M. , , 107–116. 21. A 2.6540-g sample of an iron ore, which is 53.51% w/w Fe, is dissolved in a small portion of concentrated HCl and diluted to volume in a 250-mL volumetric flask. A spectrophotometric determination of the concentration of Fe in this solution yields results of 5840, 5770, 5650, and 5660 ppm. Determine whether there is a significant difference between the experimental mean and the expected value at \(\alpha = 0.05\). 22. Horvat and co-workers used atomic absorption spectroscopy to determine the concentration of Hg in coal fly ash. Of particular interest to the authors was developing an appropriate procedure for digesting samples and releasing the Hg for analysis. As part of their study they tested several reagents for digesting samples. Their results using HNO and using a 1 + 3 mixture of HNO and HCl are shown here. All concentrations are given as ppb Hg sample. Determine whether there is a significant difference between these methods at \(\alpha = 0.05\). The data in this problem are from Horvat, M.; Lupsina, V.; Pihlar, B. , , 71–79. 23, Lord Rayleigh, John William Strutt (1842-1919), was one of the most well known scientists of the late nineteenth and early twentieth centuries, publishing over 440 papers and receiving the Nobel Prize in 1904 for the discovery of argon. An important turning point in Rayleigh’s discovery of Ar was his experimental measurements of the density of N . Rayleigh approached this experiment in two ways: first by taking atmospheric air and removing O and H ; and second, by chemically producing N by decomposing nitrogen containing compounds (NO, N O, and NH NO ) and again removing O and H . The following table shows his results for the density of N , as published in , , 340 (publication 210); all values are the grams of gas at an equivalent volume, pressure, and temperature. Explain why this data led Rayleigh to look for and to discover Ar. You can read more about this discovery here: Larsen, R. D. , , 925–928. 24. Gács and Ferraroli reported a method for monitoring the concentration of SO in air. They compared their method to the standard method by analyzing urban air samples collected from a single location. Samples were collected by drawing air through a collection solution for 6 min. Shown here is a summary of their results with SO concentrations reported in \(\mu \text{L/m}^3\). Using an appropriate statistical test, determine whether there is any significant difference between the standard method and the new method at \(\alpha = 0.05\). The data in this problem are from Gács, I.; Ferraroli, R. , , 177–185. 25. One way to check the accuracy of a spectrophotometer is to measure absorbances for a series of standard dichromate solutions obtained from the National Institute of Standards and Technology. Absorbances are measured at 257 nm and compared to the accepted values. The results obtained when testing a newly purchased spectrophotometer are shown here. Determine if the tested spectrophotometer is accurate at \(\alpha = 0.05\). 26. Maskarinec and co-workers investigated the stability of volatile organics in environmental water samples. Of particular interest was establishing the proper conditions to maintain the sample’s integrity between its collection and its analysis. Two preservatives were investigated—ascorbic acid and sodium bisulfate—and maximum holding times were determined for a number of volatile organics and water matrices. The following table shows results for the holding time (in days) of nine organic compounds in surface water. Determine whether there is a significant difference in the effectiveness of the two preservatives at \(\alpha = 0.10\). The data in this problem are from Maxkarinec, M. P.; Johnson, L. H.; Holladay, S. K.; Moody, R. L.; Bayne, C. K.; Jenkins, R. A. , , 1665–1670. 27. Using X-ray diffraction, Karstang and Kvalhein reported a new method to determine the weight percent of kaolinite in complex clay minerals using X-ray diffraction. To test the method, nine samples containing known amounts of kaolinite were prepared and analyzed. The results (as % w/w kaolinite) are shown here. Evaluate the accuracy of the method at \(\alpha = 0.05\). The data in this problem are from Karstang, T. V.; Kvalhein, O. M. , , 767–772. 28. Mizutani, Yabuki and Asai developed an electrochemical method for analyzing -malate. As part of their study they analyzed a series of beverages using both their method and a standard spectrophotometric procedure based on a clinical kit purchased from Boerhinger Scientific. The following table summarizes their results. All values are in ppm. The data in this problem are from Mizutani, F.; Yabuki, S.; Asai, M. , ,145–150. 29. Alexiev and colleagues describe an improved photometric method for determining Fe based on its ability to catalyze the oxidation of sulphanilic acid by KIO . As part of their study, the concentration of Fe in human serum samples was determined by the improved method and the standard method. The results, with concentrations in \(\mu \text{mol/L}\), are shown in the following table. Determine whether there is a significant difference between the two methods at \(\alpha = 0.05\). The data in this problem are from Alexiev, A.; Rubino, S.; Deyanova, M.; Stoyanova, A.; Sicilia, D.; Perez Bendito, D. , , , 211–219. 30. Ten laboratories were asked to determine an analyte’s concentration of in three standard test samples. Following are the results, in \(\mu \text{g/ml}\). Determine if there are any potential outliers in Sample 1, Sample 2 or Sample 3. Use all three methods—Dixon’s -test, Grubb’s test, and Chauvenet’s criterion—and compare the results to each other. For Dixon’s -test and for the Grubb’s test, use a significance level of \(\alpha = 0.05\). The data in this problem are adapted from Steiner, E. H. “Planning and Analysis of Results of Collaborative Tests,” in , Association of Official Analytical Chemists: Washington, D. C., 1975. 31.When copper metal and powdered sulfur are placed in a crucible and ignited, the product is a sulfide with an empirical formula of Cu S. The value of is determined by weighing the Cu and the S before ignition and finding the mass of Cu S when the reaction is complete (any excess sulfur leaves as SO ). The following table shows the Cu/S ratios from 62 such experiments (note that the values are organized from smallest-to-largest by rows). (a) Calculate the mean, the median, and the standard deviation for this data. (b) Construct a histogram for this data. From a visual inspection of your histogram, do the data appear normally distributed? (c) In a normally distributed population 68.26% of all members lie within the range \(\mu \pm 1 \sigma\). What percentage of the data lies within the range \(\overline{X} \pm 1 \sigma\)? Does this support your answer to the previous question? (d) Assuming that \(\overline{X}\) and \(s^2\) are good approximations for \(\mu\) and for \(\sigma^2\), what percentage of all experimentally determined Cu/S ratios should be greater than 2? How does this compare with the experimental data? Does this support your conclusion about whether the data is normally distributed? (e) It has been reported that this method of preparing copper sulfide results in a non-stoichiometric compound with a Cu/S ratio of less than 2. Determine if the mean value for this data is significantly less than 2 at a significance level of \(\alpha = 0.01\). See Blanchnik, R.; Müller, A. “The Formation of Cu S From the Elements I. Copper Used in Form of Powders,” , , , 31-52 for a discussion of some of the factors affecting the formation of non-stoichiometric copper sulfide. The data in this problem were collected by students at DePauw University. 32. Real-time quantitative PCR is an analytical method for determining trace amounts of DNA. During the analysis, each cycle doubles the amount of DNA. A probe species that fluoresces in the presence of DNA is added to the reaction mixture and the increase in fluorescence is monitored during the cycling. The cycle threshold, \(C_t\), is the cycle when the fluorescence exceeds a threshold value. The data in the following table shows \(C_t\) values for three samples using real-time quantitative PCR. Each sample was analyzed 18 times. Examine this data and write a brief report on your conclusions. Issues you may wish to address include the presence of outliers in the samples, a summary of the descriptive statistics for each sample, and any evidence for a difference between the samples. The data in this problem is from Burns, M. J.; Nixon, G. J.; Foy, C. A.; Harris, N. , ( ).

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There are a number of chemical reactions that require ammonia. In order to carry out a reaction efficiently, we need to know how much ammonia we have for stoichiometric purposes. Using gas laws, we can determine the number of moles present in the tank if we know the volume, temperature, and pressure of the system. The combined gas law shows that the pressure of a gas is inversely proportional to volume and directly proportional to temperature. Avogadro's Law shows that volume or pressure is directly proportional to the number of moles of gas. Putting these laws together gives us the following equation: \[\frac{P_1 \times V_1}{T_1 \times n_1} = \frac{P_2 \times V_2}{T_2 \times n_2}\nonumber \] As with the other gas laws, we can also say that \(\frac{\left( P \times V \right)}{\left( T \times n \right)}\) is equal to a constant. The constant can be evaluated provided that the gas being described is considered to be ideal. The is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable \(R\) for the constant, the equation becomes: \[\frac{P \times V}{T \times n} = R\nonumber \] The ideal gas law is conveniently rearranged to look this way, with the multiplication signs omitted: \[PV = nRT\nonumber \] The variable \(R\) in the equation is called the . The value of \(R\), the ideal gas constant, depends on the units chosen for pressure, temperature, and volume in the ideal gas equation. It is necessary to use Kelvin for the temperature and it is conventional to use the unit of liters for the volume. However, pressure is commonly measured in one of three units: \(\text{kPa}\), \(\text{atm}\), or \(\text{mm} \: \ce{Hg}\). Therefore, \(R\) can have three different values. We will demonstrate how \(R\) is calculated when the pressure is measured in \(\text{kPa}\). Recall that the volume of \(1.00 \: \text{mol}\) of any gas at is measured to be \(22.414 \: \text{L}\). We can substitute \(101.325 \: \text{kPa}\) for pressure, \(22.414 \: \text{L}\) for volume, and \(273.15 \: \text{K}\) for temperature into the ideal gas equation and solve for \(R\). \[R = \frac{PV}{nT} = \frac{101.325 \: \text{kPa} \times 22.414 \: \text{L}}{1.000 \: \text{mol} \times 273.15 \: \text{K}} = 8.314 \: \text{kPa} \cdot \text{L/K} \cdot \text{mol}\nonumber \] This is the value of \(R\) that is to be used in the ideal gas equation when the pressure is given in \(\text{kPa}\). The table below shows a summary of this and the other possible values of \(R\). It is important to choose the correct value of \(R\) to use for a given problem. Notice that the unit for \(R\) when the pressure is in \(\text{kPa}\) has been changed to \(\text{J/K} \cdot \text{mol}\). A kilopascal multiplied by a liter is equal to the SI unit for energy, a joule \(\left( \text{J} \right)\). What volume is occupied by \(3.760 \: \text{g}\) of oxygen gas at a pressure of \(88.4 \: \text{kPa}\) and a temperature of \(19^\text{o} \text{C}\)? Assume the oxygen is an ideal gas. In order to use the ideal gas law, the number of moles of \(\ce{O_2}\) \(\left( n \right)\) must be found from the given mass and the molar mass. Then, use \(PV = nRT\) to solve for the volume of oxygen. \[3.760 \: \text{g} \times \frac{1 \: \text{mol} \: \ce{O_2}}{32.00 \: \text{g} \: \ce{O_2}} = 0.1175 \: \text{mol} \: \ce{O_2}\nonumber \] Rearrange the ideal gas law and solve for \(V\). \[V = \frac{nRT}{P} = \frac{0.1175 \: \text{mol} \times 8.314 \: \text{J/K} \cdot \text{mol} \times 292 \: \text{K}}{88.4 \: \text{kPa}} = 3.23 \: \text{L} \: \ce{O_2}\nonumber \] The number of moles of oxygen is far less than one mole, so the volume should be fairly small compared to molar volume \(\left( 22.4 \: \text{L/mol} \right)\) since the pressure and temperature are reasonably close to standard. The result has three significant figures because of the values for \(T\) and \(P\). Since a joule \(\left( \text{J} \right) = \text{kPa} \cdot \text{L}\), the units cancel out correctly, leaving a volume in liters.

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Even before the atomic bomb had been produced, scientists and engineers had begun to think about the possibility of using the energy released by the fission process for the production of electrical energy. Shortly after World War II confident predictions were made that human beings would soon depend almost entirely on atomic energy for electricity. Alas, we are now years into the future from then and no such miracle has occurred. In the United States only 8.46 percent of the electrical energy in 2008 was produced by this method . The proportion is a little higher in some other countries, notably Great Britain, but nowhere is nuclear power even on the verge of replacing the fossil fuels. The unfortunate truth is that producing power from atomic fission has turned out to be much more expensive than was previously expected. Even in these days of high prices for the fossil fuels it is still only barely competitive. A schematic diagram of a typical nuclear reactor is given in Figure \(\Page {1}\). The uranium is present in the form of pellets of the oxide U O enclosed in long steel tubes about 2 cm in diameter. The uranium is mainly, slightly enriched with the fissionable . The rate of fission can be regulated by inserting or withdrawing control rods made of cadmium, which is a very efficient neutron absorber. In addition a moderator such as graphite or water must be present to slow down the neutrons, since slow neutrons are more efficient at causing fission than fast ones. The energy released by the fission of the uranium is carried off by a coolant, usually superheated steam at about 320°C. This steam cannot be used directly since it becomes slightly radioactive. Instead it is passed through a heat exchanger so as to produce further steam which can then be used to power a conventional steam turbine. The whole system is enclosed in a strong containment vessel which is not shown in the figure. This vessel prevents the spread of radioactivity in case of a serious accident. Nuclear power plants have two advantages over conventional power plants using fossil fuels. First, for a given energy output they consume much less fuel. Second, they produce far smaller quantities of toxic effluents. Fossil-fueled plants produce sulfur dioxide, oxides of nitrogen, and smoke particles, all of which are injurious to health. Despite the much lower cost for fuel, nuclear power plants are very expensive to build. This is largely because of their chief disadvantage, the extremely dangerous nature of the radioactive products of nuclear fission. Fission products consist of a great many neutron-rich, unstable nuclei, ranging in atomic number from 25 to 60. Particularly dangerous are the long-lived isotopes , , and the shorter-lived , all of which can be incorporated into the human body. Extreme precautions must be taken against accidental release of even traces of these materials into the environment. The worst case scenario is a reactor meltdown, of which the best known is the Chernobyl disaster, occurring in 1986 in Ukraine, then still a part of the Soviet Union. Recent study on the event estimates eventual deaths caused by the accident in the higher exposed populations as 4000 . The overall cost of the disaster is of course difficult to quantify in terms of health, psychological, economic and environmental effects. It should be realized that there are significant differences between a Hiroshima-type bomb and this sort of meltdown, primarily a difference in destructive explosive power. These differences arises because fuel used in nuclear reactors is not rich enough in for the chain reaction to produce an atomic bomb-like explosion. It should also be noted that the safety deficiencies which caused the Chernobyl did not reflect the safety precautions taken in in nuclear plants today, or in other plants at the time. Regardless, the Chernobyl event, along with the earlier Three Mile Island incident in 1979, helped to start a decline in the use and building of nuclear power plants worldwide, and many people remain wary of nuclear power. Interest in nuclear power has renewed with discussions of national and world energy strategies, however. Even if fission products are handled successfully during normal operation of a nuclear plant, there still remains the difficulty of their eventual disposal. Although many of the unstable nuclei produced by fission are short-lived, some, like (25 years) and (30 years), have quite long half-lives. Accordingly these wastes must be stored for many hundreds of years before enough nuclei decompose to reduce their radioactivity to a safe level. At the present time, spent nuclear fuel is stored in spent fuel water pools at reaction sites, which shields the surrounding environment. Spent fuel may also be placed in dry cask storage after fuel has been stored in a spent fuel pool for at least a year. The dry casks contain inert gas, are made of steel, and may be surrounded by steel and concrete to prevent leakage of radiation . Both spent fuel pools and dry cask storage are only short term solutions. One proposed long term solution is to store spent nuclear waste safely in a geological repository deep underground. In 1982, the Nuclear Waste Policy Act tasked the United States Department of Energy with finding and constructing a national repository for long term storage of nuclear waste. In 1987, the Department of Energy was directed to focus solely on Yucca Mountain, Nevada, as the site to develop a national repository. Originally set to start accepting waste in 1998, the opening of the repository has been greatly delayed, often due to a large amount of unresolved debate on the topic.. Nevertheless, in 2002, the Department of Energy determined the site acceptable, and began application to use the site . The Yucca Mountain Repository is currently out of favor. With shrinking budgets along with safety concerns about the site, the Department of Energy motioned to withdraw the Yucca Mountain site as a long term repository for nuclear waste in 2010 . The debacle over Yucca Mountain highlights both the scientific and political difficulties in finding a long term solution to spent nuclear fuels, as well as the difficulties involved in using nuclear power in general.

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. In addition to these individual basis; please contact What information can you obtain by studying the chemical kinetics of a reaction? Does a balanced chemical equation provide the same information? Why or why not? Kinetics gives information on the reaction rate and reaction mechanism; the balanced chemical equation gives only the stoichiometry of the reaction. If you were tasked with determining whether to proceed with a particular reaction in an industrial facility, why would studying the chemical kinetics of the reaction be important to you? Studying chemical kinetics determines whether to proceed with a reaction as it measures the rate of a reaction. Reactions conducted in an industrial facility mix compounds together, heating and stirring them for a while, before moving to the next process. When the compounds are then moved to the next phase of the process it is important to know how long to hold the reaction at one stage before continuing, to make sure a reaction has finished before starting the next one What is the relationship between each of the following factors and the reaction rate: reactant concentration, temperature of the reaction, physical properties of the reactants, physical and chemical properties of the solvent, and the presence of a catalyst? Reaction rate and the concentration of reactants have a direct relationship; as concentration goes up, so does reaction rate. This is because a higher concentration causes the reactants to have a higher probability to collide with a another reactant particle, possibly inducing the chemical reaction. This comes from the Collision theory. Reaction temperature: Reaction rate and the reaction temperature also have a direct relationship; as temperature goes up, so does reaction rate. A higher temperature makes the particles move at a faster speed so when two or more reactant particles collide, they collide with more energy and are more likely to reach the activation energy threshold, starting a chemical reaction. Physical properties of the reactants: If the reactants have the same physical properties, they are more likely to react, increasing the reaction rate, because the reactants mix and are more likely to collide another one of the reactants. Physical and chemical properties of the solvent: The properties of the solvent also affect the rates of a reaction. High viscosity means that particles more slowly than in low viscosity solvents, so reaction rates are slower in high viscosity solvents than in low viscosity solvents. This indicates an inverse relationship between viscosity and reaction rate. Presence of a catalyst: The chemical definition of a catalyst in a substance that increases the rate of a reaction without being used up in the reaction. Therefore, if a catalyst is present, reaction rate increases. A slurry is a mixture of a finely divided solid with a liquid in which it is only sparingly soluble. As you prepare a reaction, you notice that one of your reactants forms a slurry with the solvent, rather than a solution. What effect will this have on the reaction rate? What steps can you take to try to solve the problem? Why does the reaction rate of virtually all reactions increase with an increase in temperature? If you were to make a glass of sweetened iced tea the old-fashioned way, by adding sugar and ice cubes to a glass of hot tea, which would you add first? Increasing the temperature increases the average kinetic energy of molecules and ions, causing them to collide more frequently and with greater energy, which increases the reaction rate. First dissolve sugar in the hot tea, and then add the ice. In a typical laboratory setting, a reaction is carried out in a ventilated hood with air circulation provided by outside air. A student noticed that a reaction that gave a high yield of a product in the winter gave a low yield of that same product in the summer, even though his technique did not change and the reagents and concentrations used were identical. What is a plausible explanation for the different yields? A very active area of chemical research involves the development of solubilized catalysts that are not made inactive during the reaction process. Such catalysts are expected to increase reaction rates significantly relative to the same reaction run in the presence of a heterogeneous catalyst. What is the reason for anticipating that the relative rate will increase? Water has a dielectric constant more than two times greater than that of methanol (80.1 for H O and 33.0 for CH OH). Which would be your solvent of choice for a substitution reaction between an ionic compound and a polar reagent, both of which are soluble in either methanol or water? Why? Explain why the reaction rate is generally fastest at early time intervals. For the second-order A + B → C, what would the plot of the concentration of C versus time look like during the course of the reaction? Reactant concentrations are highest at the beginning of a reaction. The plot of [C] versus t is a curve with a slope that becomes steadily less positive. Explain the differences between a differential rate law and an integrated rate law. What two components do they have in common? Which form is preferred for obtaining a reaction order and a rate constant? Why? Diffusion-controlled reactions have rates that are determined only by the reaction rate at which two reactant molecules can diffuse together. These reactions are rapid, with second-order rate constants typically on the order of 10 L/(mol·s). Are the reactions expected to be faster or slower in solvents that have a low viscosity? Why? Consider the reactions H O + OH → 2H O and H O + N(CH ) → H O + HN(CH ) in aqueous solution. Which would have the higher rate constant? Why? Faster in a less viscous solvent because the rate of diffusion is higher; the H O /OH reaction is faster due to the decreased relative size of reactants and the higher electrostatic attraction between the reactants. What information can be obtained from the reaction order? What correlation does the reaction order have with the stoichiometry of the overall equation? During the hydrolysis reaction A + H O → B + C, the concentration of A decreases much more rapidly in a polar solvent than in a nonpolar solvent. How would this effect be reflected in the overall reaction order? The reaction rate of a particular reaction in which A and B react to make C is as follows: \(\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=\dfrac{1}{2}\left ( \dfrac{\Delta[\textrm C]}{\Delta t} \right )\) Write a reaction equation that is consistent with this rate law. What is the rate expression with respect to time if 2A are converted to 3C? While commuting to work, a person drove for 12 min at 35 mph, then stopped at an intersection for 2 min, continued the commute at 50 mph for 28 min, drove slowly through traffic at 38 mph for 18 min, and then spent 1 min pulling into a parking space at 3 mph. What was the average rate of the commute? What was the instantaneous rate at 13 min? at 28 min? Why do most studies of chemical reactions use the initial rates of reaction to generate a rate law? How is this initial rate determined? Given the following data, what is the reaction order? Estimate. Predict how the reaction rate will be affected by doubling the concentration of the first species in each equation. Cleavage of C H to produce two CH · radicals is a gas-phase reaction that occurs at 700°C. This reaction is first order, with = 5.46 × 10 s . How long will it take for the reaction to go to 15% completion? to 50% completion? 298 s; 1270 s Three chemical processes occur at an altitude of approximately 100 km in Earth’s atmosphere. \[\mathrm{N_2^+}+\mathrm{O_2}\xrightarrow{k_1}\mathrm{N_2}+\mathrm{O_2^+}\] \[\mathrm{O_2^+}+\mathrm{O}\xrightarrow{k_2}\mathrm{O_2}+\mathrm{O^+}\] \[(\mathrm{O^+}+\mathrm{N_2}\xrightarrow{k_3}\mathrm{NO^+}+\mathrm{N}\] Write a rate law for each elementary reaction. If the rate law for the overall reaction were found to be rate = [N ,O ], which one of the steps is rate limiting? The oxidation of aqueous iodide by arsenic acid to give I and arsenous acid proceeds via the following reaction: Write an expression for the initial rate of decrease of [I ], Δ[I ]/Δt. When the reaction rate of the forward reaction is equal to that of the reverse reaction: k /k = [H AsO ,I ]/[H AsO ,I ] [H ] . Based on this information, what can you say about the nature of the rate-determining steps for the reverse and the forward reactions? What are the characteristics of a zeroth-order reaction? Experimentally, how would you determine whether a reaction is zeroth order? Predict whether the following reactions are zeroth order and explain your reasoning. In a first-order reaction, what is the advantage of using the integrated rate law expressed in natural logarithms over the rate law expressed in exponential form? If the reaction rate is directly proportional to the concentration of a reactant, what does this tell you about (a) the reaction order with respect to the reactant and (b) the overall reaction order? The reaction of NO with O is found to be second order with respect to NO and first order with respect to O . What is the overall reaction order? What is the effect of doubling the concentration of each reagent on the reaction rate? Iodide reduces Fe(III) according to the following reaction: \[2Fe^{3+}(soln) + 2I^−(soln) → 2Fe^{2+}(soln) + I_2(soln)\] Experimentally, it was found that doubling the concentration of Fe(III) doubled the reaction rate, and doubling the iodide concentration increased the reaction rate by a factor of 4. What is the reaction order with respect to each species? What is the overall rate law? What is the overall reaction order? First order in Fe ; second order in I ; third order overall; rate = [Fe ,I ] . Benzoyl peroxide is a medication used to treat acne. Its rate of thermal decomposition at several concentrations was determined experimentally, and the data were tabulated as follows: What is the reaction order with respect to benzoyl peroxide? What is the rate law for this reaction? The general rate law is: rate = k [Benzoyl Peroxide] . In order to find the reaction order with respect to benzoyl peroxide, divide two rate laws and solve for m: \(\dfrac{rate_2}{rate_1}=\dfrac{k_2}{k_1} \left ( \dfrac{[Benzoyl Peroxide]_2}{[Benzoyl Peroxide]_1} \right )^m\) \(\dfrac{1.64 × 10^-4 \dfrac{M}{s}}{2.22 × 10^-4 \dfrac{M}{s}}=( \dfrac{0.070 M}{1.00 M})^m\) \(0.738=(0.7)^m\) \(m=0.85\) rate law: rate = k [Benzoyl Peroxide] 1-Bromopropane is a colorless liquid that reacts with S O according to the following reaction: \[C_3H_7Br + S_2O_3^{2−} → C_3H_7S_2O_3^− + Br^−\] The reaction is first order in 1-bromopropane and first order in S O , with a rate constant of 8.05 × 10 M ·s . If you began a reaction with 40 mmol/100 mL of C H Br and an equivalent concentration of S O , what would the initial reaction rate be? If you were to decrease the concentration of each reactant to 20 mmol/100 mL, what would the initial reaction rate be? 1.29 × 10 M/s; 3.22 × 10 M/s The experimental rate law for the reaction 3A + 2B → C + D was found to be Δ[C]/Δ = [A] [B] for an overall reaction that is third order. Because graphical analysis is difficult beyond second-order reactions, explain the procedure for determining the rate law experimentally. Although an increase in temperature results in an increase in kinetic energy, this increase in kinetic energy is not sufficient to explain the relationship between temperature and reaction rates. How does the activation energy relate to the chemical kinetics of a reaction? Why does an increase in temperature increase the reaction rate despite the fact that the average kinetic energy is still less than the activation energy? Activation energy is required for a collision between molecules to result in a chemical reaction, as well it is related to the rate due to the Arrhenius equation Activation energy is the threshold of energy needed in order for a reaction to occur. Reactant particles must collide with enough energy to be able to break chemical bonds, that will then allow the creation of new bonds. If the particles do not have enough energy, when they collide they will simply bounce off one another. k=Ae . The increase in temperature increases the rate of reaction despite the fact the average kinetic is less than the activation energy since it increase the rate of collision between molecules. With an increase in temperature, there is a greater distribution of kinetic energy among reactant particles. This increase of temperature allows the rate in which particles collide with one another to increase. Although the average kinetic energy is still lower than the activation energy, the increase of collisions among particles increases the chance of particles that contain enough energy to overcome the energy barrier to collide. Thus, the reaction rate increases due to this increased rate of collisions. Additionally, there is a higher amount of molecules that have sufficient kinetic energy to overcome the energy barrier, despite the average energy of all these molecules still being lower than the activation energy. For any given reaction, what is the relationship between the activation energy and each of the following? 1.) Electrostatic repulsion: Electrostatic repulsion is the unfavorable interaction between two species of like charge. Activation energy is the minimum amount of energy needed for a reaction to occur. Reacting molecules must have enough energy to overcome electrostatic repulsion, and a minimum amount of energy is required to break chemical bonds so that new ones may be formed. 2.) Bond formation in the activated complex: An activated complex is an intermediate state that is formed during the conversion of reactants into products. Bond breaking can increase activation energy as breaking bonds requires energy. 3.) Nature of the activated complex: The activation energy of a chemical reaction is the difference between the energy of the activated complex and the energy of the reactants. If the structure has a high steric hindrance the activation energy will be higher. If you are concerned with whether a reaction will occur rapidly, why would you be more interested in knowing the magnitude of the activation energy than the change in potential energy for the reaction? The product C in the reaction A + B → C + D can be separated easily from the reaction mixture. You have been given pure A and pure B and are told to determine the activation energy for this reaction to determine whether the reaction is suitable for the industrial synthesis of C. How would you do this? Why do you need to know the magnitude of the activation energy to make a decision about feasibility? Above , molecules collide with enough energy to overcome the energy barrier for a reaction. Is it possible for a reaction to occur at a temperature less than that needed to reach ? Explain your answer. What is the relationship between , , and ? How does an increase in affect the reaction rate? Of two highly exothermic reactions with different values of , which would need to be monitored more carefully: the one with the smaller value or the one with the higher value? Why? What happens to the approximate rate of a reaction when the temperature of the reaction is increased from 20°C to 30°C? What happens to the reaction rate when the temperature is raised to 70°C? For a given reaction at room temperature (20°C), what is the shape of a plot of reaction rate versus temperature as the temperature is increased to 70°C? The reaction rate will approximately double: 20°C to 30°C, the reaction rate increases by about 2 = 2; 20°C to 70°C, the reaction rate increases by about 2 = 32-fold. A plot of reaction rate versus temperature will give an exponential increase: rate ∝ 2 . Acetaldehyde, used in silvering mirrors and some perfumes, undergoes a second-order decomposition between 700 and 840 K. From the data in the following table, would you say that acetaldehyde follows the general rule that each 10 K increase in temperature doubles the reaction rate? Bromoethane reacts with hydroxide ion in water to produce ethanol. The activation energy for this reaction is 90 kJ/mol. If the reaction rate is 3.6 × 10 M/s at 25°C, what would the reaction rate be at the following temperatures? An enzyme-catalyzed reaction has an activation energy of 15 kcal/mol. How would the value of the rate constant differ between 20°C and 30°C? If the enzyme reduced the \(E_a\) from 25 kcal/mol to 15 kcal/mol, by what factor has the enzyme increased the reaction rate at each temperature? The data in the following table are the rate constants as a function of temperature for the dimerization of 1,3-butadiene. What is the activation energy for this reaction? 100 kJ/mol The reaction rate at 25°C is 1.0 × 10 M/s. Increasing the temperature to 75°C causes the reaction rate to increase to 7.0 × 10 M/s. Estimate \(E_a\) for this process. If \(E_a\) were 25 kJ/mol and the reaction rate at 25°C is 1.0 × 10 M/s, what would be the reaction rate at 75°C? How does the term molecularity relate to elementary reactions? How does it relate to the overall balanced chemical equation? What is the relationship between the reaction order and the molecularity of a reaction? What is the relationship between the reaction order and the balanced chemical equation? When you determine the rate law for a given reaction, why is it valid to assume that the concentration of an intermediate does not change with time during the course of the reaction? If you know the rate law for an overall reaction, how would you determine which elementary reaction is rate determining? If an intermediate is contained in the rate-determining step, how can the experimentally determined rate law for the reaction be derived from this step? Give the rate-determining step for each case. Before being sent on an assignment, an aging James Bond was sent off to a health farm where part of the program’s focus was to purge his body of radicals. Why was this goal considered important to his health? Free radicals are uncharged molecules with an unpaired valence electron. The reason these are so dangerous is because they like to grab electrons from other atoms to fill their own outer shell. This allows them to impair protein function because free radicals readily oxidize proteins and cell membrane which could lead to a loss of function. It was important to purge James Bond of radicals because radicals set off chain reactions of continuously pulling electrons from molecules, which in turn, can damage cells in the body. Cyclopropane, a mild anesthetic, rearranges to propylene via a collision that produces and destroys an energized species. The important steps in this rearrangement are as follows: cyclopropane cyclopropane Above approximately 500 K, the reaction between NO and CO to produce CO and NO follows the second-order rate law Δ[CO ]/Δ = [NO ,CO]. At lower temperatures, however, the rate law is Δ[CO ]/Δ = ′[NO ] , for which it is known that NO is an intermediate in the mechanism. Propose a complete low-temperature mechanism for the reaction based on this rate law. Which step is the slowest? Given that NO is an intermediate in the low temperature reaction mechanism, we automatically know two things: 1) , and thus, 2) . We can also assume that since we're given an intermediate from the problem, this is the only intermediate (so we won't have to dream up any other compounds that might exist in the series of reactions.) These things being said, the will look like this: \[2NO_2(g) \rightarrow NO_3(g) + NO(g) \tag{1}\] \[CO(g) + NO_3(g) \rightarrow CO_2(g) + NO_2(g) \tag{2}\] And the overall reaction will look like this (notice how NO is not present): \[CO(g) + NO_2(g) \rightarrow CO_2(g) + NO(g) \tag{overall reaction}\] Now that we have the reaction mechanism written out, we can go about determining which step is the slowest. It would be pretty tricky to do this if we weren't given any further information, however, we know two more things: 1) (given that the overall reaction is also equal to this rate) and 2) These two things being said, we've both confirmed that our proposed low-temperature reaction mechanism is in fact two steps, and that . By using the "guess-and-check" method we can label each step reaction one at a time as the and see if the rate matches up with the rate given to us. (see above) By using we can determine that the rate of the reaction must be in terms of its reactants, which follows: \[\textrm{k}\textrm{[CO,NO}_3\textrm{]}\] ... We can't have the overall reaction rate in terms of an intermediate. By looking at the 1st reaction, we can determine that we can sub in "[NO ] /[NO]" for "[NO ]" since by writing the full reaction rate of the first step and solving for [NO ] this is equivalent. So, we now have the overall rate of the mechanism as the following given the 2nd reaction is the "slow reaction:" \[\textrm{k}\dfrac{\textrm{[CO,NO}_2\textrm{]}^2}{\textrm{[NO}\textrm{]}}\] \[\textrm{k}\textrm{[NO}_2\textrm{]}^2\] What do you know... the rates are equal! Nitramide (O NNH ) decomposes in aqueous solution to N O and H O. What is the experimental rate law (Δ[N O]/Δ ) for the decomposition of nitramide if the mechanism for the decomposition is as follows? Assume that the rates of the forward and reverse reactions in the first equation are equal. \[rate_1 = {k_1}{[O_2NNH_2]}\] \[rate_{-1} = {k_{-1}}{[O_2NNH^-]}{[H^+]}\] Since these two rates produce and consume the same amount of \(O_2NNH^{-}\) over the same time period, we can set them equal to each other and solve for the intermediate \[rate_1 = rate_{-1}\] \[{k_1}{[O_2NNH_2]} = {k_{-1}}{[O_2NNH^-]}{[H^+]}\] \[{[O_2NNH^-]} = \frac{k_1{[O_2NNH_2]}}{k_{-1}{[H^+]}}\] Substituting this equation back into our original equation gives us \[rate=rate_2=k_2\frac{k_1[O_2NNH_2]}{k_{-1}[H^+]}\] With all of the rate constants (k), we can clean up our equation a little bit by saying \[k = \frac{k_2{k_1}}{k_{-1}}\] Leaving us with \[rate = \frac{k[O_2NNH_2]}{[H^+]}\] \(\textrm{rate}=k_2\dfrac{k_1[\mathrm{O_2NNH_2}]}{k_{-1}[\mathrm{H^+}]}=k\dfrac{[\mathrm{O_2NNH_2}]}{[\mathrm{H^+}]}\) The following reactions are given: \[\mathrm{A+B}\overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}}\mathrm{C+D}\] \[\mathrm{D+E}\xrightarrow{k_2}\mathrm F\] What is the relationship between the relative magnitudes of \(k_{−1}\) and \(k_2\) if these reactions have the following rate law? \[\dfrac{Δ[F]}{Δt} = k\dfrac{[A,B,E]}{[C]}\] How does the magnitude of \(k_1\) compare to that of \(k_2\)? Under what conditions would you expect the rate law to be \[\dfrac{Δ[F]}{Δt} =k′[A,B]?\] Assume that the rates of the forward and reverse reactions in the first equation are equal. First, because we have broken the equations down into we can write the rate laws for each step. Step1: \[A+B\xrightarrow[]{k_{1}} C+D\] \[rate=k_{1}[A,B]\] Step 2: \[C+D \xrightarrow[]{k_{-1}} A+B\] \[rate=k_{-1}[C,D]\] Step 3: \[D+E \xrightarrow[]{k_{2}} F\] \[rate=k_{2}[D,E]\] If we add a these steps together we see that we get overall reaction \[A+B+E \rightarrow C+F\] we can see that [D] is an intermediate and \[k_{1}=k_{-1}\] Since we are not told which steps are fast or slow we need to use Steady State Approximation. If the second step is the slower step ( >> then our rate determining step would be \[rate=k_{2}[D,E]\] Since we can only write rate laws in terms of products and reactants we have to rewrite this so that we are not including an . Assume: rate of [D] formation = rate of its disappearance \[k_{1}[A,B]=k_{-1}[C,D]+k_{2}[D,E]\] \[k_{1}[A,B]=[D](k_{-1}[C]+k_{2}[E])\] Solving for [D] we find that \[[D]= \frac{k_{1}[A,B]}{(k_{-1}[C]+k_{2}[E])}\] now we can use this to substitute the intermediate [D] in the rate law to get an appropriate rate law. \[rate=\frac{k_{2} k_{1}[A,B,E]}{(k_{-1}[C]+k_{2}[E])}\] because we had already established >> \[k_{-1}[C]+k_{2}[E]\approx k_{-1}[C]\] this would give us the observed rate law \[\frac{\Delta [F]}{\Delta t}=\frac{k_{2}k_{1}[A,B,E]}{k_{-1}[C]}\] to make this clearer we can set \[k=\frac{(k_{2})(k_{1})}{(k_{-1})}\] and we can then simplify it down to \[\dfrac{Δ[F]}{Δt} = k\dfrac{[A,B,E]}{[C]}\] we can see that all of these rate constants are related by this ratio / Since is our rate determining step >> = then we can see that >> We would expect the rate law to be \[\dfrac{Δ[F]}{Δt} =k′[A,B]?\] if the aka the slowest step is that corresponding to (step 1) since the rate law for this step is [A,B] and this is the exact the same as the rate law that we were given. What effect does a catalyst have on the activation energy of a reaction? What effect does it have on the frequency factor ( )? What effect does it have on the change in potential energy for the reaction? A catalyst lowers the activation energy of a reaction. Some catalysts can also orient the reactants and thereby increase the frequency factor. Catalysts have no effect on the change in potential energy for a reaction. How is it possible to affect the product distribution of a reaction by using a catalyst? A heterogeneous catalyst works by interacting with a reactant in a process called . What occurs during this process? Explain how this can lower the activation energy. In adsorption, a reactant binds tightly to a surface. Because intermolecular interactions between the surface and the reactant weaken or break bonds in the reactant, its reactivity is increased, and the activation energy for a reaction is often decreased. What effect does increasing the surface area of a heterogeneous catalyst have on a reaction? Does increasing the surface area affect the activation energy? Explain your answer. Identify the differences between a heterogeneous catalyst and a homogeneous catalyst in terms of the following. An area of intensive chemical research involves the development of homogeneous catalysts, even though homogeneous catalysts generally have a number of operational difficulties. Propose one or two reasons why a homogenous catalyst may be preferred. Catalysts are compounds that, when added to chemical reactions, reduce the activation energy and increase the reaction rate. The amount of a catalyst does not change during a reaction, as it is not consumed as part of the reaction process. Catalysts lower the energy required to reach the transition state of the reaction, allowing more molecular interactions to achieve that state. However, catalysts do not affect the degree to which a reaction progresses. In other words, though catalysts affect reaction kinetics, the equilibrium state remains unaffected. Catalysts can be classified into two types: homogenous and heterogeneous. Homogenous catalysts are those which exist in the same phase (gas or liquid) as the reactants, while heterogeneous catalysts are not in the same phase as the reactants. Typically, heterogeneous catalysis involves the use of solid catalysts placed in a liquid reaction mixture. For this question, we will be discussing homogenous catalysts. Most of the times, homogeneous catalysis involves the introduction of an aqueous phase catalyst into an aqueous solution of reactants. One reason why homogeneous catalysts are preferred over heterogeneous catalysts because homogeneous catalysts mix well in chemical reactions in comparison to heterogeneous catalysts. However, homogeneous catalyst is often irrecoverable after the reaction has run to completion. Consider the following reaction between cerium(IV) and thallium(I) ions: \[\ce{2Ce^{4+} + Tl^+ → 2Ce^{3+} + Tl^{3+}}\] This reaction is slow, but Mn catalyzes it, as shown in the following mechanism: \[\ce{Ce^{4+} + Mn^{2+} → Ce^{3+} + Mn^{3+}}\] \[\ce{Ce^{4+} + Mn^{3+} → Ce^{3+} + Mn^{4+}}\] \[\ce{Mn^{4+} + Tl^{+ }→ Tl^{3+} + Mn^{2+}}\] In what way does \(Mn^{2+}\) increase the reaction rate? The Mn ion donates two electrons to Ce , one at a time, and then accepts two electrons from Tl . Because Mn can exist in three oxidation states separated by one electron, it is able to couple one-electron and two-electron transfer reactions. The text identifies several factors that limit the industrial applications of enzymes. Still, there is keen interest in understanding how enzymes work for designing catalysts for industrial applications. Why? Enzymes are expensive to make, denature and fail at certain temperatures, are not that stable in a solution, and are very specific to the reaction it was made for. However, scientists can use the observations from enzymes to create catalysts that are more effective in aiding the reaction and cost less to produce. Overall, catalysts still play a large part in lowering the activation energy for reactions. Creating new catalysts can help in the improvement of areas such as medical, ecological, and even commercial products. Most enzymes have an optimal pH range; however, care must be taken when determining pH effects on enzyme activity. A decrease in activity could be due to the effects of changes in pH on groups at the catalytic center or to the effects on groups located elsewhere in the enzyme. Both examples are observed in chymotrypsin, a digestive enzyme that is a protease that hydrolyzes polypeptide chains. Explain how a change in pH could affect the catalytic activity due to (a) effects at the catalytic center and (b) effects elsewhere in the enzyme. ( : remember that enzymes are composed of functional amino acids.) At some point during an enzymatic reaction, the concentration of the activated complex, called an enzyme–substrate complex (\(ES\)), and other intermediates involved in the reaction is nearly constant. When a single substrate is involved, the reaction can be represented by the following sequence of equations: \[\text {enzyme (E) + substrate (S)} \rightleftharpoons \text{enzyme-substrate complex (ES)} \rightleftharpoons \text{enzyme (E) + product (P)}\] This can also be shown as follows: \[E + S \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} ES \underset{k_{-2}}{\stackrel{k_{2}}{\rightleftharpoons}} E+P\] Using molar concentrations and rate constants, write an expression for the rate of disappearance of the enzyme–substrate complex. Typically, enzyme concentrations are small, and substrate concentrations are high. If you were determining the rate law by varying the substrate concentrations under these conditions, what would be your apparent reaction order? A particular reaction was found to proceed via the following mechanism: What is the overall reaction? Is this reaction catalytic, and if so, what species is the catalyst? Identify the intermediates A particular reaction has two accessible pathways (A and B), each of which favors conversion of to a different product ( and , respectively). Under uncatalyzed conditions pathway A is favored, but in the presence of a catalyst pathway B is favored. Pathway B is reversible, whereas pathway A is not. Which product is favored in the presence of a catalyst? without a catalyst? Draw a diagram illustrating what is occurring with and without the catalyst. In both cases, the product of pathway A is favored. All of the produced in the catalyzed reversible pathway B will eventually be converted to as is converted irreversibly to by pathway A. The kinetics of an enzyme-catalyzed reaction can be analyzed by plotting the reaction rate versus the substrate concentration. This type of analysis is referred to as a Michaelis–Menten treatment. At low substrate concentrations, the plot shows behavior characteristic of first-order kinetics, but at very high substrate concentrations, the behavior shows zeroth-order kinetics. Explain this phenomenon.

https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Acid_Rain/Electricity_Generation

Electricity is produced at a an electric power plant. Some fuel source, such as coal, oil, natural gas, or nuclear energy produces heat, which is used to boil water to create steam. The steam under high pressure is then used to spin a turbine that interacts with a system of magnets to produce electricity. The electricity is transmitted as moving electrons through a series of wires to homes and business. This is a typical electric power plant located in Shawville, Pennsylvania. Notice the large pile of coal on the left side of the plant and the three smokestacks, each one taller than the previous. The tallest stack was built to cut down on the local air pollution, where the sulfur oxides are emitted higher into the atmosphere. This has not proven to be a solution to the probelm. As a result the sulfur oxides now travel great distances before coming down in the form of acid rain. have a number of components in common and are an interesting study in the various forms and changes of energy necessary to produce electricity. If a magnetic field can create a current then we have a means of generating electricity. Experiments showed that a magnetic just sitting next to a wire produced no current flow through that wire. However, if the magnet is moving, a current is induced in the wire. The faster the magnet moves, the greater the induced current. This is the principal behind simple electric generators in which a wire loop is rotated between to stationary magnetics. This produces a continuously varying voltage which in turn produces an alternating current . Diagram of a simple electric generator is shown above. To generate electricty then, some (mechanical) mechanism is used to turn a crank that rotates a loop of wire between stationary magnets. The faster the crank turns, the more current that is generated. In hydroelectric, the falling water turns the turbine. The wind can also turn the turbine. In fossil fuel plants and nuclear plants, water is heated to steam which turns the turbine.  

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Fermentation_in_Food_Chemistry/01%3A_Modules/1.05%3A_Intro_to_Microbial_Metabolism

Oxygen (O ) is essential for organisms growing by aerobic respiration (previous worksheet). Many organisms are unable to carry out aerobic respiration because of one or more of the following circumstances: usually refers to anaerobic processes in which organisms do not use molecular oxygen in respiration. Some microbes are ; they contain all the genes required to use either aerobic or anaerobic respiration pathways and they will use aerobic respiration unless there is no oxygen available. However, many prokaryotes are permanently incapable of respiration, even in the presence of oxygen because they lack enzymes or complexes to complete either TCA cycle or electron transport. These are . One important fermentation process is lactic acid fermentation. This process is common in lactobacilli bacteria (and many others). If respiration does not occur through oxidative phosphorylation, NADH must be re-oxidized to NAD for reuse in glycolysis through the EMP pathway (covered earlier). NAD is a catalyst in these reactions. Facultative microbes, particularly bacteria, often use pyruvate as a final electron acceptor. Lactic acid fermentation regenerates NAD but does not directly produce additional ATP. When lactic acid is the only fermentation product, the process is said to be ; such is the case for and used in yogurt production. However, many bacteria perform utilize the pentose phosphate pathway to produce a mixture of lactic acid and ethanol. More detail on this pathway follows. One important heterolactic fermenter is , which is used for souring vegetables like cucumbers and cabbage, producing pickles and sauerkraut, respectively. The has three primary roles in metabolism (human and prokaryotic). There are two phases to these pathways: oxidative phase and non-oxidative phase. Ribulose-5-phosphate (the product of the oxidative stage) is the precursor to the sugar that makes up DNA and RNA. In the , there are different options that depend on the cell’s needs. The ribose-5-phosphate from step 3 is combined with another molecule of ribose-5-phosphate to make one, 10-carbon molecule. Excess ribose-5-phosphate, which may not be needed for nucleotide biosynthesis, is converted into other sugars that can be used by the cell for metabolism. Ribulose-5-phosphate (the product of the oxidative stage) is the precursor to the sugar that makes up DNA and RNA. Of interest for heterolactic fermentation, ribose-5-phosphate is converted to glyceraldehyde-3- phosphate which enters the glycolysis pathway to be converted to pyruvate and then lactic acid. The first step is a simple epimerization alpha to the carbonyl to convert ribose-5-phosphate to xylulose-5-phosphate. Propose a mechanism for this interconversion. The second step is a reaction of xylulose-5-phosphate with a ribose-5-phosphate to prepare a 7- carbon sugar and the glyceraldehyde-3-phosphate. Draw curved arrows for this mechanism. The glyceraldehyde-3-phosphate is then converted to lactic acid. This is a repeat of glycolysis and homolactic acid fermentation. Draw out the pathway to convert glyceraldehyde-3-phosphate to lactic acid. The next steps follow a similar pathway to produce other length sugars and more glyceraldehyde-3-phosphate. In heterolactive fermentation, xylulose-5-phosphate can also be converted directly to glyceraldehyde-3-phosphate and acetyl phosphate. Draw curved arrows for this mechanism. Acetyl phosphate can then be converted to ethanol. Suggest some steps for this conversion (HINT: Look at the ethanol fermentation pathway in yeast). Some bacteria often utilize the Entner-Doudoroff (ED) Glycolytic Pathway rather than the classic glycolysis pathway. du Toit, Englebrecht, Lerm, & Krieger-Weber, Lactobacillus: The Next Generation of Malolactic Acid Fermentation Starter Cultures, . , , 876-906.

https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Atmospheric_Chemistry/Atmosphere

The atmospheric chemistry studies the chemical composition of the natural atmosphere, the way gases, liquids, and solids in the atmosphere interact with each other and with the earth's surface and associated biota, and how human activities may be changing the chemical and physical characteristics of the atmosphere. It is interesting to note that the 1995 Nobel Prize in Chemistry 1995 was awarded to the atmospheric scientists P. Crutzen, M. Molina and F. S. Rowland. For convenience of study, atmospheric scientists divide the atmosphere as if it consists of four layers. The division is mainly due to temperature variations as the altitude increases. The four layers according to the variation of temperature are. Above 100 km is the thermosphere and ionosphere where the temperature increases from 200 K at 100 km to 500 K at 300 km. The temperature goes even higher as the altitude increases. activity as the altitude decrease. In the outer space, most particles consist of single atoms, H, He, and O etc. At lower altitude (200 - 100 km), diatomic molecules N , O , NO etc are present. The ionosphere is full of electrically charged ions. The UV rays ionizes these gases. The major reactions are In the ionosphere: \[O + h v \rightarrow O^+ + e^- \label{18.1.1}\] \[N + h v \rightarrow N^+ + e^- \label{18.1.2}\] In the neutral thermosphere: \[N + O_2 \rightarrow NO + O \label{18.1.3}\] \[N + NO \rightarrow N_2 + O \label{18.1.4}\] \[O + O \rightarrow O_2 \label{18.1.5}\] Beyond the neutral thermosphere is the ionosphere and exosphere. These layers are of course interesting for space explorations and environmental concerns and space sciences. The atmosphere in the outer space is more like a plasma than a gas. Below the thermosphere is the mesosphere (100 - 50 km) in which the temperature decreases as the altitude increase. In this region, OH, H, NO, HO , O , and O are common, and the most prominent chemical reactions are: \[H_2O + h\nu \rightarrow OH + H \label{18.1.6}\] \[H_2O_2 + O \rightarrow OH + OH. \label{18.1.7}\] Below the mesosphere is the , in which the temperature increases as the altitude increase from 10 km to 50 km. In this region, the following reactions are common: \[NO_2 \rightarrow NO + O \label{18.1.8}\] \[N_2O \rightarrow N_2 + O \label{18.1.9}\] \[H_2 + O \rightarrow OH + H \label{18.1.10}\] \[CH_4 + O \rightarrow OH + CH_3 \label{18.1.11}\] Air flow is horizontal in the stratosphere. A thin ozone layer in the upper stratosphere has a high concentration of ozone. This layer is primarily responsible for absorbing the ultraviolet radiation from the sun. The ozone is generated by these reactions: \[O_2 + h\nu \rightarrow O + O \label{18.1.12}\] \[O_2 + O \rightarrow O_3 \label{18.1.13}\] The troposphere is where all weather takes place; it is the region of rising and falling packets of air. The air pressure at the top of the troposphere is only 10% of that at sea level (0.1 atmospheres). There is a thin buffer zone between the troposphere and the next layer called the tropopause. The major components in the region close to the surface of the Earth are N (78%), O (21%), Ar (1%) with variable amounts of H O, CO , CH , NO , NO , CO, N O, and O . The ozone concentration in this layer is low, about 8% of the total ozone in the atmosphere is in the troposphere. From the atmospheric science viewpoint, interactions of all gasses among themselves and their interaction with the environmental elements are of interest. However, for identification purposes, we need to identify the gases produced by man-made process (industry). Some of the gases due to human activities are: Water vapor is also considered a greenhouse gas, but it is also generated by nature continuously due to radiation from the Sun. Of course, when water vapor condense into a liquid, much energy is released in the exothermal process. Condensation of water vapor causes storms and many of the weather phenomena. Skill - Skill - Describe all the details of ozone? Discussion - At the top of the world highest mountain, ~10 km in altitude, the atmosphere is only 0.1 of that at sea level. This is the top of the troposphere. Skill - Explain the chemistry taken place in the thermosphere? Discussion - The aurora is related to the ions in the atmosphere. Skill - Describe the chemistry in the ionosphere. Discussion - Decomposition of ozone releases O, OOH, OH radicals and they are harmful to many living organisms. Discussion - Ozone in the stratosphere absorbs harmful UV C and UV B, which are harmful to humans and plants. Skill - Give your opinion please on an issue.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Reactive_Intermediates/Carbynes

Carbyne is often a general term for any compound whose molecular structure includes an electrically neutral carbon atom with three non-bonded electrons, connected to another atom by a single bond. A carbyne has the general formula R-C3•, where R is any monovalent group and the superscript 3• indicates the three unbounded valences. Carbynes are named after the simplest such compound, HC3•, the methylidyne radical or (unsubstituted) carbyne. Carbyne molecules are generally found to be in electronic doublet states: the nonbonding electrons on carbon are arranged as one radical (unpaired electron) and one electron pair, leaving a vacant atomic orbital, rather than being a tri-radical (the quartet state). The simplest case is the CH radical, which has an electron configuration 1σ 2σ 3σ 1π. Here the 1σ molecular orbital is essentially the carbon 1s atomic orbital, and the 2σ is the C-H bonding orbital formed by overlap of a carbon s-p hybrid orbital with the hydrogen 1s orbital. The 3σ is a carbon non-bonding orbital pointing along the C-H axis away from the hydrogen, while there are two non-bonding 1π orbitals perpendicular to the C-H axis. However the 3σ is an s-p hybrid which has lower energy than the 1π orbital which is pure p, so the 3σ is filled before the 1π. The CH radical is in fact isoelectronic with the nitrogen atom which does have three unpaired electrons in accordance with of maximum multiplicity. However the N atom has three degenerate p orbitals, in contrast to the CH radical where hybridization of one orbital (the 3σ) leads to an energy difference.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/10%3A_Fundamentals_of_Acids_and_Bases/10.06%3A_Types_of_Acids_and_Bases

You will already have noticed that not every compound that contains hydrogen atoms is acidic; ammonia NH , for example, gives an alkaline aqueous solution. Similarly, some compounds containing the group -OH are basic, but others are acidic. An important part of understanding chemistry is being able to recognize what substances will exhibit acidic and basic properties in aqueous solution. Fortunately, most of the common acids and bases fall into a small number of fairly well-defined groups, so this is not particularly difficult. Strictly speaking, the term refers to ionic compounds of hydrogen with the electropositive metals of ; these contain the , H , and are often referred to as "true" hydrides. However, the term is often used in its more general sense to refer to any binary compound MH in which M stands for any element. The hydride ion is such a strong base that it cannot exist in water, so salts such as sodium hydride react with water to yield hydrogen gas and an alkaline solution: \[NaH + H_2O \rightarrow Na^+ + OH^– + H_2\] The more electronegative elements form covalent hydrides which generally react as acids, a well-known example being hydrogen chloride, a gas which dissolves readily in water to give the solution we know as hydrochloric acid \[HCl_{(g)} + H_2O_{(l)} \rightarrow H_3O^+ + Cl^–\] Most of the covalent hydrogen compounds are weak acids— in some cases, such as methane, CH , so weak that their acidic properties are rarely evident. Many, such as H O and NH , are amphiprotic. The latter compound, ammonia, is a weaker acid than H O, so it exhibits basic properties in water \[NH_3 + H_2O \rightarrow NH_4^+ + OH^–\] but behaves as an acid in non-aqueous solvents such as liquid ammonia itself: \[NH_3 + NH_3 \rightarrow NH_4^+ + NH_2^–\] In general, the acidity of the non-metallic hydrides increases with the atomic number of the element to which it is connected. Thus as the element M moves from left to right across the periodic table or down within a group, the acids MH become stronger, as indicated by the acid dissociation constants shown at the right. Note that Attempts to explain these trends in terms of a single parameter such as the electronegativity of M tend not to be very useful. The reason is that acid strengths depend on a number of factors such as the strength of the M-H bond and the energy released when the resultant ions become hydrated in solution. This last factor plays a major role in making HF something of an anomaly amongst the strong acids of Group 17. is such a weak acid that its conjugate base, amide ion NH , cannot exist in water. In aqueous solution, NH acts as a weak base, accepting a proton from water and leaving a OH ion. An aqueous solution of NH is sometimes called “ammonium hydroxide”. This misnomer reflects the pre-Brønsted view that all bases contain –OH units that yield hydroxide ions on dissociation according to the Arrhenius scheme \[NH_4OH \rightleftharpoons NH_4^+ + OH^–\] A solution of ammonia in water is more correctly referred to as "aqueous ammonia" and represented by the formula NH ( ). There is no physical evidence for the existence of NH OH, but the name seems to remain forever etched on reagent bottles in chemical laboratories and in the vocabularies of chemists. Compounds containing the –OH constitute the largest category of acids, especially if the organic acids (discussed separately farther on) are included. M–OH compounds also include many of the most common bases. Whether a compound of the general type M–O–H will act as an acid or a base depends is influenced by the relative tendencies of the M–O and the O–H bonds to break apart in water. If the M–O bond cleaves more readily, then the –OH part will tend to retain its individuality and with its negative charge will become a hydroxide ion. If the O–H bond breaks, the MO-part of the molecule will remain intact as an MO and release of the proton will cause the MOH compound to act as an acid. This is not solely a matter of the relative strengths of the two bonds; the energy change that occurs when the resulting ions interact with water molecules is also an important factor. In general, if M is a metallic element, the metal hydroxide compound \(\ce{MOH}\) will be basic. The case of the highly electropositive elements of Groups 1 and 2 is somewhat special in that their solid MOH compounds exist as interpenetrating lattices of metal cations and OH ions, so those that can dissolve readily in water form strongly alkaline solutions; KOH and NaOH are well known examples of strong bases. From the Brønsted standpoint, these different “bases” are really just different sources for the single strong base OH . As one moves into Group 2 of the periodic table the M-OH compounds become less soluble; thus a saturated solution of Ca(OH) (commonly known as ) is only weakly alkaline. Hydroxides of the metallic elements of the p-block and of the transition metals are so insoluble that their solutions are not alkaline at all. Nevertheless these solids dissolve readily in acidic solutions to yield a salt plus water, so they are formally bases. The acidic character of hydroxy compounds of the nonmetals, known collectively as , is attributed to the displacement of negative charge from the hydroxylic oxygen atom by the electronegative central atom. The net effect is to make the oxygen slightly more positive, thus easing the departure of the hydrogen as H . The presence of other electron-attracting groups on the central atom has a marked effect on the strength of an oxyacid. Of special importance is the doubly-bonded oxygen atom. With the exception of the halogen halides, all of the common strong acids contain one or more of these oxygens, as in sulfuric acid SO (OH), nitric acid NO (OH) and phosphoric acid PO(OH) . In general the strengths of these acids depends more on the number of oxygens than on any other factor, so periodic trends are not so important. Most of the halogen elements form more than one oxyacid. Fluorine is an exception; being more electronegative than oxygen, no oxyacids of this element are known. Chlorine is the only halogen for which all four oxyacids are known, and the values for this series show how powerfully the Cl–O oxygen atoms affect the acid strength. Binary oxides that contain no hydrogen atoms can exhibit acid-base behavior when they react with water. The division between acidic and basic oxygen oxides largely parallels that between the hydroxy compounds. The oxygen compounds of the highly electropositive metals of Groups 1-2 actually contain the O . This ion is another case of a proton acceptor that is stronger than OH , and thus cannot exist in aqueous solution. Ionic oxides therefore tend to give strongly alkaline solutions: \[\ce{O^{-} + H2O -> 2OH^{-} (aq)}\] In some cases, such as that of MgO, the solid is so insoluble that little change in pH is noticed when it is placed in water. CaO, however, which is known as , is sufficiently soluble to form a strongly alkaline solution with the evolution of considerable heat; the result is the slightly-soluble , Ca(OH) . Oxygen compounds of the transition metals are generally insoluble solids having rather complex extended structures. Although some will dissolve in acids, they display no acidic properties in water. The oxides and hydroxides of the metals of Group 3 and higher tend to be only weakly basic, and most display an amphoteric nature. Most of these compounds are so slightly soluble in water that their acidic or basic character is only obvious in their reactions with strong acids or bases. In general, these compounds tend to be more basic than acidic; thus the oxides and hydroxides of aluminum, iron, and zinc all dissolve in mildly acidic solutions, whereas they require treatment with concentrated hydroxide ion solutions to react as acids. The product ions in the second column are known as aluminate, zincate, and ferrate. Other products, in which only some of the –OH groups of the parent hydroxides are deprotonated, are also formed, so there are actually whole series of these oxyanions for most metals. An substance is one that can act as either an acid or a base. An substance can act as either a proton donor or a proton acceptor. So all amphiprotic compounds are also amphoteric. An example of an amphoteric compound that is not amphiprotic is ZnO, which can act as an acid even though it has no protons to donate: \[\ce{ZnO(s) + 4 OH^{-} (aq) -> Zn(OH)^{2-}4 (aq)}\] As a base, it "accepts" protons but does not retain them: \[\ce{ZnO(s) + 2H^{+} <=> Zn^{2+} + H_2O}\] The same remarks can be made about the other compounds shown in the table above. For most practical purposes, the distinction between amphiprotic and amphoteric is not worth worrying about. The binary oxygen compounds of the non-metallic elements tend to produce acidic solutions when they are added to water. Such compounds are sometimes referred to as (“acids without water”.) \[CO_2 + H_2O \rightarrow H_2CO_3\] \[SO_2 + H_2O \rightarrow [H_2SO_3]\] \[SO_3 + H_2O \rightarrow H_2SO_4\] \[P_4O_{10} + 6 H_2O \rightarrow 4 H_3PO_4\] In some cases, the reaction involves more than simply incorporating the elements of water. Thus nitrogen dioxide, used in the commercial preparation of nitric acid, is not an anhydride in the strict sense: \[3 NO_2 + H_2O \rightarrow 2 HNO+3 + NO\] When sodium chloride is dissolved in pure water, the pH remains unchanged because neither ion reacts with water. However, a solution of magnesium chloride will be faintly acidic, and a solution of iron(III) chloride FeCl will be distinctly so. How can this be? Since none of these cations contains hydrogen, we can only conclude that the protons come from the water. The water molecules in question are those that find themselves close to any cation in aqueous solution; the positive field of the metal ion interacts with the polar H O molecule through ion-dipole attraction, and at the same time increases the acidity of these loosely-bound waters by making facilitating the departure H ion. In general, the smaller and more highly charged the cation, the more acidic will it be; the acidity of the alkali metals and of ions like Ag (aq) is negligible, but for more highly-charged ions such as Mg , Pb and Al , the effect is quite noticeable. Most of the transition-metal cations form organized in which four or six H O molecules are chemically bound to the metal ion where they are well within the influence of the coulombic field of the cation, and thus subject to losing a proton. Thus an aqueous solution of "Fe " is really a solution of the ion , whose first stage of "dissociation" can be represented as \[Fe(H_2O)_6^{3+} + H_2O \rightarrow Fe(H_2O)_5(OH)^{2+} + H_3O^+\] As a consequence of this reaction, a solution of FeCl turns out to be a stronger acid than an equimolar solution of acetic acid. A solution of FeCl , however, will be a much weaker acid; the +2 charge is considerably less effective in easing the loss of the proton. It should be possible for a hydrated cation to lose more than one proton. For example, an Al(H O) ion should form, successively, the following species: AlOH(H O) → Al(OH) (H O) → Al(OH) (H O) → Al(OH) (H O) → Al(OH) (H O) → Al(OH) However, removal of protons becomes progressively more difficult as the charge decreases from a high positive value to a negative one; the last three species have not been detected in solution. In dilute solutions of aluminum chloride the principal species are actually Al(H O) (commonly represented simply as Al ) and AlOH(H O) ("AlOH "). When salts dissolve in water, they yield solutions of anions and cations, so their effects on the pH of the solution will depend on the properties of the particular pair of ions. For a salt such as sodium chloride, the solution wil remain neutral because sodium ions have no acidic properties and chloride ions, being conjugate to the strong acid HCl have negligible proton-accepting tendencies. Ions of this kind are often referred to as "strong" ions (that is, derived from a strong acid and a strong base— HCl and NaOH in the case of NaCl.) The possible outcomes for the other three possibilities are shown below. The reactions that cause salt solutions to have non-neutral pH values are sometimes still referred to by the older term (“water splitting”)— a reminder of times before the concept of proton transfer acid-base reactions had developed. The –CO(OH) is the characteristic functional group of the organic acids. The acidity of the carboxylic hydrogen atom is due almost entirely to electron-withdrawal by the non-hydroxylic oxygen atom; if it were not present, we would have an alcohol –COH whose acidity is smaller even than that of H O. This partial electron withdrawal from one atom can affect not only a neighboring atom, but that atom’s neighbor as well. Thus the strength of a carboxylic acid will be affected by the bonding environment of the carbon atom to which it is connected. This propagation of partial electron withdrawal through several adjacent atoms is known as the and is extremely important in organic chemistry. A very good example of the inductive effect produced by chlorine (another highly electronegative atom) is seen by comparing the strengths of acetic acid and of the successively more highly substituted chloroacetic acids: CH –COOH acetic acid 1.8 × 10 Cl CH–COOH dichloroacetic acid 0.055 Cl C–COOH trichloroacetic acid 0.63 The acidic character of the carboxyl group is really a consequence of the enhanced acidity of the –OH group as influenced by the second oxygen atom that makes up the –COOH group. The benzene ring has a similar although weaker electron-withdrawing effect, so hydroxyl groups that are attached to benzene rings also act as acids. The most well known example of such an acid is phenol, C H OH, also known as . Compared to carboxylic acids, phenolic acids are quite very weak, as indicated by the acid dissociation constants listed below: We have already discussed organic acids, so perhaps a word about organic bases would be in order. The –OH group, when bonded to carbon, is acidic rather than basic, so alcohols are not the analogs of the inorganic hydroxy compounds. The amines, consisting of the –NH group bonded to a carbon atom, are the most common class of organic bases. Amines give weakly alkaline solutions in water: \[CH_3NH_2 + H_2O \rightarrow CH_3NH_3^+ + OH^–\] Amines are end products of the bacterial degradation of nitrogenous organic substances such as proteins. They tend to have rather unpleasant “rotten fish” odors. This is no coincidence, since seafood contains especially large amounts of nitrogen-containing compounds which begin to break down very quickly. CH NH , being a gas at room temperature, is especially apt to make itself known to us. Addition of lemon juice or some other acidic substance to fish will convert the methylamine to the methylaminium ion CH NH . Because ions are not volatile they have no odor.

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   " by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact Application Problems Problems marked with a ♦ involve multiple concepts. 1. ♦ Cadmium selenide (CdSe) is a semiconductor used in photoconductors and photoelectric cells that conduct electricity when illuminated. In a related process, a CdSe crystal can absorb enough energy to excite electrons from the valence band to the conduction band, and the excited electrons can return to the valence band by emitting light. The relative intensity and peak wavelength of the emitted light in one experiment are shown in the following table: a. Explain why the emitted light shifts to longer wavelength at higher temperatures. (Hint: consider the expansion of the crystal and the resulting changes in orbital interactions when heated.) b. Why does the relative intensity of the emitted light decrease as the temperature increases? 2. A large fraction of electrical energy is currently lost as heat during transmission due to the electrical resistance of transmission wires. How could superconducting technology improve the transmission of electrical power? What are some potential drawbacks of this technology? 3. Light-emitting diodes (LEDs) are semiconductor-based devices that are used in consumer electronics products ranging from digital clocks to fiber-optic telephone transmission lines. The color of the emitted light is determined in part by the band gap of the semiconductor. Electrons can be promoted to the conduction band and return to the valence band by emitting light or by increasing the magnitude of atomic vibrations in the crystal, which increases its temperature. If you wanted to increase the efficiency of an LED display, and thereby the intensity of the emitted light, would you increase or decrease the operating temperature of the LED? Explain your answer. 4. ♦ Strips of pure Au and Al are often used in close proximity to each other on circuit boards. As the boards become warm during use, however, the metals can diffuse, forming a purple alloy known as “the purple plague” between the strips. Because the alloy is electrically conductive, the board short-circuits. A structural analysis of the purple alloy showed that its structure contained a face-centered cubic (fcc) lattice of atoms of one element, with atoms of the other element occupying tetrahedral holes. What type of alloy is this? Which element is most likely to form the fcc lattice? Which element is most likely to occupy the tetrahedral holes? Explain your answers. What is the empirical formula of the “purple plague”? 5. ♦ Glasses are mixtures of oxides, the main component of which is silica (SiO ). Silica is called the glass former, while additives are referred to as glass modifiers. The crystalline lattice of the glass former breaks down during heating, producing the random atomic arrangements typical of a liquid. Adding a modifier and cooling the melt rapidly produces a glass. How does the three-dimensional structure of the glass differ from that of the crystalline glass former? Would you expect the melting point of a glass to be higher or lower than that of pure SiO ? Lead glass, a particular favorite of the Romans, was formed by adding lead oxide as the modifier. Would you expect lead glass to be more or less dense than soda-lime glass formed by adding sodium and potassium salts as modifiers? 6. Many glasses eventually crystallize, rendering them brittle and opaque. Modifying agents such as TiO are frequently added to molten glass to reduce their tendency to crystallize. Why does the addition of small amounts of TiO stabilize the amorphous structure of glass? 7. ♦ The carbon–carbon bond distances in polyacetylene (–CH=CH–) alternate between short and long, resulting in the following band structure: a. Is polyacetylene a metal, a semiconductor, or an insulator? b. Based on its band structure, how would treating polyacetylene with a potent oxidant affect its electrical conductivity? What would be the effect of treating polyacetylene with small amounts of a powerful reductant? Explain your answers. 8. Enkephalins are pentapeptides, short biopolymers that are synthesized by humans to control pain. Enkephalins bind to certain receptors in brain cells, which are also known to bind morphine and heroin. One enkephalin has the structure tyrosine–glycine–glycine–phenylalanine–methionine. Draw its structure. 9. A polymerization reaction is used to synthesize Saran, a flexible material used in packaging film and seat covers. The monomeric unit for Saran is 1,1-dichloroethylene (CH =CCl ), also known as vinylidene chloride. Draw a reasonable structure for the polymer. Why do pieces of Saran “cling” to one another when they are brought in contact? 10. Polymers are often amorphous solids. Like other materials, polymers can also undergo phase changes. For example, many polymers are flexible above a certain temperature, called the glass-transition temperature (Tg). Below the glass transition temperature, the polymer becomes hard and brittle. Biomedical devices that replace or augment parts of the human body often contain a wide variety of materials whose properties must be carefully controlled.

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A carbene is a molecule containing a neutral carbon atom with a valence of two and two unshared valence electrons. The general formula is R-(C:)-R' or R=C:. The term "carbene" may also refer to the specific compound H2C:, also called methylene, the parent hydride from which all other carbene compounds are formally derived. Carbenes are classified as either singlets or triplets depending upon their electronic structure. Most carbenes are very short lived, although persistent carbenes are known. One well studied carbene is Cl2C:, or dichlorocarbene, which can be generated in situ from chloroform and a strong base. The two classes of carbenes are singlet and triplet carbenes. Singlet carbenes are spin-paired. In the language of valence bond theory, the molecule adopts an sp2 hybrid structure. Triplet carbenes have two unpaired electrons. They may be either linear or bent, i.e. sp or sp2 hybridized, respectively. Most carbenes have a nonlinear triplet ground state, except for those with nitrogen, oxygen, or sulfur atoms, and halides directly bonded to the divalent carbon. Figure 1: from Carbenes are called singlet or triplet depending on the electronic spins they possess. Triplet carbenes are paramagnetic and may be observed by electron spin resonance spectroscopy if they persist long enough. The total spin of singlet carbenes is zero while that of triplet carbenes is one (in units of [\hbar] ). Bond angles are 125-140° for triplet methylene and 102° for singlet methylene (as determined by EPR). Triplet carbenes are generally stable in the gaseous state, while singlet carbenes occur more often in aqueous media. For simple hydrocarbons, triplet carbenes usually have energies 8 kcal/mol (33 kJ/mol) lower than singlet carbenes (see also Hund's rule of maximum multiplicity), thus, in general, triplet is the more stable state (the ground state) and singlet is the excited state species.Substituents that can donate electron pairs may stabilize the singlet state by delocalizing the pair into an empty p-orbital. If the energy of the singlet state is sufficiently reduced it will actually become the ground state. No viable strategies exist for triplet stabilization. The carbene called 9-fluorenylidene has been shown to be a rapidly equilibrating mixture of singlet and triplet states with an approximately 1.1 kcal/mol (4.6 kJ/mol) energy difference.[3] It is, however, debatable whether diaryl carbenes such as the fluorene carbene are true carbenes because the electrons can delocalize to such an extent that they become in fact biradicals. In silico experiments suggest that triplet carbenes can be thermodynamically stabilized with electropositive heteroatoms such as in silyl and silyloxy carbenes, especially trifluorosilyl carbenes.[4] Singlet and triplet carbenes exhibit divergent reactivity. Singlet carbenes generally participate in as either or . Singlet carbenes with unfilled p-orbital should be electrophilic. Triplet carbenes can be considered to be , and participate in stepwise radical additions. Triplet carbenes have to go through an with two unpaired electrons whereas singlet carbene can react in a single step. Due to these two modes of reactivity, reactions of singlet methylene are whereas those of triplet methylene are . This difference can be used to probe the nature of a carbene. For example, the reaction of methylene generated from of with - or with - each give a single diastereomer of the 1,2-dimethylcyclopropane product: from and from , which proves that the methylene is a singlet. If the methylene were a triplet, one would not expect the product to depend upon the starting alkene geometry, but rather a nearly identical mixture in each case. Reactivity of a particular carbene depends on the groups. Their reactivity can be affected by . Some of the reactions carbenes can do are , skeletal rearrangements, and additions to double bonds. Carbenes can be classified as nucleophilic, electrophilic, or ambiphilic. For example, if a substituent is able to donate a pair of electrons, most likely carbene will not be electrophilic. carbenes insert much more selectively than methylene, which does not differentiate between primary, secondary, and tertiary C-H bonds. Carbenes add to double bonds to form cyclopropanes. A concerted mechanism is available for singlet carbenes. Triplet carbenes do not retain stereochemistry in the product molecule. Addition reactions are commonly very fast and exothermic. The slow step in most instances is generation of carbene. A well-known reagent employed for alkene-to-cyclopropane reactions is Simmons-Smith reagent. This reagent is a system of copper, zinc, and iodine, where the active reagent is believed to be iodomethylzinc iodide. Reagent is complexed by hydroxy groups such that addition commonly happens syn to such group. Carbene insertion are another common type of carbene reactions. The carbene basically interposes itself into an existing bond. The order of preference is commonly: 1. X–H bonds where X is not carbon 2. C–H bond 3. C–C bond. Insertions may or may not occur in single step. insertion reactions present new synthetic solutions. Generally, rigid structures favor such insertions to happen. When an intramolecular insertion is possible, no insertions are seen. In flexible structures, five-membered ring formation is preferred to six-membered ring formation. Both inter- and intramolecular insertions are amendable to asymmetric induction by choosing chiral ligands on metal centers. Alkylidene carbenes are alluring in that they offer formation of moieties. To generate an alkylidene carbene a ketone can be exposed to . Carbenes and carbenoid precursors can undergo dimerization reactions to form alkenes. While this is often an unwanted side reaction, it can be employed as a synthetic tool and a direct metal carbene dimerization has been used in the synthesis of polyalkynylethenes. Persistent carbenes exist in equilibrium with their respective dimers. This is known as the Wanzlick equilibrium. In species, metal complexes with the formulae L MCRR' are often described as carbene complexes. Such species do not however react like free carbenes and are rarely generated from carbene precursors, except for the persistent carbenes. The can be classified according to their reactivity, with the first two classes being the most clearly defined:

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In efforts to reduce gas consumption from oil, ethanol is often added to regular gasoline. It has a high octane rating and burns more slowly than regular gas. This "gasohol" is widely used in many countries. It produces somewhat lower carbon monoxide and carbon dioxide emissions, but does increase air pollution from other materials. Many chemical reactions are combustion reactions. It is often important to know the energy produced in such a reaction so that we can determine which fuel might be the most efficient for a given purpose. The is the heat released when one mole of a substance is completely burned. Typical combustion reactions involve the reaction of a carbon-containing material with oxygen to form carbon dioxide and water as products. If methanol is burned in air, we have: \[\ce{CH_3OH} + \ce{O_2} \rightarrow \ce{CO_2} + 2 \ce{H_2O} \: \: \: \: \: He = 890 \: \text{kJ/mol}\nonumber \] In this case, one mole of oxygen reacts with one mole of methanol to form one mole of carbon dioxide and two moles of water. It should be noted that inorganic substances can also undergo a form of combustion reaction: \[2 \ce{Mg} + \ce{O_2} \rightarrow 2 \ce{MgO}\nonumber \] In this case, there is no water and no carbon dioxide formed. For the purposes of this chapter, these reactions are generally not considered in the discussion of combustion reactions. Heats of combustion are usually determined by burning a known amount of the material in a bomb calorimeter with an excess of oxygen. By measuring the temperature change, the heat of combustion can be determined. A 1.55 gram sample of ethanol is burned and produced a temperature increase of \(55^\text{o} \text{C}\) in 200 grams of water. Calculate the molar heat of combustion. Amount of ethanol used: \[\frac{1.55 \: \text{g}}{46.1 \: \text{g/mol}} = 0.0336 \: \text{mol}\nonumber \] Energy generated: \[4.184 \: \text{J/g}^\text{o} \text{C} \times 200 \: \text{g} \times 55^\text{o} \text{C} = 46024 \: \text{J} = 46.024 \: \text{kJ}\nonumber \] Molar heat of combustion: \[\frac{46.024 \: \text{kJ}}{0.0336 \: \text{mol}} = 1370 \: \text{kJ/mol}\nonumber \] The burning of ethanol produces a significant amount of heat.

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E2 reactions are typically seen with secondary and tertiary alkyl halides, but a hindered base is necessary with a primary halide. The mechanism by which it occurs is a single step reaction with one transition state. The rate at which this mechanism occurs is second order kinetics, and depends on both the base and alkyl halide. A good leaving group is required because it is involved in the rate determining step. The leaving groups must be coplanar in order to form a pi bond; carbons go from sp to sp hybridization states. In this reaction Ba represents the base and X represents a leaving group, typically a halogen. There is one transition state that shows the concerted reaction for the base attracting the hydrogen and the halogen taking the electrons from the bond. The product be both eclipse and staggered depending on the transition states. Eclipsed products have a transition states, while staggered products have transition states. Staggered conformation is usually the major product because of its lower energy confirmation. An E2 reaction has certain requirements to proceed: Reaction Coordinate 1. 2. What is the major product and why? 3. What is the major prodcut when 2-bromo-2-methylbutane reacts with with sodium ethoxide? 4. 5.

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Polysaccharides are carbohydrate polymers consisting of tens to hundreds to several thousand monosaccharide units. All of the common polysaccharides contain glucose as the monosaccharide unit. Polysaccharides are synthesized by plants, animals, and humans to be stored for food, structural support, or metabolized for energy. Glycogen is the storage form of glucose in animals and humans which is analogous to the starch in plants. Glycogen is synthesized and stored mainly in the liver and the muscles. Structurally, glycogen is very similar to amylopectin with alpha acetal linkages, however, it has even more branching and more glucose units are present than in amylopectin. Various samples of glycogen have been measured at 1,700-600,000 units of glucose. The structure of glycogen consists of long polymer chains of glucose units connected by an linkage. The graphic on the left shows a very small portion of a glycogen chain. All of the monomer units are alpha-D-glucose, and all the alpha acetal links connect C # 1 of one glucose to C # 4 of the next glucose. The branches are formed by linking C #1 to a C #6 through an acetal linkages. In glycogen, the branches occur at intervals of 8-10 glucose units, while in amylopectin the branches are separated by 12-20 glucose units. Carbon # 1 is called the and is the center of an acetal functional group. A carbon that has two ether oxygens attached is an acetal. The is defined as the ether oxygen being on the opposite side of the ring as the C # 6. In the chair structure this results in a . This is the same definition as the -OH in a hemiacetal. Plants make starch and cellulose through the photosynthesis processes. Animals and human in turn eat plant materials and products. Digestion is a process of hydrolysis where the starch is broken ultimately into the various monosaccharides. A major product is of course glucose which can be used immediately for metabolism to make energy. The glucose that is not used immediately is converted in the liver and muscles into glycogen for storage by the process of glycogenesis. Any glucose in excess of the needs for energy and storage as glycogen is converted to fat.

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Write the following isotopes in hyphenated form (e.g., “carbon-14”) Write the following isotopes in nuclide notation (e.g., " \(\ce{^{14}_6C}\) ") For the following isotopes that have missing information, fill in the missing information to complete the notation For each of the isotopes in Question 21.2.3, determine the numbers of protons, neutrons, and electrons in a neutral atom of the isotope. Write the nuclide notation, including charge if applicable, for atoms with the following characteristics: Calculate the density of the \(\ce{^{24}_{12}Mg}\) nucleus in g/mL, assuming that it has the typical nuclear diameter of 1 × 10 cm and is spherical in shape. What are the two principal differences between nuclear reactions and ordinary chemical changes? The mass of the atom \(\ce{^{23}_{11}Na}\) is 22.9898 amu. Which of the following nuclei lie within the band of stability? Which of the following nuclei lie within the band of stability? Write a brief description or definition of each of the following: Which of the various particles (α particles, β particles, and so on) that may be produced in a nuclear reaction are actually nuclei? Complete each of the following equations by adding the missing species: Complete each of the following equations: Write a balanced equation for each of the following nuclear reactions: Technetium-99 is prepared from Mo. Molybdenum-98 combines with a neutron to give molybdenum-99, an unstable isotope that emits a β particle to yield an excited form of technetium-99, represented as Tc . This excited nucleus relaxes to the ground state, represented as Tc, by emitting a γ ray. The ground state of Tc then emits a β particle. Write the equations for each of these nuclear reactions. The mass of the atom \(\ce{^{19}_9F}\) is 18.99840 amu. For the reaction \(\ce{^{14}_6C ⟶ ^{14}_7N +\, ?}\), if 100.0 g of carbon reacts, what volume of nitrogen gas (N ) is produced at 273 K and 1 atm? What are the types of radiation emitted by the nuclei of radioactive elements? What changes occur to the atomic number and mass of a nucleus during each of the following decay scenarios? What is the change in the nucleus that results from the following decay scenarios? Many nuclides with atomic numbers greater than 83 decay by processes such as electron emission. Explain the observation that the emissions from these unstable nuclides also normally include α particles. Why is electron capture accompanied by the emission of an X-ray? Explain how unstable heavy nuclides (atomic number > 83) may decompose to form nuclides of greater stability (a) if they are below the band of stability and (b) if they are above the band of stability. Which of the following nuclei is most likely to decay by positron emission? Explain your choice. The following nuclei do not lie in the band of stability. How would they be expected to decay? Explain your answer. The following nuclei do not lie in the band of stability. How would they be expected to decay? Predict by what mode(s) of spontaneous radioactive decay each of the following unstable isotopes might proceed: Write a nuclear reaction for each step in the formation of \(\ce{^{218}_{84}Po}\) from \(\ce{^{238}_{92}U}\), which proceeds by a series of decay reactions involving the step-wise emission of α, β, β, α, α, α, α particles, in that order. Write a nuclear reaction for each step in the formation of \(\ce{^{208}_{82}Pb}\) from \(\ce{^{228}_{90}Th}\), which proceeds by a series of decay reactions involving the step-wise emission of α, α, α, α, β, β, α particles, in that order. Define the term half-life and illustrate it with an example. A 1.00 × 10 -g sample of nobelium, \(\ce{^{254}_{102}No}\), has a half-life of 55 seconds after it is formed. What is the percentage of \(\ce{^{254}_{102}No}\) remaining at the following times? Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the Pu present today will be present in 1000 y? The isotope Tl undergoes β decay with a half-life of 3.1 min. If 1.000 g of \(\ce{^{226}_{88}Ra}\) produces 0.0001 mL of the gas \(\ce{^{222}_{86}Rn}\) at STP (standard temperature and pressure) in 24 h, what is the half-life of Ra in years? The isotope \(\ce{^{90}_{38}Sr}\) is one of the extremely hazardous species in the residues from nuclear power generation. The strontium in a 0.500-g sample diminishes to 0.393 g in 10.0 y. Calculate the half-life. Technetium-99 is often used for assessing heart, liver, and lung damage because certain technetium compounds are absorbed by damaged tissues. It has a half-life of 6.0 h. Calculate the rate constant for the decay of \(\ce{^{99}_{43}Tc}\). What is the age of mummified primate skin that contains 8.25% of the original quantity of C? A sample of rock was found to contain 8.23 mg of rubidium-87 and 0.47 mg of strontium-87. A laboratory investigation shows that a sample of uranium ore contains 5.37 mg of \(\ce{^{238}_{92}U}\) and 2.52 mg of \(\ce{^{206}_{82}Pb}\). Calculate the age of the ore. The half-life of \(\ce{^{238}_{92}U}\) is 4.5 × 10 yr. Plutonium was detected in trace amounts in natural uranium deposits by Glenn Seaborg and his associates in 1941. They proposed that the source of this Pu was the capture of neutrons by U nuclei. Why is this plutonium not likely to have been trapped at the time the solar system formed 4.7 × 10 years ago? A \(\ce{^7_4Be}\) atom (mass = 7.0169 amu) decays into a \(\ce{^7_3Li}\) atom (mass = 7.0160 amu) by electron capture. How much energy (in millions of electron volts, MeV) is produced by this reaction? A \(\ce{^8_5B}\) atom (mass = 8.0246 amu) decays into a \(\ce{^8_4Be}\) atom (mass = 8.0053 amu) by loss of a β particle (mass = 0.00055 amu) or by electron capture. How much energy (in millions of electron volts) is produced by this reaction? Isotopes such as Al (half-life: 7.2 × 10 years) are believed to have been present in our solar system as it formed, but have since decayed and are now called extinct nuclides. Write a balanced equation for each of the following nuclear reactions: Write a balanced equation for each of the following nuclear reactions: transuranium How does nuclear fission differ from nuclear fusion? Why are both of these processes exothermic? Both fusion and fission are nuclear reactions. Why is a very high temperature required for fusion, but not for fission? Cite the conditions necessary for a nuclear chain reaction to take place. Explain how it can be controlled to produce energy, but not produce an explosion. Describe the components of a nuclear reactor. In usual practice, both a moderator and control rods are necessary to operate a nuclear chain reaction safely for the purpose of energy production. Cite the function of each and explain why both are necessary. Describe how the potential energy of uranium is converted into electrical energy in a nuclear power plant. The mass of a hydrogen atom \(\ce{(^1_1H)}\) is 1.007825 amu; that of a tritium atom \(\ce{(^3_1H)}\) is 3.01605 amu; and that of an α particle is 4.00150 amu. How much energy in kilojoules per mole of \(\ce{^4_2He}\) produced is released by the following fusion reaction: \(\ce{^1_1H + ^3_1H ⟶ ^4_2He}\). How can a radioactive nuclide be used to show that the equilibrium: \[\ce{AgCl}(s)⇌\ce{Ag+}(aq)+\ce{Cl-}(aq)\] is a dynamic equilibrium? Technetium-99m has a half-life of 6.01 hours. If a patient injected with technetium-99m is safe to leave the hospital once 75% of the dose has decayed, when is the patient allowed to leave? Iodine that enters the body is stored in the thyroid gland from which it is released to control growth and metabolism. The thyroid can be imaged if iodine-131 is injected into the body. In larger doses, I-131 is also used as a means of treating cancer of the thyroid. I-131 has a half-life of 8.70 days and decays by β emission. If a hospital were storing radioisotopes, what is the minimum containment needed to protect against: Based on what is known about Radon-222’s primary decay method, why is inhalation so dangerous? Given specimens uranium-232 (\(t_{1/2} = \mathrm{68.9 \;y}\)) and uranium-233 (\(t_{1/2} = \mathrm{159,200\; y}\)) of equal mass, which one would have greater activity and why? A scientist is studying a 2.234 g sample of thorium-229 ( = 7340 y) in a laboratory. Given specimens neon-24 (\(t_{1/2} = \mathrm{3.38\; min}\)) and bismuth-211 (\(t_{1/2} = \mathrm{2.14\; min}\)) of equal mass, which one would have greater activity and why?

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Rotation in substituted ethanes is an example of general rotation about single, sigma, carbon-carbon bonds. The different conformations resulting from these single bond rotations are energetically very similar, making such rotations both common and important in understanding how alkanes act. Often a particular conformation of the molecule is required for a reaction to proceed, and can explain the product stereochemistry, so understanding how to visualize the different forms and how they interconvert is essential to understanding complicated reaction mechanisms. While the carbon-carbon single bonds of ethane must adhere to certain physical limitations (both carbons are sp hybridized, all bond angles are approximately 109.5°, etc), these criteria still allow a lot movement within the molecule. With a small amount of activation energy, substituents are free to rotate around the bond, like the blades on a fan rotate about the hub. The most common and simplest way of depicting rotations about substituted ethanes is a . This simplifies the molecule into a 2-D sketch that can readily show the differences between conformations. (A conformation is a particular arrangement of the substituents of one carbon in a bond, with respect to the substituents of the other carbon in the bond) There are two different classifications for the conformations: - the substituents, as drawn in the Newman Projection, all have 60 degrees between them. These conformers are always the lowest points in relative energy diagrams. - the substituents are overlapping, when viewed as in the Newman projection. With a model kit, this is the form where all the substituents are aligned and the ones on the carbon farthest from your eyes are hidden behind the substituents of the foremost carbon. Depending on which groups are overlapping, eclipsed forms can have varying energies. They do not have specific names, like anti vs gauche. The eclipsed conformer with the largest group on each carbon overlapping with each other will always be the highest energy conformer of that bond. To convert between eclipsed and staggered conformations (or vice versa), one carbon, with all of its substituents, is rotated 60 degrees. You can imagine the projection like a combination lock. The back carbon is like the lock itself and the front carbon like the knob you turn to enter your combination. "grasp" the front carbon, and turn in 60 degrees. The placement of the back substituents will not change. (In reality, all the substituents are rotating simultaneously.) Repeating this process 6 times will produce all staggered and eclipsed conformations of the bond in question, and bring you back to the conformation you began with. Any conformations between eclipsed and staggered are called skew conformations, but as there are an infinite number of these for each bond, (depending on how much or little the bond is rotated) skew conformations are rarely represented. (See attached "Newman Projections" to view Newman Projections of 3,4-dimethylhexane) : Only has two conformers! Because all the substituents are exactly the same (all hydrogens), there is only one eclipsed and one anti conformation, each. The eclipsed formation is 3 kcal/mol higher in energy - 1 kcal/mol for each eclipsed hydrogen. : Substituted ethanes have slightly greater differences in energy between conformations because of steric hindrance. Essentially, the larger groups bump up against each other during the transition state, so it takes more energy to move them around eachother. The large substituents are like beach balls that knock into each other while moving around, as opposed to hydrogens, which are smaller, and more compact (imagine trying the move two volleyballs past each other in a 16 inch box, versus two golf balls in the same volume). This hindrance raises the energy of all conformations of the molecule (when compared to unsubstituted ethane, or less-substituted bonds). (See bottom of attached "Newman Projections" for Ethane Conformations)

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/The_Phenyl_Group

The terms phenyl and phenol, along with benzene and benzyl, can confuse beginning organic chemistry students. Figure 1 shows what the four terms mean. : The top two structures in Figure 1 are . The -ene suffix of might indicate that it is similar to an alkene. The -ol suffix of indicates that it has an -OH group. The lower two structures in Figure 1 show . The line extending off without anything connected is the line that shows this is a group, which should be attached to something. For example, one might have phenyl chloride (C H Cl, also called chlorobenzene) or one might have benzyl chloride (C H CH Cl). (The structures of these two compounds are shown below in Figure 2.) The group is based simply on benzene, with one H removed. The group is based on methylbenzene (toluene), with one H removed from the methyl group. Figure 2 shows the benzyl chloride and phenyl chloride; these are based on the groups discussed above. If you have the misfortune to come across the word benzol, be forewarned that this is an old German word for benzene. Similarly, toluol is a German word for toluene. Benzin is a German word for gasoline; it is related to the uncommon term benzine, used for some types of gasoline. >Robert Bruner ( )

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.15%3A_Ethers

In alcohols, one of the two bonds to the oxygen atom involves hydrogen and one involves carbon. When two or more carbon atoms are present, however, isomeric structures in which oxygen is bonded to two different carbons become possible. Such compounds are called ethers. For example, dimethyl ether is isomeric with ethanol, and methyl ethyl ether is isomeric with propanol:   The general formula for an ether is R—O—R′, where R′ signifies that both R groups need not be the same. Draw projection formulas and name all the isomers which correspond to the molecular formula C H O. The formula C H would correspond to an alkane. The extra oxygen atom might be added between two carbons, giving ether, or it might be added between a carbon and a hydrogen, giving an alcohol. The alcohol molecules might have the hydroxyl group at the end of the three-carbon chain or on the second carbon atom: In an ether there are no hydrogen atoms connected to a highly electronegative neighbor, and so, unlike alcohols, ether molecules cannot hydrogen bond among themselves. Each C—O bond is polar, but the bonds are at approximately the tetrahedral angle. The polarity of one partially cancels the polarity of the other. Consequently the forces between two ether molecules are not much greater than the London forces between alkane molecules of comparable size. The boiling point of dimethyl ether, for example, is –23°C, slightly above that of propane (–42°C), but well below that of ethanol (78.5°C). All three molecules contain 26 electrons and are about the same size. In the we see this trend again, this time with compounds containing 32 or 34 electrons. The chemical reactivity of ethers is also closer to that of the alkanes than that of the alcohols. Ethers undergo few characteristic reactions other than combustion, and so they are commonly used as solvents. Diethyl ether is also used as an anesthetic, although the flammability of its vapor requires that precautions be taken to prevent fires.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.18%3A_Esters

Cellulose acetate is a complicated example of another group of organic compounds, , which can be made by combining with . A simpler case is the reaction of ethanol with acetic acid to give ethyl acetate: The general formula for an ester can be written   . In the case of ethyl acetate, R is CH CH and R′ is CH . When this notation is used, esters are named based on the number of carbon atoms present in the alcohol and carboxylic acid groups that helped to form it. The term from the alcohol is given the "-yl" suffix, and is followed by the acid term with the suffix "-ate". As an example, the ester formed by the condensation reaction between methanol and butanoic acid would be called "methyl butanoate". The synthesis of nitroglycerin, was also an example of ester formation, but in that case an inorganic acid, HNO , was combined with an alcohol. Formation of an ester is an example of an important class of reactions called condensations. In a a pair of molecules join together, giving off a small, very stable molecule like H O or HCl. In both ethyl acetate and nitroglycerin synthesis, this small molecule is H O. A condensation can often be undone if large numbers of the small molecules are added to the product. In the case of an ester, addition of large quantities of H O causes (literally, “splitting by means of water“): This is just the reverse of ethyl acetate condensation. Although the ester functional group has a polar carbonyl, it contains no hydrogen atoms suitable for hydrogen bonding. Therefore esters have low boiling points relative to most . In many cases, even though its molecules are almost twice as large as those of the constituent alcohol and acid, an ester is found to have a lower boiling point than either. Ethyl acetate, for example, boils at 77.1°C, lower than ethanol (78.5°C) or acetic acid (117.9°C). By contrast to acids and alcohols which have unpleasant and rather weak odors, respectively, esters usually smell good. The odors of many fruits and flowers are due to esters. Ethyl acetate, for example, is the most important factor in the flavor of pineapples.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Carbohydrates/Monosaccharides/Fructose

is more commonly found together with glucose and sucrose in honey and fruit juices. Fructose, along with glucose are the monosaccharides found in disaccharide, sucrose. Fructose is classified as a monosaccharide, the most important , a hexose, and is a reducing sugar. An older common name for fructose is levulose, after its levorotatory property of rotating plane polarized light to the left (in contrast to glucose which is dextrorotatory). Bees gather nectar from flowers which contains sucrose. They then use an enzyme to hydrolyze or break apart the sucrose into its component parts of glucose and fructose. High Fructose Corn Syrup The chair form of fructose follows a similar pattern as that for glucose with a few exceptions. Since fructose has a ketone functional group, the ring closure occurs at carbon # 2. In the case of fructose a five membered ring is formed. The -OH on carbon #5 is converted into the ether linkage to close the ring with carbon #2. This makes a 5 member ring - four carbons and one oxygen. The ring structure is written with the orientation depicted on the left for the monosaccharide and is consistent with the way the glucose is depicted. The is the center of a functional group. A carbon that has both an ether oxygen and an alcohol group (and is attached to two other carbons is a hemiketal. The is defined as the -OH being on the same side of the ring as the C # 6. In the ring structure this results in a for the -OH on carbon # 2. The is defined as the -OH being on the opposite side of the ring as the C # 6. In the ring structure this results in a for the -OH on carbon # 2. The alpha and beta label is not applied to any other carbon - only the anomeric carbon, in this case # 2. Figure: Compare Alpha and Beta Fructose The six member ring and the position of the -OH group on the from the -OH on C # 4 in a down projection in the ). is recognized by having a five member ring and having six carbons, a hexose. Both glucose and fructose may be either alpha or beta on the anomeric carbon, so this is not distinctive between them. Which carbon in the structure on the in Figure 2 is the anomeric carbon?

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Alkene_Stereoisomers

The carbon-carbon double bond is formed between two sp hybridized carbons, and consists of two occupied , a sigma orbital and a pi orbital. Rotation of the end groups of a double bond relative to each other destroys the p-orbital overlap that creates the pi orbital or bond. Because the pi bond has a of roughly 60 kcal/mole, this resistance to rotation stabilizes the planar configuration of this functional group. As a result, certain disubstituted alkenes may exist as a pair of configurational stereoisomers, often designated cis and trans. (one may be hydrogen). This is illustrated by the following general formulas. In the first example, the left-hand double bond carbon has two identical substituents ( ) so stereoisomerism about the double bond is not possible (reversing substituents on the right-hand carbon gives the same configuration). In the next two examples, each double bond carbon atom has two different substituent groups and stereoisomerism exists, regardless of whether the two substituents on one carbon are the same as those on the other. Some examples of this configurational stereoisomerism (sometimes called geometric isomerism) are shown below. Note that cycloalkenes smaller than eight carbons cannot exist in a stable trans configuration due to ring strain. A similar restriction holds against cycloalkynes smaller than ten carbons. Since alkynes are linear, there is no stereoisomerism associated with the carbon-carbon triple bond. Configurational stereoisomers of the kind shown above need an additional nomenclature prefix added to the IUPAC name, in order to specify the spatial orientations of the groups attached to the double bond. Thus far, the prefixes cis- and trans- have served to distinguish stereoisomers; however, it is not always clear which isomer should be called cis and which trans. For example, consider the two compounds on the right. Both compound A (1-bromo-1-chloropropene) and compound B ( 1-cyclobutyl-2-ethyl-3-methyl-1-butene) can exist as a pair of configurational stereoisomers (one is shown). How are we to name these stereoisomers so that the configuration of each is unambiguously specified? Assignment of a cis or trans prefix to any of these isomers can only be done in an arbitrary manner, so a more rigorous method is needed. A completely unambiguous system, based on a set of group priority rules, assigns a (German, zusammen for together) or (German, entgegen for opposite) to designate the stereoisomers. In the isomers illustrated above, for which cis-trans notation was adequate, is equivalent to cis and is equivalent to trans. The higher the atomic number of the immediate substituent atom, the higher the priority. For example, H– < C– < N– < O– < Cl–. (priority increases left to right) (Different isotopes of the same element are assigned a priority according to their atomic mass.) If two substituents have the same immediate substituent atom, move to the next atom (away from the double bond) until a difference is found. For example, CH – < C H – < ClCH – < BrCH – < CH O–. Once the relative priorities of the two substituents on each of the double bond carbons has been determined, a cis orientation of the higher priority pair is designated , and a trans orientation is termed . Applying these rules to the isomers of compounds A and B shown above, we assign the configuration of the 1-bromo-1-chloropropene isomer as (Br has higher priority than Cl, and CH a higher priority than H). The configuration of the 1-cyclobutyl-2-ethyl-3-methyl-1-butene isomer is determined to be (C H has higher priority than H, and the isopropyl group has higher priority than an ethyl group). The following example elaborates the priority determination for a more complex case. The line formula is expanded to give the structural formula in the center. The root name is heptene (the longest chain incorporating both carbons of the double bond), and the substituents (in red) are added to give the IUPAC name. In order to assign a configurational prefix the priority order of substituents at each double bond carbon must be determined. For the immediate substituent atoms are a chlorine and a carbon. The chlorine has a higher atomic number and therefore has higher priority (colored green and numbered 1). The more remote bromine atom does not figure in this choice. For the immediate substituent atoms are both carbons (colored orange). As a result, we must look at the next higher atomic number atoms in the substituent chain. These are also carbon, but the isopropyl group has two carbons (also orange) whereas the propyl group has only one. The priority order is therefore isopropyl ( ) > propyl ( ). Since the two higher priority groups ( ) are on the same side of the double bond, this configuration is ( ).

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/13%3A_Kinetic_Methods/13.02%3A_Chemical_Kinetics

The earliest analytical methods based on chemical kinetics—which first appear in the late nineteenth century—took advantage of the catalytic activity of enzymes. In a typical method of that era, an enzyme was added to a solution that contained a suitable substrate and their reaction was monitored for a fixed time. The enzyme’s activity was determined by the change in the substrate’s concentration. Enzymes also were used for the quantitative analysis of hydrogen peroxide and carbohydrates. The development of chemical kinetic methods continued in the first half of the twentieth century with the introduction of nonenzymatic catalysts and noncatalytic reactions. Despite the diversity of chemical kinetic methods, by 1960 they no longer were in common use. The principal limitation to their broader acceptance was a susceptibility to significant errors from uncontrolled or poorly controlled variables—temperature and pH are two such examples—and the presence of interferents that activate or inhibit catalytic reactions. By the 1980s, improvements in instrumentation and data analysis methods compensated for these limitations, ensuring the further development of chemical kinetic methods of analysis [Pardue, H. L. , , 69–107]. Every chemical reaction occurs at a finite rate, which makes it a potential candidate for a chemical kinetic method of analysis. To be effective, however, the chemical reaction must meet three necessary conditions: (1) the reaction must not occur too quickly or too slowly; (2) we must know the reaction’s rate law; and (3) we must be able to monitor the change in concentration for at least one species. Let’s take a closer look at each of these requirements. The material in this section assumes some familiarity with chemical kinetics, which is part of most courses in general chemistry. For a review of reaction rates, rate laws, and integrated rate laws, see the material in . The of the chemical reaction—how quickly the concentrations of reactants and products change during the reaction—must be fast enough that we can complete the analysis in a reasonable time, but also slow enough that the reaction does not reach equilibrium while the reagents are mixing. As a practical limit, it is not easy to study a reaction that reaches equilibrium within several seconds without the aid of special equipment for rapidly mixing the reactants. We will consider two examples of instrumentation for studying reactions with fast kinetics later in this chapter. The second requirement is that we must know the reaction’s —the mathematical equation that describes how the concentrations of reagents affect the rate—for the period in which we are making measurements. For example, the rate law for a reaction that is first order in the concentration of an analyte, , is \[\text { rate }=-\frac{d[A]}{d t}=k[A] \label{13.1}\] where is the reaction’s . Because the concentration of decreases during the reactions, [ ] is negative. The minus sign in Equation \ref{13.1} makes the rate positive. If we choose to follow a product, , then [ ] is positive because the product’s concentration increases throughout the reaction. In this case we omit the minus sign. An often is a more useful form of the rate law because it is a function of the analyte’s initial concentration. For example, the integrated rate law for Equation \ref{13.1} is \[\ln{[A]_t} = \ln{[A]_0} - kt \label{13.2}\] or \[[A]_{t}=[A]_{0} e^{-k t} \label{13.3}\] where [ ] is the analyte’s initial concentration and [ ] is the analyte’s concentration at time . Unfortunately, most reactions of analytical interest do not follow a simple rate law. Consider, for example, the following reaction between an analyte, , and a reagent, , to form a single product, \[A + R \rightleftharpoons P \nonumber\] where is the rate constant for the forward reaction, and is the rate constant for the reverse reaction. If the forward and the reverse reactions occur as single steps, then the rate law is \[\text { rate }=-\frac{d[A]}{d t}=k_{f}[A,R]-k_{r}[P] \label{13.4}\] The first term, [ , ] accounts for the loss of as it reacts with to make , and the second term, [ ] accounts for the formation of as converts back to and to . Although we know the reaction’s rate law, there is no simple integrated form that we can use to determine the analyte’s initial concentration. We can simplify Equation \ref{13.4} by restricting our measurements to the beginning of the reaction when the concentration of product is negligible. Under these conditions we can ignore the second term in Equation \ref{13.4}, which simplifies to \[\text { rate }=-\frac{d[A]}{d t}=k_{f}[A,R] \label{13.5}\] The integrated rate law for Equation \ref{13.5}, however, is still too complicated to be analytically useful. We can further simplify the kinetics by making further adjustments to the reaction conditions [Mottola, H. A. , , 279–287]. For example, we can ensure pseudo-first-order kinetics by using a large excess of so that its concentration remains essentially constant during the time we monitor the reaction. Under these conditions Equation \ref{13.5} simplifies to \[\text { rate }=-\frac{d[A]}{d t}=k_{f}[A,R]_{0}=k^{\prime}[A] \label{13.6}\] where ′ = [ ] . The integrated rate law for Equation \ref{13.6} then is \[\ln{[A]_t} = \ln{[A]_0} - k’t \label{13.7}\] or \[[A]_{t}=[A]_{0} e^{-k^{\prime} t} \label{13.8}\] It may even be possible to adjust the conditions so that we use the reaction under pseudo-zero-order conditions. \[\text { rate }=-\frac{d[A]}{d t}=k_{f}[A]_{0}[R]_{0}=k^{\prime \prime} t \label{13.9}\] \[[A]_{t}=[A]_{0}-k^{\prime \prime} t \label{13.10}\] where \(k^{\prime \prime}\) = [ ] [ ] . To say that the reaction is pseudo-first-order in means the reaction behaves as if it is first order in and zero order in even though the underlying kinetics are more complicated. We call \(k^{\prime}\) a pseudo-first-order rate constant. To say that a reaction is pseudo-zero-order means the reaction behaves as if it is zero order in and zero order in even though the underlying kinetics are more complicated. We call \(k^{\prime \prime}\) the pseudo-zero-order rate constant. The final requirement is that we must be able to monitor the reaction’s progress by following the change in concentration for at least one of its species. Which species we choose to monitor is not important: it can be the analyte, a reagent that reacts with the analyte, or a product. For example, we can determine the concentration of phosphate by first reacting it with Mo(VI) to form 12-molybdophosphoric acid (12-MPA). \[\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+6 \mathrm{Mo}(\mathrm{VI})(a q) \longrightarrow 12-\mathrm{MPA}(a q)+9 \mathrm{H}^{+}(a q) \label{13.11}\] Next, we reduce 12-MPA to heteropolyphosphomolybdenum blue, PMB. The rate of formation of PMB is measured spectrophotometrically, and is proportional to the concentration of 12-MPA. The concentration of 12-MPA, in turn, is proportional to the concentration of phosphate [see, for example, (a) Crouch, S. R.; Malmstadt, H. V. , , 1084–1089; (b) Crouch, S. R.; Malmstadt, H. V. , , 1090–1093; (c) Malmstadt, H. V.; Cordos, E. A.; Delaney, C. J. , , 26A–41A]. We also can follow reaction 13.11 spectrophotometrically by monitoring the formation of the yellow-colored 12-MPA [Javier, A. C.; Crouch, S. R.; Malmstadt, H. V. , , 239–243]. Reaction \ref{13.11} is, of course, unbalanced; the additional hydrogens on the reaction’s right side come from the six Mo(VI) that appear on the reaction’s left side where Mo(VI) is thought to be present as the molybdate dimer HMo O . Figure 13.2.1 provides one useful scheme for classifying chemical kinetic methods of analysis. Methods are divided into two broad categories: direct- computation methods and curve-fitting methods. In a direct-computation method we calculate the analyte’s initial concentration, [ ] , using the appropriate rate law. For example, if the reaction is first-order in analyte, we can use Equation \ref{13.2} to determine [ ] given values for , , and [ ] . With a curve-fitting method, we use regression to find the best fit between the data—for example, [ ] as a function of time—and the known mathematical model for the rate law. If the reaction is first-order in analyte, then we fit Equation \ref{13.2} to the data using and [ ] as adjustable parameters. A direct-computation integral method uses the integrated form of the rate law. In a , for example, we determine the analyte’s concentration at a single time and calculate the analyte’s initial concentration, [ ] , using the appropriate integrated rate law. To determine the reaction’s rate constant, , we run a separate experiment using a standard solution of analyte. Alternatively, we can determine the analyte’s initial concentration by measuring [ ] for several standards that contain known concentrations of analyte and construct a calibration curve. The concentration of nitromethane, CH NO , is determined from the kinetics of its decomposition reaction. In the presence of excess base the reaction is pseudo-first-order in nitromethane. For a standard solution of 0.0100 M nitromethane, the concentration of nitromethane after 2.00 s is \(4.24 \times 10^{-4}\) M. When a sample that contains an unknown amount of nitromethane is analyzed, the concentration of nitromethane remaining after 2.00 s is \(5.35 \times10^{-4}\) M. What is the initial concentration of nitromethane in the sample? First, we determine the value for the pseudo-first-order rate constant, \(k^{\prime}\). Using Equation \ref{13.7} and the result for the standard, we find its value is \[k^{\prime} = \frac {\ln{[A]_0} - \ln{[A]_t}} {t} = \frac {\ln{(0.0100)} - \ln{(4.24 \times 10^{-4})}} {2.00 \text{ s}} = 1.58 \text{ s}^{-1} \nonumber\] Next we use Equation \ref{13.8} to calculate the initial concentration of nitromethane in the sample. \[[A]_0 = \frac {[A]_t} {e^{-k^{\prime}t}} = \frac {5.35 \times 10^{-4} \text{ M}} {e^{-(1.58 \text{ s}^{-1})(2.00 \text{ s})}} = 0.0126 \text{ M} \nonumber\] Equation \ref{13.7} and Equation \ref{13.8} are equally appropriate integrated rate laws for a pseudo-first-order reaction. The decision to use Equation \ref{13.7} to calculate \(k^{\prime}\) and Equation \ref{13.8} to calculate [ ] is a matter of convenience. In a separate determination for nitromethane, a series of external standards gives the following concentrations of nitromethane after a 2.00 s decomposition under pseudo-first-order conditions. Analysis of a sample under the same conditions gives a nitromethane concentration of \(2.21 \times 10^{-3}\) M after 2 s. What is the initial concentration of nitromethane in the sample? The calibration curve and the calibration equation for the external standards are shown below. Substituting \(2.21 \times 10^{-3}\) M for [CH NO ] gives [CH NO ] as \(5.21 \times 10^{-2}\) M. In we determine the analyte’s initial concentration by measuring the amount of analyte that has not reacted. Sometimes it is more convenient to measure the concentration of a reagent that reacts with the analyte, or to measure the concentration of one of the reaction’s products. We can use a one-point fixed-time integral method if we know the reaction’s stoichiometry. For example, if we measure the concentration of the product, , in the reaction \[A+R \rightarrow P \nonumber\] then the concentration of the analyte at time is \[[A]_{t}=[A]_{0}-[P]_{t} \label{13.12}\] because the stoichiometry between the analyte and product is 1:1. If the reaction is pseudo-first-order in , then substituting Equation \ref{13.12} into Equation \ref{13.7} gives \[\ln \left([A]_{0}-[P]_{t}\right) = \ln{[A]_{0}} - k^{\prime} t \label{13.13}\] which we simplify by writing in exponential form. \[[A]_0 - [P]_t = [A]_0 e^{-k^{\prime}t} \label{13.14}\] Finally, solving Equation \ref{13.14} for [ ] gives the following equation. \[[A]_{0}=\frac{[P]_{t}}{1-e^{-k^{\prime}t}} \label{13.15}\] The concentration of thiocyanate, SCN , is determined from the pseudo-first-order kinetics of its reaction with excess Fe to form a reddish-colored complex of Fe(SCN) . The reaction’s progress is monitored by measuring the absorbance of Fe(SCN) at a wavelength of 480 nm. When using a standard solution of 0.100 M SCN , the concentration of Fe(SCN) after 10 s is 0.0516 M. The concentration of Fe(SCN) in a sample that contains an unknown amount of SCN is 0.0420 M after 10 s. What is the initial concentration of SCN in the sample? First, we must determine a value for the pseudo-first-order rate constant, \(k^{\prime}\). Using Equation \ref{13.13}, we find that its value is \[k^{\prime} = \frac{\ln{[A]_{0}} - \ln \left([A]_{0} - [P]_{1}\right)}{t}= \frac {\ln(0.100) - \ln(0.100 - 0.0516)} {10.0 \text{ s}} = 0.0726 \text{ s}^{-1} \nonumber\] Next, we use Equation \ref{13.15} to determine the initial concentration of SCN in the sample. \[[A]_{0}=\frac{[P]_{t}}{1-e^{-k^{\prime} t}}=\frac{0.0420 \mathrm{M}}{1-e^{-\left(0.0726 \text{ s}^{-1}\right)(10.0 \text{ s})}}=0.0868 \mathrm{M} \nonumber\] In a separate determination for SCN , a series of external standards gives the following concentrations of Fe(SCN) after a 10.0 s reaction with excess Fe under pseudo-first-order conditions. Analysis of a sample under the same conditions gives an Fe(SCN) concentration of \(3.52 \times 10^{-2}\) M after 10 s. What is the initial concentration of SCN in the sample? The calibration curve and the calibration equation for the external standards are shown below. Substituting \(3.52 \times 10^{-2}\) M for [Fe(SCN) ] gives [SCN ] as \(6.87 \times 10^{-2}\) M. A one-point fixed-time integral method has the advantage of simplicity because we need only a single measurement to determine the analyte’s initial concentration. As with any method that relies on a single determination, a one-point fixed-time integral method can not compensate for a constant determinate error. In a we correct for constant determinate errors by making measurements at two points in time and use the difference between the measurements to determine the analyte’s initial concentration. Because it affects both measurements equally, the difference between the measurements is independent of a constant de- terminate error. For a pseudo-first-order reaction in which we measure the analyte’s concentration at times and , we can write the following two equations. \[[A]_{t_{1}}=[A]_{0} e^{-k^{\prime} t_1} \label{13.16}\] \[[A]_{t_{2}}=[A]_{0} e^{-k^{\prime} t_2} \label{13.17}\] Subtracting Equation \ref{13.17} from Equation \ref{13.16} and solving for [ ] leaves us with \[[A]_{0}=\frac{[A]_{t_1}-[A]_{t_2}}{e^{-k^{\prime} t_{1}}-e^{-k^{\prime} t_{2}}} \label{13.18}\] To determine the rate constant, \(k^{\prime}\), we measure \([A]_{t_1}\) and \([A]_{t_2}\) for a standard solution of analyte. Having obtained a value for \(k^{\prime}\), we can determine [ ] by measuring the analyte’s concentration at and . We also can determine the analyte’s initial concentration using a calibration curve consisting of a plot of (\([A]_{t_1}\) – \([A]_{t_2}\)) versus [ ] . A fixed-time integral method is particularly useful when the signal is a linear function of concentration because we can replace the reactant’s concentration with the corresponding signal. For example, if we follow a reaction spectrophotometrically under conditions where the analyte’s concentration obeys Beer’s law \[(A b s)_{t}=\varepsilon b[A]_{t} \nonumber\] then we can rewrite Equation \ref{13.8} and Equation \ref{13.18} as \[(A b s)_{t}=[A]_{0} e^{-k^{\prime}} \varepsilon b=c[A]_{0} \nonumber\] \[[A]_t = \frac {(Abs)_{t_1} - (Abs)_{t_2}} {e^{-k^{\prime}t_1} - e^{-k^{\prime}t_2}} \times (\epsilon b)^{-1} = c^{\prime}[(Abs)_{t_1} - (Abs)_{t_2}] \nonumber\] where ( ) is the absorbance at time , and and \(c^{\prime}\) are constants. In a we measure the total time, \(\Delta_t\), needed to effect a specific change in concentration for one species in the chemical reaction. One important application is the quantitative analysis of catalysts, which takes advantage of the catalyst’s ability to increase the rate of reaction. As the concentration of catalyst increased, \(\Delta_t\) decreases. For many catalytic systems the relationship between \(\Delta_t\) and the catalyst’s concentration is \[\frac {1} {\Delta t} = F_{cat}[A]_0 + F_{uncat} \label{13.19}\] where [ ] is the catalyst’s concentration, and and are constants that account for the rate of the catalyzed and uncatalyzed reactions [Mark, H. B.; Rechnitz, G. A. , Interscience: New York, 1968]. Sandell and Kolthoff developed a quantitative method for iodide based on its ability to catalyze the following redox reaction [Sandell, E. B.; Kolthoff, I. M. , , 1426]. \[\mathrm{As}^{3+}(a q)+2 \mathrm{Ce}^{4+}(a q) \longrightarrow \mathrm{As}^{\mathrm{5+}}(a q)+2 \mathrm{Ce}^{3+}(a q) \nonumber\] An external standards calibration curve was prepared by adding 1 mL of a KI standard to a mixture of 2 mL of 0.05 M As , 1 mL of 0.1 M Ce , and 1 mL of 3 M H SO , and measuring the time for the yellow color of Ce to disappear. The following table summarizes the results for one analysis. What is the concentration of I in a sample if \(\Delta_t\) is 3.2 min? Figure 13.2.2 shows the calibration curve and the calibration equation for the external standards based on Equation \ref{13.19}. Substituting 3.2 min for \(\Delta_t\) gives the concentration of I in the sample as 1.4 μg/mL. In a we use the differential form of the rate law—Equation \ref{13.1} is one example of a differential rate law—to determine the analyte’s concentration. As shown in Figure 13.2.3 , the rate of a reaction at time , ( ) , is the slope of a line tangent to a curve that shows the change in concentration as a function of time. For a reaction that is first-order in analyte, the rate at time is \[(r a t e)_{t}=k[A]_{t} \nonumber\] Substituting in Equation \ref{13.3} leaves us with the following equation relating the rate at time to the analyte’s initial concentration. \[(\text {rate})_{t}=k[A]_{0} e^{-k t} \nonumber\] If we measure the rate at a fixed time, then both and are constant and we can use a calibration curve of ( ) versus [ ] for the quantitative analysis of the analyte. There are several advantages to using the reaction’s ( = 0). First, because the reaction’s rate decreases over time, the initial rate provides the greatest sensitivity. Second, because the initial rate is measured under nearly pseudo-zero-order conditions, in which the change in concentration with time effectively is linear, it is easier to determine the slope. Finally, as the reaction of interest progresses competing reactions may develop, which complicating the kinetics: using the initial rate eliminates these complications. One disadvantage of the initial rate method is that there may be insufficient time to completely mix the reactants. This problem is avoided by using an intermediate rate measured at a later time ( > 0). As a general rule (see Mottola, H. A. “Kinetic Determinations of Reactants Utilizing Uncatalyzed Reactions,” , , 279–287), the time for measuring a reaction’s initial rate should result in the consumption of no more than 2% of the reactants. The smaller this percentage, the more linear the change in concentration as a function of time. The concentration of norfloxacin, a commonly prescribed antibacterial agent, is determined using the initial rate method. Norfloxacin is converted to an N-vinylpiperazine derivative and reacted with 2,3,5,6-tetra-chloro-1,4-benzoquinone to form an N-vinylpiperazino-substituted ben-zoquinone derivative that absorbs strongly at 625 nm [Darwish, I. A.; Sultan, M. A.; Al-Arfaj, H. A. , , 1383–1388]. The initial rate of the reaction—as measured by the change in absorbance as a function of time (AU/min)—is pseudo-first order in norfloxacin. The following data were obtained for a series of external norfloxacin standards. To analyze a sample of prescription eye drops, a 10.00-mL portion is extracted with dichloromethane. The extract is dried and the norfloxacin reconstituted in methanol and diluted to 10 mL in a volumetric flask. A 5.00-mL portion of this solution is diluted to volume in a 100-mL volumetric flask. Analysis of this sample gives an initial rate of 0.0394 AU/min. What is the concentration of norfloxacin in the eye drops in mg/mL? Figure 13.2.4 shows the calibration curve and the calibration equation for the external standards. Substituting 0.0394 AU/min for the initial rate and solving for the concentration of norfloxacin gives a result of 152 μg/mL. This is the concentration in a diluted sample of the extract. The concentration in the extract before dilution is \[\frac{152 \: \mu \text{g}}{\mathrm{mL}} \times \frac{100.0 \: \mathrm{mL}}{5.00 \: \mathrm{mL}} \times \frac{1 \:\mathrm{mg}}{1000 \: \mu \mathrm{g}}=3.04 \: \mathrm{mg} / \mathrm{mL} \nonumber\] Because the dried extract was reconstituted using a volume identical to that of the original sample, the concentration of norfloxacin in the eye drops is 3.04 mg/mL. In a direct-computation method we determine the analyte’s concentration by solving the appropriate rate equation at one or two discrete times. The relationship between the analyte’s concentration and the measured response is a function of the rate constant, which we determine in a separate experiment using a single external standard (see or ), or a calibration curve (see or ). In a we continuously monitor the concentration of a reactant or a product as a function of time and use a regression analysis to fit the data to an appropriate differential rate law or integrated rate law. For example, if we are monitoring the concentration of a product for a reaction that is pseudo-first-order in the analyte, then we can fit the data to the following rearranged form of Equation \ref{13.15} \[[P]_{t}=[A]_{0}\left(1-e^{-k^{\prime} t}\right) \nonumber\] using [ ] and \(k^{\prime}\) as adjustable parameters. Because we use data from more than one or two discrete times, a curve-fitting method is capable of producing more reliable results. The data shown in the following table were collected for a reaction that is known to be pseudo-zero-order in analyte. What is the initial concentration of analyte in the sample and the rate constant for the reaction? From Equation \ref{13.10} we know that for a pseudo-zero-order reaction a plot of [ ] versus time is linear with a slope of \(-k^{\prime \prime}\) and a -intercept of [ ] . Figure 13.2.5 shows a plot of the kinetic data and the result of a linear regression analysis. The initial concentration of analyte is 0.0986 mM and the rate constant is 0.00677 M s . The best way to appreciate the theoretical and the practical details discussed in this section is to carefully examine a typical analytical method. Although each method is unique, the following description of the determination of creatinine in urine provides an instructive example of a typical procedure. The description here is based on Diamandis, E. P.; Koupparis, M. A.; Hadjiioannou, T. P. “Kinetic Studies with Ion Selective Electrodes: Determination of Creatinine in Urine with a Picrate Ion Selective Electrode,” , , 74–76. Creatine is an organic acid in muscle tissue that supplies energy for muscle contractions. One of its metabolic products is creatinine, which is excreted in urine. Because the concentration of creatinine in urine and serum is an important indication of renal function, a rapid method for its analysis is clinically important. In this method the rate of reaction between creatinine and picrate in an alkaline medium is used to determine the concentration of creatinine in urine. Under the conditions of the analysis the reaction is first order in picrate, creatinine, and hydroxide. \[\text { rate }=k[\text { picrate },\text { creatinine }]\left[\mathrm{OH}^{-}\right] \nonumber\] The reaction is monitored using a picrate ion selective electrode. Prepare a set of external standards that contain 0.5–3.0 g/L creatinine using a stock solution of 10.00 g/L creatinine in 5 mM H SO , diluting each standard to volume using 5 mM H SO . Prepare a solution of \(1.00 \times 10^{-2}\) M sodium picrate. Pipet 25.00 mL of 0.20 M NaOH, adjusted to an ionic strength of 1.00 M using Na SO , into a thermostated reaction cell at 25 C. Add 0.500 mL of the \(1.00 \times 10^{-2}\) M picrate solution to the reaction cell. Suspend a picrate ion selective in the solution and monitor the potential until it stabilizes. When the potential is stable, add 2.00 mL of a creatinine external standard and record the potential as a function of time. Repeat this procedure using the remaining external standards. Construct a calibration curve of \(\Delta E / \Delta t\) versus the initial concentration of creatinine. Use the same procedure to analyze samples, using 2.00 mL of urine in place of the external standard. Determine the concentration of creatinine in the sample using the calibration curve. 1. The analysis is carried out under conditions that are pseudo-first order in picrate. Show that under these conditions the change in potential as a function of time is linear. The potential, , of the picrate ion selective electrode is given by the Nernst equation \[E=K-\frac{R T}{F} \ln{[\text { picrate }]} \nonumber\] where is a constant that accounts for the reference electrodes, the junction potentials, and the ion selective electrode’s asymmetry potential, is the gas constant, is the temperature, and is Faraday’s constant. We know from Equation \ref{13.7} that for a pseudo-first-order reaction, the concentration of picrate at time is \[\ln {[\text{picrate}]_t}=\ln{[\text {picrate}]}_{0}-k^{\prime} t \nonumber\] where \(k^{\prime}\) is the pseudo-first-order rate constant. Substituting this integrated rate law into the ion selective electrode’s Nernst equation leaves us with the following result. \[E_{t} = K - \frac{R T} {F} \left( \ln{[\text {picrate}]}_{0} - k^{\prime} t\right) \nonumber\] \[E_{t} = K - \frac{R T} {F} \ln{[\text {picrate}]}_{0} + \frac{R T} {F} k^{\prime}t \nonumber\] Because and ( / )ln[picrate] are constants, a plot of E versus is a straight line with a slope of \(\frac{R T} {F} k^{\prime}\). 2. Under the conditions of the analysis, the rate of the reaction is pseudo-first-order in picrate and pseudo-zero-order in creatinine and OH . Explain why it is possible to prepare a calibration curve of \(\Delta E / \Delta t\) versus the concentration of creatinine. The slope of a plot of versus is \(\Delta E / \Delta t = RTk^{\prime}/F\) = ′/ (see the previous question). Because the reaction is carried out under conditions where it is pseudo-zero-order in creatinine and OH , the rate law is \[\text{rate} = k[\text{picrate},\text{creatinine}]_0[\text{OH}^-]_0 = k^{\prime}[\text{picrate}] \nonumber\] The pseudo-first-order rate constant, \(k^{\prime}\), is \[k^{\prime}=k[\text { creatinine }]_{0}\left[\mathrm{OH}^{-}\right]_{0}=c[\text {creatinine}]_{0} \nonumber\] where is a constant equivalent to [OH ] . The slope of a plot of versus , therefore, is linear function of creatinine’s initial concentration \[\frac{\Delta E}{\Delta t}=\frac{R T k^{\prime}}{F}=\frac{R T c}{F}[\text {creatinine}]_{0} \nonumber\] and a plot of \(\Delta E / \Delta t\) versus the concentration of creatinine can serve as a calibration curve. 3. Why is it necessary to thermostat the reaction cell? The rate of a reaction is temperature-dependent. The reaction cell is thermostated to maintain a constant temperature to prevent a determinate error from a systematic change in temperature, and to minimize indeterminate errors from random fluctuations in temperature. 4. Why is it necessary to prepare the NaOH solution so that it has an ionic strength of 1.00 M? The potential of the picrate ion selective electrode actually responds to the activity of the picrate anion in solution. By adjusting the NaOH solution to a high ionic strength we maintain a constant ionic strength in all standards and samples. Because the relationship between activity and concentration is a function of ionic strength, the use of a constant ionic strength allows us to write the Nernst equation in terms of picrate’s concentration instead of its activity. When using to determine the concentration of creatinine in urine, we follow the reactions kinetics using an ion selective electrode. In principle, we can use any of the analytical techniques in Chapters 8–12 to follow a reaction’s kinetics provided that the reaction does not proceed to an appreciable extent during the time it takes to make a measurement. As you might expect, this requirement places a serious limitation on kinetic methods of analysis. If the reaction’s kinetics are slow relative to the analysis time, then we can make a measurement without the analyte undergoing a significant change in concentration. If the reaction’s rate is too fast—which often is the case—then we introduce a significant error if our analysis time is too long. One solution to this problem is to stop, or quench the reaction by adjusting experimental conditions. For example, many reactions show a strong dependence on pH and are quenched by adding a strong acid or a strong base. Figure 13.2.6 shows a typical example for the enzymatic analysis of -nitrophenylphosphate, which uses the enzyme wheat germ acid phosphatase to hydrolyze the analyte to -nitrophenol. The reaction has a maximum rate at a pH of 5. Increasing the pH by adding NaOH quenches the reaction and converts the colorless -nitrophenol to the yellow-colored -nitrophenolate, which absorbs at 405 nm. An additional problem when the reaction’s kinetics are fast is ensuring that we rapidly and reproducibly mix the sample and the reagents. For a fast reaction, we need to make our measurements within a few seconds—or even a few milliseconds—of combining the sample and reagents. This presents us with a problem and an advantage. The problem is that rapidly and reproducibly mixing the sample and the reagent requires a dedicated instrument, which adds an additional expense to the analysis. The advantage is that a rapid, automated analysis allows for a high throughput of samples. Instruments for the automated kinetic analysis of phosphate using reaction \ref{13.11}, for example, have sampling rates of approximately 3000 determinations per hour. A variety of instruments have been developed to automate the kinetic analysis of fast reactions. One example, which is shown in Figure 13.2.7 , is the . The sample and the reagents are loaded into separate syringes and precisely measured volumes are dispensed into a mixing chamber by the action of a syringe drive. The continued action of the syringe drive pushes the mixture through an observation cell and into a stopping syringe. The back pressure generated when the stopping syringe hits the stopping block completes the mixing, after which the reaction’s progress is monitored spectrophotometrically. With a stopped-flow analyzer it is possible to complete the mixing of sample and reagent, and initiate the kinetic measurements in approximately 0.5 ms. By attaching an autosampler to the sample syringe it is possible to analyze up to several hundred samples per hour. Another instrument for kinetic measurements is the , a partial cross section of which is shown in Figure 13.2.8 . The sample and the reagents are placed in separate wells, which are oriented radially around a circular transfer disk. As the centrifuge spins, the centrifugal force pulls the sample and the reagents into the cuvette where mixing occurs. A single optical source and detector, located below and above the transfer disk’s outer edge, measures the absorbance each time the cuvette passes through the optical beam. When using a transfer disk with 30 cuvettes and rotating at 600 rpm, we can collect 10 data points per second for each sample. The ability to collect lots of data and to collect it quickly requires appropriate hardware and software. Not surprisingly, automated kinetic analyzers developed in parallel with advances in analog and digital circuitry—the hardware—and computer software for smoothing, integrating, and differentiating the analytical signal. For an early discussion of the importance of hardware and software, see Malmstadt, H. V.; Delaney, C. J.; Cordos, E. A. “Instruments for Rate Determinations,” , , 79A–89A. Chemical kinetic methods of analysis continue to find use for the analysis of a variety of analytes, most notably in clinical laboratories where automated methods aid in handling the large volume of samples. In this section we consider several general quantitative applications. are highly specific catalysts for biochemical reactions, with each enzyme showing a selectivity for a single reactant, or . For example, the enzyme acetylcholinesterase catalyzes the decomposition of the neurotransmitter acetylcholine to choline and acetic acid. Many enzyme–substrate reactions follow a simple mechanism that consists of the initial formation of an enzyme–substrate complex, , which subsequently decomposes to form product, releasing the enzyme to react again. \[E + S \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} ES \underset{k_{-2}}{\stackrel{k_2}{\rightleftharpoons}} E + P \label{13.20}\] where , , , and are rate constants. If we make measurement early in the reaction, the concentration of products is negligible and we can ignore the step described by the rate constant . Under these conditions the reaction’s rate is \[\text { rate }=\frac{d[P]}{d t}=k_{2}[E S] \label{13.21}\] To be analytically useful we need to write Equation \ref{13.21} in terms of the concentrations of the enzyme, , and the substrate, . To do this we use the steady-state approximation, in which we assume the concentration of remains essentially constant. Following an initial period, during which the enzyme–substrate complex first forms, the rate at which forms \[\frac{d[E S]}{d t}=k_{1}[E,S]=k_{1}\left([E]_{0}-[E S]\right)[S] \label{13.22}\] is equal to the rate at which it disappears \[-\frac{d[E S]}{d t}=k_{-1}[E S]+k_{2}[E S] \label{13.23}\] where [ ] is the enzyme’s original concentration. Combining Equation \ref{13.22} and Equation \ref{13.23} gives \[k_{1}\left([E]_{0}-[E S]\right)[S]=k_{-1}[E S]+k_{2}[E S] \nonumber\] which we solve for the concentration of the enzyme–substrate complex \[[E S]=\frac{[E]_{0}[S]}{\frac{k_{-1}+k_{2}}{k_{1}}+[S]}=\frac{[E]_{0}[S]}{K_{m}+[S]} \label{13.24}\] where is the . Substituting Equation \ref{13.24} into Equation \ref{13.21} leaves us with our final rate equation. \[\frac{d[P]}{d t}=\frac{k_{2}[E]_{0}[S]}{K_{m}+[S]} \label{13.25}\] A plot of Equation \ref{13.25}, as shown in Figure 13.2.9 , helps us define conditions where we can use the rate of an enzymatic reaction for the quantitative analysis of an enzyme or a substrate. For high substrate concentrations, where [ ] >> , Equation \ref{13.25} simplifies to \[\frac{d[P]}{d t}=\frac{k_{2}[E]_{0}[S]}{K_{m}+[S]} \approx \frac{k_{2}[E]_{0}[S]}{[S]}=k_{2}[E]_{0}=V_{\max } \label{13.26}\] where is the maximum rate for the catalyzed reaction. Under these conditions the reaction is pseudo-zero-order in substrate, and we can use to calculate the enzyme’s concentration, typically using a variable-time method. At lower substrate concentrations, where [ ] << , Equation \ref{13.25} becomes \[\frac{d[P]}{d t}=\frac{k_{2}[E]_{0}[S]}{K_{m}+[S]} \approx \frac{k_{2}[E]_{0}[S]}{K_{m}}=\frac{V_{\max }[S]}{K_{m}} \label{13.27}\] Because the reaction is first-order in substrate we can use the reaction’s rate to determine the substrate’s concentration using a fixed-time method. Chemical kinetic methods have been applied to the quantitative analysis of a number of enzymes and substrates [Guilbault, G. G. Handbook of Enzymatic Methods of Analysis, Marcel Dekker: New York, 1976]. One example, is the determination of glucose based on its oxidation by the enzyme glucose oxidase \[\text{glucose}(aq) + \text{H}_2\text{O}(g) \xrightarrow{\text{glucose oxidase}} \text{gluconolactone}(aq) + \text{H}_2\text{O}_2(aq) \nonumber\] under conditions where Equation \ref{13.20} is valid. The reaction is monitored by following the rate of change in the concentration of dissolved O using an appropriate voltammetric technique. One method for measuring the concentration of dissolved O is the Clark amperometric sensor described in . The variable-time method also is used to determine the concentration of nonenzymatic catalysts. One example uses the reduction of H O by thiosulfate, iodide, or hydroquinone, a reaction catalyzed by trace amounts of selected metal ions. For example the reduction of H O by I \[2 \mathrm{I}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{H}_{3} \mathrm{O}^{+}(a q) \longrightarrow 4 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{I}_{2}(a q) \nonumber\] is catalyzed by Mo(VI), W(VI), and Zr(IV). A variable-time analysis is conducted by adding a small, fixed amount of ascorbic acid to each solution. As I is produced it rapidly oxidizes the ascorbic acid and is reduced back to I . Once all the ascorbic acid is consumed, the presence of excess I provides a visual endpoint. Chemical kinetic methods are not as common for the quantitative analysis of analytes in noncatalytic reactions. Because they lack the enhancement of reaction rate that a catalyst affords, a noncatalytic method generally is not useful for determining small concentrations of analyte. Noncatalytic methods for inorganic analytes usually are based on a complexation reaction. One example is the determination of aluminum in serum by measuring the initial rate for the formation of its complex with 2-hydroxy-1-napthaldehyde -methoxybenzoyl-hydrazone [Ioannou. P. C.; Piperaki, E. A. , , 1481–1483]. The greatest number of noncatalytic methods, however, are for the quantitative analysis of organic analytes. For example, the insecticide methyl parathion has been determined by measuring its rate of hydrolysis in alkaline solutions [Cruces Blanco, C.; Garcia Sanchez, F. , , 513–523]. Chemical kinetic methods also find use in determining rate constants and in elucidating reaction mechanisms. Two examples from the kinetic analysis of enzymes illustrate these applications. The value of and for an enzymatic reaction are of significant interest in the study of cellular chemistry. For an enzyme that follows the mechanism in reaction \ref{13.20}, is equivalent to 2 \(\times\) [ ] , where [ ] is the enzyme’s concentration and is the enzyme’s turnover number. An enzyme’s turnover number is the maximum number of substrate molecules converted to product by a single active site on the enzyme, per unit time. A turnover number, therefore, provides a direct indication of the active site’s catalytic efficiency. The Michaelis constant, , is significant because it provides an estimate of the substrate’s intracellular concentration [(a) Northup, D. B. , , 1153–1157; (b) Zubay, G. , Macmillan Publishing Co.: New York, 2nd Ed., p 269]. An enzyme’s turnover number also is know as and is equal to /[ ] . For the mechanism in reaction \ref{13.20}, is equivalent to . For more complicated mechanisms, is a function of additional rate constants. As shown in , we can find values for and by measuring the reaction’s rate for small and for large concentrations of the substrate. Unfortunately, this is not always practical as the substrate’s limited solubility may prevent us from using the large substrate concentrations needed to determine . Another approach is to rewrite Equation \ref{13.25} by taking its reciprocal \ [\frac{1}{d[P] / d t}=\frac{1}{v}=\frac{K_{m}}{V_{\max }} \times \frac{1}{[S]}+\frac{1}{V_{\max }} \label{13.28}\] where is the reaction’s rate. As shown in Figure 13.2.10 , a plot of 1/ versus 1/[S], which is called a double reciprocal or Lineweaver–Burk plot, is a straight line with a slope of / , a -intercept of 1/ , and an -intercept of –1/ . In we noted that when faced with a nonlinear model—and Equation \ref{13.25} is one example of a nonlinear model—it may be possible to rewrite the equation in a linear form. This is the strategy used here. Linearizing a nonlinear model is not without limitations, two of which deserve a brief mention. First, because we are unlikely to have data for large substrate concentrations, we will not have many data points for small values of 1/[S]. As a result, our determination of the -intercept’s value relies on a significant extrapolation. Second, taking the reciprocal of the rate distorts the experimental error in a way that may invalidate the assumptions of a linear regression. Nonlinear regression provides a more rigorous method for fitting Equation \ref{13.25} to experimental data. The details are beyond the level of this textbook, but you may consult Massart, D. L.; Vandeginste, B. G. M.; Buydens, L. M. C. De Jong, S.; Lewi, P. J.; Smeyers-Verbeke, J. “Nonlinear Regression,” which is Chapter 11 in , Elsevier: Amsterdam, 1997, for additional details. The simplex algorithm described in of this text also can be used to fit a nonlinear equation to experimental data. The reaction between nicotineamide mononucleotide and ATP to form nicotineamide–adenine dinucleotide and pyrophosphate is catalyzed by the enzyme nicotinamide mononucleotide adenylyltransferase [(a) Atkinson, M. R.; Jackson, J. F.; Morton, R. K. , , 318–323; (b) Wilkinson, G. N. , , 324–332]. The following table provides typical data obtained at a pH of 4.95. The substrate, , is nicotinamide mononucleotide and the initial rate, , is the μmol of nicotinamide–adenine dinucleotide formed in a 3-min reaction period. Determine values for and . Figure 13.2.11 shows the Lineweaver–Burk plot for this data and the result-ng regression equation. Using the -intercept, we calculate as \[V_{\max }=\frac{1}{y\text {-intercept }}=\frac{1}{1.708 \: \mu \mathrm{mol}^{-1}}=0.585 \: \mu \mathrm{mol} \nonumber\] and using the slope we find that is \[K_{m} = \text {slope} \times V_{\max}=0.7528 \: \mu \mathrm{mol}^{-1} \mathrm{mM} \times 0.585 \: \mu \mathrm{mol}=0.440 \mathrm{ mM} \nonumber\] The following data were collected during the oxidation of catechol (the substrate) to -quinone by the enzyme -diphenyl oxidase. The reaction was followed by monitoring the change in absorbance at 540 nm. The data in this exercise is adapted from . The figure below shows the Lineweaver–Burk plot and the equation for the data. The -intercept of 9.974 min/(\(\Delta\)AU is equivalent to 1/ ; thus, is 0.10 \(\Delta\) AU/min. The slope of 11.89 min/(\Delta\)AU•mM is equivalent to / ; thus, is 1.2 mM. When an interacts with an enzyme it decreases the enzyme’s catalytic efficiency. An irreversible inhibitor binds covalently to the enzyme’s active site, producing a permanent loss in catalytic efficiency even if we decrease the inhibitor’s concentration. A reversible inhibitor forms a noncovalent complex with the enzyme, resulting in a temporary decrease in catalytic efficiency. If we remove the inhibitor, the enzyme’s catalytic efficiency returns to its normal level. There are several pathways for the reversible binding of an inhibitor and an enzyme, as shown in Figure 13.2.12 . In the substrate and the inhibitor compete for the same active site on the enzyme. Because the substrate cannot bind to an enzyme–inhibitor complex, , the enzyme’s catalytic efficiency for the substrate decreases. With the substrate and the inhibitor bind to different active sites on the enzyme, forming an enzyme–substrate–inhibitor, or complex. The formation of an complex decreases catalytic efficiency because only the enzyme–substrate complex reacts to form the product. Finally, in the inhibitor binds to the enzyme–substrate complex, forming an inactive complex. We can identify the type of reversible inhibition by observing how a change in the inhibitor’s concentration affects the relationship between the rate of reaction and the substrate’s concentration. As shown in Figure 13.2.13 , when we display kinetic data using as a Lineweaver-Burk plot it is easy to determine which mechanism is in effect. For example, an increase in slope, a decrease in the -intercept, and no change in the -intercept indicates competitive inhibition. Because the inhibitor’s binding is reversible, we can still obtain the same maximum velocity—thus the constant value for the -intercept—by adding enough substrate to completely displace the inhibitor. Because it takes more substrate, the value of increases, which explains the increase in the slope and the decrease in the -intercept’s value. provides kinetic data for the oxidation of catechol (the substrate) to -quinone by the enzyme -diphenyl oxidase in the absence of an inhibitor. The following additional data are available when the reaction is run in the presence of -hydroxybenzoic acid, PBHA. Is PBHA an inhibitor for this reaction and, if so, what type of inhibitor is it? The data in this exercise are adapted from . Figure 13.2.14 shows the resulting Lineweaver–Burk plot for the data in Exercise 13.2.3 and Example 13.2.7 . Although the two -intercepts are not identical in value—the result of uncertainty in measuring the rates—the plot suggests that PBHA is a competitive inhibitor for the enzyme’s reaction with catechol. Exercise 13.2.3 provides kinetic data for the oxidation of catechol (the substrate) to -quinone by the enzyme -diphenyl oxidase in the absence of an inhibitor. The following additional data are available when the reaction is run in the presence of phenylthiourea. Is phenylthiourea an inhibitor for this reaction and, if so, what type of inhibitor is it? The data in this exercise are adapted from . The figure below shows the Lineweaver–Burk plots for the two sets of data. The nearly identical -intercepts suggests that phenylthiourea is a noncompetitive inhibitor. The detection limit for a chemical kinetic method ranges from minor components to ultratrace components, and is determined by two factors: the rate of the reaction and the instrumental technique used to monitor the rate. Because the signal is directly proportional to the reaction’s rate, a faster reaction generally results in a lower detection limit. All other factors being equal, detection limits are smaller for catalytic reactions than for noncatalytic reactions. Not surprisingly, some of the earliest chemical kinetic methods took advantage of catalytic reactions. For example, ultratrace levels of Cu (<1 ppb) are determined by measuring its catalytic effect on the redox reaction between hydroquinone and H O . In the absence of a catalyst, most chemical kinetic methods for organic compounds use reactions with relatively slow rates, which limits the analysis to minor and to higher concentration trace analytes. Noncatalytic chemical kinetic methods for inorganic compounds that use metal–ligand complexation reactions may be fast or slow, with detection limits ranging from trace to minor analyte. The second factor that influences a method’s detection limit is the instrumentation used to monitor the reaction’s progress. Most reactions are monitored spectrophotometrically or electrochemically. The scale of operation for these techniques are discussed in Chapter 10 and Chapter 11. As noted earlier, a chemical kinetic method potentially is subject to larger errors than an equilibrium method due to the effect of uncontrolled or poorly controlled variables, such as temperature or pH. Although a direct-computation chemical kinetic method can achieve moderately accurate results (a relative error of 1–5%), the accuracy often is much worse. Curve-fitting methods provide significant improvements in accuracy because they use more data. In one study, for example, accuracy was improved by two orders of magnitude—from errors of 500% to 5%—by replacing a direct-computation analysis with a curve-fitting analysis [Pauch, J. B.; Margerum, D. W. , , 226–232]. Although not discussed in this chapter, data analysis methods that include the ability to compensate for experimental errors can lead to a significant improvement in accuracy [(a) Holler, F. J.; Calhoun, R. K.; MClanahan, S. F. , , 755–761; (b) Wentzel, P. D.; Crouch, S. R. , , 2851–2855; (c) Wentzel, P. D.; Crouch, S. R. , , 2855–2858]. The precision of a chemical kinetic method is limited by the signal-to-noise ratio of the instrumentation used to monitor the reaction’s progress. When using an integral method, a precision of 1–2% is routinely possible. The precision for a differential method may be somewhat poorer, particularly if the signal is noisy. We can improve the sensitivity of a one-point fixed-time integral method by making measurements under conditions where the concentration of the monitored species is as large as possible. When monitoring the analyte’s concentration—or the concentration of any other reactant—we want to take measurements early in the reaction before its concentration decreases. On the other hand, if we choose to monitor one of the reaction’s products, then it is better to take measurements at longer times. For a two-point fixed-time integral method, we can improve sensitivity by increasing the difference between times and . As discussed earlier, the sensitivity of a rate method improves when we choose to measure the initial rate. The analysis of closely related compounds, as discussed in earlier chapters, often is complicated by their tendency to interfere with each other. To overcome this problem we usually need to separate the analyte and the interferent before completing the analysis. One advantage of a chemical kinetic method is that it often is possible adjust the reaction conditions so that the analyte and the interferent have different reaction rates. If the difference in their respective rates is large enough, then one species will react completely before the other species has a chance to react. The need to analyze multiple analytes in complex mixtures is, of course, one of the advantages of the separation techniques covered in . Kinetic techniques provide an alternative approach for simple mixtures. We can use the appropriate integrated rate laws to find the conditions necessary to separate a faster reacting species from a more slowly reacting species. Let’s consider a system that consists of an analyte, , and an interferent, , both of which show first-order kinetics with a common reagent. To avoid an interference, the relative magnitudes of their rate constants must be sufficiently different. The fractions, , of and that remain at any point in time, , are defined by the following equations \[\left(f_{A}\right)_{t}=\frac{[A]_{t}}{[A]_{0}} \label{13.29}\] \[\left(f_{B}\right)_{t}=\frac{[B]_{t}}{[B]_{0}} \label{13.30}\] where [ ] and [ ] are the initial concentrations of and , respectively. Rearranging Equation \ref{13.2} and substituting in Equation \ref{13.29} or Equation \ref{13.30} leaves use with the following two equations. \[\ln \frac{[A]_{t}}{[A]_{0}}=\ln \left(f_{A}\right)_{t}=-k_{A} t \label{13.31}\] \[\ln \frac{[B]_{t}}{[B]_{0}}=\ln \left(f_{B}\right)_{t}=-k_{B} t \label{13.32}\] where and are the rate constants for and for . Dividing Equation \ref{13.31} by Equation \ref{13.32} leave us with \[\frac{k_{A}}{k_{B}}=\frac{\ln \left(f_{\mathcal{A}}\right)_{t}}{\ln \left(f_{B}\right)_{t}} \nonumber\] Suppose we want 99% of to react before 1% of reacts. The fraction of that remains is 0.01 and the fraction of that remains is 0.99, which requires that \[\frac{k_{A}}{k_{B}}=\frac{\ln \left(f_{A}\right)_{t}}{\ln \left(f_{B}\right)_{t}}=\frac{\ln (0.01)}{\ln (0.99)}=460 \nonumber\] the rate constant for must be at least 460 times larger than that for . When this condition is met we can determine the analyte’s concentration before the interferent begins to react. If the analyte has the slower reaction, then we can determine its concentration after we allow the interferent to react to completion. This method of adjusting reaction rates is useful if we need to analyze an analyte in the presence of an interferent, but is impractical if both and are analytes because the condition that favors the analysis of will not favor the analysis of . For example, if we adjust conditions so that 99% of reacts in 5 s, then 99% of must react within 0.01 s if it has the faster kinetics, or in 2300 s if it has the slower kinetics. The reaction of is too fast or too slow to make this a useful analytical method. What do we do if the difference in the rate constants for and are not significantly different? We still can complete an analysis if we can simultaneously monitor both species. Because both and react at the same time, the integrated form of the first-order rate law becomes \[C_{t}=[A]_{t}+[B]_{t}=[A]_{0} e^{-k_{A}t}+[B]_{0} e^{-k_{B}t} \label{13.33}\] where is the total concentration of and at time, . If we measure at times and , we can solve the resulting pair of simultaneous equations to determine values [ ] and [ ] . The rate constants and are determined in separate experiments using standard solutions of and . Equation \ref{13.33} can also serve as the basis for a curve-fitting method. As shown in Figure 13.2.15 , a plot of ln( ) as a function of time consists of two regions. At shorter times the plot is curved because and react simultaneously. At later times, however, the concentration of the faster reacting component, , decreases to zero, and Equation \ref{13.33} simplifies to \[C_{t} \approx[B]_{t}=[B]_{0} e^{-k_{B}t} \nonumber\] Under these conditions, a plot of ln( ) versus time is linear. Extrapolating the linear portion to = 0 gives [ ] , with [ ] determined by difference. Use the data in Figure 13.2.15 to determine the concentrations of and in the original sample. Extrapolating the linear part of the curve back to = 0 gives ln[ ] as –2.3, or a [ ] of 0.10 M. At = 0, ln[ ] is –1.2, which corresponds to a [ ] of 0.30 M. Because [ ] = [ ] + [ ] , the concentration of in the original sample is 0.20 M. An automated chemical kinetic method of analysis provides a rapid means for analyzing samples, with throughputs ranging from several hundred to several thousand determinations per hour. The initial start-up costs may be fairly high because an automated analysis requires a dedicated instrument designed to meet the specific needs of the analysis. When measurements are handled manually, a chemical kinetic method requires routinely available equipment and instrumentation, although the sample throughput is much lower than with an automated method.

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Write the chemical formula and Lewis structure of the following, each of which contains five carbon atoms: There are several sets of answers; one is: (a) C H (b) C H (c) C H What is the difference between the hybridization of carbon atoms’ valence orbitals in saturated and unsaturated hydrocarbons? On a microscopic level, how does the reaction of bromine with a saturated hydrocarbon differ from its reaction with an unsaturated hydrocarbon? How are they similar? Both reactions result in bromine being incorporated into the structure of the product. The difference is the way in which that incorporation takes place. In the saturated hydrocarbon, an existing C–H bond is broken, and a bond between the C and the Br can then be formed. In the unsaturated hydrocarbon, the only bond broken in the hydrocarbon is the π bond whose electrons can be used to form a bond to one of the bromine atoms in Br (the electrons from the Br–Br bond form the other C–Br bond on the other carbon that was part of the π bond in the starting unsaturated hydrocarbon). On a microscopic level, how does the reaction of bromine with an alkene differ from its reaction with an alkyne? How are they similar? Explain why unbranched alkenes can form geometric isomers while unbranched alkanes cannot. Does this explanation involve the macroscopic domain or the microscopic domain? Unbranched alkanes have free rotation about the C–C bonds, yielding all orientations of the substituents about these bonds equivalent, interchangeable by rotation. In the unbranched alkenes, the inability to rotate about the \(\mathrm{C=C}\) bond results in fixed (unchanging) substituent orientations, thus permitting different isomers. Since these concepts pertain to phenomena at the molecular level, this explanation involves the microscopic domain. Explain why these two molecules are not isomers: Explain why these two molecules are not isomers: They are the same compound because each is a saturated hydrocarbon containing an unbranched chain of six carbon atoms. How does the carbon-atom hybridization change when polyethylene is prepared from ethylene? Write the Lewis structure and molecular formula for each of the following hydrocarbons: (a) C H (b) C H (c) C H (d) C H (e) C H (f) C H Write the chemical formula, condensed formula, and Lewis structure for each of the following hydrocarbons: Give the complete IUPAC name for each of the following compounds: (a) 2,2-dibromobutane; (b) 2-chloro-2-methylpropane; (c) 2-methylbutane; (d) 1-butyne; (e) 4-fluoro-4-methyl-1-octyne; (f) -1-chloropropene; (g) 5-methyl-1-pentene Give the complete IUPAC name for each of the following compounds: Butane is used as a fuel in disposable lighters. Write the Lewis structure for each isomer of butane. Write Lewis structures and name the five structural isomers of hexane. Write Lewis structures for the isomers of \(\mathrm{CH_3CH=CHCl}\). Write structures for the three isomers of the aromatic hydrocarbon xylene, C H (CH ) . Isooctane is the common name of the isomer of C H used as the standard of 100 for the gasoline octane rating: (a) 2,2,4-trimethylpentane; (b) 2,2,3-trimethylpentane, 2,3,4-trimethylpentane, and 2,3,3-trimethylpentane: Write Lewis structures and IUPAC names for the alkyne isomers of C H . Write Lewis structures and IUPAC names for all isomers of C H Cl. Name and write the structures of all isomers of the propyl and butyl alkyl groups. Write the structures for all the isomers of the –C H alkyl group. In the following, the carbon backbone and the appropriate number of hydrogen atoms are shown in condensed form: Write Lewis structures and describe the molecular geometry at each carbon atom in the following compounds: Benzene is one of the compounds used as an octane enhancer in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: \[\ce{3C2H2 ⟶ C6H6}\] Draw Lewis structures for these compounds, with resonance structures as appropriate, and determine the hybridization of the carbon atoms in each. In acetylene, the bonding uses hybrids on carbon atoms and orbitals on hydrogen atoms. In benzene, the carbon atoms are hybridized. Teflon is prepared by the polymerization of tetrafluoroethylene. Write the equation that describes the polymerization using Lewis symbols. Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures. (a) \(\mathrm{CH=CCH_2CH_3 + 2I_2⟶ CHI_2CI_2CH_2CH_3}\) (b) \(\ce{CH3CH2CH2CH2CH3 + 8O2 ⟶ 5CO2 + 6H2O}\) Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures. What mass of 2-bromopropane could be prepared from 25.5 g of propene? Assume a 100% yield of product. 65.2 g Acetylene is a very weak acid; however, it will react with moist silver(I) oxide and form water and a compound composed of silver and carbon. Addition of a solution of HCl to a 0.2352-g sample of the compound of silver and carbon produced acetylene and 0.2822 g of AgCl. Ethylene can be produced by the pyrolysis of ethane: \(\ce{C2H6⟶C2H4 + H2}\) How many kilograms of ethylene is produced by the pyrolysis of 1.000 × 10 kg of ethane, assuming a 100.0% yield? 9.328 × 10 kg Why do the compounds hexane, hexanol, and hexene have such similar names? Write condensed formulas and provide IUPAC names for the following compounds: (a) ethyl alcohol, ethanol: CH CH OH; (b) methyl alcohol, methanol: CH OH; (c) ethylene glycol, ethanediol: HOCH CH OH; (d) isopropyl alcohol, 2-propanol: CH CH(OH)CH ; (e) glycerine, l,2,3-trihydroxypropane: HOCH CH(OH)CH OH Give the complete IUPAC name for each of the following compounds: (a) (b) (c) Give the complete IUPAC name and the common name for each of the following compounds: (a) (b) (c) (a) 1-ethoxybutane, butyl ethyl ether; (b) 1-ethoxypropane, ethyl propyl ether; (c) 1-methoxypropane, methyl propyl ether Write the condensed structures of both isomers with the formula C H O. Label the functional group of each isomer. Write the condensed structures of all isomers with the formula C H O . Label the functional group (or groups) of each isomer. HOCH CH OH, two alcohol groups; CH OCH OH, ether and alcohol groups Draw the condensed formulas for each of the following compounds: MTBE, Methyl -butyl ether, CH OC(CH ) , is used as an oxygen source in oxygenated gasolines. MTBE is manufactured by reacting 2-methylpropene with methanol. (a) (b) 4.593 × 10 L Write two complete balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures. (a) propanol is converted to dipropyl ether (b) propene is treated with water in dilute acid. Write two complete balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures. (a) \(\mathrm{CH_3CH=CHCH_3+H_2O⟶CH_3CH_2CH(OH)CH_3}\) ; (b) \(\mathrm{CH_3CH_2OH⟶CH_2=CH_2+H_2O}\) Order the following molecules from least to most oxidized, based on the marked carbon atom: Predict the products of oxidizing the molecules shown in this problem. In each case, identify the product that will result from the minimal increase in oxidation state for the highlighted carbon atom: (a) (b) (c) (a) (b) (c) Predict the products of reducing the following molecules. In each case, identify the product that will result from the minimal decrease in oxidation state for the highlighted carbon atom: (a) (b) (c) Explain why it is not possible to prepare a ketone that contains only two carbon atoms. A ketone contains a group bonded to two additional carbon atoms; thus, a minimum of three carbon atoms are needed. How does hybridization of the substituted carbon atom change when an alcohol is converted into an aldehyde? An aldehyde to a carboxylic acid? Fatty acids are carboxylic acids that have long hydrocarbon chains attached to a carboxylate group. How does a saturated fatty acid differ from an unsaturated fatty acid? How are they similar? Since they are both carboxylic acids, they each contain the –COOH functional group and its characteristics. The difference is the hydrocarbon chain in a saturated fatty acid contains no double or triple bonds, whereas the hydrocarbon chain in an unsaturated fatty acid contains one or more multiple bonds. Write a condensed structural formula, such as CH CH , and describe the molecular geometry at each carbon atom. Write a condensed structural formula, such as CH CH , and describe the molecular geometry at each carbon atom. (a) CH CH(OH)CH : all carbons are tetrahedral; (b) \(\ce{CH3C(==O)CH3}\): the end carbons are tetrahedral and the central carbon is trigonal planar; (c) CH OCH : all are tetrahedral; (d) CH COOH: the methyl carbon is tetrahedral and the acid carbon is trigonal planar; (e) CH CH CH CH(CH )CHCH : all are tetrahedral except the right-most two carbons, which are trigonal planar The foul odor of rancid butter is caused by butyric acid, CH CH CH CO H. Write the two-resonance structures for the acetate ion. Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures: Write two complete balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures. (a) \(\ce{CH3CH2CH2CH2OH + CH3C(O)OH⟶CH3C(O)OCH2CH2CH2CH3 + H2O}\): (b) \(\ce{2CH3CH2COOH + CaCO3⟶(CH3CH2COO)2Ca + CO2 + H2O}\): Yields in organic reactions are sometimes low. What is the percent yield of a process that produces 13.0 g of ethyl acetate from 10.0 g of CH CO H? Alcohols A, B, and C all have the composition C H O. Molecules of alcohol A contain a branched carbon chain and can be oxidized to an aldehyde; molecules of alcohol B contain a linear carbon chain and can be oxidized to a ketone; and molecules of alcohol C can be oxidized to neither an aldehyde nor a ketone. Write the Lewis structures of these molecules. Write the Lewis structures of both isomers with the formula C H N. What is the molecular structure about the nitrogen atom in trimethyl amine and in the trimethyl ammonium ion, (CH ) NH ? What is the hybridization of the nitrogen atom in trimethyl amine and in the trimethyl ammonium ion? Trimethyl amine: trigonal pyramidal, ; trimethyl ammonium ion: tetrahedral, Write the two resonance structures for the pyridinium ion, C H NH . Draw Lewis structures for pyridine and its conjugate acid, the pyridinium ion, C H NH . What are the geometries and hybridizations about the nitrogen atoms in pyridine and in the pyridinium ion? Write the Lewis structures of all isomers with the formula C H ON that contain an amide linkage. Write two complete balanced equations for the following reaction, one using condensed formulas and one using Lewis structures. \[\ce{CH3NH2 + H3O+ ⟶CH3NH3+ + H2O}\] Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures. Ethylammonium chloride is added to a solution of sodium hydroxide. Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in . CH CH = CHCH ( ) + Cl \(⟶\) CH CH(Cl)H(Cl)CH ( ); 2C H ( ) + 15O \(⟶\) 12CO ( ) + 6H O Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in . Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in . the carbon in CO , initially at , changes hybridization to in CO

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A close examination of the ephedrine and pseudoephedrine isomers suggests that another stereogenic center, the nitrogen, is present. As noted earlier, single-bonded nitrogen is pyramidal in shape, with the non-bonding electron pair pointing to the unoccupied corner of a tetrahedral region. Since the nitrogen in these compounds is bonded to three different groups, its configuration is chiral. The non-identical mirror-image configurations are illustrated in the following diagram (the remainder of the molecule is represented by R, and the electron pair is colored yellow). If these configurations were stable, there would be four additional stereoisomers of ephedrine and pseudoephedrine. However, pyramidal nitrogen is normally not configurationally stable. It rapidly inverts its configuration (equilibrium arrows) by passing through a planar, sp -hybridized transition state, leading to a mixture of interconverting R and S configurations. If the nitrogen atom were the only chiral center in the molecule, a 50:50 (racemic) mixture of R and S configurations would exist at equilibrium. If other chiral centers are present, as in the ephedrin isomers, a mixture of diastereomers will result. In any event, ),

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Solutions are homogeneous. Dissolved molecules as large as 1000 pm never separate as a result of gravitational forces, even in an ultracentrifuge. When suspended particles reach µm size (10 pm), they separate readily under gravity, and we classify the mixture as definitely heterogeneous. Surprisingly, suspensions of particles between these sizes, in the range 5 000 - 200 000 pm, never settle under gravity or centrifugation, yet the mixtures are definitely heterogeneous because beams of light passing though them are visible. If a laser pointer beam passes through a solution, it is invisible, but if it passes through a it is easily seen. would be invisible if it weren't for the colloidal smoke or fog that renders the beams visible. Colloids are stable suspensions of a medium, like the solid particles in smoke or fat particles in milk, in a , like air or water. The particles and medium may be any combination of solid, liquid, and gas (except that they both can't be gases, because gasses are completely miscible), as the following table shows: For reference, the table is arranged so that the suspended particles are in the column while the medium the particles are suspended in are on the row. For example, solid foam's medium is solid (row) while the colloid particles are gaseous (column). Average water droplets in fogs and clouds are around 0.01 mm in diameter, so they are ten times as big as typical colloids. Clouds "float" mostly because of air updrafts, and the hardly noticeable (~0.3 cm/s) terminal falling velocity of small droplets . When white light passes through a colloidal suspension, light scattering is highly wavelength dependent, so colloids may appear to be brightly colored. This is the case with the gold colloid that is responsible for the deep red color in most stained glass. Colloidal gold is easily made in the laboratory by reducing a gold salt dissolved in water to give elemental gold clusters. As the clusters change in size, the color of the colloid may change through virtually every color of the rainbow. Ed Vitz (Kutztown University), (University of

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Photosynthesis is a complex series of reactions carried out by algae, phytoplankton, and the leaves in plants, which utilize the energy from the sun. The simplified version of this chemical reaction is to utilize carbon dioxide molecules from the air and water molecules and the energy from the sun to produce a simple sugar such as glucose and oxygen molecules as a by product. The simple sugars are then converted into other molecules such as starch, fats, proteins, enzymes, and DNA/RNA i.e. all of the other molecules in living plants. All of the "matter/stuff" of a plant ultimately is produced as a result of this photosynthesis reaction. An important summary statement is that during photosynthesis plants use carbon dioxide and produce oxygen. Combustion occurs when any organic material is reacted (burned) in the presence of oxygen to give off the products of carbon dioxide and water and ENERGY. The organic material can be any fossil fuel such as natural gas (methane), oil, or coal. Other organic materials that combust are wood, paper, plastics, and cloth. Organic materials contain at least carbon and hydrogen and may include oxygen. If other elements are present they also ultimately combine with oxygen to form a variety of pollutant molecules such as sulfur oxides and nitrogen oxides. Metabolism occurs in animals and humans after the ingestion of organic plant or animal foods. In the cells a series of complex reactions occurs with oxygen to convert for example glucose sugar into the products of carbon dioxide and water and ENERGY. This reaction is also carried out by bacteria in the decomposition/decay of waste materials on land and in the water. An important summary statement is that during combustion/metabolism oxygen is used and carbon dioxide is a product. The whole purpose of both processes is to convert chemical energy into other forms of energy such as heat. Carbon dioxide is slightly soluble and is absorbed into bodies of water such as the ocean and lakes. It is not overly soluble as evidenced by what happens when a can of carbonated soda such as Coke is opened. Some of the dissolved carbon dioxide remains in the water, the warmer the water the less carbon dioxide remains in the water. Some carbon dioxide is used by algae and phytoplankton through the process of photosynthesis. In other marine ecosystems, some organisms such as coral and those with shells take up carbon dioxide from the water and convert it into calcium carbonate. As the shelled organisms die, bits and pieces of the shells fall to the bottom of the oceans and accumulate as sediments. The carbonate sediments are constantly being formed and redissolved in the depths of the oceans. Over long periods of time, the sediments may be raised up as dry land or into mountains. This type of sedimentary rock is called limestone. The carbonates can redissolve releasing carbon dioxide back to the air or water. In the natural carbon cycle, there are two main processes which occur: photosynthesis and metabolism. During photosynthesis, plants use carbon dioxide and produce oxyge and during metabolism oxygen is used and carbon dioxide is a product. Humans impact the carbon cycle during the combustion of any type of fossil fuel, which may include oil, coal, or natural gas. Fossil Fuels were formed very long ago from plant or animal remains that were buried, compressed, and transformed into oil, coal, or natural gas. The carbon is said to be "fixed" in place and is essentially locked out of the natural carbon cycle. Humans intervene during by burning the fossil fuels. During combustion in the presence of air (oxygen), carbon dioxide and water molecules are released into the atmosphere. The question becomes as to what happens to this extra carbon dioxide that is released into the atmosphere. This is the subject of considerable debate and about it possible effect in enhancing the greenhouse ffect which may than result in global warming.  

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Catalysts are substances that increase the reaction rate of a chemical reaction without being consumed in the process. A catalyst, therefore, does not appear in the overall stoichiometry of the reaction it catalyzes, but it must appear in at least one of the elementary reactions in the mechanism for the catalyzed reaction. The catalyzed pathway has a lower , but the net change in energy that results from the reaction (the difference between the energy of the reactants and the energy of the products) is not affected by the presence of a catalyst ( e \(\Page {1}\)). Nevertheless, because of its lower , the reaction rate of a catalyzed reaction is faster than the reaction rate of the uncatalyzed reaction at the same temperature. Because a catalyst decreases the height of the energy barrier, its presence increases the reaction rates of both the forward and the reverse reactions by the same amount. In this section, we will examine the three major classes of catalysts: heterogeneous catalysts, homogeneous catalysts, and enzymes. A catalyst affects , not Δ . In , the catalyst is in a different phase from the reactants. At least one of the reactants interacts with the solid surface in a physical process called adsorption in such a way that a chemical bond in the reactant becomes weak and then breaks. Poisons are substances that bind irreversibly to catalysts, preventing reactants from adsorbing and thus reducing or destroying the catalyst’s efficiency. An example of heterogeneous catalysis is the interaction of hydrogen gas with the surface of a metal, such as Ni, Pd, or Pt. As shown in part (a) in \(\Page {2}\) the hydrogen–hydrogen bonds break and produce individual adsorbed hydrogen atoms on the surface of the metal. Because the adsorbed atoms can move around on the surface, two hydrogen atoms can collide and form a molecule of hydrogen gas that can then leave the surface in the reverse process, called desorption. Adsorbed H atoms on a metal surface are substantially more reactive than a hydrogen molecule. Because the relatively strong H–H bond (dissociation energy = 432 kJ/mol) has already been broken, the energy barrier for most reactions of H is substantially lower on the catalyst surface. \(\Page {2}\) shows a process called , in which hydrogen atoms are added to the double bond of an alkene, such as ethylene, to give a product that contains C–C single bonds, in this case ethane. Hydrogenation is used in the food industry to convert vegetable oils, which consist of long chains of alkenes, to more commercially valuable solid derivatives that contain alkyl chains. Hydrogenation of some of the double bonds in polyunsaturated vegetable oils, for example, produces margarine, a product with a melting point, texture, and other physical properties similar to those of butter. Several important examples of industrial heterogeneous catalytic reactions are in \(\Page {1}\). Although the mechanisms of these reactions are considerably more complex than the simple hydrogenation reaction described here, they all involve adsorption of the reactants onto a solid catalytic surface, chemical reaction of the adsorbed species (sometimes via a number of intermediate species), and finally desorption of the products from the surface. In , the catalyst is in the same phase as the reactant(s). The number of collisions between reactants and catalyst is at a maximum because the catalyst is uniformly dispersed throughout the reaction mixture. Many homogeneous catalysts in industry are transition metal compounds ( \(\Page {2}\)), but recovering these expensive catalysts from solution has been a major challenge. As an added barrier to their widespread commercial use, many homogeneous catalysts can be used only at relatively low temperatures, and even then they tend to decompose slowly in solution. Despite these problems, a number of commercially viable processes have been developed in recent years. High-density polyethylene and polypropylene are produced by homogeneous catalysis. Enzymes, catalysts that occur naturally in living organisms, are almost all protein molecules with typical molecular masses of 20,000–100,000 amu. Some are homogeneous catalysts that react in aqueous solution within a cellular compartment of an organism. Others are heterogeneous catalysts embedded within the membranes that separate cells and cellular compartments from their surroundings. The reactant in an enzyme-catalyzed reaction is called a . Because enzymes can increase reaction rates by enormous factors (up to 10 times the uncatalyzed rate) and tend to be very specific, typically producing only a single product in quantitative yield, they are the focus of active research. At the same time, enzymes are usually expensive to obtain, they often cease functioning at temperatures greater than 37 °C, have limited stability in solution, and have such high specificity that they are confined to turning one particular set of reactants into one particular product. This means that separate processes using different enzymes must be developed for chemically similar reactions, which is time-consuming and expensive. Thus far, enzymes have found only limited industrial applications, although they are used as ingredients in laundry detergents, contact lens cleaners, and meat tenderizers. The enzymes in these applications tend to be proteases, which are able to cleave the amide bonds that hold amino acids together in proteins. Meat tenderizers, for example, contain a protease called papain, which is isolated from papaya juice. It cleaves some of the long, fibrous protein molecules that make inexpensive cuts of beef tough, producing a piece of meat that is more tender. Some insects, like the bombadier beetle, carry an enzyme capable of catalyzing the decomposition of hydrogen peroxide to water ( \(\Page {3}\)). cause a decrease in the reaction rate of an enzyme-catalyzed reaction by binding to a specific portion of an enzyme and thus slowing or preventing a reaction from occurring. Irreversible inhibitors are therefore the equivalent of poisons in heterogeneous catalysis. One of the oldest and most widely used commercial enzyme inhibitors is aspirin, which selectively inhibits one of the enzymes involved in the synthesis of molecules that trigger inflammation. The design and synthesis of related molecules that are more effective, more selective, and less toxic than aspirin are important objectives of biomedical research. Catalysts participate in a chemical reaction and increase its rate. They do not appear in the reaction’s net equation and are not consumed during the reaction. Catalysts allow a reaction to proceed via a pathway that has a lower activation energy than the uncatalyzed reaction. In heterogeneous catalysis, catalysts provide a surface to which reactants bind in a process of adsorption. In homogeneous catalysis, catalysts are in the same phase as the reactants. Enzymes are biological catalysts that produce large increases in reaction rates and tend to be specific for certain reactants and products. The reactant in an enzyme-catalyzed reaction is called a substrate. Enzyme inhibitors cause a decrease in the reaction rate of an enzyme-catalyzed reaction.

https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Aquatic_Chemistry/Natural_Water

Water is the most important resource. Without water life is not possible. From a chemical point of view, water, H O, is a pure compound, but in reality, you seldom drink, see, touch or use pure water. Water from various sources contains dissolved gases, minerals, organic and inorganic substances. The total water system surrounding the planet Earth is called the . It includes freshwater systems, oceans, atmosphere vapour, and biological waters. The Arctic, Atlantic, Indian, and Pacific oceans cover 71% of the Earth surface, and contain 97% of all water. Less than 1% is fresh water, and 2-3 % is ice caps and glaciers. The Antarctic Ice Sheet is almost the size of North America continent. These waters dominate our weather and climate, directly and indirectly affecting our daily lives. They cover 3.35x10 km . The four oceans have a total volume of 1.35x10 km . are steps by which water cycles on the planet Earth. These processes include sublimation of ice, evaporation of liquid, transportation of moisture by air, rain, snow, river, lake, and ocean currents. All these processes are related to the physical and chemical properties of water, and many government agencies are set up to study and record phenomena related to them. The study of these processes is called Among the planets, Earth is the only one in which there are solid, liquid and gaseous waters. These conditions are just right for life, for which water is a vital part. Water is the most abundant substance in the biosphere of Earth. Groundwater is an important part of the water system. When vapor is cooled, clouds and rain develop. Some of the rain percolate through the soil and into the underlying rocks. The water in the rocks is , which moves slowly. A body of rock, which contains appreciable quantities of water, is called an . Below the , the aquifier is filled (or saturated) with water. Above the water table is the unsaturated zone. Some regions have two or more water tables. These zones are usually separated by water-impermeable material such as boulder and clay. Groundwater can be brought to the surface by drilling below the water table, and pumped out. The amount of water that can be pumped out depends on the structure of the aquifer. Little water is stored in tight granite layers, but large quantities of water are stored in limestone aquifier layers. In some areas, there are under ground rivers. Hydrology is also the study of how solids and solute interact in, and with, water. In this link, the compositions of seawater, composition of the atmosphere, compositions of rain and snow, and compositions of river waters and lake waters are given in details. Table \(\Page {1}\) list the major ions present in seawater. The composition does vary, depending on region, depth, latitude, and water temperature. Waters at the river mouths contain less salt. If the ions are utilized by living organism, its contents vary according to the populations of organisms. Dust particles and ions present in the air are nucleation center of water drops. Thus, waters from rain and snow also contain such ions: Ca , Mg , Na , K , NH . These cations are balanced by anions, HCO , SO , NO , Cl , and NO . The pH of rain is between 5.5 and 5.6. Rain and snow waters eventually become river or lake waters. When the rain or snow waters fall, they interact with vegetation, top soil, bed rock, river bed and lake bed, dissolving whatever is soluble. Bacteria, algae, and water insects also thrive. Solubilities of inorganic salts are governed by the kinetics and equilibria of dissolution. The most common ions in lake and river waters are the same as those present in rainwater, but at higher concentrations. The pH of these waters depends on the river bed and lake bed. Natural waters contain dissolved minerals. Waters containing Ca and Mg ions are usually called . Minerals usually dissolve in natural water bodies such as lakes, rivers, springs, and underground waterways (ground waters). Calcium carbonate, CaCO , is one of the most common inorganic compounds in the Earth crust. It is the ingredient for both calcite and aragonite. These two minerals have different crystal structures and appearance. This photograph shows crystals of typical Calcite. Calcium-carbonate minerals dissolve in water, with a solubility product as shown below. \[CaCO_3 \rightleftharpoons Ca^{2+} + CO_3^{2-} \;\;\; K_{sp} = 5 \times 10^{-9}\] From the solubility product, we can (see example 1) evaluate the molar solubility to be 7.1x10 M or 7.1 mg/L (7.1 ppm of CaCO in water). The solubility increases as the pH decrease (increase acidity). This is compounded when the water is saturated with carbon dioxide, CO . Saturated CO solution contains carbonic acid, which help the dissolution due to the reaction: \[H_2O + CO_2 \rightleftharpoons H_2CO_3\] \[CaCO_3 + H_2CO_3 \rightleftharpoons Ca^{2+} + 2 HCO_3^-\] Because of these reactions, some natural waters contain more than 300 ppm calcium carbonates or its equivalents. The carbon dioxide in natural water creates an interesting phenomenon. Rainwater is saturated with CO , and it dissolves limestones. When CO is lost due to temperature changes or escaping from water drops, the reverse reaction takes place. The solid formed, however, may be a less stable phase called aragonite, which has the same chemical formula as, but a different crystal structure than that of calcite. The rain dissolves calcium carbonate by the two reactions shown above. The water carries the ions with it, sips through the crack of the rocks. When it reached the ceiling of a cave, the drop dangles there for a long time before fallen. During this time, the carbon dioxide escapes and the pH of the water increases. Calcium carbonate crystals begin to appear. Calcite, aragonite, stalactite, and stalagmite are four common solids found in the formation of caves. Natural waters contain metal ions. Water containing calcium, magnesium and their counter anions are called . Hard waters need to be treated for the following applications. Due to the reversibility of the reaction, \[CaCO_{3(s)} + H_2CO_3 \rightleftharpoons Ca^{2+} + 2 HCO_3^-\] water containing Ca , Mg and CO ions is called , because the hardness can be removed by boiling. Boiling drives the reverse reaction, causing deposit in pipes and scales in boilers. The deposits lower the efficiency of heat transfer in boilers, and diminish flow rates of water in pipes. Thus, temporary hard water has to be softened before it enters the boiler, hot-water tank, or a cooling system. The amount of metal ions that can be removed by boiling is called After boiling, metal ions remain due to presence of chloride ions, sulfate ions, nitrate ions, and a rather high solubility of MgCO . Amount of metal ions that can not be removed by boiling is called is the sum of temporary hardness and permanent hardness. Hardness is often expressed as equivalence of amount of calcium ions in the solution. Thus, water conditioning is an important topic. The value of water treatment market has been estimated to be worth $30 billion. Lime-soda softening is the removal of temporary hardness by adding a calculated amount of hydrated lime, Ca(OH) : \[Ca^{2+} + 2 HCO_3^- + Ca(OH)_{2(s)} \rightarrow 2 CaCO_{3(s)} + 2 H_2O\] Adding more lime causes the pH of water to increase, and as a result, magnesium ions are removed by the reaction: \[Mg^{2+} + Ca(OH)_{2(s)} \rightarrow Mg(OH)_{2(s)} + Ca^{2+}\] The extra calcium ions can be removed by the addition of sodium carbonate. \[Na_2CO_3 \rightarrow 2 Na^+ + CO_3^{2-}\] \[Ca^{2+} + CO_3^{2-} + \rightarrow CaCO_{3(s)}\] In this treatment, the amount of Ca(OH) required is equivalent to the temporary hardness plus the magnesium hardness. The amount of sodium carbonate required is equivalent to the permanent hardness. Thus, lime-soda softening is effective if both the temporary and total hardness have been determined. The sodium ion will remain in the water after the treatment. The pH of the water is also rather high depending on the amount of lime and sodium carbonates used. Addition of complexing reagent to form soluble complexes with Ca and Mg prevents the formation of solid. One of the complexing agents is sodium triphosphate Na PO , which is marketed as Calgon, etc. The phosphate is the complexing agent. Other complexing agents such as Na H EDTA can also be used, but the complexing agent EDTA forms strong complexes with transition metals. This causes corrosion problem, unless the pipes of the system are made of stainless steel. Today, most water softeners are using zeolites and employing ion exchange technique to soften hard water. Zeolites are a group of hydrated crystalline aluminosilicates found in certain volcanic rocks. The tetrahedrally coordinated aluminum and silicon atoms form AlO and SiO tetrahedral groups. They interconnect to each other sharing oxygen atoms forming cage-type structures as shown on the right. This diagram and the next structural diagram are taken from an introduction to zeolites There are many kinds of zeolites, some newly synthesized. Whatever kind, the crystal structure of zeolites contains large cages. The cages are connected to each other forming a framework with many cavities and channels. Both positive and negative ions can be trapped in these cavities and channels as shown below. For each oxygen that is not shared in the AlO and SiO tetrahedral groups, a negative charge is left on the group. These negative charges are balanced by trapping alkali metal and alkaline earth metal ions. When more cations are trapped, hydroxide and chloride ions will remain in the cavities and channels of the zeolites. To prepare a zeolite for water treatment, they are soaked in concentrated NaCl solution. The cavities trap as many sodium ions as they can accommodate. After the treatment, the zeolite is designated as Na-zeolite. Then the salt solution is drained, and the zeolite is washed with water to eliminate the extra salt. When hard water flow through them, calcium and magnesium ions will be trapped by the Na-zeolite. For every Ca or Mg trapped, two Na ions are released. The treated water contains a rather high concentration of Na ions, but low concentrations of Mg and Ca . Thus, zeolite ion exchange convert hard water into soft water. In most cases, the resins are polystyrene with functional -SO H groups attached to the polymer chain for cation exchange resin, and with functional group -N(CH ) attached to the chain for anion exchange resin. To prepare the resin for making , the cation resin is regenerated with HCl so that the groups are really -SO H. The anion resin is regenerated with NaOH, so that the functional groups are -N(CH ) (OH). When water containing any metal ion M and anion A passes through the ion exchange resins in two stages, the following reactions take place, \[\ce{M+ + -SO3H \rightarrow H+ + -SO3M}\] \[\ce{A^{-} + -N(CH3)3(OH) \rightarrow OH^{-} + -N(CH3)3A}\] \[\ce{H^{+} + OH- <=> H2O}\] Thus, ion exchange provides to meet laboratory requirement. This method can also be used to prepare water for domestic and laboratory applications. This method has been discussed in The following is a list of companies selling magnetic devices for . All devices are based on results of some ressearch indicating that when water runs through a magnetic field, the calcium carbonate will precipitate as aragonite rather than the usual calcite. For example, K.J. Kronenberg has published an article in , (Vol. Mag-21, No. 5, September 1985, pages 2059-2061). and stated the following: Many companies have made various devices for magnetic conditioning of water, and they claim that their devices will clean up the pipes and boilers at little or no cost. I have yet to test one of these devices for its claim, but my preliminary tests shows that permanent magnet has little effect on the calcium carbonate deposit of temporary hard water. The cleaning effect they have claimed is probably much overstated. From the solubility product shown for the dissolution of calcium carbonate, \[CaCO_3 \rightleftharpoons Ca^{2+} + CO_3^{2-} \;\;\; K_{sp} = 5 \times 10^{-9}\] Evaluate the molar solubility of Ca in saturated solution. From the definition of solubility product, we have \[[Ca^{2+}] [CO_3^{2-}] = 5 \times 10^{-9}\] Thus, \[[Ca^{2+}] = [CO_3^{2-}] = 7.1 \times 10^{-5}\; M\] The concentration of 7.1x10 M is equivalent to 7.1 mg/L (7.1 ppm of CaCO in water). There may be other ions present in the system and other equilibria conditions in addition to the equilibrium mentioned here. Problems are more complexed in the real world. Boiling of 1.0 L of water produced 10 mg of CaCO3 solid. What is the temporary hardness of the water? 10 ppm

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Photoreceptors/Photoreceptor_Proteins

Photoreceptor proteins are light-sensitive proteins involved in the sensing and response to light in a variety of organisms.[1] Photoreceptor proteins can be find in both animals and plants. Human eye retina is a good example of photoreceptor protein. Many bacteria, such as halohodospira halophila, an extremophile bacterium contain Photoactive Yellow Protein. Photoreceptor proteins are optimally suited to study the role of dynamical alterations in protein structure in relation to their function. First, such proteins can be triggered with (laser) flash illumination, and therefore excellent time-resolution is achievable in studies of the dynamical alterations in their structure. Second, because they are signal-transduction proteins, one may anticipate large conformational transitions to be involved in their signaling state formation (and its subsequent decay), which is indeed borne out by the experiments. Third, the (changing) color of these proteins often is an excellent indicator as to which time scale is relevant to resolve structural transitions. Significant and unsurpassed insight along these lines has been obtained for a number of different photoreceptor proteins. Hence, they can, indeed, be considered as “star actors” in the pursuit to understand, in general terms, the atomic details of the dynamics of functional conformational transitions [i.e., (partial) un/folding] in these proteins required for their functioning.[4] Photoreceptor protein contains two parts, the protein part, and non-protein, chromophore part. The non-protein part can response to light throught photoisomerization, or photoexcitation. [2] The figure above show protein part in secondary structure, and chromophorepart in line structure. Plants are important living organisms on earth. As autotrophs, part of plants absorb sunlight through photosynthesis to convert water and Carbon dioxide to oxygen and other chemicals. The part of plants that responsible for absorb and use sunlight is photoreceptor. There are many types of photoreceptors in plants, such as Chlorophyll. PHYTOCHROMES BLUE-LIGHT RECEPTORS characterization of two photolyase-like genes of Synechocystis sp. PCC6803. Nucleic Acids Res. 2000, 28, 2353-2362.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/03%3A_Criteria_for_Selection_of_the_Synthetic_Route

Once a Target Molecule is chosen for synthesis, one could sit down and device several routes for its synthesis. On what criteria do you select a Target and how do you arrive at a synthetic route? The answer depends on the overall goal of your project. For a natural product chemist, he might have isolated and determined the structure of a new molecule. He may need to synthesize the molecule to prove the structure. While working on structure elucidation, the route chosen for synthesis should unambiguously establish the part of the structure you are working on. Each step is chosen on this structure-criterion alone. Here the length of the route and the cost of the chemicals are not important. An elegant (enantiopure) synthesis of some complex structure is the dream of an academician working in the university laboratory. He is often more concerned with developing new routes, new reactions and new mechanistic principles. His concern is to develop new horizons and give good training to the young chemists. He is seldom worried about the cost of his research. He has time at his command and hopefully enough money to pursue his passion. He is judged by the quality (and quantity) of his research publications and the quality of training he has imparted to the students. Having a patent is an added feather to his cap. A pharmaceutical chemist and a material chemist are more interested in developing versatile and fast synthetic routes for a chosen molecule. Their efforts are directed towards the synthesis of a large number of closely related molecules, within a short time. Such a chemist is looking into Structure-Activity Studies, aimed at developing new drug molecules or molecules with special properties. He is often judged by the number of such active molecules that he has discovered and patents held in his name and not just by the number of new molecules synthesized by him or the elegance of the synthetic route. A publication to his credit is an added feather to his cap. In his endeavours, the cost and the efficiency of the synthetic routes are not the criteria for research. He believes that once the ‘right molecule’ is discovered, more efficient routes could always be generated at a suitable date. His art is directed towards fast discovery of molecules with the right properties. An industrial chemist is most concerned with the ‘cost’ of synthesis of the molecule. His efforts are directed towards development of economical synthetic procedures, which includes not only the cost of the chemicals but also the cost of waste treatment, recycling and environmental cleaning. He selects the molecules on the basis of their economic value (net profit for his company). He should be concerned about eco-friendly reactions and procedures. In general, a development chemist in an industrial R&D laboratory looks at very large-scale (typically several kilogram batch) reactions, their reproducibility, safety and cost parameters. His focus is on the commercial value of his product, the profitability and the patents. Therefore, the target molecule and the route chosen depend of the hat the chemist is wearing. We do have examples of amazing chemists who efficiently juggle with more than one hat at the same time. We take our proverbial hats off for those versatile and multifaceted chemists. This is because, for an efficient operation, different tasks demand different skill-sets and use of different sets of databases. Nonetheless, the underlying chemistry is same in all these activities. Of course, occasionally one could be creative, versatile and cost-effective at the same time. In the following pages we would look into the broad principles governing the art of organic syntheses. We have already discussed some ‘rules’ that govern synthesis. We would now explore guidelines that concern the logic of organic syntheses. We would then discuss several interesting syntheses to illustrate these principles.

https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Aquatic_Chemistry/Water_Physics

Chemical and physical properties of water are often discussed together. These properties are fundamentals of many disciplines such as hydrology, environmental studies, chemical engineering, environmental engineering, civil engineering etc. They are of interest to chemists and physicists of course. Here are some highlights of the physical properties of water. Pure liquid water has a high heat capacity of 4.182 J K g ; it is a good heat conductor, but a poor electric conductor. It is a solvent for dissolving ionic and polar substances, but interact with non-polar substance weakly. Surface tension of water is rather high, and little quantities aggregate into drops rather than spread out as thin layers. Hydrogen bonding contributes to many of the physical and chemical properties, such as the unusual but normally known freezing point of 273.15 K, and 373.15 K respectively. The critical temperature and pressure are 647.3 K and 220.5 bar (22050 kPa) and critical volume = 0.056 m kmol . Because of the many applications of water, some more details on the properties of water are desirable. Thus, keeping track of water properties is of national interest. For example, the American Society of Mechanical Engineers (ASME) is such an organization. International co-operation on research and information exchange are more economical, and for thermal properties, The International Association for the Properties of Water and Steam (IAPWS) is set up. Canada is a member of this organization. From the application point of view, the variations of the following properties as functions of temperature and pressure are required. We will discuss some of these to illustrate the point, but not all of them. is the mass per unit volume. The density of water is usually taken as 1.0 g/mL or 1.000e3 kg m at 277 K. This suggests that the density varies with temperature and water density is the highest at 277 K, and the density between 273 and 281 K from the are given in the Table here. These data are calculated from experimental data for pure water based on the standard at 276.98 K. The same source gives the density of ordinary water as 1.000000 g/mL at 277 K. The volume occupied by one mole of substance is called the . The molar volume of liquid water is 18.016/ . At 277 K, the molar volume is 18.016 mL. For liquid water, the molar volumes of liquid water increase to 18.03 mL at both 269 K and 285 K. The density of ice is 0.917 at 273 K, and the molar volume is 19.65 mL, 9% more than the molar of liquid. Thus, 9% of an ice cube containing no air bubble float above the surface, and 91% of it is below the waterline. The density makes behavior of icebergs interesting. are major tourist attractions in Newfoundland and Labrador, Canada. Charged ions interact with each other due to electrostatic attraction or repulsion. The force between two charge particles with charges , and separated by a distance is q1 q2 F = ------------ 4 p eo r2 Uncharged molecules still interact with each other, not due to electrostatic interaction, but due to electric dipole interaction. The electric dipole moment is a vector due to uneven distribution of unlike charges. In diatomic systems, the magnitude of the electric dipole moment can be estimated as the difference between the Pauling electronegativities of the two atoms. For convenience, let us assume that centres of positive and negatives of Na-Cl are separated by a distance l, then the electric dipole moment, \(\mu\) is m = q l m = q l = (1.60e-19) (240e-12 m) = 3.84e-29 C m. 1 D = 1e-18 esu cm = 3.355e-30 C m (from the calculation above) m = 11.5 D for ideallized NaCl gas molecule but mobserved = 9 D in NaCl gas =3e-29 C m. Let us assume the H-F molecule as composed of two ions, a positive and a negative ion. What is the distance that separating these two ions in order to give a dipole moment of 1.8 D?

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Nucleic_Acids/DNA/DNA%3A_Double_Helix

The secondary structure of DNA is actually very similar to the secondary structure of proteins. The protein single alpha helix structure held together by hydrogen bonds was discovered with the aid of X-ray diffraction studies. The X-ray diffraction patterns for DNA show somewhat similar patterns. In addition, chemical studies by E. Chargaff indicate several important clues about the structure of DNA. In the DNA of all organisms: The consists of two right-handed polynucleotide chains that are coiled about the same axis. The heterocyclic amine bases project inward toward the center so that the base of one strand interacts or pairs with a base of the other strand. According to the chemical and X-ray data and model building exercises, only specific heterocyclic amine bases may be paired. The Base Pairing Principle is that The base pairing is called complementary because there are specific geometry requirements in the formation of hydrogen bonds between the heterocylic amines. Heterocyclic amine base pairing is an application of the . In the structures for the complementary base pairs given in the graphic on the left, notice that the thymine - adenine pair interacts through two hydrogen bonds represented as (T=A) and that the cytosine-guanine pair interacts through three hydrogen bonds represented as (C G). Although other base pairing-hydrogen bonding combinations may be possible, they are not utilized because the bond distances do not correspond to those given by the base pairs already cited. The diameter of the helix is 20 Angstroms. The double-stranded helical model for DNA is shown in the graphic on the left. The easiest way to visualize DNA is as an immensely long rope ladder, twisted into a cork-screw shape. The sides of the ladder are alternating sequences of deoxyribose and phosphate (backbone) while the rungs of the ladder (bases) are made in two parts with each part firmly attached to the side of the ladder. The parts in the rung are heterocyclic amines held in position by hydrogen bonding. Although most DNA exists as open ended double helices, some bacterial DNA has been found as a cyclic helix. Occasionally, DNA has also been found as a single strand. QUES. Describe the structure of the double helix of DNA in your own words including the terms: backbone, heterocyclic amines, complementary base pairings, hydrogen bonding, deoxyribose, phosphate.

https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Aquatic_Chemistry/Water_Biology

Not finished yet, but the major ideas are here now! Chemical and physical properties of water discussed in other pages are essential considerations for water biology. Natural water also contains biological matters as well as living creatures. In the discusssion of biology and water, you must feel pleasant, because they make each other grow more lovely as this picture from water lily cottage. The water pages of National Wildlife Federation, offers some interesting reading too. All living things have a cycle of life. A cycle involves all or some of these process: birth, growth, mature, reproduction, matamorphosis, and death. There are millions of living organisms on Earth, ranging from single-cell amebas, bacteria, to the complicated homosapiens. There are also viruses which are fragment of DNA or RNA that depend on host cells for their reproduction. They are not cells. Living things usually have cells that isolate their systems so that the cells contain unique materials to sustain the lives of cells. Cells regulate their contents (homeostatic), and carry out their metabolsims. They divide or making copies of themselves. Many reproduction process involve two individuals and future populations are subject to a greater diversity. Mutation is a fact of life, and many adopt to their changing environment. How life started? Let the research and debate continue by not giving any clusive statements here. A physical geography course suggests that the marine invertebrates began their life 600 million years ago, and they are followed by fish, land plants, amphibians, reptiles, mommals, and then flowering plants, in this order. All these started more than a hundred million years ago, and hominid (primate) line began its evolution 20 to 15 million years ago. There are strong evidences that life on earth appeared in a body of water. Only the planet Earth has three states of water, and it offers a suitable environment for life to began, among all nine solar planets. Since all life forms involve water. Water is seen as the source, matrix, and mother of life. Water is important, because water is required for life, and some people even consider water as life blood. Since water supports life, living organisms also modify their environment, changing the nature of the water in which they live. Biology of water pollution, lists the syllabus on a course including a laboratory section. Water and biology interweave into an entangled maze waiting for explorers and curious minds. Water dissolves or emulsifies other life-supporting substances and transport them to intercellular and intracellular fluids. It is also a medium in which reactions take place. Reactions provide energy (non-matter) for living. Energy causes changes, and manifestation of changes is at least related to, if not the whole, life. An organized and systematized set of reactions is essential in each life. Many living organisms live their lives entirely in water as shown here in this photo from a job center talking about work in marine biology. Aquatic living organisms extract neutrients from water, yet maintaining a balance of electrolyte and nurrishment concentration in their cells. For living things not living in water, they extract water from their environment by whatever mechanism they can. Cells in their body are surrounded by body fluid, and all cells maintain constant concentrations of electrolytes, neutrints, and metabolites. The process of maintaining constant concentrations is called homeostasis. Certainly, some active transport mechanisms are involved in this balance. The rooting of every type of plants is unique. Generally speaking, plants having extensive roots are able to extract water under harsh conditions. On the other hand, some plants such as and juniper have little roots, but their leaves have a layer of wax that prevents water from evaporation. Water conserving plants tolerate draught, and they survive under harsh conditions. The picture shown here is a jade plant from the above link. Lately, some pumpkin growers harvested squash weighing almost 500 kg. At the peak of the growing season, the squash grows almost 0.5 kg a day. That is equivalent to 25 moles of water collected by the roots, discounting the water evaporated through the leaves. The growth is particularly good during a hot and wet day, but during a hot sunny after noon, the temperature of the leaves and fruits get very hot. In addition to water, many inorganic substances or minerals are essential to life. These substances ionize in water to form ions and their solutions conduct electricity. Therefore, they are called . Because most of these substances are already dissoved in natural water, we list ions instead of the mineral they come from. When ions dissove, they form complexes with water molecules. For most metals, the first sphere of coordination usually involve 6 water molecules. For example, when sodium chloride dissolve, we have \[\ce{NaCl + 12 H2O <=> Na(H2O)6+ + Cl(H2O)6^{-}}\] \[\ce{FeCl2 + 18 H2O <=> Fe(H2O)62+ + 2 Cl(H2O)6^{-}}\] Formation of complexes are due to the high dipole moment of water, and the dissolution can be attribute to the high dielectric constant (80). However, in most publications, we ignore the water molecules in the complexes, and simply consider them as ions. In the following, we describe some essential ions or salts as electrolytes. The eletrolytes listed above are present in significant amount in water, or fluids of organisms. There are some metals present in very minute quantities in biological systems, and these are not listed above. Metal ions also interact with proteins. An is usually a very large protein molecule, and it folds into a kidney shape enclosing one or more metal ions forming a complex. The metal is usually responsible for the enzyme activity. Cobalt, copper, iron, molebdenium, nickel, and zinc have groups of enzymes each, and further discussion can be found in The Prosthetic groups and Metal Ions in Protein Active Sites Database (PROMISE). A general discussion is called and this site has an extensive General references on Bioinorganic Chemistry. Electrolyte balance are maintained by . Diffusion process tends to make the concentration all the same throughout the entire fluid, but active or selective transport moves ions to special compartment. For example, the active transport of sodium and potasium by an enzyme called is usually known as . This process pumps potassium ions inside a cell while removing sodium ions from the cells. Thus, a high concentration of potassium is maintained inside cells. Energy is required in active transport, and cellular metabolism provides the energy and the necessary molecular motions to facilitate the process. Hormons are produced by special cells, and they are responsible for the communication between various part of the body. Some complicate harmon actions regulates the rate of transport and balance the ion concentrations depending on the portion of the tissue and the need. This is generally called the following the suggestion of . Gibbs-Donnan effect considers the equilibrium in compartments that are separated by memberances or cell walls. There will be no net change when the products of concentrations of say [Na ] , [Cl ] are the same for compartments 1 and 2. [Na+]1 [Cl-]1 = [Na+]2 [Cl-]2 the subscript 1 and 2 refer to the two compartments. When no other components are present, we have [Na+]1 = [Cl-]1 = [Na+]2 = [Cl-]2 But if compartment 2 has a sodium salt with other anions, this salt ionize to give Na too. The above condition will not be maintained, in this case. In other words, thermodynamic will be a force to adjust the concentrations. In general, the cations should be balanced with anions. Otherwise, the solution will be charged. In human, water in the tissue and body fluid is mostly free, but some fraction may be bounded in pockets of hydrophilic compartments. Body fluids have many electrolytes and neutrients dissolve in them. by J.M. Orten and O.W. Neuhaus (1982), 10th Ed. suggests that about 70% of human body weight is water, most found in three major compartments: 70% intracellular fluid, 20% interstitial fluid, and 7% blood plasma, and only 3 % in intestinal lumen, cerebrospinal fluid and other compartments. However, also suggests that blood makes up about 8% of the total body weight. For a person weighing 50 kg (110 lb), what is the weight of the blood? From the distribution given above, Amount of blood = 50 kg * 0.08 = 4 kg. This is a lot of blood, and donating 0.5 L of blood will not affect the normal function of the blood. Water in human comes from ingestion. Aside from drinking water, there is other beverages. Much of the food also contain water. When food is oxidized in the cells, all hydrogen in food converts to water, which is called metabolic water. Water is excreted via urine, feces, skin, and expiration. A typical daily water balance is shown in a table here. Water balance is maintained between cells and fluid, and the output depends on kidney functions and body insensible perspiration (Expired air from the lung is saturated with water vapor, and evaporation from the skin). Drinking water affects health. An Excite search using the phrase "drinking water" came up with 57890 documents. Drinking Water Resources gives annotated links to web sites that provide information about the drinking water. A rather recent book by S.D. Faust and O.M. Aly, 2nd Ed. (1998) [TD433 F38 1998] addresses the standards for drinking water in the first Chapter. The standards have changed over the years, as we better understand the science. Safe drinking water is a suitable combination of minerals and electrolytes. Usually, one should not drink water softened by water softeners. Using distilled water for beverages and cooking may not reach your set goals. Hard water with calcium and magnesium ions is good for drinking. Usually, a government set up a non-profit organization to provide rules for safe drinking water. This organization has an infrastructure to monitor drinking water systems, and it shall also carry out research to improve the quality of drinking water. Regarding making rules, reliable tests should be developed to determine the electrolyte we have mentioned, plus others such as lead ions, Pb , mercury Hg , methylmercury, arsenic, radioactivity, etc. Bacteria tests should be carried out regularly. This organization should also have a communication channel to release relavant message. The Environmental Protection Agency of the U.S. gives a list of comtaminants. The list has suggested limits, and it divides the contaminants into Among inorganic substances, limits are given to contents of antimony, arsenic, asbestos, barium, beryllum, cadmium, chromium, copper, mercury, nitrate, nitrite, selenium, and thallium. More than 50 organic compounds are on the list, and some familar ones are: acrylamide, benzene, carbon tetrachloride, chlorobenzene, 2 4 D, dichlorobenzen, dioxin, polychlorinated biphenyls (PCBs), toluene, and vinyl chloride. Many of these have a zero limit. In terms of microorganisms, , and , are checked. Furthermore, viruses, turbidity, total coliforms, and heterotrophic plate should be checked. The secondary standard lists most electrolytes as shown in this table on the right. The Secondary Drinking Water Standards are non-enforceable guidelines regulating contaminants that may cause cosmetic effects (such as skin or tooth discoloration) or aesthetic effects (such as taste, odor, or color) in drinking water. There are many brands of bottled drinking water, they have been very popular only in recent years. Do we know the bottle procedure? Is the industry regulated? Is the water quality reliable? Are all bottled water the same? Is there a brand bottled water for real good health? Do we know what should be in healthy drinking water? There is an opinion expressed in (OCWA). Check it out. The magnesium web site gave the following news release Oct. 4, 1999. This is a good summary from the report. Sports talk gives information about Sports Drinks. This is science, art, testing, and myth. However, some fundamentals should be considered. Our bodies are mostly water, about 70%. The body fluid has many different things dissolved in it, particularly salt. the salinity - varies somewhat with where you take the sample of water to measure. Don't worry about that). I recall the concentration as being 0.5%, but I'm not sure. Doctors call this "normal saline". Now if you put a human, or other animal, cell in saltwater that is the same concentration as the saltwater inside the cell, the cell pretty much just sits there. If you put it in distilled water, the cell absorbs the water through its cell membrane - called diffusion - until it eventually pops. If you put the cell in concentrated saltwater, the cell looses water. The water diffuses out of the cell through the membrane, leaving a small, shriveled up cell. What does this have to do with sports drinks? If you give someone distilled water, it seems like they would absorb the water faster because of what I just described. On the other hand, during sweating, you're loosing sodium, potassium and small quantities of other electrolytes. If you're exercising particularly long or hard, you need to replace those electrolytes. Researchers found that adding some salt to water replaced the salt lost through sweating and helped the body to get water to the cells. If you look at a label on a Gatorade or other drink, you'll find that the main electrolyte is simple salt. But if you put too much of the electrolytes in the water, the cells shrivel up just like the way I described above. I hope that helps you understand what happens. This stuff on how cells swell up or shrivel up is in high school biology books, and maybe even in something you can find in your school's library. Taste and order are sensations, and thus hard to quantify and systematize. Often, the Weber Fechner law is used. This law expresses the taste or order sensations as being proportional to the logarithm of the stimulus , with the proportional constant , S = K log R For some common substances, the minimum amounts detected by or gives the sensation as a threshold, below which no taste or order was detected. However, orderous sensation of water may be reported as threshold order number (TON). If mL of ordorous sample is diluted with mL of order-free water to be "just detectable" to the , the TON is defined as A + B TON = ------------- A Similarly, a flavour threshold number (FTN) can be defined in the same manner. A + B FTN = ------------- A except that and are volumes of samples and taste free water used. These formulations show a method of defining some quantities that are, otherwise, very diffcult to quantify. There are other methods of reporting orders and taste, and some bottling companies have their standard methods of comparison. Of course, the source of order and taste comes are organic, inorganic compounds as well as bacteria, algae. For example, mercaptans such as C H SH and ammonia offers disagreeable smell and taste. A sample of water was tested by 10 expert noses, and only 5 of them can detect an order. Thus, this is "just detected". What is the TON for this sample? Since no dilution was used, TON = A / A = 1. If equal amount of orderless water is required to dilute it so that the order is "just detected", then the test order number is (1+1)/1 = 2.

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The biosynthesis of RNA, called , proceeds in much the same fashion as the replication of DNA and also follows the base pairing principle. Again, a section of DNA double helix is uncoiled and only one of the DNA strands serves as a template for RNA polymerase enzyme to guide the synthesis of RNA. After the synthesis is complete, the RNA separates from the DNA and the DNA recoils into its helix. The differences in the composition of RNA and DNA have already been noted. In addition, RNA is not usually found as a double helix but as a single strand. However, the single polynucleotide strand may fold back on itself to form portions which have a double helix structure like the tertiary structure of proteins. The transcription of a single RNA strand is illustrated in the graphic on the left. One major difference is that the heterocyclic amine, adenine, on DNA codes for the incorporation of uracil in RNA rather than thymine as in DNA. Remember that thymine is not found in RNA and do not confuse the replacement of uracil in RNA for thymine in DNA in the transcription process. For example, thymine in DNA still codes for adenine on RNA not uracil, while the adenine on DNA codes for uracil in RNA. Note that the new RNA (red) is identical to non coding DNA with the exception of uracil where thymine was located in DNA. There are three major types of RNA which will be fully explained in a later section. Although RNA is synthesized in the nucleus, it migrates out of the nucleus into the cytoplasm where it is used in the synthesis of proteins. The RNA transcription process occurs in three stages: initiation, chain elongation, and termination. The first stage occurs when the binds to the promoter gene in the DNA. This also allows for the finding of the start sequence for the RNA polymerase. The promoter enzyme will not work unless the sigma protein is present (shown in blue in graphic). Specific sequences on the non coding strand of DNA are recognized as the signal to start the unwinding process. The recognition sequences are as follows: Non-coding DNA -5' recognition sections in bold GGCCGC AAAGTGTTAAATTGTGC Once the process has been initiated, then the RNA polymerase elongation enzyme takes over and is described in the next panel. The elongation begins when the RNA polymerase "reads" the template DNA. Only one strand of the DNA is read for the base sequence. The RNA which is synthesized is the complementary strand of the DNA. The RNA (top strand) and DNA (bottom strand) sequences in the model are: 5' -GACCAGGCA-3' 3'-TCTGGTCCGTAAA-5' In the graphic, the magenta color is the template DNA, while the green is the RNA strand. In the next reaction step, uracil triphosphate (UTP) is the next to be added to the RNA by bind and pairing with the adenine (A) nucleotide on the template DNA strand. A phosphodiester bond is formed; the RNA chain is than elongated to 10 nucleotides; and diphosphate left over would dissociate. Note: The coordinates used in this display have only the alpha carbons of the proteins RNA Polymerase.

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Most plastics crumble into ever-tinier fragments as they are exposed to sunlight and the elements. Except for the small amount that's been incinerated–and it's a very small amount–every bit of plastic ever made still exists, unless the material's molecular structure is designed to favor biodegradation. Unfortunately, cleaning up the garbage patch is not a realistic option, and unless we change our disposal and recycling habits, it will undoubtedly get bigger. One sensible solution would require manufacturers to use natural biodegradable packaging materials whenever possible, and consumers to conscientiously dispose of their plastic waste. Thus, instead of consigning all plastic trash to a land fill, some of it may provide energy by direct combustion, and some converted for reuse as a substitute for virgin plastics. The latter is particularly attractive since a majority of plastics are made from petroleum, a diminishing resource with a volatile price. The energy potential of plastic waste is relatively significant, ranging from 10.2 to 30.7MJ kg Ð, suggesting application as an energy source and temperature stabilizer in municipal incinerators, thermal power plants and cement kilns. The use of plastic waste as a fuel source would be an effective means of reducing landfill requirements while recovering energy. This, however, depends on using appropriate materials. Inadequate control of combustion, especially for plastics containing chlorine, fluorine and bromine, constitutes a risk of emitting toxic pollutants. Whether used as fuels or a source of recycled plastic, plastic waste must be separated into different categories. To this end, an identification coding system was developed by the Society of the Plastics Industry (SPI) in 1988, and is used internationally. This code, shown on the right, is a set of symbols placed on plastics to identify the polymer type, for the purpose of allowing efficient separation of different polymer types for recycling. The abbreviations of the code are explained in the following table. polyethylene terephthalate high density polyethylene polyvinyl chloride low density polyethylene polypropylene polystyrene polyesters, acrylics polyamides, teflon etc. Despite use of the recycling symbol in the coding of plastics, there is consumer confusion about which plastics are readily recyclable. In most communities throughout the United States, PETE and HDPE are the only plastics collected in municipal recycling programs. However, some regions are expanding the range of plastics collected as markets become available. (Los Angeles, for example, recycles all clean plastics numbered 1 through 7) In theory, most plastics are recyclable and some types can be used in combination with others. In many instances, however, there is an incompatibility between different types that necessitates their effective separation. Since the plastics utilized in a given manufacturing sector (e.g. electronics, automotive, etc.) is generally limited to a few types, effective recycling is often best achieved with targeted waste streams. ),

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Make sure you thoroughly understand the following essential ideas which have been presented below. The electrons constituting a chemical bond are simultaneously attracted by the electrostatic fields of the nuclei of the two bonded atoms. In a homonuclear molecule such as O the bonding electrons will be shared equally by the two atoms. In general, however, differences in the sizes and nuclear charges of the atoms will cause one of them to exert a greater attraction on the bonding pair, causing the electron cloud to be displaced toward the more strongly-attracting atom. The electronegativity of an atom denotes its relative electron-attracting power in a chemical bond. It is important to understand that electronegativity is not a measurable property of an atom in the sense that ionization energies and electron affinities are, although it can be correlated with both of these properties. The actual electron-attracting power of an atom depends in part on its chemical environment (that is, on what other atoms are bonded to it), so tabulated electronegativities should be regarded as no more than predictors of the behavior of electrons, especially in more complicated molecules. There are several ways of computing electronegativities, which are expressed on an arbitrary scale. The concept of electronegativity was introduced by Linus Pauling and his 0-4 scale continues to be the one most widely used. The 0-4 electronegativity scale of Pauling is the best known of several arbitrary scales of this kind. Electronegativity values are not directly observable, but are derived from measurable atomic properties properties such as ionization energy and electron affinity. The place of any atom on this scale provides a good indication of its ability to compete with another atom in attracting a shared electron pair to it, but the presence of bonds to other atoms, and of multiple- or nonbonding electron pairs may make predictions about the nature a given bond less reliable. An atom that has a small electronegativity is said to be . As the diagram shows, the metallic elements are generally electropositive. The position of hydrogen in this regard is worth noting; although physically a nonmetal, much of its chemistry is metal-like. When non-identical atoms are joined in a covalent bond, the electron pair will be attracted more strongly to the atom that has the higher electronegativity. As a consequence, the electrons will not be shared equally; the center of the negative charges in the molecule will be displaced from the center of positive charge. Such bonds are said to be and to possess , and they may confer a polar nature on the molecule as a whole. A polar molecule acts as an which can interact with electric fields that are created artificially or that arise from nearby ions or polar molecules. Dipoles are conventionally represented as arrows pointing in the direction of the negative end. The magnitude of interaction with the electric field is given by the of the molecule. The dipole moment corresponding to an individual bond (or to a diatomic molecule) is given by the product of the quantity of charge displaced and the bond length : \[μ = q \times r\] In SI units, is expressed in coulombs and in meters, so has the dimensions of \(C \cdot m\). If one entire electron charge is displaced by 100 pm (a typical bond length), then \[μ = (1.6022 \times 10^{–19}\; C) \times (10^{–10}\; m) = 1.6 \times 10^{–29}\; C \cdot m = 4.8 \;D\] The quantity denoted by D, the unit, is still commonly used to express dipole moments. It was named after (1884-1966), the Dutch-American physicist who pioneered the study of dipole moments and of electrical interactions between particles; he won the Nobel Prize for Chemistry in 1936. When a solution of polar molecules is placed between two oppositely-charged plates, they will tend to align themselves along the direction of the field. This process consumes energy which is returned to the electrical circuit when the field is switched off, an effect known as electrical . Measurement of the capacitance of a gas or solution is easy to carry out and serves as a means of determining the magnitude of the dipole moment of a substance. Estimate the percent ionic character of the bond in the hydrogen fluoride molecule from the experimental data shown at the right. In molecules containing more than one polar bond, the molecular dipole moment is just the vector combination of what can be regarded as individual "bond dipole moments". Being vectors, these can reinforce or cancel each other, depending on the geometry of the molecule; it is therefore not uncommon for molecules containing polar bonds to be nonpolar overall, as in the example of carbon dioxide: The zero dipole moment of CO is one of the simplest experimental methods of demonstrating the linear shape of this molecule. H O, by contrast, has a very large dipole moment which results from the two polar H–O components oriented at an angle of 104.5°. The nonbonding pairs on oxygen are a contributing factor to the high polarity of the water molecule. In molecules containing nonbonding electrons or multiple bonds, the elecronegativity difference may not correctly predict the bond polarity. A good example of this is carbon monoxide, in which the partial negative charge resides on the carbon, as predicted by its negative formal charge (below.) Electron densities in a molecule (and the dipole moments that unbalanced electron distributions can produce) are now easily calculated by molecular modeling programs. In this example, for methanol CH OH, the blue area centered on hydrogen represents a positive charge, the red area centered where we expect the lone pairs to be located represents a negative charge, while the light green around methyl is approximately neutral. The manner in which the individual bonds contribute to the dipole moment of the molecule is nicely illustrated by the series of chloromethanes: (Bear in mind that all four positions around the carbon atom are equivalent in this tetrahedral molecule, so there are only four chloromethanes.) Although the total number of valence electrons in a molecule is easily calculated, there is not always a simple and unambiguous way of determining how many reside in a particular bond or as non-bonding pairs on a particular atom. For example, one can write valid Lewis octet structures for carbon monoxide showing either a double or triple bond between the two atoms, depending on how many nonbonding pairs are placed on each: and (see Problem Example 3 below). The choice between structures such as these is usually easy to make on the principle that the more electronegative atom tends to surround itself with the greater number of electrons. In cases where the distinction between competing structures is not all that clear, an arbitrarily-calculated quantity known as the can often serve as a guide. The formal charge on an atom is the electric charge it would have if all bonding electrons were shared equally with its bonded neighbors. The formal charge on an atom is calculated by the following formula: in which the core charge is the electric charge the atom would have if all its valence electrons were removed. In simple cases, the formal charge can be worked out visually directly from the Lewis structure, as is illustrated farther on. Find the formal charges of all the atoms in the sulfuric acid structure shown here. The atoms here are hydrogen, sulfur, and double- and single-bonded oxygens. Remember that a double bond is made up of two electron-pairs. The general rule for choosing between alternative structures is that the one involving the formal charges is most favored, although the following example shows that this is not always the case. Write out some structures for carbon monoxide CO, both those that do and do not obey the octet rule, and select the "best" on the basis of the formal charges. For Carbon: 4 – 2 – 3 = ; Oxygen: 6 – 2 – 3 = For :C:O::: Carbon: 4 – 2 – 1 = ; Oxygen: 6 – 6 – 1 = For :C::O:: Carbon: 4 – 2 –2 = ; Oxygen: 6 – 4 – 2 = : All three structures are acceptable (because the formal charges add up to zero for this neutral molecule) and contribute to the overall structure of carbon monoxide, although not equally. Both experiment and more advanced models show that the triple-bonded form predominates. Formal charge, which is no more than a bookkeeping scheme for electrons, is by itself unable to predict this fact. In a species such as the thiocyanate ion \(SCN^-\) in which two structures having the same minimal formal charges can be written, we would expect the one in which the negative charge is on the more electronegative atom to predominate. The electrons in the structures of the top row are the valence electrons for each atom; an additional electron (purple) completes the nitrogen octet in this negative ion. The electrons in the bottom row are divided equally between the bonded atoms; the difference between these numbers and those above gives the formal charges. Formal charge can also help answer the question “where is the charge located?” that is frequently asked about polyatomic ions. Thus by writing out the Lewis structure for the ammonium ion NH , you should be able to convince yourself that the nitrogen atom has a formal charge of +1 and each of the hydrogens has 0, so we can say that the positive charge is localized on the central atom. This is another arbitrary way of characterizing atoms in molecules. In contrast to formal charge, in which the electrons in a bond are assumed to be shared equally, is the electric charge an atom would have if the bonding electrons were assigned exclusively to the more electronegative atom. Oxidation number serves mainly as a tool for keeping track of electrons in reactions in which they are exchanged between reactants, and for characterizing the “combining power” of an atom in a molecule or ion. The following diagram compares the way electrons are assigned to atoms in calculating formal charge and oxidation number in carbon monoxide. The introduced by G.N. Lewis showed how chemical bonds could form in the absence of electrostatic attraction between oppositely-charged ions. As such, it has become the most popular and generally useful model of bonding in all substances other than metals. A chemical bond occurs when electrons are simultaneously attracted to two nuclei, thus acting to bind them together in an energetically-stable arrangement. The is formed when two atoms are able to share a pair of electrons: In general, however, different kinds of atoms exert different degrees of attraction on their electrons, so in most cases the sharing will not be equal. One can even imagine an extreme case in which the sharing is so unequal that the resulting "molecule" is simply a pair of ions: The resulting substance is sometimes said to contain an . Indeed, the properties of a number of compounds can be adequately explained using the ionic model. But does this mean that there are really two kinds of chemical bonds, ionic and covalent? According to the ionic electrostatic model, solids such as NaCl consist of positive and negative ions arranged in a crystal lattice. Each ion is attracted to neighboring ions of opposite charge, and is repelled by ions of like charge; this combination of attractions and repulsions, acting in all directions, causes the ion to be tightly fixed in its own location in the crystal lattice. Since electrostatic forces are nondirectional, the structure of an ionic solid is determined purely by geometry: two kinds of ions, each with its own radius, will fall into whatever repeating pattern will achieve the lowest possible potential energy. Surprisingly, there are only a small number of possible structures; one of the most common of these, the lattice of NaCl, is shown here. When two elements form an ionic compound, is an electron really lost by one atom and transferred to the other one? In order to deal with this question, consider the data on the ionic solid LiF. The average radius of the neutral Li atom is about 2.52Å. Now if this Li atom reacts with an atom of F to form LiF, what is the average distance between the Li nucleus and the electron it has “lost” to the fluorine atom? The answer is 1.56Å; the electron is now closer to the lithium nucleus than it was in neutral lithium! So the answer to the above question is both yes and no: yes, the electron that was now in the 2 orbital of Li is now within the grasp of a fluorine 2 orbital, but no, the electron is now even closer to the Li nucleus than before, so how can it be “lost”? The one thing that is inarguably true about LiF is that there are more electrons closer to positive nuclei than there are in the separated Li and F atoms. But this is just the rule we stated at the beginning of this unit: It is obvious that the electron-pair bond brings about this situation, and this is the reason for the stability of the covalent bond. What is not so obvious (until you look at the numbers such as are quoted for LiF above) is that the “ionic” bond results in the same condition; even in the most highly ionic compounds, both electrons are close to both nuclei, and the resulting mutual attractions bind the nuclei together. This being the case, is there really any fundamental difference between the ionic and covalent bond? The answer, according to modern chemical thinking is probably “no”; in fact, there is some question as to whether it is realistic to consider that these solids consist of “ions” in the usual sense. The preferred picture that seems to be emerging is one in which the electron orbitals of adjacent atom pairs are simply skewed so as to place more electron density around the “negative” element than around the “positive” one. This being said, it must be reiterated that the ionic model of bonding is a useful one for many purposes, and there is nothing wrong with using the term “ionic bond” to describe the interactions between the atoms in the very small class of “ionic solids” such as LiF and NaCl. If there is no such thing as a “completely ionic” bond, can we have one that is completely covalent? The answer is yes, if the two nuclei have . This situation is guaranteed to be the case with homonuclear diatomic molecules-- molecules consisting of two identical atoms. Thus in Cl , O , and H , electron sharing between the two identical atoms must be exactly even; in such molecules, the center of positive charge corresponds exactly to the center of negative charge: halfway between the two nuclei. In most covalent bonds, we think of the electron pair as having a dual parentage, one electron being contributed by each atom. There are, however, many cases in which both electrons come from only one atom. This can happen if the donor atom has a non-bonding pair of electrons and the acceptor atom has a completely empty orbital that can accommodate them. These are called dative or coordinate covalent bonds. This is the case, for example, with boron trifluoride and ammonia. In BF , one the 2p orbitals is unoccupied and can accommodate the lone pair on the nitrogen atom of ammonia. The electron acceptor, BF , acts as a here, and NH is the . Bonds of this type (sometimes known as or bonds) tend to be rather weak (usually 50-200kJ/mol); in many cases the two joined units retain sufficient individuality to justify writing the formula as a molecular or .

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/20%3A_Organic_Chemistry/20.0%3A_Prelude_to_Organic_Chemistry

All living things on earth are formed mostly of carbon compounds. The prevalence of carbon compounds in living things has led to the epithet “carbon-based” life. The truth is we know of no other kind of life. Early chemists regarded substances isolated from (plants and animals) as a different type of matter that could not be synthesized artificially, and these substances were thus known as . The widespread belief called vitalism held that organic compounds were formed by a vital force present only in living organisms. The German chemist Friedrich was one of the early chemists to refute this aspect of vitalism, when, in 1828, he reported the synthesis of urea, a component of many body fluids, from nonliving materials. Since then, it has been recognized that organic molecules obey the same natural laws as inorganic substances, and the category of organic compounds has evolved to include both natural and synthetic compounds that contain carbon. Some carbon-containing compounds are classified as organic, for example, carbonates and cyanides, and simple oxides, such as and CO . Although a single, precise definition has yet to be identified by the chemistry community, most agree that a defining trait of organic molecules is the presence of carbon as the principal element, bonded to hydrogen and other carbon atoms. Today, organic compounds are key components of plastics, soaps, perfumes, sweeteners, fabrics, pharmaceuticals, and many other substances that we use every day. The value to us of organic compounds ensures that organic chemistry is an important discipline within the general field of chemistry. In this chapter, we discuss why the element carbon gives rise to a vast number and variety of compounds, how those compounds are classified, and the role of organic compounds in representative biological and industrial settings.  

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/06%3A_Properties_of_Gases/6.02%3A_Ideal_Gas_Model_-_The_Basic_Gas_Laws

Make sure you thoroughly understand the following essential ideas which have been presented above, and be able to state them in your own words. The "pneumatic" era of chemistry began with the discovery of the around 1650 which clearly established that gases are a form of matter. The ease with which gases could be studied soon led to the discovery of numerous empirical (experimentally-discovered) laws that proved fundamental to the later development of chemistry and led indirectly to the atomic view of matter. These laws are so fundamental to all of natural science and engineering that everyone learning these subjects needs to be familiar with them. Robert Boyle (1627-91) showed that the volume of air trapped by a liquid in the closed short limb of a J-shaped tube decreased in exact proportion to the pressure produced by the liquid in the long part of the tube. The trapped air acted much like a spring, exerting a force opposing its compression. Boyle called this effect " ", and published his results in a pamphlet of that title. Some of Boyle's actual data are shown in Table \(\Page {1}\). Boyle's law can be expressed as \[PV = \text{constant} \nonumber\] or, equivalently, \[P_1V_1 = P_2V_2\] These relations hold true if the number of molecules and the temperature are constant. This is a relation of ; any change in the pressure is exactly compensated by an opposing change in the volume. As the pressure decreases toward zero, the volume will increase without limit. Conversely, as the pressure is increased, the volume decreases, but can never reach zero. There will be a separate plot for each temperature; a single plot is therefore called an . Shown here are some isotherms for one mole of an ideal gas at several different temperatures. Each plot has the shape of a — the locus of all points having the property = , where is a constant. You will see later how the value of this constant ( =25 for the 300K isotherm shown here) is determined. It is very important that you understand this kind of plot which governs any relationship of inverse proportionality. You should be able to sketch out such a plot when given the value of any one ( ) pair. A related type of plot with which you should be familiar shows the product as a function of the pressure. You should understand why this yields a straight line, and how this set of plots relates to the one immediately above. In an industrial process, a gas confined to a volume of 1 L at a pressure of 20 atm is allowed to flow into a 12-L container by opening the valve that connects the two containers. What will be the final pressure of the gas? The final volume of the gas is (1 + 12)L = 13 L. The gas expands in inverse proportion two volumes \[P_2 = (20 \,atm) (1 \,L ÷ 13 \,L) = 1.5 \,atm \nonumber\] All matter expands when heated, but gases are special in that their degree of expansion is independent of their composition. The French scientists Jacques Charles (1746-1823) and Joseph Gay-Lussac (1778-1850) independently found that if the pressure is held constant, the volume of any gas changes by the same fractional amount (1/273 of its value) for each C° change in temperature. The air pressure in a car tire is 30 psi (pounds per square inch) at 10°C. What will be pressure be after driving has raised its temperature to 45°C ? (Assume that the volume remains unchanged.) The gas expands in direct proportion to the ratio of the absolute temperatures: \[P_2 = (30\, psi) × (318\,K ÷ 283\,K) = 33.7\, psi \nonumber\] The relation between the temperature of a gas and its volume has long been known. In 1702, Guillaume Amontons (1163-1705), who is better known for his early studies of friction, devised a thermometer that related the temperature to the volume of a gas. Robert Boyle had observed this inverse relationship in 1662, but the lack of any uniform temperature scale at the time prevented them from establishing the relationship as we presently understand it. Jacques Charles discovered the law that is named for him in the 1780s, but did not publish his work. John Dalton published a form of the law in 1801, but the first thorough published presentation was made by Gay-Lussac in 1802, who acknowledged Charles' earlier studies. The buoyancy that lifts a hot-air balloon into the sky depends on the difference between the density (mass ÷ volume) of the air entrapped within the balloon's envelope, compared to that of the air surrounding it. When a balloon on the ground is being prepared for flight, it is first partially inflated by an external fan, and possesses no buoyancy at all. Once the propane burners are started, this air begins to expand according to Charles' law. After the warmed air has completely inflated the balloon, further expansion simply forces excess air out of the balloon, leaving the weight of the diminished mass of air inside the envelope smaller than that of the greater mass of cooler air that the balloon displaces. Jacques Charles collaborated with the Montgolfier brothers whose hot-air balloon made the world's first manned balloon flight in June, 1783. Ten days later, Charles himself co-piloted the first hydrogen-filled balloon. Gay-Lussac, who had a special interest in the composition of the atmosphere, also saw the potential of the hot-air balloon, and in 1804 he ascended to a then-record height of 6.4 km. In the same 1808 article in which Gay-Lussac published his observations on the thermal expansion of gases, he pointed out that when two gases react, they do so in volume ratios that can always be expressed as small whole numbers. This came to be known as the . These "small whole numbers" are of course the same ones that describe the "combining weights" of elements to form simple compounds, as described in the lesson dealing with the simplest formulas. The Italian scientist Amedeo Avogadro (1776-1856) drew the crucial conclusion: these volume ratios must be related to the relative numbers of molecules that react, and thus the famous "E.V.E.N principle": thus predicts a relation between the number of moles of a gas and its volume. This relationship, originally known as , was crucial in establishing the formulas of simple molecules at a time (around 1811) when the distinction between atoms and molecules was not clearly understood. In particular, the existence of diatomic molecules of elements such as H , O , and Cl was not recognized until the results of combining-volume experiments such as those depicted below could be interpreted in terms of the E.V.E.N. principle. Early chemists made the mistake of assuming that the formula of water is HO. This led them to miscalculate the molecular weight of oxygen as 8 (instead of 16). If this were true, the reaction H + O → HO would correspond to the following combining volumes results according to the E.V.E.N principle: But a similar experiment on the formation of hydrogen chloride from hydrogen and chlorine yielded twice the volume of HCl that was predicted by the the assumed reaction H + Cl → HCl. This could be explained only if hydrogen and chlorine were diatomic molecules: This made it necessary to re-visit the question of the formula of water. The experiment immediately confirmed that the correct formula of water is H O: This conclusion was also seen to be consistent with the observation, made a few years earlier by the English chemists Nicholson and Carlisle that the reverse of the above reaction, brought about by the electrolytic decomposition of water, yields hydrogen and oxygen in a 2:1 volume ratio. If the variables and (the number of moles) have known values, then a gas is said to be in a definite , meaning that all other physical properties of the gas are also defined. The relation between these is known as an . By combining the expressions of Boyle's, Charles', and Avogadro's laws (you should be able to do this!) we can write the very important equation of state \[PV=NRT\] where the proportionality constant is known as the . This is one of the few equations you commit to memory in this course; you should also know the common value and units of \(R\). An is defined as a hypothetical substance that obeys the ideal gas equation of state. Take note of the word "hypothetical" here. No real gas (whose molecules occupy space and interact with each other) can behave in a truly ideal manner. But we will all behave more and more like an ideal gas as the pressure approaches zero. A pressure of only 1 atm is sufficiently close to zero to make this relation useful for most gases at this pressure. Many textbooks show formulas, such as \(P_1V_1 = P_2V_2\) for Boyle's law. ; if you really understand the meanings of these laws as stated above, you can easily derive them on the rare occasions when they are needed. In order to depict the relations between the three variables and we need a three-dimensional graph. A biscuit made with baking powder has a volume of 20 mL, of which one-fourth consists of empty space created by gas bubbles produced when the baking powder decomposed to CO . What weight of \(\ce{NaHCO3}\) was present in the baking powder in the biscuit? Assume that the gas reached its final volume during the baking process when the temperature was 400°C. (Baking powder consists of sodium bicarbonate mixed with some other solid that produces an acidic solution on addition of water, initiating the reaction \[\ce{NaHCO3(s) + H^{+} → Na^+ + H2O + CO2} \nonumber\] Use the ideal gas equation to find the number of moles of CO gas; this will be the same as the number of moles of NaHCO (84 g mol ) consumed : \[n=\frac{(1 \mathrm{atm})(0.005 \mathrm{L})}{\left(.082 \mathrm{L} \mathrm{atm} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)(673 \mathrm{K})}=9.1 \times 10^{-6} \,\mathrm{mol} \nonumber\] \[9.1 \mathrm{E}-6 \mathrm{mol} \times 84 \mathrm{g} \mathrm{mol}^{-1}=0.0076 \mathrm{g} \nonumber\]

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Ethers/Reactivity_of_Ethers/Cleaving_Ethers

The most common reaction of ethers is cleavage of the C–O bond by strong acids. This may occur by S 1 or E1 mechanisms for 3º-alkyl groups or by an S 2 mechanism for 1º-alkyl groups. Some examples are shown in the following diagram. The conjugate acid of the ether is an intermediate in all these reactions, just as conjugate acids were intermediates in certain alcohol reactions. The first two reactions proceed by a sequence of S 2 steps in which the iodide or bromide anion displaces an alcohol in the first step, and then converts the conjugate acid of that alcohol to an alkyl halide in the second. Since S 2 reactions are favored at least hindered sites, the methyl group in example #1 is cleaved first. The 2º-alkyl group in example #3 is probably cleaved by an S 2 mechanism, but the S 1 alternative cannot be ruled out. The phenol formed in this reaction does not react further, since S 2, S 1 and E1 reactions do not take place on . The last example shows the cleavage of a 3º-alkyl group by a strong acid. Acids having poorly nucleophilic conjugate bases are often chosen for this purpose so that E1 products are favored. The reaction shown here (#4) is the reverse of the tert-butyl ether preparation . Ethers in which oxygen is bonded to 1º- and 2º-alkyl groups are subject to peroxide formation in the presence of air (gaseous oxygen). This reaction presents an additional hazard to the use of these flammable solvents, since peroxides decompose explosively when heated or struck. The mechanism of peroxide formation is believed to be free radical in nature (note that molecular oxygen has two unpaired electrons). \[\ce{R–O–CH(CH3)2 + O2 }\rightarrow \underset{\text{a peroxide}}{\ce{R–O–C(CH3)2–O–O–H }}\] Because of their chemical stability, ethers may be used to protect hydroxyl functions from undergoing unwanted reactions.

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Quantifying_Nature/Temperature_Basics

The concept of temperature may seem familiar to you, but many people confuse temperature with heat. is a measure of how hot or cold an object is relative to another object (its thermal energy content), whereas is the flow of thermal energy between objects with different temperatures. Three different scales are commonly used to measure temperature: Fahrenheit (expressed as °F), Celsius (°C), and Kelvin (K). Thermometers measure temperature by using materials that expand or contract when heated or cooled. Mercury or alcohol thermometers, for example, have a reservoir of liquid that expands when heated and contracts when cooled, so the liquid column lengthens or shortens as the temperature of the liquid changes. The Fahrenheit temperature scale was developed in 1717 by the German physicist Gabriel Fahrenheit, who designated the temperature of a bath of ice melting in a solution of salt as the zero point on his scale. Such a solution was commonly used in the 18th century to carry out low-temperature reactions in the laboratory. The scale was measured in increments of 12; its upper end, designated as 96°, was based on the armpit temperature of a healthy person—in this case, Fahrenheit’s wife. Later, the number of increments shown on a thermometer increased as measurements became more precise. The upper point is based on the boiling point of water, designated as 212° to maintain the original magnitude of a Fahrenheit degree, whereas the melting point of ice is designated as 32°. The Celsius scale was developed in 1742 by the Swedish astronomer Anders Celsius. It is based on the melting and boiling points of water under normal atmospheric conditions. The current scale is an inverted form of the original scale, which was divided into 100 increments. Because of these 100 divisions, the Celsius scale is also called the . Lord Kelvin, working in Scotland, developed the Kelvin scale in 1848. His scale uses molecular energy to define the extremes of hot and cold. Absolute zero, or 0 K, corresponds to the point at which molecular energy is at a minimum. The Kelvin scale is preferred in scientific work, although the Celsius scale is also commonly used. Temperatures measured on the Kelvin scale are reported simply as K, not °K. A Comparison of the Fahrenheit, Celsius, and Kelvin Temperature Scales. Because the difference between the freezing point of water and the boiling point of water is 100° on both the Celsius and Kelvin scales, the size of a degree Celsius (°C) and a kelvin (K) are precisely the same. In contrast, both a degree Celsius and a kelvin are 9/5 the size of a degree Fahrenheit (°F). The kelvin is the same size as the Celsius degree, so measurements are easily converted from one to the other. The freezing point of water is 0°C = 273.15 K; the boiling point of water is 100°C = 373.15 K. The Kelvin and Celsius scales are related as follows: Degrees on the Fahrenheit scale, however, are based on an English tradition of using 12 divisions, just as 1 ft = 12 in. The relationship between degrees Fahrenheit and degrees Celsius is as follows:where the coefficient for degrees Fahrenheit is exact. (Some calculators have a function that allows you to convert directly between °F and °C.) There is only one temperature for which the numerical value is the same on both the Fahrenheit and Celsius scales: −40°C = −40°F. The relationship between the scales are as follows: °C = (5/9)*(°F-32) °F = (9/5)*(°C)+32 Convert the temperature of the surface of the sun (5800 K) and the boiling points of gold (3080 K) and liquid nitrogen (77.36 K) to °C and °F.A student is ill with a temperature of 103.5°F. What is her temperature in °C and K?

https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Basics/Valence_and_the_Periodic_Table

Many chemists were interested in knowing the "chemical equivalents" of different substances. For instance, what mass of acid A neutralizes base B? These numbers were easy to measure and practically useful. But what about elements? For instance, what mass of element A reacts with 1 g of element B? Knowing these numbers could be very useful, but it was hard to relate the equivalent masses to the actual masses of the atoms because they didn't know the formulas. There were two good ways available to figure this out. One was Avogadro's hypothesis based on Gay-Lussac's law, which allowed chemists to relate the equivalent masses to equivalent volumes. The other was Faraday's law, which measured masses of elements produced by a set amount of current. However, Berzelius, who was working hardest on this problem, didn't believe either Avogadro or Faraday. Berzelius' beliefs held up the progress of science for about 50 years. Eventually, Cannizzaro revived Avogadro's hypothesis at a big meeting of chemists in 1860. His paper explaining how to calculate molecular weights was distributed to everyone, including Julius Lothar Meyer, who wrote that when he read it "doubts disappeared and a feeling of quiet certainty took their place". Avogadro's hypothesis let chemists figure out atomic weights and formulas together. Once Cannizzaro convinced most chemists to accept it, chemists were able to study the actual atomic weights and formulas. Scientists soon observed patterns in the of the different elements. Valence is the number of connections an atom tends to form. H is defined to have a valence of 1. For instance: By the 1860s, ~60 elements were known. Using Cannizzaro's atomic weights, Mendeleev and Lothar Meyer made a great discovery, the periodic law: If you arrange the elements by their atomic weights, there is a periodic repetition in properties such as valence. The modern version of this periodic arrangement is the . Also, within a group (sharing a valence) properties like density, boiling point, heat capacity, etc follow a simple progression. Mendeleev used this to predict the properties of undiscovered elements. Here's a smaller version of the periodic table that leaves out the elements Mendeleev and Meyer found most problematic (transition metals, rare earths) and the group that hadn't been discovered yet (noble gases). Notice how the the valences repeat every 7 elements when they are arranged according to atomic mass. But there were some problems with the table. For instance, tellurium (Te) was clearly a chalcogen, in the oxygen family, and iodine (I) was clearly a halogen, based on their properties, but the weights were wrong. (Check the table!) Mendeleev said that the atomic weights must not have been determined correctly, but they were correct.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Synthesis_of_Alkanes/Wurtz_reaction

Wurtz reaction is coupling of haloalkanes using sodium metal in solvent like dry ether \[2R-X + 2Na \rightarrow R-R + 2Na^+X^−\] The reaction consists of a halogen-metal exchange involving the free radical species R• (in a similar fashion to the formation of a Grignard reagent and then the carbon-carbon bond formation in a nucleophilic substitution reaction.) One electron from the metal is transferred to the halogen to produce a metal halide and an alkyl radical. \[R-X + M → R^{\cdot} + M^+X^−\] The alkyl radical then accepts an electron from another metal atom to form an alkyl anion and the metal becomes cationic. This intermediate has been isolated in a several cases. \[R^{\cdot} + M → R^−M^+\] The nucleophilic carbon of the alkyl anion then displaces the halide in an S 2 reaction, forming a new carbon-carbon covalent bond. \[R^−M^+ + R-X → R-R + M^+X^−\]

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.11%3A_Proteins

The ability to serve a variety of functions is characteristic of most biomolecules. Nowhere is this versatility better exemplified than by the proteins. Perhaps because of their many functions, proteins are the most abundant organic molecules in living cells, constituting more than 50 percent of the mass once water is removed. It is estimated that the human body contains well over a million different kinds of protein, and even a single-cell organism contains thousands. Each of these is a polymer of amino acids which has a highly specific composition, a unique molecular weight (usually in the range from 6000 to 1 000 000) and its own sequence of different amino acids along the polymer chain. Proteins may be divided into two major classes on the basis of their behavior when reacted with water. The products obtained upon hydrolysis of are all amino acids. In the case of other organic and/or inorganic substances are obtained. The non-amino acid portions of conjugated proteins may consist of metals, lipids, sugars, phosphate, or other types of molecules. These components are referred to as . Proteins may also be subdivided on the basis of their molecular shape or conformation. In the fibrous proteins long polymer chains are arranged parallel or nearly parallel to one another to give long fibers or sheets. This arrangement results in physically tough materials which do not dissolve in water. The fibrous proteins are fundamental components of structural tissues such as tendons, bone, hair, horn, leather, claws, and feathers. By contrast, polymer chains of the fold back on themselves to produce compact, nearly spherical shapes. Most globular proteins are water soluble and hence are relatively mobile within a cell. Some examples are enzymes, antibodies, hormones, toxins, and substances such as hemoglobin whose function is to transport simple molecules or even electrons from one place to another. The enzyme trypsin, is a typical globular protein. Another class of proteins are the , which, as the name would suggest, reside in a cell's lipid bilayer membrane. Such proteins can act as channels for ions or other molecules unable to pass through the lipid bilayer; as signal transducers, able to respond to signal molecules on one side of a membrane to begin a molecular response on the other side of the membrane; or as anchors of other molecules to the cell membrane, to name a few exemplars of membrane protein function. Because these proteins interface with non-polar portions of the lipid bilayer, they do no maintain function and structure in an aqueous solution, making them far more difficult to study than globular proteins or fibrous proteins. The enzymes are the most extensive and highly specialized class of proteins. The structure and mode of action of trypsin, a typical enzyme, were described in enzymes. Most enzyme-catalyzed processes occur from 10 to 10 times faster than the uncatalyzed reactions. Specificity is so great in some cases that only one particular molecule can serve as the enzyme’s substrate. Even closely related structures are unable to fit the active site, and so their reactions cannot be sped up. Often a enzyme will be catalytically inactive until the correct substrates enter the active site and induce a conformational switch of the enzyme to its active form. Enzymes are of crucial importance to living organisms because they catalyze nearly every important reaction in cell metabolism. In cases where it is necessary to carry out a nonspontaneous reaction, enzymes are capable of coupling a reaction having a negative free-energy change to the desired one. When a living system reproduces, grows, or repairs damage caused by the external environment, enzymes catalyze the necessary reactions. Some enzymes are even stimulated or inhibited by the presence of smaller molecules. This permits regulation of the concentrations of certain substances because the latter can turn off enzymes which initiate their synthesis. On a molecular level enzymes are the means by which the mechanism of a cell is kept running.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/17%3A_Electrochemical_Cells/17.08%3A_Galvanic_Cells

In an cell electrical energy is consumed and an otherwise is reversed. A galvanic cell, on the other hand, produces electrical energy as a result of a spontaneous redox process. The electron transfer characteristic of such a process is made to occur in two separate half-cells. Electrons released during an oxidation half-equation must flow through a wire or other external circuit before they can be accepted in a reduction half-equation. Consequently an electrical current is made to flow. A typical galvanic cell, the , was used to power telegraphs 100 years ago. This cell is based on the spontaneous redox reaction \[\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)\label{1} \] (You can verify that this reaction is spontaneous by dipping a piece of zinc metal in a copper sulfate solution. In a short time the surface of the zinc will become plated with red-brown copper metal.) The half-equations \[\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + \text{2}e^{-}\label{2} \] \[\text{Cu}^{2+}(aq) + \text{2}e^{-} \rightarrow \text{Cu}(s)\label{3} \] indicate that for each mole of zinc which is oxidized and goes into solution as zinc ions, 2 mol electrons are transferred to copper ions, converting them to copper atoms. To produce electrical current we must prevent the Zn( ) from contacting the Cu ( ) ions and transferring the electrons directly. This is done in the Daniell cell by pouring a concentrated copper sulfate solution into the bottom of a glass jar and then carefully pouring a layer of less concentrated zinc sulfate solution above it. Because it contains less solute per unit volume, the zinc sulfate solution is less dense. It floats on the copper sulfate and does not mix with it. Therefore a copper electrode placed in the bottom of the jar contacts only Cu ( ) ions, and a zinc electrode suspended in the zinc sulfate solution contacts only Zn ( ) ions. In the laboratory it is more convenient to set up a cell based on the Zn + Cu reaction, as shown in Figure \(\Page {1}\). Two electrodes or half-cells are separated by a . This contains an electrolyte, KCl, so that current can flow from one half-cell to the other, but the contents of the two half-cells cannot mix. The left-hand electrode in Figure \(\Page {1}\) is a Zn rod dipping in a solution of ZnSO . Thus both components of the Zn /Zn redox couple are present, and the metal electrode can conduct electrons produced by Eq. \(\ref{2}\) to the wire in the external circuit. Since of Zn to Zn occurs at the left-hand electrode, this electrode is the . The right-hand electrode is a strip of Cu dipping in a solution of CuSO . Here both components of the Cu /Cu redox couple are present, and Eq. \(\ref3\) can occur. Electrons supplied by the external circuit are conducted through Cu to the electrode surface, where they combine with Cu ions to produce more Cu. Since occurs at this right-hand electrode, this electrode is the . The net effect of the two half-cells is that electrons are forced into the external circuit at the anode and withdrawn from it at the cathode. This will cause current to flow, or, if current is prevented from flowing by a device such as the voltmeter in Figure \(\Page {1}\), it will cause an electrical potential difference (voltage) to build up. The components of the redox couples at the electrodes in a galvanic cell need not always be a solid and a species in solution. This is evident from Figure \(\Page {2}\). In this case the spontaneous redox reaction \[\text{2Fe}^{2+}(aq) + \text{Cl}_2(g) \rightarrow \text{2Fe}^{3+}(aq) + \text{2Cl}^{-}(aq)\label{4} \] is involved. The oxidation half-equation at the anode is \[\text{Fe}^{3+}(aq) \rightarrow \text{Fe}^{2+}(aq) + e^{-}\label{5} \] Thus at the right-hand electrode in Figure \(\Page {2}\) both components of the redox couple are in aqueous solution. Reaction \(\ref{5}\) occurs at the surface of the platinum wire, which conducts the released electrons to the external circuit. The left-hand electrode in Figure \(\Page {2}\) is a . It consists of a platinum strip dipping in a solution which contains chloride ions. The electrode is surrounded by a glass tube through which chlorine gas can be pumped. At this electrode the reaction is a reduction: \[\text{Cl}_2(g) + \text{2}e^{-} \rightarrow \text{2Cl}^{-}(aq)\label{6} \] Therefore the left-hand electrode is the cathode. Since electrons are forced into the external circuit at the anode and withdrawn at the cathode, electrons flow from right to left in this cell.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/11%3A_Solutions_and_Colloids/11.1%3A_The_Dissolution_Process

An earlier chapter of this text introduced , defined as homogeneous mixtures of two or more substances. Often, one component of a solution is present at a significantly greater concentration, in which case it is called the solvent. The other components of the solution present in relatively lesser concentrations are called solutes. Sugar is a covalent solid composed of sucrose molecules, \(\ce{C12H22O11}\). When this compound dissolves in water, its molecules become uniformly distributed among the molecules of water: \[\ce{C12H22O11 (s) ⟶ C12H22O11 (aq) } \label{Eq1} \] The subscript “aq” in the equation signifies that the sucrose molecules are solutes and are therefore throughout the (water is the solvent). Although sucrose molecules are heavier than water molecules, they remain dispersed throughout the solution; gravity does not cause them to “settle out” over time. Potassium dichromate, \(\ce{K_2Cr_2O_7}\), is an ionic compound composed of colorless potassium ions, \(\mathrm{K^+}\), and orange dichromate ions, \(\ce{Cr_2O_7^{2−}}\). When a small amount of solid potassium dichromate is added to water, the compound dissolves and dissociates to yield potassium ions and dichromate ions uniformly distributed throughout the mixture (Figure \(\Page {1}\)), as indicated in this equation: \[\ce{K2Cr2O7(s) ⟶ 2K^{+} (aq) + Cr2O7^{2-} (aq)} \label{Eq2} \] As with the mixture of sugar and water, this mixture is also an aqueous solution. Its solutes, potassium and dichromate ions, remain individually dispersed among the solvent (water) molecules. Water is used so often as a solvent that the word solution has come to imply an aqueous solution to many people. However, almost any gas, liquid, or solid can act as a solvent. Many are solid solutions of one metal dissolved in another; for example, five-cent coins contain nickel dissolved in copper. Air is a gaseous solution, a homogeneous mixture of nitrogen, oxygen, and several other gases. Oxygen (a gas), alcohol (a liquid), and sugar (a solid) all dissolve in water (a liquid) to form liquid solutions. Table \(\Page {1}\) gives examples of several different solutions and the phases of the solutes and solvents. Solutions exhibit these defining traits: The formation of a solution is an example of a , a process that occurs under specified conditions without the requirement of energy from some external source. Sometimes we stir a mixture to speed up the dissolution process, but this is not necessary; a homogeneous solution would form if we waited long enough. The topic of spontaneity is critically important to the study of chemical thermodynamics and is treated more thoroughly in a later chapter of this text. For purposes of this chapter’s discussion, it will suffice to consider two criteria that , but do not guarantee, the spontaneous formation of a solution: In the process of dissolution, an internal energy change often, but not always, occurs as heat is absorbed or evolved. An increase in disorder always results when a solution forms. When the strengths of the intermolecular forces of attraction between solute and solvent species in a solution are no different than those present in the separated components, the solution is formed with no accompanying energy change. Such a solution is called an . A mixture of ideal gases (or gases such as helium and argon, which closely approach ideal behavior) is an example of an ideal solution, since the entities comprising these gases experience no significant intermolecular attractions. When containers of helium and argon are connected, the gases spontaneously mix due to diffusion and form a solution (Figure \(\Page {2}\)). The formation of this solution clearly involves an increase in disorder, since the helium and argon atoms occupy a volume twice as large as that which each occupied before mixing. Ideal solutions may also form when structurally similar liquids are mixed. For example, mixtures of the alcohols methanol (CH OH) and ethanol (C H OH) form ideal solutions, as do mixtures of the hydrocarbons pentane, \(\ce{C5H12}\), and hexane, \(\ce{C6H14}\). Placing methanol and ethanol, or pentane and hexane, in the bulbs shown in Figure \(\Page {2}\) will result in the same diffusion and subsequent mixing of these liquids as is observed for the He and Ar gases (although at a much slower rate), yielding solutions with no significant change in energy. Unlike a mixture of gases, however, the components of these liquid-liquid solutions do, indeed, experience intermolecular attractive forces. But since the molecules of the two substances being mixed are structurally very similar, the intermolecular attractive forces between like and unlike molecules are essentially the same, and the dissolution process, therefore, does not entail any appreciable increase or decrease in energy. These examples illustrate how diffusion alone can provide the driving force required to cause the spontaneous formation of a solution. In some cases, however, the relative magnitudes of intermolecular forces of attraction between solute and solvent species may prevent dissolution. Three types of intermolecular attractive forces are relevant to the dissolution process: solute-solute, solvent-solvent, and solute-solvent. As illustrated in Figure \(\Page {3}\), the formation of a solution may be viewed as a stepwise process in which energy is consumed to overcome solute-solute and solvent-solvent attractions (endothermic processes) and released when solute-solvent attractions are established (an exothermic process referred to as ). The relative magnitudes of the energy changes associated with these stepwise processes determine whether the dissolution process overall will release or absorb energy. In some cases, solutions do not form because the energy required to separate solute and solvent species is so much greater than the energy released by solvation. For example, cooking oils and water will not mix to any appreciable extent to yield solutions (Figure \(\Page {4}\)). Hydrogen bonding is the dominant intermolecular attractive force present in liquid water; the nonpolar hydrocarbon molecules of cooking oils are not capable of hydrogen bonding, instead being held together by dispersion forces. Forming an oil-water solution would require overcoming the very strong hydrogen bonding in water, as well as the significantly strong dispersion forces between the relatively large oil molecules. And, since the polar water molecules and nonpolar oil molecules would not experience very strong intermolecular attraction, very little energy would be released by solvation. On the other hand, a mixture of ethanol and water will mix in any proportions to yield a solution. In this case, both substances are capable of hydrogen bonding, and so the solvation process is sufficiently exothermic to compensate for the endothermic separations of solute and solvent molecules. As noted at the beginning of this module, spontaneous solution formation is favored, but not guaranteed, by exothermic dissolution processes. While many soluble compounds do, indeed, dissolve with the release of heat, some dissolve endothermically. Ammonium nitrate (NH NO ) is one such example and is used to make instant cold packs for treating injuries like the one pictured in Figure \(\Page {5}\). A thin-walled plastic bag of water is sealed inside a larger bag with solid NH NO . When the smaller bag is broken, a solution of NH NO forms, absorbing heat from the surroundings (the injured area to which the pack is applied) and providing a cold compress that decreases swelling. Endothermic dissolutions such as this one require a greater energy input to separate the solute species than is recovered when the solutes are solvated, but they are spontaneous nonetheless due to the increase in disorder that accompanies formation of the solution. A solution forms when two or more substances combine physically to yield a mixture that is homogeneous at the molecular level. The solvent is the most concentrated component and determines the physical state of the solution. The solutes are the other components typically present at concentrations less than that of the solvent. Solutions may form endothermically or exothermically, depending upon the relative magnitudes of solute and solvent intermolecular attractive forces. Ideal solutions form with no appreciable change in energy.

https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Lewis_Bonding_Theory/Polarity_of_Chemical_Bonds

means separation, in this case of electrical charge. If a bonding pair of electrons are pulled more toward one atom and away from the other, this will cause the first atom to be partially negatively charged, and the second to be partially positively charged. This will create an electric dipole moment, such as the the dipole moment in water that makes water so good as a solvent. Although covalent and ionic substances might seem really different, with Lewis theory we can think of them as being basically similar. At one extreme, we have a complete transfer of an electron, such as in CsF, making a positive and negative ion that are then attracted to each other. At the other extreme, we have a completely equally shared pair, such as in F . In between, we have bonds with unequal sharing. Either way, usually all atoms will have a , either by sharing or by losing electrons. In addition to average polarity of a bond, we can have temporary polarity. The electrons move around, and sometimes it will happen that both of them move toward one atom. In general, the more electrons an atom has, the looser they are held. For example, iodine has 53 electrons, which is a lot! They can move around pretty easily. Even though I has a non-polar bond on average, because the atoms are the same, it can easily become polar because the electrons are held loosely. In solution, I can split a little bit into I and I . To keep these ideas separate, means the permanent average separation of charge, and means the ability to become polarized temporarily. In a non-polar polarizable molecule like I , the average polarity is 0, but if we take many precise measurements of the instantaneous (very short time) polarity, many of them will be far from 0. In a polar, non-polarizable molecule like HF, all the instantaneous measurements will be very similar, but the average will not be 0. AgI is both polar and polarizable. Both of these have important effects on properties of materials, such as reactivity, solubility, and boiling point, that we will talk about later.

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When a crystalline solid melts, it loses its rigid form and adopts the shape of its container. At the same time there is usually an increase in volume of a few percent. On the molecular level we can interpret this as a breakdown in the regular structure of the solid. As temperature rises toward the melting point, the molecules vibrate more and more strongly. Above the melting point, these vibrations are so energetic that they overcome the forces holding the molecules in the crystal lattice. In the two animations below from the , one can compare solids and liquids. The animation on the left is that of a solid, compact and held tightly together by intermolecular forces. Pressing the play button on the bottom of the screen has little effect. On the other hand, pressing the play button on the liquid animation has a huge effect. The molecules in the liquid begin to move around rapidly, in stark contrast to the stability of the solid. The molecules no longer vibrate around an average position but begin to slide past each other. The regular arrangement of the crystal disappears, but the molecules have not escaped each other’s attractive influence (as can be seen by the dotted lines representing attractive forces in the animation). The very small volume change which occurs on melting shows that the molecules have moved apart to only a very limited extent and that there can be only a few gaps caused by the less-regular packing. This view is confirmed by the experimental fact that liquids, as opposed to gases, are very difficult to compress. Even at the bottom of the deepest oceans, under pressures of thousands of atmospheres, the density of water is only minutely larger than at the surface.

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The postulates of the kinetic molecular theory of gases ignore both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive. In reality, however, all gases have nonzero molecular volumes. Furthermore, the molecules of real gases interact with one another in ways that depend on the structure of the molecules and therefore differ for each gaseous substance. In this section, we consider the properties of real gases and how and why they differ from the predictions of the ideal gas law. We also examine liquefaction, a key property of real gases that is not predicted by the kinetic molecular theory of gases. For an ideal gas, a plot of \(PV/nRT\) versus \(P\) gives a horizontal line with an intercept of 1 on the \(PV/nRT\) axis. Real gases, however, show significant deviations from the behavior expected for an ideal gas, particularly at high pressures (part (a) in Figure \(\Page {1}\)). Only at relatively low pressures (less than 1 atm) do real gases approximate ideal gas behavior (part (b) in Figure \(\Page {1}\)). Real gases also approach ideal gas behavior more closely at higher temperatures, as shown in Figure \(\Page {2}\) for \(N_2\). Why do real gases behave so differently from ideal gases at high pressures and low temperatures? Under these conditions, the two basic assumptions behind the ideal gas law—namely, that gas molecules have negligible volume and that intermolecular interactions are negligible—are no longer valid. Because the molecules of an ideal gas are assumed to have zero volume, the volume available to them for motion is always the same as the volume of the container. In contrast, the molecules of a real gas have small but measurable volumes. At low pressures, the gaseous molecules are relatively far apart, but as the pressure of the gas increases, the intermolecular distances become smaller and smaller (Figure \(\Page {3}\)). As a result, the volume occupied by the molecules becomes significant compared with the volume of the container. Consequently, the total volume occupied by the gas is greater than the volume predicted by the ideal gas law. Thus at very high pressures, the experimentally measured value of / is greater than the value predicted by the ideal gas law. Moreover, all molecules are attracted to one another by a combination of forces. These forces become particularly important for gases at low temperatures and high pressures, where intermolecular distances are shorter. Attractions between molecules reduce the number of collisions with the container wall, an effect that becomes more pronounced as the number of attractive interactions increases. Because the average distance between molecules decreases, the pressure exerted by the gas on the container wall decreases, and the observed pressure is than expected. Thus as shown in Figure \(\Page {2}\), at low temperatures, the ratio of (PV/nRT\) is lower than predicted for an ideal gas, an effect that becomes particularly evident for complex gases and for simple gases at low temperatures. At very high pressures, the effect of nonzero molecular volume predominates. The competition between these effects is responsible for the minimum observed in the \(PV/nRT\) versus \(P\) plot for many gases. Nonzero molecular volume makes the actual volume than predicted at high pressures; intermolecular attractions make the pressure than predicted. At high temperatures, the molecules have sufficient kinetic energy to overcome intermolecular attractive forces, and the effects of nonzero molecular volume predominate. Conversely, as the temperature is lowered, the kinetic energy of the gas molecules decreases. Eventually, a point is reached where the molecules can no longer overcome the intermolecular attractive forces, and the gas liquefies (condenses to a liquid). The Dutch physicist Johannes van der Waals (1837–1923; Nobel Prize in Physics, 1910) modified the ideal gas law to describe the behavior of real gases by explicitly including the effects of molecular size and intermolecular forces. In his description of gas behavior, the so-called equation, \[ \left(P + \dfrac{an^2}{V^2}\right) (V − nb)=nRT \label{6.9.1}\] and are empirical constants that are different for each gas. The values of \(a\) and \(b\) are listed in Table \(\Page {1}\) for several common gases. The pressure term in Equation \(\ref{6.9.1}\) —\(P + (an^2/V^2\))—corrects for intermolecular attractive forces that tend to reduce the pressure from that predicted by the ideal gas law. Here, \(n^2/V^2\) represents the concentration of the gas (\(n/V\)) squared because it takes two particles to engage in the pairwise intermolecular interactions of the type shown in Figure \(\Page {4}\). The volume term—\(V − nb\)—corrects for the volume occupied by the gaseous molecules. The correction for volume is negative, but the correction for pressure is positive to reflect the effect of each factor on and , respectively. Because nonzero molecular volumes produce a measured volume that is than that predicted by the ideal gas law, we must subtract the molecular volumes to obtain the actual volume available. Conversely, attractive intermolecular forces produce a pressure that is than that expected based on the ideal gas law, so the / term must be added to the measured pressure to correct for these effects. You are in charge of the manufacture of cylinders of compressed gas at a small company. Your company president would like to offer a 4.00 L cylinder containing 500 g of chlorine in the new catalog. The cylinders you have on hand have a rupture pressure of 40 atm. Use both the ideal gas law and the van der Waals equation to calculate the pressure in a cylinder at 25°C. Is this cylinder likely to be safe against sudden rupture (which would be disastrous and certainly result in lawsuits because chlorine gas is highly toxic)? volume of cylinder, mass of compound, pressure, and temperature safety Use the molar mass of chlorine to calculate the amount of chlorine in the cylinder. Then calculate the pressure of the gas using the ideal gas law. Obtain and values for Cl from Table \(\Page {1}\). Use the van der Waals equation to solve for the pressure of the gas. Based on the value obtained, predict whether the cylinder is likely to be safe against sudden rupture. We begin by calculating the amount of chlorine in the cylinder using the molar mass of chlorine (70.906 g/mol): \[n=\dfrac{m}{M}=\rm\dfrac{500\;g}{70.906\;g/mol}=7.052\;mol\] Using the ideal gas law and the temperature in kelvins (298 K), we calculate the pressure: \[P=\dfrac{nRT}{V}=\rm\dfrac{7.052\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L}=43.1\;atm\] If chlorine behaves like an ideal gas, you have a real problem! Now let’s use the van der Waals equation with the and values for Cl from Table \(\Page {1}\). Solving for gives \[\begin{split}P&=\dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}​\\&=\rm\dfrac{7.052\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L-7.052\;mol\times0.0542\dfrac{L}{mol}}-\dfrac{6.260\dfrac{L^2atm}{mol^2}\times(7.052\;mol)^2}{(4.00\;L)^2}\\&=\rm28.2\;atm\end{split}\] This pressure is well within the safety limits of the cylinder. The ideal gas law predicts a pressure 15 atm higher than that of the van der Waals equation. A 10.0 L cylinder contains 500 g of methane. Calculate its pressure to two significant figures at 27°C using the a. 77 atm; b. 67 atm Liquefaction of gases is the condensation of gases into a liquid form, which is neither anticipated nor explained by the kinetic molecular theory of gases. Both the theory and the ideal gas law predict that gases compressed to very high pressures and cooled to very low temperatures should still behave like gases, albeit cold, dense ones. As gases are compressed and cooled, however, they invariably condense to form liquids, although very low temperatures are needed to liquefy light elements such as helium (for He, 4.2 K at 1 atm pressure). Liquefaction can be viewed as an extreme deviation from ideal gas behavior. It occurs when the molecules of a gas are cooled to the point where they no longer possess sufficient kinetic energy to overcome intermolecular attractive forces. The precise combination of temperature and pressure needed to liquefy a gas depends strongly on its molar mass and structure, with heavier and more complex molecules usually liquefying at higher temperatures. In general, substances with large van der Waals \(a\) coefficients are relatively easy to liquefy because large coefficients indicate relatively strong intermolecular attractive interactions. Conversely, small molecules with only light elements have small coefficients, indicating weak intermolecular interactions, and they are relatively difficult to liquefy. Gas liquefaction is used on a massive scale to separate O , N , Ar, Ne, Kr, and Xe. After a sample of air is liquefied, the mixture is warmed, and the gases are separated according to their boiling points. A large value of indicates the presence of relatively strong intermolecular attractive interactions. The ultracold liquids formed from the liquefaction of gases are called cryogenic liquids, from the Greek , meaning “cold,” and , meaning “producing.” They have applications as refrigerants in both industry and biology. For example, under carefully controlled conditions, the very cold temperatures afforded by liquefied gases such as nitrogen (boiling point = 77 K at 1 atm) can preserve biological materials, such as semen for the artificial insemination of cows and other farm animals. These liquids can also be used in a specialized type of surgery called , which selectively destroys tissues with a minimal loss of blood by the use of extreme cold. Moreover, the liquefaction of gases is tremendously important in the storage and shipment of fossil fuels (Figure \(\Page {5}\)). Liquefied natural gas (LNG) and liquefied petroleum gas (LPG) are liquefied forms of hydrocarbons produced from natural gas or petroleum reserves. LNG consists mostly of methane, with small amounts of heavier hydrocarbons; it is prepared by cooling natural gas to below about −162°C. It can be stored in double-walled, vacuum-insulated containers at or slightly above atmospheric pressure. Because LNG occupies only about 1/600 the volume of natural gas, it is easier and more economical to transport. LPG is typically a mixture of propane, propene, butane, and butenes and is primarily used as a fuel for home heating. It is also used as a feedstock for chemical plants and as an inexpensive and relatively nonpolluting fuel for some automobiles. No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Deviations from ideal gas behavior can be seen in plots of / versus at a given temperature; for an ideal gas, / versus = 1 under all conditions. At high pressures, most real gases exhibit larger / values than predicted by the ideal gas law, whereas at low pressures, most real gases exhibit / values close to those predicted by the ideal gas law. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the , which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces. If the temperature of a gas is decreased sufficiently, occurs, in which the gas condenses into a liquid form. Liquefied gases have many commercial applications, including the transport of large amounts of gases in small volumes and the uses of ultracold .

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A typical example of an acid is hydrogen chloride gas, HCl( ). When it dissolves in water, HCl reacts to form hydronium ions and chloride ions: \[\text{HCl}(g) + \text{H}_{2}\text{O}(l) \rightarrow \text{H}_{3}\text{O}^{+}(aq) + \text{Cl}^{-}(aq) \nonumber \] Thus the concentration of hydronium ions is increased above the value of 1.00 × 10 mol/L characteristic of pure water. Other acids, such as nitric acid, HNO behave in the same way: \[\text{HNO}_{3}(l) + \text{H}_{2}\text{O}(l) \rightarrow \text{H}_{3}\text{O}^{+}(aq) + \text{NO}_{3}^{-}(aq) \nonumber \] Thus the characteristic properties of solutions of acids are due to the presence of hydronium ions (or hydrogen ions). Whenever the concentration of hydronium ions exceeds 1.00 × 10 mol/L, an aqueous solution is said to be . In 1884 a Swedish chemist, Svante Arrhenius (1859 to 1927), first recognized the importance of hydrogen ions. He defined an acid as any substance which increases the concentration of hydrogen (or hydronium) ions in aqueous solution. The formation of a hydronium ion involves transfer of a proton from an acid molecule to a water molecule (as seen in the submicroscopic images above). This process is immediate―there are no free protons in solution which have left an acid molecule but have not yet attached themselves to a water molecule. To put it another way, a proton transfer is like a quarterback hand-off as opposed to a forward pass in foot- ball. The proton is always under the control and influence of one molecule or another. In the case of HCl we can indicate the transfer as Write balanced equations to describe the proton transfer which occurs when each of the following acids is dissolved in H O: Although a free proton is never actually produced in solution, it is often convenient to break the proton-transfer process into two hypothetical steps: (1) the loss of a proton by the acid, and (2) the gain of a proton by H O. ) When HClO loses a proton, H , the valence electron originally associated with the H atom is left behind, producing a negative ion, ClO . The proton can then be added to a water molecule in the second hypothetical step. Summing the two steps gives the overall proton transfer: \(\text{HClO}_{4} \rightarrow \cancel{\text{H}^{+}} + \text{ClO}_{4}^{-} \) step 1 \(\cancel{\text{H}^{+}} +\text{H}_{2}\text{O} \rightarrow \text{H}_{3}\text{O}^{+}\) step 2 __________________________ \( \text{HClO}_{4} +\text{H}_{2}\text{O} \rightarrow \text{H}_{3}\text{O}^{+} + \text{ClO}_{4}^{-}\) overall ) Proceeding as in part , we have \(\text{HBr} \rightarrow \cancel{\text{H}^{+}} + \text{Br}^{-}\) step 1 \(\cancel{\text{H}^{+}} +\text{H}_{2}\text{O} \rightarrow \text{H}_{3}\text{O}^{+}\) step 2 ______________________ \(\text{HBr} +\text{H}_{2}\text{O} \rightarrow \text{H}_{3}\text{O}^{+} + \text{Br}^{-}\) overall With practice, you should be able to write overall proton transfers without having to write steps 1 and 2 every time. Another point to note about proton transfers is that in any equation involving ions, the sum of the ionic charges on the left side must equal the sum of the ionic charges on the right. For example, the last overall equation in Example 1 has HBr and H O on the left. Neither is an ion, and so the sum of the ionic charges is zero. On the right we have H O and Br , which satisfy the rule because +1 + (–1) = 0. An equation which does not satisfy this rule of charge balance will involve creation or destruction of one or more electrons and therefore cannot be valid. For example, the equation \[\text{2HBr} \rightarrow \text{2H}^{+} +\text{Br}_{2} \nonumber \] cannot describe a valid proton transfer because the charges sum to zero on the left but +2 (because 2H ions) on the right. Careful examination reveals that there are 16 valence electrons (two octets in 2HBr) on the left but only 14 valence electrons (none in 2H and 14 in ) on the right. Two electrons have been destroyed—something which cannot happen. Therefore the equation must be incorrect. Because hydronium ions can be formed by transferring protons to water molecules, it is convenient when dealing with aqueous solutions to define an a . This definition was first proposed in 1923 by the Danish chemist Johannes Brönsted (1879 to 1947) and the English chemist Thomas Martin Lowry (1874 to 1936). It is called the of an acid, and we will use it for the majority of this site. The Brönsted-Lowry definition has certain advantages over Arrhenius’ idea of an acid as a producer of H O ( ). This is especially true when acid strengths are compared, a subject we shall come to a bit later. Consequently, when we speak of an acid, we will mean a proton donor, unless some qualification, such as Arrhenius acid, is used.

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Each day, hundreds of weather balloons are launched. Made of a synthetic rubber and carrying a box of instruments, each helium-filled balloon rises up into the sky. As a balloon gains altitude, the atmospheric pressure becomes less and the balloon expands. At some point the balloon bursts due to the expansion; the instruments drop (aided by a parachute) to be retrieved and studied for information about the weather. Robert Boyle (1627-1691), an English chemist, is widely considered to be one of the founders of the modern experimental science of chemistry. He discovered that doubling the pressure of an enclosed sample of gas, while keeping its temperature constant, caused the volume of the gas to be reduced by half. states that the volume of a given mass of gas varies inversely with the pressure when the temperature is kept constant. An inverse relationship is described in this way. As one variable increases in value, the other variable decreases. Physically, what is happening? The gas molecules are moving and are a certain distance apart from one another. An increase in pressure pushes the molecules closer together, reducing the volume. If the pressure is decreased, the gases are free to move about in a larger volume. Mathematically, Boyle's law can be expressed by the equation: \[P \times V = k\nonumber \] The \(k\) is a constant for a given sample of gas and depends only on the mass of the gas and the temperature. The table below shows pressure and volume data for a set amount of gas at a constant temperature. The third column represents the value of the constant \(\left( k \right)\) for this data and is always equal to the pressure multiplied by the volume. As one of the variables changes, the other changes in such a way that the product of \(P \times V\) always remains the same. In this particular case, that constant is \(500 \: \text{atm} \cdot \text{mL}\). A graph of the data in the table further illustrates the inverse relationship nature of Boyle's Law (see figure below). Volume is plotted on the \(x\)-axis, with the corresponding pressure on the \(y\)-axis. Boyle's Law can be used to compare changing conditions for a gas. We use \(P_1\) and \(V_1\) to stand for the initial pressure and initial volume of a gas. After a change has been made, \(P_2\) and \(V_2\) stand for the final pressure and volume. The mathematical relationship of Boyle's Law becomes: \[P_1 \times V_1 = P_2 \times V_2\nonumber \] This equation can be used to calculate any one of the four quantities if the other three are known. A sample of oxygen gas has a volume of \(425 \: \text{mL}\) when the pressure is equal to \(387 \: \text{kPa}\). The gas is allowed to expand into a \(1.75 \: \text{L}\) container. Calculate the new pressure of the gas. Use Boyle's Law to solve for the unknown pressure \(\left( P_2 \right)\). It is important that the two volumes (\(V_1\) and \(V_2\)) are expressed in the same units, so \(V_2\) has been converted to \(\text{mL}\). First, rearrange the equation algebraically to solve for \(P_2\). \[P_2 = \frac{P_1 \times V_1}{V_2}\nonumber \] Now substitute the known quantities into the equation and solve. \[P_2 = \frac{387 \: \text{kPa} \times 425 \: \text{mL}}{1750 \: \text{mL}} = 94.0 \: \text{kPa}\nonumber \] The volume has increased to slightly over 4 times its original value and so the pressure is decreased by about one fourth. The pressure is in \(\text{kPa}\) and the value has three significant figures. Note that any pressure or volume units can be used as long as they are consistent throughout the problem.

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An important industrial application of electrolysis is the plating of one metal on top of another. A typical example is the bumper of a car. This is made from steel and then plated with a thin layer of chromium to make it resistant to rusting and scratching. Many other metal objects, such as pins, screws, watchbands, and doorknobs, are made of one metal with another plated on the surface. An electroplating cell works in much the same way as the cell used to . The object to be plated is used as the cathode, and the electrolyte contains some ionic compound of the metal to be plated. As current flows, this compound is reduced to the metal and deposits on the surface of the cathode. In chromium plating, for instance, the electrolyte is usually a solution of potassium dichromate, K Cr O , in fairly concentrated sulfuric acid. In this very acidic solution CrO ions are completely protonated, and so the reduction half-equation is \[\text{H}_2\text{Cr}_2\text{O}_7(aq) + \text{12H}^{+}(aq) + \text{12}e^{-} \rightarrow \text{2Cr}(s) + \text{7 H}_2\text{O}(l) \nonumber \] Other metals which are often electroplated are silver, nickel, tin, and zinc. In the case of silver the electrolyte must contain the polyatomic ion Ag(CN) rather than Ag . Otherwise the solid silver will be deposited as jagged crystals instead of a shiny uniform layer.

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You know that phase changes usually depend on temperature, which determines the kinetic energy of atoms and molecules. We mentioned before that they also depend on pressure. In the section on we said that the boiling point is the place where is the same at the external pressure, so clearly boiling point depends on pressure! Melting temperature also depends on pressure (usually the density of solid and liquid are different, so it makes sense) but not nearly as much as boiling point, since the volume changes are smaller. We use to show how the transition temperatures depend on temperature and pressure both. Look at the diagram. Notice that the gas phase is on the bottom, where the pressure is low. Solid is on the left, where the temperature is low. Liquid is in between. The red line shows the sublimation point: along this line, a low pressure, solid turns directly into gas without going through liquid. The point where liquid become stable is called the triple point, where all three phases (solid, liquid and gas) are all in equilibrium. The blue line is the boiling point. Notice that the boiling temperature changes a lot with a change in pressure. The solid green line shows the melting point of most liquids. Notice that the melting point doesn't depend on pressure nearly as much as the boiling point (which makes sense, because the change in volume from solid to liquid is small). Most liquids are less dense than the solid phase, so higher pressure increase the melting point. The dotted green line shows the melting point for water. Water is denser as a liquid, so higher pressures decrease the melting temperature. The second red point in the diagram is the . The dotted black lines show the area where a exists. This is the high-temperature, high-pressure part of the diagram. Because the temperature is high, the molecules have lots of kinetic energy, so a liquid form isn't really stable because the intermolecular forces aren't strong enough to hold such energetic molecules together. However, the pressure is so high that the molecules can't really get away from each other either, so they bump into each other a lot, and feel some attractions, and don't really act like a normal gas (certainly not an ideal gas!). Past the critical point, there's no distinct liquid or gas, just a supercritical fluid with some special properties. Supercritical fluids can make good solvents. For instance, supercritical CO is commonly used because it is a safe, inert, inexpensive non-polar solvent. Most non-polar solvents are not very safe (toxic and flammable), and disposing of them is expensive; supercritical CO avoids these problems.

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The is equivalent to the law of conservation of energy, which was described previously . However, instead of describing a system in which energy changes form (KE to PE and back) but the total amount doesn't change, now we will describe a system in which energy can move in and out. The 2 ways it can do this are . Imagine we have a system that is approximately isolated, like a thermos of water, and we do work on it in various ways. First we use the energy from a falling weight to mix it very vigorously, and we see what happens to the temperature of the system. Then we run some electric current through a resistor dipped in the system, and see what happens to the temperature of the system. And we observe that the same amount of work always raises the temperature the same amount. If we do the same experiments on a cup of water that isn't isolated like the thermos (but is otherwise the same), we find that the temperature doesn't increase as much when we do the same amount of work without the insulating thermos. As the system got warmer, some energy moved from the system to the surroundings. Without the thermos, heat leaves the system when it gets warmer than the surroundings, so we have to do more work to get the same increase in temperature. Energy is the capacity to do work. Any system can do work. Perhaps it can heat a gas, causing it to expand against a force, or fall, compressing a spring, etc. But an isolated system's capacity to do work won't change. For instance, if we use a system to do work, then wait a long time, the system won't regain its original ability to do work. More concretely, if we use a falling weight to do some other work, like lifting another weight or driving a motor, the weight won't be able to do more work until we lift it to its original height. It won't float back up by itself, ready to fall again. If we use a very hot block of metal to boil water and drive motors, the block will get cooler. And if we leave it isolated and wait, it won't heat back up again by itself so we can boil more water. The capacity of a system to do work is called its . The internal energy of a system can change if work is done on or by the system, or if heat enters or leaves the system. If no work is done and no heat flows, then the internal energy of the system can't change. We can write this as an equation: \[\Delta E=q+w\] where ΔE is the change in internal energy (some people use U instead), q is heat, and w is work. Heat q is positive when it flows into the system, and negative when it flows out. Work w is positive when it is done to the system and negative when it is done by the system. (Actually, some people use the opposite signs for work, in which case the equation is E=q-w) This should make sense. The internal energy of the system increases when we put energy in; it decreases when we take energy out. Molecularly, internal energy means all the kinetic and potential energy of each particle in the system. Internal energy is a , like temperature and pressure. Work and heat are not state functions: they depend on processes. You can't look at an object and determine how much work or heat it has because that doesn't make sense, but you can measure what its temperature is, what its volume is, and what its internal energy is.

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Bases have characteristics opposite those of , and bases can be neutralized by acids. Therefore it is logical, in the Brönsted-Lowry scheme, to define a as a , that is, a species which can incorporate an extra proton into its molecular or ionic structure. For example, when barium oxide, BaO, dissolves in water, oxide ions accept protons from water molecules according to the equation Thus the general properties of solutions of bases are due to the presence of hydroxide ions [OH ( )]. Any aqueous solution which contains a concentration of hydroxide ions greater than the 1.00 × 10 mol/L characteristic of pure water is said to be . Unlike the hydronium ion, which forms very few solid compounds, hydroxide ions are often present in solid crystal lattices (like NaOH, seen below). Therefore it is possible to raise the hydroxide-ion concentration above 1.00 × 10 mol/L by dissolving compounds such as NaOH, KOH, or Ba(OH) . Hydroxide ions can accept protons from water molecules, but of course such a proton transfer has no net effect because the hydroxide ion itself becomes a water molecule: Write a balanced equation to describe the proton transfer which occurs when the base sodium hydride, NaH, is added to water. NaH consists of Na and H ions. Since positive ions repel protons, the H ion is the only likely base. Again it may be useful to use two hypothetical steps: (l) donation of a proton by an H O molecule, and (2) acceptance of a proton by the base. As in Example 11.6, we can then sum the steps \(\text{H}^{-} + \text{H}_{2}\text{O} \rightarrow \text{H}_{2} + \text{OH}^{-}\) overall

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The numerical values we deal with in science (and in many other aspects of life) represent measurements whose values are never known exactly. Our pocket-calculators or computers don't know this; they treat the numbers we punch into them as "pure" mathematical entities, with the result that the operations of arithmetic frequently yield answers that are physically ridiculous even though mathematically correct. The purpose of this unit is to help you understand why this happens, and to show you what to do about it. Consider the two statements shown below: Which of these would you be justified in dismissing immediately? Certainly not the second one, because it probably comes from a database which contains one record for each voter, so the number is found simply by counting the number of records. The first statement cannot possibly be correct. Even if a city’s population could be defined in a precise way (Permanent residents? Warm bodies?), how can we account for the minute-by minute changes that occur as people are born and die, or move in and move away? What is the difference between the two population numbers stated above? The first one expresses a quantity that cannot be known exactly — that is, it carries with it a degree of uncertainty. It is quite possible that the last census yielded precisely 157,872 records, and that this might be the “population of the city” for legal purposes, but it is surely not the “true” population. To better reflect this fact, one might list the population (in an atlas, for example) as or even . These two quantities have been rounded off to four and three significant figures, respectively, and the have the following meanings: Which of these two values we would report as “the population” will depend on the degree of confidence we have in the original census figure; if the census was completed last week, we might round to four significant digits, but if it was a year or so ago, rounding to three places might be a more prudent choice. In a case such as this, there is no really objective way of choosing between the two alternatives. This illustrates an important point: the concept of has less to do with mathematics than with our confidence in a measurement. This confidence can often be expressed numerically (for example, the height of a liquid in a measuring tube can be read to ±0.05 cm), but when it cannot, as in our population example, we must depend on our personal experience and judgment. So, what is a significant digit? According to the usual definition, it is all the numerals in a measured quantity (counting from the left) whose values are considered as known exactly, plus one more whose value could be one more or one less: Although , what we are getting rid of can be considered to be “numeric noise” that does not contribute to the quality of the measurement. The purpose in rounding off is to avoid expressing a value to a greater degree of precision than is consistent with the uncertainty in the measurement. If you know that a balance is accurate to within 0.1 mg, say, then the uncertainty in any measurement of mass carried out on this balance will be ±0.1 mg. Suppose, however, that you are simply told that an object has a length of 0.42 cm, with no indication of its precision. In this case, all you have to go on is the number of digits contained in the data. Thus the quantity “0.42 cm” is specified to 0.01 unit in 0 42, or one part in 42 . The implied relative uncertainty in this figure is 1/42, or about 2%. The precision of any numeric answer calculated from this value is therefore limited to about the same amount. It is important to understand that the number of significant digits in a value provides only a rough indication of its precision, and that information is lost when rounding off occurs. Suppose, for example, that we measure the weight of an object as 3.28 g on a balance believed to be accurate to within ±0.05 gram. The resulting value of 3.28±.05 gram tells us that the true weight of the object could be anywhere between 3.23 g and 3.33 g. The absolute uncertainty here is 0.1 g (±0.05 g), and the relative uncertainty is 1 part in 32.8, or about 3 percent. How many significant digits should there be in the reported measurement? Since only the left most “3” in “3.28” is certain, you would probably elect to round the value to 3.3 g. So far, so good. But what is someone else supposed to make of this figure when they see it in your report? The value “3.3 g” suggests an of 3.3±0.05 g, meaning that the true value is likely between 3.25 g and 3.35 g. This range is 0.02 g below that associated with the original measurement, and so rounding off has introduced a bias of this amount into the result. Since this is less than half of the ±0.05 g uncertainty in the weighing, it is not a very serious matter in itself. However, if several values that were rounded in this way are combined in a calculation, the rounding-off errors could become significant. The standard rules for rounding off are well known. Before we set them out, let us agree on what to call the various components of a numeric value. Students are sometimes told to increment the least significant digit by 1 if it is odd, and to leave it unchanged if it is even. One wonders if this reflects some idea that even numbers are somehow “better” than odd ones! (The ancient superstition is just the opposite, that only the odd numbers are "lucky".) In fact, you could do it equally the other way around, incrementing only the even numbers. If you are only rounding a single number, it doesn't really matter what you do. However, when you are rounding a series of numbers that will be used in a calculation, if you treated each first nonsignificant 5 in the same way, you would be over- or understating the value of the rounded number, thus accumulating round-off error. Since there are equal numbers of even and odd digits, incrementing only the one kind will keep this kind of error from building up. You could do just as well, of course, by flipping a coin! Suppose that an object is found to have a weight of 3.98 ± 0.05 g. This would place its true weight somewhere in the range of g to g. In judging how to round this number, you count the number of digits in “3.98” that are known exactly, and you find none! Since the “4” is the left most digit whose value is uncertain, this would imply that the result should be rounded to one significant figure and reported simply as 4 g. An alternative would be to bend the rule and round off to two significant digits, yielding 4.0 g. How can you decide what to do? In a case such as this, you should look at the implied uncertainties in the two values, and compare them with the uncertainty associated with the original measurement. Clearly, rounding off to two digits is the only reasonable course in this example. Observed values should be rounded off to the number of digits that most accurately conveys the uncertainty in the measurement. When carrying out calculations that involve multiple steps, you should avoid doing any rounding until you obtain the final result. Suppose you use your calculator to work out the area of a rectangle: Your calculator is of course correct as far as the pure numbers go, but you would be wrong to write down "1.57676 cm " as the answer. Two possible options for rounding off the calculator answer are shown at the right. It is clear that neither option is entirely satisfactory; rounding to 3 significant digits overstates the precision of the answer, whereas following the rule and rounding to the two digits in ".42" has the effect of throwing away some precision. In this case, it could be argued that rounding to three digits is justified because the implied relative uncertainty in the answer, 0.6%, is more consistent with those of the two factors. The "rules" for rounding off are generally useful, convenient guidelines, but they do not always yield the most desirable result. When in doubt, it is better to rely on relative implied uncertainties. In operations involving significant figures, the answer is reported in such a way that it reflects the reliability of the operation. An answer is no more precise that the least precise number used to get the answer. When adding or subtracting, we go by the number of (i.e., the number of digits on the right side of the decimal point) rather than by the number of significant digits. Identify the quantity having the smallest number of decimal places, and use this number to set the number of decimal places in the answer. The result must contain the same number of significant figures as in the value having the least number of significant figures. If a number is expressed in the form × 10 ("scientific notation") with the additional restriction that the coefficient is no less than 1 and less than 10, the number is in its form. Express the base-10 logarithm of a value using the same number of significant figures as is present in the of that value. Similarly, for antilogarithms (numbers expressed as powers of 10), use the same number of significant figures as are in that power. The following examples will illustrate the most common problems you are likely to encounter in rounding off the results of calculations. They deserve your careful study! A certain book has a thickness of 117 mm; find the height of a stack of 24 identical books: The last of the examples shown above represents the very common operation of converting one unit into another. There is a certain amount of ambiguity here; if we take "9 in" to mean a distance in the range 8.5 to 9.5 inches, then the implied uncertainty is ±0.5 in, which is 1 part in 18, or about ± 6%. The relative uncertainty in the answer must be the same, since all the values are multiplied by the same factor, 2.54 cm/in. In this case we are justified in writing the answer to two significant digits, yielding an uncertainty of about ±1 cm; if we had used the answer "20 cm" (one significant digit), its implied uncertainty would be ±5 cm, or ±25%. When the appropriate number of significant digits is in question, calculating the relative uncertainty can help you decide.

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1. Identify the anode and the cathode for the following electrochemical cells, and identify the oxidation or the reduction reaction at each electrode. (a) Pt| FeCl ( , 0.015), FeCl ( , 0.045) || AgNO ( , 0.1) | Ag (b) Ag | AgBr( ), NaBr ( , 1.0) || CdCl ( , 0.05) | Cd (c) Pb | PbSO ( ), H SO ( , 1.5) || H SO ( , 2.0), PbSO ( ) | PbO 2. Calculate the potential for each electrochemical cell in problem 1. The values in parentheses are the activities of the associated species. 3. Calculate the activity of KI, , in the following electrochemical cell if the potential is +0.294 V. Ag | AgCl ( ), NaCl ( , 0.1) || KI (aq, ), I ( ) | Pt 4. What reaction prevents us from using Zn as an electrode of the first kind in an acidic solution? Which other metals do you expect to behave in the same manner as Zn when immersed in an acidic solution? 5. Creager and colleagues designed a salicylate ion-selective electrode using a PVC membrane impregnated with tetraalkylammonium salicylate [Creager, S. E.; Lawrence, K. D.; Tibbets, C. R. , , 274–276]. To determine the ion-selective electrode’s selectivity coefficient for benzoate, they prepared a set of salicylate calibration standards in which the concentration of benzoate was held constant at 0.10 M. Using the following data, determine the value of the selectivity coefficient. What is the maximum acceptable concentration of benzoate if you plan to use this ion-selective electrode to analyze a sample that contains as little as 10 M salicylate with an accuracy of better than 1%? 6. Watanabe and co-workers described a new membrane electrode for the determination of cocaine, a weak base alkaloid with a p of 8.64 [Watanabe, K.; Okada, K.; Oda, H.; Furuno, K.; Gomita, Y.; Katsu, T. , , 371–375]. The electrode’s response for a fixed concentration of cocaine is independent of pH in the range of 1–8, but decreases sharply above a pH of 8. Offer an explanation for this pH dependency. 7. shows a schematic diagram for an enzyme electrode that responds to urea by using a gas-sensing NH electrode to measure the amount of ammonia released following the enzyme’s reaction with urea. In turn, the NH electrode uses a pH electrode to monitor the change in pH due to the ammonia. The response of the urea electrode is given by . Beginning with , which gives the potential of a pH electrode, show that for the urea electrode is correct. 8. Explain why the response of an NH -based urea electrode ( and ) is different from the response of a urea electrode in which the enzyme is coated on the glass membrane of a pH electrode ( and ). 9. A potentiometric electrode for HCN uses a gas-permeable membrane, a buffered internal solution of 0.01 M KAg(CN) , and a Ag S ISE electrode that is immersed in the internal solution. Consider the equilibrium reactions that take place within the internal solution and derive an equation that relates the electrode’s potential to the concentration of HCN in the sample. To check your work, search on-line for US Patent 3859191 and consult Figure 2. 10. Mifflin and associates described a membrane electrode for the quantitative analysis of penicillin in which the enzyme penicillinase is immobilized in a polyacrylamide gel coated on the glass membrane of a pH electrode [Mifflin, T. E.; Andriano, K. M.; Robbins, W. B. , , 638–639]. The following data were collected using a set of penicillin standards. (a) Over what range of concentrations is there a linear response? (b) What is the calibration curve’s equation for this concentration range? (c) What is the concentration of penicillin in a sample that yields a potential of 142 mV? 11. An ion-selective electrode can be placed in a flow cell into which we inject samples or standards. As the analyte passes through the cell, a potential spike is recorded instead of a steady-state potential. The concentration of K in serum has been determined in this fashion using standards prepared in a matrix of 0.014 M NaCl [Meyerhoff, M. E.; Kovach, P. M. , , 766–768]. A 1.00-mL sample of serum is diluted to volume in a 10-mL volumetric flask and analyzed, giving a potential of 51.1 (arbitrary units). Report the concentration of K in the sample of serum. 12. Wang and Taha described an interesting application of potentiometry, which they call batch injection [Wang, J.; Taha, Z. , , 215–221]. As shown in the figure below, an ion-selective electrode is placed in an inverted position in a large volume tank, and a fixed volume of a sample or a standard solution is injected toward the electrode’s surface using a micropipet. The response of the electrode is a spike in potential that is proportional to the analyte’s concentration. The following data were collected using a pH electrode and a set of pH standards. Determine the pH of the following samples given the recorded peak potentials: tomato juice, 167 mV; tap water, –27 mV; coffee, 122 mV. 13. The concentration of \(\text{NO}_3^-\) in a water sample is determined by a one-point standard addition using a \(\text{NO}_3^-\) ion-selective electrode. A 25.00-mL sample is placed in a beaker and a potential of 0.102 V is measured. A 1.00-mL aliquot of a 200.0-mg/L standard solution of \(\text{NO}_3^-\) is added, after which the potential is 0.089 V. Report the mg \(\text{NO}_3^-\)/L in the water sample. 14. In 1977, when I was an undergraduate student at Knox College, my lab partner and I completed an experiment to determine the concentration of fluoride in tap water and the amount of fluoride in toothpaste. The data in this problem are from my lab notebook. (a) To analyze tap water, we took three 25.0-mL samples and added 25.0 mL of TISAB to each. We measured the potential of each solution using a F ISE and an SCE reference electrode. Next, we made five 1.00-mL additions of a standard solution of 100.0 ppm F to each sample, and measured the potential after each addition, recording the potential three times. Report the parts-per-million of F in the tap water. (b) To analyze the toothpaste, we measured 0.3619 g into a 100-mL volumetric flask, added 50.0 mL of TISAB, and diluted to volume with distilled water. After we ensured that the sample was thoroughly mixed, we transferred three 20.0-mL portions into separate beakers and measured the potential of each using a F ISE and an SCE reference electrode. Next, we made five 1.00-mL additions of a standard solution of 100.0 ppm F to each sample, and measured the potential after each addition, recording the potential three times. Report the parts-per-million F in the toothpaste. 15. You are responsible for determining the amount of KI in iodized salt and decide to use an I ion-selective electrode. Describe how you would perform this analysis using external standards and how you would per-form this analysis using the method of standard additions. 16. Explain why each of the following decreases the analysis time in controlled-potential coulometry: a larger surface area for the working electrode; a smaller volume of solution; and a faster stirring rate. 17. The purity of a sample of picric acid, C H N O , is determined by controlled-potential coulometry, converting picric acid to triaminophenol, C H N O. A 0.2917-g sample of picric acid is placed in a 1000-mL volumetric flask and diluted to volume. A 10.00-mL portion of this solution is transferred to a coulometric cell and sufficient water added so that the Pt cathode is immersed. An exhaustive electrolysis of the sample re-quires 21.67 C of charge. Report the purity of the picric acid. 18. The concentration of H S in the drainage from an abandoned mine is determined by a coulometric titration using KI as a mediator and \(\text{I}_3^-\) as the titrant. \[\text{H}_{2}\text{S}(a q)+\ \mathrm{I}_{3}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+3 \mathrm{I}^{-}(a q)+\mathrm{S}(s) \nonumber\] A 50.00-mL sample of water is placed in a coulometric cell, along with an excess of KI and a small amount of starch as an indicator. Electrolysis is carried out at a constant current of 84.6 mA, requiring 386 s to reach the starch end point. Report the concentration of H S in the sample in μg/mL. 19. One method for the determination of a given mass of H AsO is a coulometric titration using \(\text{I}_3^-\) as a titrant. The relevant standard-state reactions and potentials are summarized here \[\begin{aligned} \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} &\rightleftharpoons \ \mathrm{H}_{3} \mathrm{AsO}_{3}(a q)+\ \mathrm{H}_{2} \mathrm{O}(l) \\ \mathrm{I}_{3}^{-}(a q)+2 \mathrm{e}^{-} &\rightleftharpoons 3 \mathrm{I}^{-}(a q) \end{aligned} \nonumber\] with standard state reduction potentials of, respectively, +0.559 V and +0.536 V. Explain why the coulometric titration is carried out in a neutral solution (pH ≈ 7) instead of in a strongly acidic solution (pH < 0). 20. The production of adiponitrile, NC(CH ) CN, from acrylonitrile, CH =CHCN, is an important industrial process. A 0.594-g sample of acrylonitrile is placed in a 1-L volumetric flask and diluted to volume. An exhaustive controlled-potential electrolysis of a 1.00-mL portion of the diluted acrylonitrile requires 1.080 C of charge. What is the value of for the reduction of acrylonitrile to adiponitrile? 21. The linear-potential scan hydrodynamic voltammogram for a mixture of Fe and Fe is shown in the figure below where and are the anodic and cathodic limiting currents. (a) Show that the potential is given by \[E = E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} - 0.05916 \log \frac {K_{\text{Fe}^{3+}}} {K_{\text{Fe}^{2+}}} - 0.05916 \log \frac {i - i_{l,a}}{i_{l,c} - i} \nonumber\] (b) What is the potential when = 0 for a solution that is 0.100 mM Fe and 0.050 mM Fe ? 22. The amount of sulfur in aromatic monomers is determined by differential pulse polarography. Standard solutions are prepared for analysis by dissolving 1.000 mL of the purified monomer in 25.00 mL of an electrolytic solvent, adding a known amount of sulfur, deaerating, and measuring the peak current. The following results were obtained for a set of calibration standards. Analysis of a 1.000-mL sample, treated in the same manner as the standards, gives a peak current of 1.77 μA. Report the mg S/mL in the sample. 23. The purity of a sample of K Fe(CN) is determined using linear-potential scan hydrodynamic voltammetry at a glassy carbon electrode. The following data were obtained for a set of external calibration standards. A sample of impure K Fe(CN) is prepared for analysis by diluting a 0.246-g portion to volume in a 100-mL volumetric flask. The limiting current for the sample is 444 μA. Report the purity of this sample of K Fe(CN) . 24. One method for determining whether an individual recently fired a gun is to look for traces of antimony in residue collected from the individual’s hands. Anodic stripping voltammetry at a mercury film electrode is ideally suited for this analysis. In a typical analysis a sample is collected from a suspect using a cotton-tipped swab wetted with 5% v/v HNO . After returning to the lab, the swab is placed in a vial that contains 5.0 mL of 4 M HCl that is 0.02 M in hydrazine sulfate. After soaking the swab, a 4.0-mL portion of the solution is transferred to an electrochemical cell along with 100 μL of 0.01 M HgCl . After depositing the thin film of mercury and the antimony, the stripping step gives a peak current of 0.38 μA. After adding a standard addition of 100 μL of \(5.00 \times 10^2\) ppb Sb, the peak current increases to 1.14 μA. How many nanograms of Sb were collected from the suspect’s hand? 25. Zinc is used as an internal standard in an analysis of thallium by differential pulse polarography. A standard solution of \(5.00 \times 10^{-5}\) M Zn and \(2.50 \times 10^{-5}\) M Tl has peak currents of 5.71 μA and 3.19 μA, respectively. An 8.713-g sample of a zinc-free alloy is dissolved in acid, transferred to a 500-mL volumetric flask, and diluted to volume. A 25.0-mL portion of this solution is mixed with 25.0 mL of \(5.00 \times 10^{-4}\) M Zn . Analysis of this solution gives peak currents of 12.3 μA and of 20.2 μA for Zn and Tl , respectively. Report the %w/w Tl in the alloy. 26. Differential pulse voltammetry at a carbon working electrode is used to determine the concentrations of ascorbic acid and caffeine in drug formulations [Lau, O.; Luk, S.; Cheung, Y. , , 1047–1051]. In a typical analysis a 0.9183-g tablet is crushed and ground into a fine powder. A 0.5630-g sample of this powder is transferred to a 100-mL volumetric flask, brought into solution, and diluted to volume. A 0.500-mL portion of this solution is then transferred to a voltammetric cell that contains 20.00 mL of a suitable supporting electrolyte. The resulting voltammogram gives peak currents of 1.40 μA and 3.88 μA for ascorbic acid and for caffeine, respectively. A 0.500-mL aliquot of a standard solution that contains 250.0 ppm ascorbic acid and 200.0 ppm caffeine is then added. A voltammogram of this solution gives peak currents of 2.80 μA and 8.02 μA for ascorbic acid and caffeine, respectively. Report the milligrams of ascorbic acid and milligrams of caffeine in the tablet. 27. Ratana-ohpas and co-workers described a stripping analysis method for determining tin in canned fruit juices [Ratana-ohpas, R.; Kanatharana, P.; Ratana-ohpas, W.; Kongsawasdi, W. , , 115–118]. Standards of 50.0 ppb Sn , 100.0 ppb Sn , and 150.0 ppb Sn were analyzed giving peak currents (arbitrary units) of 83.0, 171.6, and 260.2, respectively. A 2.00-mL sample of lychee juice is mixed with 20.00 mL of 1:1 HCl/HNO . A 0.500-mL portion of this mixture is added to 10 mL of 6 M HCl and the volume adjusted to 30.00 mL. Analysis of this diluted sample gave a signal of 128.2 (arbitrary units). Report the parts-per-million Sn in the original sample of lychee juice. 28. Sittampalam and Wilson described the preparation and use of an amperometric sensor for glucose [Sittampalam, G.; Wilson, G. S. , , 70–73]. The sensor is calibrated by measuring the steady-state current when it is immersed in standard solutions of glucose. A typical set of calibration data is shown here. A 2.00-mL sample is diluted to 10 mL in a volumetric flask and a steady-state current of 23.6 (arbitrary units) is measured. What is the concentration of glucose in the sample in mg/100 mL? 29. Differential pulse polarography is used to determine the concentrations of lead, thallium, and indium in a mixture. Because the peaks for lead and thallium, and for thallium and indium overlap, a simultaneous analysis is necessary. Peak currents (in arbitrary units) at –0.385 V, –0.455 V, and –0.557 V are measured for a single standard solution, and for a sample, giving the results shown in the following table. Report the mg/mL of Pb , Tl and In in the sample. 30. Abass and co-workers developed an amperometric biosensor for \(\text{NH}_4^+\) that uses the enzyme glutamate dehydrogenase to catalyze the following reaction \[2 \text { - oxyglutarate }(a q)+ \ \mathrm{NH}_{4}^{+}(a q)+\mathrm{NADH}(a q)\rightleftharpoons\text { glutamate }(a q)+\ \mathrm{NAD}^{+}(a q)+\ \mathrm{H}_{2} \mathrm{O}(l) \nonumber\] where NADH is the reduced form of nicotinamide adenine dinucleotide [Abass, A. K.; Hart, J. P.; Cowell, D. C.; Chapell, A. , , 1–8]. The biosensor actually responds to the concentration of NADH, however, the rate of the reaction depends on the concentration of \(\text{NH}_4^+\). If the initial concentrations of 2-oxyglutarate and NADH are the same for all samples and standards, then the signal is proportional to the concentration of \(\text{NH}_4^+\). As shown in the following table, the sensitivity of the method is dependent on pH. Two possible explanations for the effect of pH on the sensitivity of this analysis are the acid–base chemistry of \(\text{NH}_4^+\) and the acid–base chemistry of the enzyme. Given that the p for \(\text{NH}_4^+\) is 9.244, explain the source of this pH-dependent sensitivity. 31. The speciation scheme for trace metals in divides them into seven operationally defined groups by collecting and analyzing two samples following each of four treatments, requiring a total of eight samples and eight measurements. After removing insoluble particulates by filtration (treatment 1), the solution is analyzed for the concentration of ASV labile metals and for the total concentration of metals. A portion of the filtered solution is then passed through an ion-exchange column (treatment 2), and the concentrations of ASV metal and of total metal are determined. A second portion of the filtered solution is irradiated with UV light (treatment 3), and the concentrations of ASV metal and of total metal are measured. Finally, a third portion of the filtered solution is irradiated with UV light and passed through an ion-exchange column (treatment 4), and the concentrations of ASV labile metal and of total metal again are determined. The groups that are included in each measurement are summarized in the following table. (a) Explain how you can use these eight measurements to determine the concentration of metals present in each of the seven groups identified in . (b) Batley and Florence report the following results for the speciation of cadmium, lead, and copper in a sample of seawater [Batley, G. E.; Florence, T. M. , , 379–388]. Determine the speciation of each metal in comment on your results. measurement treatement: ASV-labile or total 32. The concentration of Cu in seawater is determined by anodic stripping voltammetry at a hanging mercury drop electrode after first releasing any copper bound to organic matter. To a 20.00-mL sample of seawater is added 1 mL of 0.05 M HNO and 1 mL of 0.1% H O . The sample is irradiated with UV light for 8 hr and then diluted to volume in a 25-mL volumetric flask. Deposition of Cu takes place at –0.3 V versus an SCE for 10 min, producing a peak current of 26.1 (arbitrary units). A second 20.00-mL sample of the seawater is treated identically, except that 0.1 mL of a 5.00 μM solution of Cu is added, producing a peak current of 38.4 (arbitrary units). Report the concentration of Cu in the seawater in mg/L. 33. Thioamide drugs are determined by cathodic stripping analysis [Davidson, I. E.; Smyth, W. F. Anal. Chem. 1977, 49, 1195–1198]. Deposition occurs at +0.05 V versus an SCE. During the stripping step the potential is scanned cathodically and a stripping peak is observed at –0.52 V. In a typical application a 2.00-mL sample of urine is mixed with 2.00 mL of a pH 4.78 buffer. Following a 2.00 min deposition, a peak current of 0.562 μA is measured. A 0.10-mL addition of a 5.00 μM solution of the drug is added to the same solution. A peak current of 0.837 μA is recorded using the same deposition and stripping conditions. Report the drug’s molar concentration in the urine sample. 34. The concentration of vanadium (V) in sea water is determined by adsorptive stripping voltammetry after forming a complex with catechol [van der Berg, C. M. G.; Huang, Z. Q. , , 2383–2386]. The catechol-V(V) complex is deposited on a hanging mercury drop electrode at a potential of –0.1 V versus a Ag/AgCl reference electrode. A cathodic potential scan gives a stripping peak that is proportional to the concentration of V(V). The following standard additions are used to analyze a sample of seawater. \(2.0 \times 10^{-8}\) Determine the molar concentration of V (V) in the sample of sea water, assuming that the standard additions result in a negligible change in the sample’s volume. 35. The standard-state reduction potential for Cu to Cu is +0.342 V versus the SHE. Given that Cu forms a very stable complex with the ligand EDTA, do you expect that the standard-state reduction potential for Cu(EDTA) is greater than +0.342 V, less than +0.342 V, or equal to +0.342 V? Explain your reasoning. 36. The polarographic half-wave potentials (versus the SCE) for Pb and for Tl in 1 M HCl are, respectively, –0.44 V and –0.45 V. In an electrolyte of 1 M NaOH, however, the half-wave potentials are –0.76 V for Pb and –0.48 V for Tl . Why does the change in electrolyte have such a significant effect on the half-wave potential for Pb , but not on the half-wave potential for Tl ? 37. The following data for the reduction of Pb were collected by normal-pulse polarography. The limiting current was 5.67 μA. Verify that the reduction reaction is reversible and determine values for and . The half-wave potentials for the normal-pulse polarograms of Pb in the presence of several different concentrations of OH are shown in the following table. Determine the stoichiometry of the Pb-hydroxide complex and its formation constant. 38. In 1977, when I was an undergraduate student at Knox College, my lab partner and I completed an experiment to study the voltammetric behavior of Cd (in 0.1 M KNO ) and Ni (in 0.2 M KNO ) at a dropping mercury electrode. The data in this problem are from my lab notebook. All potentials are relative to an SCE reference electrode. The limiting currents for Cd was 4.8 μA and that for Ni was 2.0 μA. Evaluate the electrochemical reversibility for each metal ion and comment on your results. 39. Baldwin and co-workers report the following data from a cyclic voltammetry study of the electrochemical behavior of -phenylenediamine in a pH 7 buffer [Baldwin, R. P.; Ravichandran, K.; Johnson, R. K. , , 820–823]. All potentials are measured relative to an SCE. The initial scan is toward more positive potentials, leading to the oxidation reaction shown here. Use this data to show that the reaction is electrochemically irreversible. A reaction may show electrochemical irreversibility because of slow electron transfer kinetics or because the product of the oxidation reaction participates in a chemical reaction that produces an nonelectroactive species. Based on the data in this problem, what is the likely source of -phenylenediamine’s electrochemical irreversibility?

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Metabolism/Catabolism/Beta-Oxidation

The best source of energy for eukaryotic organisms are fats. Glucose offers a ratio 6.3 moles of ATP per carbon while saturated fatty acids offer 8.1 ATP per carbon. Also the complete oxidation of fats yields enormous amounts of water for those organisms that do not have adequate access to drinkable water. Camels and killer whales are good example of this, they obtain their water requirements from the complete oxidation of fats. There are four distinct stages in the oxidation of fatty acids. Fatty acid degradation takes place within the mitochondria and requires the help of several different enzymes. In order for fatty acids to enter the mitochondria the assistance of two carrier proteins is required, Carnitine acyltransferase I and II. It is also interesting to note the similarities between the four steps of beta-oxidation and the later four steps of the . Most fats stored in eukaryotic organisms are stored as triglycerides as seen below. In order to enter into beta-oxidation bonds must be broken usually with the use of a Lipase. The end result of these broken bonds are a glycerol molecule and three fatty acids in the case of triglycerides. Other lipids are capable of being degraded as well. Key molecules in beta-oxidation: (left) A molecule, (middle) Glycerol, (right) Fatty Acids (unsaturated) A fatty acyl-CoA is oxidized by Acyl-CoA dehydrogenase to yield a trans alkene. This is done with the aid of an [FAD] prosthetic group. The trans alkene is then with the help of Enoyl-CoA hydratase The alcohol of the hydroxyacly-CoA is then by NAD to a carbonyl with the help of Hydroxyacyl-CoA dehydrogenase. NAD is used to oxidize the alcohol rather then [FAD] because NAD is capable of the alcohol while [FAD] is not. Finally acetyl-CoA is cleaved off with the help of Thiolase to yield an Acyl-CoA that is two carbons shorter than before. The cleaved acetyl-CoA can then enter into the TCA and ETC because it is already within the mitochondria.

https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Key_Elements_of_Green_Chemistry_(Lucia)/03%3A_Hazards/3.02%3A_Hazard_Concepts

To best classify hazards, the concept of GHS (Globally Harmonized System) has been introduced. Figure \(\Page {1}\) demonstrates the categories of hazards: Notice that each logo attempts to demonstrate pictorially the type of hazard represented by a specific chemical or material. For example, notice Figure \(\Page {2}\) below. This particular chemical’s hazard level could be listed as “gases under pressure” and “explosives”. The intensity of its peculiar hazards is evaluated according to the pressure and the explosive nature of the gas. Our overall sensitivity to hazards, again, depends on a number of factors that not only include the typical factors of levels, LDs, type of toxin, modality, etc., but also to our idiosyncratic immunity or responsiveness. We have been accustomed to a number of wonderful creature comforts in life that have only come about because of the power, versatility, and creativity inherent in the chemical enterprise. For example, phosgene, a notorious chemical warfare agent in the Great War, is now used as a precursor for the manufacture of a number of items including polyurethane. Shown below in Figure \(\Page {3}\) is a representation of phosgene: Lab safety is an essential part of working with hazardous materials. It is important to follow good lab practice techniques and to know how to deal with situations should they occur. In the following video by Chemistry crash courses, some good lab techniques and safety guidelines are discussed. Lab Techniques and safety by crash course: Reference: Green, H. Crashcourse. (2013, July 8). Lab Techniques & Safetey: Crash Course Chemistry#21. Retrieved from (LD50) is the amount of any ingested or interfering substance that kills 50% of a test 58 sample. It is expressed in mg/kg, or milligrams of substance per kilogram of body weight. In toxicology, it is also referred to as the median lethal dose that refers specifically to a toxin, radiation, or pathogen. The lower the LD50, the more toxic is the item being measured for toxicity. It can be considered a pragmatic approach to toxicity exposure levels because in general, toxicity does NOT always scale with body mass. The choice of 50% lethality as the gold standard avoids ambiguity because it measures in the extremes and reduces the amount of testing. However, such a fact also means that LD50 is not the lethal dose for all subjects; in other words, some may be killed by much less. Measures such as “LD1” and “LD99” (dosages required to kill 1% or 99%, respectively, of the test population) are sometimes used. Shown below is a chart of sample LD50 taken from (en. .org/wiki/Median_lethal_dose) with active links to allow further investigation of the substances whose measurements are given. Substance LD (LC ) LD : g/kg (LC g/L) standardized Water >90 g/kg >90 6,860 mg/kg 6.86 22 µg/kg 22-30 mg/kg 0.000022 0.02 Sources for toxicity information: Notice that the venom of the Brazilian Wandering Spider is particularly potent. A ten thousandth of a g/kg or ~5mg would kill a normal sized female. Luckily the venom discharged per bite is quite small. Nevertheless, the point is that the vector (or venom) represents a toxin that is of sufficient lethality that it can cause great harm or injury to a human being.