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https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/3_d-Block_Elements/Group_05%3A_Transition_Metals/Group_5_Elemental_Properties
Like the group 4 elements, all group 5 metals are normally found in nature as oxide ores that contain the metals in their highest oxidation state (+5). Because of the lanthanide contraction, the chemistry of Nb and Ta is so similar that these elements are usually found in the same ores. Three-fourths of the vanadium produced annually is used in the production of steel alloys for springs and high-speed cutting tools. Adding a small amount of vanadium to steel results in the formation of small grains of V C , which greatly increase the strength and resilience of the metal, especially at high temperatures. The other major use of vanadium is as V O , an important catalyst for the industrial conversion of SO to SO in the contact process for the production of sulfuric acid. In contrast, Nb and Ta have only limited applications, and they are therefore produced in relatively small amounts. Although niobium is used as an additive in certain stainless steels, its primary application is in superconducting wires such as Nb Zr and Nb Ge, which are used in superconducting magnets for the magnetic resonance imaging of soft tissues. Because tantalum is highly resistant to corrosion, it is used as a liner for chemical reactors, in missile parts, and as a biologically compatible material in screws and pins for repairing fractured bones. The chemistry of the two heaviest group 5 metals (Nb and Ta) is dominated by the +5 oxidation state. The chemistry of the lightest element (V) is dominated by lower oxidation states, especially +4. As indicated in Table 1.1.1, the trends in properties of the group 5 metals are similar to those of group 4. Only vanadium, the lightest element, has any tendency to form compounds in oxidation states lower than +5. For example, vanadium is the only element in the group that forms stable halides in the lowest oxidation state (+2). All three metals react with excess oxygen, however, to produce the corresponding oxides in the +5 oxidation state (M O ), in which polarization of the oxide ions by the high-oxidation-state metal is so extensive that the compounds are primarily covalent in character. Vanadium–oxygen species provide a classic example of the effect of increasing metal oxidation state on the protonation state of a coordinated water molecule: vanadium(II) in water exists as the violet hydrated ion [V(H O) ] ; the blue-green [V(H O) ] ion is acidic, dissociating to form small amounts of the [V(H O) (OH)] ion and a proton; and in water, vanadium(IV) forms the blue vanadyl ion [(H O) VO] , which contains a formal V=O bond (Figure 1.1.2). Consistent with its covalent character, V O is acidic, dissolving in base to give the vanadate ion ([VO ] ), whereas both Nb O and Ta O are comparatively inert. Oxides of these metals in lower oxidation states tend to be nonstoichiometric. Because vanadium ions with different oxidation states have different numbers of d electrons, aqueous solutions of the ions have different colors: in acid V(V) forms the pale yellow [VO ] ion; V(IV) is the blue vanadyl ion [VO] ; and V(III) and V(II) exist as the hydrated V (blue-green) and V (violet) ions, respectively. Although group 5 metals react with the heavier chalcogens to form a complex set of binary chalcogenides, the most important are the dichalcogenides (MY ), whose layered structures are similar to those of the group 4 dichalcogenides. The elements of group 5 also form binary nitrides, carbides, borides, and hydrides, whose stoichiometries and properties are similar to those of the corresponding group 4 compounds. One such compound, tantalum carbide (TiC), has the highest melting point of any compound known (3738°C); it is used for the cutting edges of high-speed machine tools.
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The electron-transfer reactions that occur within and between proteins typically involve prosthetic groups separated by distances that are often greater than 10 Å. When we consider these distant electron transfers, an explicit expression for the electronic factor is required. In the nonadiabatic limit, the rate constant for reaction between a donor and acceptor held at fixed distance and orientation is: \[k_{et} = \bigg[ \frac{H_{AB} \;^{2}}{\hbar}\left(\dfrac{\pi}{\lambda RT}\right)^{1/2} \bigg]^{\frac{-(\lambda + \Delta G^{o})^{2}}{4 \lambda RT}}\ldotp \tag{6.27}\] The electronic (or tunneling) matrix element H is a measure of the electronic coupling between the reactants and the products at the transition state. The magnitude of H depends upon donor-acceptor separation, orientation, and the nature of the intervening medium. Various approaches have been used to test the validity of Equation (6.27) and to extract the parameters H and \(\lambda\). Driving-force studies have proven to be a reliable approach, and such studies have been emphasized by many workers. In the nonadiabatic limit, the probability is quite low that reactants will cross over to products at the transition-state configuration. This probability depends upon the electronic hopping frequency (determined by H ) and upon the frequency of motion along the reaction coordinate. In simple models, the electronic-coupling strength is predicted to decay exponentially with increasing donor-acceptor separation (Equation 6.28): \[H_{AB} = (H_{AB}^{o})^{\frac{- \beta}{2} (\textbf{d} - \textbf{d}^{o})} \tag{6.28} \] In Equation (6.28), H ° is the electronic coupling at close contact ( °), and \(\beta\) is the rate of decay of coupling with distance ( ). Studies of the distance dependence of electron-transfer rates in donor-acceptor complexes, and of randomly oriented donors and acceptors in rigid matrices, have suggested 0.8 \(\leq \beta \leq\) 1.2 Å . Analysis of a large number of intramolecular electron-transfer rates has suggested a \(\beta\) value of 1.4 Å for protein reactions (Figure 6.24). Assigning a single protein \(\beta\) implies that the intervening medium is homogenous. At best this is a rough approximation, because the medium separating two redox sites in a protein is a heterogenous array of bonded and nonbonded interactions. Beratan and Onuchic have developed a formalism that describes the medium in terms of "unit blocks" connected together to form a tunneling pathway. A unit block may be a covalent bond, a hydrogen bond, or a through-space jump, each with a corresponding decay factor. Dominant tunneling pathways in proteins are largely composed of bonded groups (e.g., peptide bonds), with less favorable through-space interactions becoming important when a through-bond pathway is prohibitively long (Figure 6.25). The tunneling pathway model has been used successfully in an analysis of the electron-transfer rates in modified cytochromes c (Section IV.D.1). Plastocyanin cycles between the Cu and Cu oxidation states, and transfers electrons from cytochrome f to the P component of photosystem I in the chloroplasts of higher plants and algae. The low molecular weight (10.5 kDa) and availability of detailed structural information have made this protein an attractive candidate for mechanistic studies, which, when taken together, point to two distinct surface binding sites (i.e., regions on the plastocyanin molecular surface at which electron transfer with a redox partner occurs). The first of these, the solvent-exposed edge of the Cu ligand His-87 (the adjacent site A in Figure 6.26), is ~6 Å from the copper atom and rather nonpolar. The second site (the remote site R in Figure 6.26) surrounds Tyr-83, and is much farther (~15 Å) from the copper center. Negatively charged carboxylates at positions 42-45 and 59-61 make this latter site an attractive one for positively charged redox reagents. Bimolecular electron-transfer reactions are typically run under pseudo-first-order conditions (e.g., with an inorganic redox reagent present in ~15-fold excess): \[Rate = k[plastocyanin,complex] = k_{obs}[plastocyanin] \ldotp \tag{6.29}\] For some reactions [e.g., Co(phen) oxidation of plastocyanin (Cu )] the expected linear plot of k vs. [complex] is not observed. Instead, the rate is observed to saturate (Figure 6.27). A "minimal" model used to explain this behavior involves the two pathways for electron transfer shown in Equation (6.30). \(\tag{6.30}\) Surprisingly, the rate ratio k /k is 7. Calculations indicate that, despite the significant differences in distances, H for the remote site is ~15 percent of H for the adjacent site. This figure is much higher than would be expected from distance alone, suggesting that the value of the decay parameter \(\beta\) in Equation (6.28) depends strongly on the structure of the intervening medium. Chemical modification of structurally characterized metalloproteins by transition-metal redox reagents has been employed to investigate the factors that control long-range electron-transfer reactions. In these semisynthetic multisite redox systems, the distance is fixed, and tunneling pathways between the donor and acceptor sites can be examined. Sperm-whale myoglobin can be reacted with (NH ) Ru(OH ) and then oxidized to produce a variety of ruthenated products, including a His-48 derivative whose Ru \(\leftrightarrow\) Fe tunneling pathway is depicted in Figure 6.28. Electrochemical data (Table 6.5) indicate that the (NH ) Ru group does not significantly perturb the heme center, and that equilibrium (i.e., k = k + k ) should be approached when a mixed-valent intermediate is produced by flash-photolysis techniques: \[(NH_{3})_{5}Ru^{3+}-Mb(Fe^{3+}) \xrightarrow[e^{-}]{fast} (NH_{3})_{5}Ru^{2+}-Mb(Fe^{3+}) \xrightleftharpoons[k_{-1}]{k_{1}} (NH_{3})_{5}Ru^{3+}-Mb(Fe^{2+}) \tag{6.31}\] This kinetic behavior was observed, and both the forward (k ) and reverse (k ) reactions were found to be markedly temperature-dependent: k = 0.019 s (25 °C), \(\Delta\)H = 7.4 kcal/mol, k = 0.041 s ) (25°C), \(\Delta\)H = 19.5 kcal/mol. X-ray crystallographic studies indicate that the axial water ligand dissociates upon reduction of the protein. This conformational change does not control the rates, since identical results were obtained when a second flash-photolysis technique was used to generate (NH ) Ru -Mb(Fe ) in order to approach the equilibrium from the other direction. Fe (NH ) Ru Cyanogen bromide has been used to modify the six-coordinate metmyoglobin heme site, causing the coordinated water ligand to dissociate. The CNBr-modified myoglobin heme site is thus five-coordinate in oxidation states. As expected, the self-exchange rate increased from ~1 M s to ~10 M s . Recent efforts in modeling biological electron transfers using chemically modified redox proteins point the way toward the design of semisynthetic redox enzymes for catalytic applications. An intriguing example, termed flavohemoglobin, was produced by reaction of hemoglobin with a flavin reagent designed to react with Cys-93 of the \(\beta\)-chain (i.e., the hemoglobin molecule was modified by two flavin moieties). The resulting derivative, unlike native hemoglobin, accepts electrons directly from NADPH and catalyzes the -hydroxylation of aniline in the presence of O and NADPH. In physiologically relevant precursor complexes, both redox centers are frequently buried in protein matrices. Characterization of such protein-protein complexes is clearly important, and several issues figure prominently: Most of our knowledge about the structures of protein-protein complexes comes from crystallographic studies of antigen-antibody complexes and multisubunit proteins; such systems generally exhibit a high degree of thermodynamic stability. On the other hand, complexes formed as a result of bimolecular collisions generally are much less stable, and tend to resist attempts to grow x-ray-quality crystals; the high salt conditions typically used in protein crystallizations often lead to dissociation of such complexes. One of the most widely studied protein-protein complexes is that formed between mammalian cytochrome b and cytochrome c. Using the known x-ray structures of both proteins, Salemme generated a static computer graphics model of this electron-transfer complex by docking the x-ray structures of the individual proteins. Two features of this model and its revision by molecular dynamics simulations (Figure 6.29 See color plate section, page C-12.) are noteworthy: (1) several Lys residues on cytochrome c and carboxylate-containing groups on cytochrome b form "salt bridges" (i.e., intermolecular hydrogen bonds); and (2) the hemes are nearly coplanar and are ~17 Å (Fe-Fe) apart. This distance was confirmed by an energy-transfer experiment in which the fluorescence of Zn-substituted cytochrome c was quenched by cytochrome b . Spectroscopic studies have verified the suggestion that these proteins form a 1:1 complex at low ionic strength (Figure 6.30). In addition, chemical modification and spectroscopic analyses are all in agreement with the suggestion that the complex is primarily stabilized by electrostatic interactions of the (-NH ••• O C—) type. The effect of ionic strength on the reduction of cytochrome c by cytochrome b is also in accord with this picture: lowering the ionic strength increases the reaction rate, as expected for oppositely charged molecules. A common experimental strategy for studying electron transfers proteins uses a metal-substituted heme protein as one of the reactants. In particular, the substitution of zinc for iron in one of the porphyrin redox centers allows facile initiation of electron transfer through photoexcitation of the zinc porphyrin (ZnP). The excited zinc porphyrin, ZnP* in Equation (6.32), may decay back (k ~ 10 s ) to the ground state or transfer an electron to an acceptor. \(\tag{6.32}\) The ZnP cation radical produced in the k step is a powerful oxidant; back electron transfer (k ) will thus occur and regenerate the starting material. The reactions shown in Equation (6.32) have been investigated in mixedmetal [Zn, Fe] hemoglobins. A hemoglobin molecule can be viewed as two independent electron-transfer complexes, each consisting of an \(\alpha_{1}\)-\(\beta_{2}\) subunit pair (Figure 6.31), since the \(\alpha_{1}\)-\(\alpha_{2}\), (\beta_{1}\)-(\beta_{2}\), and \(\alpha_{1}\)-(\beta_{1}\) distances are prohibitively long (> 30 Å). Both [\(\alpha\)(Zn), \(\beta\)(Fe)] and [\(\alpha\)(Fe), \(\beta\)(Zn)] hybrids have been studied. The ZnP and FeP are nearly parallel, as in the cytochrome b -cytochrome c model complex. Long-range electron transfer ( ZnP* → Fe ) between the \(\alpha_{1}\) and \(\beta_{2}\) subunits has been observed (the heme-edge/heme-edge distance is ~20 Å). The driving force for the forward electron-transfer step is ~0.8 eV, and k (see Equation 6.32) is ~100 s at room temperature, but decreases to ~9 s in the low-temperature region (Figure 6.32). Below 140-160 K the vibrations that induce electron transfer "freeze out"; nuclear tunneling is usually associated with such slow, temperature-independent rates. A complete analysis of the full temperature dependence of the rate requires a quantum-mechanical treatment of \(\lambda_{i}\) rather than that employed in the Marcus theory. It is interesting to note that the heme b vinyl groups (see Figure 6.6) for a given [\(\alpha_{1}\)(Fe), \(\beta_{2}\)Zn)] hybrid point toward each other and appear to facilitate electron transfer.
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Argon is it is colorless, tasteless and odorless noble gas that is located in Group 18 on the Periodic Table. It was discovered by Henry Cavendish in 1785 and was named Argon, which is derived from the Greek word "argos" meaning inactive. Cavendish formed oxides of nitrogen by passing electric currents through air, then dissolved them in water to get nitric acid, but was unable to get all of the air to react. He suspected that there was a then unidentified gas component of air; Ramsay and Rayleigh went on to isolate this component in 1894, and the new found element was thus named Argon.
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Band Theory was developed with some help from the knowledge gained during the quantum revolution in science. In 1928, Felix Bloch had the idea to take the quantum theory and apply it to solids. In 1927, Walter Heitler and Fritz London discovered bands- very closely spaced orbitals with not much difference in energy. In this image, orbitals are represented by the black horizontal lines, and they are being filled with an increasing number of electrons as their amount increases. Eventually, as more orbitals are added, the space in between them decreases to hardly anything, and as a result, a band is formed where the orbitals have been filled. Different metals will produce different combinations of filled and half filled bands. Sodium's bands are shown with the rectangles. Filled bands are colored in blue. As you can see, bands may overlap each other (the bands are shown askew to be able to tell the difference between different bands). The lowest unoccupied band is called the conduction band, and the highest occupied band is called the valence band. Bands will follow a trend as you go across a period: The probability of finding an electron in the conduction band is shown by the equation: \[ P= \dfrac{1}{e^{ \Delta E/RT}+1} \] The ∆E in the equation stands for the change in energy or energy gap. t stands for the temperature, and R is a bonding constant. That equation and this table below show how the bigger difference in energy is, or gap, between the valence band and the conduction band, the less likely electrons are to be found in the conduction band. This is because they cannot be excited enough to make the jump up to the conduction band. Metals are conductors. There is no band gap between their valence and conduction bands, since they overlap. There is a continuous availability of electrons in these closely spaced orbitals. In insulators, the band gap between the valence band the the conduction band is so large that electrons cannot make the energy jump from the valence band to the conduction band. Semiconductors have a small energy gap between the valence band and the conduction band. Electrons can make the jump up to the conduction band, but not with the same ease as they do in conductors. There are two different kinds of semiconductors: and . An intrinsic semiconductor is a semiconductor in its pure state. For every electron that jumps into the conduction band, the missing electron will generate a hole that can move freely in the valence band. The number of holes will equal the number of electrons that have jumped. In extrinsic semiconductors, the band gap is controlled by purposefully adding small impurities to the material. This process is called . Doping, or adding impurities to the lattice can change the electrical conductivity of the lattice and therefore vary the efficiency of the semiconductor. In extrinsic semiconductors, the number of holes will not equal the number of electrons jumped. There are two different kinds of extrinsic semiconductors, p-type (positive charge doped) and n-type (negative charge doped). Jim Clark ( )
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This page describes the collision theory of reaction rates, concentrating on the key factors that determine whether a particular collision will result in a reaction—in particular, the energy of the collision, and the orientation of the collision. Reactions in which a single species falls apart are simpler because the orientation of the molecule is unimportant. Reactions involving collisions between more than two species are extremely uncommon (see below). Two species can only react with each other if they come into contact with each other and then they may react. However, it is not sufficient for the two species to just collide: The chances of this happening for more than two particles at a time are remote. All three (or more) particles would have to arrive at exactly the same point in space at the same time, with everything lined up exactly right, and with enough energy to react. Such circumstances are rarely observed in nature. e, \(\ce{CH2=CH2}\), a \[ \ce{H_2C=CH_2 + HCl \rightarrow CH_3CH_2Cl} \nonumber \] As a result of the collision between the two molecules, the double bond in ethene is converted into a single bond. A hydrogen atom is now attached to one of the carbons and a chlorine atom to the other. The reaction can only happen if the hydrogen end of the \(\ce{H-Cl}\) bond approaches the carbon-carbon double bond. No other collision between the two molecules produces the same effect. The two simply bounce off each other. Of the collisions shown in Figure \(\Page {2}\) only collision 1 may possibly lead on to a reaction. With no knowledge of the reaction mechanism, one might wonder why collision 2 would be unsuccessful. The double bond has a high concentration of negative charge around it due to the electrons in the bonds. The approaching chlorine atom is also partially negative due to dipole created by the electronegativity difference between it and hydrogen. The repulsion simply causes the molecules to bounce off each other. In any collision involving unsymmetrical species, the way they hit each other is important in determining whether a reaction occurs. Even if the species are orientated properly, a reaction will not take place unless the particles collide with a certain minimum energy called the activation energy of the reaction. Activation energy is the minimum energy required to make a reaction occur. This can be illustrated on an energy profile for the reaction. An energy profile for a simple exothermic reaction is given in Figure \(\Page {2}\). If the particles collide with less energy than the activation energy, nothing interesting happens. They bounce apart. The activation energy can be thought of as a barrier to the reaction. Only those collisions with energies equal to or greater than the activation energy result in a reaction. Any chemical reaction results in the breaking of some bonds (which requires energy) and the formation of new ones (which releases energy). Some bonds must be broken before new ones can be formed. Activation energy is involved in breaking some of the original bonds. If a collision is relatively gentle, there is insufficient energy available to initiate the bond-breaking process, and thus the particles do not react. Because of the key role of activation energy in deciding whether a collision will result in a reaction, it is useful to know the proportion of the particles present with high enough energies to react when they collide. In any system, the particles present will have a very wide range of energies. For gases, this can be shown in Figure \(\Page {3}\) and is called the Maxwell-Boltzmann distribution, a plot showing the number of particles with each particular energy. The area under the curve measures of the total number of particles present. Remember that for a reaction to occur, particles must collide with energies equal to or greater than the activation energy for the reaction. The activation energy is marked on the with a green line. Notice that the large majority of the particles have insufficient energy to react when they collide. To enable them to react, either the shape of the curve must be altered, or the activation energy shifted further to the left. This is described on other pages.
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Because energy takes many forms, only some of which can be seen or felt, it is defined by its effect on matter. For example, microwave ovens produce energy to cook food, but we cannot see that energy. In contrast, we can see the energy produced by a light bulb when we switch on a lamp. In this section, we describe the forms of energy and discuss the relationship between energy, heat, and work. The forms of energy include thermal energy, radiant energy, electrical energy, nuclear energy, and chemical energy (Figure \(\Page {1}\)). Thermal energy results from atomic and molecular motion; the faster the motion, the greater the thermal energy. The temperature of an object is a measure of its thermal energy content. Radiant energy is the energy carried by light, microwaves, and radio waves. Objects left in bright sunshine or exposed to microwaves become warm because much of the radiant energy they absorb is converted to thermal energy. Electrical energy results from the flow of electrically charged particles. When the ground and a cloud develop a separation of charge, for example, the resulting flow of electrons from one to the other produces lightning, a natural form of electrical energy. Nuclear energy is stored in the nucleus of an atom, and chemical energy is stored within a chemical compound because of a particular arrangement of atoms. Electrical energy, nuclear energy, and chemical energy are different forms of potential energy (PE), which is energy stored in an object because of the relative positions or orientations of its components. A brick lying on the windowsill of a 10th-floor office has a great deal of potential energy, but until its position changes by falling, the energy is contained. In contrast, kinetic energy (KE) is energy due to the motion of an object. When the brick falls, its potential energy is transformed to kinetic energy, which is then transferred to the object on the ground that it strikes. The electrostatic attraction between oppositely charged particles is a form of potential energy, which is converted to kinetic energy when the charged particles move toward each other. Energy can be converted from one form to another (Figure \(\Page {2}\)) or, as we saw with the brick, transferred from one object to another. For example, when you climb a ladder to a high diving board, your body uses chemical energy produced by the combustion of organic molecules. As you climb, the chemical energy is converted to to overcome the force of gravity. When you stand on the end of the diving board, your potential energy is greater than it was before you climbed the ladder: the greater the distance from the water, the greater the potential energy. When you then dive into the water, your potential energy is converted to kinetic energy as you fall, and when you hit the surface, some of that energy is transferred to the water, causing it to splash into the air. Chemical energy can also be converted to radiant energy; one common example is the light emitted by fireflies, which is produced from a chemical reaction. Although energy can be converted from one form to another, . This is known as the : The kinetic energy of an object is related to its mass \(m\) and velocity \(v\): For example, the kinetic energy of a 1360 kg (approximately 3000 lb) automobile traveling at a velocity of 26.8 m/s (approximately 60 mi/h) is \[\begin{align} KE &=\dfrac{1}{2}(1360\, kg)(26.8 \, ms)^2 \label{12.1.5} \\[4pt] &= 4.88 \times 10^5 g \cdot m^2\nonumber \end{align} \] Because all forms of energy can be interconverted, energy in any form can be expressed using the same units as kinetic energy. The SI unit of energy, the joule (J), is defined as 1 kilogram·meter /second (kg·m /s ). Because a joule is such a small quantity of energy, chemists usually express energy in kilojoules (1 kJ = 10 J). For example, the kinetic energy of the 1360 kg car traveling at 26.8 m/s is 4.88 × 10 J or 4.88 × 10 kJ. It is important to remember that , whether thermal, radiant, chemical, or any other form. Because heat and work result in changes in energy, their units must also be the same. To demonstrate, let’s calculate the potential energy of the same 1360 kg automobile if it were parked on the top level of a parking garage 36.6 m (120 ft) high. Its potential energy is equivalent to the amount of work required to raise the vehicle from street level to the top level of the parking garage, which is \[w = F\,d.\] The force (\(F\)) exerted by gravity on any object is equal to its mass (\(m\), in this case, 1360 kg) times the acceleration (\(a\)) due to gravity ( , 9.81 m/s at Earth’s surface). The distance (\(d\)) is the height ( ) above street level (in this case, 36.6 m). Thus the potential energy of the car is as follows: \[ \begin{align} PE &= F\;d = m\,a\;d = m\,g\,h \label{12.1.6} \\[4pt] &=(1360\, Kg)\left(\dfrac{9.81\, m}{s^2}\right)(36.6\;m)\nonumber \\[4pt] &= 4.88 \times 10^5\; \frac{Kg \cdot m^2}{s^2}\nonumber \\[4pt] &=4.88 \times 10^5 J = 488\; kJ\nonumber \end{align} \] The units of potential energy are the same as the units of kinetic energy. Notice that in this case the potential energy of the stationary automobile at the top of a 36.6 m high parking garage is the same as its kinetic energy at 60 mi/h. If the vehicle fell from the roof of the parking garage, its potential energy would be converted to kinetic energy, and it is reasonable to infer that the vehicle would be traveling at 60 mi/h just before it hit the ground, neglecting air resistance. After the car hit the ground, its potential and kinetic energy would both be zero. Potential energy is usually defined relative to an arbitrary standard position (in this case, the street was assigned an elevation of zero). As a result, we usually calculate only differences in potential energy: in this case, the difference between the potential energy of the car on the top level of the parking garage and the potential energy of the same car on the street at the base of the garage. The units of energy are the same for all forms of energy. Energy can also be expressed in the non-SI units of calories (cal), where 1 cal was originally defined as the amount of energy needed to raise the temperature of exactly 1 g of water from 14.5 °C to 15.5 °C. The name is derived from the Latin , meaning “heat.” Although energy may be expressed as either calories or joules, calories were defined in terms of heat, whereas joules were defined in terms of motion. Because calories and joules are both units of energy, however, the calorie is now defined in terms of the joule: \[ \begin{align} 1 \;cal &\equiv 4.184 \;J \; \quad \text{(exactly)} \label{12.1.7a} \\[4pt] 1 \;J &= 0.2390\; cal \label{12.1.7b} \end{align}\] In this text, we will use the SI units—joules (J) and kilojoules (kJ)—exclusively, except when we deal with nutritional information. mass and velocity or height kinetic and potential energy Use Equation \(\ref{12.1.4}\) to calculate the kinetic energy and Equation \(\ref{12.1.6}\) to calculate the potential energy, as appropriate. The kinetic energy of the baseball is therefore (via Equation \(\ref{12.1.4}\)) \[\begin{align*} KE &= \dfrac{1}{2} 149 \;\cancel{g} \left(\dfrac{1\; kg}{1000 \;\cancel{g}} \right) \left(\dfrac{44.7 \;m}{s} \right)^2 \\[4pt] &= 1.49 \times 10^2 \dfrac{kg⋅m^2}{s^2} \\[4pt] &= 1.49 \times10^2\; J \end{align*}\] 3.10 × 10 J 65 J ​​​​​​ To study the flow of energy during a chemical reaction, we need to distinguish between a , the small, well-defined part of the universe in which we are interested (such as a chemical reaction), and its , the rest of the universe, including the container in which the reaction is carried out (Figure \(\Page {3}\)). In the discussion that follows, the mixture of chemical substances that undergoes a reaction is always the system, and the flow of heat can be from the system to the surroundings or vice versa. Three kinds of systems are important in chemistry. An can exchange both matter and energy with its surroundings. A pot of boiling water is an open system because a burner supplies energy in the form of heat, and matter in the form of water vapor is lost as the water boils. A can exchange energy but not matter with its surroundings. The sealed pouch of a ready-made dinner that is dropped into a pot of boiling water is a closed system because thermal energy is transferred to the system from the boiling water but no matter is exchanged (unless the pouch leaks, in which case it is no longer a closed system). An exchanges neither energy nor matter with the surroundings. Energy is always exchanged between a system and its surroundings, although this process may take place very slowly. A truly isolated system does not actually exist. An insulated thermos containing hot coffee approximates an isolated system, but eventually the coffee cools as heat is transferred to the surroundings. In all cases, the amount of heat lost by a system is equal to the amount of heat gained by its surroundings and vice versa. That is, , which must be true if . The is a complete description of a system at a given time, including its temperature and pressure, the amount of matter it contains, its chemical composition, and the physical state of the matter. A is a property of a system whose magnitude depends on only the present state of the system, not its previous history. Temperature, pressure, volume, and potential energy are all state functions. The temperature of an oven, for example, is independent of however many steps it may have taken for it to reach that temperature. Similarly, the pressure in a tire is independent of how often air is pumped into the tire for it to reach that pressure, as is the final volume of air in the tire. Heat and work, on the other hand, are not state functions because they are . For example, a car sitting on the top level of a parking garage has the same potential energy whether it was lifted by a crane, set there by a helicopter, driven up, or pushed up by a group of students (Figure \(\Page {4}\)). The amount of work expended to get it there, however, can differ greatly depending on the path chosen. If the students decided to carry the car to the top of the ramp, they would perform a great deal more work than if they simply pushed the car up the ramp (unless, of course, they neglected to release the parking brake, in which case the work expended would increase substantially!). The potential energy of the car is the same, however, no matter which path they choose. is a branch of chemistry that qualitatively and quantitatively describes the energy changes that occur during chemical reactions. is the capacity to do work. is the amount of energy required to move an object a given distance when opposed by a force. is due to the random motions of atoms, molecules, or ions in a substance. The of an object is a measure of the amount of thermal energy it contains. is the transfer of thermal energy from a hotter object to a cooler one. Energy can take many forms; most are different varieties of , energy caused by the relative position or orientation of an object. is the energy an object possesses due to its motion. Energy can be converted from one form to another, but the states that energy can be neither created nor destroyed. The most common units of energy are the , defined as 1 (kg·m )/s , and the , defined as the amount of energy needed to raise the temperature of 1 g of water by 1°C (1 cal = 4.184 J).
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The physical properties of haloalkanes are much as one might expect. Volatility decreases: (a) with increasing molecular weight along a homologous series, (b) with increasing atomic number of the halogen, and (c) with the structure of the alkyl group in the order such that \(<\) \(<\) for isomeric halides. These trends are apparent from the physical properties listed in Table 14-1, which includes data for simple halogen derivatives of alkanes, alkenes, alkynes, and arenes. The boiling points of many halogen compounds are similar to hydrocarbons of the same molecular weight, but there are some conspicuous exceptions. lodomethane, for example, has about the same molecular weight as decane (MW 142), but the boiling point of iodomethane is \(132^\text{o}\) than that of decane. Likewise, fluorocarbons (e.g., tetrafluoromethane, \(\ce{CF_4}\), MW 88, bp \(-129^\text{o}\)) are far more volatile than hydrocarbons of similar weights (e.g., hexane, \(\ce{C_6H_{14}}\), MW 86, bp \(69^\text{o}\)). In general, halogen compounds are insoluble in water but are readily soluble in organic solvents and, with the exception of some fluoro and monochloro compounds, they are more dense than water. Aryl halides are fairly pleasant smelling liquids, but arylmethyl (benzylic) halides of structure \(\ce{ARCH_2X}\) are irritating to the eyes, skin, and nasal passages. Toxicity varies, but the chlorinated hydrocarbons such as \(\ce{CCl_4}\) (“carbon tet”) and \(\ce{CHCl_2-CHCl_2}\) are quite toxic and should be used with care. and (1977)
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Potassium is a metal, abbreviated as K on the . In its pure form, potassium has a white-sliver color, but quickly oxidizes upon exposure to air and tarnishing in minutes if it is not stored under oil or grease. Potassium is essential to several aspects of plant, animal, and human life and is thus mined, manufactured, and consumed in huge quantities around the world. The seventh most abundant element, potassium was discovered and isolated in 1807 by Sir Humphry Davy. Important compounds of potassium include potassium hydroxide (used in some drain cleaners), potassium superoxide, \(KO_2\), which is used in respiratory equipment and potassium nitrate, used in fertilizers and pyrotechnics. Potassium, like sodium, melts below the boiling point of water (63 °C) and is less dense than water also. Like most of the alkali metals, potassium compounds impart a characteristic color to flames. In the case of the 19th element, the color is pale lavender. Like sodium ions, the presence of potassium ions in the body is essential for the correct function of many cells. Potassium reacts so violently with water that it bursts into flame. The silvery white metal is very soft and reacts rapidly with the oxygen in air. Its chemical symbol is derived from the Latin word kalium which means "alkali". Its English name is from potash which is the common name for a compound containing it. Potassium has a 2.6% abundance by mass in the earth's crust and is found mostly in mineral form as part of feldspars (groups of minerals) and clays. Potassium easily leaches out of these minerals over time and thus has a relatively high concentration in sea water as well (0.75g/L). Today, most of the world's potassium is mined in Canada, the U.S., and Chile but was originally monopolized by Germany. Plants, animals and humans all depend on potassium for survival and good health. The element is part of many bodily fluids and assists related functions of the human body. Most notably, potassium aids nerve functions and is found in several cell types (including skeletal cells, smooth muscle cells, endocrine cells, cardiac cells, and central neurons). Plants depend on potassium for healthy growth. Potassium found in animal excretions and dead plants easily binds to clay in the soil they fall on and is thus utilized by plants. The element helps maintain osmotic pressure and cell size and plays a role in photosynthesis and energy production. 95% of manufactured potassium is used in fertilizers and the rest is used to produce specific compounds of potassium, such as potassium hydroxide (\(KOH\)), which can then be turned into potassium carbonate (\(K_2CO_3\)). Potassium carbonate is used in glass manufacturing and potassium hydroxide is found in liquid soaps and detergents. Potassium chloride is used in many pharmaceuticals and other salts of potassium are used in baking, photography, tanning leather, and iodized salt. In these cases, potassium is utilized for its negative anion. Potassium can be obtained through various known reactions, all of which require heat treatment: \[K_2CO_3+2C \overset{\Delta}{\longrightarrow} 3CO+2K \label{1}\] \[2KCl+CaC_2 \overset{\Delta}{\longrightarrow} CaCl_2+2C+K \label{2}\] \[2KN_3 \overset{\Delta}{\longrightarrow} 3N_2+2K \label{3}\] Due to expenses, these processes are not commercially adaptable. Therefore the element is commonly obtained through reduction at elevated heats (i.e., ). Sodium is often combined with \(KCl\), \(KOH\), or \(K_2CO_3\) to produce potassium sodium alloys and in the 1950's the Mine Safety Appliances Company developed a reduction process that yields high purity potassium: \[KCl+Na \overset{\Delta}{\longrightarrow} K+NaCl \label{4}\] The reaction is heated in a special device equipped with a furnace, heat-exchanger tubes, a fractionating column, a \(KCl\) feed, a waste removal system, and a vapor condensing system. Because the reaction attains equilibrium quickly, potassium can be removed continuously as a product in order to shift equilibrium to the right and produce even more potassium in its place. of potassium include \(NaK\) (Sodium) and \(KLi\) (Lithium). Both of these alloys produce metals of low vapor pressure and melting points.
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In the late 1600's minerals which we now know contain fluorine were used in etching glass. The discovery of the element was prompted by the search for the chemical substance which was able to attack glass (it is HF, a weak acid). The early history of the isolation and work with fluorine and hydrogen fluoride is filled with accidents since both are extremely dangerous. Eventually, electrolysis of a mixture of KF and HF (carefully ensuring that the resulting hydrogen and fluorine would not come in contact) in a platinum apparatus yielded the element. Fluorine was discovered in 1530 by Georgius Agricola. He originally found it in the compound Fluorspar, which was used to promote the fusion of metals. It was under this application until 1670, when Schwanhard discovered its usefulness in etching glass. Pure fluorine (from the Latin fluere, for "flow") was was not isolated until 1886 by Henri Moissan, burning and even killing many scientists along the way. It has many uses today, a particular one being used in the Manhattan project to help create the first nuclear bomb. Fluorine is the most electronegative element on the periodic table, which means that it is a very strong oxidizing agent and accepts other elements' electrons. Fluorine is the most electronegative element because it has 5 electrons in it's 2P shell. The optimal electron configuration of the 2P orbital contains 6 electrons, so since Fluorine is so close to ideal electron configuration, the electrons are held very tightly to the nucleus. The high electronegativity of fluorine explains its small radius because the positive protons have a very strong attraction to the negative electrons, holding them closer to the nucleus than the bigger and less electronegative elements. Because of its reactivity, elemental fluorine is never found in nature and no other chemical element can displace fluorine from its compounds. elf (\(F_2\)), or s (\(ClF\), \(ClF_3\), \(ClF_5\)). It will \(OH^-\). \[3F_2+2H_2O \rightarrow O_2+4HF \tag{1}\] \[HF+H_2O \rightarrow H_3O^++F^- \tag{2}\] \(CF_4\), \(C_2F_6\), an \(C_5F_{12}\). \[C_{(s)} + F_{2(g)} \rightarrow CF_{4(g)} + C_2F_6 + C_5F_{12} \tag{3}\] \(OF_2\) be \[2F_2 + O_2 \rightarrow 2OF_2 \tag{4}\] \(XeF_2\). \[Xe + F_2 \rightarrow XeF_2 \tag{5}\] \[F_2 + 2NaOH \rightarrow O_2 + 2NaF +H_2 \tag{6}\] \[4F_2 + HCl + H_2O \rightarrow 3HF + OF_2 + ClF_3 \tag{7}\] \[F_2 + 2HNO_3 \rightarrow 2NO_3F + H_2 \tag{8}\] Compounds of fluorine are present in fluoridated toothpaste and in many municipal water systems where they help to prevent tooth decay. And, of course, fluorocarbons such as Teflon have made a major impact on life in the 20th century. Fluorine can either be found in nature or produced in a lab. To make it in a lab, compounds like Potassium Fluoride are put through with Hydrofluoric acid to create pure Fluorine and other compounds. It can be carried out with a variety of compounds, usually ionic ones involving Fluorine and a metal. Fluorine can also be found in nature in various minerals and compounds. The two main compounds it can be found in are Fluorspar (\(CaF_2\)) and Cryolite (\(Na_3AlF_6\)). 2. Q. Is Fluorine usually oxidized or reduced? explain. A. 3. Q. What are some common uses of Fluorine? A. 4. Q. Does Fluorine form compounds with nonmetals? if so, give two examples, one of them being of an oxide. A. 5. Q. What group is Fluorine in? (include name of group and number) A.
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In classical physics, studying the behavior of a physical system is often a simple task due to the fact that several physical qualities can be measured simultaneously. However, this possibility is absent in the quantum world. In 1927 the German physicist Werner Heisenberg described such limitations as the Heisenberg Uncertainty Principle, or simply the Uncertainty Principle, stating that it is not possible to measure both the momentum and position of a particle simultaneously. In order to understand the conceptual background of the Heisenberg Uncertainty Principle it is important to understand how physical values are measured. In almost any measurement that is made, light is reflected off the object that is being measured and processed. The shorter the wavelength of light used, or the higher its frequency and energy, the more accurate the results. For example, when attempting to measure the speed of a tennis ball as it is dropped off of a ledge, photons(measurement of light) are shot off the tennis ball, reflected, and then processed by certain equipment. Because the tennis ball is so large compared to the photons, it is unaffected by the efforts of the observer to measure its physical quantities. However, if a photon is shot at an electron, the minuscule size of the electron and its unique wave-particle duality introduces consequences that can be ignored when taking measurements of macroscopic objects. Heisenberg himself encountered such limitations as he attempted to measure the position of an electron with a microscope. As noted, the accuracy of any measurement is limited by the wavelength of light illuminating the electron. Therefore, in principle, one can determine the position as accurately as one wishes by using light of very high frequency, or short wave-lengths. However, the collision between such high energy photons of light with the extremely small electron causes the momentum of the electron to be disturbed. Thus, increasing the energy of the light (and increasing the accuracy of the electron's position measurement), increases such a deviation in momentum. Conversely, if a photon has low energy the collision does not disturb the electron, yet the position cannot be accurately determined. Heisenberg concluded in his famous 1927 paper on the topic, "At the instant of time when the position is determined, that is, at the instant when the photon is scattered by the electron, the electron undergoes a discontinuous change in momentum. This change is the greater the smaller the wavelength of the light employed, i.e., the more exact the determination of the position. At the instant at which the position of the electron is known, its momentum therefore can be known only up to magnitudes which correspond to that discontinuous change; thus, the more precisely the position is determined, the less precisely the momentum is known..." (Heisenberg, 1927, p. 174-5). Heisenberg realized that since both light and particle energy are quantized, or can only exist in discrete energy units, there are limits as to how small, or insignificant, such an uncertainty can be. As proved later in this text, that bound ends up being expressed by Planck's Constant, h = 6.626*10 J*s. It is important to mention that The Heisenberg Principle should not be confused with the observer effect. The observer effect is generally accepted to mean that the act of observing a system will influence that which is being observed. While this is important in understanding the Heisenberg Uncertainty Principle, the two are not interchangeable. The error in such thinking can be explained using the wave-particle duality of electromagnetic waves, an idea first proposed by Louis de Broglie. Wave-particle duality asserts that any energy exhibits both particle- and wave-like behavior. As a consequence, in quantum mechanics, a particle cannot have both a definite position and momentum. Thus, the limitations described by Heisenberg are a natural occurrence and have nothing to do with any limitations of the observational system. It is mathematically possible to express the uncertainty that, Heisenberg concluded, always exists if one attempts to measure the momentum and position of particles.First, we must define the variable “x” as the position of the particle, and define “p” as the momentum of the particle. The momentum of a photon of light is known to simply be its frequency, expressed by the ratio h/λ, where h represents Planck’s constant and λ represents the wavelength of the photon. The position of a photon of light is simply its wavelength, \lambda\).. In order to represent finite change in quantities, the Greek uppercase letter delta, or Δ, is placed in front of the quantity. Therefore, \[\Delta{p}=\dfrac{h}{\lambda}\] \[\Delta{x}= \lambda\] By substituting \(\Delta{x}\) for \(\lambda\) in the first equation, we derive \[\Delta{p}=\dfrac{h}{\Delta{x}}\] or, \[\Delta{p}\Delta{x}=h\] Note, we can derive the same formula by assuming the particle of interest is behaving as a particle, and not as a wave. Simply let Δp=mu, and Δx=h/mu (from De Broglie’s expression for the wavelength of a particle). Substituting in Δp for mu in the second equation leads to the very same equation derived above-ΔpΔx=h. This equation was refined by Heisenberg and his colleague Niels Bohr, and was eventually rewritten as \[\Delta{p}\Delta{x} \ge \dfrac{h}{4\pi}\] What this equation reveals is that the more accurately a particle’s position is known, or the smaller Δx is, the less accurately the momentum of the particle Δp is known. Mathematically, this occurs because the smaller Δx becomes, the larger Δp must become in order to satisfy the inequality. However, the more accurately momentum is known the less accurately position is known. It is hard for most people to accept the uncertainty principle, because in classical physics the velocity and position of an object can be calculated with certainty and accuracy. However, in quantum mechanics, the wave-particle duality of electrons does not allow us to accurately calculate both the momentum and position because the wave is not in one exact location but is spread out over space. A "wave packet" can be used to demonstrate how either the momentum or position of a particle can be precisely calculated, but not both of them simultaneously. An accumulation of waves of varying wavelengths can be combined to create an average wavelength through an interference pattern: this average wavelength is called the "wave packet". The more waves that are combined in the "wave packet", the more precise the position of the particle becomes and the more uncertain the momentum becomes because more wavelengths of varying momenta are added. Conversely, if we want a more precise momentum, we would add less wavelengths to the "wave packet" and then the position would become more uncertain. Therefore, there is no way to find both the position and momentum of a particle simultaneously. Several scientists have debated the Uncertainty Principle, including Einstein. Einstein created a slit experiment to try and disprove the Uncertainty Principle. He had light passing through a slit, which causes an uncertainty of momentum because the light behaves like a particle and a wave as it passes through the slit. Therefore, the momentum is unknown, but the initial position of the particle is known. Here is a video that demonstrates particles of light passing through a slit and as the slit becomes smaller, the final possible array of directions of the particles becomes wider. As the position of the particle becomes more precise when the slit is narrowed, the direction, or therefore the momentum, of the particle becomes less known as seen by a wider horizontal distribution of the light. Heisenberg’s Uncertainty Principle not only helped shape the new school of thought known today as quantum mechanics, but it also helped discredit older theories. Most importantly, the Heisenberg Uncertainty Principle made it obvious that there was a fundamental error in the Bohr model of the atom. Since the position and momentum of a particle cannot be known simultaneously, Bohr’s theory that the electron traveled in a circular path of a fixed radius orbiting the nucleus was obsolete. Furthermore, Heisenberg’s uncertainty principle, when combined with other revolutionary theories in quantum mechanics, helped shape wave mechanics and the current scientific understanding of the atom. 1.) The Heisenberg Uncertainty Principle discredits the aspect of the Bohr atom model that an electron is constrained to a one-dimensional orbit of a fixed radius around the nucleus. 2.) The Observer Effect means the act of observing a system will influence what is being observed, whereas the Heisenberg Uncertainty Principle has nothing to do with the observer or equipment used during observation. It simply states that a particle behaves both as a wave and a particle and therefore cannot have both a definite momentum and position. 3.) Uncertainty principle: ΔxΔp≥h/4Π Can be written (x)(m)(v)(%)=h/4Π (position)(mass of electron)(velocity)(percent accuracy)=(Plank's Constant)/4Π (.05*10 m)(9.11*10 kg)(v)(.01)=(6.626*10 J*s)/4Π v=1*10 m/s 4.) Uncertainty principle: ΔxΔp≥h/4Π (10*10 m)(Δp)≥(6.626*10 J*s)/4Π Δp≥5*10 (kg*m)/s 5.) Uncertainty principle: ΔxΔp≥h/4Π Can be written (x)(m)(v)(%)=h/4Π (position)(mass of electron)(velocity)(percent accuracy)=h/4Π (radius)(9.11*10 kg)(2*10 )(.03)=(6.626*10 J*s)/4Π r=9*10 m r=9*10 nm
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Helium is at the top of the (which also contains neon, argon, krypton, xenon, and radon) and is the least reactive element. Helium has many interesting characteristics, such as making balloons float and raising the pitch of one's voice; these applications are discussed below. Helium is the second most abundant element in the universe, next to hydrogen. Helium is colorless, odorless, and tasteless. It has a very low boiling point, and is monatomic. Helium is small and extremely light, and is the least reactive of all elements; it does not react with any other elements or ions, so there are no helium-bearing minerals in nature. Helium was first observed by studying the sun, and was named after the Greek word for the sun, . Helium is one of the most abundant elements in the universe. Large quantities are produced in the energy-producing fusion reactions in stars. Previously, helium was rarely used, because only .0004% of Earth's atmosphere is helium—that equates to one helium molecule for every 200,000 air molecules, including oxygen, hydrogen, and nitrogen. However, the discovery of helium-rich wells in Texas, Russia, Poland, Algeria, China, and Canada has made helium more accessible. Helium is produced in minerals through radioactive decay. Helium is extracted from natural gas deposits, which often contain as much as 10% helium. These natural gas reserves are the only industrially-available source of helium. The total world helium resources theoretically add up to 25.2 billion cubic meters; the United States contains 11.1 billion cubic meters. The extracted gas is subjected to chemical pre-purification, using an alkaline wash to remove carbon dioxide and hydrogen sulfide. The remaining gas is cooled to -200°C, where all materials, except helium gas, are liquefied. in 1868 Jenssen, who was studying the chromosphere of the Sun during a solar eclipse. He used a spectrometer to resolve the light into its spectrum, in which each color represents a different gaseous element. He observed a new yellow light, concluding that it indicated the presence of an element not previously known. In 1895, the existence of helium on Earth was proved by Sir William Ramsay. Heating cleveite (a radioactive mineral) released an inert gas, which was found to be helium; this helium is a by-product of the natural decay of radioactive elements. The chemists Norman Lockyer and Edward Frankland confirmed helium as an element and named it after the Greek word for the Sun. Helium has a number of applications due to its inert nature. Liquefied helium has cryogenic properties, and is used to freeze biological materials for long term storage and later use. Twenty percent of industrial helium use is in wielding and industrial applications. Helium protects the heated parts of metals such as aluminum and titanium from air. Mixtures of helium and oxygen are used in tanks for underwater breathing devices: due to its low density, helium gas allows oxygen to stream easily through the lungs. Because helium remains a gas, even at temperatures low enough to liquefy hydrogen, it is used as pressure gas to move liquid hydrogen into rocket engines. Its inert nature also makes helium useful for cooling nuclear power plants. The most commonly known characteristic of helium is that it is lighter than air. It can levitate balloons during parties and fly blimps over sports stadiums. Helium has 92% of the lifting power of hydrogen; however, it is safer to use because it is noncombustible and has lower rate of diffusion than that of hydrogen gas. The famous Hindenburg disaster is an example of the hazards of using combustible gas like hydrogen. Because helium was previously very expensive only available from natural gas reserves in U.S., Nazi Germany had only hydrogen gas at its disposal. The consequences were devastating, as shown below: Currently, helium is found in other natural gas reserves around the world. The cost of helium has decreased from $2500/ft in 1915 to $0.15/ft in 1989. Helium is what keeps the Goodyear blimps afloat over stadiums. Helium is often inhaled from balloons to produce a high, squeaky voice. This practice can be very harmful. Inhaling helium can lead to loss of consciousness and cerebral arterial gas embolism, which can temporarily lead to complete blindness. This occurs when blood vessels in the lungs rupture, allowing the gas to gain access to the pulmonary vasculature and subsequently the brain. Helium is naturally found in the gas state. Helium is the second least reactive element and noble gas (after neon). Its low atomic mass, thermal conductivity, specific heat, and sound speed are greatest after hydrogen. Due to the small size of helium atoms, the diffusion rate through solids is three times greater than that of air and 65% greater than that of hydrogen. The element is inert, monatomic in standard conditions, and the least water soluble gas. At normal ambient temperatures, helium has a negative . Thus, upon free expansion, helium naturally heats up. However, below its Joule Thomson inversion temperature (32-50 K at 1 atm), it cools when allowed to freely expand. Once cooled, helium can be liquefied through expansion cooling. Helium is commonly found throughout the universe as plasma, a state in which electrons are not bound to nuclei. Plasmas have high electrical conductivities and are highly influenced by magnetic and electric fields. Helium is the only element that cannot be solidified by lowering the temperature at ordinary pressures; this must be accompanied by a pressure increase. The volume of solid helium, He and He, can be decreased by more than 30% by applying pressure. Solid helium has a projected density of 0.187 ± 0.009 g/mL at 0 K and 25 bar. Solid helium also has a sharp melting point and a crystalline structure. There are two forms of liquid helium: He I and He II. Helium I is formed when temperature falls below 4.22 K and above the lambda point of 2.1768 K. It is a clear liquid that boils when heat is applied and contracts when temperature is lowered. Below the lambda point, helium does not boil, but expands. Helium I has a gas-like index of refraction of 1.026 which makes its surface difficult to see. It has a very low viscosity and a density 1/8th that of water. This property can be explained with quantum mechanics. Both helium I and II are quantum fluids, displaying atomic properties on a macroscopic scale due to the fact that the boiling point of helium is so close to absolute zero. At 2.174 K, helium I forms into helium II. Its properties are very unusual, and the substance is described as superfluid. Superfluid is a quantum-mechanical state of matter; the two-fluid model for helium II explains why one portion of helium atoms exists in a ground state, flowing with zero viscosity, and another portion is in an excited state, behaving like an ordinary fluid. The viscosity of He II is so low that there is no internal friction. He II can conduct heat 300 times more effectively than silver, making it the best heat conductor known. Its thermal conductivity is a million times that of helium I and several hundred times that of copper. The conductivity and viscosity of helium II do not obey classical rules, but are consistent with the rules of quantum mechanics. When temperature is lowered, helium II expands in volume. It cannot be boiled, but evaporates directly to gas when heated. In this superfluid state, liquid helium can flow through thin capillaries or cracks much faster than helium gas. It also exhibits a creeping effect, moving along the surface seemingly against gravity. Helium II creeps along the sides of a open vessel until it reaches a warmer region where it evaporates. As a result of the creeping behavior and the ability to leak rapidly through tiny openings, helium II is very difficult to confine. Helium II also exhibits a fountain effect. Suppose a chamber allows a reservoir of helium II to filter superfluid and non-superfluid helium. When the interior of the container is heated, superfluid helium converts to non-superfluid helium to maintain equilibrium. This creates intense pressure on the superfluid helium, causing the liquid to fountain out of the container. Helium has eight known isotopes but only two are stable: He and He. He is found in only very small quantities compared to He. is produced in trace amounts by the beta decay of tritium. This form is found in abundance in stars, as a product of nuclear fusion. Extraplanetary materials have trace amounts of He from solar winds. He is produced by the alpha decay of heavier radioactive elements on Earth. It is an unusually stable isotope because its nucleons are arranged in complete shells. a). nothing b). the cigarette is incinerated before touching the cylinder c). the cylinder explodes d). the cylinder becomes a flame thrower a). It starts boiling. b). It starts "creeping" over the divider, soon filling up the other sections of the dish. c). It evaporates and soon leaves the dish. d.) It solidifies and expands, breaking the dividers of the petri dish, and filling up the whole dish.
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. If a substance has relatively strong intermolecular forces, a high melting point, and is not easily compressed, will this substance be a solid, liquid, or gas? The substance is a solid because solids have relatively stronger intermolecular forces than liquids or gases. These strong intermolecular forces strongly hold the molecules in the solid together which makes it hard to compress a solid. Solids also have higher melting points because you need to put a lot more energy into solids to break apart the intermolecular forces to make the solid change phases. An unknown sample at \(25^oC\) had a volume of \(3.00\times 10^{14} \mu m^3\) and a mass of \(3.20\times 10^3\ g\). We need to compute the density of the liquid first: \[\left(\dfrac{3.2 \times 10^3 g\;}{3.0 \times 10^{14} \cancel{µm^{3}}} \right)\left(\dfrac{1.0 \times 10^{12}\cancel{µm^{3}}\;}{\textrm{1} \textrm{ cm}^{3}\;}\right )=1.07 \times 10^{1}\textrm{g cm}^{-3}\nonumber \] This material is known to be condensed because its density is higher than that of water. Calculate the molar volume by: \[\left(\dfrac{\textrm{1} \textrm{ cm}^{3}\;}{\textrm{10.7} \cancel{g}\;}\right)\left(\dfrac{127\cancel{g}\;}{\textrm{1} \textrm{ mol}\;}\right )=1.19\times10^{1}\textrm{cm}^{3} {mol}^{-1}\nonumber \] Cooling a sample of matter from 130° to 50° at a constant pressure causes its volume to increase from 769.1 to 930.1 cm . Classify the material as nearly ideal gas, a non ideal gas, or condensed. Charles' Law - Ideal Gases \[\dfrac{V_{1}}{T_{1}}=\dfrac{V_{2}}{T_{2}}\nonumber \] \[0.00191 \dfrac{L}{K} = 0.00288 \dfrac{L}{K}\nonumber \] The material is a non-ideal gas. The similarity of the molar volumes of solid and liquid forms of the same substance suggests that the distance between neighboring molecules of the substance in solid phase is approximately the same as the distance between neighboring molecules in liquid phase. The great difference of the molar volumes of solid/liquid and gas forms of the same substance suggests that the distance between neighboring molecules of the substance in gas phase is greater the distance between neighboring molecules in solid/liquid phase. This would explain why solids and liquids are called condensed states of matter, because the particles in these two phases are very close together, hence, solids and liquids have definite volumes. While the particles in the gas phase are so far away from each other that they can move freely at high speed, allowing the gas substance to assume the shape and volume of its container. Suppose you have to choose between solid \(\ce{I_2}\) and solid \(\ce{NH_4Cl}\) to fill your cushion and you prefer softer cushions, which one would you choose and why? Non-directional ion-ion interaction is the dominant intermolecular interaction solid \(\ce{NH_4Cl}\) has, and weaker London dispersion forces are the main intermolecular interaction \(\ce{I_2}\) has, while the feeling of softness (indentation that breaks bonds in the solid) of the two solids depends on their strength of intermolecular interactions; in this case \(\ce{I_2}\) has weaker intermolecular interactions and will feel softer. The diffusion constant is defined as the amount of substance that when diffusing from a region of high concentration to that of a low concentration goes through each unit of cross section per unit time. What happens to the diffusion constant, at constant temperature, as density of a liquid decreases, as the density of a solid increases, and as the density of a gas decreases? Explain the phenomena. As the density of a liquid decreases, the diffusion constant will increase. This is because as density of the liquid decreases, there is more space in between the liquid molecules, and hence more movement is possible for the liquid molecules. As density of a solid increases, diffusion constant will decrease. Similarly, in concept for the liquid molecules, if the density of a solid increases, there is even less space between the solid molecules than there was before. This further restricts the movements of the solid molecules and decreases diffusion. Note: solids also undergo diffusion; however, it occurs at an extremely slow rate. As the density of gas decreases, diffusion constant will increase. As the density of a gas decreases, collision between gas molecules will occur less frequently, thus increasing freedom of movement for gas molecules and increasing the diffusion constant. For the following chemicals list attractive intermolecular forces that you expect to see with each chemical. Of the forces you listed for each example, rank the intermolecular forces from strongest to weakest. Which of the the following will be most strongly attracted to a lithium ion: A fluoride ion will be most strongly attracted to a lithium ion. The attraction between unlike charges such as lithium and fluoride are much stronger than the ion dipole attraction between \(\ce{Li^+}\) and \(\ce{HF}\) and the ion-induced dipole attraction between \(\ce{Li^+}\) and \(\ce{Ar}\). Estimate the bond length of \(\ce{He2}\), \(\ce{Ar2}\), and \(\ce{Xe2}\) from the potential energy curves below. For each interacting pair, identify the attractive and repulsive distances. What are the intermolecular forces exist in each of the pair of molecules? Order the potential interactions in term of strength of intermolecular forces. Repulsion is when R less than R at equilibrium. Attraction is when R is bigger than R at equilibrium. Increasing order of intermolecular forces: \(\ce{He2}\) < \(\ce{Ar2}\) < \(\ce{Xe2}\). The intermolecular force that the molecules have is London Dispersion Force (Van der Waals force). Larger molecules tend to have greater polarizability because they have more electrons and their electrons are further away from the nucleus. Therefore, LDF tend to get stronger when the molecule becomes larger. Based on the graph, the higher the potential energy, the stronger the interaction is. Arrange the following substances in order of decreasing normal boiling points and explain the rationale. \[\ce{CH_3CH_2CH_2CH_2OH > CH_3CH_2CH_2CH_2CH_3 > CH_3CH(CH_3)CH_2CH_3 > CH_4 }\nonumber \] 1-Butanol has the greatest boiling point due to strong hydrogen bonding with the alcohol group. Pentane has the second strongest boiling point due to its systematical structure and large London Dispersion force. Although 2-methylbutane has the same molecular mass as pentane, its geometrical structure hinders stacking which leads to lower intermolecular force. Therefore, 2-methylbutane has the lowest boiling point. Like a normal human being, Kris enjoys breathing. But unfortunately for Kris, she is clinically paranoid, and feels as though she's breathing in a little too much ethanol. To calm Kris down, you say that as a vapor, ethanol exists, to an extent, as a dimer, (\(\ce{(CH3CH2OH)2}\)), in which two \(\ce{CH3CH2OH}\) molecules are held together by hydrogen bonds. Propose and draw a reasonable structure for this dimer to help Kris deal with her problems. Hydrogen bonds are a type of intermolecular forces. It is the bond between a hydrogen atom and a high electronegative atom such as N, O, and F. The only hydrogen bonding which can take place between the ethanol molecules is between one's oxygen and the other's hydrogen bonded onto the other's oxygen. So any two membered structure of two ethanol molecukes with this criteria indicated is acceptable. The diagram above shows a reasonable structure of a acetic acid dimer, where the dash lines represent hydrogen bonds. Hypochlorous acid (\(\ce{HOCl}\)) is a similar compound to \(\ce{HOF}\), which is the simplest possible compound that allows comparison between fluoride and oxygen in their abilities to form hydrogen bonds. Although \(\ce{Cl}\) attracts electrons more strongly than \(\ce{O}\), solid \(\ce{HOCl}\) cannot form \(\ce{H-Cl}\) bonds. Draw the proposed structure for chains of \(\ce{HOCl}\) molecules in the crystalline state. The bond angle for \(\ce{HOCl}\) is \(103^o\). How do the boiling points of hydrogen halides differ from that of hydrogen fluoride (\(\ce{HF}\)). Explain your reasoning. Answer: \(\ce{HF}\) has a much higher boiling point (\(20^oC\)) compared to the other hydrogen halides. This is because \(\ce{HF}\) is capable of forming hydrogen bonding within its compounds. Fluorine is capable of doing hydrogen bonding as opposed to the other halogens in the periodic table. Since fluorine is the element with the highest electronegativity, the hydrogen-fluorine bond in hydrogen fluoride is highly polarized, creating a partial positive charge (δ+) on the hydrogen and, a negative charge (δ-) on the fluorine atom. Furthermore, the lone pairs situated on the fluorine atom are in the second energy level, thus they are very close and the negative charge on the atom is very concentrated, thus it has a strong attraction force. Hydrogen bonds form between the positively charged H atoms and the lone pair of the F atom. The other halogens are not capable of hydrogen bonding because they are not as electronegative as fluorine and they are larger in size. Thus the bond within the molecule is not as polarized and the lone pairs on the halogen atom are not as concentrated. Thus, the stronger the intermolecular forcers of the liquid, the harder it is for a molecule to gain enough energy to overcome the intermolecular forces that bond it in a liquid. The more energy is required to enter the gas phase. The higher the boiling point. Fluorine has higher intermolecular forces, thus it has a higher boiling point compared to the other hydrogen halides. Oxygen is stored at a temperature of 105 K and a pressure of 3.356 atm. If the volume of the container is 2.5 L, calculate the number of moles of oxygen in the container. Compare this number of moles to the number of moles in the same container at standard pressure and 298 K. Is it fewer, more, or unchanged? This is an ideal gas law problem, so the formula \(PV=nRT\) should be used. \(P=3.356 \, \text{atm}\)   \(V=2.5 \, \text{L}\)   \(n=?\)   \(R=0.0821 \dfrac{\text{L} \ \text{atm}}{\text{mol} \ \text{K}}\)  \(T=105 K\) \[n= \dfrac{3.356 \times 2.5}{0.0821 \times 105} \, \text{mol} \nonumber \] \[n= 0.9733 \, \text{mol} \nonumber \] Standard pressure = 1 atm \[n=\dfrac{1\times 2.5}{0.0821\times 298}\nonumber \] \[n= 0.102 \, \text{mol} < 0.9733 \, \text{mol} \nonumber \] Therefore, under standard conditions, there are fewer moles of oxygen in the container. Consider the reaction at 25ºC. \[\ce{CaC2(s) + 2H2O(l) -> C2H2(g) + Ca(OH)2(s)} \nonumber \] If the total pressure is 0.9124 atm and the vapor pressure of water at this temperature is 0.0313 atm. What is the mass of acetylene (\(\ce{C2H2}\)) per liter of "wet" acetylene collected by this reaction (i.e., collected over a pool of water)? Assume all gases behave ideally. Partial pressures in gases are additive via . \[P_T-P_{\ce{H2O}}=P_{\ce{C2H2}} = 0.8811 atm \nonumber\] Using the Ideal Gas Law, moles per liter of acetylene can be found by rearranging the equation: \[\ce\dfrac{n}{V}=\dfrac{P_{\ce{C2H2}}}{RT}\nonumber \] \[\dfrac {0.8811}{0.082057 \times 298.15} =\dfrac{0.036 mol}{1 L}\nonumber \] The question asks for grams per liter, so: \[\left (\dfrac{0.036 \: \cancel{mol \; C_2H_2}}{1 L} \right ) \left (\dfrac{26.036\; g C_2H_2}{1\: \cancel{mol\; C_2H_2}} \right ) = 0.9376\; \dfrac{g}{L}\nonumber \] sea-level pressure on According to the graph, a location where the boiling temperature is 80 C correlates approximately to a fraction of 0.5 atm pressure. Use the phase diagram of water to determine whether water is a solid, a liquid, or a gas at each of the following combinations of temperature and pressure. Using the phase diagram of water, What phase does candle wax exist in? How about butter? Matter can exist in different phases depending on the temperature of the substance. For example, candle wax and butter can exist as a solid at room temperature, but when enough heat is applied to these substances, they can exist in a liquid or a gaseous phase. r is pla \[PV=nRT\nonumber \] \[(0.02168atm)(180m^{3})(\dfrac{1000L}{m^{3}})=(n)(0.08206\frac{L\ atm}{mol\ K})(292.15K)\nonumber \] \[n=162.77mols\nonumber \] \[m=162.77mols\times \dfrac{18.02g}{mol}=2930g\nonumber \] A cooling bath is a mixture that is primarily liquid which is kept at a constant, low temperature. While in the lab, Po’lah accidently severed its hand after trying to make some mac and cheese. As a result it had to immediately transfer its hand to a cooling bath to sufficiently preserve it but not so that it could be of some use later on. In order to achieve this, the cooling bath had to be kept at around \(0^{\circ}C\). The only liquids it had access to were water and methanol, both at room temperature, and the only cooling agents that it had were liquid nitrogen and dry ice (solid carbon dioxide). What combination did Po’lah use and why? This question is a bit tricky. The coolants Liquid Nitrogen and Dry ice, (\(\ce{CO2 (s)}\)), both turn into gasses at very low temperatures. After they turn into gaseous, they are no longer “useful”, since they cannot be added into the bath. Therefore, after adding sufficient amounts of either of the coolants to either of the liquids, the temperature of the bath would decrease way past the freezing point of either liquid, resulting in a cooling “block” rather than a bath. So what is the deal? Is this problem unsolvable? Fortunately, it isn’t. Think about a glass of ice water. If enough ice is put into the water, then eventually the water and ice would equilibrate at \(0^{\circ}C\) which is the equilibrium point of the water where both its solid and liquid states can exit. Unfortunately, while the ice is not available at this time, the same concept can be applied. Remember, if enough of either coolant is used, then the entire cup of water will freeze over. However, if just enough of either coolant is put in so that the cooling bath remains at \(0^{\circ}C\), a cooling bath that is somewhat solid and somewhat liquid will form, giving an adequate place for Po’lah to put its hand. Water is perfect for this because it remains a liquid at \(0^{\circ}C\), and cannot exist as such at any lower temperature. Since Ethanol freezes at \(-143.7^{\circ}C\), adding dry ice or liquid nitrogen would bring the cooling bath much lower than \(0^{\circ}C\), which would destroy Po’lah’s hand Thus, it doesn’t matter what cooling agent is used, either can bring water down to \(0^{\circ}C\). However, methane cannot be used. Add a little of either coolant to bring the water down to \(0^{\circ}C\), and it will remain at both a liquid and a solid which creates a \(0^{\circ}C\) cooling bath. Methanol cannot be used. The melting points of the chlorides of third period elements are: \(\ce{NaCl} = 1074 \,K\), \(\ce{MgCl_2} = 987\,K\), \(\ce{AlCl_3} =465\,K\), \(\ce{SiCl_4} = 204.4\, K\), \(\ce{PCl_3} = 179\, K\). Based on the applicable intermolecular forces present in each compound, explain this trend in melting points. s the chloride compounds make their way across the third period elements, the melting points gradually decrease. Both (\(\ce{NaCl}\) and \(\ce{MgCl_2}\)) are both "ionic solids" with strong ion-ion bonding and hence have the highest melting temperatures in the comparative series. We would expect the \(\ce{Na+}\) ions in (\(\ce{NaCl}\) to have weaker Coulombic interactions than the \(\ce{Mg^{2+}}\) ions in \(\ce{MgCl_2}\) since the latter are smaller and double the charge (i.e., a higher charge density). That would make the lattice energy higher because it would generate stronger ionic bonds (lattice energy of \(\ce{NaCl}\) and \(\ce{MgCl2}\) are 876 and 2526 kJ/mol). However, this ionic perspective is insufficient to explain the observed melting points with (\(\ce{NaCl} > \ce{MgCl_2}\)), which we ascribe to bonding never being 100% ionic. As expected for teh difference in electronegativity of chlorine, magnesium, and sodium, the percent ionic character of the \(\ce{Na-Cl}\) bond is 71% and for \(\ce{Mg-Cl}\) is even lower, at 58%. Hence, the \(\ce{Na-Cl}\) bond is 29% covalent and the \(\ce{Mg-Cl}\) bond is 42% covalent. The covalent character reduces the melting point of \(\ce{MgCl2}\), even though \(\ce{Mg+}\) has a higher charge-to-size ratio than \(\ce{Na+}\). \(\ce{AlCl_3}\) is a molecular framework solid which means the solid is an extended covalent molecule and not individual \(\ce{AlCl_3}\) molecules bound by intermolecular forces. Both \(\ce{SiCl_4}\) and \(\ce{PCl_4}\) solids are molecular in nature. To understand the strength of intermolecular forces as play this properly, we need to apply basic VESPR rules to identify the structures of these molecules. \(\ce{SiCl_4}\) is a tetrahedral molecule with no dipole moment (nor quadrupole moment either). \(\ce{PCl_4}\) is trigonal pyramidal and since \(\ce{P}\) has a different electronegativity than \(\ce{Cl}\), it has a dipole moment (0.97D). From this argument, we would think that \(\ce{PCl_4}\) would have a higher melting point, however both molecules have dispersive forces (London) too. Since the magnitude of dispersive forces increased wtih larger molecules (with more electrons to shift around), this is higher in \(\ce{PCl_4}\), which beats out the weak dipole-dipole interactions in \(\ce{PCl_4}\).
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How long does it take for a chemical reaction to occur under a given set of conditions? As with many "simple" questions, no meaningful answer can be given without being more precise. In this case, A reaction is "completed" when it has reached equilibrium — that is, when concentrations of the reactants and products are no longer changing. If the equilibrium constant is quite large, then the answer reduces to a simpler form: the reaction is completed when the concentration of a reactant falls to zero. In the interest of simplicity, we will assume that this is the case in the remainder of this discussion. If the reaction takes place very slowly, the time it takes for every last reactant molecule to disappear may be too long for the answer to be practical. In this case, it might make more sense to define "completed" when a reactant concentration has fallen to some arbitrary fraction of its initial value — 90%, 70%, or even only 20%. The particular fraction one selects depends on the cost of the reactants in relation to the value of the products, balanced against the cost of operating the process for a longer time or the inconvenience of waiting for more product. This kind of consideration is especially important in industrial processes in which the balances of these costs affect the profitability of the operation. Instead of trying to identify the time required for the reaction to become completed, it is far more practical to specify the the time required for the concentration of a reactant to fall to half of its initial value. This is known as the (or half-time) of the reaction. )
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Fertilizers, Formulas and Ecological Stoichiometry When you shop for fertilizers, you will find them labeled according to the "NPK" (nitrogen/phosphorus/potassium) analysis. For example, one fertilizer might be labeled "18-51-20". Deciphering this label requires understanding of chemical formulas. The nitrogen value (18) is the actual percent nitrogen in the fertilizer, but the phosphorus and potassium values are not. The phosphorus content is actually expressed as if the phosphorus compound were P O . So the "51" in the label means 51% P O . But what percent of this is actually phosphorus itself? We can determine that from the formula. Determine the percent phosphorus and the percent oxygen in P O . The formula indicates the relative amounts (in mol) of phosphorus and oxygen, so 1 mol of the compound contains 2 mol of P and 5 mol of O. The percent phosphorus is \(\text{ }\!\!%\!\!\text{ P = }\frac{m_{\text{P}}}{m_{\text{compound}}}\text{ }\times \text{ 100 }\!\!%\!\!\) 1 mol of P O has a mass of (2 mol P * 31 g/mol P + 5 mol O * (16.0 g/mol O) = 142 g, and it contains 2 mol of P, weighing 2 mol P * 31.0 g/mol = 62.0 g, so... \(\text{ }\!\!%\!\!\text{ P = }\frac{m_{\text{P}}}{m_{\text{P}_{\text{2}}\text{O}_{\text{5}}}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{62}\text{.0 g}}{\text{142}\text{.0 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 43}\text{.7 }\!\!%\!\!\text{ }\) Similarly, %O = 100% x (5 mol O * 16 g/mol) / (2 mol P * 31 g/mol P + 5 mol O * (16.0 g/mol O) % O = 56.3% If the fertilizer contains 51% P O , it contains only 43.7% of that as pure phosphorus, or 0.437 * 51 = 22.3% P. How does an analytical laboratory determine which phosphorus compound is present? First, the compound must be separated from the others and purified, possibly by dissolving it in a solvent that does not dissolve the others, then evaporating the solvent until the compound "recrystallizes". Then techniques of are used to determine the percentage by mass of each element in the compound. Such data are usually reported as the . When 10.0 g of phosphorus burns in oxygen, 22.9 g of a pure compound is formed. Calculate the percent composition from these experimental data. The percentage of phosphorus is the mass of phosphorus divided by the total mass of compound times 100 percent: \(\text{ }\!\!%\!\!\text{ P = }\frac{m_{\text{P}}}{m_{\text{compound}}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{10}\text{.0 g}}{\text{22}\text{.9 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 43}\text{.7 }\!\!%\!\!\text{ }\) The remainder of the compound (22.9 g – 10 g = 12.9 g) is oxygen: \(\text{ }\!\!%\!\!\text{ O = }\frac{m_{\text{O}}}{m_{\text{compound}}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{12}\text{.9 g}}{\text{22}\text{.9 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 56}\text{.3 }\!\!%\!\!\text{ }\) As a check, verify that the percentages add to 100: 56.3% + 43.7% = 100% The potassium (K) content of a fertilizer is actually expressed as if the potassium compound were K O. So the "20" in the label (above) means 20% K O, which is 17% oxygen and 83% elemental potassium: % K = 100% x m / m = (2 mol K * 39.1 g/mol) / (2 mol K * 39.1 g/mol + 1 mol O * 16.0 g/mol) So the true percentage of potassium in the fertilizer is 0.83 x % K2O = 0.83 x 20% = 16.6% K So 18−51−20 fertilizer actually contains (by weight) 18% elemental (N), 22% elemental (P), and 16% elemental (K). is an approach to ecology which studies how the chemical elements in a population of plants or animals matches the elements in the food supply for the population. Most plants and animals (including us) show a high degree of , that is, they have a characteristic chemical composition regardless of their environment, or food supply. They need food which supplies the elements in the proper ratios. If we want plants to grow well, we need to supply elements in the proper ratios. One claim of the practitioners of ecological stoichiometry is that fast growth requires high concentrations of RNA, which in turn requires a lot of phosphorus. This is the "Growth Rate Hypothesis" . The nitrogen to phosphorus ratio in RNA is similar to that in ATP, which is one of its constituents (along with UTP, CTP and GTP). Suppose that this N:P ratio establishes the requirements for the plant. We can then calculate the optimal N:P food ratio by calculating the N:P ratio in ATP, whose structure is:   From the formula, C H N O P ,we see that % N = 100* (5*14.0 / 507.2 ) = 13.8% %P = 100* (3*31 / 507.2) = 18.3% P So the optimal N/P ratio is 13.8/18.3 = 0.753:1 Is the 18−51−20 fertilizer optimal for ATP synthesis in the plant? The N:P ratio is 18/51 = 0.35, much less than that required. This particular fertilizer is designed for soils that are depleted in phosphorus. focuses on the chemical requirements of each trophic level, in addition to energy requirements. Proponents say that "Ecological stoichiometry recognizes that organisms themselves are outcomes of chemical reactions and thus their growth and reproduction can be constrained by supplies of key chemical elements [especially carbon (C), nitrogen (N) and phosphorus (P)]". . Organisms in each trophic level of an ecosystem must have evolved to utilize the nutrient ratios provided by the lower trophic levels. To obtain the formula from percent-composition data for a phosphorus oxide, we must find how many phosphorus atoms there are per oxygen atom. On a macroscopic scale this corresponds to the ratio of the amount of phosphorus to the amount of oxygen. If the formula is P O , it not only indicates that there are five oxygen for every two phosphorus , it also says that there are 5 of oxygen atoms for each 2 of phosphorus atoms. That is, the of oxygen is 2.5 times the of phosphorus. The numbers in the ratio of the amount of phosphorus to the amount of oxygen (2:5) are the subscripts of phosphorus and oxygen in the formula. Phosphorus forms several oxides, including phosphorus pentoxide, P O , phosphorus trioxide, P O , P O , P O , P O , PO and P O . Determine the formula for the compound whose percent phosphorus is 49.2%. For convenience, assume that we have 100 g of the compound. Of this, 49.2 g (49.2%) is phosphorus and 50.8 g is oxygen (100% - 49.2%). Each mass can be converted to an amount of substance \(\begin{align} & n_{\text{P}}=\text{49}\text{.2 g}\times \frac{\text{1 mol P}}{\text{31}\text{.0 g}}=\text{1}\text{.59 mol P} \\ & n_{\text{O}}=\text{50}\text{.8 g}\times \frac{\text{1 mol O}}{\text{16}\text{.0 g}}=\text{3}\text{.18 mol O} \\ \end{align}\) Dividing the larger amount by the smaller, we have \(\frac{n_{\text{O}}}{n_{\text{P}}}=\frac{\text{3}\text{.18 mol O}}{\text{1}\text{.59 mol P}}=\frac{\text{2}\text{.00 mol O}}{\text{1 mol P}}\) The ratio 2.0 mol O to 1 mol P also implies that there are 2.00 O atoms per 1 P atom. Therefore we write the formula as PO . This is the "Empirical Formula", and it gives the smallest ratio of atoms. The "molecular formula" could be twice as large, P O , or some other multiple of the empirical formula. In this case, the known oxide with the 1:2 ratio is PO O . Occasionally the ratio of amounts is not a whole number. Aspirin contains 60.0% C, 4.48% H, and 35.5% O. What is its empirical formula? \(\begin{align} & n_{\text{H}}=\text{14}\text{.4 g}\times \frac{\text{1 mol H}}{\text{1}\text{.008 g}}=\text{14}\text{.3 mol H} \\ & n_{\text{C}}=\text{85}\text{.6 g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{7}\text{.13 mol C} \\ & n_{\text{O}}=\text{35}\text{.5 g}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g}}=\text{2}\text{.22 mol O} \\ \end{align}\) Divide all three by the smallest amount of substance \(\begin{align} & \frac{n_{\text{C}}}{n_{\text{O}}}=\frac{\text{5}\text{.00 mol C}}{\text{2}\text{.22 mol O}}=\frac{\text{2}\text{.25 mol H}}{\text{1 mol O}} \\ & \frac{n_{\text{H}}}{n_{\text{O}}}=\frac{\text{4}\text{.44 mol H}}{\text{2}\text{.22 mol O}}=\frac{\text{2}\text{.00 mol H}}{\text{1 mol O}} \\ \end{align}\) Clearly there are twice as many H atoms as O atoms, but the ratio of C to O is not so obvious. We must convert 2.25 to a ratio of small whole numbers. This can be done by changing the figures after the decimal point to a fraction. In this case, .25 becomes ¼. Thus 2.25 = 2¼ = \(\tfrac{\text{9}}{\text{4}}\), and \(\frac{n_{\text{C}}}{n_{\text{O}}}=\frac{\text{2}\text{.25 mol C}}{\text{1 mol O}}=\frac{\text{9 mol C}}{\text{4 mol O}}\) We can also write \(\frac{n_{\text{H}}}{n_{\text{O}}}=\frac{\text{2 mol H}}{\text{1 mol O}}=\frac{\text{8 mol H}}{\text{4 mol O}}\) Thus the empirical formula is C H O . Once someone has determined a formula–empirical or molecular—it is possible for someone else to do the reverse calculation. Finding the weight-percent composition from the formula often proves quite informative, as the following example shows. In order to replenish nitrogen removed from the soil when plants are harvested, the compounds NaNO (sodium nitrate), NH NO (ammonium nitrate), and NH (ammonia) are used as fertilizers. If a farmer could buy each at the same cost per gram, which would be the best bargain? In other words, which compound contains the largest percentage of nitrogen? We will show the detailed calculation only for the case of NH NO . 1 mol NH NO contains 2 mol N, 4 mol H, and 3 mol O. The molar mass is thus = (2 × 14.01 + 4 × 1.008 + 3 × 16.00) g mol = 80.05 g mol A 1-mol sample would weigh 80.05 g. The mass of 2 mol N it contains is \(m_{\text{N}}\text{ = 2 mol N }\times \text{ }\frac{\text{14}\text{.01 g}}{\text{1 mol N}}\text{ = 28}\text{.02 g}\) Therefore the percentage of N is \(\text{ }\!\!%\!\!\text{ N = }\frac{m_{\text{N}}}{m_{\text{NH}_{\text{4}}\text{NO}_{\text{3}}}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{28}\text{.02 g}}{\text{80}\text{.05 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 35}\text{.00 }\!\!%\!\!\text{ }\) The percentages of H and O are easily calculated as \(\begin{align} & m_{\text{H}}\text{ = 4 mol H }\times \text{ = }\frac{\text{1}\text{.008 g}}{\text{1 mol H}}\text{ = 4}\text{.032 g} \\ & \text{ }\!\!%\!\!\text{ H = }\frac{\text{4}\text{.032 g}}{\text{80}\text{.05 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 5}\text{.04 }\!\!%\!\!\text{ } \\ & m_{\text{O}}\text{ = 3 mol O }\times \text{ = }\frac{\text{16}\text{.00 g}}{\text{1 mol O}}\text{ = 48}\text{.00 g} \\ & \text{ }\!\!%\!\!\text{ O = }\frac{\text{48}\text{.00 g}}{\text{80}\text{.05 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 59}\text{.96 }\!\!%\!\!\text{ } \\ \end{align}\) Though not strictly needed to answer the problem, the latter two percentages provide a check of the results. The total 35.00 + 5.04% + 59.96% = 100.00% as it should. Similar calculations for NaNO and NH yield 16.48% and 82.24% nitrogen, respectively. The farmer who knows chemistry chooses ammonia!
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One of the most interesting and basic problems connected with the synthesis of proteins in living cells is how the component amino acids are induced to link together in the sequences that are specific for each type of protein. There also is the related problem of how the information as to the amino-acid sequences is perpetuated in each new generation of cells. We now know that the substances responsible for genetic control in plants and animals are present in and originate from the chromosomes of cell nuclei. Chemical analysis of the chromosomes has revealed them to be composed of giant molecules of deoxyribonucleoproteins, which are deoxyribonucleic acids (DNA) bonded to proteins. Since it is known that DNA rather than the protein component of a nucleoprotein contains the genetic information for the biosynthesis of enzymes and other proteins, we shall be interested mainly in DNA and will first discuss its structure. Part or perhaps all of a particular DNA is the chemical equivalent of the Mendelian gene - the unit of inheritance. The role of DNA in living cells is analogous to that of a punched tape used for controlling the operation of an automatic turret lathe - DNA supplies the information for the development of the cells, including synthesis of the necessary enzymes and such replicas of itself as are required for reproduction by cell division. Obviously, we would not expect the DNA of one kind of organism to be the same as DNA of another kind of organism. It is therefore impossible to be very specific about the structure of DNA without being specific about the organism from which it is derived. Nonetheless, the basic structural features of DNA are the same for many kinds of cells, and we mainly shall be concerned with these basic features in the following discussion. In the first place, DNA molecules are quite large, sufficiently so to permit them to be seen individually in photographs taken with electron microscopes. The molecular weights vary considerably, but values of 1,000,000 to 4,000,000,000 are typical. X-ray diffraction indicates that DNA is made up of two long-chain molecules twisted around each other to form a double-stranded helix about \(20 \: \text{Å}\) in diameter. The arrangement is shown schematically in \(12\): As we shall see, the components of the chains are such that the strands can be held together efficiently by hydrogen bonds. In agreement with this structure, it has been found that, when DNA is heated to about \(80^\text{o}\) under proper conditions, the strands of the helix unwind and dissociate into two randomly coiled fragments. Furthermore, when the dissociated material is allowed to cool slowly under the proper conditions, the fragments recombine and regenerate the helical structure. Chemical studies show that the strands of DNA have the structure of a long-chain polymer made of alternating phosphate and sugar residues carrying nitrogen bases, \(13\): The sugar is \(D\)-2-deoxyribofuranose, \(14\), and each sugar residue is bonded to two phosphate groups by way of ester links involving the 3- and 5-hydroxyl groups: The backbone of DNA is thus a Each of the sugar residues of DNA is bonded at the 1-position of one of four bases: cytosine, \(15\); thymine, \(16\); adenine, \(17\); and guanine, \(18\). The four bases are derivatives of either or , both of which are heterocyclic nitrogen bases: Unlike phenols ( ), structural analysis of many of the hydroxy-substituted aza-aromatic compounds is complicated by isomerism of the keto-enol type, sometimes called isomerism. For 2-hydroxypyrimidine, \(19\), these isomers are \(19a\) and \(19b\), and the lactam form is more stable, as also is true for cytosine, \(15\), thymine, \(16\), and the pyrimidine ring of guanine, \(18\). For the sake of simplicity in illustrating \(\ce{N}\)-glycoside formation in DNA, we shall show the type of bonding involved for the sugar and base components only (i.e., the deoxyribose nucleoside structure). Attachment of 2-deoxyribose is through a \(\ce{NH}\) group to form the \(\beta\)-\(\ce{N}\)-deoxyribofuranoside ( ): Esterification of the 5'-hydroxyl group of deoxyribose , such as cytosine deoxyriboside, with phosphoric acid gives the corresponding :\(^{11}\) The number of nucleotide units in a DNA chain varies from about 3,000 to 10,000,000 Although the sequence of the purine and pyrimidine bases in the chains are not known, there is a striking equivalence between the numbers of certain of the bases regardless of the origin of DNA. Thus the number of adenine (A) groups equals the number of thymine (T) groups, and the number of guanine (G) groups equals the number of cytosine (C) groups: A = T and G = C. The bases of DNA therefore are half purines and half pyrimidines. Furthermore, although the ratios of A to G and T to C are constant for a given species, they vary widely from one species to another. The equivalence between the purine and pyrimidine bases in DNA was accounted for by J. D. Watson and F. Crick (1953) through the suggestion that the two strands are constructed so that, when twisted together in the helical structure, hydrogen bonds are formed involving adenine in one chain and thymine in the other, or cytosine in one chain and guanine in the other. Thus each adenine occurs paired with a thymine and each cytosine with a guanine and the strands are said to have complementary structures. The postulated hydrogen bonds are shown in Figure 25-22, and the relationship of the bases to the strands in Figure 25-23. It is now well established that DNA provides the genetic recipe that determines how cells reproduce. In the process of cell division, the DNA itself also is reproduced and thus perpetuates the information necessary to regulate the synthesis of specific enzymes and other proteins of the cell structure. In replicating itself prior to cell division, the DNA double helix evidently separates at least partly into two strands (see Figure 25-24). Each of the separated parts serves as a guide (template) for the assembly of a complementary sequence of nucleotides along its length. Ultimately, new DNA double strands are formed, each of which contains one strand from the parent DNA. The genetic information inherent in DNA depends on the arrangement of the bases (A, T, G, and C) along the phosphate-carbohydrate backbone - that is, on the arrangement of the four nucleotides specific to DNA. Thus the sequence A-G-C at a particular point conveys a different message than the sequence G-A-C. It is quite certain that the code involves a particular sequence of nucleotides for each amino acid. Thus the sequence A-A-A codes for lysine, and U-C-G codes for serine. The sequences or for all twenty amino acids are known. It is clear that DNA does not play a direct role in the synthesis of proteins and enzymes because most of the protein synthesis takes place outside of the cell nucleus in the cellular cytoplasm, which does not contain DNA. Furthermore, it has been shown that protein synthesis can occur in the absence of a cell nucleus or, equally, in the absence of DNA. Therefore the genetic code in DNA must be passed on selectively to other substances that carry information from the nucleus to the sites of protein synthesis in the cytoplasm. These other substances are (RNA), which are polymeric molecules similar in structure to DNA, except that \(D\)-2-deoxyribofuranose is replaced by \(D\)-ribofuranose and the base thymine is replace by uracil, as shown in Figure 25-25. RNA also differs from DNA in that there are not the same regularities in the overall composition of its bases and it usually consists of a single polynucleotide chain. There are different types of RNA, which fulfill different functions. About \(80\%\) of the RNA in a cell is located in the cytoplasm in clusters closely associated with proteins. These ribonucleoprotein particles specifically are called ribosomes, and the ribosomes are the sites of most of the protein synthesis in the cell. In addition to the (rRNA), there are ribonucleic acids called (mRNA), which convey instructions as to what protein to make. In addition, there are ribonucleic acids called (tRNA), which actually guide the amino acids into place in protein synthesis. Much is now known about the structure and function of tRNA. The principal structural features of tRNA molecules are shown schematically in Figure 25-26. Some of the important characteristics of tRNA molecules are summarized as follows. 1. There is at least one particular tRNA for each amino acid. 2. The tRNA molecules have single chains with 73-93 ribonucleotides. Most of the tRNA bases are adenine (A), cytosine (C), guanine (G), and uracil (U). There also are a number of unusual bases that are methylated derivatives of A, C, G, and U. 3. The clover-leaf pattern of Figure 25-26 shows the general structure of tRNA. There are regions of the chain where the bases are complementary to one another, which causes it to fold into two double-helical regions. The chain has three bends or separating the helical regions. 4. The 5'-terminal residue usually is a guanine nucleotide; it is phosphorylated at the 5'-\(\ce{OH}\). The terminus at the 3' end has the same sequence of three nucleotides in all tRNA's, namely, CCA. The 3'-\(\ce{OH}\) of the adenosine in this grouping is the point of attachment of the tRNA to its specific amino acid: With this information on the structure of tRNA, we can proceed to a discussion of the essential features of biochemical protein synthesis. The information that determines amino-acid sequence in a protein to be synthesized is contained in the DNA of a cell nucleus as a particular sequence of nucleotides derived from adenine, guanine, thymine, and cytosine. For each particular amino acid there is a sequence of nucleotides called a . The information on protein structure is transmitted from the DNA in the cell nucleus to the cytoplasm where the protein is assembled by messenger RNA. This messenger RNA, or at least part of it, is assembled in the nucleus with a base sequence that is complementary to the base sequence in the parent DNA. The assembly mechanism is similar to DNA replication except that thymine (T) is replaced by uracil (U). The uracil is complementary to adenine in the DNA chain (see Figure 25-27). After the mRNA is assembled, it is transported to the cytoplasm where it becomes attached to the ribosomes. The amino acids in the cytoplasm will not form polypeptides unless activated by ester formation with appropriate tRNA molecules. The ester linkages are through the 3'-\(\ce{OH}\) of the terminal adenosine nucleotide (Equation 25-9) and are formed only under the influence of a synthetase enzyme that is specific for the particular amino acid. The energy for ester formation comes from ATP hydrolysis ( and ). The product is called an . The aminoacyl-tRNA's form polypeptide chains in the order specified by codons of the mRNA bound to the ribosomes (see Figure 25-28). The order of incorporation of the amino acids depends on the recognition of a codon in mRNA by the corresponding anticodon in tRNA by a complementary base-pairing of the type A \(\cdots\) U and C \(\cdots\) G. The first two bases of the codon recognize only their complementary bases in the anticodon, but there is some flexibility in the identity of the third base. Thus phenylalanine tRNA has the anticodon A-A-G and responds to the codons U-U-C and U-U-U, but not U-U-A or U-U-G: The codons of the mRNA on the ribosomes are read from the 5' to the 3' end. Thus the synthetic polynucleotide (5')A-A-A-(A-A-A)\(_n\)-A-A-C(3') contains the code for lysine (A-A-A) and asparagine (A-A-C); the actual polypeptide obtained using this mRNA in a cell-free system was Lys-(Lys)\(_n\)-Asn, and not Asn-(Lys)\(_n\)-Lys. The start of protein synthesis is signaled by specific codon-anticodon interactions. Termination is also signaled by a codon in the mRNA, although the stop signal is not recognized by tRNA, but by proteins that then trigger the hydrolysis of the completed polypeptide chain from the tRNA. Just how the secondary and tertiary structures of the proteins are achieved is not yet clear, but certainly the mechanism of protein synthesis, which we have outlined here, requires little modification to account for preferential formation of particular conformations. \(^{11}\)The positions on the sugar ring are primed to differentiate them from the positions of the nitrogen base. and (1977)
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The best energies obtained at the Hartree-Fock level are still not accurate, because they use an average potential for the electron-electron interactions. Configuration interaction (CI) methods help to overcome this limitation. Because electrons interact and repel each other, their motion in atoms is correlated. When one electron is close to the nucleus, the other tends to be far away. When one is on one side, the other tends to be on the other side. This motion is related to that of two people playing tag around a house. As we said before, the exact wavefunction must depend upon the coordinates of both electrons simultaneously. We have shown that it is a reasonable approximation in calculating energies to neglect this correlation and use wavefunctions that only depend upon the coordinates of one electron, which assumes the electrons move independently. This "orbital approximation" is similar to playing tag without keeping track of the other person. This independent-electron approximation gives reasonable, even good values, for the energy, and correlation can be taken into account to improve this description even more. The method for taking correlation into account is called . In describing electrons in atoms, it is not necessary to be restricted to only a single orbital configuration given by a . We developed the Slater determinant as a way to create correctly antisymmetrized product wavefunctions that approximate the exact multi-electron function for an atom. By using more than one configuration and putting electrons in different orbitals, spatial correlations in the electron motion can be taken into account. This procedure is called Configuration Interaction (CI). For example, for the two-electron Slater determinant wavefunction of helium, we could write \[\psi (r_1, r_2) = \underbrace{c_1Det | \varphi _{1s} (r_1) \varphi _{1s} (r_2) |}_{\text{ground state: }1s^2} + \underbrace{c_2 Det | \varphi _{1s} (r_1) \varphi _{2s} (r_2) | }_{\text{excited state: }1s^12s^1} \label{9-66}\] where \(c_1\) and \(c_2) are coefficients (that can be varied in variational method). This wavefunction adds the excited (higher energy) configuration 1s 2s to the ground (lowest energy) configuration 1s . The lowest energy configuration corresponds to both electrons being in the same region of space at the same time; the higher energy configuration represents one electron being close to the nucleus and the other electron being further away. This makes sense because the electrons repel each other. When one electron is in one region of space, the other electron will be in another region. Configuration interaction is a way to account for this correlation. Write a paragraph without using any equations that describes the essential features of configuration interaction.
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The 3-dimensional structure of \(\ce{BF3}\) is different from \(\ce{PF3}\), and this is difficult to comphrend by considering their formulas alone. However, the Lewis dot structures for them are different, and the electron pair in \(\textrm{:PF}_3\) is the reason for its structure being different from \(\ce{BF3}\) (no lone pair). Three-dimensional arrangements of atoms or bonds in molecules are important properties as are bond lengths, bond angles and bond energies. The Lewis dot symbols led us to the non-bonding electron pairs, whose role in determining the shape of a molecule was examined by N.V. Sidgwick and H.E. Powell in 1940, and later by R.S. Nyholm and R.J. Gillespie. They have developed an extensive rationale called valence-shell electron-pair repulsion (VSEPR) model of molecular geometry. The Valence-Shell Electron-Pair Repulsion (VSEPR) models consider the unshared pairs (or lone electron pairs) and the bonding electrons. These considerations of lone and bonding electron pairs give an excellent explanation about the molecular shapes. The VSEPR model counts both bonding and nonbonding (lone) electron pairs, and calls the total number of pairs the ( ). If the element A has atoms bonded to it and nonbonding pairs, then \[SN = m + n\] is useful for predicting shapes of molecules. If \(\ce{X}\) is any atom bonded to \(\ce{A}\) (in single, double, or triple bond), a molecule may be represented by \(\mathrm{AX_mE_n}\) where \(\ce{E}\) denotes a lone electron pair. This formula enables us to predict its geometry. The common , descriptor, and examples are given in the table on the right. Note that the is also called the or . The VSEPR model has another general rule: The geometry of the molecules with their SNs equal to 2 to 6 are given in the Table 1. The first line for each is the shape including the lone electron pair(s). If the lone electron pairs are ignored, the geometry of the molecule is given by another descriptor. To get an idea about the shapes of molecules and ions, three dimensional models are the best to use. However, good computer graphics sometimes also illustrate very well. The link VSEPR Illustration: View and manipulate molecular models gives excellent graphics, and you may enjoy seeing some of the graphics of the molecules. Hint: Carbon is the central atom. Identifying the central atom is a good start to drawing a structure: Hint: Sulfur has 4 valence electrons. Give the number of valence electrons of any main-group element. Hint: There is no lone electron pair. Draw a Lewis dot structure, and express it in the form \(\ce{CCl2=O}\). This carbonyl chloride is also called phosgene, a colourless, chemically reactive, highly toxic gas having an odour like that of musty hay. It was used during World War I against troops. Hint: Sulfur is the central atom. Identifying the central atom is a good start to drawing a structure: Hint: Sulfur has 6 valence electrons. Give the number of valence electrons of any main-group element. Hint: There is one lone electron pair. Draw a Lewis dot structure, and express in the form \(\textrm{:SCl}_2\ce{=O}\) or \(\ce{SCl2=OE}\). Hint: There are two lone electron pairs. The structure is Hint: There is no lone electron pair for carbon dioxide. From the structure of \(\ce{O=C=O}\), figure out the number of lone electron pairs. \(\ce{CO2}\) is linear; so is \(\ce{H-CN}\). Hint: The molecule has a T shape. Predict the shape of a molecule \(\textrm{::ClF}_3\) or \(\ce{ClF3E2}\). Hint: The angle for ideal tetrahedral molecule is 109.5 degrees. Calculate bond angles from the geometry. Hint: The geometrical shape is octahedral. Apply terms linear, T shape, seesaw, trigonal planar, trigonal bipyramidal, tetrahedral, octahedral etc. to describe the geometric shapes of molecules.
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Bromine is a reddish-brown fuming liquid at room temperature with a very disagreeable chlorine-like smell. In fact its name is derived from the Greek bromos or "stench". It was first isolated in pure form by Balard in 1826. It is the only non-metal that is a liquid at normal room conditions. Bromine on the skin causes painful burns that heal very slowly. It is an element to be treated with the utmost respect in the laboratory. Most bromine is produced by displacement from ordinary sea water. Chlorine (which is more active) is generally used to dislodge the bromine from various compounds in the water. Before leaded gasolines were removed from the market, bromine was used in an additive to help prevent engine "knocking". Production now is chiefly devoted to dyes, disinfectants and photographic chemicals.
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The presence of a phenyl group in a compound can be ascertained with a fair degree of certainty from its infrared spectrum. For example, in Figure 22-1 we see the infrared spectra of methylbenzene, and of 1,2-, 1,3-, and 1,4-dimethylbenzene. That each spectrum is of a benzene derivative is apparent from certain common features. The two bands near \(1600 \: \text{cm}^{-1}\) and \(1500 \: \text{cm}^{-1}\), although of variable intensity, have been correlated with the stretching vibrations of the carbon-carbon bonds of the aromatic ring; also, the sharp bands near \(3030 \: \text{cm}^{-1}\) are characteristic of aromatic \(\ce{C-H}\) bonds. Other bands in the spectra, especially those between \(1650 \: \text{cm}^{-1}\) and \(2000 \: \text{cm}^{-1}\), between \(1225 \: \text{cm}^{-1}\) and \(950 \: \text{cm}^{-1}\), and below \(900 \: \text{cm}^{-1}\), have been correlated with the number and position of ring substituents. Although we shall not document all these various bands in detail, each of the spectra in Figure 22-1 is marked to show some of the correlations that have been made. Compared to straight-chain conjugated polyenes, aromatic compounds have relatively complex absorption spectra with several bands in the ultraviolet region. Benzene and the alkylbenzenes show two bands in which we shall be primarily interested, one near \(200 \: \text{nm}\) and the other near \(260 \: \text{nm}\). The \(200\)-\(\text{nm}\) band is of fairly high intensity and corresponds to excitation of a \(\pi\) electron of the conjugated system to a \(\pi^*\) orbital (i.e., a \(\pi \rightarrow \pi^8*\) transition). The excited state has significant contributions from dipolar structures such as \(1\): This is analogous to the absorption bands of conjugated dienes ( ) except that the wavelength of absorption of benzenes is shorter. In fact, the \(200\)-\(\text{nm}\) absorptions of benzene and the alkylbenzenes are just beyond the range of most commercial quartz spectrometers. However, these absorptions (which we say arise from the benzene \(^2\)) are intensified and shifted to longer wavelengths when the conjugated system is extended by replacement of the ring hydrogens by unsaturated groups (e.g., \(\ce{-CH=CH_2}\), \(\ce{-C \equiv CH}\), \(\ce{-CH=O}\), and \(\ce{-C \equiv N}\); see Table 22-2). The delocalized \(\pi\)-electron system of the absorbing chromophore now includes the electrons of the unsaturated substituent as well as those of the ring. In the specific case of ethenylbenzene the excited state is a hybrid structure composite of \(2a\) and \(2b\) and other related dipolar structures: Similar effects are observed for benzene derivatives in which the substituent has unshared electron pairs that can be conjugated with the benzene ring (e.g., \(\ce{-NH_2}\), \(\ce{-OH}\), \(\ce{-Cl}\)). An unshared electron pair is to some extent delocalized to become a part of the aromatic \(\pi\)-electron system in both the ground and excited states, but more importantly in the excited state. This is illustrated for benzenamine (aniline) by the following structures, which contribute to the hybrid structure: The data of Table 22-3 show the effect on the benzene chromophore of this type of substituent - the substituent often being called an .\(^2\) This term means that, although the substituent itself is not responsible for the absorption band, it shifts the absorption of the chromophoric group, in this case the benzene ring, toward wavelengths. The auxochromic groups usually increase the intensity of the absorption also. The benzenoid band corresponds to a low-energy \(\pi \rightarrow \pi^*\) transition of the benzene molecules. The absorption intensity is weak because the \(\pi^*\) state involved has the same electronic symmetry as the ground state of benzene, and transitions between symmetrical states usually are forbidden. The transitions are observed in this case only because the vibrations of the ring cause it to be slightly distorted at given instants. In the valence-bond treatment this excited state of benzene is an antibonding state of the \(\pi\) electrons. The electronic spectra of polynuclear aromatic hydrocarbons such as naphthalene and anthracene, in which aromatic rings are fused together in a linear manner, resemble the spectrum of benzene except that the bands are shifted to longer wavelengths. In fact, with the four linearly connected rings of naphthacene, the benzenoid band is shifted far enough to longer wavelengths to be in the visible region of the spectrum (see Table 22-4). Accordingly, naphthacene is yellow. The next higher member, pentacene, is blue. Compounds such as phenanthrene, chrysene, and pyrene, in which the aromatic rings are fused in an angular manner, have complex electronic spectra with considerable fine structure. The \(\lambda_\text{max}\) values normally are at shorter wavelengths than those of their linear isomers. The chemical shifts of arene protons (\(6.5 \: \text{ppm}\) to \(8.0 \: \text{ppm}\)) characteristically are toward lower magnetic fields than those of protons attached to ordinary double bonds (\(4.6 \: \text{ppm}\) to \(6.9 \: \text{ppm}\)). The difference of about \(2 \: \text{ppm}\) cannot be easily explained because the hydrogens in both types of systems are bonded to carbon through \(sp^2\)-\(\sigma\) bonds ( and ). At least part of the chemical-shift difference between arene protons and alkene protons is the result of the special property of \(\pi\) electrons in aromatic systems of circulating freely above and below the plane of the carbon nuclei, as shown in Figure 22-4. When a molecule such as benzene is subjected to a magnetic field that has a component to the plane of the ring, the electrons circulate around the ring in such a way as to produce a local magnetic dipole in the direction to the applied field. This effect acts to reduce the applied field in the center of the ring. Therefore, if a proton could be located in the center of the ring, the applied field would have to be higher than normal to counteract the local diamagnetic field and bring the proton into resonance. A proton outside the ring is affected in the opposite way ( effect) because, as can be seen from the diagram, such protons are located in the return path of the lines of force associated with the local field and thus are in a field greater than that arising from the external magnet alone. When the plane of the molecule is oriented parallel to the field, the diamagnetic circulation is cut off. As a result, as the molecules tumble over and over in the liquid the component of magnetization perpendicular to the plane of the ring varies rapidly. Nonetheless, a substantial paramagnetic effect is experienced by the ring hydrogens. The resonance line positions therefore are shifted to lower magnetic fields. Strong evidence in confirmation of the above explanation of the chemical shifts of aromatic hydrogens is provided by a study of the cyclic conjugated polyene [18]annulene, which has hydrogens both "inside" and "outside" the ring: The hydrogens are strongly deshielded, coming at \(1.9 \: \text{ppm}\) from tetramethylsilane, while the hydrogens are deshielded and come at \(8.8 \: \text{ppm}\) from TMS. As we shall see, the ring current effect is quite general and constitutes a widely used test for aromatic character in conjugated polyene ring systems. In general, the spin-spin splittings observed between the five protons of a phenyl group can be extremely complex. An example is afforded by nitrobenzene (Figure 22-5), which has different chemical shifts for its ortho, meta, and para hydrogens and six different spin-spin interaction constants: \(J_{23}\), \(J_{24}\), \(J_{25}\), \(J_{26}\), \(J_{34}\), \(J_{35}\), (the subscripts correspond to position numbers of the protons): Such a spectrum is much too complex to be analyzed by any simple procedure. Nonetheless, nuclear magnetic resonance can be useful in assigning structures to aromatic derivatives, particularly in conjunction with integrated line intensities and approximate values of the coupling constants between the ring hydrogens, as shown below: \(^2\)A chromophore is a grouping of atoms in an organic molecule that gives rise to color, or has the potential of doing so when other groups called are present (also see ). and (1977)
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Francium is the last of the known alkali metals and does not occur to any significant extent in nature. All known isotopes are radioactive and have short half-lives (22 minutes is the longest). The existence of Francium was predicted by Dmitri Mendeleev in the 1870's and he presumed it would have chemical and physical properties similar to cesium. That may well be, but not enough francium has been isolated to test. Numerous historical claims to the discovery of element 87 were made resulting in the names russium, virginium, and moldavium. However, the confirmed discovery is credited to Marguerite Perey who was an assistant to Marie Curie at the Radium Institute in Paris. She named the element after her native country.
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The solubility of many compounds depends strongly on the pH of the solution. For example, the anion in many sparingly soluble salts is the conjugate base of a weak acid that may become protonated in solution. In addition, the solubility of simple binary compounds such as oxides and sulfides, both strong bases, is often dependent on pH. In this section, we discuss the relationship between the solubility of these classes of compounds and pH. We begin our discussion by examining the effect of pH on the solubility of a representative salt, \(\ce{M^{+}A^{−}}\), where \(\ce{A^{−}}\) is the conjugate base of the weak acid \(\ce{HA}\). When the salt dissolves in water, the following reaction occurs: \[\ce{MA (s) \rightleftharpoons M^{+} (aq) + A^{-} (aq)} \label{17.13a} \nonumber\] with \[K_{sp} = [\ce{M^{+}},\ce{A^{−}}] \label{17.13b} \nonumber\] The anion can also react with water in a hydrolysis reaction: \[\ce{A^{-} (aq) + H2O (l) \rightleftharpoons OH^{-} (aq) + HA (aq)} \label{17.14}\] Because of the reaction described in , the predicted solubility of a sparingly soluble salt that has a basic anion such as S , PO , or CO is increased. If instead a strong acid is added to the solution, the added H will react essentially completely with A to form HA. This reaction decreases [A ], which decreases the magnitude of the ion product \[Q = [\ce{M^{+}},\ce{A^{-}}]\] According to , more MA will dissolve until \(Q = K_{sp}\). Hence an acidic pH dramatically increases the solubility of virtually all sparingly soluble salts whose anion is the conjugate base of a weak acid. In contrast, pH has little to no effect on the solubility of salts whose anion is the conjugate base of a stronger weak acid or a strong acid, respectively (e.g., chlorides, bromides, iodides, and sulfates). For example, the hydroxide salt Mg(OH) is relatively insoluble in water: \[Mg(OH)_{2(s)} \rightleftharpoons Mg^{2+} (aq) + 2OH^− (aq) \label{17.15a}\] with \[K_{sp} = 5.61 \times 10^{−12} \label{17.15b}\] When acid is added to a saturated solution that contains excess solid Mg(OH) , the following reaction occurs, removing OH from solution: \[H^+ (aq) + OH^− (aq) \rightarrow H_2O (l) \label{17.16}\] The overall equation for the reaction of Mg(OH) with acid is thus \[Mg(OH)_{2(s)} + 2H^+ (aq) \rightleftharpoons Mg^{2+} (aq) + 2H_2O (l) \label{17.17}\] As more acid is added to a suspension of Mg(OH) , the equilibrium shown in is driven to the right, so more Mg(OH) dissolves. Such pH-dependent solubility is not restricted to salts that contain anions derived from water. For example, CaF is a sparingly soluble salt: \[CaF_{2(s)} \rightleftharpoons Ca^{2+} (aq) + 2F^− (aq) \label{17.18a}\] with \[K_{sp} = 3.45 \times 10^{−11} \label{17.18b}\] When strong acid is added to a saturated solution of CaF , the following reaction occurs: \[H^+ (aq) + F^− (aq) \rightleftharpoons HF (aq) \label{17.19}\] Because the forward reaction decreases the fluoride ion concentration, more CaF dissolves to relieve the stress on the system. The net reaction of CaF with strong acid is thus \[CaF_{2(s)} + 2H^+ (aq) \rightarrow Ca^{2+} (aq) + 2HF (aq) \label{17.20}\] Example \(\Page {1}\) shows how to calculate the solubility effect of adding a strong acid to a solution of a sparingly soluble salt. Sparingly soluble salts derived from weak acids tend to be more soluble in an acidic solution. Lead oxalate (PbC O ), lead iodide (PbI ), and lead sulfate (PbSO ) are all rather insoluble, with values of 4.8 × 10 , 9.8 × 10 , and 2.53 × 10 , respectively. What effect does adding a strong acid, such as perchloric acid, have on their relative solubilities? values for three compounds relative solubilities in acid solution Write the balanced chemical equation for the dissolution of each salt. Because the strongest conjugate base will be most affected by the addition of strong acid, determine the relative solubilities from the relative basicity of the anions. The solubility Equilibria for the three salts are as follows: \[PbC_2O_{4(s)} \rightleftharpoons Pb^{2+} (aq) + C_2O^{2−}_{4(aq)} \nonumber\] \[PbI_{2(s)} \rightleftharpoons Pb^{2+} (aq) + 2I^− (aq) \nonumber\] \[PbSO_{4(s)} \rightleftharpoons Pb^{2+} (aq) + SO^{2−}_{4(aq)} \nonumber\] The addition of a strong acid will have the greatest effect on the solubility of a salt that contains the conjugate base of a weak acid as the anion. Because HI is a strong acid, we predict that adding a strong acid to a saturated solution of PbI will not greatly affect its solubility; the acid will simply dissociate to form H (aq) and the corresponding anion. In contrast, oxalate is the fully deprotonated form of oxalic acid (HO CCO H), which is a weak diprotic acid (p = 1.23 and p = 4.19). Consequently, the oxalate ion has a significant affinity for one proton and a lower affinity for a second proton. Adding a strong acid to a saturated solution of lead oxalate will result in the following reactions: \[C_2O^{2−}_{4(aq)} + H^+ (aq) \rightarrow HO_2CCO^−_{2(aq)} \nonumber\] \[HO_2CCO^−_{2(aq)} + H^+ (aq) \rightarrow HO_2CCO_2H (aq) \nonumber\] These reactions will decrease [C O ], causing more lead oxalate to dissolve to relieve the stress on the system. The p of HSO (1.99) is similar in magnitude to the p of oxalic acid, so adding a strong acid to a saturated solution of PbSO will result in the following reaction: \[ SO^{2-}_{4(aq)} + H^+ (aq) \rightleftharpoons HSO^-_{4(aq)} \nonumber\] Because HSO has a pKa of 1.99, this reaction will lie largely to the left as written. Consequently, we predict that the effect of added strong acid on the solubility of PbSO will be significantly less than for PbC O . Which of the following insoluble salts—AgCl, Ag CO , Ag PO , and/or AgBr—will be substantially more soluble in 1.0 M HNO than in pure water? Ag CO and Ag PO Solubility Products and pH: Caves and their associated pinnacles and spires of stone provide one of the most impressive examples of pH-dependent solubility Equilbria(part (a) in ). Perhaps the most familiar caves are formed from limestone, such as Carlsbad Caverns in New Mexico, Mammoth Cave in Kentucky, and Luray Caverns in Virginia. The primary reactions that are responsible for the formation of limestone caves are as follows: \[\ce{CO2(aq) + H2O (l) \rightleftharpoons H^{+} (aq) + HCO^{−}3(aq)} \label{17.21}\] \[\ce{HCO^{−}3(aq) \rightleftharpoons H^{+} (aq) + CO^{2-}3(aq)} \label{17.22}\] \[\ce{Ca^{2+} (aq) + CO^{2−}3(aq) \rightleftharpoons CaCO3(s)} \label{17.23}\] Limestone deposits that form caves consist primarily of CaCO from the remains of living creatures such as clams and corals, which used it for making structures such as shells. When a saturated solution of CaCO in CO -rich water rises toward Earth’s surface or is otherwise heated, CO gas is released as the water warms. CaCO then precipitates from the solution according to the following equation (part (b) in ): \[Ca^{2+} (aq) + 2HCO^−_{3(aq)} \rightleftharpoons CaCO_{3(s)} + CO_{2(g)} + H_2O (l) \label{17.24}\] The forward direction is the same reaction that produces the solid called scale in teapots, coffee makers, water heaters, boilers, and other places where hard water is repeatedly heated. When groundwater-containing atmospheric CO ( finds its way into microscopic cracks in the limestone deposits, CaCO dissolves in the acidic solution in the reverse direction of . The cracks gradually enlarge from 10–50 µm to 5–10 mm, a process that can take as long as 10,000 yr. Eventually, after about another 10,000 yr, a cave forms. Groundwater from the surface seeps into the cave and clings to the ceiling, where the water evaporates and causes the equilibrium in to shift to the right. A circular layer of solid CaCO is deposited, which eventually produces a long, hollow spire of limestone called a stalactite that grows down from the ceiling. Below, where the droplets land when they fall from the ceiling, a similar process causes another spire, called a stalagmite, to grow up. The same processes that carve out hollows below ground are also at work above ground, in some cases producing fantastically convoluted landscapes like that of Yunnan Province in China ( ). One of the earliest classifications of substances was based on their solubility in acidic versus basic solution, which led to the classification of oxides and hydroxides as being either basic or acidic. and hydroxides either react with water to produce a basic solution or dissolve readily in aqueous acid. or hydroxides either react with water to produce an acidic solution or are soluble in aqueous base. There is a clear correlation between the acidic or the basic character of an oxide and the position of the element combined with oxygen in the periodic table. Oxides of metallic elements are generally basic oxides, and oxides of nonmetallic elements are acidic oxides. Compare, for example, the reactions of a typical metal oxide, cesium oxide, and a typical nonmetal oxide, sulfur trioxide, with water: \[Cs_2O (s) + H_2O (l) \rightarrow 2Cs^+ (aq) + 2OH^− (aq) \label{17.25}\] \[SO_{3(g)} + H_2O (l) \rightarrow H_2SO_{4(aq)} \label{17.26}\] Cesium oxide reacts with water to produce a basic solution of cesium hydroxide, whereas sulfur trioxide reacts with water to produce a solution of sulfuric acid—very different behaviors indeed Metal oxides generally react with water to produce basic solutions, whereas nonmetal oxides produce acidic solutions. The difference in reactivity is due to the difference in bonding in the two kinds of oxides. Because of the low electronegativity of the metals at the far left in the periodic table, their oxides are best viewed as containing discrete M cations and O anions. At the other end of the spectrum are nonmetal oxides; due to their higher electronegativities, nonmetals form oxides with covalent bonds to oxygen. Because of the high electronegativity of oxygen, however, the covalent bond between oxygen and the other atom, E, is usually polarized: E –O . The atom E in these oxides acts as a Lewis acid that reacts with the oxygen atom of water to produce an oxoacid. Oxides of metals in high oxidation states also tend to be acidic oxides for the same reason: they contain covalent bonds to oxygen. An example of an acidic metal oxide is MoO , which is insoluble in both water and acid but dissolves in strong base to give solutions of the molybdate ion (MoO ): \[MoO_{3(s)} + 2OH^− (aq) \rightarrow MoO^{2−}_{4(aq)} + H_2O (l) \label{17.27}\] As shown in Figure \(\Page {3}\), there is a gradual transition from basic metal oxides to acidic nonmetal oxides as we go from the lower left to the upper right in the periodic table, with a broad diagonal band of oxides of intermediate character separating the two extremes. Many of the oxides of the elements in this diagonal region of the periodic table are soluble in both acidic and basic solutions; consequently, they are called (from the Greek ampho, meaning “both,” as in amphiprotic). Amphoteric oxides either dissolve in acid to produce water or dissolve in base to produce a soluble complex. As shown in for example, mixing the amphoteric oxide Cr(OH) (also written as Cr O •3H O) with water gives a muddy, purple-brown suspension. Adding acid causes the Cr(OH) to dissolve to give a bright violet solution of Cr (aq), which contains the [Cr(H O) ] ion, whereas adding strong base gives a green solution of the [Cr(OH) ] ion. The chemical equations for the reactions are as follows: \[\mathrm{Cr(OH)_3(s)}+\mathrm{3H^+(aq)}\rightarrow\underset{\textrm{violet}}{\mathrm{Cr^{3+}(aq)}}+\mathrm{3H_2O(l)} \label{17.28}\] Aluminum hydroxide, written as either Al(OH) or Al O •3H O, is amphoteric. Write chemical equations to describe the dissolution of aluminum hydroxide in (a) acid and (b) base. amphoteric compound dissolution reactions in acid and base Using and \(\ref{17.29}\) as a guide, write the dissolution reactions in acid and base solutions. \[Al(OH)_{3(s)} + 3H^+ (aq) \rightarrow Al^{3+} (aq) + 3H_2O (l) \nonumber\] In aqueous solution, Al forms the complex ion [Al(H O) ] . \[Al(OH)_{3(s)} + OH^− (aq) \rightarrow [Al(OH)_4]^− (aq) \nonumber\] Copper(II) hydroxide, written as either Cu(OH) or CuO•H O, is . Write chemical equations that describe the dissolution of cupric hydroxide both in an acid and in a base. \[Cu(OH)_{2(s)} + 2H^+ (aq) \rightarrow Cu^{2+} (aq) + 2H_2O (l) \nonumber\] \[Cu(OH)_{2(s)} + 2OH^− (aq) \rightarrow [Cu(OH)_4]^2_{−(aq)} \nonumber\] Many dissolved metal ions can be separated by the selective precipitation of the cations from solution under specific conditions. In this technique, pH is often used to control the concentration of the anion in solution, which controls which cations precipitate. The concentration of anions in solution can often be controlled by adjusting the pH, thereby allowing the selective precipitation of cations. Suppose, for example, we have a solution that contains 1.0 mM Zn and 1.0 mM Cd and want to separate the two metals by selective precipitation as the insoluble sulfide salts, ZnS and CdS. The relevant solubility equilbria can be written as follows: \[ZnS (s) \rightleftharpoons Zn^{2+} (aq) + S^{2−} (aq) \label{17.30a}\] with \[K_{sp}= 1.6 \times 10^{−24} \label{17.30b}\] and \[CdS (s) \rightleftharpoons Cd^{2+} (aq) + S^{2−} (aq) \label{17.31a}\] with \[K_{sp} = 8.0 \times 10^{−27} \label{17.31b}\] Because the S ion is quite basic and reacts extensively with water to give HS and OH , the solubility equilbria are more accurately written as \(MS (s) \rightleftharpoons M^{2+} (aq) + HS^− (aq) + OH^−\) rather than \(MS (s) \rightleftharpoons M^{2+} (aq) + S^{2−} (aq) \). Here we use the simpler form involving S , which is justified because we take the reaction of S with water into account later in the solution, arriving at the same answer using either equilibrium equation. The sulfide concentrations needed to cause \(ZnS\) and \(CdS\) to precipitate are as follows: \[K_{sp} = [Zn^{2+},S^{2−}] \label{17.32a}\] \[1.6 \times 10^{−24} = (0.0010\; M)[S^{2−}]\label{17.32b}\] \[1.6 \times 10^{−21}\; M = [S^{2−}]\label{17.32c}\] and \[K_{sp} = [Cd^{2+},S^{2−}] \label{17.33a}\] \[8.0 \times 10^{−27} = (0.0010\; M)[S^{2−}]\label{17.33b}\] \[8.0 \times 10^{−24}\; M = [S^{2−}] \label{17.33c}\] Thus sulfide concentrations between 1.6 × 10 M and 8.0 × 10 M will precipitate CdS from solution but not ZnS. How do we obtain such low concentrations of sulfide? A saturated aqueous solution of H S contains 0.10 M H S at 20°C. The p for H S is 6.97, and p corresponding to the formation of [S ] is 12.90. The equations for these reactions are as follows: \[H_2S (aq) \rightleftharpoons H^+ (aq) + HS^− (aq) \label{17.34a}\] with \[pK_{a1} = 6.97 \; \text{and hence} \; K_{a1} = 1.1 \times 10^{−7} \label{17.34b}\] \[HS^− (aq) \rightleftharpoons H^+ (aq) + S^{2−} (aq) \label{17.34c}\] with \[pK_{a2} = 12.90 \; \text{and hence} \; K_{a2} = 1.3 \times 10^{−13} \label{17.34d}\] We can show that the concentration of S is 1.3 × 10 by comparing and and recognizing that the contribution to [H ] from the dissociation of HS is negligible compared with [H ] from the dissociation of H S. Thus substituting 0.10 M in the equation for for the concentration of H S, which is essentially constant regardless of the pH, gives the following: Substituting this value for [H ] and [HS ] into the equation for , \[K_{\textrm{a2}}=1.3\times10^{-13}=\dfrac{[\mathrm{H^+},\mathrm{S^{2-}}]}{[\mathrm{HS^-}]}=\dfrac{(1.1\times10^{-4}\textrm{ M})x}{1.1\times10^{-4}\textrm{ M}}=x=[\mathrm{S^{2-}}]\] Although [S ] in an H S solution is very low (1.3 × 10 M), bubbling H S through the solution until it is saturated would precipitate both metal ions because the concentration of S would then be much greater than 1.6 × 10 M. Thus we must adjust [S ] to stay within the desired range. The most direct way to do this is to adjust [H ] by adding acid to the H S solution (recall ), thereby driving the equilibrium in to the left. The overall equation for the dissociation of H S is as follows: \[H_2S (aq) \rightleftharpoons 2H^+ (aq) + S^{2−} (aq) \label{17.36}\] Now we can use the equilibrium constant for the overall reaction, which is the product of and , and the concentration of H S in a saturated solution to calculate the H concentration needed to produce [S ] of 1.6 × 10 M: \[K=K_{\textrm{a1}}K_{\textrm{a2}}=(1.1\times10^{-7})(1.3\times10^{-13})=1.4\times10^{-20}=\dfrac{[\mathrm{H^+}]^2[\mathrm{S^{2-}}]}{[\mathrm{H_2S}]} \label{17.37}\] Thus adding a strong acid such as HCl to make the solution 0.94 M in H will prevent the more soluble ZnS from precipitating while ensuring that the less soluble CdS will precipitate when the solution is saturated with H S. A solution contains 0.010 M Ca and 0.010 M La . What concentration of HCl is needed to precipitate La (C O ) •9H O but not Ca(C O )•H O if the concentration of oxalic acid is 1.0 M? values are 2.32 × 10 for Ca(C O ) and 2.5 × 10 for La (C O ) ; p = 1.25 and p = 3.81 for oxalic acid. concentrations of cations, values, and concentration and p values for oxalic acid concentration of HCl needed for selective precipitation of La (C O ) Because the salts have different stoichiometries, we cannot directly compare the magnitudes of the solubility products. Instead, we must use the equilibrium constant expression for each solubility product to calculate the concentration of oxalate needed for precipitation to occur. Using ox for oxalate, we write the solubility product expression for calcium oxalate as follows: \[K_{sp} = [Ca^{2+},ox^{2−}] = (0.010)[ox^{2−}] = 2.32 \times 10^{−9} \nonumber\] \[[ox^{2−}] = 2.32 \times 10^{−7}\; M \nonumber\] The expression for lanthanum oxalate is as follows: \[K_{sp} = [La^{3+}]^2[ox^{2−}]^3 = (0.010)^2[ox^{2−}]^3 = 2.5 \times 10^{−27} \nonumber\] \[[ox^{2−}] = 2.9 \times 10^{−8}\; M \nonumber\] Thus lanthanum oxalate is less soluble and will selectively precipitate when the oxalate concentration is between \(2.9 \times 10^{−8} M\) and \(2.32 \times 10^{−7} M\). To prevent Ca from precipitating as calcium oxalate, we must add enough H to give a maximum oxalate concentration of 2.32 × 10 M. We can calculate the required [H ] by using the overall equation for the dissociation of oxalic acid to oxalate: \[HO_2CCO_2H (aq) \rightleftharpoons 2H^+ (aq) + C_2O^{2−}_{4(aq)}\] = = (10 )(10 ) = 10 = 8.7×10 Substituting the desired oxalate concentration into the equilibrium constant expression, \[\begin{align}8.7\times10^{-6}=\dfrac{[\mathrm{H^+}]^2[\mathrm{ox^{2-}}]}{[\mathrm{HO_2CCO_2H}]} &= \dfrac{[\mathrm{H^+}]^2(2.32\times10^{-7})}{1.0} \\ [\mathrm{H^+}] &=\textrm{6.1 M}\end{align} \nonumber\] Thus adding enough HCl to give [H ] = 6.1 M will cause only La (C O ) •9H O to precipitate from the solution. A solution contains 0.015 M Fe and 0.015 M Pb . What concentration of acid is needed to ensure that Pb precipitates as PbS in a saturated solution of H S, but Fe does not precipitate as FeS? values are 6.3 × 10 for FeS and 8.0 × 10 for PbS. 0.018 M H The anion in many sparingly soluble salts is the conjugate base of a weak acid. At low pH, protonation of the anion can dramatically increase the solubility of the salt. Oxides can be classified as acidic oxides or basic oxides. Acidic oxides either react with water to give an acidic solution or dissolve in strong base; most acidic oxides are nonmetal oxides or oxides of metals in high oxidation states. Basic oxides either react with water to give a basic solution or dissolve in strong acid; most basic oxides are oxides of metallic elements. Oxides or hydroxides that are soluble in both acidic and basic solutions are called amphoteric oxides. Most elements whose oxides exhibit amphoteric behavior are located along the diagonal line separating metals and nonmetals in the periodic table. In solutions that contain mixtures of dissolved metal ions, the pH can be used to control the anion concentration needed to selectively precipitate the desired cation.
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In the field of inorganic chemistry, color is commonly associated with d–d transitions. If this is the case, why is it that some transition metal complexes show intense color in solution, but possess no d electrons? In transition metal complexes a change in electron distribution between the metal and a ligand gives rise to charge transfer (CT) bands when performing Ultraviolet-visible spectroscopy experiments. For complete understanding, a brief introduction to electron transfer reactions and is necessary. Electron transfer reactions(charge transfer) fall into two categories: and . Inner-sphere mechanisms involve electron transfer occurring via a covalently bound bridging ligand (Figure \(\Page {1}\)). Outer-sphere mechanisms involve electron transfer occurring without a covalent linkage between reactants, e.g., \[\ce{[ML6]^{2+} + [ML6]^{3+} -> [ML6]^{3+} + [ML6]^{2+}} \label{eq1}\] Here, we focus on outer sphere mechanisms. In a self-exchange reaction the reactant and product side of a reaction are the same. No chemical reaction takes place and only an electron transfer is witnessed. This reductant-oxidant pair involved in the charge transfer is called the precursor complex. The states that a molecular electronic transition occurs much faster than a molecular vibration. Consider the reaction described in Equation \ref{1}. This process has a Franck-Condon restriction: Electron transfer can only take place when the \(\ce{M–L}\) bond distances in the \(\ce{ML(II)}\) and \(\ce{ML(III)}\) states are the same. This means that vibrationally excited states with equal bonds lengths must be formed in order to allow electron transfer to occur. This would mean that the \(\ce{[ML6]^{2+}}\) bonds must be compressed and \(\ce{[ML6]^{3+}}\) bonds must be elongated for the reaction to occur. Self exchange rate constants vary, because the activation energy required to reach the vibrational states varies according to the system. The greater the changes in bond length required to reach the precursor complex, the slower the rate of charge transfer. Marcus-Hush theory relates kinetic and thermodynamic data for two self-exchange reactions with data for the cross-reaction between the two self-exchange partners. This theory determines whether an outer sphere mechanism has taken place. This theory is illustrated in the following reactions Self exchange 1: [ ML ] + [ML ] → [ ML ] + [ ML ] ∆G = 0 Self exchange 2: [ L ] + [ L ] → [ L ] + [ L ] ∆G = 0 Cross Reaction: [ ML ] + [ L ] → [ ML ] + [ L ] The Gibbs free energy of activation \(\Delta G^{\mp }\) is represented by the following equation: \[\Delta G^{\mp } = \Delta_{w} G^{\mp } + \Delta _{o}G^{\mp } + \Delta _{s}G^{\mp }+ RT \ln ( k’T / hZ) \nonumber \] The rate constant for the self-exchange is calculated using the following reaction \[k = \kappa Z e^{-\Delta G^{\mp }/RT} \nonumber \] where \(\kappa\) is the transmission coefficient ~1. The Marcus-Hush equation is given by the following expression \[k_{12}=(k_{11}k_{22}K_{12}f_{12})^{1/2} \label{marcus}\] where: \[\log f_{12}=\frac{(\log K_{12})^{2}}{4\log(\frac{k_{11}k_{22}}{Z^{2}})} \nonumber \] The following equation is an approximate from of the Marcus-Hush equation (Equation \ref{marcus}): \[\log k_{12}\approx 0.5\log k_{11}+0.5\log \log \nonumber \] since \(f\approx 1\) and \(\log f \approx 0\). How is the Marcus-Hush equation used to determine if an outer sphere mechanism is taking place? If an outer sphere mechanism is taking place the calculated values of \(k_{12}\) will match or agree with the experimental values. If these values do not agree, this would indicate that another mechanism is taking place. d- d transitions are forbidden by the . d-d transitions result in weak absorption bands and most d-block metal complexes display low intensity colors in solution (exceptions d and d complexes). The low intensity colors indicate that there is a low probability of a d-d transition occurring. Ultraviolet-visible (UV/Vis) spectroscopy is the study of the transitions involved in the rearrangements of valence electrons. In the field of inorganic chemistry, UV/Vis is usually associated with d – d transitions and colored transition metal complexes. The color of the transition metal complex solution is dependent on: the metal, the metal oxidation state, and the number of metal d-electrons. For example iron(II) complexes are green and iron(III) complexes are orange/brown. If color is dependent on d-d transitions, why is it that some transition metal complexes are intensely colored in solution but possess no d electrons? In transition metal complexes a change in electron distribution between the metal and a ligand give rise to charge transfer (CT) bands. CT absorptions in the UV/Vis region are intense (ε values of 50,000 L mole cm or greater) and selection rule allowed. The intensity of the color is due to the fact that there is a high probability of these transitions taking place. Selection rule forbidden d-d transitions result in weak absorptions. For example octahedral complexes give ε values of 20 L mol cm or less. A charge transfer transition can be regarded as an internal oxidation-reduction process. Ligands possess \(σ\), \(σ^*\), \(π\), \(π^*\), and nonbonding (\(n\)) molecular orbitals. If the ligand molecular orbitals are full, charge transfer may occur from the ligand molecular orbitals to the empty or partially filled metal d-orbitals. The absorptions that arise from this process are called ligand-to-metal charge-transfer (LMCT) bands (Figure 3). LMCT transitions result in intense bands. Forbidden d-d transitions may also take place giving rise to weak absorptions. Ligand to metal charge transfer results in the reduction of the metal. If the metal is in a low oxidation state (electron rich) and the ligand possesses low-lying empty orbitals (e.g., \(\ce{CO}\) or \(\ce{CN^{-}}\)) then a metal-to-ligand charge transfer (MLCT) transition may occur. MLCT transitions are common for coordination compounds having π-acceptor ligands. Upon the absorption of light, electrons in the metal orbitals are excited to the ligand \(π^*\) orbitals. Figure 4 illustrates the metal to ligand charge transfer in a d octahedral complex. MLCT transitions result in intense bands. Forbidden d–d transitions may also occur. This transition results in the oxidation of the metal. The position of the CT band is reported as a transition energy and depends on the solvating ability of the solvent. A shift to lower wavelength (higher frequency) is observed when the solvent has high solvating ability. Polar solvent molecules align their dipole moments maximally or perpendicularly with the ground state or excited state dipoles. If the ground state or excited state is polar an interaction will occur that will lower the energy of the ground state or excited state by solvation. The effect of solvent polarity on CT spectra is illustrated in the following example. You are preparing a sample for a UV/Vis experiment and you decide to use a polar solvent. Is a shift in wavelength observed when: When both the ground state and the excited state are neutral a shift in wavelength is not observed. No change occurs. Like dissolves like and a polar solvent won’t be able to align its dipole with a neutral ground and excited state. If the excited state is polar, but the ground state is neutral the solvent will only interact with the excited state. It will align its dipole with the excited state and lower its energy by solvation. This interaction will lower the energy of the polar excited state. (increase wavelength, decrease frequency, decrease energy) If the ground state is polar the polar solvent will align its dipole moment with the ground state. Maximum interaction will occur and the energy of the ground state will be lowered. (increased wavelength, lower frequency, and lower energy) The dipole moment of the excited state would be perpendicular to the dipole moment of the ground state, since the polar solvent dipole moment is aligned with the ground state. This interaction will raise the energy of the polar excited state. (decrease wavelength, increase frequency, increase energy) If the ground state is polar the polar solvent will align its dipole moment with the ground state. Maximum interaction will occur and the energy of the ground state will be lowered. (increased wavelength, lower frequency, and lower energy). If the excited state is neutral no change in energy will occur. Like dissolves like and a polar solvent won’t be able to align its dipole with a neutral excited state. Overall you would expect an increase in energy (Illustrated below), because the ground state is lower in energy (decrease wavelength, increase frequency, increase energy). CT absorptions are selection rule allowed and result in intense (ε values of 50,000 L mole cm or greater) bands in the UV/Vis region. Selection rule forbidden d-d transitions result in weak absorptions. For example octahedral complexes give ε values of 20 L mol cm or less. CT bands are easily identified because they: \(\ce{KMnO4}\) dissolved in water exhibits intense CT Bands. The one LMCT band in the visible is observed around 530 nm. The band at 528 nm gives rise to the deep purple color of the solution. An electron from a “oxygen lone pair” character orbital is transferred to a low lying \(\ce{Mn}\) orbital. Tris(bipyridine)ruthenium(II) dichloride (\(\ce{[Ru(bpy)3]Cl2}\)) is a coordination compound that exhbits a CT band is observed (Figure \(\Page {6}\)) A d electron from the ruthenium atom is excited to a bipyridine anti-bonding orbital. The very broad absorption band is due to the excitation of the electron to various vibrationally excited states of the π* electronic state.
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In 1964 researchers in the Soviet Union at Dubna announced their discovery of element 104. A similar claim was made by researchers at the University of California at Berkeley. The Soviet scientists claimed to have bombarded a target of Pu-242 with Ne-22, resulting in a nucleus with 104 protons and a mass number of 260. \[\ce{^{22}_{10}Ne + ^{242}_{94}Pu \rightarrow ^{260}_104}Rf\] The Berkeley team used a Cf-249 target and isotopes of carbon for projectiles, resulting in isotopes of 104 with mass numbers of 257 and 259. Several other isotopes were also prepared by the American team. \[\ce{^{12}_{6}C + ^{249}_{98}Cf \rightarrow ^{257}_{104}Rf + 4^1_0n}\] The naming and priority of discovery controversy has raged ever since and Rutherfordium is the approved name selected by the IUPAC in August 1997.
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Rubidium (Latin: rubidius = red) is similar in physical and chemical characteristics to potassium, but much more reactive. It is the seventeenth most abundant element and was discovered by its red spectral emission in 1861 by Bunsen and Kirchhoff. Its melting point is so low you could melt it in your hand if you had a fever (39°C). But that would not be a good idea because it would react violently with the moisture in your skin. Rubidium was once thought to be quite rare but recent discoveries of large deposits indicate that there is plenty to use. However at present it finds only limited application in the manufacture of cathode ray tubes.
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The oxygen family, also called the chalcogens, consists of the elements found in Group 16 of the periodic table and is considered among the main group elements. It consists of the elements , , , tellurium and polonium. These can be found in nature in both free and combined states. are intimately related to life. We need oxygen all the time throughout our lives. Did you know that sulfur is also one of the essential elements of life? It is responsible for some of the protein structures in all living organisms. Many industries utilize sulfur, but emission of sulfur compounds is often seen more as a problem than the natural phenomenon. The metallic properties increase as the atomic number increases. The element polonium has no stable isotopes, and the isotope with mass number 209 has the longest half life of 103 years. Properties of oxygen are very different from other elements of the group, but they all have 2 elections in the outer s orbital, and 4 electrons in the p orbitals, usually written as s p The electron configurations for each element are given below: Polonium can be written as [Xe] 6s 4f 5d 6p The trends of their properties in this group are interesting. Knowing the trend allows us to predict their reactions with other elements. Most trends are true for all groups of elements, and the group trends are due mostly to the size of the atoms and number of electrons per atom. The trends are described below: Metallic character increases down the group, with tellurium classified as a metalloid and polonium as a metal. Melting point, boiling point, density, atomic radius, and ionic radius all increase down the group. Ionization energy decreases down the group. The most common oxidation state is -2; however, sulfur can also exist at a +4 and +6 state, and +2, +4, and +6 oxidation states are possible for Se, Te, and Po. Oxygen is a gas at room temperature and 1 atm, and is colorless, odorless, and tasteless. It is the most abundant element by mass in both the Earth's crust and the human body. It is second to nitrogen as the most abundant element in the atmosphere. There are many commercial uses for oxygen gas, which is typically obtained through fractional distillation. It is used in the manufacture of iron, steel, and other chemicals. It is also used in water treatment, as an oxidizer in rocket fuel, for medicinal purposes, and in petroleum refining. Oxygen has two O and O . In general, O (or dioxygen) is the form referred to when talking about the elemental or molecular form because it is the most common form of the element. The O bond is very strong, and oxygen can also form strong bonds with other elements. However, compounds that contain oxygen are considered to be more thermodynamically stable than O . The latter allotrope, ozone, is a pale-blue poisonous gas with a strong odor. It is a very good oxidizing agent, stronger than dioxygen, and can be used as a substitute for chlorine in purifying drinking water without giving the water an odd taste. However, because of its unstable nature it disappears and leaves the water unprotected from bacteria. Ozone at very high altitudes in the atmosphere is responsible for protecting the Earth's surface from ultraviolet radiation; however, at lower altitudes it becomes a major component of smog. Oxygen's primary oxidation states are -2, -1, 0, and -1/2 (in O ), but is the most common. Typically, compounds with oxygen in this oxidation state are called . When oxygen reacts with metals, it forms oxides that are mostly ionic in nature. These can dissolve in water and react to form hydroxides; they are therefore called or . Nonmetal oxides, which form covalent bonds, are simple molecules with low melting and boiling points. Compounds with oxygen in an oxidation state of -1 are referred to as . Examples of this type of compound include \(Na_2O_2\) and \(BaO_2\). When oxygen has an oxidation state of -1/2, as in \(O_2^-\), the compound is called a . Oxygen is rarely the central atom in a structure and can never bond with more than 4 elements due to its small size and its inability to create an expanded valence shell. Oxygen reacts with hydrogen to form water, which is extensively hydrogen-bonded, has a large dipole moment, and is considered an universal solvent. There are a wide variety of oxygen-containing compounds, both organic and inorganic: oxides, peroxides and superoxides, alcohols, phenols, ethers, and carbonyl-containing compounds such as aldehydes, ketones, esters, amides, carbonates, carbamates, carboxylic acids and anhydrides. Sulfur is a solid at room temperature and 1 atm pressure. It is usually yellow, tasteless, and nearly odorless. It is the sixteenth most abundant element in Earth's crust. It exists naturally in a variety of forms, including elemental sulfur, sulfides, sulfates, and organosulfur compounds. Since the 1890s, sulfur has been mined using the Frasch process, which is useful for recovering sulfur from deposits that are under water or quicksand. Sulfur produced from this process is used in a variety of ways including in vulcanizing rubber and as fungicide to protect grapes and strawberries. Sulfur is unique in its ability to form a wide range of allotropes, more than any other element in the periodic table. The most common state is the solid S ring, as this is the most thermodynamically stable form at room temperature. Sulfur exists in the gaseous form in five different forms (S, S , S , S , and S ). In order for sulfur to convert between these compounds, sufficient heat must be supplied. Two common oxides of sulfur are sulfur dioxide (SO ) and sulfur trioxide (SO ). Sulfur dioxide is formed when sulfur is combusted in air, producing a toxic gas with a strong odor. These two compounds are used in the production of sulfuric acid, which is used in a variety of reactions. Sulfuric acid is one of the top manufactured chemicals in the United States, and is primarily used in the manufacture of fertilizers. Sulfur also exhibits a wide range of oxidation states, with values ranging from -2 to +6. It is often the central ion in a compound and can easily bond with up to 6 atoms. In the presence of hydrogen it forms the compound hydrogen sulfide, H S, a poisonous gas incapable of forming hydrogen bonds and with a very small dipole moment. Hydrogen sulfide can easily be recognized by its strong odor that is similar to that of rotten eggs, but this smell can only be detected at low, nontoxic concentrations. This reaction with hydrogen epitomizes how differently oxygen and sulfur act despite their common valence electron configuration and common nonmetallic properties. A variety of sulfur-containing compounds exist, many of them organic. The prefix in from of the name of an oxygen-containing compound means that the oxygen atom has been substituted with a sulfur atom. General categories of sulfur-containing compounds include thiols (mercaptans), thiophenols, organic sulfides (thioethers), disulfides, thiocarbonyls, thioesters, sulfoxides, sulfonyls, sulfamides, sulfonic acids, sulfonates, and sulfates. Selenium appears as a red or black amorphous solid, or a red or grey crystalline structure; the latter is the most stable. Selenium has properties very similar to those of sulfur; however, it is more metallic though it is still classified as a nonmetal. It acts as a semiconductor and therefore is often used in the manufacture of rectifiers, which are devices that convert alternating currents to direct currents. Selenium is also photoconductive, which means that in the presence of light the electrical conductivity of selenium increases. It is also used in the drums of laser printers and copiers. In addition, it has found increased use now that lead has been removed from plumbing brasses. It is rare to find selenium in its elemental form in nature; it must typically be removed through a refining process, usually involving copper. It is often found in soils and in plant tissues that have bioaccumulated the element. In large doses, the element is toxic; however, many animals require it as an essential micronutrient. Selenium atoms are found in the glutathione peroxidase, which destroys lipid-damaging peroxides. In the human body it is an essential cofactor in maintaining the function of the thyroid gland. In addition, some research has shown a correlation between selenium-deficient soils and an increased risk of contracting the HIV/AIDS virus. Tellurium is the metalloid of the oxygen family, with a silvery white color and a metallic luster similar to that of tin at room temperature. Like selenium, it is also displays photoconductivity. Tellurium is an extremely rare element, and is most commonly found as a telluride of gold. It is often used in metallurgy in combination with copper, lead, and iron. In addition, it is used in solar panels and memory chips for computers. It is not toxic or carcinogenic; however, when humans are exposed to too much of it they develop a garlic-like smell on their breaths. Polonium is a very rare, radioactive metal. There are 33 different isotopes of the element and all of the isotopes are radioactive. It exists in a variety of states, and has two metallic allotropes. It dissolves easily into dilute acids. Polonium does not exist in nature in compounds, but it can form synthetic compounds in the laboratory. It is used as an alloy with beryllium to act as a neutron source for nuclear weapons. Polonium is a highly toxic element. The radiation it emits makes it very dangerous to handle. It can be immediately lethal when applied at the correct dosage, or cause cancer if chronic exposure to the radiation occurs. Methods to treat humans who have been contaminated with polonium are still being researched, and it has been shown that agents could possibly be used to decontaminate humans.
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Previously, we saw that it is the of the entropy changes of the system and surroundings that determines whether a process will occur spontaneously. In chemical thermodynamics we prefer to focus our attention on the system rather than the surroundings, and would like to avoid having to calculate the entropy change of the surroundings explicitly. In this unit we introduce a new thermodynamic function, the , which turns out to be the single most useful criterion for predicting the direction of a chemical reaction and the composition of the system at equilibrium. However, the term "free energy", although still widely used, is rather misleading, so we will often refer to it as "Gibbs energy." The free energy enables us to do this for changes that occur at a constant temperature and pressure (the ) or constant temperature and volume (the .) The Gibbs energy (also known as the or ) is defined as \[G = H – T S \label{23.4.1}\] in which \(S\) refers to the entropy of the . Since \(H\), \(T\) and \(S\) are all state functions, so is \(G\). Thus for any change in state (under constant temperature), we can write the extremely important relation \[ΔG = ΔH – T ΔS \label{23.4.2}\] How does this simple equation encompass the entropy change of the world \(ΔS_{total}\), which we already know is the sole criterion for spontaneous change from the second law of thermodynamics? Starting with the definition \[ΔS_{total} = ΔS_{surr} + ΔS_{sys} \label{23.4.3}\] we would first like to get rid of \(ΔS_{surr}\). How can a chemical reaction (a change in the ) affect the entropy of the ? Because most reactions are either exothermic or endothermic, they are accompanied by a flow of heat across the system boundary. The enthalpy change of the reaction \(ΔH\) is defined as the flow of heat into the system from the surroundings when the reaction is carried out at constant pressure, so the heat withdrawn from the surroundings will be \(–q_p\) which will cause the entropy of the surroundings to change by \(–q_p / T = –ΔH/T\). We can therefore rewrite Equation \(\ref{23.4.3}\) as \[ΔS_{total} = \dfrac{- ΔH}{T} + ΔS_{sys} \label{23.4.4}\] Multiplying each side by \(-T\), we obtain \[-TΔS_{total} = ΔH - TΔS_{sys} \label{23.4.5}\] which expresses the entropy change of the world in terms of thermodynamic properties of the exclusively. If \(-TΔS_{total}\) is denoted by \(ΔG\), then we have Equation \(\ref{23.4.2}\) which defines the change for the process. From the foregoing, you should convince yourself that \(G\) will decrease in any process occurring at constant temperature and pressure which is accompanied by an overall increase in the entropy. The constant temperature is a consequence of the temperature and the enthalpy appearing in the preceding Equation \(\ref{23.4.5}\). Since most chemical and phase changes of interest to chemists take place under such conditions, the Gibbs energy is the most useful of all the thermodynamic properties of a substance, and (as we shall see in the lesson that follows this one) it is closely linked to the equilibrium constant. Some textbooks and teachers say that the free energy, and thus the spontaneity of a reaction, depends on both the enthalpy and entropy changes of a reaction, and they sometimes even refer to reactions as "energy driven" or "entropy driven" depending on whether \(ΔH\) or the \(TΔS\) term dominates. This is technically correct, but misleading because it disguises the important fact that \(ΔS_{total}\), which this equation expresses in an indirect way, is the criterion of spontaneous change. We will deal only with the Gibbs energy in this course. The is of interest mainly to chemical engineers (whose industrial-scale processes are often confined to tanks and reactors of fixed volume) and some geochemists whose interest is centered on the chemistry that occurs deep within the earth's surface. Remember that \(ΔG\) is meaningful only for changes in which the . These are the conditions under which most reactions are carried out in the laboratory; the system is usually open to the atmosphere (constant pressure) and we begin and end the process at room temperature (after any heat we have added or which is liberated by the reaction has dissipated.) The importance of the Gibbs function can hardly be over-stated: it serves as the single master variable that determines whether a given chemical change is thermodynamically possible. Thus if the free energy of the reactants is greater than that of the products, the entropy of the world will increase when the reaction takes place as written, and so the reaction will tend to take place spontaneously. Conversely, if the free energy of the products exceeds that of the reactants, then the reaction will not take place in the direction written, but it will tend to proceed in the reverse direction. \(ΔG\) serves as the single master variable that determines whether a given chemical change is thermodynamically possible. Moreover, it determines the direction and extent of chemical change. In a spontaneous change, Gibbs energy always decreases and never increases. This of course reflects the fact that the entropy of the world behaves in the exact opposite way (owing to the negative sign in the \(TΔS\) term). \[\ce{H_2O(l) \rightarrow H2O(s)} \label{23.5.6}\] water below its freezing point undergoes a decrease in its entropy, but the heat released into the surroundings more than compensates for this, so the entropy of the world increases, the free energy of the H O diminishes, and the process proceeds spontaneously. In a spontaneous change, Gibbs energy decreases and never increases. An important consequence of the one-way downward path of the free energy is that once it reaches its minimum possible value, all net change comes to a halt. This, of course, represents the state of chemical equilibrium. These relations are nicely summarized as follows: This might seem strange, given the key importance \(ΔG\) in determining whether or not a reaction will take place in a given direction. It turns out, however, that it is almost never necessary to explicitly evaluate \(ΔG\). As we will show in the lesson that follows this one, it is far more convenient to work with the equilibrium constant of a reaction, within which \(ΔG\) is "hidden". This is just as well, because for most reactions (those that take place in solutions or gas mixtures) the value of \(ΔG\) depends on the of the various reaction components in the mixture; it is not a simple sum of the "products minus reactants" type, as is the case with \(ΔH\). Recalling the condition for spontaneous change \[ΔG = ΔH – TΔS < 0\] it is apparent that the temperature dependence of Δ depends almost entirely on the entropy change associated with the process. (We say "almost" because the values of \(ΔH\) and \(ΔS\) are themselves slightly temperature dependent; both gradually increase with temperature). In particular, notice that in the above equation For any given reaction, the sign of \(ΔH\) can also be positive or negative. This means that there are four possibilities for the influence that temperature can have on the spontaneity of a process: Both enthalpic \(\Delta H\) and entropic \(-T\Delta S\) terms will be negative, so \(ΔG\) will be negative regardless of the temperature. An exothermic reaction whose entropy increases will be spontaneous at all temperatures. If the reaction is sufficiently exothermic it can force \(ΔG\) negative only at temperatures below which \(|TΔS| < |ΔH|\). This means that there is a temperature \(T = ΔH / ΔS\) at which the reaction is at equilibrium; the reaction will only proceed spontaneously below this temperature. The freezing of a liquid or the condensation of a gas are the most common examples of this condition. This is the reverse of the previous case; the entropy increase must overcome the handicap of an endothermic process so that \(TΔS > ΔH\). Since the effect of the temperature is to "magnify" the influence of a positive \(ΔS\), the process will be spontaneous at temperatures above \(T = ΔH / ΔS\). (Think of melting and boiling.) With both \(ΔH\) and \(ΔS\) working against it, this kind of process will not proceed spontaneously at any temperature. Substance A always has a greater number of accessible energy states, and is therefore always the preferred form. The plots above are the important ones; do not try to memorize them, but make sure you understand and can explain or reproduce them for a given set of Δ and Δ . The other two plots on each diagram are only for the chemistry-committed. You have already been introduced to the terms such as \(ΔU^o\) and \(ΔH^o\) in which the \(^o\) sign indicates that all components (reactants and products) are in their . This concept of standard states is especially important in the case of the free energy, so let's take a few moments to review it. More exact definitions of the conventional standard states can be found in most physical chemistry textbooks. In specialized fields such as biochemistry and oceanography, alternative definitions may apply. For example, the "standard pH" of zero (corresponding to \([H^{+}] = 1\,M\)) is impractical in biochemistry, so pH = 7 is commonly employed. For most practical purposes, the following definitions are good enough: To make use of Gibbs energies to predict chemical changes, we need to know the free energies of the individual components of the reaction. For this purpose we can combine the standard enthalpy of formation and the standard entropy of a substance to get its \[ΔG_f^o = ΔH_f^o – TΔS_f^o \label{23.4.7}\] Recall that the symbol ° refers to the of a substance measured under the conditions of 1 atm pressure or an effective concentration of 1 mol L and a temperature of 298 K. Then determine the standard Gibbs energy of the reaction according to \[ ΔG^o = \sum ΔG_f^o \;(\text{products})– \sum ΔG_f^o \;(\text{reactants}) \label{24.4.8}\] As with standard heats of formation, the standard free energy of a substance represents the free energy change associated with the formation of the substance from the elements in their most stable forms as they exist under the standard conditions of 1 atm pressure and 298 K. Standard Gibbs free energies of formation are normally found directly from tables. Once the values for all the reactants and products are known, the standard Gibbs energy change for the reaction is found by Equation \(\ref{23.4.7}\). Most tables of thermodynamic values list \(ΔG_f^o\) values for common substances (e.g., ), which can, of course, always be found from values of and . Find the standard Gibbs energy change for the reaction \[\ce{CaCO3(s) \rightarrow CaO (s) + CO2(g)} \nonumber\] The \(ΔG_f^o°\) values for the three components of this reaction system are \(\ce{CaCO3(s)}\): –1128 kJ mol , CaO : –603.5 kJ mol , CO : –137.2 kJ mol . Substituting into Equation \(\ref{23.4.7}\), we have \[ΔG^o = (–603.5 –137.2) – (–1128) kJ\, mol^{–1} = +130.9\, kJ\, mol^{–1} \nonumber \] This indicates that the process is not spontaneous under standard conditions (i.e., solid calcium carbone will not form solid calcium oxide and CO at 1 atm partial pressure at 25° C). This reaction is carried out on a huge scale to manufacture cement, so it is obvious that the process can be spontaneous under different conditions. The practical importance of the Gibbs energy is that it allows us to make predictions based on the properties (Δ values) of the reactants and products themselves, eliminating the need to experiment. But bear in mind that while thermodynamics always correctly predicts whether a given process take place (is spontaneous in the thermodynamic sense), it is unable to tell us if it take place at an observable rate. When thermodynamics says "no", it means exactly that. When it says "yes", it means "maybe". The reaction \[\ce{ 1/2 O2(g) + H2(g) → H2O(l)} \nonumber\] is used in fuel cells to produce an electrical current. The reaction can also be carried out by direct combustion. : molar entropies in J mol K : O (g) 205.0; H (g)130.6; H O(l) 70.0; H O(l) Δ ° = –285.9 kJ mol . Use this information to find First, we need to find \(ΔH^o\) and \(ΔS^o\) for the process. Recalling that the standard enthalpy of formation of the elements is zero, \[\begin{align*} ΔH^o &= ΔH^p_f(\text{products}) – ΔH^°_f(\text{reactants}) \\[4pt] &= –285.9\, kJ\, mol^{–1} – 0 \\[4pt] &= –285.9 \,kJ \,mol^{–1} \end{align*}.\] Similarly, \[\begin{align*} ΔS^o &= S^o_f(\text{products}) – S^o_f(\text{reactants}) \\[4pt] &= (70.0) – (½ \times 205.0 + 130.6) \\[4pt] &= –163\, J\, K^{–1}mol^{–1} \end{align*}\] The foregoing example illustrates an important advantage of fuel cells. Although direct combustion of a mole of hydrogen gas yields more energy than is produced by the same net reaction within the fuel cell, the latter, in the form of electrical energy, can be utilized at nearly 100-percent energy efficiency by a motor or some other electrical device. If the thermal energy released by direct combustion were supplied to a heat engine, second-law considerations would require that at least half of this energy be "wasted" to the surroundings. Δ refer to in which all components (reactants and products) are in their . The \(ΔG_f^o\) of a substance, like \(ΔH_f^o\), refers to the reaction in which that substance is formed from the elements as they exist in their most stable forms at 1 atm pressure and (usually) 298 K. Both of these terms are by definition zero for the elements in their standard states. There are only a few common cases in which this might create some ambiguity: Ions in aqueous solution are a special case; their standard free energies are relative to the hydrated hydrogen ion \(\ce{H^{+}(aq)}\) which is assigned \(ΔG_f^o = 0\). \(ΔG\) is very different from ΔG°. The distinction is nicely illustrated in Figure \(\Page {5}\) in which Δ is plotted on a vertical axis for two hypothetical reactions having opposite signs of Δ . The horizontal axis schematically expresses the relative concentrations of reactants and products at any point of the process. Note that the origin corresponds to the composition at which half of the reactants have been converted into products. Take careful note of the following: The important principle you should understand from this is that a negative Δ does not mean that the reactants will be completely transformed into products. By the same token, a positive Δ does not mean that no products are formed at all. It should now be clear from the discussion above that a given reaction carried out under standard conditions is characterized by a of . The reason for the Gibbs energy minimum at equilibrium relates to the increase in entropy when products and reactants coexist in the same phase. As seen in the plot, even a minute amount of "contamination" of products by reactants reduces the free energy below that of the pure products. In contrast, composition of a chemical reaction system undergoes continual change until the equilibrium state is reached. So the a single reaction can have an infinite number of Δ values, reflecting the infinite possible compositions between the extremes of pure reactants (zero extent of reaction) and pure products (unity extent of reaction). In the example of a reaction A → B, depicted in the above diagram, the standard free energy of the products is smaller than that of the reactants , so the reaction will take place spontaneously. T . For reactions in which products and reactants occupy a single phase (gas or solution), the meaning of "spontaneous" is that the equilibrium composition will correspond to an extent of reaction greater than 0.5 but smaller than unity.Note, however, that for values in excess of about ±50 kJ mol , the equilibrium composition will be negligibly different from zero or unity extent-of-reaction. The physical meaning of Δ is that it tells us how far the free energy of the system has changed from ° of the pure reactants . As the reaction proceeds to the right, the composition changes, and Δ begins to fall. When the composition reaches , Δ reaches its minimum value and further reaction would cause it to rise. But because free energy can only decrease but never increase, this does not happen. The composition of the system remains permanently at its equilibrium value. A . extent-of-reaction diagram for a non-spontaneous reaction can be interpreted in a similar way; the equilibrium composition will correspond to an extent of reaction greater than zero but less than 0.5. In this case, the minimum at reflects the increase in entropy when the reactants are "contaminated" by a small quantity of products. If all this detail about Δ seems a bit overwhelming, do not worry: it all gets hidden in the equilibrium constant and reaction quotient that we discuss in the next lesson! Although it is \(ΔG\) \(ΔG^o\) that serves as a criterion for spontaneous change at constant temperature and pressure, \(ΔG^o\) values are so readily available that they are often used to get a rough idea of whether a given chemical change is possible. This is practical to do in some cases, but not in others: It generally works for reactions such as \[\ce{4 NH_3(g) + 5 O_2(g) → 4 NO(g) + 6 H_2O(g)} \nonumber\] with \(ΔG^o = –1,010\, kJ\). (industrially important for the manufacture of nitric acid) because \(ΔG^o\) is so negative that the reaction will be spontaneous and virtually complete under just about any reasonable set of conditions. The following reaction expresses the fact that the water molecule is thermodynamically stable: \[\ce{2 H_2(g) + 1/2 O_2(g)→ H_2O(l)} \nonumber\] with \(ΔG^o = –237.2 \,kJ\). Note that this refers to water (the standard state of H O at 25°). If you think about it, a negative standard Gibbs energy of formation (of which this is an example) can in fact be considered a definition of molecular stability. Similarly, dissociation of dihydrogen into its atoms is highly unlikely under standard conditions: \[\ce{H_2O(g) → 2 H(g) + O(g)} \nonumber\] with \(ΔG^o = +406.6\, kJ\). Again, an analogous situation would apply to any stable molecule. Now consider the dissociation of dinitrogen tetroxide \[\ce{N_2O_4(g) → 2 NO_2(g)} \nonumber\] with \(ΔG^o = +2.8 kJ\). in which the positive value of tells us that N O at 1 atm pressure will not change into two moles of NO at the same pressure, but owing to the small absolute value of , we can expect the spontaneity of the process to be quite sensitive to both the temperature (as shown in the table below) and to the pressure in exactly the way the Le Chatelier principle predicts. For reactions involving dissolved ions, one has to be quite careful. Thus for the dissociation of the weak hydrofluoric acid \[\ce{HF(aq) → H^+(aq) + F^–(aq)} \nonumber\] with \(ΔG^o = –317 \,kJ\). it is clear that a 1 mol/L solution of HF will not dissociate into 1M ions, but this fact is not very useful because if the HF is added to water, the initial concentration of the fluoride ion will be zero (and that of H very close to zero), and the Le Chatelier principle again predicts that dissociation will be spontaneous. It is common knowledge that dissociation of water into hydrogen- and hydroxyl ions occurs only very sparingly: \[\ce{H_2O(l) → H^+(aq) + OH^–(aq) } \nonumber\] with \(ΔG^o = 79.9 \,kJ\). which correctly predicts that the water will not form 1M (effective concentration) of the ions, but this is hardly news if you already know that the product of these ion concentrations can never exceed 10 at 298K. Finally, consider this most familiar of all phase change processes, the vaporization of liquid water: \[\ce{H_2O(l) → H_2O(g)} \nonumber\] with \(ΔG^o = 8.58 \,kJ \). Conversion of liquid water to its vapor at 1 atm partial pressure does not take place at 25° C, at which temperature the equilibrium partial pressure of the vapor (the "vapor pressure") is only 0.031 atm (23.8 torr.) Gaseous H O at a pressure of 1 atm can only exist at 100° C. Of course, water left in an open container at room temperature will spontaneously evaporate if the partial pressure of water vapor in the air is less than 0.031 atm, corresponding to a relative humidity of under 100% A reaction is in its equilibrium state when \[ΔG = ΔH – TΔS = 0 \label{23.4.1a}\] The temperature at which this occurs is given by \[T = \dfrac{ΔH}{TΔS} \label{23.4.1b}\] If we approximate \(ΔH\) by \(ΔH^o\) and \(ΔS\) by \(ΔS^o\), so Equation \ref{23.4.1a} would be \[ΔG \approx ΔH^o – TΔS^o = 0 \label{23.4.1aa}\] We can then estimate the normal boiling point of a liquid. From the following thermodynamic data for water: Because ΔH° values are normally expressed in kilojoules while ΔS° is given in joules, a very common student error is to overlook the need to express both in the same units. We find that liquid water is in equilibrium with water vapor at a partial pressure of 1 atm when the temperature is \[T = \dfrac{44,100\, J}{118.7\, J\, K^{–1}} = 371.5\, K\] But " ", you say? Very true. The reason we are off here is that both Δ  ° and Δ  ° have their own temperature dependencies; we are using the "standard" 25° values without correcting them to 100° C. Nevertheless, if you think about it, the fact that we can estimate the boiling point of a liquid from a table of thermodynamic data should be rather impressive! Of course, the farther one gets from 298 K, the more unreliable will be the result. Thus for the dissociation of dihydrogen into its atoms, All one can say here is that H will break down at something over 3000 K or so. (You may already know that molecules will dissociate into their atoms at high temperatures.) We tend to think of high temperatures as somehow "forcing" molecules to dissociate into their atoms, but this is wrong. In order to get the H–H bond to vibrate so violently through purely thermal excitation that the atoms would fly apart, a temperature more like 30,000 K would be required. The proper interpretation is at the temperature corresponding to Δ sufficient to overcome the H-H bond strength. The \(T\Delta S\) term interacts with the \(ΔH\) term in \(\Delta G\) to determine whether the reaction can take place at a given temperature. This can be more clearly understood by examining plots of \(TΔS^o\) and \(ΔH^o\) as functions of the temperature for some actual reactions. Of course these parameters refer to standard states that generally do not correspond to the temperatures, pressures, or concentrations that might be of interest in an actual case. Nevertheless, these quantities are easily found and they can usefully predict the way that temperature affects these systems. \[\ce{C(graphite) + O_2(g) → CO_2(g)} \nonumber \] This , like most such reactions, is . The positive entropy change is due mainly to the greater mass of \(\ce{CO2}\) molecules compared to those of \(\ce{O2}\). \[\ce{3 H_2 + N_2 → 2 NH_3(g) } \nonumber\] The decrease in moles of gas in the drives the entropy change negative, making the reaction . Thus higher T, which speeds up the reaction, also reduces its extent. \[ \ce{N_2O_4(g) → 2 NO_2(g)} \nonumber\] are typically endothermic with positive entropy change, and are therefore . \[\ce{ 1/2 N_2 (g) + O_2 (g)→ NO_2(g)} \nonumber\] This reaction is , meaning that . But because the reverse reaction is kinetically inhibited, NO can exist indefinitely at ordinary temperatures even though it is thermodynamically unstable. The appellation “free energy” for \(G\) has led to so much confusion that many scientists now refer to it simply as the . The “free” part of the older name reflects the steam-engine origins of thermodynamics with its interest in converting heat into work: \(ΔG\) is the maximum amount of energy, which can be “freed” from the system to perform useful work. By "useful", we mean work other than that which is associated with the expansion of the system. This is most commonly in the form of electrical work (moving electric charge through a potential difference), but other forms of work (osmotic work, increase in surface area) are also possible. A much more serious difficulty with the Gibbs function, particularly in the context of chemistry, is that although \(G\) has the of energy (joules, or in its intensive form, J mol ), it lacks one of the most important attributes of energy in that it is not . Thus, although the free energy always falls when a gas expands or a chemical reaction takes place spontaneously, there need be no compensating in energy anywhere else. Referring to \(G\) as an energy also reinforces the false but widespread notion that a fall in energy must accompany any change. But if we accept that energy is conserved, it is apparent that the only necessary condition for change (whether the dropping of a weight, expansion of a gas, or a chemical reaction) is the of energy. he increase in the entropy. The quotient \(–ΔG/T\) is in fact identical with \(ΔS_{total}\), the entropy change of the world, whose increase is the primary criterion for an \(G\) differs from the thermodynamic quantities H and S in another significant way: it has no physical reality as a property of matter, whereas \(H\) and \(S\) can be related to the quantity and distribution of energy in a collection of molecules. The free energy is simply a useful construct that serves as a criterion for change and makes calculations easier.
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If we know the probabilities of the possible outcomes of a trial, we can calculate the probabilities for combinations of outcomes. These calculations are based on two rules, which we call . If we partition the outcomes into exhaustive and mutually exclusive events, the laws of probability also apply to events. Since, as we define them, “events” is a more general term than “outcomes,” we call them the law of the and the law of the . These laws are valid so long as three conditions are satisfied. We have already discussed the first two of these conditions, which are that the outcomes possible in any individual trial must be exhaustive and mutually exclusive. The third condition is that, if we make more than one trial, the outcomes must be ; that is, the outcome of one trial must not be influenced by the outcomes of the others. We can view the laws of probability as rules for inferring information about combinations of events. The law of the probability of alternative events applies to events that belong to the same distribution. The law of the probability of compound events applies to events that can come from one or more distributions. An important special case occurs when the compound events are \(N\) successive samplings of a given distribution that we identify as the . If the random variable is a number, and we average the numbers that we obtain from \(N\) successive samplings of the parent distribution, these “averages-of-\(N\)” themselves constitute a distribution. If we know certain properties of the parent distribution, we can calculate corresponding properties of the “distribution of averages-of-\(N\) values obtained by sampling the parent distribution.” These calculations are specified by the , which we discuss in . In general, when we combine events from two distributions, we can view the result as an event that belongs to a third distribution. At first encounter, the idea of combining events and distributions may seem esoteric. A few examples serve to show that what we have in mind is very simple. Since an event is a set of outcomes, an event occurs whenever any of the outcomes in the set occurs. Partitioning the outcomes of tossing a die into “even outcomes” and “odd outcomes” illustrates this idea. The event “even outcome” occurs whenever the outcome of a trial is \(2\), \(4,\) or \(6\). The probability of an event can be calculated from the probabilities of the underlying outcomes. We call the rule for this calculation the law of the probabilities of alternative events. (We create the opportunity for confusion here because we are illustrating the idea of alternative events by using an example in which we call the alternatives “alternative outcomes” rather than “alternative events.” We need to remember that “event” is a more general term than “outcome.” One possible partitioning is that which assigns every outcome to its own event.) We discuss the probabilities of alternative events further below. To illustrate the idea of compound events, let us consider a first distribution that comprises “tossing a coin” and a second distribution that comprises “drawing a card from a poker deck.” The first distribution has two possible outcomes; the second distribution has \(52\) possible outcomes. If we combine these distributions, we create a third distribution that comprises “tossing a coin and drawing a card from a poker deck.” The third distribution has \(104\) possible outcomes. If we know the probabilities of the outcomes of the first distribution and the probabilities of the outcomes of the second distribution, and these probabilities are independent of one another, we can calculate the probability of any outcome that belongs to the third distribution. We call the rule for this calculation the law of the probability of compound events. We discuss it further below. A similar situation occurs when we consider the outcomes of tossing two coins. We assume that we can tell the two coins apart. Call them coin \(1\) and coin \(2\). We designate heads and tails for coins \(1\) and \(2\) as \(H_1\), \(T_1\), \(H_2\), and \(T_2\), respectively. There are four possible outcomes in the distribution we call “tossing two coins:” \(H_1H_2\), \(H_1T_2\), \(T_1H_2\), and \(T_1T_2\). (If we could not tell the coins apart, \(H_1T_2\) would be the same thing as \(T_1H_2\); there would be only three possible outcomes.) We can view the distribution “tossing two coins” as being a combination of the two distributions that we can call “tossing coin \(1\)” and “tossing coin\(\ 2\).” We can also view the distribution “tossing two coins” as a combination of two distributions that we call “tossing a coin a first time” and “tossing a coin a second time.” We view the distribution “tossing two coins” as being equivalent to the distribution “tossing one coin twice.” This is an example of , which is a frequently encountered type of distribution. In general, Chapter 19 considers the distribution of outcomes when a trial is repeated many times. Understanding the properties of such distributions is the single most essential element in understanding the theory of statistical thermodynamics. The central limit theorem relates properties of the repeated-trials distribution to properties of the parent distribution. If we know the probability of each of two mutually exclusive events that belong to an exhaustive set, the probability that one or the other of them will occur in a single trial is equal to the sum of the individual probabilities. Let us call the independent events and , and represent their probabilities as \(P(A)\) and \(P(B)\), respectively. The probability that one of these events occurs is the same thing as the probability that either occurs or occurs. We can represent this probability as \(P(A\ or\ B)\). The probability of this combination of events is the sum: \(P(A)+P(B)\). That is, \[P\left(A\ or\ B\right)=P\left(A\right)+P(B)\] Above we define as the event that a single toss of a die comes up either \(1\) or \(3\). Because each of these outcomes is one of six, mutually-exclusive, equally-likely outcomes, the probability of either of them is \({1}/{6}\): \(P\left(tossing\ a\ 1\right)=P\left(tossing\ a\ 3\right)\)\(={1}/{6}\). From the law of the probability of alternative events, we have \[\begin{align*} P\left(event\ Y\right) &=(tossing\ a\ 1\ or\ tossing\ a\ 3) \\[4pt] &=P\left(tossing\ a\ 1\right)\ or P\left(tossing\ a\ 3\right) \\[4pt] &= {1}/{6}+{1}/{6} \\[4pt] &={2}/{6} \end{align*}\] We define \(X\) as the event that a single toss of a die comes up even. From the law of the probability of alternative events, we have \[\begin{align*} P\left(event\ X\right) &=P\left(tossing\ 2\ or\ 4\ or\ 6\right) \\[4pt] &=P\left(tossing\ a\ 2\right)+P\left(tossing\ a\ 4\right)+P\left(tossing\ a\ 6\right) \\[4pt] &={3}/{6} \end{align*}\] We define \(Z\) as the event that a single toss comes up \(5\). \[P\left(event\ Z\right)=P\left(tossing\ a\ 5\right)=1/6\] If there are \(\omega\) independent events (denoted \(E_1,E_2,\dots ,E_i,\dots ,E_{\omega }\)), the law of the probability of alternative events asserts that the probability that one of these events will occur in a single trial is \[ \begin{align*} P\left(E_1\ or\ E_2\ or\dots E_i\dots or\ E_{\omega }\right) &=P\left(E_1\right)+P\left(E_2\right)+\dots +P\left(E_i\right)+\dots +P\left(E_{\omega }\right) \\[4pt] &=\sum^{\omega }_{i=1} P\left(E_i\right) \end{align*}\] If these \(\omega\) independent events encompass all of the possible outcomes, the sum of their individual probabilities must be unity. Let us now suppose that we make two trials in circumstances where event \(A\) is possible in the first trial and event \(B\) is possible in the second trial. We represent the probabilities of these events by \(P\left(A\right)\) and \(P(B)\) and stipulate that they are independent of one another; that is, the probability that \(B\) occurs in the second trial is independent of the outcome of the first trial. Then, the probability that \(A\) occurs in the first trial and \(B\) occurs in the second trial, \(P(A\ and\ B)\), is equal to the product of the individual probabilities. \[P\left(A\ and\ B\right)=P\left(A\right)\times P(B)\] To illustrate this using outcomes from die-tossing, let us suppose that event \(A\) is tossing a \(1\) and event \(B\) is tossing a \(3\). Then, \(P\left(A\right)={1}/{6}\) and \(P\left(B\right)={1}/{6}\). The probability of tossing a 1 in a first trial and tossing a \(3\) in a second trial is then \[\begin{align*} P\left( \text{tossing a 1 first and tossing a 3 second}\right) &=P\left(\text{tossing a 1}\right)\times P\left(\text{tossing a 3}\right) \\[4pt] &={1}/{6}\times {1}/{6} \\[4pt] &={1}/{36} \end{align*}\] If we want the probability of getting one \(1\) and one \(3\) in two tosses, we must add to this the probability of tossing a \(3\) first and a \(1\) second. If there are \(\omega\) independent events (denoted \(E_1,E_2,\dots ,E_i,\dots ,E_{\omega }\)), the law of the probability of compound events asserts that the probability that \(E_1\) will occur in a first trial, and \(E_2\) will occur in a second trial, , is \[\begin{align*} P\left(E_1\ and\ E_2\ and\dots E_i\dots and\ E_{\omega }\right) &=P\left(E_1\right)\times P\left(E_2\right)\times \dots \times P\left(E_i\right)\times \dots \times P\left(E_{\omega }\right)\\[4pt] &=\prod^{\omega }_{i=1}{P(E_i)} \end{align*}\]
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The Maxwell-Boltzmann equation, which forms the basis of the kinetic theory of gases, defines the distribution of speeds for a gas at a certain temperature. From this distribution function, the most probable speed, the average speed, and the root-mean-square speed can be derived. The is used to determine the motion of a molecule of an ideal gas under a certain set of conditions. However, when looking at a mole of , it is impossible to measure the velocity of each molecule at every instant of time. Therefore, the Maxwell-Boltzmann distribution is used to determine how many molecules are moving between velocities \(v\) and \(v + dv\). Assuming that the one-dimensional distributions are independent of one another, that the velocity in the and directions does not affect the velocity, for example, the Maxwell-Boltzmann distribution is given by \[ \dfrac{dN}{N}= \left(\frac{m}{2\pi k_BT} \right)^{1/2}e^{\frac{-mv^2}{2k_BT}}dv \label{1} \] where Additionally, the function can be written in terms of the scalar quantity speed instead of the vector quantity velocity. This form of the function defines the distribution of the gas molecules moving at different speeds, between \(c_1\) and \(c_2\), thus \[f(c)=4\pi c^2 \left (\frac{m}{2\pi k_BT} \right)^{3/2}e^{\frac{-mc^2}{2k_BT}} \label{2} \] Finally, the Maxwell-Boltzmann distribution can be used to determine the distribution of the kinetic energy of for a set of molecules. The distribution of the kinetic energy is identical to the distribution of the speeds for a certain gas at any temperature. Figure 1 shows the Maxwell-Boltzmann distribution of speeds for a certain gas at a certain temperature, such as nitrogen at 298 K. The speed at the top of the curve is called the most probable speed because the largest number of molecules have that speed. Figure 2 shows how the Maxwell-Boltzmann distribution is affected by temperature. At lower temperatures, the molecules have less energy. Therefore, the speeds of the molecules are lower and the distribution has a smaller range. As the temperature of the molecules increases, the distribution flattens out. Because the molecules have greater energy at higher temperature, the molecules are moving faster. Figure 3 shows the dependence of the Maxwell-Boltzmann distribution on molecule mass. On average, heavier molecules move more slowly than lighter molecules. Therefore, heavier molecules will have a smaller speed distribution, while lighter molecules will have a speed distribution that is more spread out. Three speed expressions can be derived from the Maxwell-Boltzmann distribution: the most probable speed, the average speed, and the root-mean-square speed. The most probable speed is the maximum value on the distribution plot. This is established by finding the velocity when the following derivative is zero \[\dfrac{df(c)}{dc}|_{C_{mp}} = 0 \nonumber \] which is \[C_{mp}=\sqrt {\dfrac {2RT}{M}} \label{3a} \] The average speed is the sum of the speeds of all the molecules divided by the number of molecules. \[C_{avg}=\int_0^{\infty} c f(c) dc = \sqrt {\dfrac{8RT}{\pi M}} \label{3b} \] The root-mean-square speed is square root of the average speed-squared. \[ C_{rms}=\sqrt {\dfrac {3RT}{M}} \label{3c} \] where It follows that for gases that follow the Maxwell-Boltzmann distribution (if thermallized) \[C_{mp}< C_{avg}< C_{rms} \label{4} \] 1. 0.00141 2. \(3.92 \times 10^{20}\) argon molecules 3. C = 194.27 m/s C = 219.21 m/s C = 237.93 m/s 4. As stated above, C is the most probable speed, thus it will be at the top of the distribution curve. To the right of the most probable speed will be the average speed, followed by the root-mean-square speed. 5. Hint: Use the related speed expressions to determine the distribution of the gas molecules: helium at 500 K. helium at at 300 K. argon at 1000 K.
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"Oxidation state" implies a description that can change: a metal can go from one oxidation state to another. For example, a Cu(I) can become a Cu(0). It does so by an electron transfer from one place to another. In the case of Cu(I)/Cu(0), an electron would have to be donated by some other species. A loss of electrons is called an "oxidation", whereas a gain of electrons is called a "reduction" (as immortalized in the mnemonic: LEO the lion goes GER). Since an electron always goes from somewhere to somewhere else, one thing is always oxidized when something else is reduced. (Note that this is a little like proton transfer reactions: a proton is always transferred from one basic site to another, and is never really by itself.) These paired processes are called "reduction-oxidation" reactions, or "redox" for short. So the reduction of Cu(I) to Cu(0) is just a "half-reaction"; it needs a corresponding oxidation to make it happen. Li could donate an electron, for example, to become Li . In a biological setting, a Fe(III) in an important protein may need an extra electron to become Fe(II) in order to do its job (your life may be at stake here). It could get the electron from a nearby Cu(I), which becomes Cu(II). Balance the following half reactions by adding the right number of electrons to one side or the other. a) Cu → Cu(I) b) Fe(III) → Fe c) Mn → Mn d) Zn → Zn e) F → F f) H → H Sometimes, redox reactions work out very neatly: one participant needs an electron or two, and the other participant has one or two electrons to give. For example, a Cu ion in need of two electrons to become a Cu atom might get them from a Zn atom, which would become a Zn ion. In other words (or other symbols): \[ Cu^{2+} + 2e^- \rightarrow Cu \] \[ Zn^{2+} \rightarrow Zn + 2 e^- \] Adding those together: \[ Cu^{2+} + Zn \rightarrow Cu + Zn^{2+} \] Note that the electrons on each side just cancel each other, much like adding the same thing to both sides of an equals sign. Other times, things may be slightly more complicated. There may be an issue of conserving matter, for instance. For example, hydrogen gas, H , can be oxidised to give proton, H . We can't have more hydrogen atoms before the reaction than afterwards; matter can't just be created or destroyed. To solve that problem, the oxidation of hydrogen gas, H , produces two protons, not just one, and so two electrons are involved as well. Alternatively, two half-reactions may actually involve different numbers of electrons, and so proportions of each half reaction need to be adjusted in order to match the number of electrons properly. Put the following pairs of half reactions together to make a full reaction in each case. Make sure you have balanced the half reactions first. a) Cu(I) → Cu(II) and Fe(III) → Fe(II) b) Cu(I) → Cu(0) and Ag(0) → Ag(I) c) F → F and Fe → Fe d) Mo → Mo and Mn → Mn In many cases, redox reactions do not just involve simple metal ions or atoms. Often, the metal atom is found within a compound or a complex ion. For example, one of the most common oxidizing agents in common use is permanganate ion, MnO , which is usually converted to manganese dioxide, MnO during a reaction. That means the half reaction here is: \[ MnO_4^- \rightarrow MnO_2 \] In order to sort out how many electrons are being traded, we need to know the oxidation state of manganese before and after the reaction. That turns out to be Mn(VII) before and Mn(IV) afterwards. That means 3 electrons are added to permanganate to produce manganese dioxide. Now we have: \[ MnO_4^- + 3 e^- \rightarrow MnO_2 \] But now we just have a mass balance problem again. There are oxygen atoms before the reaction that have just disappeared after the reaction. Where could those oxygen atoms have gotten to? On this planet, the simplest answer to that question is always water. So maybe 2 waters were produced as part of the reaction. That gives us: \[ MnO_4^- + 3 e^- \rightarrow MnO_2 + 2 H_2O \] Only now we have more problems. First of all, now we have some hydrogen atoms on the right that we didn't have on the left. Where did these things come from? Also, there is this niggling problem of negative charges that we had before the reaction that we don't have after the reaction. Charge, like matter, doesn't just appear or disappear. It has to go someplace, and we have to explain where. We'll kill two birds with one stone. Maybe some protons were added to the reaction at the beginning, giving us those hydrogen atoms for the water and balancing out the charge. We are left with: \[ MnO_4^- + 3 e^- + 4 H^+ \rightarrow MnO_2 + 2H_2O \] It all works out. There are four negative charges on the left, and four plus charges, so no charge overall. There are no charges on the right. There is one manganese on the left and on the right. There are four oxygens on the left and on the right. There are four hydrogens on the left and on the right. Now let's take a little detour. We're going to go back in time, to the point where we realized we had a problem with our oxygen atoms. Now we have: \[ MnO_4^- + 3 e^- \rightarrow MnO_2 \] But now we just have a mass balance problem again. There are oxygen atoms before the reaction that have just disappeared after the reaction. Where could those oxygen atoms have gotten to? And while we're at it, there are four negative charges on the left and none on the right. Charge doesn't just appear or disappear during a reaction. It has to be balanced. One solution for this problem involves the production of hydroxide ions, HO , during the reaction. Hydroxide ions are pretty common; there are a few in every glass of water. The charge in the reaction would be balanced if four hydroxide ions were produced, and it would explain where those oxygen atoms went. That gives us: \[ MnO_4^- + 3 e^- \rightarrow MnO_2 + 4 OH^- \] Now the charge is balanced! But the oxygen atoms aren't. And where did those hydrogen atoms come from? Well, on this planet, a good source of hydrogen and oxygen atoms is water. Maybe the reaction needs water. That means: \[ MnO_4^- + 3 e^- + 2 H_2O \rightarrow MnO_2 + 4 OH^- \] One manganese on each side. Four hydrogens on each side. Six oxygens on each side. Four negative charges on each side. Nothing has appeared or disappeared during the reaction. We know where everything went. There is actually a shortcut to get to this solution. If we already know how to balance the reaction in acididc media, we just add enough hydroxides to neutralize the acid (H + OH = H O). But we have to add those hydroxides . Some of the waters will then cancel out to leave the balanced reaction. Start with acid \[ MnO_4^- + 3 e^- + 4 H^+ \rightarrow MnO_2 + 2 H_2 O \] Add base to both sides \[ MnO_4^- + 3 e^- + 4 H^+ + 4OH^- \rightarrow MnO_2 + 2 H_2O + 4 ^- OH \] Neutralize \[ MnO_4^- + 3 e^- + 4 H_2O \rightarrow MnO_2 + 2 H_2O + 4 OH^- \] Cancel the extra waters \[ MnO_4^- + 3 e^- + 2 H_2O \rightarrow MnO_2 + 4 OH^- \] Now we're back. We have seen two different outcomes to this problem. The moral of the story is that sometimes there is more than one right answer. In the case of redox reactions, sometimes things happen a little differently depending on whether things are occurring under acidic conditions (meaning, in this context, that there are lots of protons around) or in basic conditions (meaning there aren't many protons around but there is hydroxide ion). Apart from helping us to keep track of things, the presence of acids (protons) and bases (hydroxide ions) in redox reactions are common in reality. For example, batteries rely on redox reactions to produce electricity; the electricity is just a current of electrons trying to get from one site to another to carry out a redox reaction. Many batteries, such as car batteries, contain acid. Other batteries, like "alkaline" batteries, for example, contain hydroxide ion. Balance the following half reactions. Assume the reactions are in acidic conditions. a)MnO → Mn(OH) b) NO → N O c)HPO → H PO d) Sn(OH) → HSnO ,
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Electron transfer is one of the most basic processes that can happen in chemistry. It simply involves the movement of an electron from one atom to another. Many important biological processes rely on electron transfer, as do key industrial transformations used to make valuable products. In biology, for example, electron transfer plays a central role in respiration and the harvesting of energy from glucose, as well as the storage of energy during photosynthesis. In society, electron transfer has been used to obtain metals from ores since the dawn of civilization. Oxidation state is a useful tool for keeping track of electron transfers. It is most commonly used in dealing with metals and especially with transition metals. Unlike metals from the first two columns of the periodic table, such as sodium or magnesium, transition metals can often transfer different numbers of electrons, leading to different metal ions. Sodium is generally found as Na and magnesium is almost always Mg , but manganese could be Mn , Mn , and so on, as far as Mn . At first glance, "oxidation state" is often synonymous with "charge on the metal". However, there is a subtle difference between the two terms. For example, in a coordination complex, a metal atom that is ostensibly an ion with a charge of +2 may have very little charge on it at all. Instead, the positive charge may be delocalized onto the ligands that are donating their own electrons to the metal. Oxidation state is used instead to describe what the charge on the metal ion would be if the coordinated ligands were removed and the metal ion left by itself. Oxidation state is commonly denoted by Roman numerals after the symbol for the metal atom. This designation can be shown either as a superscript, as in Mn , or in parentheses, as in Mn(II); both of these descriptions refer to a Mn ion, or what might have been a Mn ion before it got into a bonding situation. Translate the following oxidation state descriptions into charges on the metal. a) Ag b) Ni(II) c) Mn d) Cr(VI) e) Cu(III) f) Fe g) Os h) Re(V) The oxidation state of a metal within a compound can be determined only after the other components of the compound have been conceptually removed. For example, metals are frequently found in nature as oxides. An oxide anion is O , so every oxygen in a compound will correspond to an additional 2- charge. In order to balance charge, the metal must have a corresponding plus charge. For example, sodium oxide has the formula Na O. If the oxygen ion is considered to have a 2- charge, and there is no charge overall, there must be a corresponding charge of +2. That means each sodium ion has a charge of +1. Determine the charge on the metal in each of the following commercially valuable ores. Note that sulfur, in the same column of the periodic table as oxygen, also has a 2- charge as an anion. a) , PbS b) , SnO c) , HgS d) , FeS e) , Fe O f) , Fe O is a common zinc ore, ZnS. However, sphalerite always has small, variable fractions of iron in place of some of the zinc in its structure. What is the likely oxidation state of the iron? Sometimes non-metals such as carbon are thought of in different oxidations states, too. For example, the coke used in smelting metal ores is roughly C, in oxidation state 0. Determine the oxidation state of carbon in each case, assuming oxygen is always 2- and hydrogen is always 1+. a) carbon monoxide, CO b) carbon dioxide, CO c) methane, CH d) formaldehyde, H CO e) oxalate, C O Sometimes it is useful to know the charges and structures of some of the earth's most common anions. Draw Lewis structures for the following anions: a) hydroxide, HO b) carbonate, CO c) sulfate, SO d) nitrate, NO e) phosphate, PO f) silicate, SiO g) inosilicate, SiO Use your knowledge of common anions to determine the oxidation states on the metals in the following ores. a) , MgCO b) , Cu CO (OH) c) , MnO(OH) d) , CaSO e) , MnCO f) , MnSiO In mixed-metal species, the presence of two different metals may make it difficult to assign oxidation states to each. For the following ores, propose one solution for the oxidation states of the metals. a) , CuFeS b) , ZnFe O c) , Be Al (SiO ) d) or peacock ore, Cu FeS e) , CuAl (PO ) (OH) are believed to make up about 60% of the earth's crust. The alkali, alkaline earth and aluminum metals in these tectosilicates are typically found in their highest possible oxidation states. What are the charges on the silicates in the following examples? a) , KAlSi O b) , CaAl Si O c) , BaAl Si O d) , NaAlSi O Frequently, minerals are solid solutions in which repeating units of different compositions are mixed together homogeneously. For example, is a variation of anorthite in which about 50% of the aluminum ions are replaced by silicon ions and about 50% of the calcium ions are replaced by sodium ions. Show that this composition would still be charge neutral overall. a) Cu → Cu(I) + e b) Fe(III) + 3 e → Fe c) Mn → Mn(III) + 3 e d) Zn(II) + 2 e → Zn e) 2 F → F + 2 e f) H → 2 H + 2 e ,
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The Ideal Gas Law is very simply expressed: \[ PV=nRT\] from which simpler gas laws such as Boyle's, Charles's, Avogadro's and Amonton's law be derived. When dealing with gas, a famous equation was used to relate all of the factors needed in order to solve a gas problem. This equation is known as the Ideal Gas Equation. As we have always known, anything ideal does not exist. In this issue, two well-known assumptions should have been made beforehand: An is a hypothetical gas dreamed by chemists and students because it would be much easier if things like intermolecular forces do not exist to complicate the simple . Ideal gases are essentially point masses moving in constant, random, straight-line motion. Its behavior is described by the assumptions listed in the . This definition of an ideal gas contrasts with the Non-Ideal Gas definition, because this equation represents how gas actually behaves in reality. For now, let us focus on the Ideal Gas. We must emphasize that this gas law is . As students, professors, and chemists, we sometimes need to understand the concepts before we can apply it, and assuming the gases are in an ideal state where it is unaffected by real world conditions will help us better understand the behavior the gases. In order for a gas to be , its behavior must follow the whereas the will deviate from this theory due to real world conditions. Before we look at the , let us state the four gas variables and one constant for a better understanding. The four gas variables are: (P), (V), (n), and (T). Lastly, the constant in the equation shown below is R, known as the the , which will be discussed in depth further later: \[ PV=nRT \] Another way to describe an ideal gas is to describe it in mathematically. Consider the following equation: \[ \dfrac{PV}{nRT}=1 \] The term \(\frac{pV}{nRT}\) is also called the and is a measure of the ideality of the gas. An ideal gas will always equal 1 when plugged into this equation. The greater it deviates from the number 1, the more it will behave like a real gas rather than an ideal. A few things should always be kept in mind when working with this equation, as you may find it extremely helpful when checking your answer after working out a gas problem. The Ideal Gas Law is simply the combination of all Simple Gas Laws (Boyle's Law, Charles' Law, and Avogadro's Law), and so learning this one means that you have learned them all. The Simple Gas Laws can always be derived from the describes the inverse proportional relationship between pressure and volume at a constant temperature and a fixed amount of gas. This law came from a manipulation of the Ideal Gas Law. \[ P \propto \dfrac{1}{V} \] or expressed from two pressure/volume points: \[ P_1V_1=P_2V_2 \] This equation would be ideal when working with problem asking for the initial or final value of pressure or volume of a certain gas when one of the two factor is missing. describes the directly proportional relationship between the volume and temperature (in Kelvin) of a fixed amount of gas, when the pressure is held constant. \[ V\propto \; T \] or express from two volume/temperature points: \[ \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2} \] This equation can be used to solve for initial or final value of volume or temperature under the given condition that pressure and the number of mole of the gas stay the same. Volume of a gas is directly proportional to the amount of gas at a constant temperature and pressure. \[ V \propto \; n\] or expressed as a two volume/number points: \[ \dfrac{V_1}{n_1}=\dfrac{V_2}{n_2} \] can apply well to problems using (see below), because of a set amount of pressure and temperature. Given a constant number of mole of a gas and an unchanged volume, pressure is directly proportional to temperature. \[ P \propto \; T\] or expressed as two pressure/temperature points: \[ \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2} \] Boyle's Law, Charles' Law, and Avogradro's Law and Amontons's Law are given under certain conditions so directly combining them will not work. Through advanced mathematics (provided in outside link if you are interested), the properties of the three simple gas laws will give you the Ideal Gas Equation. The table below lists the different units for each property. Use the following table as a reference for pressure. *note: This is the SI unit for pressure Here comes the tricky part when it comes to the , R. Value of R change when dealing with different unit of pressure and volume (Temperature factor is overlooked because temperature will always be in Kelvin instead of Celsius when using the Ideal Gas equation). Only through appropriate value of R will you get the correct answer of the problem. It is simply a constant, and the different values of R correlates accordingly with the units given. When choosing a value of R, choose the one with the appropriate units of the given information (sometimes given units must be converted accordingly). Here are some commonly used values of R: *note: This is the SI unit for the gas constant So, which value of R should I use? Because of the various value of R you can use to solve a problem. It is crucial to match your units of Pressure, Volume, number of mole, and Temperature with the units of R. How do you know the Ideal Gas Equation is the correct equation to use? Use the Ideal Gas Equation to solve a problem when the and . There are various type of problems that will require the use of the Ideal Gas Equation. Other things to keep in mind: Know what Standard Temperature and Pressure (STP) values are. Know how to do . Know your basic equations. Take a look at the problems below for examples of each different type of problem. Attempt them initially, and if help is needed, the solutions are right below them. Remark: The units must cancel out to get the appropriate unit; knowing this will help you double check your answer. 5.0 g of neon is at 256 mm Hg and at a temperature of 35º C. What is the volume? Pressure: \( 256 \; \rm{mmHg} \times (1 \; \rm{atm/} 760 \; \rm{mmHg}) = 0.3368 \; \rm{atm} \) Moles: \( 5.0 \; \rm{g}\; Ne \times (1 \; \rm{mol} / 20.1797\; \rm{g}) = 0.25 \; \rm{mol}\; \rm{Ne} \) Temperature: \(35º C + 273 = 308 \; \rm{K} \) \[ V = (nRT/P) \] \[ V = \dfrac{(0.25\; \rm{mol})(0.08206\; \rm{L atm}/\rm{K mol})(308\; \rm{K})}{(0.3368\; \rm{atm})}] \] \[ V = 19\; \rm{L}\] What is a gas’s temperature in Celsius when it has a volume of 25 L, 203 mol, 143.5 atm? \[T = \dfrac{PV}{nR}\] \[T = \dfrac{(143.5\; \rm{atm})(25\; \rm{L})}{(203 \; \rm{mol})(0.08206 L•atm/K mol)}\] \[T = 215.4\; \rm{K}\] \[215.4 K - 273 = -57.4º C\] What is the density of nitrogen gas (\(N_2\)) at 248.0 Torr and 18º C? \[(248 \; \rm{Torr}) \times \dfrac{1 \; \rm{atm}}{760 \; \rm{Torr}} = 0.3263 \; \rm{atm}\] \[18ºC + 273 = 291 K\] *Write down all known equations: \[PV = nRT\] \[\rho=\dfrac{m}{V}\] \[m=M \times n\] *Now take the density equation. \[\rho=\dfrac{m}{V}\] *Keeping in mind \(m=M \times n\)...replace \((M \times n)\) for \(mass\) within the density formula. \[\rho=\dfrac{M \times n}{V}\] \[\dfrac{\rho}{M} = \dfrac{n}{V}\] *Now manipulate the Ideal Gas Equation \(PV = nRT\) \[\dfrac{n}{V} = \dfrac{P}{RT}\] *\((n/V)\) is in both equations. \[\dfrac{n}{V} = \dfrac{\rho}{M}\] \[\dfrac{n}{V} = \dfrac{P}{RT}\] *Now combine them please. \[\dfrac{\rho}{M} = \dfrac{P}{RT}\] *Isolate density. \[\rho = \dfrac{PM}{RT}\] \[\rho = \dfrac{PM}{RT}\] \[\rho = \dfrac{(0.3263\; \rm{atm})(2*14.01 \; \rm{g/mol})}{(0.08206 L atm/K mol)(291 \; \rm{K})}\] \[\rho = 0.3828 \; g/L\] Find the volume, in mL, when 7.00 g of \(O_2\) and 1.50 g of \(Cl_2\) are mixed in a container with a pressure of 482 atm and at a temperature of 22º C. \[n_{total} = n_{O_2}+ n_{Cl_2}\] \[= \left[7.0 \; \rm{g} \; O_2 \times \dfrac{1 \; \rm{mol} \; O_2}{32.00 \; \rm{g} \; O_2}\right] + \left[1.5 \; \rm{g}\; Cl_2 \times \dfrac{1 \; \rm{mol} \; Cl_2}{70.905 \; \rm{g} \; Cl_2}\right]\] \[= 0.2188 \; \rm{mol} \; O_2 + 0.0212 \; \rm{mol} \; Cl_2\] \[= 0.24 \; \rm{mol}\] \[V= \dfrac{nRT}{P}\] \[V= \dfrac{(0.24\; \rm{mol})(0.08206 L atm/K mol)(295\; \rm{K})}{(482\; \rm{atm})}\] \[V= 0.0121\; \rm{L}\] \[0.0121\; \rm{L} \times \dfrac{1000\; \rm{ml}}{1\; \rm{L}} = 12.1\; \rm{mL}\] A 3.00 L container is filled with \(Ne_{(g)}\) at 770 mmHg at 27 C. A \(0.633\;\rm{g}\) sample of \(CO_2\) vapor is then added. What is the partial pressure of \(CO_2\) and \(Ne\) in atm? What is the total pressure in the container in atm? , Before: Other Unknowns: \(n_{CO_2}\)= ? \[n_{CO_2} = 0.633\; \rm{g} \;CO_2 \times \dfrac{1 \; \rm{mol}}{44\; \rm{g}} = 0.0144\; \rm{mol} \; CO_2\] \[n_{Ne} = \dfrac{PV}{RT}\] \[n_{Ne} = \dfrac{(1.01\; \rm{atm})(3.00\; \rm{L})}{(0.08206\;atm\;L/mol\;K)(300\; \rm{K})}\] \[n_{Ne} = 0.123 \; \rm{mol}\] Because the pressure of the container before the \(CO_2\) was added contained only \(Ne\), that is your partial pressure of \(Ne\). After converting it to atm, you have already answered part of the question! \[P_{Ne} = 1.01\; \rm{atm}\] Step 3: Now that have pressure for Ne, you must find the partial pressure for \(CO_2\). Use the ideal gas equation. \[ \dfrac{P_{Ne}V}{n_{Ne}RT} = \dfrac{P_{CO_2}V}{n_{CO_2}RT}\] but because both gases share the same Volume (\(V\)) and Temperature (\(T\)) and since the Gas Constant (\(R\)) is constants, all three terms cancel and can be removed them from the equation. \[\dfrac{P}{n_{Ne}} = \dfrac{P}{n_{CO_2}}\] \[\dfrac{1.01 \; \rm{atm}}{0.123\; \rm{mol} \;Ne} = \dfrac{P_{CO_2}}{0.0144\; \rm{mol} \;CO_2} \] \[P_{CO_2} = 0.118 \; \rm{atm}\] \(CO_2\). \[P_{total}= P_{Ne} + P_{CO_2}\] \[P_{total}= 1.01 \; \rm{atm} + 0.118\; \rm{atm}\] \[P_{total}= 1.128\; \rm{atm} \approx 1.13\; \rm{atm} \; \text{(with appropriate significant figures)} \]
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Elemental iodine is a dark grey solid with a faint metallic luster. When heated at ordinary air pressures it sublimes to a violet gas. The name iodine is taken from the Greek ioeides which means "violet colored". It was discovered in 1811 by Courtois. Commercially iodine is recovered from seaweed and brines. It is an important trace element in the human diet, required for proper function of the thyroid gland. Thus iodine is added to table salt ("iodized") to insure against iodine deficiencies. Radioactive isotopes of iodine are used in medical tracer work involving the thyroid and also to treat diseases of that gland.
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This page describes the structures of the Period 3 elements from sodium to argon, and shows how these structures can be used to explain the physical properties of the elements. In a moment we shall explain all the ups and downs in this graph. Sodium, magnesium and aluminum are all good conductors of electricity. Silicon is a semiconductor. None of the rest conduct electricity. These trends are explained below. Sodium, magnesium and aluminum all have metallic structures, which accounts for their electrical conductivity and relatively high melting and boiling points. Melting and boiling points rise across the three metals because of the increasing number of electrons which each atom can contribute to the delocalized "sea of electrons". The atoms also get smaller and have more protons as you go from sodium to magnesium to aluminum. The attractions and therefore the melting and boiling points increase because: Silicon is a non-metal, and has a giant covalent structure exactly the same as carbon in diamond - hence the high melting point. You have to break strong covalent bonds in order to melt it. There are no obviously free electrons in the structure, and although it conducts electricity, it doesn't do so in the same way as metals. Silicon is a semiconductor. Phosphorus, sulfur, chlorine and argon are simple molecular substances with only van der Waals attractions between the molecules. Their melting or boiling points will be lower than those of the first four members of the period which have giant structures. The presence of individual molecules prevents any possibility of electrons flowing, and so none of them conduct electricity. The sizes of the melting and boiling points are governed entirely by the sizes of the molecules: Argon molecules consist of single argon atoms. Jim Clark ( )
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The Shroud of Turin is a linen wrapping cloth that appears to possess the image of Jesus Christ. Some people believe this to be the cloth that he was wrapped in following his crucifixion. In 1988, several groups of scientists were allowed samples of the shroud to subject these samples to C dating. The carbon-14 to carbon-12 ratio was found to be 92% of that in living organisms. On the above graph, which depicts the decay curve for carbon-14, you can draw a line from 1988 up to the curve and then from this intersection over to the percent value on the Y axis. The value obtained from the curve is about 80%. This means that the Shroud of Turin may be younger than was previously thought. The age of the shroud can be obtained from the graph by starting at 92% and drawing a line to the curve. Draw a line from this intersection down to the years and the value obtained is about 1000 AD, which means that the Shroud of Turin was probably created in the Middle Ages.
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Electrons with more than one atom, such as Helium (He), and Nitrogen (N), are referred to as multi-electron atoms. Hydrogen is the only atom in the periodic table that has one electron in the orbitals under ground state. We will learn how additional electrons behave and affect a certain atom. Let's review some basic concepts. First, electrons repel against each other. Particles with the same charge repel each other, while oppositely charged particles attract each other. For example, a proton, which is positively charged, is attracted to electrons, which are negatively charged. However, if we put two electrons together or two protons together, they will repel one another. Since neutrons lack a charge, they will neither repel nor attract protons or electrons. a) The two electrons are placed together and repel each other because of the same charge. b) The two protons are repelling each other for the same reason. c) When oppositely charged particles, an electron and a proton, are placed together, they attract each other. Protons and neutrons are located in an atom's nucleus. Electrons float around the atom in energy levels. Energy levels consist of orbitals and sub-orbitals. The lower the energy level the electron is located at, the closer it is to nucleus. As we go down and to the right of the periodic table, the number of electrons that an element has increases. Since there are more electrons, the atom experiences greater repulsion and electrons will tend to stay as far away from each other as possible. Our main focus is what effects take place when more electrons surround the nucleus. To better understand the following concepts it is a good idea to first review . With more protons in the nucleus, the attractive force for electrons to the nucleus is stronger. Thus, the orbital energy becomes more negative (less energy). Orbital energy also depends on the type of l orbital an electron is located in. The lower the number of l, the closer it is to the nucleus. For example, l=0 is the s orbital. S orbitals are closer to the nucleus than the p orbitals (l=1) that are closer to the nucleus than the d orbitals (l=2) that are closer to the f orbitals (l=3). More electrons create the or screening effect. means the electrons closer to the nucleus block the outer valence electrons from getting close to the nucleus. See figure 2. Imagine being in a crowded auditorium in a concert. The enthusiastic fans are going to surround the auditorium, trying to get as close to the celebrity on the stage as possible. They are going to prevent people in the back from seeing the celebrity or even the stage. This is the shielding or screening effect. The stage is the nucleus and the celebrity is the protons. The fans are the electrons. Electrons closest to the nucleus will try to be as close to the nucleus as possible. The outer/valence electrons that are farther away from the nucleus will be shielded by the inner electrons. Therefore, the inner electrons prevent the outer electrons from forming a strong attraction to the nucleus. The degree to which the electrons are being screened by inner electrons can be shown by ns<np<nd<nf where n is the energy level. The inner electrons will be attracted to the nucleus much more than the outer electrons. Thus, the attractive forces of the valence electrons to the nucleus are reduced due to the shielding effects. That is why it is easier to than the inner electrons. It also reduces the nuclear charge of an atom. is the ability of an electron to get close to the nucleus. The penetration of ns > np > nd > nf. Thus, the closer the electron is to the nucleus, the higher the penetration. Electrons with higher penetration will shield outer electrons from the nucleus more effectively. The s orbital is closer to the nucleus than the p orbital. Thus, electrons in the s orbital have a higher penetration than electrons in the p orbital. That is why the s orbital electrons shield the electrons from the p orbitals. Electrons with higher penetration are closer to the nucleus than electrons with lower penetration. Electrons with lower penetration are being shielded from the nucleus more. is a type of probability to find where an electron is mostly likely going to be in an atom. The higher the penetration, the higher probability of finding an electron near the nucleus. As shown by the graphs, electrons of the s orbital are found closer to the nucleus than the p orbital electrons. Likewise, the lower the energy level an electron is located at, the higher chance it has of being found near the nucleus. The smaller the energy level (n) and the orbital angular momentum quantum number (l) of an electron is, the more likely it will be near the nucleus. As electrons get to higher and higher energy levels, the harder it is to locate it because the radius of the sphere is greater. Thus, the probability of locating an electron will be more difficult. Radial probability distribution can be found by multiplying By using the radial probability distribution equation, we can get a better understanding about an electron's behavior, as shown on Figures 3.1-3.3. Figures 3.1, 3.2, and 3.3 show that the lower the energy level, the higher the probability of finding the electron close to the nucleus. Also, the lower momentum quantum number gets, the closer it is to the nucleus.
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We have seen that the strengths of acids and bases vary over many orders of magnitude. In this section, we explore some of the structural and electronic factors that control the acidity or basicity of a molecule. In general, the stronger the \(\ce{A–H}\) or \(\ce{B–H^+}\) bond, the less likely the bond is to break to form \(H^+\) ions and thus the less acidic the substance. This effect can be illustrated using the hydrogen halides: The trend in bond energies is due to a steady decrease in overlap between the 1s orbital of hydrogen and the valence orbital of the halogen atom as the size of the halogen increases. The larger the atom to which H is bonded, the weaker the bond. Thus the bond between H and a large atom in a given family, such as I or Te, is weaker than the bond between H and a smaller atom in the same family, such as F or O. As a result, acid strengths of binary hydrides increase as we go down a column of the periodic table. For example, the order of acidity for the binary hydrides of elements is as follows, with \(pK_a\) values in parentheses: \[H_2O (14.00 = pK_w) < H_2S (7.05) < H_2Se (3.89) < H_2Te (2.6) \label{1}\] Whether we write an acid–base reaction as \(AH \rightleftharpoons A^−+H^+\) or as \(BH^+ \rightleftharpoons B+H^+\), the conjugate base (\(A^−\) or \(B\)) contains one more lone pair of electrons than the parent acid (\(AH\) or \(BH^+\)). Any factor that stabilizes the lone pair on the conjugate base favors dissociation of \(H^+\) and makes the parent acid a stronger acid. Let’s see how this explains the relative acidity of the binary hydrides of the elements in the second row of the periodic table. The observed order of increasing acidity is the following, with pKa values in parentheses: \[CH_4 (~50) \ll NH_3 (~36) < H_2O (14.00) < HF (3.20) \label{2}\] Consider, for example, the compounds at both ends of this series: methane and hydrogen fluoride. The conjugate base of \(CH_4\) is \(CH_3^−\), and the conjugate base of \(HF\) is \(F^−\). Because fluorine is much more electronegative than carbon, fluorine can better stabilize the negative charge in the \(F^−\) ion than carbon can stabilize the negative charge in the CH3− ion. Consequently, \(\ce{HF}\) has a greater tendency to dissociate to form \(H^+\) and \(F^−\) than does methane to form \(H^+\) and \(CH_3^−\), making HF a much stronger acid than \(CH_4\). The same trend is predicted by analyzing the properties of the conjugate acids. For a series of compounds of the general formula \(HE\), as the electronegativity of E increases, the E–H bond becomes more polar, favoring dissociation to form \(E^−\) and \(H^+\). Due to both the increasing stability of the conjugate base and the increasing polarization of the E–H bond in the conjugate acid, acid strengths of binary hydrides increase as we go from left to right across a row of the periodic table. Acid strengths of binary hydrides increase as we go down a column or from left to right across a row of the periodic table. The stornger acid, the weaker the covalent bond to a hydrogen atom. So the strongest acid possible is the molecule with the weakest bond. That is the hydrohelium (1+) cation, \(\ce{HeH^{+}}\), which is a positively charged ion formed by the reaction of a proton with a helium atom in the gas phase. It was first produced in the laboratory in 1925 and is isoelectronic with molecular hydrogen (\ce{H2}}). It is the strongest known acid, with a proton affinity of 177.8 kJ/mol. \(\ce{HeH^{+}}\) cannot be prepared in a condensed phase, as it would protonate any anion, molecule or atom with which it were associated. However it is possible to estimate a aqueous acidity using : A free energy change of dissociation of −360 kJ/mol is equivalent to a p of −63. It has been suggested that \(\ce{HeH^{+}}\) should occur naturally in the interstellar medium, but it has not yet been detected. Atoms or groups of atoms in a molecule other than those to which H is bonded can induce a change in the distribution of electrons within the molecule. This is called an inductive effect, and, much like the coordination of water to a metal ion, it can have a major effect on the acidity or basicity of the molecule. For example, the hypohalous acids (general formula HOX, with X representing a halogen) all have a hydrogen atom bonded to an oxygen atom. In aqueous solution, they all produce the following equilibrium: \[ HOX_{(aq)} \rightleftharpoons H^+_{(aq)} + OX^−{(aq)} \label{3}\] The acidities of these acids vary by about three orders of magnitude, however, due to the difference in electronegativity of the halogen atoms: As the electronegativity of \(X\) increases, the distribution of electron density within the molecule changes: the electrons are drawn more strongly toward the halogen atom and, in turn, away from the H in the O–H bond, thus weakening the O–H bond and allowing dissociation of hydrogen as \(H^+\). The acidity of oxoacids, with the general formula \(HOXO_n\) (with \(n\) = 0−3), depends strongly on the number of terminal oxygen atoms attached to the central atom \(X\). As shown in Figure \(\Page {1}\), the \(K_a\) values of the oxoacids of chlorine increase by a factor of about \(10^4\) to \(10^6\) with each oxygen as successive oxygen atoms are added. The increase in acid strength with increasing number of terminal oxygen atoms is due to both an inductive effect and increased stabilization of the conjugate base. Any inductive effect that withdraws electron density from an O–H bond increases the acidity of the compound. Because oxygen is the second most electronegative element, adding terminal oxygen atoms causes electrons to be drawn away from the O–H bond, making it weaker and thereby increasing the strength of the acid. The colors in Figure \(\Page {1}\) show how the electrostatic potential, a measure of the strength of the interaction of a point charge at any place on the surface of the molecule, changes as the number of terminal oxygen atoms increases. In Figure \(\Page {1}\) and Figure \(\Page {2}\), blue corresponds to low electron densities, while red corresponds to high electron densities. The oxygen atom in the O–H unit becomes steadily less red from \(HClO\) to \(HClO_4\) (also written as \(HOClO_3\), while the H atom becomes steadily bluer, indicating that the electron density on the O–H unit decreases as the number of terminal oxygen atoms increases. The decrease in electron density in the O–H bond weakens it, making it easier to lose hydrogen as \(H^+\) ions, thereby increasing the strength of the acid. At least as important, however, is the effect of delocalization of the negative charge in the conjugate base. As shown in Figure \(\Page {2}\), the number of resonance structures that can be written for the oxoanions of chlorine increases as the number of terminal oxygen atoms increases, allowing the single negative charge to be delocalized over successively more oxygen atoms. Electron delocalization in the conjugate base increases acid strength. The electrostatic potential plots in Figure \(\Page {2}\) demonstrate that the electron density on the terminal oxygen atoms decreases steadily as their number increases. The oxygen atom in ClO− is red, indicating that it is electron rich, and the color of oxygen progressively changes to green in \(ClO_4^+\), indicating that the oxygen atoms are becoming steadily less electron rich through the series. For example, in the perchlorate ion (\(ClO_4^−\)), the single negative charge is delocalized over all four oxygen atoms, whereas in the hypochlorite ion (\(OCl^−\)), the negative charge is largely localized on a single oxygen atom (Figure \(\Page {2}\)). As a result, the perchlorate ion has no localized negative charge to which a proton can bind. Consequently, the perchlorate anion has a much lower affinity for a proton than does the hypochlorite ion, and perchloric acid is one of the strongest acids known. As the number of terminal oxygen atoms increases, the number of resonance structures that can be written for the oxoanions of chlorine also increases, and the single negative charge is delocalized over more oxygen atoms. As these electrostatic potential plots demonstrate, the electron density on the terminal oxygen atoms decreases steadily as their number increases. As the electron density on the oxygen atoms decreases, so does their affinity for a proton, making the anion less basic. As a result, the parent oxoacid is more acidic. Similar inductive effects are also responsible for the trend in the acidities of oxoacids that have the same number of oxygen atoms as we go across a row of the periodic table from left to right. For example, \(H_3PO_4\) is a weak acid, \(H_2SO_4\) is a strong acid, and \(HClO_4\) is one of the strongest acids known. The number of terminal oxygen atoms increases steadily across the row, consistent with the observed increase in acidity. In addition, the electronegativity of the central atom increases steadily from P to S to \(Cl\), which causes electrons to be drawn from oxygen to the central atom, weakening the \(\ce{O–H}\) bond and increasing the strength of the oxoacid. Careful inspection of the data in Table \(\Page {1}\) shows two apparent anomalies: carbonic acid and phosphorous acid. If carbonic acid \((H_2CO_3\)) were a discrete molecule with the structure \(\ce{(HO)_2C=O}\), it would have a single terminal oxygen atom and should be comparable in acid strength to phosphoric acid (\(H_3PO_4\)), for which pKa1 = 2.16. Instead, the tabulated value of \(pK_{a1}\) for carbonic acid is 6.35, making it about 10,000 times weaker than expected. As we shall see, however, \(H_2CO_3\) is only a minor component of the aqueous solutions of \(CO_2\) that are referred to as carbonic acid. Similarly, if phosphorous acid (\(H_3PO_3\)) actually had the structure \((HO)_3P\), it would have no terminal oxygen atoms attached to phosphorous. It would therefore be expected to be about as strong an acid as \(HOCl\) (pKa = 7.40). In fact, the \(pK_{a1}\) for phosphorous acid is 1.30, and the structure of phosphorous acid is \(\ce{(HO)_2P(=O)H}\) with one H atom directly bonded to P and one \(\ce{P=O}\) bond. Thus the pKa1 for phosphorous acid is similar to that of other oxoacids with one terminal oxygen atom, such as \(H_3PO_4\). Fortunately, phosphorous acid is the only common oxoacid in which a hydrogen atom is bonded to the central atom rather than oxygen. Inductive effects are also observed in organic molecules that contain electronegative substituents. The magnitude of the electron-withdrawing effect depends on both the nature and the number of halogen substituents, as shown by the pKa values for several acetic acid derivatives: \[pK_a CH_3CO_2H 4.76< CH_2ClCO_2H 2.87<CHCl_2CO_2H 1.35<CCl_3CO_2H 0.66<CF_3CO_2H 0.52 \nonumber\] As you might expect, fluorine, which is more electronegative than chlorine, causes a larger effect than chlorine, and the effect of three halogens is greater than the effect of two or one. Notice from these data that inductive effects can be quite large. For instance, replacing the \(\ce{–CH_3}\) group of acetic acid by a \(\ce{–CF_3}\) group results in about a 10,000-fold increase in acidity! Arrange the compounds of each series in order of increasing acid or base strength. The structures are shown here. : series of compounds : relative acid or base strengths : Use relative bond strengths, the stability of the conjugate base, and inductive effects to arrange the compounds in order of increasing tendency to ionize in aqueous solution. : Although both sulfuric acid and sulfurous acid have two –OH groups, the sulfur atom in sulfuric acid is bonded to two terminal oxygen atoms versus one in sulfurous acid. Because oxygen is highly electronegative, sulfuric acid is the stronger acid because the negative charge on the anion is stabilized by the additional oxygen atom. In comparing sulfuric acid and fluorosulfonic acid, we note that fluorine is more electronegative than oxygen. Thus replacing an –OH by –F will remove more electron density from the central S atom, which will, in turn, remove electron density from the S–OH bond and the O–H bond. Because its O–H bond is weaker, \(FSO_3H\) is a stronger acid than sulfuric acid. The predicted order of acid strengths given here is confirmed by the measured pKa values for these acids:  \[pKa H_2SO_3 1.85<H_2SO_4^{−2} < FSO_3H−10 \nonumber\] The structures of both trifluoramine and hydroxylamine are similar to that of ammonia. In trifluoramine, all of the hydrogen atoms in NH3 are replaced by fluorine atoms, whereas in hydroxylamine, one hydrogen atom is replaced by OH. Replacing the three hydrogen atoms by fluorine will withdraw electron density from N, making the lone electron pair on N less available to bond to an \(H^+\) ion. Thus \(NF_3\) is predicted to be a much weaker base than \(NH_3\). Similarly, because oxygen is more electronegative than hydrogen, replacing one hydrogen atom in \(NH_3\) by \(OH\) will make the amine less basic. Because oxygen is less electronegative than fluorine and only one hydrogen atom is replaced, however, the effect will be smaller. The predicted order of increasing base strength shown here is confirmed by the measured \(pK_b\) values:   \[pK_bNF_3—<<NH_2OH 8.06<NH_3 4.75 \nonumber\] Trifluoramine is such a weak base that it does not react with aqueous solutions of strong acids. Hence its base ionization constant has never been measured. Arrange the compounds of each series in order of \(HClO-3 > CH_3PO_3H_2 > H_3PO_4\) \(CF_3S^− < CH_3S^− < OH^−\)​​ Inductive effects and charge delocalization significantly influence the acidity or basicity of a compound. The acid–base strength of a molecule depends strongly on its structure. The weaker the A–H or B–H+ bond, the more likely it is to dissociate to form an \(H^+\) ion. In addition, any factor that stabilizes the lone pair on the conjugate base favors the dissociation of \(H^+\), making the conjugate acid a stronger acid. Atoms or groups of atoms elsewhere in a molecule can also be important in determining acid or base strength through an inductive effect, which can weaken an \(\ce{O–H}\) bond and allow hydrogen to be more easily lost as \(H^+\) ions.
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This page is a basic introduction to the electromagnetic spectrum sufficient for chemistry students interested in UV-visible absorption spectroscopy. If you are looking for any sort of explanations suitable for physics courses, then I'm afraid this isn't the right place for you. Any wave is essentially just a way of shifting energy from one place to another - whether the fairly obvious transfer of energy in waves on the sea or in the much more difficult-to-imagine waves in light. In waves on water, the energy is transferred by the movement of water molecules. But a particular water molecule doesn't travel all the way across the Atlantic - or even all the way across a pond. Depending on the depth of the water, water molecules follow a roughly circular path. As they move up to the top of the circle, the wave builds to a crest; as they move down again, you get a trough. The energy is transferred by relatively small local movements in the environment. With water waves it is fairly easy to draw diagrams to show this happening with real molecules. With light it is more difficult. The energy in light travels because of local fluctuating changes in electrical and magnetic fields - hence "electromagnetic" radiation. If you draw a beam of light in the form of a wave (without worrying too much about what exactly is causing the wave!), the distance between two crests is called the wavelength of the light. (It could equally well be the distance between two troughs or any other two identical positions on the wave.) You have to picture these wave crests as moving from left to right. If you counted the number of crests passing a particular point per second, you have the frequency of the light. It is measured in what used to be called "cycles per second", but is now called Hertz, Hz. Cycles per second and Hertz mean exactly the same thing. Orange light, for example, has a frequency of about 5 x 10 Hz (often quoted as 5 x 10 MHz - megahertz). That means that 5 x 10 wave peaks pass a given point every second. Light has a constant speed through a given substance. For example, it always travels at a speed of approximately 3 x 10 meters per second in a vacuum. This is actually the speed that all electromagnetic radiation travels - not just visible light. There is a simple relationship between the wavelength and frequency of a particular color of light and the speed of light: . . . and you can rearrange this to work out the wavelength from a given frequency and vice versa: \[ \lambda = \dfrac{c}{\nu}\] \[ \nu= \dfrac{c}{\lambda}\] These relationships mean that if you increase the frequency, you must decrease the wavelength. Compare this diagram with the similar one above. . . and, of course, the opposite is true. If the wavelength is longer, the frequency is lower. It is really important that you feel comfortable with the relationship between frequency and wavelength. If you are given two figures for the wavelengths of two different colors of light, you need to have an immediate feel for which one has the higher frequency. For example, if you were told that a particular color of red light had a wavelength of 650 nm, and a green had a wavelength of 540 nm, it is important for you to know which has the higher frequency. (It's the green - a shorter wavelength means a higher frequency. Don't go on until that feels right!) Each particular frequency of light has a particular energy associated with it, given by another simple equation: You can see that the higher the frequency, the higher the energy of the light. Light which has wavelengths of around 380 - 435 nm is seen as a sequence of violet colours. Various red colours have wavelengths around 625 - 740 nm. Which has the highest energy? The light with the highest energy will be the one with the highest frequency - that will be the one with the smallest wavelength. In other words, violet light at the 380 nm end of its range. The diagram shows an approximation to the spectrum of visible light. The main colour regions of the spectrum are approximately: Don't assume that there is some clear cut-off point between all these colours. In reality, the colours just merge seamlessly into one another - much more seamlessly than in my diagram! The electromagnetic spectrum doesn't stop with the colors you can see. It is perfectly possible to have wavelengths shorter than violet light or longer than red light. On the spectrum further up the page, I have shown the ultra-violet and the infra-red, but this can be extended even further into x-rays and radio waves, amongst others. The diagram shows the approximate positions of some of these on the spectrum. Once again, don't worry too much about the exact boundaries between the various sorts of electromagnetic radiation - because there are no boundaries. Just as with visible light, one sort of radiation merges into the next. Just be aware of the general pattern. Also be aware that the energy associated with the various kinds of radiation increases as the frequency increases.
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We now turn our attention to acid–base reactions to see how the concepts of chemical equilibrium and equilibrium constants can deepen our understanding of this kind of chemical behavior. We begin with a qualitative description of acid–base equilibria in terms of the Brønsted–Lowry model and then proceed to a quantitative description the following sections. Owing to the overwhelming excess of \(H_2O\) molecules in aqueous solutions, a bare hydrogen ion has no chance of surviving in water. The hydrogen ion in aqueous solution is no more than a proton, a bare nucleus. Although it carries only a single unit of positive charge, this charge is concentrated into a volume of space that is only about a hundred-millionth as large as the volume occupied by the smallest atom. (Think of a pebble sitting in the middle of a sports stadium!) The resulting extraordinarily high of the proton strongly attracts it to any part of a nearby atom or molecule in which there is an excess of negative charge. In the case of water, this will be the lone pair (unshared) electrons of the oxygen atom; the tiny proton will be buried within the lone pair and will form a shared-electron (coordinate) bond with it, creating a , \(H_3O^+\). In a sense, \(H_2O\) is acting as a base here, and the product \(H_3O^+\) is the conjugate acid of water: Although other kinds of dissolved ions have water molecules bound to them more or less tightly, the interaction between H and \(H_2O\) is so strong that writing “H ” hardly does it justice, although it is formally correct. The formula \(H_3O^+\) more adequately conveys the sense that it is both a molecule in its own right, and is also the conjugate acid of water. However, the equation \[\ce{HA → H^{+} + A^{–}} \nonumber\] is so much easier to write that chemists still use it to represent acid-base reactions in contexts in which the proton donor-acceptor mechanism does not need to be emphasized. Thus it is permissible to talk about “hydrogen ions” and use the formula H in writing chemical equations as long as you remember that they are not to be taken literally in the context of aqueous solutions. Interestingly, experiments indicate that the proton does not stick to a single \(H_2O\) molecule, but changes partners many times per second. This molecular promiscuity, a consequence of the uniquely small size and mass the proton, allows it to move through the solution by rapidly hopping from one \(H_2O\) molecule to the next, creating a new \(H_3O^+\) ion as it goes. The overall effect is the same as if the \(H_3O^+\) ion itself were moving. Similarly, a hydroxide ion, which can be considered to be a “proton hole” in the water, serves as a landing point for a proton from another \(H_2O\) molecule, so that the OH ion hops about in the same way. The ion is an important factor when dealing with chemical reactions that occur in aqueous solutions. Because and hydroxide ions can “move without actually moving” and thus without having to plow their way through the solution by shoving aside water molecules as do other ions, solutions which are acidic or alkaline have extraordinarily high . The ion has a pyramidal geometry and is composed of three hydrogen atoms and one oxygen atom. There is a lone pair of electrons on the oxygen giving it this shape. The bond angle between the atoms is 113 degrees. As H ions are formed, they bond with \(H_2O\) molecules in the solution to form \(H_3O^+\) (the ion). This is because hydrogen ions do not exist in aqueous solutions, but take the form the ion, \(H_3O^+\). A reversible reaction is one in which the reaction goes both ways. In other words, the water molecules dissociate while the OH ions combine with the H ions to form water. Water has the ability to attract H ions because it is a polar molecule. This means that it has a partial charge, in this case the charge is negative. The partial charge is caused by the fact that oxygen is more electronegative than hydrogen. This means that in the bond between hydrogen and oxygen, oxygen "pulls" harder on the shared electrons thus causing a partial negative charge on the molecule and causing it to be attracted to the positive charge of H to form . Another way to describe why the water molecule is considered polar is through the concept of dipole moment. The electron geometry of water is tetrahedral and the molecular geometry is bent. This bent geometry is asymmetrical, which causes the molecule to be polar and have a dipole moment, resulting in a partial charge. As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water ( ). The concentrations of these ions in a solution are often critical determinants of the solution’s properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions. A common means of expressing quantities, the values of which may span many orders of magnitude, is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where “X” is the quantity of interest and “log” is the base-10 logarithm: \[\mathrm{pX=−\log X} \label{\(\Page {1}\)}\] The pH of a solution is therefore defined as shown here, where [H O ] is the molar concentration of hydronium ion in the solution: \[\mathrm{pH=-\log[H_3O^+]}\label{\(\Page {2}\)}\] Rearranging this equation to isolate the hydronium ion molarity yields the equivalent expression: \[\mathrm{[H_3O^+]=10^{−pH}}\label{\(\Page {3}\)}\] Likewise, the hydroxide ion molarity may be expressed as a p-function, or pOH: \[\mathrm{pOH=-\log [OH^−]}\label{\(\Page {4}\)}\] or \[\mathrm{[OH^-]=10^{−pOH}} \label{\(\Page {5}\)}\] Finally, the relation between these two ion concentration expressed as p-functions is easily derived from the \(K_w\) expression: \[ \begin{align*} K_\ce{w} &=\ce{[H_3O^+,OH^- ]} \\[4pt] -\log K_\ce{w} &= -\log([H_3O^+,OH^−]) \\[4pt] &=-\log[H_3O^+] + -\log[OH^-] \\[4pt] p \mathit{K}_w &=pH + pOH \end{align*}\] At 25 °C, the value of \(K_w\) is \(1.0 \times 10^{−14}\) and so: \[\mathrm{14.00=pH + pOH} \label{\(\Page {9}\)}\] The hydronium ion molarity in pure water (or any neutral solution) is \( 1.0 \times 10^{-7}\; M\) at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore: \[\mathrm{pH=-\log[H_3O^+]=-\log(1.0\times 10^{−7}) = 7.00} \label{\(\Page {1}\)0}\] \[\mathrm{pOH=-\log[OH^−]=-\log(1.0\times 10^{−7}) = 7.00} \label{\(\Page {1}\)1}\] And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than \( 1.0 \times 10^{-7}\; M\) and hydroxide ion molarities less than \( 1.0 \times 10^{-7}\; M\) (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than \( 1.0 \times 10^{-7}\; M\) and hydroxide ion molarities greater than \( 1.0 \times 10^{-7}\; M\) (corresponding to pH values greater than 7.00 and pOH values less than 7.00). Since the autoionization constant \(K_w\) is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than 25 °C. For example, the hydronium molarity of pure water at 80 °C is 4.9 × 10 , which corresponds to pH and pOH values of: \[ \begin{align*} \mathrm{pH} &= -\log[H_3O^+] \\[4pt] &= -\log(4.9\times 10^{−7}) \\[4pt] &=6.31 \label{\(\Page {1}\)2} \end{align*}\] \[ \begin{align*} \mathrm{pOH} &= -\log[OH^-] \\[4pt] &= -\log(4.9\times 10^{−7}) \\[4pt] &=6.31 \label{\(\Page {1}\)3} \end{align*}\] At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at nonstandard temperatures, such as enzyme reactions in warm-blooded organisms. Unless otherwise noted, references to pH values are presumed to be those at standard temperature (25 °C) (Table \(\Page {1}\)). Figure \(\Page {2}\) shows the relationships between [H O ], [OH ], pH, and pOH, and gives values for these properties at standard temperatures for some common substances. What is the pH of stomach acid, a solution of HCl with a hydronium ion concentration of \(1.2 \times 10^{−3}\; M\)? \[ \begin{align*} \mathrm{pH} &= -\log [H_3O^+] \\[4pt] &= -\log(1.2 \times 10^{−3}) \\[4pt] &=−(−2.92) =2.92 \end{align*} \] Water exposed to air contains carbonic acid, H CO , due to the reaction between carbon dioxide and water: \[\ce{CO}_{2(aq)}+\ce{H_2O}_{(l)} \rightleftharpoons \ce{H_2CO}_{3(aq)}\nonumber\] Air-saturated water has a hydronium ion concentration caused by the dissolved \(\ce{CO_2}\) of \(2.0 \times 10^{−6}\; M\), about 20-times larger than that of pure water. Calculate the pH of the solution at 25 °C. 5.70 Calculate the hydronium ion concentration of blood, the pH of which is 7.3. \[ \begin{align*} \mathrm{pH} =-\log [\ce{H_3O^+}] &=7.3 \\[4pt] [\ce{H_3O^+}] &=10^{−7.3} \\[4pt] [\ce{H_3O^+}] &=5\times 10^{−8}\;M \end{align*}\] (On a calculator take the antilog, or the “inverse” log, of −7.3, or calculate 10 .) Calculate the hydronium ion concentration of a solution with a pH of −1.07. 12 M This uses the definition of pH that we commonly use: \[pH=-\log[H_3O^+] \nonumber\] So for this solution: \[ \begin{align*} \mathrm{pH} =-\log [\ce{H_3O^+}] &= -1.07 \\[4pt] [\ce{H_3O^+}] & = 10^{+1.07} \\[4pt] &= 12\, (\text{with significant figure}) \end{align*}\] However, at this high concentration, the solution will be non-ideal and we have to use the proper definition in terms of hydronium activities \[pH=-\log a\{H_3O^+\} \nonumber \] See this module for more details. The ionization constant of water \(\ce{K_w}\) at 37 °C is \(2.42 \times 10^{-14}\). What is the pH for a neutral solution at this human physiological temperature? Is the water acidic, basic or neutral? \[ \ce{K_w} = \ce{[H3O^{+}] [OH^{-}]} = 2.42 \times 10^{-14}\nonumber\] and \[ \begin{align*} \ce{[H3O^{+}]} = \ce{[OH^{-}]} &= \sqrt{2.42 \times 10^{-14} } \\[4pt] &= 1.55 \times 10^{-7} \end{align*} \] \[pH=-\log[H_3O^+] = -\log 1.55 \times 10^{-7} = 6.81 \] If we use the definition of acidic systems like in Figure \(\Page {2}\), then we would (incorrectly) argue the solution is acidic. However, since \(\ce{[OH^{-}]} = \ce{[H3O^{+}]}\), the solution is still neutral. This is only a strange idea, if one ignores the temperature dependence of \(\ce{K_w}\). Normal rainwater has a pH between 5 and 6 due to the presence of dissolved CO which forms carbonic acid: \[\ce{H_2O}_{(l)}+\ce{CO}_{2(g)}⟶\ce{H_2CO}_{3(aq)} \label{\(\Page {1}\)4}\] \[\ce{H_2CO}_{3(aq)} \rightleftharpoons \ce{H^+}_{(aq)} + \ce{HCO^-}_{3(aq)} \label{\(\Page {1}\)5}\] Acid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including CO , SO , SO , NO, and NO being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here: \[\ce{H_2O}_{(l)} + \ce{SO}_{3(g)} ⟶\ce{H_2SO}_{4(aq)} \label{\(\Page {1}\)6}\] \[\ce{H_2SO}_{4(aq)} ⟶ \ce{H^+}_{(aq)} + \ce{HSO}_{4(aq)}^- \label{\(\Page {1}\)7}\] Carbon dioxide is naturally present in the atmosphere because we and most other organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or when we burn wood or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also stems from burning fossil fuels, which have traces of sulfur, and from the process of “roasting” ores of metal sulfides in metal-refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine. Figure \(\Page {3}\): (a) Acid rain makes trees more susceptible to drought and insect infestation, and depletes nutrients in the soil. (b) It also is corrodes statues that are carved from marble or limestone. (credit a: modification of work by Chris M Morris; credit b: modification of work by “Eden, Janine and Jim”/Flickr) Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure \(\Page {3}\)). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India. The acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure \(\Page {4}\)). Figure \(\Page {4}\): (a) A research-grade pH meter used in a laboratory can have a resolution of 0.001 pH units, an accuracy of ± 0.002 pH units, and may cost in excess of $1000. (b) A portable pH meter has lower resolution (0.01 pH units), lower accuracy (± 0.2 pH units), and a far lower price tag. (credit b: modification of work by Jacopo Werther). What are the pOH and the pH of a 0.0125-M solution of potassium hydroxide, KOH at 25 °C? Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding [OH ] = 0.0125 M: \[\mathrm{pOH=-\log[OH^− ]=-\log 0.0125}\] \[=−(−1.903)=1.903\] The pH can be found from the \(\ce{pOH}\): \[\mathrm{pH+pOH=14.00}\] \[\mathrm{pH=14.00−pOH=14.00−1.903=12.10}\] The hydronium ion concentration of vinegar is approximately \(4 \times 10^{−3}\; M\) at 25 °C. What are the corresponding values of pOH and pH? pOH = 11.6, pH = 2.4 The pH of a solution may also be visually estimated using colored indicators (Figure \(\Page {5}\)). Figure \(\Page {5}\): (a) A universal indicator assumes a different color in solutions of different pH values. Thus, it can be added to a solution to determine the pH of the solution. The eight vials each contain a universal indicator and 0.1-M solutions of progressively weaker acids: HCl (pH = l), CH CO H (pH = 3), and NH Cl (pH = 5), deionized water, a neutral substance (pH = 7); and 0.1-M solutions of the progressively stronger bases: KCl (pH = 7), aniline, C H NH (pH = 9), NH (pH = 11), and NaOH (pH = 13). (b) pH paper contains a mixture of indicators that give different colors in solutions of differing pH values. (credit: modification of work by Sahar Atwa). The concentration of hydronium ion in a solution of an acid in water is greater than \( 1.0 \times 10^{-7}\; M\) at 25 °C. The concentration of hydroxide ion in a solution of a base in water is greater than \( 1.0 \times 10^{-7}\; M\) at 25 °C. The concentration of H O in a solution can be expressed as the pH of the solution; \(\ce{pH} = -\log \ce{H3O+}\). The concentration of OH can be expressed as the pOH of the solution: \(\ce{pOH} = -\log[\ce{OH-}]\). In pure water, pH = 7.00 and pOH = 7.00   ).
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Electrons in an atom reside in shells characterized by a particular value of n, the Principal Quantum Number. Within each shell an electron can occupy an orbital which is further characterized by an Orbital Quantum Number, \(l\), where \(l\) can take all values in the range: \[l = 0, 1, 2, 3, ... , (n-1),\] traditionally termed s, p, d, f, etc. orbitals. Each orbital has a characteristic shape reflecting the motion of the electron in that particular orbital, this motion being characterized by an angular momentum that reflects the angular velocity of the electron moving in its orbital. A quantum mechanics approach to determining the energy of electrons in an element or ion is based on the results obtained by solving the Schrödinger Wave Equation for the H-atom. The various solutions for the different energy states are characterized by the three quantum numbers, n, and m . In summary then, each electron in an orbital is characterized by four quantum numbers (Table 1). The ways in which the angular momenta associated with the orbital and spin motions in many-electron-atoms can be combined together are many and varied. In spite of this seeming complexity, the results are frequently readily determined for simple atom systems and are used to characterize the electronic states of atoms. The interactions that can occur are of three types. There are two principal coupling schemes used: In the Russell Saunders scheme (named after Henry Norris Russell, 1877-1957 a Princeton Astronomer and Frederick Albert Saunders, 1875-1963 a Harvard Physicist and published in ) it is assumed that: spin-spin coupling > orbit-orbit coupling > spin-orbit coupling. This is found to give a good approximation for first row transition series where spin-orbit (J) coupling can generally be ignored, however for elements with atomic number greater than thirty, spin-orbit coupling becomes more significant and the j-j coupling scheme is used. S - the resultant spin quantum number for a system of electrons. The overall spin S arises from adding the individual m together and is as a result of coupling of spin quantum numbers for the separate electrons. L - the total orbital angular momentum quantum number defines the energy state for a system of electrons. These states or term letters are represented as follows: Coupling occurs between the resultant spin and orbital momenta of an electron which gives rise to J the total angular momentum quantum number. Multiplicity occurs when several levels are close together and is given by the formula (2S+1). The Russell Saunders term symbol that results from these considerations is given by: L=2 and the Ground Term is written as D The Russell Saunders term symbols for the other free ion configurations are given in the Table below. Note that d gives the same terms as d The Ground Terms are deduced by using . The two rules are: A simple graphical method for determining just the ground term alone for the free-ions uses a "fill in the boxes" arrangement. To calculate , simply sum the electrons using a value of ½ for each. To calculate , use the labels for each column to determine the value of for that box, then add all the individual box values together. For a d configuration, then: Total value of is therefore +4 +2 +0 -1 -2 or =3. Note that for 5 electrons with 1 electron in each box then the total value of is 0. This is why for a d configuration is the same as for a d . The other thing to note is the idea of the "hole" approach. A d configuration can be treated as similar to a d configuration. In the first case there is 1 electron and in the latter there is an absence of an electron i.e., a hole. The overall result shown in the Table above is that: The effect of a crystal field on the different orbitals (s, p, d, etc.) will result in splitting into subsets of different energies, depending on whether they are in an octahedral or tetrahedral environment. The magnitude of the d orbital splitting is generally represented as a fraction of Δ or 10Dq. The ground term energies for free ions are also affected by the influence of a crystal field and an analogy is made between orbitals and ground terms that are related due to the angular parts of their electron distribution. The effect of a crystal field on different orbitals in an octahedral field environment will cause the d orbitals to split to give t and e subsets and the D ground term states into T and E , (where upper case is used to denote states and lower case orbitals). f orbitals are split to give subsets known as t , t and a . By analogy, the F ground term when split by a crystal field will give states known as T , T , and A . Note that it is important to recognize that the F ground term here refers to states arising from d orbitals and not f orbitals and depending on whether it is in an octahedral or tetrahedral environment the lowest term can be either A or T . Note that, for simplicity, spin multiplicities are not included in the table since they remain the same for each term. The table above shows that the Mulliken symmetry labels, developed for atomic and molecular orbitals, have been applied to these states but for this purpose they are written in CAPITAL LETTERS. For splitting in a tetrahedral crystal field the components are similar, except that the symmetry label g (gerade) is absent. The ground term for first-row transition metal ions is either D, F or S which in high spin octahedral fields gives rise to A, E or T states. This means that the states are either non-degenerate, doubly degenerate or triply degenerate.
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In the early days of organic chemistry, alkenes were described as "unsaturated" because, in contrast to the "saturated" alkanes, they were found to react readily with substances such as halogens, hydrogen halides, oxidizing agents, and so on. Therefore, the "chemical affinity" of alkenes was regarded as unsatisfied or "unsaturated". (Also see .) One reason alkenes and alkynes react more readily than alkanes is because the carbon-carbon bonds of a multiple bond are individually weaker than normal carbon-carbon single bonds. Consider the bond energies involved. According to Table 4-3, the strengths of carbon-carbon single, double, and triple bonds are \(83\), \(146\), and \(200 \: \text{kcal}\), respectively. From these values we can calculate that cleavage of one-half of a carbon-carbon double bond should require \(63 \: \text{kcal}\) and cleavage of one-third of a carbon-carbon triple bond should require \(54 \: \text{kcal}\): As a result, addition reactions to multiple bonds are expected to be about \(20\)-\(30 \: \text{kcal}\) more exothermic than the corresponding cleavage reactions of carbon-carbon single bonds, as estimated here for reaction with bromine: The substantial difference in the heats of reaction of ethane, ethene, and ethyne with bromine is reflected in a very important practical consideration in handling ethyne (acetylene), namely its thermodynamic stability relative to solid carbon and hydrogen gas. Unlike ethane, both ethene and ethyne can be shown from bond energies to be unstable with respect to formation of solid carbon and gaseous hydrogen: Although this does not seem to offer particular problems with ethene, an explosive decomposition of ethyne to carbon and hydrogen may occur if the gas is compressed to \(10\)-\(20 \: \text{kg cm}^{-2}\). Even liquid ethyne (bp \sim 83^\text{o}\)) must be handled with care. Ethyne is not used commercially under pressure unless it is mixed with an inert gas and handled in rugged equipment. Ethyne burns with pure oxygen to give a very hot flame that is widely used for welding. For this purpose, the gas is dissolved under about \(15 \: \text{kg cm}^{-2}\) in 2-propanone (acetone, , bp \(56.5^\text{o}\)) and contained in cylinders packed with diatomaceous earth. Why is ethyne so much less stable than ethene or ethane? First, \(\ce{C-C}\) bonds are not as strong as \(\ce{C-H}\) bonds. Therefore a gain in stability usually is to be expected when \(\ce{C-H}\) bonds are made at the expense of \(\ce{C-C}\) bonds; ethene and ethane each have more \(\ce{C-H}\) bonds than ethyne has. Second, ethyne has six electrons held between the two carbons and these electrons experience considerable mutual interelectronic repulsion. This accounts for the fact that the \(\ce{C-C}\) bond strength for the triple bond of an alkyne is \(200/3 = 67 \: \text{kcal}\), compared to \(146/2 = 73 \: \text{kcal}\) for the double bond of an alkene and \(83 \: \text{kcal}\) for a normal single bond of an alkane. and (1977)
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FTIR spectrometers (Fourier Transform Infrared Spectrometer) are widely used in organic synthesis, polymer science, petrochemical engineering, pharmaceutical industry and food analysis. In addition, since FTIR spectrometers can be hyphenated to chromatography, the mechanism of chemical reactions and the detection of unstable substances can be investigated with such instruments. FTIR spectrometers are the third generation infrared spectrometer. FTIR spectrometers have several prominent advantages: (1) The signal-to-noise ratio of spectrum is significantly higher than the previous generation infrared spectrometers. (2) The accuracy of wavenumber is high. The error is within the range of ± 0.01 cm . (3) The scan time of all frequencies is short (approximately 1 s). (4) The resolution is extremely high (0.1 ~ 0.005 cm ). (5) The scan range is wide (1000 ~ 10 cm ). (6) The interference from stray light is reduced. Due to these advantages, FTIR Spectrometers have replaced dispersive IR spectrometers. have some background information of dispersive IR Spectrometer. The basic components of a dispersive IR spectrometer include a radiation source, monochromator, and detector. The common IR radiation sources are inert solids that are heated electrically to promote thermal emission of radiation in the infrared region of the electromagnetic spectrum. The monochromator is a device used to disperse or separate a broad spectrum of IR radiation into individual narrow IR frequencies. Generally, dispersive spectrometers have a double-beam design with two equivalent beams from the same source passing through the sample and reference chambers as independent beams. These reference and sample beams are alternately focused on the detector by making use of an optical chopper, such as, a sector mirror. One beam will proceed, traveling through the sample, while the other beam will pass through a reference species for analytical comparison of transmitted photon wavefront information. After the incident radiation travels through the sample species, the emitted wavefront of radiation is dispersed by a monochromator (gratings and slits) into its component frequencies. A combination of prisms or gratings with variable-slit mechanisms, mirrors, and filters comprise the dispersive system. Narrower slits gives better resolution by distinguishing more closely spaced frequencies of radiation and wider slits allow more light to reach the detector and provide better system sensitivity. The emitted wavefront beam (analog spectral output) hits the detector and generates an electrical signal as a response. Detectors are devices that convert the analog spectral output into an electrical signal. These electrical signals are further processed by the computer using mathematical algorithm to arrive at the final spectrum. The detectors used in IR spectrometers can be classified as either photon/quantum detectors or thermal detectors. It is the absorption of IR radiation by the sample, producing a change of IR radiation intensity, which gets detected as an off-null signal (e.g. different from reference signal). This change is translated into the recorder response through the actions of synchronous motors. Each frequency that passes through the sample is measured individually by the detector which consequently slows the process of scanning the entire IR region. A block diagram of a classic dispersive IR spectrometer is shown in Figure \(\Page {1}\). \[\delta =2\Delta \label{1} \] , constructive interference occurs because crests overlap with crests, troughs with troughs. As a result, a maximum intensity signal is observed by the detector. This situation can be described by the following equation: \[\delta =n\lambda \label{2} \] \[\delta =(n+\dfrac{1}{2})\lambda \label{3} \] with n = 0,1,2,3... Since the mirror moves back and forth, the intensity of the signal increases and decreases which gives rise to a cosine wave. The plot is defined as an interferogram. When detecting the radiation of a broad band source rather than a single-wavelength source, a peak at ZPD is found in the interferogram. At the other distance scanned, the signal decays quickly since the mirror moves back and forth. Figure \(\Page {4}\)(a) shows an interferogram of a broad band source. The interferogram is a function of time and the values outputted by this function of time are said to make up the time domain. The time domain is Fourier transformed to get a frequency domain, which is deconvolved to product a spectrum. Figure \(\Page {4}\) shows the Fast Fourier transform from an interferogram of polychromatic light to its spectrum. Advantages of Fourier Transform over Continuous-Wave Spectrometry Fourier transform, named after the French mathematician and physicist Jean Baptiste Joseph Fourier, is a mathematical method to transform a function into a new function. The following equation is a common form of the Fourier transform with unitary normalization constants: The following equation is another form of the Fourier transform(cosine transform) which applies to real, even functions: The following equation shows how f(t) is related to F(v) via a Fourier transform: The math description of the Fourier transform can be tedious and confusing. An alternative explanation of the Fourier transform in FTIR spectrometers is provided here before we jump into the math description to give you a rough impression which may help you understand the math description. The wave functions of the reflected and transmitted beams may be represented by the general form of: \[E_{1}=rtc E_{m}\times \cos(\nu t-2\pi kx) \label{7} \] and \[E_{1}=rtc E_{m}\times \cos[\nu t-2\pi k(\nu x+\Delta d)] \label{8} \] where \[E=E_{1}+E_{2}=2(r\times t\times c\times E_{m})\times \cos(\nu t-2\pi kx)\cos(\pi k\Delta d) \label{9} \] where E , ν, and k are the amplitude, frequency and wave number of the IR radiation source. The intensity (\(I\)) detected is the time average of \(E_2\) and is written as \[I=4r^{2}t^{2}c^{2}E_{m}^{2}cos^{2}(\nu t-2\pi kx)\cos^{2}(\pi k\Delta d) \label{10} \] Since the time average of the first cosine term is just ½, then \[I=2I(k)\cos^{2}(\pi k\Delta d) \label{11} \] and \[I(\Delta d)=I(k)[1+\cos(2\pi k\Delta d)]\label{12} \] ere \(I(k)\) is a constant that depends only upon \(k\) and \(I(∆d)\) is From \(I(∆d)\) we can get \(I(k)\) using Fourier transform as follows: \[I(\Delta d)-I(\infty)=\int_{0}^{k_{m}}I(k)\cos(2\Pi k\Delta d)dk \label{13} \] Letting K ∞, we can write \[I(k)=\int_{0}^{\infty}[I(\Delta d)-I(\infty)]\cos(2\Pi k\Delta d)d\Delta d \label{14} \] The physically measured information recorded at the detector produces an interferogram, which provides information about a response change over time within the mirror scan distance. Therefore, the interferogram obtained at the detector is a time domain spectrum. This procedure involves sampling each position, which can take a long time if the signal is small and the number of frequencies being sampled is large. In terms of ordinary frequency, \(\nu\), the Fourier transform of this is given by (angular frequency \(\omega= s\pi \nu\)): \[f(\nu )=\int_{-\infty}^{\infty}f(t)e^{-i2\Pi \nu t}dt \label{15} \] The inverse Fourier transform is given by: \[f(\nu )=\int_{-\infty}^{\infty}f(t)e^{+i2\pi \nu t}dt \label{16} \] The interferogram is transformed into IR absorption spectrum (Figure \(\Page {5}\)) that is commonly recognizable with absorption intensity or % transmittance plotted against the wavelength or wavenumber. The ratio of radiant power transmitted by the sample (I) relative to the radiant power of incident light on the sample (I ) results in quantity of Transmittance, (T). Absorbance (A) is the logarithm to the base 10 of the reciprocal of the transmittance (T): \[A=log_{10}\dfrac{1}{T}=-log_{10}T=-log_{10}\dfrac{I}{I_{0}} \label{17} \] The first step is sample preparation The standard method to prepare solid sample for FTIR spectrometer is to use KBr. About 2 mg of sample and 200 mg KBr are dried and ground. The particle size should be unified and less than two micrometers. Then, the mixture is squeezed to form transparent pellets which can be measured directly. For liquids with high boiling point or viscous solution, it can be added in between two NaCl pellets. Then the sample is fixed in the cell by skews and measured. For volatile liquid sample, it is dissolved in CS or CCl to form 10% solution. Then the solution is injected into a liquid cell for measurement. Gas sample needs to be measured in a gas cell with two KBr windows on each side. The gas cell should first be vacuumed. Then the sample can be introduced to the gas cell for measurement. The second step is getting a background spectrum by collecting an interferogram and its subsequent conversion to frequency data by inverse Fourier transform. We obtain the background spectrum because the solvent in which we place our sample will have traces of dissolved gases as well as solvent molecules that contribute information that are not our sample. The background spectrum will contain information about the species of gases and solvent molecules, which may then be subtracted away from our sample spectrum in order to gain information about just the sample. Figure \(\Page {6}\) shows an example of an FTIR background spectrum. The background spectrum also takes into account several other factors related to the instrument performance, which includes information about the source, interferometer, detector, and the contribution of ambient water (note the two irregular groups of lines at about 3600 cm and about 1600 cm in Figure \(\Page {6}\)) and carbon dioxide (note the doublet at 2360 cm and sharp spike at 667 cm in Figure \(\Page {6}\)) present in the optical bench. Next, we collect a single-beam spectrum of the sample, which will contain absorption bands from the sample as well as the background (gaseous or solvent). The ratio between the single-beam sample spectrum and the single beam background spectrum gives the spectrum of the sample (Figure \(\Page {7}\)). Data analysis is done by assigning the observed absorption frequency bands in the sample spectrum to appropriate normal modes of vibrations in the molecules. Despite of the powerfulness of traditional FTIR spectrometers, they are not suitable for real-time monitoring or field use. So various portable FTIR spectrometers have been developed. Below are two examples. Ahonen developed a portable, real-time FTIR spectrometer as a gas analyzer for industrial hygiene use. The instrument consists of an operational keyboard, a control panel, signal and control processing electronics, an interferometer, a heatable sample cell and a detector. All the components were packed into a cart. To minimize the size of the instrument, the resolution of FTIR spectrometer was sacraficed. But it is good enough for the use of industrial hygiene. The correlation coefficient of hygienic effect between the analyzer and adsorption tubes is about 1 mg/m . Korb developed a portable FTIR spectrometer which only weighs about 12.5 kg so that it can be held by hand. Moreover, the energy source of the instrument is battery so that the mobility is significantly enhanced. Besides, the instrument can function well within the temperature range of 0 to 45 C and the humidity range of 0 to 100%. Additionally, this instrument resists vibration. It works well in an operating helicopter. Consequently, this instrument is excellent for the analysis of radiation from the surface and atmosphere of the Earth. The instrument is also very stable. After a three-year operation, it did not lose optical alignment. The reduction of size was implemented by a creative design of optical system and accessory components. Two KBr prisms were used to constitute the interferometer cavity. Optical coatings replaced the mirrors and beam splitter in the interferometer. The optical path is shortened with a much more compact packaging of components. A small, low energy consuming interferometer drive was designed. It is also mass balanced to resist vibration. The common He-Ne tube was replaced by a smaller laser diode.
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Hydrogen is a colorless, odorless and tasteless gas that is the most abundant element in the known universe. It is also the lightest (in terms of atomic mass) and the simplest, having only one proton and one electron (and no neutrons in its most common isotope). It is all around us. It is a component of water (H O), fats, petroleum, table sugar (C H O ), ammonia (NH ), and hydrogen peroxide (H O )—things essential to life, as we know it. Hydrogen comes from Greek meaning “water producer” (“hydro” =water and “gennao”=to make). First isolated and identified as an element by Cavendish in 1766, hydrogen was believed to be many different things. Cavendish himself thought that it was "inflammable air from metals", owing to its production by the action of acids on metals. Before that, Robert Boyle and Paracelsus both used reactions of iron and acids to produce hydrogen gas and Antoine Lavoisier gave hydrogen its name because it produced water when ignited in air. Others thought it was pure phlogiston because of its flammability. Hydrogen is among the ten most abundant elements on the planet, but very little is found in elemental form due to its low density and reactivity. Much of the terrestrial hydrogen is locked up in water molecules and organic compounds like hydrocarbons. Hydrogen is a nonmetal and is placed above group in the periodic table because it has ns electron configuration like the alkali metals. However, it varies greatly from the alkali metals as it forms cations (H ) more reluctantly than the other alkali metals. Hydrogen‘s ionization energy is 1312 kJ/mol, while lithium (the alkali metal with the highest ionization energy) has an ionization energy of 520 kJ/mol. Because hydrogen is a nonmetal and forms H (hydride anions), it is sometimes placed above the halogens in the periodic table. Hydrogen also forms H dihydrogen like halogens. However, hydrogen is very different from the halogens. Hydrogen has a much smaller electron affinity than the halogens. H dihydrogen or molecular hydrogen is non-polar with two electrons. There are weak attractive forces between H molecules, resulting in low boiling and melting points. However, H has very strong intramolecular forces; H reactions are generally slow at room temperature due to strong H—H bond. H is easily activated by heat, irradiation, or catalysis. Activated hydrogen gas reacts very quickly and exothermically with many substances. Hydrogen also has an ability to form covalent bonds with a large variety of substances. Because it makes strong O—H bonds, it is a good reducing agent for metal oxides. Example: CuO(s) + H (g) → Cu(s) + H O(g) H (g) passes over CuO(s) to reduce the Cu to Cu(s), while getting oxidized itself. Hydrogen's low ionization energy makes it act like an alkali metal: \[H_{(g)} \rightarrow H^+_{(g)} + e^-\] However, it half-filled valence shell (with a \(1s^1\) configuration) with one \(e^-\) also causes hydrogen to act like a halogen non-metal to gain noble gas configuration by adding an additional electron \[H_{(g)} + e^- \rightarrow H^-_{(g)}\] Hydrogen accepts e- from an active metal to form ionic hydrides like LiH. By forming an ion with -1 charge, the hydrogen behaves like a halogen. \[2M_{(s)}+H_{2(g)} \rightarrow 2MH_{(s)}\] with \(M\) representing Examples: \[M_{(s)}+H_{2(g)} \rightarrow MH_{2(s)}\] with \(M\) representing Example: Unlike metals forming ionic bonds with nonmetals, hydrogen forms polar covalent bonds. Despite being electropositive like the active metals that form with nonmetals, hydrogen is much less electropositive than the active metals, and forms covalent bonds. Hydrogen + Halogen → Hydrogen Halide \[H_{2(g)}+ Cl_{2(g)} \rightarrow HCl_{(g)}\] Hydrogen gas reacting with oxygen to produce water and a large amount of heat: Hydrogen + Oxygen → Water \[(H_{2(g)}+O_{2(g)} \rightarrow H_2O_{(g)}\] Reactions of hydrogen with (Group 3-12) form metallic hydrides. There is no fixed ratio of hydrogen atom to metal because the hydrogen atoms fill holes between metal atoms in the crystalline structure. The vast majority of hydrogen produced industrially today is made either from treatment of methane gas with steam or in the production of "water gas" from the reaction of coal with steam. Most of this hydrogen is used in the Haber process to manufacture ammonia. Hydrogen is also used for vegetable oils, turning them into margarine and shortening, and some is used for liquid rocket fuel. Liquid hydrogen (combined with liquid oxygen) is a major component of rocket fuel (as mentioned above combination of hydrogen and oxygen relapses a huge amount of energy). Because hydrogen is a good reducing agent, it is used to produce metals like iron, copper, nickel, and cobalt from their ores. Because one cubic feet of hydrogen can lift about 0.07 lbs, hydrogen lifted airships or Zeppelins became very common in the early 1900s.However, the use of hydrogen for this purpose was largely discontinued around World War II after the explosion of ; this prompted greater use of inert helium, rather than flammable hydrogen for air travel. Showing the explosion of . ( from Youtube) Recently, due to the fear of fossil fuels running out, extensive research is being done on hydrogen as a source of energy.Because of their moderately high energy densities liquid hydrogen and compressed hydrogen gas are possible fuels for the future.A huge advantage in using them is that their combustion only produces water (it burns “clean”). However, it is very costly, and not economically feasible with current technology. Combustion of fuel produces energy that can be converted into electrical energy when energy in the steam turns a turbine to drive a generator. However, this is not very efficient because a great deal of energy is lost as heat. The production of electricity using voltaic cell can yield more electricity (a form of usable energy). Voltaic cells that transform chemical energy in fuels (like H and CH ) are called fuel cells. These are not self-contained and so are not considered batteries. The hydrogen cell is a type of fuel cell involving the reaction between H (g) with O (g) to form liquid water; this cell is twice as efficient as the best internal combustion engine. In the cell (in basic conditions), the oxygen is reduced at the cathode, while the hydrogen is oxidized at the anode. Reduction: O (g)+2H O(l)+4e Oxidation: H (g) + 2OH (aq) Overall: 2H (g) + O (g) → 2H O(l) E cell= Reduction- Oxidation= E - E = 0.401V – (-0.828V) = +1.23 However, this technology is far from being used in everyday life due to its great costs. Hydrogen is the fuel for reactions of the Sun and other stars (fusion reactions). Hydrogen is the lightest and most abundant element in the universe. About 70%- 75% of the universe is composed of hydrogen by mass. All stars are essentially large masses of hydrogen gas that produce enormous amounts of energy through the fusion of hydrogen atoms at their dense cores. In smaller stars, hydrogen atoms collided and fused to form helium and other light elements like nitrogen and carbon(essential for life). In the larger stars, fusion produces the lighter and heavier elements like calcium, oxygen, and silicon. On Earth, hydrogen is mostly found in association with oxygen; its most abundant form being water (H O). Hydrogen is only .9% by mass and 15% by volume abundant on the earth, despite water covering about 70% of the planet. Because hydrogen is so light, there is only 0.5 ppm (parts per million) in the atmosphere, which is a good thing considering it is EXTREMELY flammable. Hydrogen gas can be prepared by reacting a dilute strong acid like hydrochloric acids with an active metal. The metal becomes oxides, while the H (from the acid) is reduced to hydrogen gas. This method is only practical for producing small amounts of hydrogen in the lab, but is much too costly for industrial production: \[Zn_{(s)} + 2H^+_{(aq)} \rightarrow Zn^{2+}_{(aq)} + H_{2(g)}\] The purest form of H (g) can come from electrolysis of H O(l), the most common hydrogen compound on this plant. This method is also not commercially viable because it requires a significant amount of energy (\(\Delta H = 572 \;kJ\)): \[2H_2O_{(l)} \rightarrow 2H_{2(g)} + O_{2(g)} \] \(H O\) is the most abundant form of hydrogen on the planet, so it seems logical to try to extract hydrogen from water without electrolysis of water. To do so, we must reduce the hydrogen with +1 oxidation state to hydrogen with 0 oxidation state (in hydrogen gas). Three commonly used reducing agents are carbon (in coke or coal), carbon monoxide, and methane. These react with water vapor form H (g): \[C_{(s)} + 2H_2O_{(g)} \rightarrow CO(g) + H_{2(g)}\] \[CO_{(g)} + 2H_2O_{(g)} \rightarrow CO2 + H_{2(g)}\] Reforming of Methane: \[CH_{4(g)} + H_2O_{(g)} \rightarrow CO(g) + 3H_{2(g)}\] These three methods are most industrially feasible (cost effective) methods of producing H (g). There are two important isotopes of hydrogen. Deuterium ( H) has an abundance of 0.015% of terrestrial hydrogen and the nucleus of the isotope contains one neutron.
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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/03%3A_Rate_Laws/3.02%3A_Reaction_Mechanisms/3.2.02%3A_Pre-equilibrium_Approximation
The pre-equilibrium approximation assumes that the reactants and intermediates of a multi-step reaction exist in dynamic equilibrium. According to the article, reactions involving many steps can be analyzed using approximations. Like the steady state method, the pre-equilibrium approximation method derives an expression for the rate of product formation with approximated concentrations. Unlike the steady state method, the pre-equilibrium approximation does so by assuming that the reactants and intermediate are in equilibrium. Although both methods are used to solve for a rate of reaction, they are used under different conditions. The steady state method can only be used if the first step of a reaction is much slower than the second step, whereas the pre-equilibrium approximation requires the first step to be faster. These opposing conditions prevent the two methods from being interchangeable. Both the steady state approximation and pre-equilibrium approximation apply to intermediate-forming , in which the product of the first step of the reaction serves as the reactant for the second step. As shown in Figure 1, there are three rate constants in the consecutive reaction involving an intermediate. These three rate constants, k , k , and k , are incredibly important for the pre-equilibrium approximation, as discussed in the next section. A consecutive reaction found in living systems is the In this type of reaction, an enzyme binds to a substrate to produce an enzyme-substrate intermediate, which then forms the final product. As shown in Figure 2, this reaction follows the basic pattern of the consecutive reaction in Figure 1. The two reactants, E (enzyme) and S (substrate), form the intermediate ES. This enzyme-substrate intermediate forms the product P, usually an essential biomolecule. The enzyme then exits the reaction unchanged and able to catalyze future reactions. As before, there are three reaction rates in this reaction: k , k , and k . The pre-equilibrium approximation uses the rate constants to solve for the rate of the reaction, indicating how quickly the reaction is likely to produce the biomolecule. The enzyme substrate reaction in Figure 2 is used below to demonstrate a rate law derivation. The overall strategy for the pre-equilibrium reaction is as follows: I. Assume reactants and intermediate are in equilibrium II. Solve for the rate of product formation 1. The general reaction used to derive a rate law is as following: \( E + S \xrightleftharpoons[k_{-1}]{\ k_1\ } ES \xrightarrow{k_2} E + P \) 2. Breaking up the overall reaction into elementary steps gives: \[E + S \rightarrow ES \nonumber \] \[ES \rightarrow E + S \nonumber \] \[ES \rightarrow E + P \nonumber \] 3. In the steady state reaction, the intermediate concentration [ES] is assumed to remain at a small constant value. The pre-equilibrium approximation takes a different approach. Because the reversible reaction \(E + S \rightarrow ES\) is much faster than the product formation of \(ES \rightarrow E + P\), E, S, and ES are considered to be in equilibrium throughout the reaction. This concept greatly simplifies the mathematics leading to the rate law. 4. Using the idea that the reactants and intermediate are in equilibrium, the rate of formation of E + P can be written as: \[\frac{d[P]}{dt}=k_2[ES]=k_2K[E,S] \nonumber \] where \[K = \frac{[ES]}{[E,S]} = \frac{k_1}{k_{-1}} \nonumber \] 5. The rate of product formation is simplified through the k: 6. The rate of product formation can now be shortened to: \[\dfrac{d[P]}{dt} = k[E,S] \nonumber \] This equation can be solved for unknown variables in consecutive reactions involving intermediates. Many reactions have at least one that must be reached in order for the reaction to go forward. There are three activation energies in a pre-equilibrium reaction: two for the reversible steps and one for the final step. The three activation energies are denoted E E , and E : Forward Reverse Product Formation The overall activation energy is given below: The overall activation energy is positive if Ea + Ea > Ea The overall activation energy is negative if Ea + Ea < Ea For the reaction \(A + B \xrightarrow[k_{-1}]{k_1} AB \xrightarrow[]{k_2} C\)(see Figure 1), derive a rate law using the pre-equilibrium approximation. \(\dfrac{d[P]}{dt} = k[A,B]\) where \(k =\dfrac{k_1 k_2}{k_{-1}}\)
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The s of most liquids have similar shapes with the vapor pressure steadily increasing as the temperature increases (Figure \(\Page {1}\)). A good approach is to find a mathematical model for the pressure increase as a function of temperature. Experiments showed that the vapor pressure \(P\) and temperature \(T\) are related, \[P \propto \exp \left(- \dfrac{\Delta H_{vap}}{RT}\right) \ \label{1}\] where \(\Delta{H_{vap}}\) is the Enthalpy (heat) of Vaporization and \(R\) is the gas constant (8.3145 J mol K ). A simple relationship can be found by integrating Equation \ref{1} between two pressure-temperature endpoints: \[\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right) \label{2}\] where \(P_1\) and \(P_2\) are the vapor pressures at two temperatures \(T_1\) and \(T_2\). Equation \ref{2} is known as the and allows us to estimate the vapor pressure at another temperature, if the vapor pressure is known at some temperature, and if the enthalpy of vaporization is known. The order of the temperatures in Equation \ref{2} matters as the Clausius-Clapeyron Equation is sometimes written with a negative sign (and switched order of temperatures): \[\ln \left( \dfrac{P_1}{P_2} \right) = - \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_1}- \dfrac{1}{T_2} \right) \label{2B} \] The vapor pressure of water is 1.0 atm at 373 K, and the enthalpy of vaporization is 40.7 kJ mol . Estimate the vapor pressure at temperature 363 and 383 K respectively. Using the Clausius-Clapeyron equation (Equation \(\ref{2B}\)), we have: \[\begin{align} P_{363} &= 1.0 \exp \left[- \left(\dfrac{40,700}{8.3145}\right) \left(\dfrac{1}{363\;K} -\dfrac{1}{373\; K}\right) \right] \nonumber \\[4pt] &= 0.697\; atm \nonumber \end{align} \nonumber\] \[\begin{align} P_{383} &= 1.0 \exp \left[- \left( \dfrac{40,700}{8.3145} \right)\left(\dfrac{1}{383\;K} - \dfrac{1}{373\;K} \right) \right] \nonumber \\[4pt] &= 1.409\; atm \nonumber \end{align} \nonumber\] Note that the increase in vapor pressure from 363 K to 373 K is 0.303 atm, but the increase from 373 to 383 K is 0.409 atm. The increase in vapor pressure is not a linear process. We can use the Clausius-Clapeyron equation to construct the entire vaporization curve. There is a deviation from experimental value, that is because the enthalpy of vaporization varies slightly with temperature. The Clausius-Clapeyron equation can be also applied to sublimation; the following example shows its application in estimating the heat of sublimation. The vapor pressures of ice at 268 K and 273 K are 2.965 and 4.560 torr respectively. Estimate the heat of sublimation of ice. The enthalpy of sublimation is \(\Delta{H}_{sub}\). Use a piece of paper and derive the Clausius-Clapeyron equation so that you can get the form: \[\begin{align} \Delta H_{sub} &= \dfrac{ R \ln \left(\dfrac{P_{273}}{P_{268}}\right)}{\dfrac{1}{268 \;K} - \dfrac{1}{273\;K}} \nonumber \\[4pt] &= \dfrac{8.3145 \ln \left(\dfrac{4.560}{2.965} \right)}{ \dfrac{1}{268\;K} - \dfrac{1}{273\;K} } \nonumber \\[4pt] &= 52,370\; J\; mol^{-1}\nonumber \end{align} \nonumber\] Note that the heat of sublimation is the sum of heat of melting (6,006 J/mol at 0°C and 101 kPa) and the heat of vaporization (45,051 J/mol at 0 °C). Show that the vapor pressure of ice at 274 K is higher than that of water at the same temperature. Note the curve of vaporization is also called the curve of evaporization. Calculate \(\Delta{H_{vap}}\) for ethanol, given vapor pressure at 40 C = 150 torr. The normal boiling point for ethanol is 78 C. Recognize that we have TWO sets of \((P,T)\) data: We then directly use these data in Equation \ref{2B} \[\begin{align*} \ln \left(\dfrac{150}{760} \right) &= \dfrac{-\Delta{H_{vap}}}{8.314} \left[ \dfrac{1}{313} - \dfrac{1}{351}\right] \\[4pt] \ln 150 -\ln 760 &= \dfrac{-\Delta{H_{vap}}}{8.314} \left[ \dfrac{1}{313} - \dfrac{1}{351}\right] \\[4pt] -1.623 &= \dfrac{-\Delta{H_{vap}}}{8.314} \left[ 0.0032 - 0.0028 \right] \end{align*}\] Then solving for \(\Delta{H_{vap}}\) \[\begin{align*} \Delta{H_{vap}} &= 3.90 \times 10^4 \text{ joule/mole} \\[4pt] &= 39.0 \text{ kJ/mole} \end{align*} \] It is important to not use the Clausius-Clapeyron equation for the solid to liquid transition. That requires the use of the more general \[\dfrac{dP}{dT} = \dfrac{\Delta \bar{H}}{T \Delta \bar{V}} \nonumber\] where \(\Delta \bar{H}\) and \(\Delta \bar{V}\) is the molar change in enthalpy (the enthalpy of fusion in this case) and volume respectively between the two phases in the transition.
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Saturated compounds such as the alkanes and their derivatives, which have normal tetrahedral angles for the bonds to carbon, can be formulated readily in terms of atomic orbitals with \(sp^3 \: \sigma\) bonds to carbon. An example is shown in Figure 6-11, which also shows how an atomic-orbital model can be drawn in abbreviated style. The lines in this drawing correspond to bonds and are labeled as \(sp^3\) with \(sp^3\) (the overlapping orbitals of the \(C-C\) bond) or as \(sp^3\) with \(s\) (the overlapping orbitals of the \(C-H\) bonds). Many important molecules such as ammonia, water, and hydrogen fluoride have atoms with unshared pairs of electrons: If we formulate each of these molecules in such a way to minimize repulsions between like charges, a basically tetrahedral arrangement will be expected because this will place the nuclei (and electron pairs) as widely separated as possible. The water molecule could be formulated this way, as in \(6\), with the oxygen at the center of the tetrahedron: The simple picture predicts that the \(H-O-H\) bond angle should be tetrahedral, \(109.5^\text{o}\). But actually it is \(104.5^\text{o}\). There are two schools of thought as to why the angle is \(104.5^\text{o}\). One idea is that the repulsion model is too simple and has to be modified to take into account that the repulsion is more severe between pairs of unshared electrons than between electrons in bonding orbitals on the same atom. This is because when a bond is formed between nuclei, the attraction of the nuclei for the electrons shrinks the orbitals available to the bonding electrons, thereby reducing their electrostatic repulsion with other pairs. The degree of repulsion between electron pairs diminishes in the sequence: unshared pairs . unshared pairs \(>\) unshared pairs . bonding pairs \(>\) bonding pairs . bonding pairs. From this, we expect that in water the \(H-O-H\) angle will be than tetrahedral, because the larger repulsion between the two unshared pairs will tend to push the bonding pairs closer together. A similar, but smaller, effect is expected for ammonia because now the repulsion is only between the one unshared pair and bonding pairs. The ammonia \(H-N-H\) angle is \(107.3^\text{o}\), which is only slightly smaller than the tetrahedral value of \(109.5^\text{o}\). The alternative point of view of why the bond angle of water is \(104.5^\text{o}\) starts with the premise that, in the simplest approximation, the angle should be \(90^\text{o}\)! To see how this comes about let us compare \(H:Be:H\) with \(H:\underset{\cdot \cdot}{\ddot{O}}:H\). You will recall that to form two bonds to \(Be\), we had to promote an electron and change the electronic configuration to the valence configuration, \(\left( 2s \right)^1 \left( 2p \right)^1\). The situation with \(H_2O\) is different in that the oxygen ground state and valence state are the same, \(\left( 2s \right)^2 \left( 2p_x \right)^1 \left( 2p_y \right)^1 \left( 2p_z \right)^1\). This means we could form two  bonds to oxygen using the \(2p_y\) and \(2p_z\) orbitals at an angle of \(90^\text{o}\) (Figure 6-12). Now, to explain why the \(H-O-H\) bond angles are \(104.5^\text{o}\) instead of \(90^\text{o}\), we can say that the repulsion between the hydrogen nuclei is expected to widen the bond angle. An argument in favor of this formulation is provided by the bond angle in \(H_2S\), which is \(92.2^\text{o}\). This is much closer to the \(90^\text{o}\) expected for \(p\)-bond orbitals and the hydrogens in \(H_2S\) would not be expected to repel each other as much as in \(H_2O\) because sulfur is a larger atom than oxygen. Both ways of formulating the orbitals used in the bonding of water molecules are in current use. Arguments can be advanced in favor of both. Highly sophisticated quantum-mechanical calculations, which we will say more about later, suggest that oxygen in water molecules uses orbitals that are \(18\% \: s\) and \(82\% \: p\) in its bonds (\(sp^{4.5}\)), and furthermore, that the unshared pairs are in orbitals [ one pair as \(\left( 2s \right)^2\) and the other as \(\left( 2p \right)^2\)]. Each of the unshared electron-pair orbitals of oxygen in water is calculated to be about \(40\% \: s\) and \(60\% \: p\) (\(sp^{1.5}\)). The results are hardly clearcut, but the bonding orbitals are considerably closer to \(sp^3\) (\(25\% \: s\) and \(75\% \: p\)) than they are to \(100\% \: p\). We recommend that the bonding orbitals of nitrogen and oxygen be considered to be \(sp^3\) and the unshared pairs designated simply as \(\left( n \right)^2\). An abbreviated atomic orbital model of methanol, \(CH_3OH\), made on this basis is shown in Figure 6-13. Recall from Chapter 2 that bond angles in compounds with carbon-carbon double bonds such as ethene are closer to \(120^\text{o}\) than to the normal tetrahedral value of \(109.5^\text{o}\). There are several way sin which a carbon-carbon double bond can be formulated in terms of atomic-orbital models. One very popular approach is to consider that ethene has two \(sp^2\)-hybridized carbons that form one carbon-carbon \(\sigma\) bond and four carbon-hydrogen \(\sigma\) bonds by overlap of the six \(sp^2\) orbitals, as shown in Figure 6-14. The remaining carbon-carbon bond is formulated as arising from overlap of the two \(p\) orbitals, one on each carbon, that are not utilized in making the \(sp^2\) hybrids. Sidewise overlap of \(p\) orbitals is called to distinguish it from the endwise of the type we have discussed previously (Figure 6-15). The resulting differs from the in that electron density is concentrated in the regions above and below the bond axis rather than along the bond axis. Formulations of ethene in this way suggests that it should be a planar molecule with \(H-C-H\) angles of \(120^\text{o}\). Ethene is indeed planar, but its \(H-C-H\) angles are found to be \(117^\text{o}\), rather than the \(120^\text{o}\) predicted for \(sp^2\) bonds. An explanation of this discrepancy using further electron-repulsion arguments will be discussed later in the chapter. The simple elegance of the \(\sigma\)-\(\pi\) model of ethene should not be taken as proving that there actually are two different kinds of bonds between the carbons. The \(\sigma\)-\(\pi\) representation of double bonds is not really unique. Given \(sp^2\) hybridization of the carbons so there are \(sp^2\)-\(\sigma\) bonds to the hydrogens, it is possible to take the \(sp^2\) and \(p\) orbitals used for the \(\sigma\) and \(\pi\) bonds, rehybridize them, and so derive a new set of overlapping orbitals for the double bond. These orbitals are called \(\tau\) ( ) and can be represented by two banana-shaped orbitals between the carbons (Figure 6-16). The result is two completely equivalent \(C-C\) bonds. The \(\tau\) model has the advantage of offering a striking parallel to ball-and-stick models, whereas the \(\sigma\)-\(\pi\) model is of particular value as a basis for quantitative calculations, as will be discussed in Chapter 21.   Using the \(\sigma\)-\(\pi\) model of double bonds, we conclude that the twisted configuration shown in Figure 6-17 should not be very stable. Here the \(p\) orbitals are not in position to overlap effectively in the \(\pi\) manner. The favored configuration is expected to have the axes of the \(p\)-\(\pi\) orbitals parallel. Because considerable energy would have to be expended to break the \(p\)-\(\pi\) double bond and to permit rotation about the remaining \(sp^2\)-\(\sigma\) bond, restricted rotation and stable cis-trans isomers are expected. Similar conclusions can be reached on the basis of the \(\tau\) model of the double bond. Ethyne, \(C_2H_2\), is an organic compound that usually is formulated with \(sp\) hybrid bonds. The carbon-hydrogen framework is built up through \(\sigma\) overlap of two \(sp\)-hybrid orbitals, one from each carbon atom, to form a \(C-C\) bond, and \(\sigma\) overlap of the remaining \(sp\) orbitals with the \(s\) orbital of two hydrogens to form \(C-H\) bonds. The remaining carbon-carbon bonds result through sidewise \(\pi\) overlap of the pure \(p\) orbitals, as shown in Figure 6-18. This model fits well with the properties of the ethyne molecule being linear (bond angles of \(180^\text{o}\). Also, the \(C-H\) bonds in ethyne are different from those in ethene or ethane, as judged by their \(C-H\) stretching and bending frequencies in the infrared (Chapter 9), their bond energies, (Table 4-6), and their acidities ( ). These differences in properties are in keeping with the different states of hybridization of the carbon orbitals that we have postulated for ethane, ethene, and ethyne. A summary of the directional character of the \(s\)-\(p\) hybrid atomic orbitals discussed so far is given in Table 6-2. By referring to this table, it usually is possible to deduce the nature of the bonding orbitals for most organic compounds from the molecular geometry, if this is known, Thus a tetrahedral molecule \(AX_4\) with four attached ligands uses \(sp^3\) hybrid orbitals localized on atom \(A\); a planar triangular molecule \(AX_3\) with three attached ligands at angles of \(120^\text{o}\) is \(sp^2\) hybridized at atom \(A\); a linear molecule \(AX_2\) with two ligands is \(sp\) hybridized at \(A\). Applying the converse of these rules, one should be able to predict molecular geometry by making reasonable assumptions as to the state of hybridization for each atom in the molecule. Obviously in doing this we have to take account of unshared electron pairs. Prediction is easy if unshared pairs are absent. Thus, four attached ligands, as in \(CH_4\), \(CCl_4\), or \(BF_4^\ominus\), imply \(sp^3\) hybridization at the central atom and therefore a tetrahedral arrangement of ligands. Three ligands, as bonded to carbon in \(CH_3^\oplus\) or to boron in \(BF_3\), imply \(sp^2\) hybridization for the central atom and a planar triangular arrangement if ligands. Two ligands, as in \(CO_2\), imply \(sp\) hybridization and linear geometry. In many of our later discussions of organic reactions, we will be concerned with cationic, radical, and anionic carbon species that are substitution products of \(CH_3^\oplus\), \(CH_3 \cdot\), and \(CH_3:^\ominus\). Because of the importance of these entities, you should know how to formulate them and related substances, such as \(^\ominus: \ddot{N}H_2\), with atomic orbitals. Perhaps the most straightforward way is to start from \(CH_4\) and see what changes in the \(C-H\) bonds we would expect as the result of the hypothetical processes: \(CH_4 \rightarrow CH_3:^\ominus + H^\oplus\), \(CH_4 \rightarrow CH_3^\oplus + H:^\ominus\), and \(CH_4 \rightarrow CH_3 \cdot + H \cdot\). Methane is tetrahedral with \(sp^3\) carbon bonding orbitals. Removal of \(H^\oplus\) gives \(CH_3:^\ominus\), which corresponds in electronic structure to \(H_3N:\) and, for the same reasons, should have a pyramidal shape with nearly tetrahedral \(H-C-H\) angles. Removal of \(H:^\ominus\) from \(CH_4\) to give \(CH_3^\oplus\) with bonding electrons, suggests a change to \(sp^2\) bonding orbitals for the carbon and planar geometry with \(H-C-H\) angles of \(120^\text{o}\). The radical, \(CH_3 \cdot\) presents a special problem. We can think of it as being formed by the loss of \(H \cdot\) from \(CH_4\), by adding an electron to \(CH_3^\oplus\), or by removing an electron from pyramidal \(CH_3:^\ominus\). We can formulate \(CH_3 \cdot\) with \(Sp^2\) orbitals for the \(C-H\) bonds and the extra electron in a \(p\) orbital, or with \(sp^3\) orbitals for the \(C-H\) bonds and the extra electron in an \(sp^3\) orbital: The actual structure of \(CH_3 \cdot\) has the hydrogens and carbons in a plane (left). Therefore it appears that the repulsions between the bonding electron pairs is greater than the repulsions between the extra electron and the bonding pairs. The actual structure corresponds to the one in which the bonding pairs are as far apart as possible. Molecules of the type \(AX_4\), which have four identical ligands on the central atom and no unshared electrons on \(A\) (e.g., \(CH_4\) and \(CCl_4\)), are expected to be, and are, tetrahedral. By the same reasoning, three electron pairs around one atom should seek a planar arrangement with \(120^\text{o}\) angles to minimize electron repulsion; accordingly, species of the type \(AX_3\), which have no unshared pairs on \(A\) (e.g., \(BF_3\) and \(CH_3^\oplus\)), have this geometry. With only two electron pairs, the preferred arrangement is linear. The bond angles of compounds with multiple bonds can be explained similarly. For example, in ethene the four electrons of the double bond occupy the region in space between the two carbon nuclei. The situation at either carbon is rather like the \(AX_3\) case, except that one of the ligands now has a double complement of bonding electrons: Therefore the carbon orbitals are expected to be directed in one plane to give bond angles that deviate somewhat from \(120^\text{o}\) because of the high density of electrons in the multiple bond. Thus the \(H-C-H\) angle shrinks to \(117^\text{o}\), whereas the \(H-C=C\) angles open up to \(122^\text{o}\), because repulsion between electrons in the \(H-C=C\) bonds is greater than between electrons in the \(H-C-H\) bonds. Electron-attracting power (or ) of the ligands also is important in determining bond angles. Thus for compounds of the type \(CH_3X\), in which \(X\) is a more electron-attracting group than carbon, the \(C-X\) bond is polarized in the sense \(H_3 \overset{\delta \oplus}{C} - - - \overset{\delta \ominus}{X}\), and the carbon then should have some of the character of \(CH_3^\oplus\). Thus the \(H-C-H\) angles are expected to be greater than \(109.5^\text{o}\), as in fact they are. In chloromethane, for example, the \(H-C-H\) angle is \(111^\text{o}\). Also, we can explain on the basis of electron repulsions why the bond angle in phosphine, \(:PH_3\) (\(93^\text{o}\)), is less than that in ammonia, \(:NH_3\) (\(107.3^\text{o}\)), and the bond angle in \(H: \underset{\cdot \cdot}{\ddot{S}} :H\) (\(92.2^\text{o}\)) is less than that in \(H: \underset{\cdot \cdot}{\ddot{O}}\) (\(104.5^\text{o}\)). The important point is that phosphorus and sulfur are larger atoms than nitrogen and oxygen. This means than the \(H-S-H\) and \(H-P-H\) bond angles can be about \(90^\text{o}\) without bringing the hydrogens and the bonding pairs as close together as they are in \(H_2O\) and \(NH_3\) where the bond angles are near to the tetrahedral value. and (1977)
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An atom is composed of a nucleus containing neutrons and protons with electrons dispersed throughout the remaining space. Electrons, however, are not simply floating within the atom; instead, they are fixed within electronic orbitals. Electronic orbitals are regions within the atom in which electrons have the highest probability of being found. There are multiple orbitals within an atom. Each has its own specific energy level and properties. Because each orbital is different, they are assigned specific : 1s, 2s, 2p 3s, 3p,4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p. The numbers, (n=1,2,3, etc.) are called and can only be positive numbers. The letters ( ) represent the ( ) and the orbital angular momentum quantum number may be 0 or a positive number, but can never be greater than n-1. Each letter is paired with a specific value: An orbital is also described by its (m ). The magnetic quantum number can range from to + . This number indicates how many orbitals there are and thus how many electrons can reside in each atom. Orbitals that have the same or identical energy levels are referred to as An example is the 2p orbital: 2p has the same energy level as 2p . This concept becomes more important when dealing with molecular orbitals. The states that no two electrons can have the same exact orbital configuration; in other words, the same quantum numbers. However, the electron can exist in spin up (m = +1/2) or with spin down (m = -1/2) configurations. This means that the s orbital can contain up to two electrons, the p orbital can contain up to six electrons, the d orbital can contain up to 10 electrons, and the f orbital can contain up to 14 electrons. As discussed in the previous section, the magnetic quantum number (m ) can range from –l to +l. The number of possible values is the number of lobes (orbitals) there are in the s, p, d, and f subshells. As shown in Table 1, the s subshell has one lobe, the p subshell has three lobes, the d subshell has five lobes, and the f subshell has seven lobes. Each of these lobes is labeled differently and is named depending on which plane the lobe is resting in. If the lobe lies along the x plane, then it is labeled with an x, as in 2p . If the lobe lies along the xy plane, then it is labeled with a xy such as d . Electrons are found within the lobes. The plane (or planes) that the orbitals do not fill are called nodes. These are regions in which there is a 0 probability density of finding electrons. For example, in the d orbital, there are nodes on planes xz and yz. This can be seen in Figure \(\Page {1}\). There are two types of nodes, angular and radial nodes. are typically flat plane (at fixed angles), like those in the diagram above. The quantum number determines the number of angular nodes in an orbital. are spheres (at fixed radius) that occurs as the principal quantum number increases. The total nodes of an orbital is the sum of angular and radial nodes and is given in terms of the \(n\) and \(l\) quantum number by the following equation: \[ N = n-l -1\] For example, determine the nodes in the 3p orbital, given that n = 3 and = 1 (because it is a p orbital). The total number of nodes present in this orbital is equal to n-1. In this case, 3-1=2, so there are 2 total nodes. The quantum number determines the number of angular nodes; there is 1 angular node, specifically on the xy plane because this is a p orbital. Because there is one node left, there must be one radial node. To sum up, the 3p orbital has 2 nodes: 1 angular node and 1 radial node. This is demonstrated in Figure 2. Another example is the 5d orbital. There are four nodes total (5-1=4) and there are two angular nodes (d orbital has a quantum number =2) on the xz and zy planes. This means there there must be two radial nodes. The number of radial and angular nodes can only be calculated if the principal quantum number, type of orbital (s,p,d,f), and the plane that the orbital is resting on (x,y,z, xy, etc.) are known. We can think of an atom like a hotel. The nucleus is the lobby where the protons and neutrons are, and in the floors above, we find the rooms (orbitals) with the electrons. The principal quantum number is the floor number, the subshell type lets us know what type of room it is (s being a closet, p being a single room, d having two adjoining rooms, and f being a suit with three rooms) , the magnetic quantum number lets us know how many beds there are in the room, and two electrons can sleep in one bed (this is because each has a different spin; -1/2 and 1/2). For example, on the first floor we have the s orbital. The s orbital is a closet and has one bed in it so the first floor can hold a total of two electrons. The second floor has the room styles s and p. The s is a closet with one bed as we know and the p room is a single with three beds in it so the second floor can hold a total of 8 electrons. Each orbital, as previously mentioned, has its own energy level associated to it. The lowest energy level electron orbitals are filled first and if there are more electrons after the lowest energy level is filled, they move to the next orbital. The order of the electron orbital energy levels, starting from least to greatest, is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p. Since electrons all have the same charge, they stay as far away as possible because of repulsion. So, if there are open orbitals in the same energy level, the electrons will fill each orbital singly before filling the orbital with two electrons. For example, the 2p shell has three p orbitals. If there are more electrons after the 1s, and 2s orbitals have been filled, each p orbital will be filled with one electron first before two electrons try to reside in the same p orbital. This is known as . The way electrons move from one orbital to the next is very similar to walking up a flight of stairs. When walking up stairs, you place one foot on the first stair and then another foot on the second stair. At any point in time, you can either stand with both feet on the first stair, or on the second stair but it is impossible to stand in between the two stairs. This is the way electrons move from one electron orbital to the next. Electrons can either jump to a higher energy level by absorbing, or gaining energy, or drop to a lower energy level by emitting, or losing energy. However, electrons will never be found in between two orbitals.
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The construction of linear combinations of the basis of atomic movements allows the vibrations belonging to irreducible representations to be investigated. The wavefunction of these symmetry equivalent orbitals is referred to as , or SALCs. SALCs (Symmetry Adapted Linear Combinations) are the linear combinations of basis sets composed of the stretching vectors of the molecule. The SALCs of a molecule can help determine binding schemes and symmetries. The procedure used to determine the SALCs of a molecule is also used to determine the LCAO of a molecule. The LCAO, Linear Combination of Atomic Orbitals, uses the basis set of atomic orbitals instead of stretching vectors. The LCAO of a molecule provides a detailed description of the molecular orbitals, including the number of nodes and relative energy levels. Symmetry adapted linear combinations are the sum over all the basis functions: \[\phi_{i} =\displaystyle\sum_{j} c_{ij} b_{j} \label{1}\] \(ϕ_i\) is the i SALC function, bj is the j basis function, and \(c_{ij}\) is a coefficient which controls how much of \(b_j\) appears in \(ϕ_i\). In method two, the projection operator is used to obtain the coefficients consistent with each irreducible representation. The SALCs of a molecule may be constructed in ways. The first method uses a basis set composed of the irreducible representation of the stretching modes of the molecule. On the other hand, the second method uses a projection operator on each stretching vector. When determining the irreducible representations of the stretching modes, the reducible representations for all the vibrational modes must first be determined. Basis vectors are assigned characters and are treated as individual objects. A In order to understand and construct SALCs, a background in is required. The identification of the point group of the molecule is essential for understanding how the application of operations affects the molecule. This allows for the determination of the nature of the stretching modes. As a review, let’s first determine the stretching modes of water together. The first step in determining stretching modes of a molecule is to add the characters contained in the x, y, and z rows to obtain the total reducible representation of the xyz coordinates, Γ . Γ can also be found by applying the symmetry operations to the three vectors (x, y, and z) of the coordinate system of the molecule. The next step involves the investigation of the atoms that remain unchanged when an operation is applied, Γ . This step refers to the unmoved atoms (UMA). Multiplying Γ and Γ gives the reducible representation for the molecule referred to as Γ . The Γ is the reducible representation for all the modes of the molecule (vibrational, rotational, and translational) and can also be determined by applying the symmetry operations to each coordinate vector (x, y, and z) on each atom. Γ is then reduced to later give the stretching modes that are unique to the molecule. First, the reduction formula is applied to decompose the reducible representation: \[a_{i}=\dfrac{1}{h}\displaystyle\sum_{R}(X^{R}X_{i}^{R}C^{R}) \label{2}\] Here, is the number of times the irreducible representation will appear in the initial reducible representation. The order of the point group is represented by is an operation of the group; is a character of the operation in the reducible representation; is a character of the operation in the irreducible representation, and is the number of members in the class to which belongs. Applying this formula and subtracting the representations obtained from the basis functions x, y, z, R , R , and R (for the translations and rotations of the molecule) gives the irreducible representation that corresponds to the vibrational states of the molecule: Γ = 2a + b A simple check can be performed to determine that the right number of modes was obtained. For linear molecules (3N-5) gives the correct number of . For molecules with any other shape otherwise known as non-linear molecules, the formula is (3N-6). represents the number of atoms in a molecule. Let’s double check the above water example: \[(3N-6) N=3 \label{3}\] \[[3(3)-6] = 3\] Water should have three vibrational modes. When the irreducible representation was obtained, it was seen that water has two a modes and a b mode for a total of three. When double checking that you have the correct number of normal modes for other molecules, remember that the irreducible representation E is doubly degenerate and counts as two normal modes. T is triply degenerate and counts for three normal modes, etc. There are multiple ways of constructing the SALCs of a molecule. The first method uses the known symmetries of the stretching modes of the molecule. To investigate this method, the construction of the SALCs of water is examined. Water has three vibrational states, 2a +b . Two of these vibrations are stretching modes. One is symmetric with the symmetry A , and the other is antisymmetric with the symmetry B . While looking for the SALC of a molecule, one uses vectors represented by as the basis set. The vectors demonstrate the irreducible representations of molecular vibrations. The SALCs of water can be composed by creating a linear combination of the stretching vectors. \[\phi(A_{1})=b_{1}+b_{2}\] and \[\phi(B_{1})=b_{1}-b_{2} \label{4}\] The final step in constructing the SALCs of water is to normalize expressions. To normalize the SALC, multiply the entire expression by the normalization constant that is the inverse of the square root of the sum of the squares of the coefficients within the expression. \[\phi_{i}=N\displaystyle\sum_{j}c_{ij}b_{j} \label{5}\] \[N=\dfrac{1}{\sqrt{\displaystyle\sum_{j=1}^{n}c_{ij}^{2}}} \label{6}\] \[\phi(A_{1})=\dfrac{1}{\sqrt{2}}(b_{1}+b_{2})\] and \[\phi(B_{1})=\dfrac{1}{\sqrt{2}}(b_{1}-b_{2})\] Normalizing the SALCs ensures that the magnitude of the SALC is unity, and therefore the dot product of any SALC with itself will equal one. The other method for constructing SALCs is the projection operator method. The SALC of a molecule can be constructed in the same manner as the LCAO, Linear Combination of Atomic Orbitals, however the basis set differs. While looking for the SALCs of a molecule, one uses vectors represented by , on the other hand, while looking for the LCAO of a molecule, one uses atomic orbitals as the basis set. The vectors demonstrate the possible vibration of the molecule. While constructing SALCs, the basis vectors can be treated as individual vectors. Let’s take a look at how to construct the SALC for water. The first step in constructing the SALC is to label all vectors in the basis set. Below are the bond vectors of water that will be used as the basis set for the SALCs of the molecule. Next, the basis vector, , is transformed by , the th symmetry operation of the molecule’s point group. As the vector of the basis set is transformed, record the vector that takes its place. Water is a member of the point group C . The Symmetry elements of the C point group are E, C , σ , σ . The i SALC function, using the vector . Once the transformations have been determined, the SALC can be constructed by taking the sum of the products of each character of a representation within the point group and the corresponding transformation. The SALCs functions are the collective transformations of the basis sets represented by ϕ where is the character of the th irreducible representation and the th symmetry operation. \[\phi_{i}=\displaystyle\sum_{j}X_{i}(j)T_{j}\nu \label{7}\] Projection Operator method for C The final step in constructing the SALCs of water is to normalize expressions. There are two SALCs for the water molecule, ϕ (A ) and ϕ (B ). This demonstrates that water has two stretching modes, one is a totally symmetric stretch with the symmetry, A , and the other is an antisymmetric stretch with the symmetry B . Both methods of construction result in the same SALCs. Only irreducible representations corresponding to the symmetries of the stretching modes of the molecule will produce a SALC that is non-zero. Method 1 only utilized the known symmetries of the vibrational modes. All irreducible representations of the point group were used, but the representations that were not vibrational modes resulted in SALCs equal to zero. Therefore, with the SALCs of a molecule given, all the symmetries of the stretching modes are identified. This allows for a clearer understanding of the spectroscopy of the molecule. Even though vibrational modes can be observed in both infrared and Raman spectroscopy, the SALCs of a molecule cannot identify the magnitude or frequency of the peak in the spectra. The normalized SALCs can, however, help to determine the relative magnitude of the stretching vectors. The magnitude can be determined by the equation below. \[a \cdot b= |a||b|cos \theta \label{8}\] The resulting A and B symmetries for the above water example are each active in both Raman and IR spectroscopies, according to the C character table. If the vibrational mode allows for a change in the dipole moment, the mode can be observed through infrared spectroscopy. If the vibrational mode allows for a change in the polarization of the molecule, the mode can be observed through Raman spectroscopy. Both stretching and bending modes are seen in the spectra, however only stretching modes are expressed in the SALCs. The SALCs of a molecule can also provide insight to the geometry of a molecule. For example, SALCs can aid in determining the differences between para-difluorobenzene and ortho-difluorobenzene. The SALCs for these two molecules are given below. \[\phi (A_{g})= \dfrac{1}{2} (b_{1} + b_{2} + b_{3} +b_{4})\] \[\phi (B_{1g})= \dfrac{1}{2} (b_{1} - b_{2} + b_{3} -b_{4})\] \[\phi (B_{2u})= \dfrac{1}{2} (b_{1} - b_{2} - b_{3} +b_{4})\] \[\phi (B_{3u})= \dfrac{1}{2} (b_{1} + b_{2} - b_{3} -b_{4})\] \[\phi _{1} (A_{1})= \dfrac{1}{\sqrt{2}} (b_{1} + b_{2} + b_{3} +b_{4})\] \[\phi _{2} (A_{1})= \dfrac{1}{\sqrt{2}} (b_{1} - b_{2} + b_{3} -b_{4})\] \[\phi _{1} (B_{1})= \dfrac{1}{\sqrt{2}} (b_{1} - b_{2} - b_{3} +b_{4})\] \[\phi _{2} (B_{1})= \dfrac{1}{\sqrt{2}} (b_{1} + b_{2} - b_{3} -b_{4})\] From the SALCs, it is seen that para-difluorobenzene has four stretching modes and ortho-difluorobenzene has only two. Therefore, it is no surprise that the vibrational spectroscopy of the para-difluorobenzene shows more peaks than the ortho-difluorobenzene. Once the LCAOs of the molecule have been determined, the expressions can be interpreted into images of the orbitals bonding. If two orbitals are of the same sign in the expression, the electrons in the orbitals are in phase with each other and are bonding. If two orbitals are of the opposite sign in the expression, the electrons in the orbitals are out of phase with each other and are antibonding. The image below shows the atomic orbitals' phases (or signs) as red or blue lobes. Any separation between two antibonding atomic orbitals is a planar node. As the number of nodes increases, so does the level of antibonding. This allows for the LCAO to place the molecular orbitals in order of increasing energy, which can be used in constructing the molecular orbital (MO) diagram of the molecule. The irreducible representation used to construct the LCAO is used to describe the MOs. The LCAOs for water are shown below with red dotted lines showing the nodes. Notice the nodes for the p orbitals are in a different plane than the s orbitals of the hydrogens, so these are degenerate and nonbonding. Information from the LCAO of water can also be used to analyze and anticipate the adsorption of water onto various surfaces. Evarestov and Bandura used this technique to identify the water adsorption on Y-doped BaZrO and TiO Applying a combination of Methods 1 and 2, the SALCs for CBr H can be determined. The point group of this molecule is C , making it similar to the determination of SALCs for water. However, the central carbon contains more than one type of attached atom; therefore, the stretching analysis must be performed in pieces. First, the C-H stretches are examined, followed by the C-Br stretches: Applying the projection operator method to C-Br and C-H stretches individually, the SALCs are obtained in the same fashion as before. The results are normalized and the following SALCs are obtained for the C molecule CBr H : \[\phi CBr(A_{1})=\frac{1}{\sqrt{2}}(b_{1}+b_{2})\] \[\phi CBr(B_{1}) =\frac{1}{\sqrt{2}} (b_{1}-b_{2})\] \[\phi CH(A_{1}) =\frac{1}{\sqrt{2}} (a_{1}+a_{2})\] \[\phi CH(B_{2}) =\frac{1}{\sqrt{2}} (a_{1}-a_{2})\] To obtain the SALCs for PtCl , the same general method is applied. However, even though the point group of the molecule is D , the cyclic subgroup C may be used (this is a more simplified character table used for spherically symmetrical molecules). Some manipulation is required in order to use this cyclic subgroup and will be discussed. Below is the C cyclic character table. Notice, there are two rows for E, each singly degenerate. To solve for the characters of E, one must take the sum and difference of the two rows. Then, a reduction can be applied to obtain the easiest possible characters by dividing each row by a common factor (removing the common factor is not necessary, but it does simplify the problem as well as remove any imaginary terms): Sum = [ (1+1) (i-i) (-1-1) (-i+i) ] = (2 0 -2 0) ÷ 2 = E (1 0 -1 0) Difference = [ (1-1) (i+i) (-1+1) (-i-i) ] = (0 2i 0 -2i) ÷ 2i = E (0 1 0 -1) Using the above cyclic group, and the newly obtained characters for , the projection operator can be applied using Method 2 for the construction of SALCs. Normalizing the sum as mentioned in Method 1, the following SALCs are obtained for the D molecule PtCl : \[\phi (A) =\frac{1}{2} (b_{1}+b_{2}+b_{3}+b_{4})\] \[\phi (B)=\frac{1}{2} (b_{1}-b_{2}+b_{3}-b_{4})\] \[\phi (E^{1})=\frac{1}{\sqrt{2}} (b_{1}-b_{3})\] \[\phi (E^{2})=\frac{1}{\sqrt{2}} (b_{2}-b_{4})\] Applying a combination of Methods 1 and 2, the SALCs for the C-H stretches of PF H can be determined. The point group of this molecule is C . The central carbon contains more than one type of attached hydrogen; therefore, the stretching analysis must be performed in pieces. First, the C-H stretches are examined, followed by the C-H stretches: Applying the projection operator method to C-H and C-H stretches individually, the SALCs are obtained in the same fashion as before. The results are normalized and the following SALCs are obtained for the C molecule PF H : \[\phi_{1} A'=\frac{1}{\sqrt{2}} (b_{1} +b_{2} )+a_{1} = \frac{1}{sqrt{3}} (b_{1} +b_{2} +a_{1})\] \[\phi_{2} A'=\frac{1}{\sqrt{2}} (b_{1} +b_{2} )-a_{1} = \frac{1}{sqrt{3}} (b_{1} +b_{2} -a_{1})\] \[\phi A''=\frac{1}{\sqrt{2}} (b_{1} -b_{2} )\] B T (b )= \frac{1}{\sqrt{2}} (b_{1} -b_{2})\) A T (b )= \frac{1}{\sqrt{2}} (b_{3} +b_{4})\) B T (b )= \frac{1}{\sqrt{2}} (b_{3} -b_{4})\) Then add and subtract (for in phase and out of phase) the individual linear combinations found by the projection operator to give the SALCs. ϕ (A ) = \frac{1}{2} (b_{1} +b_{2} +b_{3} +b_{4} )\) ϕ (A ) = \frac{1}{2} (b_{1} +b_{2} -b_{3} -b_{4} )\) ϕ (B ) = \frac{1}{2} (b_{1} -b_{2} +b_{3} -b_{4} )\) ϕ (B ) = \frac{1}{2} (b_{1} -b_{2} -b_{3} +b_{4} )\). 2. AT (b )= \frac{1}{\sqrt{3}} (b_{1} +b_{2} +b_{3} )\) E T (b )= \frac{1}{\sqrt{6}} (2b_{1} -b_{2} -b_{3} )\) E T (b )= \frac{1}{\sqrt{2}} (b_{2} -b_{3} )\) 3.
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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/11%3A_Quantum_Mechanics_and_Atomic_Structure/11.11%3A_Many-Electron_Atoms_and_the_Periodic_Table
Quantum mechanics can account for the periodic structure of the elements, by any measure a major conceptual accomplishment for any theory. Although accurate computations become increasingly more challenging as the number of electrons increases, the general patterns of atomic behavior can be predicted with remarkable accuracy. Figure \(\Page {1}\) shows a schematic representation of a helium atom with two electrons whose coordinates are given by the vectors \(r_1\) and \(r_2\). The electrons are separated by a distance \(r_{12} = |r_1-r_2|\). The origin of the coordinate system is fixed at the nucleus. As with the hydrogen atom, the nuclei for multi-electron atoms are so much heavier than an electron that the nucleus is assumed to be the center of mass. Fixing the origin of the coordinate system at the nucleus allows us to exclude translational motion of the center of mass from our quantum mechanical treatment. The Schrödinger equation operator for the hydrogen atom serves as a reference point for writing the Schrödinger equation for atoms with more than one electron. Start with the same general form we used for the hydrogen atom Hamiltonian \[ (\hat {T} + \hat {V}) \psi = E \psi \label {9-1}\] Include a kinetic energy term for each electron and a potential energy term for the attraction of each negatively charged electron for the positively charged nucleus and a potential energy term for the mutual repulsion of each pair of negatively charged electrons. The He atom Schrödinger equation is \[ \left( -\dfrac {\hbar ^2}{2m_e} (\nabla ^2_1 + \nabla ^2_2) + V (r_1) + V (r_2) + V (r_{12}) \right) \psi = E \psi \label {9-2}\] where \[ V(r_1) = -\dfrac {2e^2}{4 \pi \epsilon _0 r_1} \label {9-3}\] \[ V(r_2) = -\dfrac {2e^2}{4 \pi \epsilon _0 r_2} \label {9-4}\] \[ V(r_{12}) = -\dfrac {e^2}{4 \pi \epsilon _0 r_{12}} \label {9-5}\] Equation \(\ref{9-2}\) can be extended to any atom or ion by including terms for the additional electrons and replacing the He nuclear charge +2 with a general charge Z; e.g. \[V(r_1) = -\dfrac {Ze^2}{4 \pi \epsilon _0 r_1} \label {9-6}\] Equation \(\ref{9-2}\) then becomes \[ \left( -\dfrac {\hbar ^2}{2m_e} \sum _i \nabla ^2_i + \sum _i V (r_i) + \sum _{i \ne j} V (r_{ij}) \right) \psi = E \psi \label {9-7}\] Given what we have learned from the previous quantum mechanical systems we’ve studied, we predict that exact solutions to the multi-electron Schrödinger equation would consist of a family of multi-electron wavefunctions, each with an associated energy eigenvalue. These wavefunctions and energies would describe the ground and excited states of the multi-electron atom, just as the hydrogen wavefunctions and their associated energies describe the ground and excited states of the hydrogen atom. We would predict quantum numbers to be involved, as well. The fact that electrons interact through their electron-electron repulsion means that an exact wavefunction for a multi-electron system would be a single function that depends upon the coordinates of all the electrons; i.e., a multi-electron wavefunction: \[\Psi (r_1, r_2, \cdots r_i) \label{8.3.4}\] Unfortunately, the electron-electron repulsion terms make it impossible to find an exact solution to the Schrödinger equation for many-electron atoms. The most basic ansatz to the exact solutions involve writing a multi-electron wavefunction as a electron wavefunctions \[\psi (r_1, r_2, \cdots , r_i) = \varphi _1 (r_1) \varphi _2 (r_2) \cdots \varphi _i(r_i) \label{8.3.5}\] Obtaining the energy of the atom in the state described by that wavefunction as the of the energies of the one-electron components. By writing the multi-electron wavefunction as a product of single-electron functions in Equation \(\ref{8.3.5}\), we conceptually transform a multi-electron atom into a collection of individual electrons located in individual orbitals whose spatial characteristics and energies can be separately identified. For atoms these single-electron wavefunctions are called atomic orbitals. For molecules, as we will see in the next chapter, they are called molecular orbitals. While a great deal can be learned from such an analysis, it is important to keep in mind that such a discrete, compartmentalized picture of the electrons is an approximation, albeit a powerful one. The specific arrangement of electrons in orbitals of an atom determines many of the chemical properties of that atom and is formulated via the principle, which means "building-up" in German. Aufbau principles determine the order in which atomic orbitals are filled as the atomic number is increased. For the hydrogen atom, the order of increasing orbital energy is given by 1 < 2 = 2 < 3 = 3 = 3 , etc. The dependence of energy on alone leads to extensive degeneracy, which is however removed for orbitals in many-electron atoms. Thus 2 lies below 2 , as already observed in helium. Similarly, 3 , 3 and 3 increase energy in that order, and so on. The 4 is lowered sufficiently that it becomes comparable to 3 . The general ordering of atomic orbitals is summarized in the following scheme: \[ 1s < 2s < 2p < 3s < 3p < 4s \sim 3d < 4p < 5s \sim 4d\\< 5p < 6s \sim 5d \sim 4f < 6p < 7s \sim 6d \sim 5f \label{4}\] and illustrated in Figure \(\Page {2}\). This provides enough orbitals to fill the ground states of all the atoms in the periodic table. For orbitals designated as comparable in energy, e.g., 4 \(\sim\) 3 , the actual order depends which other orbitals are occupied. The energy of atomic orbitals increases as the principal quantum number, \(n\), increases. In any atom with two or more electrons, the repulsion between the electrons makes energies of subshells with different values of \(l\) differ so that the energy of the orbitals increases within a shell in the order < < < Figure \(\Page {2}\) depicts how these two trends in increasing energy relate. The 1 orbital at the bottom of the diagram is the orbital with electrons of lowest energy. The energy increases as we move up to the 2 and then 2 , 3 , and 3 orbitals, showing that the increasing value has more influence on energy than the increasing value for small atoms. However, this pattern does not hold for larger atoms. The 3 orbital is higher in energy than the 4 orbital. Such overlaps continue to occur frequently as we move up the chart. Electrons in successive atoms on the periodic table tend to fill low-energy orbitals first. The arrangement of electrons in the orbitals of an atom is called the of the atom. We describe an electron configuration with a symbol that contains three pieces of information ( Figure \(\Page {3}\)): For example, the notation 2 (read "two–p–four") indicates four electrons in a subshell ( = 1) with a principal quantum number ( ) of 2. The notation 3 (read "three–d–eight") indicates eight electrons in the subshell (i.e., = 2) of the principal shell for which = 3. To determine the electron configuration for any particular atom, we can “build” the structures in the order of atomic numbers. Beginning with hydrogen, and continuing across the periods of the periodic table, we add one proton at a time to the nucleus and one electron to the proper subshell until we have described the electron configurations of all the elements. This procedure is called the , from the German word (“to build up”). Each added electron occupies the subshell of lowest energy available (in the order shown in Figure \(\Page {4}\)), subject to the limitations imposed by the allowed quantum numbers according to the Pauli exclusion principle. Electrons enter higher-energy subshells only after lower-energy subshells have been filled to capacity. Figure \(\Page {3}\) illustrates the traditional way to remember the filling order for atomic orbitals. We will now construct the ground-state electron configuration and orbital diagram for a selection of atoms in the first and second periods of the periodic table. are pictorial representations of the electron configuration, showing the individual orbitals and the pairing arrangement of electrons. We start with a single hydrogen atom (atomic number 1), which consists of one proton and one electron. Referring to either Figure \(\Page {4}\), we would expect to find the electron in the 1 orbital. By convention, the \(m_s=+\dfrac{1}{2}\) value is usually filled first. The electron configuration and the orbital diagram are: Following hydrogen is the noble gas helium, which has an atomic number of 2. The helium atom contains two protons and two electrons. The first electron has the same four quantum numbers as the hydrogen atom electron ( = 1, = 0, = 0, \(m_s=+\dfrac{1}{2}\)). The second electron also goes into the 1 orbital and fills that orbital. The second electron has the same , , and quantum numbers, but must have the opposite spin quantum number, \(m_s=−\dfrac{1}{2}\). This is in accord with the Pauli exclusion principle: No two electrons in the same atom can have the same set of four quantum numbers. For orbital diagrams, this means two arrows go in each box (representing two electrons in each orbital) and the arrows must point in opposite directions (representing paired spins). The electron configuration and orbital diagram of helium are: The = 1 shell is completely filled in a helium atom. The next atom is the alkali metal lithium with an atomic number of 3. The first two electrons in lithium fill the 1 orbital and have the same sets of four quantum numbers as the two electrons in helium. The remaining electron must occupy the orbital of next lowest energy, the 2 orbital (Figure \(\Page {4}\) ). Thus, the electron configuration and orbital diagram of lithium are: An atom of the alkaline earth metal beryllium, with an atomic number of 4, contains four protons in the nucleus and four electrons surrounding the nucleus. The fourth electron fills the remaining space in the 2 orbital. An atom of boron (atomic number 5) contains five electrons. The = 1 shell is filled with two electrons and three electrons will occupy the = 2 shell. Because any subshell can contain only two electrons, the fifth electron must occupy the next energy level, which will be a 2 orbital. There are three degenerate 2 orbitals ( = −1, 0, +1) and the electron can occupy any one of these orbitals. When drawing orbital diagrams, we include empty boxes to depict any empty orbitals in the same subshell that we are filling. Carbon (atomic number 6) has six electrons. Four of them fill the 1 and 2 orbitals. The remaining two electrons occupy the 2 subshell. We now have a choice of filling one of the 2 orbitals and pairing the electrons or of leaving the electrons unpaired in two different, but degenerate, orbitals. The orbitals are filled as described by : the lowest-energy configuration for an atom with electrons within a set of degenerate orbitals is that having the maximum number of unpaired electrons. Thus, the two electrons in the carbon 2 orbitals have identical , , and quantum numbers and differ in their quantum number (in accord with the Pauli exclusion principle). The electron configuration and orbital diagram for carbon are: Nitrogen (atomic number 7) fills the 1 and 2 subshells and has one electron in each of the three 2 orbitals, in accordance with Hund’s rule. These three electrons have unpaired spins. Oxygen (atomic number 8) has a pair of electrons in any one of the 2 orbitals (the electrons have opposite spins) and a single electron in each of the other two. Fluorine (atomic number 9) has only one 2 orbital containing an unpaired electron. All of the electrons in the noble gas neon (atomic number 10) are paired, and all of the orbitals in the = 1 and the = 2 shells are filled. The electron configurations and orbital diagrams of these four elements are: Similarly, the abbreviated configuration of lithium can be represented as [He]2 , where [He] represents the configuration of the helium atom, which is identical to that of the filled inner shell of lithium. Writing the configurations in this way emphasizes the similarity of the configurations of lithium and sodium. Both atoms, which are in the alkali metal family, have only one electron in a valence subshell outside a filled set of inner shells. \[\ce{Li:[He]}\,2s^1\\ \ce{Na:[Ne]}\,3s^1\] The alkaline earth metal magnesium (atomic number 12), with its 12 electrons in a [Ne]3 configuration, is analogous to its family member beryllium, [He]2 . Both atoms have a filled subshell outside their filled inner shells. Aluminum (atomic number 13), with 13 electrons and the electron configuration [Ne]3 3 , is analogous to its family member boron, [He]2 2 . Beginning with the transition metal scandium (atomic number 21), additional electrons are added successively to the 3 subshell. This subshell is filled to its capacity with 10 electrons (remember that for = 2 [ orbitals], there are 2 + 1 = 5 values of , meaning that there are five orbitals that have a combined capacity of 10 electrons). The 4 subshell fills next. Note that for three series of elements, scandium (Sc) through copper (Cu), yttrium (Y) through silver (Ag), and lutetium (Lu) through gold (Au), a total of 10 electrons are successively added to the ( – 1) shell next to the shell to bring that ( – 1) shell from 8 to 18 electrons. For two series, lanthanum (La) through lutetium (Lu) and actinium (Ac) through lawrencium (Lr), 14 electrons ( = 3, 2 + 1 = 7 values; thus, seven orbitals with a combined capacity of 14 electrons) are successively added to the ( – 2) shell to bring that shell from 18 electrons to a total of 32 electrons. What is the electron configuration and orbital diagram for a phosphorus atom? What are the four quantum numbers for the last electron added? The atomic number of phosphorus is 15. Thus, a phosphorus atom contains 15 electrons. The order of filling of the energy levels is 1 , 2 , 2 , 3 , 3 , 4 , . . . The 15 electrons of the phosphorus atom will fill up to the 3 orbital, which will contain three electrons: The last electron added is a 3 electron. Therefore, = 3 and, for a -type orbital, = 1. The value could be –1, 0, or +1. The three orbitals are degenerate, so any of these values is correct. For unpaired electrons, convention assigns the value of \(+\dfrac{1}{2}\) for the spin quantum number; thus, \(m_s=+\dfrac{1}{2}\). Identify the atoms from the electron configurations given: (a) Mn (b) Xe For an atom or an ion with only a single electron, we can calculate the potential energy by considering only the electrostatic attraction between the positively charged nucleus and the negatively charged electron. When more than one electron is present, however, the total energy of the atom or the ion depends not only on attractive electron-nucleus interactions but also on repulsive electron-electron interactions. When there are two electrons, the repulsive interactions depend on the positions of electrons at a given instant, but because we cannot specify the exact positions of the electrons, it is impossible to exactly calculate the repulsive interactions. Consequently, we must use approximate methods to deal with the effect of electron-electron repulsions on orbital energies. If an electron is far from the nucleus (i.e., if the distance between the nucleus and the electron is large), then at any given moment, most of the other electrons will be that electron and the nucleus. Hence the electrons will cancel a portion of the positive charge of the nucleus and thereby decrease the attractive interaction between it and the electron farther away. As a result, the electron farther away experiences an effective nuclear charge ( ) that is than the actual nuclear charge Z (Figure \(\Page {6}\)). This effect is called electron shielding. As the distance between an electron and the nucleus approaches infinity, approaches a value of 1 because all the other ( − 1) electrons in the neutral atom are, on the average, between it and the nucleus. If, on the other hand, an electron is very close to the nucleus, then at any given moment most of the other electrons are farther from the nucleus and do not shield the nuclear charge. At ≈ 0, the positive charge experienced by an electron is approximately the full nuclear charge, or ≈ . At intermediate values of , the effective nuclear charge is somewhere between 1 and : 1 ≤ ≤ . Thus the actual experienced by an electron in a given orbital depends not only on the spatial distribution of the electron in that orbital but also on the distribution of all the other electrons present. This leads to large differences in for different elements, as shown in Figure 2.5.1 for the elements of the first three rows of the periodic table. Notice that only for hydrogen does = , and only for helium are and comparable in magnitude. Because of the effects of shielding and the different radial distributions of orbitals with the same value of but different values of , the different subshells are not degenerate in a multielectron atom. For a given value of , the orbital is always lower in energy than the orbitals, which are lower in energy than the orbitals, and so forth. As a result, some subshells with higher principal quantum numbers are actually lower in energy than subshells with a lower value of ; for example, the 4 orbital is lower in energy than the 3 orbitals for most atoms. Except for the single electron containing hydrogen atom, in every other element \(Z_{eff}\) is always less than \(Z\).   Ionization energy is the energy required to remove an electron from a neutral atom in its gaseous phase. Conceptually, ionization energy is the opposite of electronegativity. The lower this energy is, the more readily the atom becomes a cation. Therefore, the higher this energy is, the more unlikely it is the atom becomes a cation. Generally, elements on the right side of the periodic table have a higher ionization energy because their valence shell is nearly filled. Elements on the left side of the periodic table have low ionization energies because of their willingness to lose electrons and become cations. Thus, ionization energy increases from left to right on the periodic table. Another factor that affects ionization energy is . Electron shielding describes the ability of an atom's inner electrons to shield its positively-charged nucleus from its valence electrons. When moving to the right of a period, the number of electrons increases and the strength of shielding increases. As a result, it is easier for valence shell electrons to ionize, and thus the ionization energy decreases down a group. Electron shielding is also known as . Some elements have several ionization energies; these varying energies are referred to as the first ionization energy, the second ionization energy, third ionization energy, etc. The first ionization energy is the energy requiredto remove the outermost, or highest, energy electron, the second ionization energy is the energy required to remove any subsequent high-energy electron from a gaseous cation, etc. Below are the chemical equations describing the first and second ionization energies: First Ionization Energy: \[ X_{(g)} \rightarrow X^+_{(g)} + e^- \] Second Ionization Energy: \[ X^+_{(g)} \rightarrow X^{2+}_{(g)} + e^- \] Generally, any subsequent ionization energies (2nd, 3rd, etc.) follow the same periodic trend as the first ionization energy. Ionization energies decrease as atomic radii increase. This observation is affected by \(n\) (the principal quantum number) and \(Z_{eff}\) (based on the atomic number and shows how many protons are seen in the atom) on the ionization energy (I). The relationship is given by the following equation: \[ I = \dfrac{R_H Z^2_{eff}}{n^2} \] The periodic structure of the elements is evident for many physical and chemical properties, including chemical valence, atomic radius, electronegativity, melting point, density, and hardness. The classic prototype for periodic behavior is the variation of the first ionization energy with atomic number, which is plotted in Figure \(\Page {8}\). The electron affinity (EA) of an element E is defined as minus the internal energy change associated with the gain of an electron by a gaseous atom, at 0 K : \[E_{(g)} + e^- → E^-_{(g)}\] Unlike ionization energies, which are always positive for a neutral atom because energy is required to remove an electron, electron affinities can be positive (energy is released when an electron is added), negative (energy must be added to the system to produce an anion), or zero (the process is energetically neutral). The periodic trends of electron affinity (Figure \(\Page {9}\)) shows that chlorine has the most positive electron affinity of any element, which means that more energy is released when an electron is added to a gaseous chlorine atom than to an atom of any other element, EA= 348.6 kJmol and the elements have the largest values overall. The addition of a second electron to an element is expected to be much less favored since there will be repulsion between the negatively charged electron and the overall negatively charged anion. For example, for O the values are: \[O_{(g)} + e^ \rightarrow O^-_{(g)} \;\;\;\; EA = +141\; kJmol^{-1}\] \[O^-_{(g)} + e^- → O^{2-}){(g)} \;\;\;\; EA = -798\; kJmol^{-1}\] ")
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Semipermiable membranes do no H O Osmosis is the diffusion of a fluid through a semipermeable membrane. When a semipermeable membrane (animal bladders, skins of fruits and vegetables) separates a solution from a solvent, then only solvent molecules are able to pass through the membrane. The osmotic pressure of a solution is the pressure difference needed to stop the flow of solvent across a semipermeable membrane. The osmotic pressure of a solution is proportional to the of the solute particles in solution. \[\Pi = i \dfrac{n}{V}RT = i M RT \label{eq1}\] where Calculate molarity of a sugar solution in water (300 K) has osmotic pressure of 3.00 atm. Since it is sugar, we know it doesn't dissociate in water, so \(i\) is 1. Then we use Equation \ref{eq1} directly \[M = \dfrac{\Pi}{RT} = \dfrac{3.00\, atm}{(0.0821\, atm.L/mol.K)(300\,K)} = 0.122\,M \nonumber\] Calculate osmotic pressure for 0.10 M \(\ce{Na3PO4}\) aqueous solution at 20°C. Since \(\ce{Na3PO4}\) ionizes into four particles (3 Na+1 + \(PO_4^{-3}\)), then \(i = 4\). We can then calculate the osmotic pressure via Equation \ref{eq1} \[\Pi = iMRT = (0.40)(0.0821)(293) = 9.6\, atm \nonumber\] Hemoglobin is a large molecule that carries oxygen in human blood. A water solution that contains 0.263 g of hemoglobin (Hb) in 10.0 mL of solution has an osmotic pressure of 7.51 torr at \(25 ^oC\). What is the molar mass of the hemoglobin? \(6.51 \times 10^4 \; g/mol\)
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Prof. Linus Pauling (1931) first developed the Hybridization state theory in order to explain the structure of molecules such as methane (CH ). This concept was developed for simple chemical systems but this one applied more widely later on and from today’s point of view it is considered an operative empirical for excusing the structures of organic and inorganic compounds along with their related problems. An innovative method proposed for the determination of hybridization state on time economic ground . We Know, hybridization is nothing but the mixing of orbital’s in different ratio to form some newly synthesized orbitals called hybrid orbitals. The mixing pattern is as follows: s + p (1:1) - sp hybrid orbital; s + p (1:2) - sp hybrid orbital ; s + p (1:3) - sp hybrid orbital Formula used for the determination of sp, sp2 and sp3 hybridization state: All single (-) bonds are σ bond, in double bond (=) there is one σ and 1π, in triple bond (≡) there is one σ and 2π. In addition to these each lone pair (LP) and can be treated as one σ bond subsequently. a. In : central atom N is surrounded by three N-H single bonds i.e. three sigma (σ) bonds and one lone pair (LP) i.e. one additional σ bond. So, in NH there is a total of four σ bonds [3 bond pairs (BPs) + 1 lone pair (LP)] around central atom N. Therefore, in this case power of the hybridization state of N = 4-1 = 3 i.e. hybridization state = sp . d. In : I and Cl both have 4 bonds and 3LPs, so, in this case power of the hybridization state of both I and Cl = 4 - 1 = 3 i.e. hybridization state of I and Cl both are sp . In case of sp d, sp d and sp d hybridization state there is a common term sp for which 4 sigma bonds are responsible. So, in addition to 4 sigma bonds, for each additional sigma, added one d orbital gradually as follows:- Eg:- a. : b. : 7 I-F single bonds i.e. 7σ bonds = 4σ bonds + 3 additional σ bonds = sp d hybridization. c. : I has 7 e s in its outermost shell, so, in this case, add one e with 7(overall charge on the compound) i.e. 07+1= 08. So, out of 08 electrons, 02 electrons form 02 I-Cl bonds i.e. 02 sigma bonds and there is 03 LPs. So, altogether there are 05σ bonds. So, 5σ bonds = 04 σ bonds + 01 additional σ bond = sp d hybridization. d. : Xe, an inert gas, consider 8 e s in its outermost shell, 04 of which form 04 Xe-F sigma bonds and there is two LPs, i.e. altogether there is 06 σ bonds = 04 σ bonds + 02 additional σ bonds = sp d hybridization. In case of determination of the hybridization state by using the above method, one must have a clear idea about the outermost electrons of different family members in the periodic table as follows: Family Outermost electrons Nitrogen family 05 Oxygen family 06 Halogen family 07 Inert gas family 08 Total number of sigma (σ) bonds Nature of Hybridization State Examples
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Hint: Number of electrons around N = 1 + (6/2) Evaluate the formal charge of any atom of any dot structure. What is the formal charge for the resonance structure having the odd electron on the oxygen atom? Four resonance structures have been given, and all indicate a bond order of 1.5 for \(\ce{[O-N=O]}\) <=> \(\ce{[O=N-O]}\). Solving real-world problems requires you to know what information to look for before the problem is solvable. This is one of those problems. In an exercise or an open-book test, we can ask you to evaluate the lattice energy of a salt, and you need to find the various data. Since the CAcT quiz is an open-book test, this type of problem should have been given, but you will not have this type of problem in this quiz. To Quizzes
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We already have alluded to the difficulties encountered in the interpretation of the structure of benzene in and . Our task here is to see what new insight the VB and MO treatments can give us about benzene, but first we will indicate those properties of benzene that are difficult to explain on the basis of simple structure theory. From x-ray diffraction and spectroscopic measurements, benzene is known to be a planar molecule with six carbons \(1.390\) Å apart in a hexagonal ring, \(5\). Six hydrogen atoms, one associated with each carbon, are located \(1.09\) Å from those carbons. All \(\ce{H-C-C}\) and \(\ce{C-C-C}\) bond angles are \(120^\text{o}\): The 1,3,5-cyclohexatriene structure, \(6\), proposed for benzene in 1866 by Kekule, has alternating single and double bonds around the ring, which would be predicted to have bond lengths of \(1.48\) Å and \(1.34\) Å, respectively (see Table 2-1): The knowledge that the bond lengths are equal in the ring in benzene is a point against the Kekule formulation, but a more convincing argument is available from a comparison of the chemistry of benzene with that of 1,3,5-hexatriene, \(7\): Benzene also is more stable by about \(36\)-\(38 \: \text{kcal mol}^{-1}\) than anticipated for the 1,3,5-cyclohexatriene structure. You will recall from earlier discussions that the heat of combustion of one mole of benzene is \(38 \: \text{kcal}\) than calculated for cyclohexatriene (see ). Also, the heat of hydrogenation of benzene is only \(49.8 \: \text{kcal mol}^{1}\), which is \(36 \: \text{kcal}\) than expected for 1,3,5-cyclohexatriene; this estimate is based on the assumption that the heat of hydrogenation of 1,3,5-cyclohexatriene (with three double bonds) would be three times that of cyclohexane (\(28.5 \: \text{kcal mol}^{-1}\), for one double bond), or \(3 \times 28.2 = 85.5 \: \text{kcal mol}^{-1}\). The extra stability of benzene relative to the hypothetical 1,3,5-cyclohexatriene can be called its . Most (but not all) of this stabilization may be ascribed to resonance or electron delocalization. In an atomic model of benzene was discussed in some detail. Each carbon in the ring was considered to form three coplanar \(sp^2\)-hybrid \(\sigma\) bonds at \(120^\text{o}\) angles. These carbon-carbon and carbon-hydrogen \(\sigma\) bonds use three of the four valence electrons of each carbon. The remaining six carbon electrons are in parallel \(p\) orbitals, one on each of the six carbons. Each of the \(\pi\) electrons can be regarded as being paired with its immediate neighbors all around the ring, as shown by \(8\): As mentioned in , delocalization of the electrons over all six centers in benzene should give a more stable electron distribution than any structure in which the electrons are localized in pairs between adjacent carbons (as in the classical 1,3,5-cyclohexatriene structure). The simple MO and VB treatments of benzene begin with the atomic-orbital model and each treats benzene as a six-electron \(\pi\)-bonding problem. The assumption is that the \(\sigma\) bonds of benzene should not be very much different from those of ethene and may be regarded as independent of the \(\pi\) system. Extension of the ideas of for the MO treatment of an electron-pair bond between two nuclei to the \(\pi\) bonding in benzene is fairly straightforward. What is very important to understand is that there must be more than one molecular orbital for the \(\pi\) electrons because there are six \(\pi\) electrons, and the Pauli principle does not allow more than two paired electrons to occupy a given orbital. In fact, combination (or mixing) os the six \(2p\) orbitals of benzene, shown in \(8\), gives \(\pi\) molecular orbitals. Without exception, . The details of the mathematics of the mixing process to give an optimum set of molecular orbitals will not be described here,\(^1\) but the results are shown in Figure 21-5. Of the six predicted molecular orbitals, three are bonding and three are antibonding. The six \(\pi\) electrons are assigned to the three bonding orbitals in pairs and are calculated to have a total \(\pi\)-electron energy of \(6 \alpha + 8 \beta\). The calculation that leads to the results shown in Figure 21-5 is not very sophisticated. It is based on the assumption that the \(\pi\) bonding between each carbon and its immediate neighbors is equal all around the ring and that bonding involving carbons more than \(2\) Å apart is unimportant. What happens if we use the MO method to calculate the \(\pi\)-electron energy of classical 1,3,5-cyclohexatriene? The procedure is exactly as for benzene, except that we decree that each carbon \(p\) orbital forms a \(\pi\) bond with only of its neighboring \(p\) orbitals. The results are shown in Figure 21-6. The \(\pi\)-electron energy turns out to be three times that of ethene, or \(6 \alpha + 6 \beta\) compared to \(6 \alpha + 8 \beta\) for benzene. The calculated for benzene is the difference between these quantities, or \(\left( 6 \alpha + 8 \beta \right) - \left( 6 \alpha + 6 \beta \right) = 2 \beta\). That is to say, the calculated delocalization energy is the difference between the energy of benzene with full \(\pi\) bonding and the energy of 1,3,5-cyclohexatriene with alternating single and double bonds. If the electron delocalization energy \(\left( 2 \beta \right)\) is equal to the stabilization energy \(\left( 38 \: \text{kcal mol}^{-1} \right)\), then \(\beta = 19 \: \text{kcal mol}^{-1}\). Whether this is a valid method for determining \(\beta\) has been a matter of dispute for many years. Irrespective of this, the results of the calculations do account for the fact that benzene is more stable than would be expected for 1,3,5-cyclohexatriene. However, do the results also account for the low reactivity toward the various reagents in Figure 214, such as those that donate \(\ce{Br}^\oplus\) to double bonds (see )? To settle this question, we have calculated the changes in \(\pi\)-electron energy that occur in each of the following reactions: This means calculating the \(\pi\)-electron energies of all four entities and assuming that the differences in the \(\sigma\)-bond energies cancel between the two reactions. The result of this rather simple calculation is that attack of \(\ce{Br}^\oplus\) on benzene is thermodynamically favorable than on 1,3,5-cyclohexatriene by about \(\beta\). If \(\beta\) is \(19 \: \text{kcal mol}^{-1}\), this is clearly a sizable energy difference, and we can conclude that the simple MO method does indeed account for the fact that benzene is attacked by \(\ce{Br}^\oplus\) far less readily than is 1,3,5-cyclohexatriene. Extension of the basic ideas of the VB treatment described in to the atomic-orbital model of benzene is straightforward. We can write VB structures that represent pairing schemes of electrons in the atomic orbitals as shown in \(9\) through \(13\): Pairing schemes \(9\) and \(10\) correspond to Kekule's structures, whereas \(11\), \(12\), and \(13\) are called "Dewar structures" because J. Dewar suggested, in 1869, that benzene might have a structure such as \(14\): The electrons are paired in the configurations represented by \(11\), \(12\), and \(13\), but these pairing schemes are not as energetically favorable as \(9\) and \(10\). The reason is that the two electrons paired according to the dashed lines in \(11\), \(12\), and \(13\) are on nuclei separated by \(2.8\) Å, which is too far apart for effective bonding. The dashed lines between the distant carbons in \(11\), \(12\), and \(13\) are significant in that they define a pairing scheme. Such lines sometimes are said to represent "formal bonds". We hope that it is clear from what we have said here and previously that the ; indeed, the energy of the actual molecule is than any one of the contributing structures. The double-headed arrow between the structures is used to indicate that they represent different electron-pairing schemes for a molecule and different forms of the molecule in equilibrium with one another. When we use the resonance method in a qualitative way, we consider that the contribution of each of the several structures is to be weighted in some way that accords with the degree of bonding each would have, it were to represent an actual molecule with the specified geometry. Thus the Kekule-type electron-pairing schemes, \(9\) and \(10\), are to be taken as contributing and to the hybrid structure of benzene - equally because they are energetically equivalent, and predominantly because they can contribute much more to the overall bonding than \(11\), \(12\), and \(13\). In using the resonance method, we assume that all the resonance structures contributing to a given resonance hybrid have the same spatial arrangements of the nuclei but different pairing schemes for the electrons. Therefore \(11\), \(12\), and \(13\) are not to be confused with bicyclo[2.2.0]-2,5-hexadiene, \(15\), because \(15\) is a known (albeit not very stable) molecule with different atom positions and therefore vastly different bond angles and bond lengths from benzene: The electron-pairing schemes \(9\) and \(10\) represent the electron pairing that \(15\) would have if it were grossly distorted, with each carbon at the corner of a regular hexagon and a formal bond in place of a carbon-carbon single bond. Thus \(9\) and \(10\) would contribute in a significant way to the resonance hybrid of \(15\). Clearly, it is inconvenient and tedious to write the structures of the contributing forms to show the structure of a resonance hybrid. A shorthand notation is therefore desirable. Frequently, dashed rather than full lines are used where the bonding electrons are expected to be delocalized over several atoms. For benzene, \(16a\) or \(16b\) is quite appropriate: Unfortunately, although these are clear and explicit renderings, they are tedious to draw. As a result, many authors use (as we will most often) a single Kekule structure to represent benzene, with the understanding that all the \(\ce{C-C}\) bonds are equivalent. Other authors choose to represent benzene as a hexagon with an inscribed circle: This is a simple notation for benzene, but is quite uninformative and even can be actively misleading with some aromatic ring systems, and thus should be used with this limitation in mind. In calculations of the resonance energy of benzene, the five electronic configurations (valence-bond structures \(9\) through \(13\)) are combined mathematically to give five hybrid states, and of these the lowest-energy state is assumed to correspond to the normal state of the molecule. Thus benzene is considered by this approach to be a resonance hybrid of the valence-bond structures \(9\) through \(13\). In this simple treatment, \(9\) and \(10\) are calculated to contribute about \(80\%\) and \(11\), \(12\), and \(13\) about \(20\%\) to the hybrid. The actual numerical VB calculations, which are much more difficult to carry through than the corresponding MO calculations, give an energy of \(Q + 2.61 J\) for benzene and \(Q + 1.50 J\) for classical 1,3,5-cyclohexatriene.\(^3\) The resonance or delocalization energy then is \(\left( Q + 2.61 J \right) - \left( Q + 1.50 J \right) = 1.11 J\), which makes \(J \sim 35 \: \text{kcal mol}^{-1}\) if the resonance energy is taken to be equal to the \(38 \: \text{kcal}\) value obtained for the stabilization energy. If one carries through a simple VB calculation of the energy change associated with attack of \(\ce{Br}^\oplus\) on benzene as compared to 1,3,5-cyclohexatriene, the value obtained is \(0.63 J\), which corresponds to \(22 \: \text{kcal}\). This is in excellent agreement with the \(19 \: \text{kcal}\) value obtained by the MO method ( ). \(^1\)There are many excellent books that cover this subject in great detail; however, the simplest introductory work is J. D. Roberts, , W. A. Benjamin, Inc., Menlo Park, Calif., 1961. \(^2\) Note that in \(9\) and also in \(10\), we show a particular way of pairing the electrons. However, just as \(1 \leftrightarrow 2\), and \(4a \leftrightarrow 4b\), we also must consider other sets that represent exchanges of electrons across the dashed lines of \(9\) and also of \(10\). \(^3\)\(Q\) and \(J\) are negative VB energy parameters that correspond roughly to the MO parameters \(\alpha\) and \(\beta\). and (1977)
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is a rare element found primarily in molten rock and saltwater in very small amounts. It is understood to be non-vital in human biological processes, although it is used in many drug treatments due to its positive effects on the human brain. Because of its reactive properties, humans have utilized lithium in batteries, nuclear fusion reactions, and thermonuclear weapons. Lithium was first identified as a component of of the mineral petalite and was discovered in 1817 by Johan August Arfwedson, but not isolated until some time later by W.T. Brande and Sir Humphry Davy. In its mineral forms it accounts for only 0.0007% of the earth's crust. It compounds are used in certain kinds of glass and porcelain products. More recently lithium has become important in dry-cell batteries and nuclear reactors. Some compounds of lithium have been used to treat manic depressives. Lithium is an alkali metal with the atomic number = 3 and an atomic mass of 6.941 g/mol. This means that lithium has 3 protons, 3 electrons and 4 neutrons (6.941 - 3 = ~4). Being an alkali metal, lithium is a soft, flammable, and highly reactive metal that tends to form hydroxides. It also has a pretty low density and under standard conditions, it is the least dense solid element. Lithium is the lightest of all metals and is named from the Greek work for stone (lithos). It is the first member of the Alkali Metal family. It is less dense than water (with which it reacts) and forms a black oxide in contact with air. In compounds lithium (like all the alkali metals) has a +1 charge. In its pure form it is soft and silvery white and has a relatively low melting point (181oC). 1s When placed in contact with water, pure lithium reacts to form lithium hydroxide and hydrogen gas. \[ 2Li (s) + 2H_2O (l) \rightarrow 2LiOH (aq) + H_2 (g)\] Out of all the group 1 metals, lithium reacts the least violently, slowly releasing the hydrogen gas which may create a bright orange flame only if a substantial amount of lithium is used. This occurs because lithium has the highest activation energy of its group - that is, it takes more energy to remove lithium's one valence electron than with other group 1 elements, because lithium's electron is closer to its nucleus. Atoms with higher activation energies will react slower, although lithium will release more total heat through the entire process. Pure lithium will form lithium hydroxide due to moisture in the air, as well as lithium nitride (\(Li_3N\)) from \(N_2\) gas, and lithium carbonate \((Li_2CO_3\)) from carbon dioxide. These compounds give the normally the silver-white metal a black tarnish. Additionally, it will combust with oxygen as a red flame to form lithium oxide. \[ 4Li (s) + O_2 (g) \rightarrow 2Li_2O \] In its mineral forms it accounts for only 0.0007% of the earth's crust. It compounds are used in certain kinds of glass and porcelain products. More recently lithium has become important in dry-cell batteries and nuclear reactors. Some compounds of lithium have been used to treat manic depressives. Lithium is able to be used in the function of a Lithium battery in which the Lithium metal serves as the anode. Lithium ions serve in lithium ion batteries (chargeable) in which the lithium ions move from the negative to positive electrode when discharging, and vice versa when charging. Lithium has the highest specific heat capacity of the solids, Lithium tends to be used as a cooler for heat transfer techniques and applications. Lithium is most commonly found combined with aluminum, silicon, and oxygen to form the minerals known as (LiAl(SiO ) ) or / (LiAlSi O ). These have been found on each of the 6 inhabited continents, but they are mined primarily in Western Australia, China, and Chile. Mineral sources of lithium are becoming less essential, as methods have now been developed to make use of the lithium salts found in saltwater. The mineral forms of lithium are heated to a high enough temperature (1200 K - 1300 K) in order to crumble them and thus allow for subsequent reactions to more easily take place. After this process, one of three methods can be applied. The lithium chloride obtained from any of the three methods undergoes an oxidation-reduction reaction in an electrolytic cell, to separate the chloride ions from the lithium ions. The chloride ions are oxidized, and the lithium ions are reduced. \[2Cl^- - 2e^- \rightarrow Cl_2 \;\; \text{(oxidation)}\] \[Li^+ + e^- \rightarrow Li \;\; \text{(reduction)}\] Saltwater naturally contains lithium chloride, which must be extracted in the form of lithium carbonate, then it is re-treated, separated into its ions, and reduced in the same electrolytic process as in extraction from lithium ores. Only three saltwater lakes in the world are currently used for lithium extraction, in Nevada, Chile, and Argentina. Saltwater is channeled into shallow ponds and over a period of a year or more, water evaporates out to leave behind various salts. Lime is used to remove the magnesium salt, so that the remaining solution contains a fairly concentrated amount of lithium chloride. The solution is then treated with sodium carbonate in order for usable lithium carbonate to precipitate out.
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Entropy is related to the number of microstates a collection of particles can occupy. As both the molecular mass and molecular structure of the particles will affect the number of available microstates, they also affect the entropy of the collection of particles. From quantum theory, we know that increasing the molecular mass of a particle decreases the energy spacing between states. For a given temperature, more states are available to be occupied, increasing the number of available microstates the system may occupy, and hence the entropy of the system. The table below shows the molar entropies for the noble gases. As the mass of increases, so does the molar entropy. The same is true for the number of atoms in a molecule. A molecule with more atoms will, in general, have a more degrees of freedom to take up energy, increasing its number of available microstates and entropy.
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Prior to the early 1920's, chemists doubted the existence of molecules having molecular weights greater than a few thousand. This limiting view was challenged by , a German chemist with experience in studying natural compounds such as rubber and cellulose. In contrast to the prevailing rationalization of these substances as aggregates of small molecules, Staudinger proposed they were made up of composed of 10,000 or more atoms. He formulated a structure for , based on a repeating isoprene unit (referred to as a monomer). For his contributions to chemistry, Staudinger received the 1953 Nobel Prize. The terms and were derived from the Greek roots (many), (one) and (part). Recognition that polymeric macromolecules make up many important natural materials was followed by the creation of synthetic analogs having a variety of properties. Indeed, applications of these materials as fibers, flexible films, adhesives, resistant paints and tough but light solids have transformed modern society. Some important examples of these substances are discussed in the following sections. ),
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The National Fire Protection Association (NFPA) provides vast fire safety information including the four-color diamond used on labels to indicate the hazard level of a chemical; see Figure 1. NFPA Health, Flammability, and Reactivity hazards are rated from 0 (none) to 4 (extreme). The Health rating is in the blue section, Flammability in red, and Reactivity in yellow. The white section is reserved for Other Specific Hazards including oxidizers, acids, corrosives, and radioactive materials. Since 1960, the NFPA labeling standard has been a simple, recognizable, and understandable labeling system that provides information regarding important hazards of a material and the severity of these hazards as they relate to handling, fire prevention, exposure, and control. The Flammability rating of an NFPA chemical label provides basic information to fire fighters and other emergency personnel, enabling them to decide whether to evacuate an area or to select the appropriate fire fighting tactics and emergency procedures. The NFPA labeling standard also provides laboratory personnel with the information necessary to select the appropriate level of personal protection when working with a material and the correct method of storage for that material. Proper use of chemical labeling should be common laboratory practice. Flash point provides a simple and convenient index for the flammability and combustibility of substances. Flash point provides valuable information to those who handle, transport, and store chemicals. Flash point is the minimum temperature at which the vapor present over a liquid forms a flammable mixture when mixed with air. The NFPA Flammability rating is determined by the flash point of the material, with ratings of 4, 3, 2, or 1 for chemicals with flash points below 73°F, below 100°F, below 200°F, or above 200°F, respectively. The NFPA Flammability rating of 0 is reserved for substances that will not burn. When assessing the safety and flammability of a material or mixture, experimental data on flash point is preferred. But, with so many new compounds being synthesized and mixtures being formulated every year, an estimate of flash point is often sufficient for a preliminary assessment of the flammability of these new materials. Estimates of the flash point of a compound can be based (i) on an assessment of its chemical structure and a comparison to the known flash point of similar chemical compounds or (ii) on a correlation to known physical properties of the material. The normal boiling point of a compound is generally known and this physical property can be used to estimate the compound’s flash point. Linear correlations between flash point and boiling point generally obtain excellent correlation coefficients (R2 between about 0.90 and 0.98) across many different functional group families and across a wide range in boiling point. This correlation is not surprising because a sufficient amount of vapor must be present to form a flammable mixture when in air, and boiling point correlates well to the minimum vapor content of the material necessary for a flammable mixture with air. Compared to the measurement of flash point, the measurement of boiling point is less risky, is inexpensive, uses simple equipment, and requires minimal training. A reliable correlation between boiling point and flash point can provide the information necessary for a preliminary assessment of the flammability of a material or mixture. Advanced QSPR models predict flash point for pure compounds and mixtures based on boiling point, specific gravity, heat of vaporization, and other strategies. The link below (Table 1) contains flash point and normal boiling point data for 248 compounds. The data has been saved in a text only-tab delimited format, which can be imported into most software packages. The data was gathered from “Yaws' 2003 Handbook of Thermodynamic and Physical Properties of Chemical Compounds” (online version available at ) and through Internet searches of publicly available safety and physical property data. Flash point is generally reported in degrees Fahrenheit, because flash point is widely used in chemical shipping and storage by people who do not generally use Celsius for measuring temperature. The normal boiling points were converted to Fahrenheit so both physical properties could be compared in a common scale. Table 2 (below) contains flash points for a series of organic compounds organized by carbon chain length in rows (methyl, ethyl, n-propyl, phenyl, etc.) and functional (end) groups in columns (methyl, hydroxymethyl, chloromethyl, aminomethyl, carboxylic acid, etc.). Table 2 can be cut-n-pasted into the analysis software that you are utilizing. If using M.S. Excel, the “Data Analysis” functions necessary for these exercises are located in the “Tools” drop down menu. If these functions are not available, follow the instructions for the “Analysis ToolPak” add-in found in the HELP menu.
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So far in our molecular orbital descriptions we have not dealt with polyatomic systems with multiple bonds. To do so, we can use an approach in which we describe σ bonding using localized electron-pair bonds formed by hybrid atomic orbitals, and π bonding using molecular orbitals formed by unhybridized atomic orbitals. We begin our discussion by considering the bonding in ethylene (C H ). Experimentally, we know that the H–C–H and H–C–C angles in ethylene are approximately 120°. This angle suggests that the carbon atoms are hybridized, which means that a singly occupied orbital on one carbon overlaps with a singly occupied orbital on each H and a singly occupied lobe on the other C. Thus each carbon forms a set of three σ bonds: two C–H ( + ) and one C–C ( + ) (part (a) in ). The hybridization can be represented as follows: After hybridization, each carbon still has one unhybridized 2 orbital that is perpendicular to the hybridized lobes and contains a single electron (part (b) in ). The two singly occupied 2 orbitals can overlap to form a π bonding orbital and a π* antibonding orbital, which produces the energy-level diagram shown in With the formation of a π bonding orbital, electron density increases in the plane between the carbon nuclei. The π* orbital lies outside the internuclear region and has a nodal plane perpendicular to the internuclear axis. Because each 2 orbital has a single electron, there are only two electrons, enough to fill only the bonding (π) level, leaving the π* orbital empty. Consequently, the C–C bond in ethylene consists of a σ bond and a π bond, which together give a C=C double bond. Our model is supported by the facts that the measured carbon–carbon bond is shorter than that in ethane (133.9 pm versus 153.5 pm) and the bond is stronger (728 kJ/mol versus 376 kJ/mol in ethane). The two CH fragments are coplanar, which maximizes the overlap of the two singly occupied 2 orbitals. As in the diatomic molecules discussed previously, the singly occupied 2 orbital in ethylene can overlap to form a bonding/antibonding pair of π molecular orbitals. The two electrons remaining are enough to fill only the bonding π orbital. With one σ bond plus one π bond, the carbon–carbon bond order in ethylene is 2. Triple bonds, as in acetylene (C H ), can also be explained using a combination of hybrid atomic orbitals and molecular orbitals. The four atoms of acetylene are collinear, which suggests that each carbon is hybridized. If one lobe on each carbon atom is used to form a C–C σ bond and one is used to form the C–H σ bond, then each carbon will still have two unhybridized 2 orbitals (a 2 pair), each with one electron (part (a) in ). The two 2 orbitals on each carbon can align with the corresponding 2 orbitals on the adjacent carbon to simultaneously form a of π bonds (part (b) in ). Because each of the unhybridized 2 orbitals has a single electron, four electrons are available for π bonding, which is enough to occupy only the bonding molecular orbitals. Acetylene must therefore have a carbon–carbon triple bond, which consists of a C–C σ bond and two mutually perpendicular π bonds. Acetylene does in fact have a shorter carbon–carbon bond (120.3 pm) and a higher bond energy (965 kJ/mol) than ethane and ethylene, as we would expect for a triple bond. (a) In the formation of the σ-bonded framework, two sets of singly occupied carbon hybrid orbitals and two singly occupied hydrogen 1 orbitals overlap. (b) In the formation of two carbon–carbon π bonds in acetylene, two singly occupied unhybridized 2 orbitals on each carbon atom overlap. With one σ bond plus two π bonds, the carbon–carbon bond order in acetylene is 3. In complex molecules, hybrid orbitals and valence bond theory can be used to describe σ bonding, and unhybridized orbitals and molecular orbital theory can be used to describe π bonding. Describe the bonding in HCN using a combination of hybrid atomic orbitals and molecular orbitals. The HCN molecule is linear. chemical compound and molecular geometry bonding description using hybrid atomic orbitals and molecular orbitals From the geometry given, predict the hybridization in HCN. Use the hybrid orbitals to form the σ-bonded framework of the molecule and determine the number of valence electrons that are used for σ bonding. Determine the number of remaining valence electrons. Use any remaining unhybridized orbitals to form π and π* orbitals. Fill the orbitals with the remaining electrons in order of increasing energy. Describe the bonding in HCN. Because HCN is a linear molecule, it is likely that the bonding can be described in terms of hybridization at carbon. Because the nitrogen atom can also be described as hybridized, we can use one hybrid on each atom to form a C–N σ bond. This leaves one hybrid on each atom to either bond to hydrogen (C) or hold a lone pair of electrons (N). Of 10 valence electrons (5 from N, 4 from C, and 1 from H), 4 are used for σ bonding: We are now left with 2 electrons on N (5 valence electrons minus 1 bonding electron minus 2 electrons in the lone pair) and 2 electrons on C (4 valence electrons minus 2 bonding electrons). We have two unhybridized 2 atomic orbitals left on carbon and two on nitrogen, each occupied by a single electron. These four 2 atomic orbitals can be combined to give four molecular orbitals: two π (bonding) orbitals and two π* (antibonding) orbitals. With 4 electrons available, only the π orbitals are filled. The overall result is a triple bond (1 σ and 2 π) between C and N. Exercise Describe the bonding in formaldehyde (H C=O), a trigonal planar molecule, using a combination of hybrid atomic orbitals and molecular orbitals. σ-bonding framework: Carbon and oxygen are hybridized. Two hybrid orbitals on oxygen have lone pairs, two hybrid orbitals on carbon form C–H bonds, and one hybrid orbital on C and O forms a C–O σ bond. π bonding: Unhybridized, singly occupied 2 atomic orbitals on carbon and oxygen interact to form π (bonding) and π* (antibonding) molecular orbitals. With two electrons, only the π (bonding) orbital is occupied. In , we used resonance structures to describe the bonding in molecules such as ozone (O ) and the nitrite ion (NO ). We showed that ozone can be represented by either of these Lewis electron structures: Although the VSEPR model correctly predicts that both species are bent, it gives no information about their bond orders. . Experimental evidence indicates that ozone has a bond angle of 117.5°. Because this angle is close to 120°, it is likely that the central oxygen atom in ozone is trigonal planar and hybridized. If we assume that the terminal oxygen atoms are also hybridized, then we obtain the σ-bonded framework shown in . Two of the three lobes on the central O are used to form O–O sigma bonds, and the third has a lone pair of electrons. Each terminal oxygen atom has two lone pairs of electrons that are also in lobes. In addition, each oxygen atom has one unhybridized 2 orbital perpendicular to the molecular plane. The σ bonds and lone pairs account for a total of 14 electrons (five lone pairs and two σ bonds, each containing 2 electrons). Each oxygen atom in ozone has 6 valence electrons, so O has a total of 18 valence electrons. Subtracting 14 electrons from the total gives us 4 electrons that must occupy the three unhybridized 2 orbitals. With a molecular orbital approach to describe the π bonding, three 2 atomic orbitals give us three molecular orbitals, as shown in . One of the molecular orbitals is a π bonding molecular orbital, which is shown as a banana-shaped region of electron density above and below the molecular plane. This region has nodes perpendicular to the O plane. The molecular orbital with the highest energy has two nodes that bisect the O–O σ bonds; it is a π* antibonding orbital. The third molecular orbital contains a single node that is perpendicular to the O plane and passes through the central O atom; it is a nonbonding molecular orbital. Because electrons in nonbonding orbitals are neither bonding nor antibonding, they are ignored in calculating bond orders. We can now place the remaining four electrons in the three energy levels shown in , thereby filling the π bonding and the nonbonding levels. The result is a single π bond holding oxygen atoms together, or 1/2 bonds per O–O. We therefore predict the overall O–O bond order to be 1 1/2 (1/2 bond plus 1 σ bond), just as predicted using resonance structures. The molecular orbital approach, however, shows that the π nonbonding orbital is localized on the O atoms, which suggests that they are more electron rich than the central O atom. The reactivity of ozone is consistent with the predicted charge localization. Resonance structures are a crude way of describing molecular orbitals that extend over more than two atoms. Describe the bonding in the nitrite ion in terms of a combination of hybrid atomic orbitals and molecular orbitals. Lewis dot structures and the VSEPR model predict that the NO ion is bent. chemical species and molecular geometry bonding description using hybrid atomic orbitals and molecular orbitals Calculate the number of valence electrons in NO . From the structure, predict the type of atomic orbital hybridization in the ion. Predict the number and type of molecular orbitals that form during bonding. Use valence electrons to fill these orbitals and then calculate the number of electrons that remain. If there are unhybridized orbitals, place the remaining electrons in these orbitals in order of increasing energy. Calculate the bond order and describe the bonding. The lone pair of electrons on nitrogen and a bent structure suggest that the bonding in NO is similar to the bonding in ozone. This conclusion is supported by the fact that nitrite also contains 18 valence electrons (5 from N and 6 from each O, plus 1 for the −1 charge). The bent structure implies that the nitrogen is hybridized. If we assume that the oxygen atoms are hybridized as well, then we can use two hybrid orbitals on each oxygen and one hybrid orbital on nitrogen to accommodate the five lone pairs of electrons. Two hybrid orbitals on nitrogen form σ bonds with the remaining hybrid orbital on each oxygen. The σ bonds and lone pairs account for 14 electrons. We are left with three unhybridized 2 orbitals, one on each atom, perpendicular to the plane of the molecule, and 4 electrons. Just as with ozone, these three 2 orbitals interact to form bonding, nonbonding, and antibonding π molecular orbitals. The bonding molecular orbital is spread over the nitrogen and both oxygen atoms. Placing 4 electrons in the energy-level diagram fills both the bonding and nonbonding molecular orbitals and gives a π bond order of 1/2 per N–O bond. The overall N–O bond order is 1 1/2 consistent with a resonance structure. Exercise Describe the bonding in the formate ion (HCO ), in terms of a combination of hybrid atomic orbitals and molecular orbitals. Like nitrite, formate is a planar polyatomic ion with 18 valence electrons. The σ bonding framework can be described in terms of hybridized carbon and oxygen, which account for 14 electrons. The three unhybridized 2 orbitals (on C and both O atoms) form three π molecular orbitals, and the remaining 4 electrons occupy both the bonding and nonbonding π molecular orbitals. The overall C–O bond order is therefore Hydrocarbons in which two or more carbon–carbon double bonds are directly linked by carbon–carbon single bonds are generally more stable than expected because of resonance. Because the double bonds are close enough to interact electronically with one another, the π electrons are shared over all the carbon atoms, as illustrated for 1,3-butadiene in . As the number of interacting atomic orbitals increases, the number of molecular orbitals increases, the energy spacing between molecular orbitals decreases, and the systems become more stable ( ). Thus as a chain of alternating double and single bonds becomes longer, the energy required to excite an electron from the highest-energy occupied (bonding) orbital to the lowest-energy unoccupied (antibonding) orbital decreases. If the chain is long enough, the amount of energy required to excite an electron corresponds to the energy of visible light. For example, vitamin A is yellow because its chain of five alternating double bonds is able to absorb violet light. Many of the colors we associate with dyes result from this same phenomenon; most dyes are organic compounds with alternating double bonds. As the number of interacting atomic orbitals increases, the energy separation between the resulting molecular orbitals steadily decreases. A derivative of vitamin A called is used by the human eye to detect light and has a structure with alternating C=C double bonds. When visible light strikes retinal, the energy separation between the molecular orbitals is sufficiently close that the energy absorbed corresponds to the energy required to change one double bond in the molecule from , where like groups are on the same side of the double bond, to , where they are on opposite sides, initiating a process that causes a signal to be sent to the brain. If this mechanism is defective, we lose our vision in dim light. Once again, a molecular orbital approach to bonding explains a process that cannot be explained using any of the other approaches we have described. To describe the bonding in more complex molecules with multiple bonds, we can use an approach that uses hybrid atomic orbitals to describe the σ bonding and molecular orbitals to describe the π bonding. In this approach, unhybridized orbitals on atoms bonded to one another are allowed to interact to produce bonding, antibonding, or nonbonding combinations. For π bonds between two atoms (as in ethylene or acetylene), the resulting molecular orbitals are virtually identical to the π molecular orbitals in diatomic molecules such as O and N . Applying the same approach to π bonding between three or four atoms requires combining three or four unhybridized orbitals on adjacent atoms to generate π bonding, antibonding, and nonbonding molecular orbitals extending over all of the atoms. Filling the resulting energy-level diagram with the appropriate number of electrons explains the bonding in molecules or ions that previously required the use of resonance structures in the Lewis electron-pair approach. What information is obtained by using the molecular orbital approach to bonding in O that is not obtained using the VSEPR model? Can this information be obtained using a Lewis electron-pair approach? How is resonance explained using the molecular orbital approach? Indicate what information can be obtained by each method: Using both a hybrid atomic orbital and molecular orbital approaches, describe the bonding in BCl and CS . Use both a hybrid atomic orbital and molecular orbital approaches to describe the bonding in CO and N .
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There are alternative ways to add hydrogen to a multiple bond besides the catalytic methods described in the previous sections. The most useful of these are homogeneous reactions utilizing diimide, \(\ce{HN=NH}\), and diborane, \(\ce{B_2H_6}\). The behavior and reactivity of diimide can be understood best by considering the thermochemistry of hydrogenation of nitrogen: The first step is strongly endothermic and is the main hurdle to overcome in the hydrogenation of nitrogen to ammonia. Conversely, the reverse reaction, which is the dehydrogenation of diimide, is strongly exothermic. Therefore we may expect that diimide will have a pronounced tendency to revert to molecular nitrogen. This is in fact so and, at normal temperatures, diimide exists only as a transient intermediate that cannot be isolated. It is extremely reactive and readily transfers hydrogen to carbon-carbon multiple bonds: In practice, diimide is generated as it is needed in the presence of the compound to be hydrogenated. There are various ways to do this, but one of the simplest methods is dehydrogenation of hydrazine with oxidizing agents such as atmospheric oxygen or hydrogen peroxide: \[\ce{H_2N-N_2} \underset{\text{or} \: \ce{O_2}}{\overset{\ce{H_2O_2}}{\longrightarrow}} \ce{HN=NH} + 2 \ce{H}^\oplus + 2 \ce{e}^\ominus\] Hydrazine actually has been used as a hydrogenating agent for over sixty years, but it was not until the 1960's that the diimide intermediate in such reactions was recognized. The hydrogenation step is stereospecific and transfers hydrogen in the suprafacial manner. For example, alkynes are converted to -alkenes: There are no detectable intermediate stages or rearrangements in diimide hydrogenation. The reaction is visualized as a six-center ( ) process in which the bonds are broken and made in a concerted fashion: An important difference between diimide hydrogenation and catalytic hydrogenation is that diimide will react only with symmetrical or nonpolar bonds \(\left( \ce{C=C}, \: \ce{C \equiv C}, \: \ce{N=N} \right)\), whereas hydrogen can add, albeit reluctantly, to polar bonds \(\left( \ce{C=O}, \: \ce{C=N} \right)\). Diimide does not attack the stronger polar bonds probably because it does not survive long enough to do so. It selfdestructs in the absence of a reactive substrate to give nitrogen and hydrazine: and (1977)
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are containing one or more in a . The solvent that makes up most of the solution, whereas a solute is the substance that is dissolved inside the solvent. Concentrations are often expressed in terms of relative unites (e.g. percentages) with three different types of percentage concentrations commonly used: \[\text{Mass/Volume Percent}= \dfrac{\text{Mass of Solute (g)}}{\text{Volume of Solution (mL)}} \times 100\% \label{3}\] For example, In the United States, alcohol content in spirits is defined as twice the percentage of alcohol by volume (v/v) called proof. What is the concentration of alcohol in spirits that is sold at 151 proof (hence the name)? It will have an alcohol content of 75.5% (w/w) as per definition of "proof". When calculating these percentages, that the units of the solute and solution should be equivalent units (and weight/volume percent (w/v %) is defined in terms of grams and mililiters). Sometimes when solutions are too dilute, their percentage concentrations are too low. So, instead of using really low percentage concentrations such as 0.00001% or 0.000000001%, we choose another way to express the concentrations. This next way of expressing concentrations is similar to cooking recipes. For example, a recipe may tell you to use 1 part sugar, 10 parts water. As you know, this allows you to use amounts such as 1 cup sugar + 10 cups water in your equation. However, instead of using the recipe's "1 part per ten" amount, chemists often use , or to describe dilute concentrations. Here is a table with the volume percent of different gases found in air. Volume percent means that for 100 L of air, there are 78.084 L Nitrogen, 20.946 L Oxygen, 0.934 L Argon and so on; Volume percent mass is different from the composition by mass or composition by amount of moles. The molarity and molality equations differ only from their denominators. However, this is a huge difference. As you may remember, volume varies with different temperatures. At higher temperatures, volumes of liquids increase, while at lower temperatures, volumes of liquids decrease. Therefore, molarities of solutions also vary at different temperatures. This creates an advantage for using molality over molarity. Using molalities rather than molarities for lab experiments would best keep the results within a closer range. Because volume is not a part of its equation, it makes molality independent of temperature. In a solution, there is 111.0 mL (110.605 g) solvent and 5.24 mL (6.0508 g) solute present in a solution. Find the mass percent, volume percent and mass/volume percent of the solute. Mass Percent =(Mass of Solute) / (Mass of Solution) x 100%| =(6.0508g) / (110.605g + 6.0508g) x 100% =(0.0518688312) x 100% =5.186883121% Volume Percent =(Volume of Solute) / (Volume of Solution) x 100% =(5.24mL) / (111.0mL + 5.24mL) x 100% =(0.0450791466) x 100% =4.507914659% Mass/Volume Percent =(Mass of Solute) / (Volume of Solution) x 100% =(6.0508g) / (111.0mL + 5.24mL) x 100% =(0.0520) x 100% =5.205% With the solution shown in the picture below, find the mole percent of substance C. Moles of C= (5 C molecules) x (1mol C / 6.022x10 C molecules) = 8.30288941x10 mol C Total Moles= (24 molecules) x (1mol / 6.022x10 molecules)= 3.98538691x10 mol total X = (8.30288941x10 mol C) / (3.98538691x10 mol) = .2083333333 Mole Percent of C = X x 100% =(o.2083333333) x 100% =20.83333333 A 1.5L solution is composed of 0.25g NaCl dissolved in water. Find its molarity. Moles of NaCl= (0.25g) / (22.99g + 35.45g) = 0.004277 mol NaCl Molarity =(Moles of Solute) / (Liters of Solution) =(0.004277mol NaCl) / (1.5L) =0.002851 M 0.88g NaCl is dissolved in 2.0L water. Find its molality. Moles of NaCl= (0.88g) / (22.99g + 35.45g) = 0.01506 mol NaCl Mass of water= (2.0L) x (1000mL / 1L) x (1g / 1mL) x (1kg / 1000g) = 2.0kg water Molality =(Moles of Solute) / (kg of Solvent) =(0.01506 mol NaCl) / (2.0kg) =0.0075290897m
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Technetium has the distinction of being the first synthetically produced element. It was the last of the six elements predicted by Mendeleev to be discovered (in 1937 by Segrè and Perrier). Trace amounts were found in a sample of molybdenum that had been bombarded with deuterons in a cyclotron. All known isotopes of Tc are radioactive and none appears to occur naturally on earth. However, it is produced commercially in kg quantities because very tiny amounts (55 parts per million) can transform iron into a corrosion resistant alloy. All applications of the metal involve closed systems, however, because of the radioactivity. Pure Tc is a silvery-gray metal similar in appearance to platinum but it is normally produced in powdered form. Chemically it resembles a cross between Mn and Re. Although of little interest, compounds have been prepared in which Tc exhibits oxidation states of +4, +6 and +7.
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In Figure 28.8.1 , the point at which we evaluate or measure \(E_a\) serves as a dividing line (also called a ) between reactants and products. At this point, we do not have \(\text{A} + \text{B}\), and we do not have \(\text{C}\). Rather, what we have is an activated complex of some kind called a between reactants and products. The value of the reaction coordinate at the transition state is denoted \(q^\ddagger\). Recall our notation \(\text{x}\) for the complete set of coordinates and momenta of all of the atoms in the system. Generally, the reaction coordinate \(q\) is a function \(q \left( \text{x} \right)\) of all of the coordinates and momenta, although typically, \(q \left( \text{x} \right)\) is a function of a subset of the coordinates and, possibly, the momenta. As an example, let us consider two atoms \(\text{A}\) and \(\text{B}\) undergoing a collision. An appropriate reaction coordinate could simply be the distance \(r\) between \(\text{A}\) and \(\text{B}\). This distance is a function of the positions \(\textbf{r}_\text{A}\) and \(\textbf{r}_\text{B}\) of the two atoms, in that \[q = r = \left| \textbf{r}_\text{A} - \textbf{r}_\text{B} \right| \label{20.24} \] When \(\text{A}\) and \(\text{B}\) are molecules, such as proteins, \(q \left( \text{x} \right)\) is a much more complicated function of \(\text{x}\). Now, recall that the mechanical energy \(\mathcal{E} \left( \text{x} \right)\) is given by \[\mathcal{E} \left( \text{x} \right) = \sum_{i=1}^N \frac{\textbf{p}_i^2}{2m_1} + U \left( \textbf{r}_1, \ldots, \textbf{r}_N \right) \label{20.25} \] and is a sum of kinetic and potential energies. Transition state theory assumes the following: Define a function \(\theta \left( y \right)\), which is \(1\) if \(y \geq 0\) and \(0\) if \(y < 0\). The function \(\theta \left( y \right)\) is known as a . We now define a of reactive trajectories \(k \left( t \right)\) using statistical mechanics \[k \left( t \right) = \frac{1}{h Q_r} \int_{q \left( \text{x} \right) = q^\ddagger} d \text{x} \: e^{-\beta \mathcal{E} \left( \text{x} \right)} \left| \dot{q} \left( \text{x} \right) \right| \theta \left( q \left( \text{x}_t \left( \text{x} \right) \right) - q^\ddagger \right) \label{20.27} \] where \(h\) is Planck’s constant. Here \(Q_r\) is the partition function of the reactants \[Q_r = \int d \text{x} \: e^{-\beta \mathcal{E} \left( \text{x} \right)} \theta \left( q^\ddagger - q \left( \text{x} \right) \right) \label{20.28} \] The meaning of Equation \(\ref{20.27}\) is an ensemble average over a canonical ensemble of the product \(\left| \dot{q} \left( \text{x} \right) \right|\) and \(\theta \left( q \left( \text{x}_t \left( \text{x} \right) \right) - q^\ddagger \right)\). The first factor in this product \(\left| \dot{q} \left( \text{x} \right) \right|\) forces the initial velocity of the reaction coordinate to be positive, i.e., toward products, and the step function \(\theta \left( q \left( \text{x}_t \left( \text{x} \right) \right) - q^\ddagger \right)\) requires that the trajectory of \(q \left( \text{x}_t \left( \text{x} \right) \right)\) be reactive, otherwise, the step function will give no contribution to the flux. The function \(k \left( t \right)\) in Equation \(\ref{20.27}\) is known as the . In the definition of \(Q_r\) the step function \(\theta \left( q^\ddagger - q \left( \text{x} \right) \right)\) measures the total number of microscopic states on the reactive side of the energy profile. A plot of some examples of reactive flux functions \(k \left( t \right)\) is shown in Figure 28.8.2 . These functions are discussed in greater detail in , 5809 (1991). These examples all show that \(k \left( t \right)\) decays at first but then finally reaches a plateau value. This plateau value is taken to be the true rate of the reaction under the assumption that eventually, all trajectories that will become reactive will have done so after a sufficiently long time. Thus, \[k = \underset{t \rightarrow \infty}{\text{lim}} k \left( t \right) \label{20.29} \] gives the true rate constant. On the other hand, a common approximation is to take the value \(k \left( 0 \right)\) as an estimate of the rate constant, and this is known as the transition to \(k\), i.e., \[\begin{align} k^\text{(TST)} &= k \left( 0 \right) \\ &= \frac{1}{Q_r} \int_{q \left( \text{x} \right) = q^\ddagger} d \text{x} \: e^{-\beta \mathcal{E} \left( \text{x} \right)} \left| \dot{q} \left( \text{x} \right) \right| \theta \left( q \left( \text{x} \right) - q^\ddagger \right) \end{align} \label{20.30} \] However, note that since we require \(\dot{q} \left( \text{x} \right)\) to initially be toward products, then by definition, at \(t = 0\), \(q \left( \text{x} \right) \geq q^\ddagger\), and the step function in the above expression is redundant. In addition, if \(\dot{q} \left( \text{x} \right)\) only depends on momenta (or velocities) and not actually on coordinates, which will be true if \(q \left( \text{x} \right)\) is not curvilinear (and is true for some curvilinear coordinates \(q \left( \text{x} \right)\)), and if \(q \left( \text{x} \right)\) only depends on coordinates, then Equation \(\ref{20.30}\) reduces to \[k^\text{(TST)} = \frac{1}{h Q_r} \int d \text{x}_\textbf{p} e^{-\beta \sum_{i=1}^N \textbf{p}_i^2/2m_i} \left| \dot{q} \left( \textbf{p}_1, \ldots, \textbf{p}_N \right) \right| \int_{q \left( \textbf{r}_1, \ldots, \textbf{r}_N \right) = q^\ddagger} d \text{x}_\textbf{r} e^{-\beta U \left( \textbf{r}_1, \ldots, \textbf{r}_N \right)} \label{20.31} \] The integral \[Z^\ddagger = \int_{q \left( \textbf{r}_1, \ldots, \textbf{r}_N \right) = q^\ddagger} d \text{x}_\textbf{r} e^{-\beta U \left( \textbf{r}_1, \ldots, \textbf{r}_N \right)} \nonumber \] counts the number of microscopic states consistent with the condition \(q \left( \textbf{r}_1, \ldots, \textbf{r}_N \right) = q^\ddagger\) and is, therefore, a kind of partition function, and is denoted \(Q^\ddagger\). On the other hand, because it is a partition function, we can derive a free energy \(\Delta F^\ddagger\) from it \[F^\ddagger \propto -k_B T \: \text{ln} \: Z^\ddagger \label{20.32} \] Similarly, if we divide \(Q_r\) into its ideal-gas and configurational contributions \[Q_r = Q_r^\text{(ideal)} Z_r \label{20.33} \] then we can take \[Z_r = e^{-\beta F_r} \label{20.34} \] where \(F_r\) is the free energy of the reactants. Finally, setting \(\dot{q} = p/\mu\), where \(\mu\) is the associated mass, and \(p\) is the corresponding momentum of the reaction coordinate, then, canceling most of the momentum integrals between the numerator and \(Q_r^\text{(ideal)}\), the momentum integral we need is \[\int_0^\infty e^{-\beta p^2/2 \mu} \frac{p}{\mu} = k_B T \label{20.35} \] which gives the final expression for the transition state theory rate constant \[k^\text{(TST)} = \frac{k_B T}{h} e^{-\beta \left( F^\ddagger - F_r \right)} = \frac{k_B T}{h} e^{-\beta \Delta F^\ddagger} \label{20.36} \] Figure 28.8.2 actually shows \(k \left( t \right)/ k^\text{(TST)}\), which must start at \(1\). As the figure shows, in addition, for \(t > 0\), \(k \left( t \right) < k^\text{(TST)}\). Hence, \(k^\text{(TST)}\) is always an upper bound to the true rate constant. Transition state theory assumes that any trajectory that initially moves toward products will be a reactive trajectory. For this reason, it overestimates the reaction rate. In reality, trajectories can cross the dividing surface several or many times before eventually proceeding either toward products back toward reactants. Figure 28.8.3 shows that one can obtain trajectories of both types. Here, the dividing surface lies at \(q = 0\). Left, toward \(q = -1\) is the reactant side, and right, toward \(q = 1\) is the product side. Because some trajectories return to reactants and never become products, the true rate is always less than \(k^\text{(TST)}\), and we can write recrossings ( )
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As noted above, a variety of other \(\alpha\)-donor or \(\pi\)-acceptor ligands will bind to the active sites of biological oxygen carriers. As documented in Table 4.2, carbon monoxide (CO) generally binds more strongly to hemoglobin than does dioxygen, hence causing carbon-monoxide poisoning. In addition to being readily available from car exhausts and tobacco smoke to convert oxyhemoglobin to carbonmonoxyhemoglobin, CO is produced in the catabolism of heme molecules. Thus under even the most favorable of conditions, about 3 percent of human hemoglobin is in the carbonmonoxy form. When CO binds to a single metal atom in nonbiological systems, it does so through the carbon atom and in a linear manner: \[Fe—C=O \leftarrow Fe^{II} +\;^{-}C=O^{+} \tag{4.40}\] Model systems for carbonmonoxy (also called carbonyl) hemoglobin show a geometry similar to that of the Fe—C\(\equiv\)O group, linear or nearly so and essentially perpendicular to the porphyrin plane. The biochemical literature is littered with reports that this is the geometry adopted by CO in binding to hemoglobins. We will return to this topic later in this chapter, since the physiological consequences are potentially important. Carbon monoxide binds weakly as a \(\sigma\)-donor ligand to four-coordinate cobalt(II) systems. Despite a bout of artifactual excitement, CO has never been observed to bind significantly to five-coordinate Co systems with a nitrogenous axial base to yield octahedral six-coordinate species. The sulfur analogue thiocarbonyl (CS), although not stable as a free entity, binds very strongly to iron-porphyrin species in a linear manner. Nitric oxide (NO) binds to hemes even more strongly than CO (and hence O ), so strongly, in fact, that the Fe—N bond is very weak and easily ruptured. Attachment to the metal is via the nitrogen atom; however, the geometry of attachment is sensitive to the \(\pi\) basicity of the metalloporphyrin, and ranges from linear to strongly bent. In binding to Co the NO ligand is effectively reduced to NO , with concomitant oxidation of Co to Co : \(\tag{4.42}\) In much the same way that cobalt-dioxygen systems are paramagnetic (S = \(\frac{1}{2}\)) and amenable to EPR studies, iron-nitric oxide (also called iron nitrosyl) species are also paramagnetic and isoelectronic with cobalt-dioxygen species. The unpaired spin is localized mostly on the NO group. In contrast to the dioxygen, carbon-monoxide, and nitric-oxide ligands, the isocyanide and nitroso functions bear an organic tail. Moreover, nitroso ligands are isoelectronic with dioxygen. \(\tag{4.42}\) Thus, in principle, not only may the steric bulk of the ligand be varied, in order to probe the dimensions of the dioxygen-binding pocket,* but also the \(\sigma\)-donor/\(\pi\)-acceptor properties of the ligands may be varied by appropriate substituents on the aryl ring. Isocyanide groups may bind to metals in a variety of ways. For 1:1 adducts (Figure 4.19), the isocyanide group is approximately linear, although some flexibility seems to exist in a bis(t-butylisocyanide)iron(Il)tetraphenylporphyrinato complex. For zerovalent metals with much electron density available for donation into ligand \(\pi\)* orbitals, the isocyanide ligand has been observed to bend at the N atom. One prediction exists that an isocyanide ligand binds in this manner to hemoglobin. For 1:1 adducts of nitroso ligands, side-on, O-, and N-ligated modes are possible (Figure 4.19). No O-nitroso complexes have been definitively characterized by diffraction methods. For hemoglobin the N-nitroso mode is likely, since this is the mode found for the nitrosoalkane in Fe(TPP)(amine)(RNO). To date isocyanide ligands have not achieved their potential as probes of the geometry of the ligand-binding pocket in hemoglobin, partly because we lack structural data on the preferred geometry of attachment of these ligands in a sterically uncongested environment. * In fact, this classic experiment of St. George and Pauling established, for the first time and before any crystallographic data were available, that the heme group and the ligand-binding site in hemoglobin reside at least partway inside the protein, rather than on the surface.
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The molecular orbital picture of dioxygen differs from the Lewis picture. Both models predict an oxygen-oxygen double bond, but one model suggests unpaired electrons whereas the other indicates an electron-paired system. Often, there is experimental evidence available to check the reliability of predictions about structure. These data include measurements of bond lengths and bond strengths as well as magnetic properties. Bond dissociation energy data tell us how difficult it is to separate one atom from another in a molecule. Bond order is one of the factors that influences bond strength. Thus, measuring a bond dissociation energy is one way to confirm that dioxygen really does contain an oxygen-oxygen double bond. First, we need something to compare it to. Peroxides (such as hydrogen peroxide, H O , or sodium peroxide, Na O ) probably contain oxygen-oxygen single bonds, according to their Lewis structures. These bonds are relatively weak, costing about 35 kcal/mol to break. In contrast, the bond in dioxygen costs about 70 kcal/mol to break. Its bond is about twice as strong; it is a double bond. Bond dissociation energies can be complicated to measure. They require a comparison of energy changes in numerous chemical reactions so that the energy change resulting from cleavage of a specific bond can be inferred. In contrast, infrared absorption frequencies are easy to measure. They simply require shining infrared light through a sample and measuring what frequencies of the light are absorbed by the material. (A related technique, Raman spectroscopy, gives similar information by measuring subtle changes in the frequency of laser light that is scattered off a sample). The frequencies absorbed depend on what bonds are present in the material. These frequencies vary according to two basic factors: the weights of the atoms at the ends of the bond, and the strength of the bond between them. The stronger the bond, the higher the absorption frequency. Peroxides absorb infrared light at around 800 cm (this unusual frequency unit is usually pronounced "wavenumbers"). Dioxygen absorbs infrared light around 1300 cm . Since the atoms at the ends of the bond in both peroxide and dioxygen are oxygens, we can be sure that this difference in frequency is not due to a difference in mass. It is due to a difference in bond strength. The bond in dioxygen is much stronger than the O-O bond in peroxide, because the former is a double bond and the latter is a single bond. A third measure of bond order is found in bond length measurements. The more strongly bound two atoms are, the closer they are together. An O=O bond should be shorter than an O-O bond. Bond lengths can be measured by microwave spectroscopy (usually for gas-phase molecules), in which frequencies absorbed depend on the distance between the molecules. Alternatively, bond lengths can be measured by x-ray crystallography. X-rays can be diffracted through crystals of solid materials. The interference pattern that is produced can be mathematically decoded to produce a three-dimensional map of where all the atoms are in the material. The distances between these atoms can be measured very accurately. The O-O bond in peroxides are about 1.49 Angstroms long (an Angstrom is 10 m; this unit is often used for bond lengths because it is a convenient size for this task. Covalent bonds are generally one to three Angstroms long). The O-O bond in dioxygen is about 1.21 A long. The O-O bond in dioxygen is shorter and stronger than in a peroxide. In addition to bond order, there is the question of electron pairing in dioxygen. The Lewis structure suggests electrons are paired in dioxygen. The molecular orbital picture suggests two unpaired electrons. Compounds with paired electrons are referred to as diamagnetic. Those with unpaired electrons are called paramagnetic. Paramagnetic substances interact strongly with magnetic fields. It turns out that oxygen does interact with a magnetic fields. A sample of liquid-phase oxygen can be held between the poles of a magnet. Oxygen has unpaired electrons. This finding is consistent with molecular orbital theory, but not with simple Lewis structures. Thus, MO theory tells us something that the Lewis picture cannot. A final important source of experimental data is photoelectron spectroscopy. Photoelectron spectroscopy gives information about the electron energy levels in an atom or compound. In this technique, gas-phase molecules are subjected to high-energy electromagnetic radiation, such as ultraviolet light or X-rays. Electrons are ejected from various energy levels in the molecule, and the binding energies of electrons in those levels is determined. Thus, photoelectron spectroscopy provides verification for exactly the sort of information that quantitative molecular orbital calculations are designed to deliver. ,
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Penetration and shielding are two underlying principles in determining the physical and chemical properties of elements. We can predict basic properties of elements by using shielding and penetration characteristics to assess basic trends. Electrons are negatively charged and are pulled pretty close to each other by their attraction to the positive charge of a nucleus. The electrons are attracted to the nucleus at the same time as electrons repel each other. The balance between attractive and repulsive forces results in \[ F=k \dfrac{q_1q_1}{r^2} \label{1}\] The force that an electron feels is dependent on the distance from the nearest charge (i.e., an electron, usually with bigger atoms and on the outer shells) and the amount of charge. More distance between the charges will result in less force, and more charge will have more force of attraction or repulsion. However, this is not the case when observing atomic behavior. When considering the core electrons (or the electrons closest to the nucleus), the nuclear charge "felt" by the electrons ( (\(Z_{eff}\))) is close to the actual nuclear charge. As you proceed from the core electrons to the outer valence electrons, \(Z_{eff}\) falls significantly. This is because of ore electrons \[ F_{electron-nucleus}=k \dfrac{Ze^2}{r^2} \label{2}\] with Penetration and shielding result in an (\(F_{eff}\)) that holds the outer electrons to the atom, akin to Equation \(\ref{2}\), but with \(Z_{eff}\) substituted for \(Z\): \[ F_{eff}=k \dfrac{Z_{eff}e^2}{r^2} \label{3}\] Penetration describes the proximity to which an electron can approach to the nucleus. In a multi-electron system, electron penetration is defined by an electron's relative electron density (probability density) near the nucleus of an atom. Electrons in different orbitals have different wavefunctions and therefore different radial distributions and probabilities (defined by quantum numbers and around the nucleus). In other words, penetration depends on the shell ( ) and subshell ( . For example, we see that since a 2s electron has more electron density near the nucleus than a 2p electron, it is penetrating the nucleus of the atom more than the 2p electron. The penetration power of an electron, in a multi-electron atom, is dependent on the values of both the shell and subshell. Within the same shell value ( ), the penetrating power of an electron follows this trend in subshells ( ): s>p>d>f And for different values of shell (n) and subshell (l), penetrating power of an electron follows this trend: 1s>2s>2p>3s>3p>4s>3d>4p>5s>4d>5p>6s>4f.... and the energy of an electron for each shell and subshell goes as follows... 1s<2s<2p<3s<3p<4s<3d<4p.... The electron probability density for s-orbitals is highest in the center of the orbital, or at the nucleus. If we imagine a dartboard that represents the circular shape of the s-orbital and if the darts landed in correlation to the probability to where and electron would be found, the greatest dart density would be at the 50 points region but most of the darts would be at the 30 point region. When considering the 1s-orbital, the spherical shell of 53 pm is represented by the 30 point ring. Electrons which experience greater penetration experience stronger attraction to the nucleus, less shielding, and therefore experience a larger (\(Z_{eff}\)), but shield other electrons more effectively. An atom (assuming its atomic number is greater than 2) has core electrons that are extremely attracted to the in the middle of the atom. However the number of protons in the nucleus are never equal to the number of core electrons (relatively) adjacent to the nucleus. The number of protons increase by one across the periodic table, but the number of core electrons change by periods. The first period has no core electrons, the second has 2, the third has 10, and etc. This number is not equal to the number of protons. So that means that the core electrons feel a stronger pull towards the nucleus than any other electron within the system. The valence electrons are farther out from the nucleus, so they experience a smaller force of attraction. Shielding refers to the core electrons repelling the outer rings and thus lowering the 1:1 ratio. Hence, the nucleus has "less grip" on the outer electrons and are shielded from them. Electrons that have greater penetration can get closer to the nucleus and effectively block out the charge from electrons that have less proximity. For example, \(Z_{eff}\) is calculated by subtracting the magnitude of shielding from the total nuclear charge. The value of \(Z_{eff}\) will provide information on how much of a charge an electron actually experiences. Because the order of electron penetration from greatest to least is s, p, d, f; the order of the amount of shielding done is also in the order s, p, d, f. Since the 2s electron has more density near the nucleus of an atom than a 2p electron, it is said to shield the 2p electron from the full effective charge of the nucleus. Therefore the 2p electron feels a lesser effect of the positively charged nucleus of the atom due to the shielding ability of the electrons closer to the nucleus than itself, (i.e. 2s electron). These electrons that are shielded from the full charge of the nucleus are said to experience an \(Z_{eff}\))of the nucleus, which is some degree less than the full nuclear charge an electron would feel in a hydrogen atom or hydrogenlike ions. The effective nuclear charge of an atom is given by the equation: \[ Z_{eff}=Z-S \label{4}\] where. We can see from this equation that the effective nuclear charge of an atom increases as the number of protons in an atom increases. Therefore as we go from left to right on the periodic table the effective nuclear charge of an atom increases in strength and holds the outer electrons closer and tighter to the nucleus. This phenomena can explain the decrease in atomic radii we see as we go across the periodic table as electrons are held closer to the nucleus due to increase in number of protons and increase in effective nuclear charge. What is the effective attraction \(Z_{eff}\) experienced by the valence electrons in the three isoelectronic species: the fluorine anion, the neutral neon atom, and sodium cation? Each species has 10 electrons, and the number of nonvalence electrons is 2 (10 total electrons - 8 valence) but the effective nuclear charge varies because each has a different \(A\). The charge \(Z\) of the nucleus of a fluorine atom is 9, but the valence electrons are screened appreciably by the core electrons (four electrons from the 1s and 2s orbitals) and partially by the 7 electrons in the 2p orbitals. So the sodium cation has the greatest effective nuclear charge, and thus the smallest radius. A radial distribution function graph describes the distribution of orbitals with the effects of shielding (Figure \(\Page {2}\)). The small peak of the 2s orbital shows that the electrons in the 2s orbital are closest to the nucleus. Therefore, it is the electrons in the 2p orbital of Be that are being shielded from the nucleus, by the electrons in the 2s orbital. The following is the radial distribution of the 1s and 2s orbitals. Notice the 1s orbital is shifted to the right, while the 2s orbital has a node.
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The halogens chlorine and bromine add rapidly to a wide variety of alkenes without inducing the kinds of structural rearrangements noted for strong acids (first example below). The stereoselectivity of these additions is strongly , as shown in many of the following examples. An important principle should be restated at this time. The alkenes shown here are all achiral, but the addition products have chiral centers, and in many cases may exist as enantiomeric stereoisomers. In the absence of chiral catalysts or reagents, reactions of this kind will always give racemic mixtures if the products are enantiomeric. On the other hand, if two chiral centers are formed in the addition the reaction will be diastereomer selective. This is clearly shown by the addition of bromine to the isomeric 2-butenes. Anti-addition to cis-2-butene gives the racemic product, whereas anti-addition to the trans-isomer gives the meso-diastereomer. We can account both for the high stereoselectivity and the lack of rearrangement in these reactions by proposing a stabilizing interaction between the developing carbocation center and the electron rich halogen atom on the adjacent carbon. This interaction, which is depicted for bromine in the following equation, delocalizes the positive charge on the intermediate and blocks halide ion attack from the syn-location. The stabilization provided by this halogen-carbocation bonding makes rearrangement unlikely, and in a few cases three-membered cyclic halonium cations have been isolated and identified as true intermediates. A resonance description of such a bromonium ion intermediate is shown below. The positive charge is delocalized over all the atoms of the ring, but should be concentrated at the more substituted carbon (carbocation stability), and this is the site to which the nucleophile will bond. The stereoselectivity described here is in large part due to a . Because they proceed by way of polar ion-pair intermediates, chlorine and bromine addition reactions are faster in polar solvents than in non-polar solvents, such as hexane or carbon tetrachloride. However, in order to prevent solvent nucleophiles from competing with the halide anion, these non-polar solvents are often selected for these reactions. In water or alcohol solution the nucleophilic solvent may open the bromonium ion intermediate to give an α-halo-alcohol or ether, together with the expected vic-dihalide. Such reactions are sensitive to pH and other factors, so when these products are desired it is necessary to modify the addition reagent. Aqueous chlorine exists as the following equilibrium, K ≈ 10 . By adding AgOH, the concentration of HOCl can be greatly increased, and the chlorohydrin addition product obtained from alkenes. \[\ce{Cl_2 + H_2O <<=> HOCl + HCl} \] The more widely used HOBr reagent, hypobromous acid, is commonly made by hydrolysis of N-bromoacetamide, as shown below. Both HOCl and HOBr additions occur in an anti fashion, and with the regioselectivity predicted by this mechanism (OH bonds to the more substituted carbon of the alkene). \[ CH_3CONHBr + H_2O \rightarrow HOBr + CH_3CONH_2\] Vicinal halohydrins provide an alternative route for the epoxidation of alkenes over that of . As illustrated in the following diagram, a base induced intramolecular forms a three-membered cyclic ether called an epoxide. Both the halohydrin formation and halide displacement reactions are stereospecific, so stereoisomerism in the alkene will be reflected in the epoxide product ( trans-2-butene forms a trans-disubstituted epoxide). A general procedure for forming these useful compounds will be discussed in the next section.
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This theory was discovered due to Arrhenius's theory having deficiencies. Arrhenius's theory states that ions exist in a solid substance and dissociated from each other once the solid dissolves. Arhennius's theory did not take into account the fact that strong electrolytes are not as great as he originally thought and the values of the van 't Hoff factor relied on the concentration of the solution. The theory of electrolyte solution was brought about by Peter Debye and Erich Huckel in 1923. Interionic Attractions are when an ion is surrounded by an ionic atmosphere which has a net charge opposite for its own. For example an anion would be completely surrounded by ions mostly composed of cations and a cation would mostly be surrounded by ions of anions. The ionic atmosphere decreases the mobility of each ion by exerting a drag on it, which in turn also decreases the magnitude of colligative properties. The ionic atmosphere cannot created nor destroyed. In solutions with weak electrolytes the number of ions is not large, therefore the effect of the interionic attraction is small. In a concentrated solution of strong electrolytes the ion count is large, and therefore the interionic attraction will be apparent. The reason behind the differences in the interionic attraction is that in concentrated solutions ions are closer together due to the large ion count, while in less concentrated solutions they are further apart.
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The composition of relatively complex mixtures of metal ions can be determined using qualitative analysis , a procedure for discovering the of metal ions present in the mixture (rather than quantitative information about their amounts). The procedure used to separate and identify more than 20 common metal cations from a single solution consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively selective until almost all of the metal ions are precipitated, as illustrated in . Most metal chloride salts are soluble in water; only Ag , Pb , and Hg form chlorides that precipitate from water. Thus the first step in a qualitative analysis is to add about 6 M HCl, thereby causing AgCl, PbCl , and/or Hg Cl to precipitate. If no precipitate forms, then these cations are not present in significant amounts. The precipitate can be collected by filtration or centrifugation. Next, the acidic solution is saturated with H S gas. Only those metal ions that form very insoluble sulfides, such as As , Bi , Cd , Cu , Hg , Sb , and Sn , precipitate as their sulfide salts under these acidic conditions. All others, such as Fe and Zn , remain in solution. Once again, the precipitates are collected by filtration or centrifugation. Ammonia or NaOH is now added to the solution until it is basic, and then (NH ) S is added. This treatment removes any remaining cations that form insoluble hydroxides or sulfides. The divalent metal ions Co , Fe , Mn , Ni , and Zn precipitate as their sulfides, and the trivalent metal ions Al and Cr precipitate as their hydroxides: Al(OH) and Cr(OH) . If the mixture contains Fe , sulfide reduces the cation to Fe , which precipitates as FeS. The next metal ions to be removed from solution are those that form insoluble carbonates and phosphates. When Na CO is added to the basic solution that remains after the precipitated metal ions are removed, insoluble carbonates precipitate and are collected. Alternatively, adding (NH ) HPO causes the same metal ions to precipitate as insoluble phosphates. At this point, we have removed all the metal ions that form water-insoluble chlorides, sulfides, carbonates, or phosphates. The only common ions that might remain are any alkali metals (Li , Na , K , Rb , and Cs ) and ammonium (NH ). We now take a second sample from the solution and add a small amount of NaOH to neutralize the ammonium ion and produce NH . (We cannot use the same sample we used for the first four groups because we added ammonium to that sample in earlier steps.) Any ammonia produced can be detected by either its odor or a litmus paper test. A flame test on another original sample is used to detect sodium, which produces a characteristic bright yellow color. As discussed in , the other alkali metal ions also give characteristic colors in flame tests, which allows them to be identified if only one is present. Metal ions that precipitate together are separated by various additional techniques, such as forming complex ions, changing the pH of the solution, or increasing the temperature to redissolve some of the solids. For example, the precipitated metal chlorides of group 1 cations, containing Ag , Pb , and Hg , are all quite insoluble in water. Because PbCl is much more soluble in hot water than are the other two chloride salts, however, adding water to the precipitate and heating the resulting slurry will dissolve any PbCl present. Isolating the solution and adding a small amount of Na CrO solution to it will produce a bright yellow precipitate of PbCrO if Pb was in the original sample ( ). As another example, treating the precipitates from group 1 cations with aqueous ammonia will dissolve any AgCl because Ag forms a stable complex with ammonia: [Ag(NH ) ] . In addition, Hg Cl in ammonia (2Hg → Hg + Hg ) to form a black solid that is a mixture of finely divided metallic mercury and an insoluble mercury(II) compound, which is separated from solution: As another example, treating the precipitates from group 1 cations with aqueous ammonia will dissolve any AgCl because Ag forms a stable complex with ammonia: [Ag(NH ) ] . In addition, Hg Cl disproportionates in ammonia (2Hg → Hg + Hg ) to form a black solid that is a mixture of finely divided metallic mercury and an insoluble mercury(II) compound, which is separated from solution: \[Hg_2Cl_{2(s)} + 2NH_{3(aq)} \rightarrow Hg_{(l)} + Hg(NH_2)Cl_{(s)} + NH^+_{4(aq)} + Cl^−_{(aq)} \tag{17.39}\] Any silver ion in the solution is then detected by adding HCl, which reverses the reaction and gives a precipitate of white AgCl that slowly darkens when exposed to light: \[[Ag(NH_3)_2]^+_{(aq)} + 2H^+_{(aq)} + Cl^−_{(aq)} \rightarrow AgCl_{(s)} + 2NH^+_{4(aq)} \tag{17.40}\] Similar but slightly more complex reactions are also used to separate and identify the individual components of the other groups. In , the identity, not the amount, of metal ions present in a mixture is determined. The technique consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively selective until almost all the metal ions are precipitated. Other additional steps are needed to separate metal ions that precipitate together. Given a solution that contains a mixture of NaCl, CuCl , and ZnCl , propose a method for separating the metal ions.
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\newcommand{\gph}{^{\gamma}}     % gamma phase superscript\)  \( \newcommand{\aphp}{^{\alpha'}}   % alpha prime phase superscript\)  \( \newcommand{\bphp}{^{\beta'}}    % beta prime phase superscript\)  \( \newcommand{\gphp}{^{\gamma'}}   % gamma prime phase superscript\)  \( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\)  \( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\)  \( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\)  \( \newcommand{\dif}{\mathop{}\!\mathrm{d}}   % roman d in math mode, preceded by space\)  \( \newcommand{\Dif}{\mathop{}\!\mathrm{D}}   % roman D in math mode, preceded by space\)  \( \newcommand{\df}{\dif\hspace{0.05em} f} % df\)  \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)  \( \newcommand{\dq}{\dBar q} % heat differential\)  \( \newcommand{\dw}{\dBar w} % work differential\)  \( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)  \( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\)  \( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\)  \( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\)  \( \newcommand{\Del}{\Delta}\)  \( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\)  \( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\)  \( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\)  \( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\)  \( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\)  \( \newcommand{\dotprod}{\small\bullet}\)  \( \newcommand{\fug}{f} % fugacity\)  \( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)  \( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)  \( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)  \( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\)  \( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\)  \( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\) \( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\)  \( \newcommand{\D}{\displaystyle} % for a line in built-up\)  \( \newcommand{\s}{\smash[b]} % use in equations with conditions of validity\)  \( \newcommand{\cond}[1]{\\[-2.5pt]{}\tag*{#1}}\)  \( \newcommand{\nextcond}[1]{\\[-5pt]{}\tag*{#1}}\)  \( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}}     % gas constant value\)  \( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \) \( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \) \( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \)  The advancement \(\xi\) of a chemical reaction in a closed system describes the changes in the amounts of the reactants and products from specified initial values of these amounts. We have seen that if the system is maintained at constant temperature and pressure, \(\xi\) changes spontaneously in the direction that decreases the Gibbs energy. The change continues until the system reaches a state of reaction equilibrium at the minimum of \(G\). The value of the advancement in this equilibrium state will be denoted \(\xi\eq\), as shown in Fig. 11.15. The value of \(\xi\eq\) depends in general on the values of \(T\) and \(p\). Thus when we change the temperature or pressure of a closed system that is at equilibrium, \(\xi\eq\) usually changes also and the reaction spontaneously to a new equilibrium position. To investigate this effect, we write the total differential of \(G\) with \(T\), \(p\), and \(\xi\) as independent variables \begin{equation} \dif G = -S\dif T + V\difp + \Delsub{r}G\dif\xi \tag{11.9.1} \end{equation} and obtain the reciprocity relations \begin{equation} \Pd{\Delsub{r}G}{T}{p, \xi} = -\Pd{S}{\xi}{T,p} \qquad \Pd{\Delsub{r}G}{p}{T, \xi} = \Pd{V}{\xi}{T,p} \tag{11.9.2} \end{equation} We recognize the partial derivative on the right side of each of these relations as a molar differential reaction quantity: \begin{equation} \Pd{\Delsub{r}G}{T}{p, \xi} = -\Delsub{r}S \qquad \Pd{\Delsub{r}G}{p}{T, \xi} = \Delsub{r}V \tag{11.9.3} \end{equation} We use these expressions for two of the coefficients in an expression for the total differential of \(\Delsub{r}G\): \begin{gather} \s{ \dif\Delsub{r}G = -\Delsub{r}S\dif T + \Delsub{r}V\difp + \Pd{\Delsub{r}G}{\xi}{T,p}\dif\xi } \tag{11.9.4} \cond{(closed system)} \end{gather} Since \(\Delsub{r}G\) is the partial derivative of \(G\) with respect to \(\xi\) at constant \(T\) and \(p\), the coefficient \(\pd{\Delsub{r}G}{\xi}{T,p}\) is the partial derivative of \(G\) with respect to \(\xi\): \begin{equation} \Pd{\Delsub{r}G}{\xi}{T,p} = \Pd{^2 G}{\xi^2}{T,p} \tag{11.9.5} \end{equation} We know that at a fixed \(T\) and \(p\), a plot of \(G\) versus \(\xi\) has a slope at each point equal to \(\Delsub{r}G\) and a minimum at the position of reaction equilibrium where \(\xi\) is \(\xi\eq\). At the minimum of the plotted curve, the slope \(\Delsub{r}G\) is zero and the second derivative is positive (see Fig. 11.15). By setting \(\Delsub{r}G\) equal to zero in the general relation \(\Delsub{r}G = \Delsub{r}H - T\Delsub{r}S\), we obtain the equation \(\Delsub{r}S = \Delsub{r}H/T\) which is valid only at reaction equilibrium where \(\xi\) equals \(\xi\eq\). Making this substitution in Eq. 11.9.4, and setting \(\dif\Delsub{r}G\) equal to zero and \(\dif\xi\) equal to \(\dif\xi\eq\), we obtain \begin{gather} \s{ 0 = -\frac{\Delsub{r}H}{T}\dif T + \Delsub{r}V\difp + \Pd{^2 G}{\xi^2}{T,p}\dif\xi\eq } \tag{11.9.6} \cond{(closed system)} \end{gather} which shows how infinitesimal changes in \(T\), \(p\), and \(\xi\eq\) are related. Now we are ready to see how \(\xi\eq\) is affected by changes in \(T\) or \(p\). Solving Eq. 11.9.6 for \(\dif\xi\eq\) gives \begin{gather} \s{ \dif\xi\eq = \frac{\displaystyle \frac{\Delsub{r}H}{T}\dif T - \Delsub{r}V\difp} { \Pd{^2 G}{\xi^2}{T,p}} } \tag{11.9.7} \cond{(closed system)} \nextcond{} \end{gather} The right side of Eq. 11.9.7 is the expression for the total differential of \(\xi\) in a closed system at reaction equilibrium, with \(T\) and \(p\) as the independent variables. Thus, at constant pressure the equilibrium shifts with temperature according to \begin{gather} \s{ \Pd{\xi\eq}{T}{p} = \frac{\Delsub{r}H} { T\Pd{^2 G}{\xi^2}{T,p}} } \tag{11.9.8} \cond{(closed system)} \nextcond{} \end{gather} and at constant temperature the equilibrium shifts with pressure according to \begin{gather} \s{ \Pd{\xi\eq}{p}{T} = - \frac{\Delsub{r}V} { \Pd{^2 G}{\xi^2}{T,p}} } \tag{11.9.9} \cond{(closed system)} \nextcond{} \end{gather} Because the partial second derivative \(\pd{^2G}{\xi^2}{T,p}\) is positive, Eqs. 11.9.8 and 11.9.9 show that \(\pd{\xi\eq}{T}{p}\) and \(\Delsub{r}H\) have the same sign, whereas \(\pd{\xi\eq}{p}{T}\) and \(\Delsub{r}V\) have opposite signs. These statements express the application to temperature and pressure changes of what is known as : When a change is made to a closed system at equilibrium, the equilibrium shifts in the direction that tends to oppose the change. Here are two examples. Another kind of change for which Le Chatelier’s principle gives an incorrect prediction is the addition of an inert gas to a gas mixture of constant volume. Adding the inert gas at constant \(V\) increases the pressure, but has little effect on the equilibrium position of a gas-phase reaction regardless of the value of \(\Delsub{r}V\). This is because the inert gas affects the activities of the reactants and products only slightly, and not at all if the gas mixture is ideal, so there is little or no effect on the value of \(Q\subs{rxn}\). (Note that the dependence of \(\xi\eq\) on \(p\) expressed by Eq. 11.9.9 does not apply to an open system.)
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The goal of equilibrium statistical mechanics is to calculate the diagonal elements of \(\rho_{e q}\) so we can evaluate average observables \(\langle A\rangle=\operatorname{Tr}\left\{A \rho_{e q}\right\}=A\) that give us fundamental relations or equations of state. Just as thermodynamics has its potentials \(\mathrm{U}, \mathrm{A}, \mathrm{H}, \mathrm{G}\) etc., so statistical mechanics has its ensembles, which are useful depending on what macroscopic variables are specified. We first consider the microcanonical ensemble because it is the one directly defined in postulate II of statistical mechanics. In the microcanonical ensemble \(U\) is fixed (Postulate I), and other constraints that are fixed are the volume \(V\) and mole number \(n\) (for a simple system), or other extensive parameters (for more complicated systems). The 'partition function' of an ensemble describes how probability is partitioned among the available microstates compatible with the constraints imposed on the ensemble. In the case of the microcanonical ensemble, the partitioning is equal in all microstates at the same energy: according to postulate II, with \(p_{i}=\rho_{i i}^{(e q)}=1 / W(U)\) for each microstate "i" at energy U. Using just this, we can evaluate equations of state and fundamental relations. 2. Calculation of thermodynamic quantities from W(U) Consider again the model system of a box with \(\mathrm{M}=\mathrm{V} / \mathrm{V}_{0}\) volume elements \(\mathrm{V}_{0}\) and \(\mathrm{N}\) particles of volume \(\mathrm{V}_{0}\), so each particle can fill one volume elements. The particles can randomly hop among unoccupied volume elements to randomly sample the full volume of the box. This is a simple model of an ideal gas. As shown in the last chapter, \[W=\dfrac{M !}{(M-N) ! N !}\] for identical particles, and we can approximate this, if \(\mathrm{M}<<\mathrm{N}\) by \[W \approx \dfrac{1}{N !}\left(\dfrac{V}{V_{0}}\right)^{N}\] since \(M ! /(M-N !) \approx M^{N}\) in that case. Assuming the hopping samples all microstates so the system reaches equilibrium, we compute the equilibrium entropy, as proved in chapter 10 from postulate III, as \[S=k_{B} \ln W \approx S_{0}+N k_{B} \ln \left(V / V_{0}\right)\] where \(S_{0}\) is independent of volume. This gives the volume dependence of the entropy of an ideal gas. Note that by taking the derivative \((\partial \mathrm{S} / \partial \mathrm{V})=k_{B} \mathrm{~N} / \mathrm{V}=\mathrm{P} / \mathrm{T}\) we can immediately derive the ideal gas law \(\mathrm{PV}=\mathrm{N} k_{B} \mathrm{~T}=\mathrm{nRT}\). The above model does not give us the energy dependence, since we did not explicitly consider the energy of the particles, other than to assume there was enough energy for them to randomly hop around. We now remedy this by considering the energy levels of particles in a box. The result will also demonstrate once more that \(\tilde{\Omega}\) increases so ferociously fast that it is equal to \(\mathrm{W}\) with incredibly high accuracy for more than a handful of particles. Let the total energy \(U\) be randomly distributed among particles in a box of volume \(L^{3}=\) \(\mathrm{V}\). The energy is given by \[U=\dfrac{1}{2 m} \sum_{i=1}^{3 N} p_{i}^{2},\] where \(\mathrm{i}=1,2,3\) are the \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) coordinates of particle #1, and so forth to \(\mathrm{i}=3 N-2,3 N-1,3 N\) are the \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) coordinates of particle \(\# N\). In quantum mechanics, the momentum of a free particle is given by \(p=h / \lambda\), where \(h\) is Planck's constant. Only certain waves \(\Psi(x)\) are allowed in the box, such that \(\Psi(x)=0\) at the boundaries of the box, as shown in the figure below. The wavelengths \(\lambda=\mathrm{L} / 2, \mathrm{~L}, 3 \mathrm{~L} / 2 \cdots\) can be inserted in the equation for total energy, yielding \[U=\dfrac{1}{2 m} \sum_{i=1}^{3 N}\left(\dfrac{h n_{i}}{2 L}\right)^{2}=\sum_{i=1}^{3 N} \dfrac{h^{2} n_{i}^{2}}{8 m L^{2}}, \quad n_{i}=1,2,3 \cdots,\] the energy for a bunch of particles in a box. W(U) is the number of states at energy U. Looking at the figure again, all the energy levels are "dots" in a \(3 \mathrm{~N}\)-dimensional cartesian space, called the "state space", or "action space" or sometimes "quantum number space." The surface of constant energy \(\mathrm{U}\) is the surface of a hypersphere of dimension \(3 \mathrm{~N}-1\) in state space. The reason is that the above equation is of the form constant \(=x^{2}+y^{2}+\cdots\) where the variables are the quantum numbers. The number of states within a thin shell of energy \(U\) at the surface of the sphere is \[W(U) \; where\; \lim _{N \rightarrow \infty} W(U)=\tilde{\Omega}\]. \(\tilde{\Omega}\) is the total number of states inside the sphere, which at a first glance would seem to be much larger than W(U), the states in the shell. In fact, for a very high dimensional hypervolume, a thin shell at the surface contains all the volume, so in fact, \(\tilde{\Omega}\) is essentially equal to W(U) and we can just calculate the former to a good approximation when \(N\) is large. If this is hard to believe, consider an analogous example of a hypercube instead of a hypersphere. Its volume is \(L^{\mathrm{m}}\), where \(m\) is the number of dimensions. The change in volume with side length \(\mathrm{L}\) is \(\partial \mathrm{V} / \partial \mathrm{L}=\mathrm{mL}^{\mathrm{m}-1}\), so \(\Delta \mathrm{V}=\mathrm{mL}^{\mathrm{m}-1} \Delta \mathrm{L}\) is the volume of a shell of width \(\Delta \mathrm{L}\) at the surface of the cube. The ratio of that volume to the total volume is \(\Delta \mathrm{V} / \mathrm{V}=\mathrm{m} \Delta \mathrm{L} / \mathrm{L}\). Let's take the example our intuition is built on, \(\mathrm{m}=3\), and assume \(\Delta \mathrm{L} / \mathrm{L}=0.001\), just a \(0.1 \%\) surface layer. Then \(\Delta \mathrm{V} / \mathrm{V}=3 \cdot 10^{-3}<<\mathrm{V}\) indeed. But now consider \(\mathrm{m}=10^{20}\), a typical number of particles in a statistical mechanical system. Now \(\Delta \mathrm{V} / \mathrm{V}=10^{20} 10^{-3}=10^{17}\). The little increment in volume is much greater than the original volume of the cube, and contains essentially all the volume of the new "slightly larger" cube. It may be "slightly" large in side length, but it is astronomically larger in volume. Now back to our hypersphere in the figure. Its volume, which is essentially equal to W(U), the number of states just at the surface of the sphere, is \[W(U)=\left(\dfrac{1}{2}\right)^{3 N} V_{\text {hypersphere }}=\left(\dfrac{1}{2}\right)^{3 N} \dfrac{\pi^{3 N / 2}}{\Gamma(3 N / 2+1)} R^{3 N}=\left(\dfrac{U}{U_{0}}\right)^{3 N / 2} .\] The \((1 / 2)^{3 \mathrm{~N}}\) is there because all quantum numbers must be greater than zero, so only the positive part of the sphere should be counted. The Gamma function \(\Gamma\) is related to the factorial function, and \(R\) is the radius of the sphere, which is given by \[R=n_{\text {max }}=\sqrt{\dfrac{8 m L^{2} U}{h^{2}}},\] the largest quantum number, if all energy is in a single mode. They key is that \(R \sim \sqrt{U}\), so \(U\) is raised to the \(3 \mathrm{~N} / 2\) power, where \(N\) is the number of particles, 3 is because there are three modes per particle, and the \(1 / 2\) is because of the energy of a free particle depends on the square of the quantum number. Thus \[S(U)=k_{B} \ln W(U)=S_{0}+\dfrac{3}{2} N k_{B} \ln U=S_{0}+\dfrac{3}{2} n R \ln U \text {, }\] where the constant \(S_{0}\) is not the same as in the previous example. We used the volume equation from the previous example to obtain an equation of state (PV=nRT), and we can obtain another equation of state here: \[\left(\dfrac{\partial S}{\partial U}\right)_{V, n}=\dfrac{1}{T}=\dfrac{3}{2} n R \dfrac{1}{U} \text { or } U=\dfrac{3}{2} n R T\] This equation relates the energy of an ideal gas to its temperature. \(3 n\) is the number of modes or degrees of freedom (3 velocities per particle time \(n\) moles of particles), whereas the factor of 2 comes directly from the particle-in-a-box energy function - in case you ever wondered where that comes from. So, for a harmonic oscillator, \(\mathrm{n} \sim U\) \(\left(E=\hbar \omega(n+1 / 2)\right.\) as you may recall) instead of \(n \sim U^{1 / 2}\), and you might expect \(U=3 n R T\) for \(3 \mathrm{~N}\) particles held together by springs into a solid crystal lattice. And indeed, that is true for an ideal lattice at high temperature (in analogy to an ideal gas at high temperature). Unlike free particles, the energy of oscillators does not have the factor of 1/2. The 'deep' reason is that an oscillator has two degrees of freedom to store energy in each direction, not just one: there's still the kinetic energy, but there's also potential energy. Example 3: A system of \(\mathrm{N}\) uncoupled spins \(\mathrm{s}_{\mathrm{z}}=\pm 1 / 2\) The Hamiltonian for this system in a magnetic field is given by \[H=\sum_{j=1}^{N} s_{z j} B+\dfrac{N B}{2} \text {, }\] where the extra term at the end is added so the energy equals zero when all the spins are pointing down. At energy \(U=0\), no spin is excited. For each excited spin, the energy increases by \(B\), so at energy \(U, U / B\) atoms are excited. These \(U / B\) excitations are indistinguishable and can be distributed in \(\mathrm{N}\) sites: \[W(U)=\dfrac{N !}{\left(N-\dfrac{U}{B}\right) !\left(\dfrac{U}{B}\right) !}=\dfrac{\Gamma(N+1)}{\Gamma\left(N+1-\dfrac{U}{B}\right) \Gamma\left(\dfrac{U}{B}+1\right)} .\] This is our usual formula for permutations; the right side is in terms of Gamma functions, which are defined even when \(U / B\) is not an integer. Gamma functions basically interpolate the factorial function for noninteger values. This formula has a potential problem built-in: clearly, when \(U\) starts out at 0 and then increases, \(W\) initially increases. But for \(U=N B\) (the maximum energy), \(W=1\) again. In fact, \(W\) reaches its maximum for \(U=N B / 2\). But if \(W(U)\) is not monotonic in \(U\), then \(S\) isn't either, violating P3 of thermodynamics. Let's see how this works out. For large \(\mathrm{N}\), and temperature neither so low that \(\dfrac{U}{B} \sim O(1)\), nor so high that \(\dfrac{U}{B} \sim O(N)\), we can use the Stirling expansion \(\ln N ! \approx N \ln N-N\), yielding \[\begin{aligned} &\dfrac{S}{k_{B}}=\ln \Omega \approx N \ln N-N-\left(N-\dfrac{U}{B}\right) \ln \left(N-\dfrac{U}{B}\right)+N \dfrac{U}{B}-\dfrac{U}{B} \ln \dfrac{U}{B}+\dfrac{U}{B} \\ &\approx N \ln N-\left(N-\dfrac{U}{B}\right) \ln \left(N-\dfrac{U}{B}\right)-\dfrac{U}{B} \ln \dfrac{U}{B}+\dfrac{U}{B} \ln N-\dfrac{U}{B} \ln N \\ &\approx-N \ln \left(1-\dfrac{U}{N B}\right)+\dfrac{U}{B} \ln \left(1-\dfrac{U}{N B}\right)-\dfrac{U}{B} \ln \left(\dfrac{U}{N B}\right) \\ &\approx\left(\dfrac{U}{B}-N\right) \ln \left(1-\dfrac{U}{N B}\right)-\dfrac{U}{B} \ln \left(\dfrac{U}{N B}\right) \end{aligned}\] after canceling terms as much as possible. We can now calculate the temperature and obtain the equation of state \(U(T)\) : \[\dfrac{1}{T}=\left(\dfrac{\partial S}{\partial U}\right)_{N} \approx \dfrac{k}{B} \ln \left(\dfrac{N B}{U}-1\right) \Rightarrow U \approx \dfrac{N B}{1+e^{B / k T}}\] In this equations, at \(T \sim 0, \quad U \rightarrow 0\); and as \(T \rightarrow \infty, U \rightarrow N B / 2\). So even at infinite temperature the energy can only go up to half the maximum value, where \(W(U)\) is monotonic. The population cannot be 'inverted' to have more spins point up than down. At most half the spins can be made to point up by heating. This should come as no surprise: if the number of microstates is maximized by having only half the spins point up when energy is added, then that's the state you will get (this is true even in the exact solution). Note that this does not mean that it is impossible to get all spins to point up. It is just not an equilibrium state at any temperature between 0 and \(\infty .\) Such nonequilibrium states with more spins up (or atoms excited) than down are called "inverted" populations. In lasers, such states are created by putting the system (like a laser crystal) far out of equilibrium. Such a state will then relax back to an equilibrium state, releasing a pulse of energy as the spins (or atoms) drop from the excited to the ground state. The heat capacity of the above example system is \[c_{v}=\left(\dfrac{\partial U}{\partial T}\right)_{N} \approx \dfrac{N B^{2}}{k T^{2}} \dfrac{e^{B / k T}}{\left(1+e^{B / k T}\right)^{2}} \text { peaked at } 4 k_{B} T / B \text {, }\] so we can calculate thermodynamic quantities as input for thermodynamic manipulations. As we shall see in detail later (actually, we saw it in the previous example!), in any real system the heat capacity must eventually approach \(c_{v}=N k_{B} / 2\), where \(N\) is the number of degrees of freedom. However, a broad peak at \(4 k_{B} T / B\) is a sign of two low-lying energy levels spaced by \(B\). Levels at higher energy will eventually contribute to \(c_{v}\), making sure it does not drop. Example 4: Let us check that T derived from \[\left(\dfrac{\partial S}{\partial U}\right)_{N}=\left(\dfrac{\partial k_{B} \ln W}{\partial U}\right)_{N}=\dfrac{1}{T}\] indeed agrees with the intuitive concept of temperature. Consider two baths within a closed system, so \(U=U_{1}+U_{2}=\) const. \(\Rightarrow d U=0 \Rightarrow d U_{1}=-d U_{2}\). If we know \[W_{i}\left(U_{i}\right) \Rightarrow d W_{i}=\dfrac{\partial W_{i}}{\partial U_{i}} d U_{i}\] for each bath, then \[\begin{aligned} W_{\text {tot }} &=W_{1} W_{2} \\ d W_{\text {tot }} &=\left(W_{1}+d W_{1}\right) \cdot\left(W_{2}+d W_{2}\right)-W_{1} W_{2}+O\left(d W^{2}\right) \\ &=W_{1} d W_{2}+W_{2} d W_{1} \\ &=\left(-W_{1} \dfrac{\partial W_{2}}{\partial U_{2}}+W_{2} \dfrac{\partial W_{1}}{\partial U_{1}}\right) d U_{1}=0 \end{aligned}\] at equilibrium because the maximum number of states is already occupied. For this to be true for any infinitesimal energy flow \(d U_{1}\), \[\begin{aligned} &\Rightarrow \dfrac{1}{W_{2}}\left(\dfrac{\partial W_{2}}{\partial U_{2}}\right)_{V, N}=\dfrac{1}{W_{1}}\left(\dfrac{\partial W_{1}}{\partial U_{2}}\right)_{V, N} \\ &\Rightarrow\left(\dfrac{\partial \ln W_{2}}{\partial U_{2}}\right)_{V, N}=\left(\dfrac{\partial \ln W_{1}}{\partial U_{2}}\right)_{V, N} \text { or }\left(\dfrac{\partial S_{2}}{\partial U_{2}}\right)_{V, N}=\dfrac{1}{T_{2}}=\left(\dfrac{\partial S_{1}}{\partial U}\right)_{V, N}=\dfrac{1}{T_{1}} \end{aligned}\] At equilibrium, the temperatures are equal, fitting out thermodynamic definition that "temperature is equalized when heat flow is allowed."
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We have seen how the uncertainty principle relates the attainable line widths in different kinds of spectroscopy to the lifetimes of the states - the the lifetime, the the spread in energy of the states and the greater the spectroscopic line width. So far we have associated short lifetimes with excited states, but this need not necessarily be so. Short lifetimes also may be associated with chemical or conformational changes. As a specific example, suppose we have a magnetic nucleus in the \(+\frac{1}{2}\) state located in a chemical environment whereby it experiences a magnetic field \(H\) such that \(H = H_0 \left( 1 - \sigma \right)\). This nucleus will have a particular magnetic energy, call it \(E\). Now suppose the nucleus has a lifetime \(\Delta t\) before it moves to a different chemical environment where it experiences a different field \(H' = H_0 \left( 1 - \sigma' \right)\) and has a different energy \(E'\). Clearly, there will be an uncertainty in the energy \(E\) depending on the lifetime of the \(+\frac{1}{2}\) nucleus in the particular chemical environment before it switches to the new environment with a different shielding and a different energy. Consider a specific example, 2,2,3,3-tetrachlorobutane. This substance can exist in three different conformations, \(1\), \(2\), and \(3\). By reference to the discussions in , you will recognize that \(1\) is achiral, whereas \(2\) and \(3\) are enantiomers: Clearly, if we could separate \(1\) from \(2\) and \(3\), the protons of its methyl groups would have chemical shifts from those of \(2\) and \(3\) (which, as enantiomers, would have their methyl proton resonances at the same frequency). Now consider a mixture of the conformations \(1\), \(2\), and \(3\) in which the before they convert are \(\Delta t\). Assuming that the lifetimes of the \(+\frac{1}{2}\) and \(-\frac{1}{2}\) magnetic states are long compared to \(\Delta t\), then uncertainty in the transition energies will depend on the lifetimes of the states (conformations) with different chemical shifts for the protons. The chemical-shift difference between \(1\) and \(2\) or \(3\) at \(-44^\text{o}\), as shown by Figure 27-3, is about \(5 \: \text{Hz}\). From Equation 27-1, we can see that \(5 \: \text{Hz}\) also will be the degree of the uncertainty in the frequency when \(\Delta t \sim 1/ \left( 2 \pi \Delta \nu \right) = 1/ \left( 2 \pi \times 5 \: \text{Hz} \right) = 0.03 \: \text{sec}\). Thus if \(1\) has a lifetime much than \(0.03 \: \text{sec}\), say \(1 \: \text{sec}\), before going to \(2\) or \(3\), it will give a resonance of its own and, of course, \(2\) and \(3\) will also. However, if \(1\), \(2\), and \(3\) have lifetimes much than \(0.03 \: \text{sec}\), say \(0.001 \: \text{sec}\), then we expect average resonance for \(1\), \(2\), and \(3\). Either condition can be realized for 2,2,3,3-tetrachlorobutane by taking the proton nmr spectrum at different temperatures (Figure 27-3). At \(-44^\text{o}\), at which \(\Delta t\) is \(1.0 \: \text{sec}\), we see the separate peaks for \(1\) and for \(2\) and \(3\). At \(-20^\text{o}\), at which \(\Delta t\) is \(0.045 \: \text{sec}\), the uncertainty is such that the lines have coalesced and we no longer can see the separate peaks. When the spectrum is taken at room temperature, at which \(\Delta t\) is about \(0.00005 \: \text{sec}\), a single very sharp line is observed. We get a sharp line at this temperature because, for practical purposes, there is no uncertainty about the chemical shift of \(1\), \(2\), and \(3\). The line width now is determined again by the lifetimes of the \(+\frac{1}{2}\) and \(-\frac{1}{2}\) magnetic states, by the lifetimes of the conformations. and (1977)
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The polarity of the carbonyl group is manifest in the physical properties of carbonyl compounds. Boiling points for the lower members of a series of aldehydes and ketones are \(50\)-\(80^\text{o}\) than for hydrocarbons of the same molecular weight; this may be seen by comparing the data of Table 16-2 (physical properties of aldehydes and ketones) with those in Table 4-1 (physical properties of alkanes). The water solubility of the lower-molecular-weight aldehydes and ketones is pronounced (see Table 16-2). This is to be expected for most carbonyl compounds of low molecular weight and is the consequence of hydrogen-bonding between the water and the electronegative oxygen of the carbonyl group: and (1977)
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We have seen that the partition function of a system gives us the key to calculate thermodynamic functions like energy or pressure as a moment of the energy distribution. We can extend this formulism to calculate the entropy of a system once its \(Q\) is known. We can start with Boltzmann's (statistical) definition of entropy: \[S = k \ln(W) \label{Boltz} \] with \[W=\frac{A!}{\prod_j{a_j}!} \nonumber \] Combining these equations, we obtain: \[S_{ensemble} = k \ln\frac{A!}{\prod_j{a_j}!} \nonumber \] Rearranging: \[S_{ensemble} = k \ln{A!}-k \sum_j{\ln{a_j!}} \nonumber \] Using : \[\begin{split}S_{ensemble} &= k A\ln{A}-k A - k \sum_j{a_j\ln{a_j}} + k \sum_j{a_j} \\ &= k A\ln{A}- k \sum_j{a_j\ln{a_j}}\end{split} \nonumber \] Since: \[\sum_j{a_j}=A \nonumber \] The probability of finding the system in state \(a_j\) is: \[p_j=\frac{a_j}{A} \nonumber \] Rearranging: \[a_j = p_jA \nonumber \] Plugging in: \[S_{ensemble}=k A\ln{A}- k \sum_j{p_jA\ln{p_jA}} \nonumber \] Rearranging: \[S_{ensemble}=k A\ln{A}- k \sum_j{p_jA\ln{p_j}}- k \sum_j{p_jA\ln{A}} \nonumber \] If \(A\) is constant, then: \[k \sum_j{p_jA\ln{A}} = k A\ln{A}\sum_j{p_j} \nonumber \] Since: \(\sum_j{p_j} = 1\) We get: \[S_{ensemble}=k A\ln{A}- k \sum_j{p_jA\ln{p_j}}- k A\ln{A} \nonumber \] The first and last term cancel out: \[S_{ensemble}=- k \sum_j{p_jA\ln{p_j}} \nonumber \] We can divide by \(A\) to get the entropy of the system: \[S_{system}=- k \sum_j{p_j\ln{p_j}} \label{eq10}\] If all the \(p_j\) are zero except for the for one, then the system is perfectly ordered and the entropy of the system is zero. The probability of being in state \(j\) is \[p_j=\frac{e^{-\beta E_j}}{Q} \label{eq15}\] Plugging Equation \ref{eq15} into Equation \ref{eq10} results in \[\begin{align*}S &= - k \sum_j{\frac{e^{-\beta E_j}}{Q}\ln{\frac{e^{-\beta E_j}}{Q}}} \\[4pt] &= - k \sum_j{\frac{e^{-\beta E_j}}{Q}\left(-\beta E_j- \ln{Q}\right)} \\[4pt] &= - \beta k \sum_j{\frac{E_je^{-\beta E_j}}{Q}}+\frac{k\ln{Q}}{Q}\sum_j{e^{-\beta E_j}} \end{align*}\] Making use of: \[\beta k=\frac{1}{T} \nonumber \] And: \[\sum{\frac{e^{-\beta E_j}}{Q}}=\sum{p_j}=1 \nonumber \] We obtain: \[S= \dfrac{U}{T} + k\ln Q \label{20.42} \]
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There are many ways in which ligand affinity may be perturbed (Figure 4.25). It is convenient to divide these into two groups, referred to as distal and proximal effects. effects are associated with the stereochemistry of the metalloporphyrinato moiety and the coordination of the axial base, and thus their influence on O and CO affinity is indirect. effects pertain to noncovalent interactions of the metal-porphyrinato skeleton and the sixth ligand (O , CO, etc.) with neighboring solvent molecules, with substituents, such as pickets or caps, on the porphyrin, and with the surrounding protein chain. The distal groups that hover over the O -binding site engender the most important distal effects. For convenience, the effects of crystal packing and the protein matrix on porphyrin conformation will be discussed among the proximal effects, although as nonbonded interactions they properly are distal effects. To a first approximation, the effects of substituents on the porphyrin ring, as transmitted through bonds to the metal center, do not perturb the ligand binding properties as much as do distal effects. Thus substituents, such as vinyl and propionic-acid groups on protoporphyrin IX and o-pivalamidophenyl pickets, are ignored; one porphyrin is much like another. At the end of this subsection the various ways ligand affinity may be modulated will be summarized in an augmented version of Figure 4.3. Few molecules have had their conformational properties characterized as exhaustively as have metalloporphyrins. The cyclic aromatic 24-atom porphyrinato skeleton offers a tightly constrained metal-binding site. The conformation of least strain is planar, and the radius of the hole of the dianion is close to 2.00 Å, leading to metal-porphyrinato nitrogen-atom separations, M-N , of 2.00 Å if the metal is centered in the square plane defined by the four porphyrinato nitrogen atoms. Small deviations from planarity are generally observed and attributed to crystal packing effects; large deviations may be induced by bulky substituents on the porphyrin skeleton, especially at the positions, by the crystal matrix, or by the highly anisotropic protein matrix. The 2.00 Å radius hole neatly accommodates low-spin (S = 0) and intermediatespin (S = 1) iron(II), low-spin (S = \(\frac{1}{2}\)) iron(III), and cobalt(II) and cobalt(III) ions. * With few exceptions the metal is centered in or above the central hole for mononuclear porphyrin species; only rarely do M-N bonds show a significant (though still small) scatter about their mean value. * In order to accommodate smaller ions, such as nickel(II), the porphyrin skeleton may contract by ruffling, with little loss of aromaticity; like a pleated skirt the pyrrole rings rotate alternately clockwise and counterclockwise about their respective M-N vectors. This distortion leaves the four porphyrinato nitrogen atoms, N , still coplanar. Alternatively, the porphyrin skeleton may buckle to give a saddle conformation; the N atoms may acquire a small tetrahedral distortion in this process. M-N bonds as short as 1.92 Å have been observed. Metals with one or two electrons in their 3d orbital have a radius larger than 2.00 Å. In order to accommodate them in the plane of the porphyrin, the porphyrin skeleton expands. M-N separations as long as 2.07 Å may occur with the metal still centered in the plane of the N atoms. For five-coordinate complexes the magnitude of the displacement of the metal from the plane of the four nitrogen atoms, M • • • porph, is a consequence of the electronic configuration of ML complexes. Of course, the effect is augmented if the 3d orbital (directed along M-N bonds, Figure 4.16) is occupied. Compare a displacement of 0.14 Å for Co(TPP)(1,2-Me Im) (no 3d occupancy) with 0.43 Å for Fe(PF)(2-MeIm) (3d occupied). For six-coordinate complexes where the two axial ligands, L and L , are different, the M • • • porph displacement usually reflects relative influences. Generally, displacement of the metal from the plane of the porphyrinatonitrogen atoms is within 0.04 Å of the displacement from the 24-atom mean plane of the entire porphyrin skeleton. On occasions this second displacement may be much larger, for example in Fe(TPP)(2-MeIm), where it is 0.15 Å larger than it is for Fe(PF)(2-MeIm). This effect is called , and it is usually attributed to crystal packing forces. Interaction of the porphyrin with protein side chains leads to considerable doming or folding of the heme in vertebrate hemoglobins. The metal-axial ligand separations, M-L (when more than one, L denotes the heterocyclic axial base), are dependent on the nature of the ligand, L. When L and L are different, the M-L separations are sensitive to the relative influences of L and L as well as to steric factors. For example, for Fe(TPP)(1-Melm) , the Fe—N bond length is 2.016(5) Å, whereas for Fe(TPP)(1-Melm)(NO) it is 2.180(4) Å. For sterically active ligands, such as 2-methylimidazole compared to 1-methylimidazole (4.34), the longer Co—N bond occurs for the 2-Melm ligand because of steric clash between the 2-methyl group and the porphyrin. It is possible that combinations of intrinsic bonding and steric factors may give rise to a double minimum and two accessible axial ligand conformations (Figure 4.26). This situation seems to occur in the solid state for Fe(PF)(2-Melm)(O )•EtOH, where a short Fe—N and a long Fe—O bond are observed both from the structure revealed by single-crystal x-ray diffraction methods and by EXAFS data. On the other hand, for solvate-free Fe(PF)(2-MeIm)(O ) and for Fe(PF)(1,2-Me Im)(O ), the EXAFS patterns are interpreted in terms of a short Fe—O and long Fe—Im bond. This parameter is the minimum angle that the plane of the axial base (e.g., pyridine, substituted imidazole, etc.) makes with a plane defined by the N , M, and L atoms (Figure 4.25). If there are two axial ligands, e.g., 1-methylimidazole and O , then, as before, the angle the axial base makes is denoted \(\phi_{1}\) and the other angle \(\phi_{2}\). For a linear CO ligand bound perpendicularly to the porphyrin plane, \(\phi_{2}\) is undefined. Note that the orientation of the second ligand is influenced by distal effects. When \(\phi\) = 0, the axial base eclipses a pair of M-N bonds; contacts with the porphyrin are maximized. When \(\phi\) = 45°, contacts are minimized. Unless the axial base has a 2-substituent, however, the contacts are not excessively close for any value of \(\phi\). With a 2-methyl substituent, the contacts are sufficiently severe that the M-N vector is no longer perpendicular to the porphyrin plane, and the imidazole group is rotated so that the M-N vector no longer approximately bisects the imidazole C—N—C bond angle, as illustrated Figure 4.25. Distal effects arise from noncovalent interactions of the coordinated dioxygen, carbon monoxide, or other ligand with its surroundings. The protein matrix, the pickets, and the caps are functionally equivalent to an anisotropic solvent matrix that contains a variety of solutes. The limits of this simplification are illustrated in the following example. The electronically similar cobalt meso-, deutero-, and protoporphyrin IX complexes bind dioxygen with similar affinities under identical solvent conditions. When they are embedded in globin, larger differences in affinity and changes in cooperativity are observed. These effects are attributed to the slightly different nestling of the porphyrin molecules in the cleft in hemoglobin or, in the generalization introduced, to slightly different solvation effects. Interaction of the coordinated O or CO molecule with solvent molecules or with the protein has a profound influence on kinetics and thermodynamics (see Figure 4.24, and Tables 4.2 and 4.5). As discussed earlier, there is accumulation of negative charge on the dioxygen ligand. The possibility then arises for stabilization of coordination through hydrogen bonding or dipolar interactions with solute molecules, porphyrin substituents (such as amide groups in the picket-fence porphyrins and some species of strapped porphyrins ), or with protein residues* (such as histidine). Destabilization of coordinated ligands and lowered affinity can result if the coordinated ligand is unable, through steric clash, to achieve its optimum stereochemistry or if the closest neighboring groups are electronegative, as are the ether and ester linkages on capped porphyrins. We will describe in detail in the next subsection (III.C) the fascinating variety of means by which ligand binding is modulated by distal amino-acid residues. * For CoMbO no change in EPR parameters occurs on substituting D O for H O. No hydrogen bond between O and a distal group comparable in strength to that in whale CoMbO was inferred. Dissimilar systems may show similar affinities for a ligand as a result of a different mix of the proximal and distal effects enumerated above. These effects are not all of equal magnitude, and an attempt is made here to show the increment in free energy that occurs if the effect is manifest in the deoxy or liganded state of Figure 4.3. Increasing the free energy of the deoxy state while holding that of the liganded state constant leads to an increase in affinity. The reference state is gaseous Fe(TPP)(1-MeIm). The magnitude and sign of these effects are shown in Figure 4.27. For the coordination of alkylisocyanide molecules to hemoglobin, the steric effects of different alkyl groups have been quantified. Lowered affinity occurs with increasing alkyl chain length, with the exception of methyl isocyanide.
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The molar Gibbs free energy of mixing (∆Gmix) for a binary solution is given by: \[\begin{align} ∆G_{mix} &= RT (\chi_1 \ln a_1 + \chi_2 \ln a_2) \\[5pt] &= RT (\chi_1 \ln \gamma_1 \chi_1 + \chi_2 \ln \gamma_2 \chi_2) \\[5pt] &= RT (\chi_1 \ln \gamma_1 + \chi_2 \ln \gamma_2) + RT (\chi_1 \ln \chi_1 + \chi_2 \ln \chi_2) \label{1} \end{align}\] where \(R\) is the ideal gas constant, \(T\) the absolute temperature, \(\chi_i\) the molar fraction of the \(i^{th}\) components, \(a_i\) its activity, and \(\gamma_i\) its activity coefficient. The ideal molar Gibbs free energy of mixing (\(∆G_{mix}^{ideal}\)) and the excess molar Gibbs free energy of mixing (\(∆G_{mix}^{excess}\)) are then defined as: \[∆G_{mix}^{ideal} = RT (\chi_1 \ln \chi_1 + \chi_2 \ln \chi_2) \label{2}\] \[∆G_{mix}^{excess} = RT (\chi_1 \ln \gamma_1 + \chi_2 \ln \gamma_2) \label{3}\] The regular solution model assumes that the molar entropy of mixing \(∆S_{mix}\) corresponds to an ideal solution, assumption that is reasonable for small molecules of similar size: \[∆S_{mix} = - R (\chi_1 \ln \chi_1 + \chi_2 \ln \chi_2). \label{4}\] Since: \[∆G_{mix} = ∆H_{mix} - T ∆S_{mix} \label{5}\] then, the molar heat of mixing \(∆H_{mix}\) for a real solution is given by: \[∆H_{mix} = RT (\chi_1 \ln \gamma_1 + \chi_2 \ln \gamma_2) = ∆G_{mix}^{excess} \label{6}\] The composition with the lowest Gibbs free energy of mixing (Equation \ref{1}) corresponds to the azeotrope. The mole fractions (\(\chi_i\)) can be calculated from the composition of the binary solution. The activity coefficients (\(\gamma_i\)) can be derived from an analysis of the composition of the distillate of the binary solution, using Raoult's and 's laws. Considering the vapor pressure of the ith component above the binary solution (\(P_i\)), Raoult's law convention states that: \[P_i = a_i P^o_i = \gamma_i \chi_i P^o_i. \label{7}\] where \(P^0_i\) is the vapor pressure of the \(i^{th}\) pure component. If the vapor phase above the solution can be assumed to be ideal, 's law states that: \[P_i = y_i P_{total} \label{8}\] where \(yi\) is the mole fraction for the ith component in the vapor phase and \(P_{total}\) is the total pressure. If \(P_{total}\), \(P^o_i\), \(x_i\) and \(y_i\) are known, then the activity coefficients in the liquid phase (gi) can be calculated by equating [7] and [8]: \[y_i P_{total} = \gamma_i \chi_i P^o_i \label{9}\] A series of binary solutions of cyclohexane and ethanol were prepared. The mole fractions of the solutions (\(\chi_i\)) can be calculated from the composition of the binary solutions, knowing the density of the components and their molar mass. The solutions were analyzed with a gas chromatograph (GC) and the peak heights were recorded (hi) for the two components. The peak height functions (Hi) may be calculated, as follows: \[ H_i = \dfrac{h_1}{h_i + h_j} \label{10}\] The solutions were boiled, the distillates were collected and analyzed using the GC under the same conditions, and the peak heights were recorded (\(h_i\)) for the two components. The peak height functions for the distillate (H'i) may be calculated following Equation \ref{10}. In order to calculate the molar fraction in the gas phase (\(y_i\)), standard curves are constructed for the two components of the binary solutions, using mole fractions (\(\chi_i\)) and peak height functions (Hi) from the liquid phase. The standard curves consist of polynomial fits of the mole fractions (xi) as a function of peak height functions (Hi). The mole fractions of the distillates (yi) are determined using the standard curves, when the peak height functions of the distillates (H'i) are used as the independent variables. The barometric pressure and the room temperature were measured. The reported value of barometric pressure includes a temperature correction. Since the vapors from the solution displace the air in the apparatus, \(P_{total}\) in Equation \ref{9} is equal to the barometric pressure and to the sum of the vapor pressures of the two components. The vapor pressure of the pure components (\(P^o_i\_) at the boiling temperature of the binary solutions are estimated using and available parameters \(A\) and \(B\) for cyclohexane and ethanol: \[ \ln P = A +\dfrac{B}{T} \label{11}\] The molar Gibbs free energy of mixing (\(∆G_{mix}\)) for the various solutions is calculated from Equation \ref{1}. Other thermodynamic terms (\(∆G_{mix}^{ideal}\), \(∆G_{mix}^{excess}\), \(∆S_{mix}\), \(∆H_{mix}\)) are also obtained. All values are calculated with their associated uncertainty, based on the uncertainty in the data. Composition of the solutions*7: GC peak heights for the solutions (cm): GC peak heights for the distillates (cm): (Note: See Sample Calculations below) 1. Calculate the mole fractions for the solutions (xi), and their uncertainties. Remember that the error of sums or subtractions is the sum of the errors of the individual terms and that the relative error of products or divisions is the sum of the relative errors of the individual terms. 2. Calculate the peak height functions for the solutions (Hi), as well as for the distillates (H'i). 3. Obtain third-order polynomial fits of xi vs Hi for cyclohexane and for ethanol. 4. Using the third-order polynomial functions, calculate the mole fractions in the distillates (ycy and yet) as a function of the peak height function of the distillates (H'cy and H'et). Assume that the associated uncertainty for the y values is the standard error of the y estimates. 5. Using the corresponding Antoine equation, calculate the vapor pressure of pure cyclohexane and ethanol (P0i) at the boiling temperatures. Calculate the uncertainty for the vapor pressures of the pure compounds, at the boiling temperatures, using the formulas: 6. Estimate the activity coefficients for the components (γi), and estimate their associated errors. 7. Calculate the molar Gibbs free energies of mixing (\(∆G_{mix}\)) at the boiling temperature, with their associated errors. Remember that \[ \Delta \ln x =dfrac{\Delta x}{x}\]. 8. Calculate ∆Hmix (or ∆Gmixexcess) and ∆Smix (or -∆Gmixideal/T) and estimate their associated errors. 9. Plot ∆Gmix vs xi. Fit the data into a third-order polynomial function. Calculate the azeotrope composition by finding the value of xi at the minimum of the ∆Gmix curve. 10. Estimate, by interpolation, the boiling temperature for the azeotrope. Sample Calculations: 1. Mole fraction for cyclohexane (xcy) in solution 2: where ni is the number of moles, ri is the density, Vi is the volume, and Mi the molar mass of the ithcomponent. Associated uncertainty for xcy in solution 2 (∆xcy): 2. Peak height functions for cyclohexane in solution 2 (Hcy and H'cy): 3. Standard curve of xi vs Hi for cyclohexane: x = a + b H + c H2 + d H3. Solution # xcy Hcy Hcy2 Hcy3 1 0.000 0.000 0.000 0.000 2 7.670E-02 1.023E-01 1.046E-02 1.070E-03 3 4 5 6 7 8 Using LINEST(known_y's,[known_x's],true,true) {Ctrl-Shift,Enter} d c b a ± Dd ± Dc ± Db ± Da 0.9996 0.0095 xcy = a + b Hcy + c Hcy2 +d Hcy3 Coefficient of determination = 0.9996 Standard error of the y estimates: 0.0095. 4. Mole fraction for cyclohexane in the distillate of solution 2: 5. Vapor pressure for pure cyclohexane in solution 2: 6. Activity coefficient for cyclohexane in solution 2: 7. Molar Gibbs free energy of mixing (∆Gmix) for solution 2; Associated uncertainty in solution 2 [∆(∆Gmix)]: 8. Other thermodynamic properties: 9. Location of the azeotrope:
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We have identified three criteria for whether a given reaction will occur spontaneously: Recall that if \(Q < K\), then the reaction proceeds spontaneously to the right as written, resulting in the net conversion of reactants to products. Conversely, if \(Q > K\), then the reaction proceeds spontaneously to the left as written, resulting in the net conversion of products to reactants. If \(Q = K\), then the system is at equilibrium, and no net reaction occurs. Table \(\Page {1}\) summarizes these criteria and their relative values for spontaneous, nonspontaneous, and equilibrium processes. Because all three criteria assess the same thing—the spontaneity of the process—it would be most surprising indeed if they were not related. In this section, we explore the relationship between the standard free energy of reaction (\(ΔG°\)) and the equilibrium constant (\(K\)). Because ΔH° and ΔS° determine the magnitude of ΔG° and because K is a measure of the ratio of the concentrations of products to the concentrations of reactants, we should be able to express K in terms of ΔG° and vice versa. "Free Energy", ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature, thereby eliminating \(ΔH\) from the equation for \(ΔG\). The general relationship can be shown as follow (derivation not shown): \[ \Delta G = V \Delta P − S \Delta T \label{18.29}\] If a reaction is carried out at constant temperature (\(ΔT = 0\)), then Equation \(\ref{18.29}\) simplifies to \[\Delta{G} = V\Delta{P} \label{18.30}\] Under normal conditions, the pressure dependence of free energy is not important for solids and liquids because of their small molar volumes. For reactions that involve gases, however, the effect of pressure on free energy is very important. Assuming ideal gas behavior, we can replace the \(V\) in Equation \(\ref{18.30}\) by nRT/P (where n is the number of moles of gas and R is the ideal gas constant) and express \(\Delta{G}\) in terms of the initial and final pressures (\(P_i\) and \(P_f\), respectively): \[\begin{align} \Delta G &=\left(\dfrac{nRT}{P}\right)\Delta P \\[4pt] &=nRT\dfrac{\Delta P}{P}=nRT\ln\left(\dfrac{P_\textrm f}{P_\textrm i}\right) \label{18.31} \end{align} \] If the initial state is the standard state with P = 1 atm, then the change in free energy of a substance when going from the standard state to any other state with a pressure P can be written as follows: \[G − G^° = nRT\ln{P}\] This can be rearranged as follows: \[G = G^° + nRT\ln {P} \label{18.32}\] As you will soon discover, Equation \(\ref{18.32}\) allows us to relate \(ΔG^o\) and \(K_p\). Any relationship that is true for \(K_p\) must also be true for \(K\) because \(K_p\) and \(K\) are simply different ways of expressing the equilibrium constant using different units. Let’s consider the following hypothetical reaction, in which all the reactants and the products are ideal gases and the lowercase letters correspond to the stoichiometric coefficients for the various species: \[aA+bB \rightleftharpoons cC+dD \label{18.33}\] Because the free-energy change for a reaction is the difference between the sum of the free energies of the products and the reactants, we can write the following expression for \(ΔG\): \[\begin{align} \Delta{G}&=\sum_m G_{products}−\sum_n G_{reactants} \\[4pt] &=(cG_C+dG_D)−(aG_A+bG_B) \label{18.34} \end{align} \] Substituting Equation \(\ref{18.32}\) for each term into Equation \(\ref{18.34}\), \[ΔG=[(cG^o_C+cRT \ln P_C)+(dG^o_D+dRT\ln P_D)]−[(aG^o_A+aRT\ln P_A)+(bG^o_B+bRT\ln P_B)]\] Combining terms gives the following relationship between \(ΔG\) and the reaction quotient \(Q\): \[\begin{align} \Delta G &=\Delta G^\circ+RT \ln\left(\dfrac{P^c_\textrm CP^d_\textrm D}{P^a_\textrm AP^b_\textrm B}\right) \\[4pt] &=\Delta G^\circ+RT\ln Q \label{18.35} \end{align}\] where \(ΔG°\) indicates that all reactants and products are in their standard states. For gases at equilibrium (\(Q = K_p\),), and as you’ve learned in this chapter, \(ΔG = 0\) for a system at equilibrium. Therefore, we can describe the relationship between \(ΔG^o\) and \(K_p\) for gases as follows: \[ \begin{align} 0 &= ΔG° + RT\ln K_p \label{18.36a} \\[4pt] ΔG° &= −RT\ln K_p \label{18.36b} \end{align} \] If the products and reactants are in their standard states and ΔG° < 0, then K > 1, and products are favored over reactants. Conversely, if ΔG° > 0, then K < 1, and reactants are favored over products. If ΔG° = 0, then \(K_p = 1\), and neither reactants nor products are favored: the system is at equilibrium. For a spontaneous process under standard conditions, \(K_{eq}\) and \(K_p\) are greater than 1. To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of ), equilibrium is established when the system’s free energy is minimized (Figure \(\Page {3}\)). If a system is present with reactants and products present in nonequilibrium amounts ( ≠ ), the reaction will proceed spontaneously in the direction necessary to establish equilibrium. \(ΔG^o\) is −32.7 kJ/mol of N for the reaction \[\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)}\nonumber\] This calculation was for the reaction under standard conditions—that is, with all gases present at a partial pressure of 1 atm and a temperature of 25°C. Calculate \(ΔG\) for the same reaction under the following nonstandard conditions: Does the reaction favor products or reactants? : balanced chemical equation, partial pressure of each species, temperature, and ΔG° : whether products or reactants are favored : : A The relationship between ΔG° and ΔG under nonstandard conditions is given in Equation \(\ref{18.35}\). Substituting the partial pressures given, we can calculate \(Q\): \[Q=\dfrac{P^2_{\textrm{NH}_3}}{P_{\textrm N_2}P^3_{\textrm H_2}}=\dfrac{(0.021)^2}{(2.00)(7.00)^3}=6.4\times10^{-7}\nonumber\] B Substituting the values of ΔG° and Q into Equation \(\ref{18.35}\), \[ \begin{align*} \Delta G &=\Delta G^\circ+RT\ln Q \\[4pt] &=-32.7\textrm{ kJ}+\left[(\textrm{8.314 J/K})(\textrm{373 K})\left(\dfrac{\textrm{1 kJ}}{\textrm{1000 J}}\right)\ln(6.4\times10^{-7})\right] \\[4pt] &=-32.7\textrm{ kJ}+(-44\textrm{ kJ}) \\[4pt] &=-77\textrm{ kJ/mol of N}_2 \end{align*}\] Because ΔG < 0 and Q < 1.0, the reaction is spontaneous to the right as written, so products are favored over reactants. Calculate \(ΔG\) for the reaction of nitric oxide with oxygen to give nitrogen dioxide under these conditions: T = 50°C, P = 0.0100 atm, \(P_{\mathrm{O_2}}\) = 0.200 atm, and \(P_{\mathrm{NO_2}}\) = 1.00 × 10 atm. The value of ΔG° for this reaction is −72.5 kJ/mol of O . Are products or reactants favored? −92.9 kJ/mol of O ; the reaction is spontaneous to the right as written, so products are favored. Calculate K for the reaction of H with N to give NH at 25°C. \(ΔG^o\) for this reaction is −32.7 kJ/mol of N . : balanced chemical equation from Example \(\Page {2}\), ΔG°, and temperature : K Substitute values for ΔG° and T (in kelvin) into Equation \(\ref{18.36b}\) to calculate \(K_p\), the equilibrium constant for the formation of ammonia. In Example 10, we used tabulated values of ΔG to calculate ΔG° for this reaction (−32.7 kJ/mol of N ). For equilibrium conditions, rearranging Equation \(\ref{18.36b}\), \[\begin{align*} \Delta G^\circ &=-RT\ln K_\textrm p \\[4pt] \dfrac{-\Delta G^\circ}{RT} &=\ln K_\textrm p \end{align*}\] Inserting the value of ΔG° and the temperature (25°C = 298 K) into this equation, \[\begin{align*}\ln K_\textrm p &=-\dfrac{(-\textrm{32.7 kJ})(\textrm{1000 J/kJ})}{(\textrm{8.314 J/K})(\textrm{298 K})}=13.2 \\[4pt] K_\textrm p &=5.4\times10^5\end{align*}\] Thus the equilibrium constant for the formation of ammonia at room temperature is favorable. However, the rate at which the reaction occurs at room temperature is too slow to be useful. Calculate K for the reaction of NO with O to give NO at 25°C. ΔG° for this reaction is −70.5 kJ/mol of O . 2.2 × 10 Although \(K_p\) is defined in terms of the partial pressures of the reactants and the products, the equilibrium constant \(K\) is defined in terms of the concentrations of the reactants and the products. We described the relationship between the numerical magnitude of \(K_ and \(K\) previously and showed that they are related: \[K_p = K(RT)^{Δn} \label{18.37}\] where \(Δn\) is the number of moles of gaseous product minus the number of moles of gaseous reactant. For reactions that involve only solutions, liquids, and solids, \(Δn = 0\), so \(K_p = K\). For all reactions that do not involve a change in the number of moles of gas present, the relationship in Equation \(\ref{18.36b}\) can be written in a more general form: \[ΔG° = −RT \ln K \label{18.38}\] Only when a reaction results in a net production or consumption of gases is it necessary to correct Equation \(\ref{18.38}\) for the difference between \(K_p\) and \(K\). Although we typically use concentrations or pressures in our equilibrium calculations, recall that equilibrium constants are generally expressed as unitless numbers because of the use of or in precise thermodynamic work. Systems that contain gases at high pressures or concentrated solutions that deviate substantially from ideal behavior require the use of fugacities or activities, respectively. Combining Equations \(\ref{18.38}\) with \(ΔG^o = ΔH^o − TΔS^o\) provides insight into how the components of ΔG° influence the magnitude of the equilibrium constant: \[ΔG° = ΔH° − TΔS° = −RT \ln K \label{18.39}\] Notice that \(K\) becomes larger as ΔS° becomes more positive, indicating that the magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder. Moreover, K increases as ΔH° decreases. Thus the magnitude of the equilibrium constant is also directly influenced by the tendency of a system to seek the lowest energy state possible. The magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder and seek the lowest energy state possible. The fact that ΔG° and K are related provides us with another explanation of why equilibrium constants are temperature dependent. This relationship is shown explicitly in Equation \(\ref{18.39}\), which can be rearranged as follows: \[\ln K=-\dfrac{\Delta H^\circ}{RT}+\dfrac{\Delta S^\circ}{R} \label{18.40}\] Assuming ΔH° and ΔS° are temperature independent, for an exothermic reaction (ΔH° < 0), the magnitude of K decreases with increasing temperature, whereas for an endothermic reaction (ΔH° > 0), the magnitude of K increases with increasing temperature. The quantitative relationship expressed in Equation \(\ref{18.40}\) agrees with the qualitative predictions made by applying Le Chatelier’s principle. Because heat is produced in an exothermic reaction, adding heat (by increasing the temperature) will shift the equilibrium to the left, favoring the reactants and decreasing the magnitude of K. Conversely, because heat is consumed in an endothermic reaction, adding heat will shift the equilibrium to the right, favoring the products and increasing the magnitude of K. Equation \(\ref{18.40}\) also shows that the magnitude of ΔH° dictates how rapidly K changes as a function of temperature. In contrast, the magnitude and sign of ΔS° affect the magnitude of K but not its temperature dependence. If we know the value of K at a given temperature and the value of ΔH° for a reaction, we can estimate the value of K at any other temperature, even in the absence of information on ΔS°. Suppose, for example, that K and K are the equilibrium constants for a reaction at temperatures T and T , respectively. Applying Equation \(\ref{18.40}\) gives the following relationship at each temperature: \[\begin{align}\ln K_1&=\dfrac{-\Delta H^\circ}{RT_1}+\dfrac{\Delta S^\circ}{R} \\[4pt] \ln K_2 &=\dfrac{-\Delta H^\circ}{RT_2}+\dfrac{\Delta S^\circ}{R}\end{align}\] Subtracting \(\ln K_1\) from \(\ln K_2\), \[\ln K_2-\ln K_1=\ln\dfrac{K_2}{K_1}=\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \label{18.41}\] Thus calculating ΔH° from tabulated enthalpies of formation and measuring the equilibrium constant at one temperature (K ) allow us to calculate the value of the equilibrium constant at any other temperature (K ), assuming that ΔH° and ΔS° are independent of temperature. Equation \(\ref{18.41}\) is often referred to as the after Dutch chemist Jacobus Henricus van 't Hoff in 1884 in his book Études de Dynamique chimique (Studies in dynamic chemistry). The equilibrium constant for the formation of NH from H and N at 25°C was calculated to be K = 5.4 × 10 in Example \(\Page {2}\). What is K at 500°C? (Use the data from Example \(\Page {1}\)) : balanced chemical equation, ΔH°, initial and final T, and K at 25°C K at 500°C : Convert the initial and final temperatures to kelvin. Then substitute appropriate values into Equation \(\ref{18.41}\) to obtain \(K_2\), the equilibrium constant at the final temperature. : The value of ΔH° for the reaction obtained using Hess’s law is −91.8 kJ/mol of N . If we set \(T_1\) = 25°C = 298 K and \(T_2\) = 500°C = 773 K, then from Equation \(\ref{18.41}\) we obtain the following: \[\begin{align*}\ln\dfrac{K_2}{K_1}&=\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \\[4pt] &=\dfrac{(-\textrm{91.8 kJ})(\textrm{1000 J/kJ})}{\textrm{8.314 J/K}}\left(\dfrac{1}{\textrm{298 K}}-\dfrac{1}{\textrm{773 K}}\right)=-22.8 \\[4pt] \dfrac{K_2}{K_1}&=1.3\times10^{-10} \\[4pt] K_2&=(5.4\times10^5)(1.3\times10^{-10})=7.0\times10^{-5}\end{align*}\] Thus at 500°C, the equilibrium strongly favors the reactants over the products. The equilibrium constant for the reaction of \(\ce{NO}\) with \(\ce{O2}\) to give \(\ce{NO2}\) at 25°C is \(K_p = 2.2 \times 10{12}\). Use the \(ΔH^o_f\) values in the exercise in Example \(\Page {4}\) to calculate \(K_p\) for this reaction at 1000°C. 5.6 × 10 For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature. If we assume ideal gas behavior, the ideal gas law allows us to express \(ΔG\) in terms of the partial pressures of the reactants and products, which gives us a relationship between ΔG and K , the equilibrium constant of a reaction involving gases, or K, the equilibrium constant expressed in terms of concentrations. If ΔG° < 0, then K or K > 1, and products are favored over reactants. If ΔG° > 0, then K or K < 1, and reactants are favored over products. If ΔG° = 0, then K or K = 1, and the system is at equilibrium. We can use the measured equilibrium constant K at one temperature and ΔH° to estimate the equilibrium constant for a reaction at any other temperature. ( )
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The laws of probability apply to events that are independent. If the result of one trial depends on the result of another trial, we may still be able to use the laws of probability. However, to do so, we must know the nature of the interdependence. If the activity associated with event precedes the activity associated with event , the probability of may depend on whether occurs. Suppose that the first activity is tossing a coin and that the second activity is drawing a card from a deck; however, the deck we use depends on whether the coin comes up heads or tails. If the coin is heads, we draw a card from an ordinary deck; if the coin is tails, we draw a coin from a deck with the face cards removed. Now we ask about the probability of drawing an ace. If the coin is heads, the probability of drawing an ace is \({4}/{52}={1}/{13}\). If the coin is tails, the probability of drawing an ace is \({4}/{40}={1}/{10}\). The combination coin is heads and card is ace has probability: \(\left({1}/{2}\right)\left({1}/{13}\right)={1}/{26}\). The combination coin is tails and card is ace has probability \(\left({1}/{2}\right)\left({1}/{10}\right)={1}/{20}\). In this case, the probability of drawing an ace depends on the modification we make to the deck based on the outcome of the coin toss. Applying the laws of probability is straightforward. An example that illustrates the application of these laws in a transparent way is provided by villages First, Second, Third, and Fourth, which are separated by rivers. (See Figure 1.) Bridges \(1\), \(2\), and \(3\) span the river between First and Second. Bridges \(a\) and \(b\) span the river between Second and Third. Bridges \(A\), \(B\), \(C\), and \(D\) span the river between Third and Fourth. A traveler from First to Fourth who is free to take any route he pleases has a choice from among \(3\times 2\times 4=24\) possible combinations. Let us consider the probabilities associated with various events: The outcomes of rolling dice, rolling provide more illustrations. If we roll two dice, we can classify the possible outcomes according to the sums of the outcomes for the individual dice. There are thirty-six possible outcomes. They are displayed in Table 1. Table 1: Outcomes from tossing two dice Let us consider the probabilities associated with various dice-throwing events: Above we looked at the number of outcomes associated with a score of \(3\) to find that the probability of this event is \({1}/{18}\). We can use another argument to get this result. The probability that two dice roll a score of three is equal to the probability that the first die shows \(1\) or \(2\) times the probability that the second die shows whatever score is necessary to make the total equal to three. This is: \[\begin{align*} P\left(first\ die\ shows\ 1\ or\ 2\right)\times \left({1}/{6}\right) &= \left[\left({1}/{6}\right)+\left({1}/{6}\right)\right]\times {1}/{6} \\[4pt] &={2}/{36} \\[4pt]& ={1}/{18} \end{align*}\] Application of the laws of probability is frequently made easier by recognizing a simple restatement of the requirement that events be mutually exclusive. In a given trial, either an event occurs or it does not. Let the probability that an event occurs be \(P\left(A\right)\). Let the probability that event does not occur be \(P\left(\sim A\right)\). Since in any given trial, the outcome must belong either to event or to event \(\sim A\), we have \[P\left(A\right)+P\left(\sim A\right)=1\] For example, if the probability of success in a single trial is \({2}/{3}\), the probability of failure is \({1}/{3}\). If we consider the outcomes of two successive trials, we can group them into four events. Using the laws of probability, we have \[ \begin{align*} 1 &=P\left(Event\ SS\right)+P\left(Event\ SF\right)+P\left(Event\ FS\right)+\ P(Event\ FF) \\[4pt] &=P_1\left(S\right)\times P_2\left(S\right)+P_1\left(S\right)\times P_2\left(F\right) +P_1(F)\times P_2(S)+P_1(F)\times P_2(F) \end{align*}\] where \(P_1\left(X\right)\) and \(P_2\left(X\right)\) are the probability of event \(X\) in the first and second trials, respectively. This situation can be mapped onto a simple diagram. We represent the possible outcomes of the first trial by line segments on one side of a unit square \(P_1\left(S\right)+P_1\left(F\right)=1\). We represent the outcomes of the second trial by line segments along an adjoining side of the unit square. The four possible events are now represented by the areas of four mutually exclusive and exhaustive portions of the unit square as shown in Figure 2.
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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$)}}\)  \( \newcommand{\dil}{\tx{(dil)}}\)  \( \newcommand{\sln}{\tx{(sln)}}\)  \( \newcommand{\mix}{\tx{(mix)}}\)  \( \newcommand{\rxn}{\tx{(rxn)}}\)  \( \newcommand{\expt}{\tx{(expt)}}\)  \( \newcommand{\solid}{\tx{(s)}}\)  \( \newcommand{\liquid}{\tx{(l)}}\)  \( \newcommand{\gas}{\tx{(g)}}\)  \( \newcommand{\pha}{\alpha}        % phase alpha\)  \( \newcommand{\phb}{\beta}         % phase beta\)  \( \newcommand{\phg}{\gamma}        % phase gamma\)  \( \newcommand{\aph}{^{\alpha}}     % alpha phase superscript\)  \( \newcommand{\bph}{^{\beta}}      % beta phase superscript\)  \( \newcommand{\gph}{^{\gamma}}     % gamma phase superscript\)  \( \newcommand{\aphp}{^{\alpha'}}   % alpha prime phase superscript\)  \( \newcommand{\bphp}{^{\beta'}}    % beta prime phase superscript\)  \( \newcommand{\gphp}{^{\gamma'}}   % gamma prime phase superscript\)  \( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\)  \( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\)  \( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\)  \( \newcommand{\dif}{\mathop{}\!\mathrm{d}}   % roman d in math mode, preceded by space\)  \( \newcommand{\Dif}{\mathop{}\!\mathrm{D}}   % roman D in math mode, preceded by space\)  \( \newcommand{\df}{\dif\hspace{0.05em} f} % df\)  \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)  \( \newcommand{\dq}{\dBar q} % heat differential\)  \( \newcommand{\dw}{\dBar w} % work differential\)  \( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)  \( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\)  \( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\)  \( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\)  \( \newcommand{\Del}{\Delta}\)  \( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\)  \( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\)  \( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\)  \( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\)  \( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\)  \( \newcommand{\dotprod}{\small\bullet}\)  \( \newcommand{\fug}{f} % fugacity\)  \( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)  \( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)  \( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)  \( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\)  \( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\)  \( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\) \( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\)  \( \newcommand{\D}{\displaystyle} % for a line in built-up\)  \( \newcommand{\s}{\smash[b]} % use in equations with conditions of validity\)  \( \newcommand{\cond}[1]{\\[-2.5pt]{}\tag*{#1}}\)  \( \newcommand{\nextcond}[1]{\\[-5pt]{}\tag*{#1}}\)  \( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}}     % gas constant value\)  \( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \) \( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \) \( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \)  Section 9.6.3 explained how we can evaluate the activity coefficient \(\g\mbB\) of a nonelectrolyte solute of a binary solution if we know the variation of the osmotic coefficient of the solution from infinite dilution to the molality of interest. A similar procedure for the mean ionic activity coefficient of an electrolyte solute was described in Sec. 10.6. The physical measurements needed to find the osmotic coefficient \(\phi_m\) of a binary solution must be directed to the calculation of the quantity \(\mu\A^*-\mu\A\), the difference between the chemical potentials of the pure solvent and the solvent in the solution at the temperature and pressure of interest. This difference is positive, because the presence of the solute reduces the solvent’s chemical potential. To calculate \(\phi_m\) from \(\mu\A^*-\mu\A\), we use Eq. 9.6.16 for a nonelectrolyte solute, or Eq. 10.6.1 for an electrolyte solute. Both equations are represented by \begin{equation} \phi_m=\frac{\mu\A^*-\mu\A}{RTM\A\nu m\B} \tag{12.2.1} \end{equation} where \(\nu\) for a nonelectrolyte is \(1\) and for an electrolyte is the number of ions per formula unit. The sequence of steps, then, is (1) the determination of \(\mu\A^*-\mu\A\) over a range of molality at constant \(T\) and \(p\), (2) the conversion of these values to \(\phi_m\) using Eq. 12.2.1, and (3) the evaluation of the solute activity coefficient by a suitable integration from infinite dilution to the molality of interest. A measurement of \(\mu\A^*-\mu\A\) also gives us the activity coefficient, based on the pure-solvent reference state, through the relation \(\mu\A=\mu\A^*+RT\ln(\g\A x\A)\) (Eq. 9.5.15). Sections 12.2.1 and 12.2.2 will describe freezing-point and osmotic-pressure measurements, two much-used methods for evaluating \(\mu\A^*-\mu\A\) in a binary solution at a given \(T\) and \(p\). The isopiestic vapor-pressure method was described in Sec. 9.6.4. The freezing-point and isopiestic vapor-pressure methods are often used for electrolyte solutions, and osmotic pressure is especially useful for solutions of macromolecules. This section explains how we can evaluate \(\mu\A^*-\mu\A\) for a solution of a given composition at a given \(T\) and \(p\) from the freezing point of the solution combined with additional data obtained from calorimetric measurements. Consider a binary solution of solvent A and solute B. We assume that when this solution is cooled at constant pressure and composition, the solid that first appears is pure A. For example, for a dilute aqueous solution the solid would be ice. The temperature at which solid A first appears is \(T\f\), the freezing point of the solution. This temperature is lower than the freezing point \(T\f^*\) of the pure solvent, a consequence of the lowering of \(\mu\A\) by the presence of the solute. Both \(T\f\) and \(T\f^*\) can be measured experimentally. Let \(T'\) be a temperature of interest that is equal to or greater than \(T\f^*\). We wish to determine the value of \(\mu\A^*(\tx{l},T')-\mu\A(\tx{sln},T')\), where \(\mu\A^*(\tx{l},T')\) refers to pure liquid solvent and \(\mu\A(\tx{sln},T')\) refers to the solution. A second method for evaluating \(\mu\A^*-\mu\A\) uses the solution property called . A simple apparatus to measure the osmotic pressure of a binary solution is shown schematically in Fig. 12.2. The system consists of two liquid phases separated by a semipermeable membrane. Phase \(\pha\) is pure solvent and phase \(\phb\) is a solution with the same solvent at the same temperature. The semipermeable membrane is permeable to the solvent and impermeable to the solute. The presence of the membrane makes this system different from the multiphase, multicomponent system of Sec. 9.2.7, used there to derive conditions for transfer equilibrium. By a modification of that procedure, we can derive the conditions of equilibrium for the present system. We take phase \(\phb\) as the reference phase because it includes both solvent and solute. In order to prevent expansion work in the isolated system, both pistons shown in the figure must be fixed in stationary positions. This keeps the volume of each phase constant: \(\dif V\aph=\dif V\bph=0\). Equation 9.2.41, expressing the total differential of the entropy in an isolated multiphase, multicomponent system, becomes \begin{equation} \dif S = \frac{T\bph-T\aph}{T\bph}\dif S\aph + \frac{\mu\A\bph-\mu\A\aph}{T\bph}\dif n\A\aph \tag{12.2.6} \end{equation} In an equilibrium state, the coefficients \((T\bph-T\aph)/T\bph\) and \((\mu\A\bph-\mu\A\aph)/T\bph\) must be zero. Therefore, in an equilibrium state the temperature is the same in both phases and the solvent has the same chemical potential in both phases. The presence of the membrane, however, allows the pressures of the two phases to be unequal in the equilibrium state. Suppose we start with both phases shown in Fig. 12.2 at the same temperature and pressure. Under these conditions, the value of \(\mu\A\) is less in the solution than in the pure liquid, and a spontaneous flow of solvent will occur through the membrane from the pure solvent to the solution. This phenomenon is called (Greek for ). If we move the right-hand piston down slightly in order to increase the pressure \(p''\) of the solution in phase \(\phb\), \(\mu\A\) increases in this phase. The of the solution, \(\varPi\), is defined as the additional pressure the solution must have, compared to the pressure \(p'\) of the pure solvent at the same temperature, to establish an equilibrium state with no flow of solvent in either direction through the membrane: \(p'' = p' + \varPi\). In practice, the membrane may not be completely impermeable to a solute. All that is required for the establishment of an equilibrium state with different pressures on either side of the membrane is that solvent transfer equilibrium be established on a short time scale compared to the period of observation, and that the amount of solute transferred during this period be negligible. The osmotic pressure \(\varPi\) is an intensive property of a solution whose value depends on the solution’s temperature, pressure, and composition. Strictly speaking, \(\varPi\) in an equilibrium state of the system shown in Fig. 12.2 refers to the osmotic pressure of the solution at pressure \(p'\), the pressure of the pure solvent. In other words, the osmotic pressure of a solution at temperature \(T\) and pressure \(p'\) is the additional pressure that would have to be exerted on the solution to establish transfer equilibrium with pure solvent that has temperature \(T\) and pressure \(p'\). A solution has the property called osmotic pressure regardless of whether this additional pressure is actually present, just as a solution has a freezing point even when its actual temperature is different from the freezing point. Because in an equilibrium state the solvent chemical potential must be the same on both sides of the semipermeable membrane, there is a relation between chemical potentials and osmotic pressure given by \begin{gather} \s{ \mu\A(p'') = \mu\A(p'+\varPi)=\mu\A^*(p') } \tag{12.2.7} \cond{(equilibrium state)} \end{gather} We can use this relation to derive an expression for \(\mu\A^*(p')-\mu\A(p')\) as a function of \(\varPi\). The dependence of \(\mu\A\) on pressure is given according to Eq. 9.2.49 by \begin{equation} \Pd{\mu\A}{p}{T,\allni} = V\A \tag{12.2.8} \end{equation} where \(V\A\) is the partial molar volume of the solvent in the solution. Rewriting this equation in the form \(\dif\mu\A=V\A\difp\) and integrating at constant temperature and composition from \(p'\) to \(p'+\varPi\), we obtain \begin{equation} \mu\A(p'+\varPi)-\mu\A(p')=\int_{p'}^{p'+\varPi}\!V\A\difp \tag{12.2.9} \end{equation} Substitution from Eq. 12.2.7 changes this to \begin{gather} \s {\mu\A^*(p')-\mu\A(p')=\int_{p'}^{p'+\varPi}\!V\A\difp } \tag{12.2.10} \cond{(constant \(T\))} \end{gather} which is the desired expression for \(\mu\A^*-\mu\A\) at a single temperature and pressure. To evaluate the integral, we need an experimental value of the osmotic pressure \(\varPi\) of the solution. If we assume \(V\A\) is constant in the pressure range from \(p'\) to \(p'+\varPi\), Eq. 12.2.10 becomes simply \begin{equation} \mu\A^*(p')-\mu\A(p') = V\A\varPi \tag{12.2.11} \end{equation}
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The 1-octanol-water partition coefficient (Kow) is the equilibrium ratio of the concentration of a material in 1-octanol to the concentration of that material in water in an immiscible mixture of 1-octanol and water. The published values of log10 of Kow, i.e. Log(Kow), range from about -4 for saccharides to +10 for phthalate esters. Differences in Log(Kow) of 1.0 indicate a ten-fold difference in the equilibrium ratio of concentrations. Log(Kow) provides a quantitative data set for introducing students to the principles of partitioning of organic materials between two immiscible solvents, which leads to an understanding of solubility, of the hydrophilic-hydrophobic nature of materials, and of the separation of chemicals by extraction. The Log(Kow) of organic compounds is a thermodynamic property which is widely used for modeling the biological activity and environmental fate of organic chemicals, including estimating soil-water partition coefficients, dissolved organic matter-water partition coefficients, lipid solubility, the formation of micelles, transport across membranes, environmental risk assessment (including bioconcentration, aqueous toxicity, and biodegradation), and as a valuable tool in drug and pesticide design. The data in Tables 1 and 2 below will be used to analyze the effect on Log(Kow) of increasing the carbon chain length by –CH2– . A commonly used rule-of-thumb is that Log(Kow) increases by 0.5 for each additional –CH2– in a homologous series. Table 1 contains the published Log(Kow) values for 196 organic compounds, primarily from the LOGKOW© database; available on-line at logkow.cisti.nrc.ca/logkow/index.jsp ; provided by : Sangster Research Laboratories, P.O. Box 49562, CSP du Musée, 5122 Cote des Neiges , Montréal, Québec, Canada , H3T 2A5. In Table 1, the Log(Kow) values are arranged with rows of increasing alkyl chain length from hydrogen through 1-decyl and with columns for the various functional (end) groups. Many entries in Table 1 are blank because no reliable published Log(Kow) data was found for these compounds, though these compounds are certainly known. Table 2 contains the difference in Log(Kow) values between 175 homologous pairs of compounds in Table 1. on the following link and save each data set on your computer prior to performing the statistical analyses in the exercise. Rearrange the data in Table 2 into a single column for data analysis. If you are using M.S. Excel, the “Data Analysis” functions are located in the “Tools” drop down menu. If these functions are not available, follow the instructions for the “Analysis ToolPak” add-in found in the HELP menu. The 95% Confidence Interval for both the overall population and the population mean depend on the standard deviation, but the Confidence Interval for the population mean also depends the number of data points. 95% Confidence Interval for the population = mean +/- (1.96)(standard deviation)
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The equilibria (relative stabilities) and equilibration (rate of interconversion) of the rotational conformations of ethane and butane were discussed in Section 5-2. If you review this material, it will be clear that forming a ring from a hydrocarbon chain will greatly reduce the number of possible staggered and eclipsed conformations. We will begin our discussion with cyclohexane because of its special importance, proceed to smaller rings, then give a brief exposition of the conformations of the larger rings. If the carbons of a cyclohexane ring were placed at the corners of a regular hexagon, all the \(\ce{C-C-C}\) bond angles would have to be \(120^\text{o}\). Because the expected normal \(\ce{C-C-C}\) bond angle should be near the tetrahedral value of \(109.5^\text{o}\), the suggested planar configuration of cyclohexane would have at each of the carbons, and would correspond to less stable cyclohexane molecules than those with more normal bond angles. The actual normal value for the \(\ce{C-C-C}\) bond angle of an open-chain \(\ce{-CH_2-CH_2-CH_2}-\) unit appears to be about \(112.5^\text{o}\), which is \(3^\text{o}\) greater than the tetrahedral value. From this we can conclude that the angle strain at each carbon of a planar cyclohexane would be \(\left( 120^\text{o} - 112.5^\text{o} \right) = 7.5^\text{o}\). Angle strain is not the whole story with regard to the instability of the planar form, because in addition to having \(\ce{C-C-C}\) bond angles different from their normal values, the planar structure also has its carbons and hydrogens in the unfavorable arrangement, as shown in Figure 12-2. If the carbon valence angles are kept near the tetrahedral value, you will find that you can construct ball-and-stick models of the cyclohexane six-carbon ring with two quite different conformations. These are known as the "chair" and "boat" conformations (Figure 12-3). It has not been possible to separate cyclohexane at room temperature into pure isomeric forms that correspond to these conformations, and actually the two forms appear to be rapidly interconverted. The chair conformation is considerably more stable and comprises more than \(99.9\%\) of the equilibrium mixture at room temperature.\(^1\) Why is the boar form less stable than the chair form, if both have normal \(\ce{C-C-C}\) bond angles? The answer is that the boat form has unfavorable nonbonded interactions between the hydrogen atoms around the ring. If we make all of the bond angles normal and orient the carbons to give the "extreme boat" conformation of Figure 12-4, a pair of 1,4 hydrogens (the so-called "flagpole" hydrogens) have to be very close together \(\left( 1.83 \: \text{Å} \right)\). Hydrogens this close together would be on the rising part of a repulsion potential energy curve, such as Figure 4-6, for hydrogen-hydrogen nonbonded interactions. This at an \(\ce{H-H}\) distance of \(1.83 \: \text{Å}\) corresponds to a repulsion energy of about \(3 \: \text{kcal mol}^{-1}\). There is still another factor that makes the extreme boat unfavorable; namely, that the eight hydrogens along the "sides" of the boat are eclipsed, which brings them substantially closer together than they would be in a staggered arrangement (about \(2.27 \: \text{Å}\) compared with \(2.50 \: \text{Å}\)). This is in striking contrast with the chair form (Figure 12-5), for which adjacent hydrogens are in staggered positions with respect to one another all around the ring. Therefore the chair form is expected to be more stable than the boat form because it has less repulsion between the hydrogens. You should make and inspect models such as those in Figure 12-3 to see the rather striking difference between the chair and boat conformations that is not obvious from the diagrams. You will find that the chair structure is quite rigid, and rotation does occur around the \(\ce{C-C}\) bonds with interconversion to the boat structure. In contrast, the boat form is quite flexible. Rotation about the \(\ce{C-C}\) bonds permits the ring to twist one way or the other from the extreme boat conformation to considerably more stable, equal-energy conformations, in which the flagpole hydrogens move farther apart and the eight hydrogens along the sides become largely but not completely staggered. These arrangements are called the (sometimes ) conformations (see Figure 12-6) and are believed to be about \(5 \: \text{kcal mol}^{-1}\) less stable than the chair form. It is possible to measure the spectral properties of the twist-boat form by a very elegant technique employed by F. A. L. Anet. Because the equilibrium constant for conversion of chair to boat increases with temperature, a considerable proportion of the molecules exist as the twist-boat form in the vapor at \(800^\text{o}\). If such vapor is allowed to impinge on a surface cooled to \(20 \: \text{K}\), the film condensate contains about \(25\%\) of the twist-boat form. At this low temperature, the twist-boat form is converted to the more stable chair form at a very slow rate. Infrared spectra can be taken of the boat-chair mixture at \(10 \: \text{K}\). If the mixture is allowed to warm to \(75 \: \text{K}\), the normal equilibrium favoring the chair form is established in a short time. The spatial arrangement (stereochemistry) of cyclohexane and other organic compounds are studied conveniently with the aid of , which are made with standard bond angles and scaled bond distances. The bonds have stainless-steel rods that make a snap-fit into stainless-steel sleeves. Rotation is smooth about the bonds and there is sufficient flexibility to accommodate some angle strain. Dreding models of the conformations of cyclohexane are shown in Figure 12-7. Notice that these models correspond closely to the sawhorse representations in Figures 12-4, 12-5, and 12-6. Figure 12-5 shows that there are two distinct kinds of hydrogen in the chair form of cyclohexane - six that are close to the "average" plane of the ring (called hydrogens) and three above and three below this average plane (called hydrogens). This raises interesting questions in connection with substituted cyclohexanes: For example, is the methyl group in methylcyclohexane equatorial or axial? Since only methylcyclohexane is known, the methyl group must be exclusively equatorial \(\left( e \right)\), exclusively axial \(\left( a \right)\), or the two forms must be interconverted so rapidly that they cannot be separated into isomeric forms. It appears that the latter circumstance prevails, with the ring changing rapidly from one chair form to another by flipping one end of the chair up and the other end down: Such a change would cause a substituent in an axial position to go to an equatorial position and . This process is called and its rate often is called the . With cyclohexane, inversion is so fast at room temperature that, on average, the molecules, flip about 100,000 times per second, over an energy barrier of about \(11 \: \text{kcal mol}^{-1}\). You will understand this flipping process if you make a model of a cyclohexane ring carrying a single substituent. By manipulating the model you can discover some of the different ways the process can occur. The simplest route is simply to flip up one corner of the ring to convert the chair into a boat and then flip down the opposite carbon: Because of the flexibility of the boat conformation, it is possible to transform it to other boat conformations whereby carbons other than the one indicated flip down and complete the interconversion. At room temperature the conformation of methylcyclohexane with the methyl equatorial is more stable than the one with the methyl axial by \(1.7 \: \text{kcal mol}^{-1}\). The same is true of all monosubstituted cyclohexanes to a greater or lesser degree. Reasons for this can be seen from space-filling models (Figure 12-8), which show that a substituent group has more room when the substituent is equatorial than when it is axial. In the axial position the substituent is considerably closer to the two axial hydrogens on the same side of the ring than to other hydrognes, even hydrogens on adjacent carbons when the substituent is in the equatorial position (Figure 12-8). For example, when the substituent is bromine, which has a \(\ce{C-Br}\) bond length of \(1.94 \: \text{Å}\), the distance from axial bromine to the axial hydrogen at \(\ce{C_3}\) or \(\ce{C_5}\) on the same side of the ring is about \(2.7 \: \text{Å}\). In contrast, the distance from equatorial bromine to any of the hydrogens on the adjacent carbons is about \(3.1 \: \text{Å}\). There is a very important general aspect of the difference between these two nonbonded \(\ce{H} \cdot \cdot \cdot \ce{Br}\) interactions at \(2.7 \: \text{Å}\) and \(3.1 \: \text{Å}\). Whenever two nonbonded atoms are brought close together, and before the massive repulsion sets in (which is so evident in Figure 4-6), there is a slight in the energy curve corresponding to .\(^2\) For nonbonded \(\ce{H} \cdot \cdot \cdot \ce{Br}\) interactions the bottom of the dip comes at about \(3.1 \: \text{Å}\) (Figure 12-9), and the resulting attraction between the atoms will provide some stabilization of the equatorial conformation relative to the axial conformation. Weak attractive forces between nonbonded atoms are called or , and are of great importance in determining the properties of liquids. They also can be expected to play a role in determining conformation equilibria whenever the distances between the atoms in the conformations correspond to the so-called . Table 12-2 shows the contribution made by various substituents to the free-energy change from the axial to the equatorial orientations of the substituent. Thus, for bromine, the free-energy change, \(\Delta G^0\), is \(-0.5 \: \text{kcal mol}^{-1}\), which means that at \(25^\text{o}\), the equilibrium constant, \(K\), for the axial \(\rightleftharpoons\) equatorial equilibrium is about 2.3 (from \(-2.303 RT \: \text{log} \: K = \Delta G^0\); see Section 4-4A). From many studies it is known that the interconversion of conformations with the substituent in the equatorial and the axial positions occurs about 100,000 times per second, which corresponds to a transition-state energy (activation energy) of about \(11 \: \text{kcal mol}^{-1}\) above the ground-state energy. The rate decreases as the temperature is lowered. If one cools chlorocyclohexane to its melting point \(\left( -44^\text{o} \right)\), the substance crystallizes to give the pure equatorial isomer. The crystals then can be cooled to \(-150^\text{o}\) and dissolved at this temperature in a suitable solvent. At \(-150^\text{o}\) it would take about 130 days for half of the equatorial form to be converted to the axial form. However, when the solution is warmed to \(-60^\text{o}\) the equatorial conformation is converted to the equilibrium mixture in a few tenths of a second. The cis-trans isomerism of cyclohexane derivatives (Section 5-1A) is complicated by conformational isomerism. For example, 4- -butylcyclohexyl chloride theoretically could exist in four stereoisomeric chair forms, \(1\), \(2\), \(3\), and \(4\). trans cis Use of \(\ce{^{13}C}\) nmr spectroscopy to determine whether a substituent is in an axial or equatorial position is well illustrated with - and -4- -butylcyclohexanols, \(5\) and \(6\): The five \(\ce{-CH_2}-\) groups of cyclopentane theoretically could form a regular planar pentagon (internal angles of \(108^\text{o}\)) with only a little bending of the normal \(\ce{C-C-C}\) bond angles. Actually, cyclopentane molecules are flat. The planar structure has completely eclipsed hydrogens, which makes it less stable by about \(10 \: \text{kcal mol}^{-1}\) than if there were no eclipsed hydrogens. The result is that each molecule assumes a puckered conformation that is the best compromise between distortion of bond angles and eclipsing of hydrogens. The best compromise conformations have the ring twisted with one or two of the \(\ce{-CH_2}-\) groups bent substantially out of a plane passed through the other carbons (Figure 12-14). The flexibility of the ring is such that these deformations move rapidly around the ring. Formation of a four-membered ring of carbon atoms can be achieved only with substantial distortion of the normal valence angles of carbon, regardless of whether the ring is planar or nonplanar. In cyclobutane, for example, if the valence bonds are assumed to lie along straight lines drawn between the carbon nuclei, each \(\ce{C-C-C}\) bond angle will be \(19.5^\text{o}\) smaller than the \(109.5^\text{o}\) tetrahedral value: The three carbon atoms of the cyclopropane ring lie in a plane. Therefore the angle strain is expected to be considerable because each \(\ce{C-C-C}\) valence angle must be deformed \(49.5^\text{o}\) from the tetrahedral value. It is likely that some relief from the strain associated with the eclipsing of the hydrogens of cyclopropane is achieved by distortion of the \(\ce{H-C-H}\) and \(\ce{H-C-C}\) bond angles: If one is willing to consider a carbon-carbon double bond as a two-membered ring, then ethene, \(\ce{C_2H_4}\), is the simplest possible cycloalkane ("cycloethane"). As such, \(\ce{C_2H_4}\) has \(\ce{C-C-C}\) valence angles of \(0^\text{o}\) and therefore an angle strain of \(109.5^\text{o}\) at each \(\ce{CH_2}\) group compared to the tetrahedral value: and (1977) \(^1\)Pioneering work on the conformations of cyclohexane and its derivatives was carried out by O. Hassel (Norway) and D. H. R. Barton (United Kingdom) for which they shared a Nobel Prize in 1969. \(^2\)The vertical scale of Figure 4-6 does not permit seeing the dip in the curve resulting from attractive forces between neon atoms. It is deepest when \(r\) is about \(3.12 \: Å\) and amounts to \(0.070 \: \text{kcal mol}^{-1}\). \(^3\)After F. London, who developed a quantum-mechanical theory of the origin of these forces and also pioneered many quantum calculations of great consequence to chemistry, including bonding in \(\ce{H_2}\), which will be discussed in Section 21-1.
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We can make the same argument for the heat along C. If we do the three processes A and B+C only to a tiny extent we can write: And now we can integrate from \(V_1\) to \(V_2\) over the reversible adiabatic work along B and from \(T_1\) to \(T_2\) for the reversible isochoric heat along C. To separate the variables we do need to bring the temperature to the right side of the equation.: The latter expression is valid for a reversible adiabatic expansion of a monatomic ideal gas (say Argon) because we used the \(C_v\) expression for such a system. We can use the gas law \(PV=nRT\) to translate this expression in one that relates pressure and volume see Eq 19.23 We can mathematically show that the temperature of a gas decreases during an adiabatic expansion. Assuming an ideal gas, the internal energy along an adiabatic path is: \[\begin{split} d\bar{U}&= \delta q+\delta w \\ &= 0-Pd\bar{V}\\ &= -Pd\bar{V} \end{split} \nonumber \] The constant volume heat capacity is defined as: \[{\bar{C}}_V=\left(\frac{\partial\bar{U}}{\partial T}\right)_V \nonumber \] We can rewrite this for internal energy: \[d\bar{U}={\bar{C}}_VdT \nonumber \] Combining these two expressions for internal energy, we obtain: \[{\bar{C}}_VdT=-Pd\bar{V} \nonumber \] Using the ideal gas law for pressure of an ideal gas: \[{\bar{C}}_VdT=-\frac{RT}{\bar{V}}d\bar{V} \nonumber \] Separating variables: \[\frac{\bar{C}_V}{T}dT=-\frac{R}{\bar{V}}d\bar{V} \nonumber \] This is an expression for an ideal path along a reversible, adiabatic path that relates temperature to volume. To find our path along a PV surface for an ideal gas, we can start in TV surface and convert to a PV surface. Let's go from (\(T_1,V_1\)) to (\(T_2,V_2\)). \[\int_{T1}^{T_2}{\frac{\bar{C}_V}{T}dT=-\int_{\bar{V}_1}^{\bar{V}_2}{\frac{R}{\bar{V}}d\bar{V}}} \nonumber \] \[\bar{C}_V\ln{\left(\frac{T_2}{T_1}\right)}=-R\ln{\left(\frac{{\bar{V}}_2}{{\bar{V}}_1}\right)}=R\ln{\left(\frac{{\bar{V}}_1}{{\bar{V}}_2}\right)} \nonumber \] \[\ln{\left(\frac{T_2}{T_1}\right)}=\frac{R}{\bar{C}_V}\ln{\left(\frac{\bar{V}_1}{\bar{V}_2}\right)} \nonumber \] \[\left(\frac{T_2}{T_1}\right)=\left(\frac{\bar{V}_1}{\bar{V}_2}\right)^{\frac{R}{\bar{C}_V}} \nonumber \] We know that: \[R={\bar{C}}_P-{\bar{C}}_V \nonumber \] \[\frac{R}{\bar{C}_V}=\frac{\bar{C}_P-\bar{C}_V}{\bar{C}_V}=\frac{\bar{C}_P}{\bar{C}_V}-1 \nonumber \] \[\frac{R}{{\bar{C}}_V}=\gamma-1 \nonumber \] Therefore: \[\left(\frac{T_2}{T_1}\right)=\left(\frac{\bar{V}_1}{\bar{V}_2}\right)^{\gamma-1} \nonumber \] This expression shows that volume and temperature are inversely related. That is, as the volume increase from \(V_1\) to \(V_2\), the temperature must decrease from \(T_1\) to \(T_2\).
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Reaction enthalpies are important, but difficult to tabulate. However, because enthalpy is a state function, it is possible to use to simplify the tabulation of reaction enthalpies. Hess’ Law is based on the addition of reactions. By knowing the reaction enthalpy for constituent reactions, the enthalpy of a reaction that can be expressed as the sum of the constituent reactions can be calculated. The key lies in the canceling of reactants and products that °Ccur in the “data” reactions but not in the “target reaction. Find \(\Delta H_{rxn}\) for the reaction \[2 CO(g) + O_2(g) \rightarrow 2 CO_2(g) \nonumber \] Given \[C(gr) + ½ O_2(g) \rightarrow CO(g) \nonumber \] with \(\Delta H_1 = -110.53 \,kJ\) \[C(gr) + O_2(g) \rightarrow CO_2(g) \nonumber \] with \(\Delta H_2 = -393.51\, kJ\) The target reaction can be generated from the data reactions. \[ {\color{red} 2 \times} \left[ CO(g) \rightarrow C(gr) + O_2(g) \right] \nonumber \] plus \[ { \color{red} 2 \times} \left[ C(gr) + 2 O_2(g) \rightarrow 2 CO_2(g) \right] \nonumber \] equals \[2 CO(g) + O_2(g) \rightarrow 2 CO_2(g) \nonumber \] so \[{ \color{red} 2 \times} \Delta H_1 = -787.02 \, kJ \nonumber \] \[{ \color{red} 2 \times} \Delta H_2 = 221.06\, kJ \nonumber \] \[ { \color{red} 2 \times} \Delta H_1 + { \color{red} 2 \times} \Delta H_2 = -565.96 \,kJ \nonumber \] One of the difficulties with many thermodynamic state variables (such as enthalpy) is that while it is possible to measure changes, it is impossible to measure an absolute value of the variable itself. In these cases, it is necessary to define a zero to the scale defining the variable. For enthalpy, the definition of a zero is that the standard enthalpy of formation of a pure element in its standard state is zero. All other enthalpy changes are defined relative to this standard. Thus it is essential to very carefully define a standard state. The standard state of a substance is the most stable form of that substance at 1 atmosphere pressure and the specified temperature. Using this definition, a convenient reaction for which enthalpies can be measured and tabulated is the . This is a reaction which forms one mole of the substance of interest in its standard state from elements in their standard states. The enthalpy of a standard formation reaction is the (\(\Delta H_{f^o}\)). Some examples are It is important to note that the standard state of a substance is . For example, the standard state of water at -10 °C is solid, whereas the standard state at room temperature is liquid. Once these values are tabulated, calculating reaction enthalpies becomes a snap. Consider the heat combustion (\(\Delta H_c\)) of methane (at 25 °C) as an example. \[CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \nonumber \] The reaction can expressed as a sum of a combination of the following standard formation reactions. \[C(gr) + 2 H_2(g) \rightarrow CH_4(g) \nonumber \] with \(\Delta H_f^o = -74.6\, kJ/mol\) \[C(gr) + O_2(g) \rightarrow CO_2(g) \nonumber \] with \(\Delta H_f^o = -393.5\, kJ/mol\) \[H_2(g) + ½ O_2(g) \rightarrow H_2O(l) \nonumber \] with \(\Delta H_f^o = -285.8 \,kJ/mol\) The target reaction can be generated from the following combination of reactions \[{ \color{red} -1 \times} \left[ C(gr) + 2 H_2(g) \rightarrow CH_4(g)\right] \nonumber \] \[CH_4(g) \rightarrow C(gr) + 2 H_2(g) \nonumber \] with \(\Delta H_f^o ={ \color{red} -1 \times} \left[ -74.6\, kJ/mol \right]= 74.6\, kJ/mol\) \[C(gr) + O_2(g) \rightarrow CO_2(g) \nonumber \] with \(\Delta H_f^o = -393.5\, kJ/mol\) \[{ \color{red} 2 \times} \left[ H_2(g) + ½ O_2(g) \rightarrow H_2O(l) \right] \nonumber \] \[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \nonumber \] with \(\Delta H_f^o = {\color{red} 2 \times} \left[ -285.8 \,kJ/mol \right] = -571.6\, kJ/mol\). \[CH_4(g) + 2 O_2(g) \rightarrow CO2_(g) + 2 H_2O(l) \nonumber \] with \(\Delta H_c^o = -890.5\, kJ/mol\) Alternately, the reaction enthalpy could be calculated from the following relationship \[\Delta H_{rxn} = \sum_{products} \nu \cdot \Delta H_f^o - \sum_{reactants} \nu \cdot \Delta H_f^o \nonumber \] where \(\nu\) is the stoichiometric coefficient of a species in the balanced chemical reaction. For the combustion of methane, this calculation is \[ \begin{align} \Delta _{rxn} & = (1\,mol) \left(\Delta H_f^o(CO_2)\right) + (2\,mol) \left(\Delta H_f^o(H_2O)\right) - (1\,mol) \left(\Delta H_f^o(CH_4)\right) \\ & = (1\,mol) (-393.5 \, kJ/mol) + (2\,mol) \left(-285.8 \, kJ/mol \right) - (1\,mol) \left(-74.6 \, kJ/mol \right) \\ & = -890.5 \, kJ/mol \end{align} \nonumber \] A note about units is in order. Note that reaction enthalpies have units of kJ, whereas enthalpies of formation have units of kJ/mol. The reason for the difference is that enthalpies of formation (or for that matter enthalpies of combustion, sublimation, vaporization, fusion, etc.) refer to specific substances and/or specific processes involving those substances. As such, the total enthalpy change is scaled by the amount of substance used. General reactions, on the other hand, have to be interpreted in a very specific way. When examining a reaction like the combustion of methane \[CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \nonumber \] with \(\Delta H_{rxn} = -890.5\, kJ\). The correct interpretation is that the reaction of one mole of CH (g) with two moles of O (g) to form one mole of CO (g) and two moles of H O(l) releases 890.5 kJ at 25 °C. Ionized species appear throughout chemistry. The energy changes involved in the formation of ions can be measured and tabulated for several substances. In the case of the formation of positive ions, the enthalpy change to remove a single electron at 0 K is defined as the . \[ M(g) \rightarrow M^+(g) + e^- \nonumber \] with \(\Delta H (0 K) \equiv 1^{st} \text{ ionization potential (IP)}\) The removal of subsequent electrons requires energies called the 2 Ionization potential, 3 ionization potential, and so on. \[M^+(g) \rightarrow M^{2+}(g) + e^- \nonumber \] with \(\Delta H(0 K) ≡ 2^{nd} IP\) \[M^{2+}(g) \rightarrow M^{3+}(g) + e^- \nonumber \] with \(\Delta H(0 K) ≡ 3^{rd} IP\) An atom can have as many ionization potentials as it has electrons, although since very highly charged ions are rare, only the first few are important for most atoms. Similarly, the can be defined for the formation of negative ions. In this case, the first electron affinity is defined by \[X(g) + e^- \rightarrow X^-(g) \nonumber \] with \(-\Delta H(0 K) \equiv 1^{st} \text{ electron affinity (EA)}\) The minus sign is included in the definition in order to make electron affinities mostly positive. Some atoms (such as noble gases) will have negative electron affinities since the formation of a negative ion is very unfavorable for these species. Just as in the case of ionization potentials, an atom can have several electron affinities. \[X^-(g) + e^- \rightarrow X^{2-}(g) \nonumber \] with \(-\Delta H(0 K) ≡ 2^{nd} EA\). \[X^{2-}(g) + e^- \rightarrow X^{3-}(g) \nonumber \] with \(-\Delta H(0 K) ≡ 3^{rd} EA\). In the absence of standard formation enthalpies, reaction enthalpies can be estimated using average bond enthalpies. This method is not perfect, but it can be used to get ball-park estimates when more detailed data is not available. A \(D\) is defined by \[XY(g) \rightarrow X(g) + Y(g) \nonumber \] \(\Delta H \equiv D(X-Y)\) In this process, one adds energy to the reaction to break bonds, and extracts energy for the bonds that are formed. \[\Delta H_{rxn} = \sum (\text{bonds broken}) - \sum (\text{bonds formed}) \nonumber \] As an example, consider the combustion of ethanol: In this reaction, five C-H bonds, one C-C bond, and one C-O bond, and one O=O bond must be broken. Also, four C=O bonds, and one O-H bond are formed. The reaction enthalpy is then given by \[ \begin{align} \Delta H_c = \, &5(413 \,kJ/mol) + 1(348\, kJ/mol) + 1(358 \,kJ/mol) \nonumber \\ & + 1(495\, kJ/mol) - 4(799 \,kJ/mol) – 2(463\, kJ/mol) \nonumber \\ =\,& -856\, kJ/mol \end{align} \nonumber \] Because the bond energies are defined for gas-phase reactants and products, this method does not account for the enthalpy change of condensation to form liquids or solids, and so the result may be off systematically due to these differences. Also, since the bond enthalpies are averaged over a large number of molecules containing the particular type of bond, the results may deviate due to the variance in the actual bond enthalpy in the specific molecule under consideration. Typically, reaction enthalpies derived by this method are only reliable to within ± 5-10%.
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A characteristic and synthetically important reaction of ethyne and 1-alkynes is salt (“acetylide”) formation with very strong bases. In such reactions the alkynes behaves as an acids in the sense that they give up protons to suitably strong bases: Water is too weak a base to accept protons from alkynes; consequently no measurable concentration of \(\ce{H_3O}^\oplus\) is expected from the ionization of alkynes in dilute aqueous solutions. Therefore we have no quantitative measure of 1-alkyne acidity in aqueous solution other than that it probably is about \(10^{10}\) times less acidic than water, as judged from measurements in other solvents to be discussed shortly. In the gas phase, however, the situation is reversed, and : This reversal is of little practical value because organic reactions ions normally are not carried out in the gas phase. However, it should alert us to the tremendous role that solvents play in determining acidities by their abilities (some much more than others) to stabilize ions by the property known as . ( .) Liquid ammonia is a more useful solvent than water for the preparation of 1-alkyne salts. A substantial amount of the alkyne can be converted to the conjugate base by amide anions (potassium or sodium amide) because a 1-alkyne is a stronger acid than ammonia. The acidity of the terminal hydrogen in 1-alkynes provides a simple and useful test for 1-alkynes. With silver-ammonia solution (\(\ce{AgNO_3}\) in aqueous ammonia), 1-alkynes give insoluble silver salts, whereas disubstituted alkynes do not: The silver “acetylides” appear to have substantially covalent carbon-metal bonds and are less ionic than sodium and potassium alkynides. Silver-ammonia solution may be used to precipitate 1-alkynes from mixtures with other hydrocarbons. The 1-alkynes are regenerated easily from the silver precipitates by treatment with strong inorganic acids. It should be noted, however, that silver alkynides may be shock sensitive and can decompose explosively, especially when dry. Some idea of the importance of solvation can be gained from the calculated \(\Delta H\) for the following process: \[ \ce{Na^+ (g) + Cl^- (g) \rightarrow Na^+ (aq) + Cl^- (aq)}\] with \(Delta H^o = -187\,kcal\). The solvation energies of ions are so large that relatively small differences for different ions can have a very large effect on equilibrium constants. Thus, the ratio between the relative acidities of ethyne and water in the gas phase and in water of \(10^{12}\) corresponds at \(25^\text{o}\) to an overall \(\Delta G^0\) difference in solvation energies of approximately \(16 \: \text{kcal}\), which is less than \(10\%\) of the total solvation energies of the ions. Further difficulties arise because of differences between solvation energies and interactions between the ions in different solvents. Thus the acidities of 1-alkynes relative to other acids have been found to change by a factor of \(10^{11}\) in different solvents. For this reason, we must be particularly careful in comparing the rates and equilibrium constants of ionic reactions to take proper account of solvation and ion interaction effects. An excellent rule of thumb is that, other things being equal, large ions are more stable than small ions in the gas phase, with the opposite being true in polar solvents, where small ions are more strongly solvated (thus more stable) than large ions. For comparison, From (1) minus (3), the solvation energy of gaseous \(\ce{Li}^\oplus\) is \(47 \: \text{kcal mol}^{-1}\) greater than \(\ce{K}^\oplus\); and from (1) minus (2), that of \(\ce{F}^\ominus\) is \(35 \: \text{kcal mol}^{-1}\) greater than that of \(\ce{I}^\ominus\). Such differences in solvation energies can have considerable effects on reactivity, and you may remember from Section 8-7E that \(\ce{F}^\ominus\) is a weaker nucleophile than \(\ce{I}^\ominus\), largely because of its greater solvation energy. If we compare acid strengths of the simple hydrocarbons, we find that ethyne is substantially more acidic than ethene or ethane in the gas phase or in solution. Why is this? The simplest explanation is that there is a direct connection between \(\ce{C-H}\) acidity and the amount of \(s\) character associated with the \(\sigma\)-bonding carbon orbital. Other things being equal, acidity increases with increasing \(s\) character in the carbon orbital. 1-Alkynes are very weak acids, hence their conjugate bases, \(\ce{RC \equiv C}^\ominus\), are quite strong bases. These anions also are reactive carbon nucleophiles, and it is this property that makes them useful for organic synthesis. Recall from Chapter 8 that one of the most generally useful organic reactions is a displacement reaction in which an anionic nucleophile, \(\ce{Nu}^\ominus\), attacks an alkyl derivative, \(\ce{RX}\), to displace \(\ce{X}^\ominus\) and form a new bond between carbon and the nucleophile: The displaced group \(\ce{X}\) often is a halide ion (chloride, bromide, or iodide), and if the entering nucleophile \(\ce{Nu}^\ominus\) is an alkynide anion, the reaction leads to formation of a carbon-carbon bond: With those \(\ce{RX}\) derivatives that undergo nucleophilic displacement readily, this is a general method of forming a \(\ce{C-C}\) bond, thereby leading to substituted alkynes. The alkynide salts generally used are those of lithium, sodium, potassium, or magnesium. Another reaction of 1-alkynes that extends the carbon chain is a coupling reaction in which the alkyne dimerizes under the influence of a cuprous salt, usually cuprous ammonium chloride: This addition of one molecule of alkyne to another is formally analogous to the dimerization of alkenes under the influence of sulfuric acid (see ), but the mechanisms are quite different. If the reaction is carried out in the presence of an oxidizing agent, such as \(\ce{O_2}\) or a cupric salt dissolved in pyridine (a weak base), a different product is obtained. Under these conditions, oxidative coupling occurs to give a conjugated diyne: Although the details of the mechanisms of these alkyne reactions are not known, it is likely that the ability of 1-alkynes to form carbon-metal bonds with metals such as copper is a key factor. Other oxidative coupling reactions occur with transition metals, and this will be discussed in detail in . and (1977)
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Discovered in 1803 by William Wollaston and named after the recently discovered asteroid Pallas, palladium is a silvery-white, soft metal similar to platinum. It is a rare metal (only about 1 part per million in the earth) but occurs commonly along with copper, silver and gold. Palladium is part of the the Platinum Group Metals (PGM) whic is located in the 5th and 6th rows of the transition metal section of the periodic table and includes , , Palladium, , , and . Common characteristics include resistance to wear, oxidation, and corrosion, high melting points, and oxidation states of +2 to +4. They are generally non-toxic. Palladium is used as an alloying agent with gold in jewelry ("white gold") and in some dental applications in place of silver or gold. Unlike the other so-called platinum metals, palladium is more susceptible to attack by acids, even hydrochloric acid. \[ \ce{Pd_{(s)} + 2HCl \rightarrow Pd^{2+} + H_{2(g)}} + \ce{2Cl_{(aq)}^-}\] Palladium has the curious ability to absorb large quantities of hydrogen gas (up to 900 times its own volume) and this has generated some interest in its alloys as a storage system for hydrogen as a portable fuel for automobiles. It was also prominent in the "cold fusion" controversy some years ago when it was said (apparently falsely) that when it was made to absorb heavy hydrogen (deuterium), the atoms would undergo fusion and release more energy than was put into the process.
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We will consider here the reverse process of photosynthesis, namely how carbohydrates, especially glucose, are converted to energy by being broken down into carbon dioxide and water. A general summary of the several stages involved is shown in Figure 20-8. Initially, the storage fuels or foodstuffs (fats, carbohydrates, and proteins) are hydrolyzed into smaller components (fatty acids and glycerol, glucose and other simple sugars, and amino acids). In the next stage, these simple fuels are degraded further to two-carbon fragments that are delivered as the \(\ce{CH_3C=O}\) group (ethanoyl, or acetyl) in the form of the thioester of coenzyme A, \(\ce{CH_3COS}\) . The structure of this compound and the manner in which fatty acids are degraded has been considered in , and amino acid metabolism is discussed briefly in . This section is concerned mainly with the pathway by which glucose is metabolized by the process known as . In the conversion of glucose to \(\ce{CH_3COS}\) , two carbons are oxidized to carbon dioxide with consumption of two oxygen molecules: \[\ce{C_6H_{12}O_6} + 2 \textbf{CoA} \ce{SH} + \left[ 2 \ce{O_2} \right] \rightarrow 2 \ce{CH_3COS} \textbf{CoA} + 4 \ce{H_2O} + 2 \ce{CO_2} \tag{20-5}\] For further oxidation to occur, the \(\ce{CH_3COS}\) must enter the next stage of metabolism, whereby the \(\ce{CH_3C=O}\) group is converted to \(\ce{CO_2}\) and \(\ce{H_2O}\). This stage is known variously as the citric acid cycle, the tricarboxylic acid cycle, or the Krebs cycle, in honor of H. A. Krebs (Nobel Prize, 1953), who first recognized its cyclic nature in 1937. We can write an equation for the process as if it involved oxygen: \[2 \ce{CH_3COS} \textbf{CoA} + \left[ 4 \ce{O_2} \right] \rightarrow 4 \ce{CO_2} + 2 \ce{H_2O} + 2 \textbf{CoA} \ce{SH} \tag{20-6}\] Notice that combination of the reactions of Equations 20-5 and 20-6, glycolysis plus the citric acid cycle, oxidizes glucose completely to \(\ce{CO_2}\) and \(\ce{H_2O}\): \[\ce{C_6H_{12}O_6} + \left[ 6 \ce{O_2} \right] \rightarrow 6 \ce{CO_2} + 6 \ce{H_2O} \tag{20-7}\] But, as you will see, none of the steps uses molecular oxygen directly. Hence there must be a stage in metabolism whereby molecular oxygen is linked to production of oxidizing agents that are consumed in glycolysis and in the citric acid cycle. The coupling of oxygen into the metabolism of carbohydrates is an extremely complex process involving transport of the oxygen to the cells by an oxygen carrier such as hemoglobin, myoglobin, or hemocyanin. This is followed by a series of reactions, among which \(\ce{NADH}\) is converted to \(\ce{NAD}^\oplus\) with associated formation of three moles of ATP from three moles of ADP and inorganic phosphate. Another electron-carrier is flavin adenine dinucleotide (\(\ce{FAD}\); ), which is reduced to \(\ce{FADH_2}\) with an associated production of two moles of ATP from two moles of ADP. These processes are known as and can be expressed by the equations: Oxidative phosphorylation resembles photophosphorylation, discussed in , in that electron transport in photosynthesis also is coupled with ATP formation. By suitably juggling Equations 20-7 through 20-9, we find that the metabolic oxidation of one mole of glucose is achieved by ten moles of \(\ce{NAD}^\oplus\) and two moles of \(\ce{FAD}\): The overall result is production of 36 moles of ATP from ADP and phosphate per mole of glucose oxidized to \(\ce{CO_2}\) and \(\ce{H_2O}\). Of these, 34 ATPs are produced according to Equation 20-10 and, as we shall see, two more come from glycolysis. Glycolysis is the sequence of steps that converts glucose into two \(\ce{C_3}\) fragments with the production of ATP. The \(\ce{C_3}\) product of interest here is 2-oxopropanoate (pyruvate): There are features in this conversion that closely resemble the dark reactions of photosynthesis, which build a \(\ce{C_6}\) chain (fructose) from \(\ce{C_3}\) chains ( ). For example, the reactants are either phosphate esters or mixed anhydrides, and the phosphorylating agent is ATP: \[\ce{ROH} + \text{ATP} \rightarrow \ce{RO-PO_3^2-} + \text{ADP} + \ce{H}^\oplus\] Furthermore, rearrangements occur that interconvert an aldose and ketose, and the cleavage of a \(\ce{C_6}\) chain into two \(\ce{C_3}\) chains is achieved by a reverse aldol condensation: Also, oxidation of an aldehyde to an acid is accomplished with \(\ce{NAD}^\oplus\). There is a related reaction in photosynthesis ( ) that accomplishes the reduction of an acid to an aldehyde and is specific for \(\ce{NADPH}\), not \(\ce{NADH}\): First, glucose is phosphorylated to glucose 6-phosphate with ATP. Then an aldose \(\rightleftharpoons\) ketose rearrangement converts glucose 6-phosphate into fructose 6-phosphate. A second phosphorylation with ATP gives fructose 1,6-diphosphate: At this stage the enzyme aldolase catalyzes the aldol cleavage of fructose 1,6-diphosphate. One product is glyceraldehyde 3-phosphate and the other is 1,3-dihydroxypropanone phosphate. Another ketose \(\rightleftharpoons\) aldose equilibrium converts the propanone into the glyceraldehyde derivative: The next step oxidizes glyceraldehyde 3-phosphate with \(\ce{NAD}^\oplus\) in the presence of phosphate with the formation of 1,3-diphosphoglycerate: The mixed anhydride of phosphoric acid and glyceric acid then is used to convert ADP to ATP and form 3-phosphoglycerate. Thereafter the sequence differs from that in photosynthesis. The next few steps accomplish the formation of pyruvate by transfer of the phosphoryl group from \(\ce{C_3}\) to \(\ce{C_2}\) followed by dehydration to phosphoenolpyruvate. Phosphoenolpyruvate is an effective phosphorylating agent that converts ADP to ATP and forms pyruvate: The net reaction at this point produces more ATP than is consumed in the phosphorylation of glucose and fructose. What happens thereafter depends on the organism. With yeast and certain other microorganisms, pyruvate is decarboxylated and reduced to ethanol. The end result of glycolysis in this instance is . In higher organisms, pyruvate can be stored temporarily as a reduction product (lactate) or it can be oxidized further to give \(\ce{CH_3COS}\) and \(\ce{CO_2}\). The \(\ce{CH_3COS}\) then enters the citric acid cycle to be oxidized to \(\ce{CO_2}\) and \(\ce{H_2O}\), as discussed in the next section: Glycolysis to the pyruvate or lactate stage liberates heat, which can help keep the organism warm and produce ATP from ADP for future conversion into energy. However, glycolysis does not directly involve oxygen and does not liberate \(\ce{CO_2}\), as we might expect from the overall process of the metabolic conversion of glucose to carbon dioxide and water (Equation 20-10). The liberation of \(\ce{CO_2}\) occurs subsequent to pyruvate formation in a process called variously, the citric acid cycle, the Krebs cycle, or the tricarboxylic acid (TCA) cycle. The initial step, which is not really part of the cycle, is conversion of pyruvate to ethanoyl (acetyl ): To achieve the oxidation of acetyl on a continuing basis, intermediates consumed in certain steps must be regenerated in others. Thus we have a situation similar to that in the Calvin cycle ( ), whereby the first stage of the cycle achieved the desired reaction (\(\ce{CO_2}\) formation) and the second stage is designed to regenerate intermediates necessary to perpetuate the cycle. The entry point is the reaction between acetyl and a four-carbon unit, 2-oxobutanedioic acid. An aldol-type addition of the \(\ce{CH_3CO}\) group to this \(\ce{C_4}\) keto acid extends the chain to a branched \(\ce{C_6}\) acid (as citric acid): Dehydration-rehydration of citrate converts it to isocitrate: From here, oxidation of the hydroxyl function with \(\ce{NAD}^\oplus\) gives a keto acid, which loses \(\ce{CO_2}\) readily ( ) and affords 2-oxopentanedioate: We now have a \(\ce{C_5}\) keto acid that can be oxidized in the same way as the \(\ce{C_3}\) keto acid, pyruvic acid, to give a butanedioyl : Two molecules of \(\ce{CO_2}\) now have been produced and the remaining part of the citric acid cycle is concerned with regeneration of the for forming acetyl from 2-oxopropanoate, and also with regenerating the 2-oxobutanedioate, which is the precursor of citrate. The steps involved are The hydrolysis of the acyl in the first step is used for energy storage by conversion of guanosine diphosphate (GDP) to guanosine triphosphate (GTP): The hydration of the -butenedioate ( ) and the final oxidation reaction ( ) have been discussed previously. There is an alternative route, called the , by which glucose enters the glycolytic sequence to pyruvate. This route achieves the oxidative decarboxylation of glucose to give ribose, as the 5-phosphate ester. The essential steps are The net result is that three pentoses are converted into two molecules of fructose and one of glyceraldehyde \(\left( 3 \ce{C_5} \rightarrow 2 \ce{C_6} + \ce{C_3} \right)\). The relationship of the pentose-phosphate pathway to glycolysis is shown in Figure 20-11. The steps involved in the pentose shunt are readily reversible, but there are several steps in glycolysis that are not. These are the phosphorylation steps (Figure 20-9). Yet, there has to be a return route from pyruvate to glucose. This route is called and, in animals, takes place in the liver. We shall not discuss the steps in gluconeogenesis except to indicate again that they are not all the reverse of glycolysis. For comparison, the steps that differ are indicated in Figure 20-9 by dashed lines. Why is lactate formed from pyruvate in the metabolism of glucose? Pyruvate \( + \: \ce{NADH} + \ce{H}^\oplus \rightarrow\) lactate \(+ \: \ce{NAD}^\oplus\) is a dead-end path, but it does furnish the \(\ce{NAD}^\oplus\) needed for glycolysis in active muscle. This route for forming \(\ce{NAD}^\oplus\) is important, because in circumstances of physical exertion, the rate of production of \(\ce{NAD}^\oplus\) from oxidative phosphorylation may be slower than the demand for \(\ce{NAD}^\oplus\), in which case a temporary supply is available from the pyruvate \(\rightarrow\) lactate reduction. The lactate so formed builds up in muscle tissue under conditions of physical exertion and is apt to cause muscles to "cramp". The excess lactate so formed ultimately is removed by being converted back to pyruvate by oxidation with \(\ce{NAD}^\oplus\). The beauty of the metabolic cycle through pyruvate, shown in summary in Figure 20-11, is the way it can be tapped at various points according to whether the organism requires ATP (from glycolysis), \(\ce{NADH}\) (from pentose shunt), or \(\ce{NAD}^\oplus\) (from the lactate siding). and (1977)
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A formal charge (FC) is the charge assigned to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity. When determining the best Lewis structure (or predominant resonance structure) for a molecule, the structure is chosen such that the formal charge on each of the atoms is as close to zero as possible. The formal charge of any atom in a molecule can be calculated by the following equation: \[{\displaystyle FC=V-N-{\frac {B}{2}}\ }\] where is the number of valence electrons of the neutral atom in isolation (in its ground state); is the number of non-bonding valence electrons on this atom in the molecule; and is the total number of electrons shared in bonds with other atoms in the molecule. Even though all three structures gave us a total charge of zero, the final structure is the superior one because there are no charges in the molecule at all. The following is equivalent: It is important to keep in mind that formal charges are just that – , in the sense that this system is a formalism. The formal charge system is just a method to keep track of all of the valence electrons that each atom brings with it when the molecule is formed. Formal charge is a tool for estimating the distribution of electric charge within a molecule. The concept of oxidation states constitutes a competing method to assess the distribution of electrons in molecules. If the formal charges and oxidation states of the atoms in carbon dioxide are compared, the following values are arrived at: The reason for the difference between these values is that formal charges and oxidation states represent fundamentally different ways of looking at the distribution of electrons amongst the atoms in the molecule. With formal charge, the electrons in each covalent bond are assumed to be split exactly evenly between the two atoms in the bond (hence the dividing by two in the method described above). The formal charge view of the CO molecule is essentially shown below: The covalent (sharing) aspect of the bonding is overemphasized in the use of formal charges, since in reality there is a higher electron density around the oxygen atoms due to their higher electronegativity compared to the carbon atom. This can be most effectively visualized in an electrostatic potential map. With the oxidation state formalism, the electrons in the bonds are "awarded" to the atom with the greater electronegativity. The oxidation state view of the CO molecule is shown below: Oxidation states overemphasize the ionic nature of the bonding; the difference in electronegativity between carbon and oxygen is insufficient to regard the bonds as being ionic in nature. In reality, the distribution of electrons in the molecule lies somewhere between these two extremes. The inadequacy of the simple Lewis structure view of molecules led to the development of the more generally applicable and accurate valence bond theory of Slater, Pauling, et al., and henceforth the molecular orbital theory developed by Mulliken and Hund.
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Krypton is one of the six Noble Gas elements ( ), which are widely known for their relative "inertness" and difficulty in forming chemical compounds with any other elements, due to these elements having full valence shells. Contrary to original thinking, however, Krypton has been made to react with the highly electronegative elements and is used in lighting and other commercial purposes. Krypton is found in the Group 18 elements, otherwise known as the Noble Gases. In 1785, Henry Cavendish suggested that air contained nonreactive gases after he was unsuccessful in getting a sample of air to react. A century later, British chemists John Rayleigh and William Ramsey began to isolate these inert gases (beginning with Argon) and seperated them in their own group on the periodic table since each of these elements had full electron valence shells. One of these gases, Krypton, was discovered along with Neon and Xenon by Rayleigh and fellow chemist Morris Travers in 1898 in a residue left from evaporating almost all components of liquid air. The name Krypton is derived from the Greek word "kryptos", meaning "hidden". However, the inert quality of these gases was disproved when Xenon compounds were created in 1962 and a Krypton compound (KrF2) was synthesized successfully a year later. This proved that this group of gases is not necessarily inert. Although both Kr and Xe have full valence shells, they are both the most easily ionized of the group. It simply took an element of high electronegativity, in this case Fluorine, to force Xe and Kr to react under high temperatures. It ranks sixth in abundance in the atmosphere. As with the other noble gases, krypton is isolated from the air by liquefaction. Although Krypton is naturally chemically nonreactive, krypton difluoride was synthesized in 1963. \[Kr_{(g)}+F_{2(g)} \rightarrow KrF_{2(g)}\] It has also been discovered that Krypton can bond with other atoms besides Fluorine, however such compounds are much more unstable than krypton difluoride. For example, KrF can bond with nitrogen when it reacts with [HC≡NH] [AsF ] under -50°C to form [HC≡N–Kr–F] . There have been other reports of successfully synthesizing additional Krypton compounds, but none have been verified. Krypton has 6 stable isotopes: Kr, Kr, Kr, Kr, Kr, and Kr. There are a total of 31 isotopes of Krypton, and the only isotope besides the six given that occur naturally is Kr which is a product of atmospheric reactions between the other natural isotopes. Krypton gas is used in various kinds of lights, from small bright flashlight bulbs to special strobe lights for airport runways. Due to Krypton's large number of spectral lines, it's ionized gas is white, which is why light bulbs that are krypton based are used in photography and studio lighting in the film industry. In neon lights, Krypton reacts with other gases to produce a bright yellow light as well. The isotope Kr can also be used in combination with phosphors to produce materials that shine in the dark due to the fact that this particular isotope of Krypton reflects off of phosphors. Krypton is also used in lasers as a control for a desired wavelength, especially in red lasers because Krypton has a much higher light density in the red spectral region than other gases such as Neon, which is why krypton-based lasers are used to produce red light in laser-light shows. Yet another important application of Krypton, specifically Kr, is in Magnetic Resonance Imaging (MRI), which is used instead of other gases because of its high spin and smaller/less polar electron cloud compared to other noble gases such as Xenon. It is used to distinguish hydrophobic and hydrophillic regions containing an airway.
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Ruthenium is a rare metal (less than 0.01 parts per million in the earth's crust). It is generally described as gray-white, hard and brittle. At room temperature it is resistant to virtually all acids. It's high melting point and brittle nature make casting difficult. Its chief commercial use is as a hardening agent in platinum jewelry. Element 44 (named from the Latin, ruthenia, for Russia) was originally discovered in 1807 by the Polish chemist Sniadecki, but the claim from a relative unknown was not accepted by a Paris commission and he withdrew it. It was not until 1828 that Gottfried Osann claimed to have found three new elements in his platinum samples that interest in the metal increased. Although the he could not substantiate the claim, he did not withdraw it. In 1844 Karl Klaus showed that two of Osann's metals were not new elements at all, but the third he was able to isolate and characterize. Palladium is part of the the Platinum Group Metals (PGM) whic is located in the 5th and 6th rows of the transition metal section of the periodic table and includes Ruthenium, , , , , and . Common characteristics include resistance to wear, oxidation, and corrosion, high melting points, and oxidation states of +2 to +4. They are generally non-toxic. Recently interest has grown in some compounds of ruthenium which possess the ability to convert visible light into a suitable energy source for splitting water into hydrogen and oxygen (for fuel). Work in this field is ongoing.
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You learned that the integrated rate law for each common type of reaction (zeroth, first, or second order in a single reactant) can be plotted as a straight line. Using these plots offers an alternative to the methods described for showing how reactant concentration changes with time and determining reaction order. We will illustrate the use of these graphs by considering the thermal decomposition of NO gas at elevated temperatures, which occurs according to the following reaction: \[\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)} \label{14.26} \] Experimental data for this reaction at 330°C are listed in \(\Page {1}\); they are provided as [NO ], ln[NO ], and 1/[NO ] versus time to correspond to the integrated rate laws for zeroth-, first-, and second-order reactions, respectively. The actual concentrations of NO are plotted versus time in part (a) in \(\Page {1}\). Because the plot of [NO ] versus is not a straight line, we know the reaction is not zeroth order in NO . A plot of ln[NO ] versus (part (b) in \(\Page {1}\)) shows us that the reaction is not first order in NO because a first-order reaction would give a straight line. Having eliminated zeroth-order and first-order behavior, we construct a plot of 1/[NO ] versus (part (c) in \(\Page {1}\)). This plot is a straight line, indicating that the reaction is second order in NO . We have just determined the reaction order using data from a single experiment by plotting the concentration of the reactant as a function of time. Because of the characteristic shapes of the lines shown in \(\Page {2}\), the graphs can be used to determine the reaction order of an unknown reaction. In contrast, the method of initial rates required multiple experiments at different NO concentrations as well as accurate initial rates of reaction, which can be difficult to obtain for rapid reactions. Dinitrogen pentoxide (N O ) decomposes to NO and O at relatively low temperatures in the following reaction: \[2N_2O_5(soln) → 4NO_2(soln) + O_2(g) \nonumber \] This reaction is carried out in a CCl solution at 45°C. The concentrations of N O as a function of time are listed in the following table, together with the natural logarithms and reciprocal N O concentrations. Plot a graph of the concentration versus , ln concentration versus , and 1/concentration versus and then determine the rate law and calculate the rate constant. balanced chemical equation, reaction times, and concentrations graph of data, rate law, and rate constant Use the data in the table to separately plot concentration, the natural logarithm of the concentration, and the reciprocal of the concentration (the vertical axis) versus time (the horizontal axis). Compare the graphs with those in \(\Page {1}\) to determine the reaction order. Write the rate law for the reaction. Using the appropriate data from the table and the linear graph corresponding to the rate law for the reaction, calculate the slope of the plotted line to obtain the rate constant for the reaction. Here are plots of [N O ] versus , ln[N O ] versus , and 1/[N O ] versus : The plot of ln[N O ] versus gives a straight line, whereas the plots of [N O ] versus and 1/[N O ] versus do not. This means that the decomposition of N O is first order in [N O ]. The rate law for the reaction is therefore rate = [N O ] Calculating the rate constant is straightforward because we know that the slope of the plot of ln[A] versus for a first-order reaction is − . We can calculate the slope using any two points that lie on the line in the plot of ln[N O ] versus . Using the points for = 0 and 3000 s, \[\begin{align*} \textrm{slope} &=\dfrac{\ln[\mathrm{N_2O_5}]_{3000}-\ln[\mathrm{N_2O_5}]_0}{3000\textrm{ s}-0\textrm{ s}} \\[4pt] &=\dfrac{(-4.756)-(-3.310)}{3000\textrm{ s}} \\[4pt] &=-4.820\times10^{-4}\textrm{ s}^{-1} \end{align*} \] Thus \(k = 4.820 \times 10^{−4}\, s^{−1}\). 1,3-Butadiene (CH =CH—CH=CH ; C H ) is a volatile and reactive organic molecule used in the production of rubber. Above room temperature, it reacts slowly to form products. Concentrations of C H as a function of time at 326°C are listed in the following table along with ln[C H ] and the reciprocal concentrations. Graph the data as concentration versus , ln concentration versus , and 1/concentration versus . Then determine the reaction order in C H , the rate law, and the rate constant for the reaction. second order in C H ; rate = [C H ] ; = 1.3 × 10 M ·s For a zeroth-order reaction, a plot of the concentration of any reactant versus time is a straight line with a slope of − . For a first-order reaction, a plot of the natural logarithm of the concentration of a reactant versus time is a straight line with a slope of − . For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of .
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Named for the mineral zircon in which it can be found, zirconium was discovered in 1789 by Klaproth and eventually isolated in 1824 by Berzelius. The metal reacts with oxygen and nitrogen in the atmosphere to form a protective coating that inhibits further corrosion. It is resistant to weak acids and even forms a low-temperature superconductor when alloyed with niobium. Zirconium finds applications in industry which suit its high corrosion resistance and strength. It is very similar to the less plentiful Hafnium (see below) and the two are very difficult to separate. Most samples of either are contaminated with small amounts of the other element.
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Lewis’ octet theory correctly predicts formulas for nearly all ionic compounds and accounts for most covalent bonding schemes for biologically important molecules. But the ones that are exceptions are among the most biologically interesting compounds. Lewis’ theory concentrates on resemblances to noble-gas valence octets. Therefore it is most successful in accounting for formulas of compounds of the representative elements (especially in periods 1 and 2), whose distinguishing electrons are also and electrons. The octet rule is much less useful in dealing with compounds of the transition elements or inner transition elements, most of which involve some participation of or orbitals in bonding. Even among the representative elements there are some interesting exceptions to the Lewis theory. These fall mainly into three categories: The typical structural formula for ATP (adenosine triphosphate) shows three single bonds and one double bond on the phosphorus atom, for a total of 5 electron pairs. This is possible because P has 3d orbitals as well as 3s and 3p orbitals to accommodate the electrons. The bonding is shown in the simple phosphate ion [PO ] : Other examples of molecules with more than an octet of electrons are phosphorus pentafluoride (PF ) and sulfur hexafluoride (SF ). Phosphorus pentafluoride is a gas at room temperature. It consists of PF molecules in which each fluorine atom is bonded to the phosphorus atom. Since each bond corresponds to a shared pair of electrons, the Lewis structure isExpansion of the valence shell is impossible for an atom in the second period because there is no such thing as a 2 orbital. The valence ( = 2) shell of nitrogen, for example, consists of the 2 and 2 subshells only. Thus nitrogen can form NF (in which nitrogen has an octet) but not NF . Phosphorus, on the other hand, forms both PF and PF , the latter involving expansion of the valence shell to include part of the 3 subshell.   Instead of an octet the phosphorus atom has 10 electrons in its valence shell. Sulfur hexafluoride (also a gas) consists of SF molecules. Its structure is   Here the sulfur atom has six electron pairs in its valence shell. Another exception to the octet rule occurs in molecules called . These molecules contain at least one unpaired electron, a clear violation of the octet rule. Free radicals play many important roles a wide range of applied chemistry fields, including biology and medicine. Nitric oxide, known as the 'endothelium-derived relaxing factor', or 'EDRF', is synthesized in the body from arginine and oxygen by various enzymes and by reduction of inorganic nitrate. It is released by the lining of blood vessels, and causes the surrounding muscle to relax, increasing blood flow. During World War I, soldiers who handled nitroglycerin explosives were found to have low blood pressure. This led to the discovery that nitroglycerine serves as a vasodilator because it is converted to nitric oxide in the body. Sildenafil, popularly known by the trade name Viagra, stimulates erections primarily by enhancing signaling through the nitric oxide pathway in the penis. Nitric oxide therapy may in some cases save the lives of infants at risk for pulmonary vascular disease. As expected for a non-octet compound, nitric oxide is highly reactive (having a lifetime of a few seconds), yet diffuses freely across membranes. Appropriate levels of NO production are important in protecting an organ such as the liver from ischemic (constriction of blood vessels) damage. | Nitrogen dioxide, NO is another free radical which is the brown, toxic component of smog. It partially dimerizes to attain a Lewis octet: Free radical scavengers may be important anti-aging biomolecules. Free radicals are generated by several natural processes in the body, and if they aren't destroyed immediately, they can attack DNA and other critical molecules. It has been claimed that α-tocopherol (Vitamin E) rapidly accepts electrons from free radicals produced by the lipid peroxidation chain reaction, becoming the | Vitamin E Free Radical. and protecting DNA. Note the similarity to the 7 electron , one of the damaging (but short lived) free radicals of the body: The Lewis structure for Oxygen usually hides the fact that it is a , containing two unpaired electrons. This is sometimes cited as a serious flaw in Lewis bond theory, and was a major impetus for development of molecular orbital theory. The paramagnetic character of oxygen was mentioned before, and is easily demonstrated by attraction of oxygen to an external magnet. The elements beryllium, boron, and aluminum are notorious for forming "electron deficient" compounds (with fewer than the typical 8 valence electrons). This gives them special properties in synthetic biochemical and organic chemistry, but they almost always react to form species that obey the Lewis octet rule in biochemical sytems. A good example is Beryllium dichloride, BeCl , which melts at 405°C and boils at 520°C. That compares with 714°C and 1412°C for magnesium chloride. Individual BeCl molecules exist only in the gas phase above 750°C: Instead of an octet, the valence shell of Be contains only electron pairs. It is reactive because it does not have the octet, and upon cooling, forms molecules that do, where Cl atoms donate a pair of their non-bonding electrons to a neighboring Be atom so that they bridge two Be atoms, satisfying the valence of both: But the extremely high toxicity of beryllium is probably due to ionic species, and the [pillsm.com/?a=7539 | Blood Beryllium Lymphocyte Proliferation Test] uses beryllium sulfate, a typical ionic compound of Be (aq), or Be(H ) : , to mimic the biological form of beryllium. Beryllium may also acquire and octet by formation of "coordinate covalent bonds", accepting both electrons in the bond from another species. Thus BeCl reacts with Cl ions or OH ions under normal conditions to form BeCl or Be(OH) :   Similar arguments can be applied to boron trichloride, BCl , which is a stable gas at room temperature. We are forced to write its structure as   in which the valence shell of boron has only three pairs of electrons. Molecules such as BeCl and BCl are referred to as because some atoms do not have complete octets. BCl reacts with NH in the following way: Again, the most stable boron compounds under normal conditions follow the Lewis rule: Boric acid, H BO is traditionally used as an insecticide, notably against ants, fleas, and cockroaches, but it is as innocuous as salt to Humans, as are borates, which contain the BO ion. Aluminum Chloride is a low melting covalent compound which exists as AlCl only in the gas phase. Because it has only 6 electrons in it's Lewis structure, it reacts with electron pair donors like water:
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The early literature concerning the addition of hydrogen bromide to unsymmetrical alkenes at best is confused. Sometimes the same alkene was reported to give addition both according to, and in opposition to, the principles discussed for electrophilic ionic addition ( ). Much of the uncertainty on the addition of hydrogen bromide was removed by the classical researches of M. S. Kharasch and F. R. Mayo (1933) who showed that there must be two reaction mechanisms, each giving a different product. Kharasch and Mayo found, in the presence of radical inhibitors, hydrogen bromide adds to propene in a rather slow reaction to give pure 2-bromopropane: With light, peroxides, radical initiators, and in the absence of radical inhibitors a rapid radical-chain addition of hydrogen bromide occurs to yield \(80\%\) or more of 1-bromopropane: Similar effects have been noted occasionally with hydrogen chloride, but never with hydrogen iodide or hydrogen fluoride. A few substances apparently add to alkenes only by radical mechanisms, and always add in the opposite way to that expected for electrophilic ionic addition. The ionic addition of hydrogen bromide was discussed in and will not be considered further at this point. Two questions with regard to the so-called will be given special attention. Why does the radical mechanism give a product of different structure than the ionic addition? Why does the radical addition occur readily with hydrogen bromide but rarely with the other hydrogen halides? The abnormal addition of hydrogen bromide is catalyzed strongly by peroxides, which have the structure \(\ce{R-O-O-R}\) and decompose thermally to give \(\ce{RO} \cdot\) radicals (see ): The \(\ce{RO} \cdot\) radicals can react with hydrogen bromide in two ways, to abstract either hydrogen atoms or bromine atoms: Clearly, the formation of \(\ce{ROH}\) and a bromine atom is energetically more favorable. The overall process of decomposition of peroxide and attack on hydrogen bromide, which results in the formation of a bromine atom, can initiate a radical-chain addition of hydrogen bromide to an alkene. The two chain-propagating steps, taken together, are exothermic by \(16 \: \text{kcal}\) and have a fairly reasonable energy balance between the separate steps. The reaction chains apparently are rather long, because the addition is strongly inhibited by radical traps and only traces of peroxide catalyst are needed. The direction of addition of hydrogen bromide to propene clearly depends on which end of the double bond the bromine atom attacks. The important question is which of the two possible carbon radicals that may be formed is the more stable, the 1-bromo-2-propyl radical, \(5\), or the 2-bromo-1-propyl radical, \(6\): From \(\ce{C-H}\) bond-dissociation energies of alkanes (see Table 5-6), the ease of formation and stabilities of the carbon radicals is seen to follow the sequence \(>\) \(>\) . By analogy, the 1-bromo-2-propyl radical, \(5\), is expected to be more stable and more easily formed than the 2-bromo-1-propyl radical, \(6\). The product of radical addition should be, and indeed is, 1-bromopropane: Other reagents, such as the halogens, also can add to alkenes and alkynes by both radical-chain and ionic mechanisms. Radical addition usually is initiated by light, whereas ionic addition is favored by low temperatures and no light. Nevertheless, it often is difficult to keep both mechanisms from operating at the same time. This is important even when the alkene is symmetrical because, although the adduct will then have the same structural formula regardless of mechanism, the stereochemical configurations may differ. Electrophilic addition of halogens generally is a stereospecific antarafacial addition, but radical-chain additions are less stereospecific. There are many reagents that add to alkenes only by radical-chain mechanisms. A number of these are listed in Table 10-3. They have in common a relatively weak bond, \(\ce{X-Y}\), that can be cleaved homolytically either by light or by chemical initiators such as peroxides. In the propagation steps, the radical that attacks the double bond does so to produce the more stable carbon radical. For addition to simple alkenes and alkynes, the more stable carbon radical is the one with the fewest hydrogens or the most alkyl groups at the radical center. The principles of radical addition reactions of alkenes appear to apply equally to alkynes, although there are fewer documented examples of radical additions to triple bonds. Two molecules of hydrogen bromide can add to propyne first to give -1-bromopropene (by antarafacial addition) and then 1,2-dibromopropane: and (1977)
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An electrolyte solution is a solution that generally contains ions, atoms or molecules that have lost or gained electrons, and is electrically conductive. For this reason they are often called ionic solutions, however there are some cases where the electrolytes are not ions. For this discussion we will only consider solutions of ions. A basic principle of electrostatics is that opposite charges attract and like charges repel. It also takes a great deal of force to overcome this electrostatic attraction. The general form of Coulomb's law describes the force of attraction between charges: \[F=k\frac{q_1mq_2}{r^2}\] However, we must make some changes to this physics formula to be able to use it for a solution of oppositely charged ions. In Coulomb's Law, the constant \[k=\frac{1}{4\pi\varepsilon_{0}}\], where \(\varepsilon_{0}\) is the permittivity of free space, such as in a vacuum. However, since we are looking at a solution, we must consider the effect that the medium (the solvent in this case) has on the electrostatic force, which is represented by the dielectric constant \(\varepsilon\): \[F=\frac{q_{1}q_{2}}{4\pi\varepsilon_{0}\varepsilon r^{2}}\] Polar substances such as water have a relatively high dielectric constant. Ions are not stable on their own, and thus no ions can ever be studied separately. Particularly in biology, all ions in a certain cell or tissue have a counterion that balances this charge. Therefore, we cannot measure the enthalpy or entropy of a single ion as we can atoms of a pure element. So we define a reference point. The \(\Delta_{f}\overline{H}^{\circ} \) of a hydrogen ion \(H^+\) is equal to zero, as are the other thermodynamic quantities. \[\Delta_{f}\overline{H}^{\circ}[H^{+}(aq)]=0\] \[\Delta_{f}\overline{G}^{\circ}[H^{+}(aq)]=0\] \[\overline{S}^{\circ}[H^{+}(aq)]=0\] When studying the formation of ionic solutions, the most useful quantity to describe is chemical potential \(\mu\), defined as the partial molar Gibbs energy of the ith component in a substance: \[\mu_{i}=\overline{G}_{i}=\left(\frac{\partial G}{\partial n_{i}}\right)_{T,P,n_{j}}=\mu_{i}^{\circ}+RT\ln x_{i}\] where \(x_{i}\) can be any unit of concentration of the component: mole fraction, molality, or for gases, the partial pressure divided by the pressure of pure component. To express the chemical potential of an electrolyte in solution in terms of molality, let us use the example of a dissolved salt such as magnesium chloride, \(MgCl_{2}\). \[MgCl_{2}\rightleftharpoons Mg^{2+}+2Cl^{-} \label{1}\] We can now write a more general equation for a dissociated salt: \[M_{\nu+}X_{\nu-}\rightleftharpoons\nu_{+}M^{z+}+\nu_{-}X^{z-} \label{2} \] where \(\nu_{\pm}\) represents the stoichiometric coefficient of the cation or anion and \(z_\pm\) represents the charge, and M and X are the metal and halide, respectively. The total chemical potential for these anion-cation pair would be the sum of their individual potentials multiplied by their stoichiometric coefficients: \[\mu=\nu_{+}\mu_{+}+\nu_{-}\mu_{-} \label{3} \] The chemical potentials of the individual ions are: \[\mu_{+} = \mu_+^{\circ}+RT\ln m_+ \label{4} \] \[\mu_{-} = \mu_-^{\circ}+RT\ln m_- \label{5} \] And the molalities of the individual ions are related to the original molality of the salt m by their stoichiometric coefficients \[m_{+}=\nu_{+}m\] Substituting Equations \(\ref{4}\) and \(\ref{5}\) into Equation \(\ref{3}\), \[ \mu=\left( \nu_+\mu_+^{\circ}+\nu_- mu_-^{\circ}\right)+RT\ln\left(m_+^{\nu+}m_-^{\nu-}\right) \label{6} \] since the total number of moles \(\nu=\nu_{+}+\nu_{-}\), we can define the mean ionic molality as the geometric average of the molarity of the two ions: \[ m_{\pm}=(m_+^{\nu+}m_-^{\nu-})^{\frac{1}{\nu}}\] then Equation \(\ref{6}\) becomes \[\mu=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+\nu RT\ln m_{\pm} \label{7} \] We have derived this equation for a ideal solution, but ions in solution exert electrostatic forces on one another to deviate from ideal behavior, so instead of molarities we must use the activity a to represent how the ion is behaving in solution. Therefore the mean ionic activity is defined as \[a_{\pm}=(a_{+}^{\nu+}+a_{-}^{\nu-})^{\frac{1}{\nu}}\] where \[a_{\pm}=\gamma m_{\pm} \label{mean}\] and \(\gamma_{\pm}\) is the , which is dependent on the substance. Substituting the mean ionic activity of \Equation \(\ref{mean}\) into Equation \(\ref{7}\), \[\mu=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+\nu RT\ln a_{\pm}=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+RT\ln a_{\pm}^{\nu}=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+RT \ln a \label{11}\] when \(a=a_{\pm}^{\nu}\). Equation \(\ref{11}\) then represents the chemical potential of a nonideal electrolyte solutions. To calculate the mean ionic activity coefficient requires the use of the Debye-Hückel limiting law, part of the . Let us now write out the chemical potential in terms of molality of the salt in our first example, \(MgCl_{2}\). First from Equation \(\ref{1}\), the stoichiometric coefficients of the ions are: \[\nu_{+} = 1,\nu_{-} = 2,\nu\; = 3 \nonumber\] The mean ionic molality is \[\begin{align*} m_{\pm} &= (m_{+}^{1}m_{-}^{2})^{\frac{1}{3}} \\[4pt] &= (\nu_{+}m\times\nu_{-}m)^{\frac{1}{3}} \\[4pt] &=m(1^{1}2^{2})^{\frac{1}{3}} \\[4pt] &=1.6\, m \end{align*}\] The expression for the chemical potential of \[MgCl_{2}\] is \[\mu_{MgCl_{2}}=\mu_{MgCl_{2}}^{\circ}+3RT\ln 1.6\m m \nonumber\]
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Secondary structure refers to the shape of a folding protein due exclusively to hydrogen bonding between its backbone amide and carbonyl groups. Secondary structure does not include bonding between the R-groups of amino acids, hydrophobic interactions, or other interactions associated with tertiary structure. The two most commonly encountered secondary structures of a polypeptide chain are α-helices and beta-pleated sheets. These structures are the first major steps in the folding of a polypeptide chain, and they establish important topological motifs that dictate subsequent tertiary structure and the ultimate function of the protein. Thumbnail: Structure of human hemoglobin. The proteins α and β subunits are in red and blue, and the iron-containing heme groups in green. (CC BY- ; ).
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In , you learned that the integrated rate law for each common type of reaction (zeroth, first, or second order in a single reactant) can be plotted as a straight line. Using these plots offers an alternative to the methods described for showing how reactant concentration changes with time and determining reaction order. We will illustrate the use of these graphs by considering the thermal decomposition of NO gas at elevated temperatures, which occurs according to the following reaction: \( 2NO_{2}\left ( g \right ) \overset{\Delta}{\rightarrow} 2NO\left ( g \right )+O_{2}\left ( g \right ) \tag{14.4.1}\) Experimental data for this reaction at 330°C are listed in ; they are provided as [NO ], ln[NO ], and 1/[NO ] versus time to correspond to the integrated rate laws for zeroth-, first-, and second-order reactions, respectively. The actual concentrations of NO are plotted versus time in part (a) in . Because the plot of [NO ] versus is not a straight line, we know the reaction is not zeroth order in NO . A plot of ln[NO ] versus (part (b) in ) shows us that the reaction is not first order in NO because a first-order reaction would give a straight line. Having eliminated zeroth-order and first-order behavior, we construct a plot of 1/[NO ] versus (part (c) in ). This plot is a straight line, indicating that the reaction is second order in NO . We have just determined the reaction order using data from a single experiment by plotting the concentration of the reactant as a function of time. Because of the characteristic shapes of the lines shown in , the graphs can be used to determine the reaction order of an unknown reaction. In contrast, the method described in required multiple experiments at different NO concentrations as well as accurate initial rates of reaction, which can be difficult to obtain for rapid reactions. Dinitrogen pentoxide (N O ) decomposes to NO and O at relatively low temperatures in the following reaction: \( 2N_{2}O_{5}\left ( soln \right ) \rightarrow 4NO_{2}\left ( soln \right )+O_{2}\left ( g \right ) \) This reaction is carried out in a CCl solution at 45°C. The concentrations of N O as a function of time are listed in the following table, together with the natural logarithms and reciprocal N O concentrations. Plot a graph of the concentration versus , ln concentration versus , and 1/concentration versus and then determine the rate law and calculate the rate constant. balanced chemical equation, reaction times, and concentrations graph of data, rate law, and rate constant Use the data in the table to separately plot concentration, the natural logarithm of the concentration, and the reciprocal of the concentration (the vertical axis) versus time (the horizontal axis). Compare the graphs with those in to determine the reaction order. Write the rate law for the reaction. Using the appropriate data from the table and the linear graph corresponding to the rate law for the reaction, calculate the slope of the plotted line to obtain the rate constant for the reaction. Here are plots of [N O ] versus , ln[N O ] versus , and 1/[N O ] versus : The plot of ln[N O ] versus gives a straight line, whereas the plots of [N O ] versus and 1/[N O ] versus do not. This means that the decomposition of N O is first order in [N O ]. The rate law for the reaction is therefore \( rate = k \left [N_{2}O_{5} \right ]\) Calculating the rate constant is straightforward because we know that the slope of the plot of ln[A] versus for a first-order reaction is − . We can calculate the slope using any two points that lie on the line in the plot of ln[N O ] versus . Using the points for = 0 and 3000 s, \( slope= \dfrac{ln\left [N_{2}O_{5} \right ]_{3000}-ln\left [N_{2}O_{5} \right ]_{0}}{3000\;s-0\;s} = \dfrac{\left [-4.756 \right ]-\left [-3.310 \right ]}{3000\;s} =4.820\times 10^{-4}\;s^{-1} \) Thus = 4.820 × 10 s . Exercise 1,3-Butadiene (CH =CH—CH=CH C H ) is a volatile and reactive organic molecule used in the production of rubber. Above room temperature, it reacts slowly to form products. Concentrations of C H as a function of time at 326°C are listed in the following table along with ln[C H ] and the reciprocal concentrations. Graph the data as concentration versus , ln concentration versus , and 1/concentration versus . Then determine the reaction order in C H , the rate law, and the rate constant for the reaction. second order in C H ; rate = [C H ] ; = 1.3 × 10 M ·s For a zeroth-order reaction, a plot of the concentration of any reactant versus time is a straight line with a slope of − . For a first-order reaction, a plot of the natural logarithm of the concentration of a reactant versus time is a straight line with a slope of − . For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of . Compare first-order differential and integrated rate laws with respect to the following. Is there any information that can be obtained from the integrated rate law that cannot be obtained from the differential rate law? In the single-step, second-order reaction 2A → products, how would a graph of [A] versus time compare to a plot of 1/[A] versus time? Which of these would be the most similar to the same set of graphs for A during the single-step, second-order reaction A + B → products? Explain. For reactions of the same order, what is the relationship between the magnitude of the rate constant and the reaction rate? If you were comparing reactions with different orders, could the same arguments be made? Why? The reaction rate increases as the rate constant increases. We cannot directly compare reaction rates and rate constants for reactions of different orders because they are not mathematically equivalent. One method of using graphs to determine reaction order is to use relative rate information. Plotting the log of the relative rate versus log of relative concentration provides information about the reaction. Here is an example of data from a zeroth-order reaction: Varying [A] does not alter the reaction rate. Using the relative rates in the table, generate plots of log(rate) versus log(concentration) for zeroth-, first- and second-order reactions. What does the slope of each line represent? The table below follows the decomposition of N O gas by examining the partial pressure of the gas as a function of time at 45°C. What is the reaction order? What is the rate constant? How long would it take for the pressure to reach 105 mmHg at 45°C?
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Tartaric acid (TA) is a byproduct of wine production. This organic acid and its salts are used in foods such as fruit jellies, preserves, jams, baked goods, and confections. TA is hardly metabolized and degraded by yeast and spoilage bacteria providing microbiological stability to foods that contain it. In addition to its use as antimicrobial and acidulant, TA and its salts are used as emulsifiers, leavening, and anticacking agents. TA is also used in the beverage industry and has non-food uses in textile coloring, galvanizing, and mirror production. The increasing popularity in wine consumption in recent years has resulted in the increase of waste from wine making practices. One liter of white wine generates the same amount of water pollution as 3 people in one day. Waste-waters from wine production contain biodegradable compounds and fruit suspended solids; their treatment is of great importance because their high pollutant activity. Moreover, waste treatment is of economic interest because the organic compounds present in waste from the wine making process can have value as additives, ingredients, and substrates in the food and pharmaceutical industries. Vintage process Grape marc Vine shoots Vine pruning Grapes and yeast Tartaric Acid separated from grape juice Tartaric Acid crystals In the method described above, addition of calcium ions prompts the precipitation of calcium tartrate. Once tartaric acid is removed from the solution as tartrate, it will partially dissociate into calcium and tartrate ions establishing the following equilibrium with the solid salt: \(Q=\text{(}c_{\text{Ca}^{2+}}\text{)(}c_{\text{C}_{4}\text{H}_{4}\text{O}_{6}^{2-}})\) (1) \(\text{Q}>{K}_{sp}\,\) precipitation occurs while if \(\text{Q}<{K}_{sp}\,\) no precipitation occurs Decide whether CaC O , calcium oxalate, will precipitate or not when (a)100 cm of 0.02 CaCl and 100 cm of 0.02 Na C O are mixed, and also when (b) 100 cm of 0.0001 CaCl and 1000 cm of 0.0001 Na C O are mixed. = 2.32 × 10 mol dm . Since is larger than (2.32 × 10 mol dm ), precipitation will occur. thus \(\begin{align}{Q}&={c}_{\text{Ca}^{2+}}\times {c}_{\text{C}_{2}\text{O}_{4}^{2-}}\\ \text{ }&=(\text{9.09} \times \text{10}^{-6}\text{mol dm}^{-3}) (\text{9.09} \times \text{10}^{-5} \text{mol dm}^{-3})\\ \text{ }&=\text{8.26} \times \text{10}^{-10}\text{mol}^{2} \text{dm}^{-6}\end{align}\) Calculate the mass of CaC O precipitated when 100 cm of 0.0200 CaCl and 100 cm of 0.0200 Na C O are mixed together. In in part of the previous example we determined that precipitation does actually occur. In order to find how much calcium oxalate is precipitated, we must concentrate on the amount of each species. Since 100 cm of 0.02 CaCl was used, we have \(n_{\text{Ca}^{2+}}=\text{0.0200 }\frac{\text{mmol}}{\text{cm}^{3}}\times \text{ 100 cm}^{3}=\text{2.00 mmol}\) Thus \({K}_{sp} = [\text{Ca}^{2+},\text{C}_{2}\text{O}_{4}^{2-}]\,\) Ed Vitz (Kutztown University), (University of
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The dependence of the extensive properties of a system (\(V\), \(E\), \(H\), \(S\), \(G\), etc.) on its composition is given by what are called . For a general extensive property \(F\) the partial molar value with respect to component \(i\) is given by \[ \bar{F}_i = \left( \dfrac{\partial F}{\partial n_i} \right)_{T,P,n_{j \neq i}} \label{1}\] In other words, \( \bar{F}_i\) equals the partial derivative of \(F\) with respect to the number of moles of \(i\) with temperature, pressure, and the number of moles of all other components held constant. If temperature and pressure are held constant, then \[ dF = \sum_i \bar{F}_i dn_i \label{2}\] In this exercise we will be looking at the volume of a two-component system consisting of water (w) and an ionic solute (s). In this case Equation \ref{2} becomes \[ dV = \bar{V}_s dn_s + \bar{V}_w dn_w \label{3}\] where \[ \bar{V}_s = \left( \dfrac{\partial F}{\partial n_s} \right)_{T,P,n_{w}} \label{4}\] and \[ \bar{V}_w = \left( \dfrac{\partial F}{\partial n_w} \right)_{T,P,n_{s}} \label{5}\] The integration of Equation \ref{3} at constant temperature and pressure from \(n_s\) and \(n_w\) equal to zero to some final values of \(n_s\) and \(n_w\), gives \[ V = \bar{V}_s n_s + \bar{V}_w n_w \label{6}\] Equation \ref{6} gives the impression that \(\bar{V}_s\) and \(\bar{V}_w\) are the volumes of one mole of the solute and water respectively; however Equations \ref{4} and \ref{5} show that this is not true. These equations show that \(\bar{V}_s\) and \(\bar{V}_w\) are the rates at which the volume of the whole solution changes as the moles of solute or of water are changed while temperature, pressure, and the moles of the other are held constant. Evidence that \(\bar{V}_s\) and \(\bar{V}_w\) are not volumes will appear as you do the exercise. Below are tables of densities and percent by mass of solute for aqueous solutions of three compounds: sodium chloride, sodium phosphate, and zinc sulfate. These data are taken from the CRC Handbook of Chemistry and Physics 66th edition (1985-1986). The data all apply to solutions at 20 ºC and 1 atm of pressure. Carry out the following steps for each of the three solutions (NOTE: sample calculations are given at the end of this exercise for the 0.4% NaCl solution): Below are the calculations relating to the 0.4% sodium chloride solution. The chosen mass of water is 1000 g. Step a: Mass of solution containing 1000 g water Step b: Mass of solute in the mass of solution calculated in step a Step c: Moles of solute corresponding to the mass calculated in step b Step d: Volume of solution containing 1000 g water
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Many of the physical properties of solutions differ significantly from those of the pure substances discussed in earlier chapters, and these differences have important consequences. For example, the limited temperature range of liquid water (0°C–100°C) severely limits its use. Aqueous solutions have both a lower freezing point and a higher boiling point than pure water. Probably one of the most familiar applications of this phenomenon is the addition of ethylene glycol (“antifreeze”) to the water in an automobile radiator. This solute lowers the freezing point of the water, preventing the engine from cracking in very cold weather from the expansion of pure water on freezing. Antifreeze also enables the cooling system to operate at temperatures greater than 100°C without generating enough pressure to explode. Changes in the freezing point and boiling point of a solution depend primarily on the number of solute particles present rather than the kind of particles. Such properties of solutions are called colligative properties (from the Latin colligatus, meaning “bound together” as in a quantity). As we will see, the vapor pressure and osmotic pressure of solutions are also colligative properties. When we determine the number of particles in a solution, it is important to remember that not all solutions with the same molarity contain the same concentration of solute particles. Consider, for example, 0.01 M aqueous solutions of sucrose, \(NaCl\), and \(\ce{CaCl_2}\). Because sucrose dissolves to give a solution of neutral molecules, the concentration of solute particles in a 0.01 M sucrose solution is 0.01 M. In contrast, both \(\ce{NaCl}\) and \(\ce{CaCl_2}\) are ionic compounds that dissociate in water to yield solvated ions. As a result, a 0.01 M aqueous solution of \(\ce{NaCl}\) contains 0.01 M Na+ ions and 0.01 M \(Cl^−\) ions, for a total particle concentration of 0.02 M. Similarly, the \(\ce{CaCl_2}\) solution contains 0.01 M \(Ca^{2+}\) ions and 0.02 M \(Cl^−\) ions, for a total particle concentration of 0.03 M.These values are correct for dilute solutions, where the dissociation of the compounds to form separately solvated ions is complete. At higher concentrations (typically >1 M), especially with salts of small, highly charged ions (such as \(Mg^{2+}\) or \(Al^{3+}\)), or in solutions with less polar solvents, dissociation to give separate ions is often incomplete. The sum of the concentrations of the dissolved solute particles dictates the physical properties of a solution. In the following discussion, we must therefore keep the chemical nature of the solute firmly in mind. Recall that the normal boiling point of a substance is the temperature at which the vapor pressure equals 1 atm. If a nonvolatile solute lowers the vapor pressure of a solvent, it must also affect the boiling point. Because the vapor pressure of the solution at a given temperature is less than the vapor pressure of the pure solvent, achieving a vapor pressure of 1 atm for the solution requires a higher temperature than the normal boiling point of the solvent. Thus the boiling point of a solution is always greater than that of the pure solvent. We can see why this must be true by comparing the phase diagram for an aqueous solution with the phase diagram for pure water (Figure \(\Page {1}\)). The vapor pressure of the solution is less than that of pure water at all temperatures. Consequently, the liquid–vapor curve for the solution crosses the horizontal line corresponding to P = 1 atm at a higher temperature than does the curve for pure water. The boiling point of a solution with a nonvolatile solute is always greater than the boiling point of the pure solvent. The magnitude of the increase in the boiling point is related to the magnitude of the decrease in the vapor pressure. As we have just discussed, the decrease in the vapor pressure is proportional to the concentration of the solute in the solution. Hence the magnitude of the increase in the boiling point must also be proportional to the concentration of the solute (Figure \(\Page {2}\)). We can define the boiling point elevation (\(ΔT_b\)) as the difference between the boiling points of the solution and the pure solvent: \[ΔT_b=T_b−T^0_b \label{eq1}\] where \(T_b\) is the boiling point of the solution and \(T^0_b\) is the boiling point of the pure solvent. We can express the relationship between \(ΔT_b\) and concentration as follows \[ΔT_b = mK_b \label{eq2}\] where m is the concentration of the solute expressed in molality, and \(K_b\) is the molal boiling point elevation constant of the solvent, which has units of °C/m. Table \(\Page {1}\) lists characteristic Kb values for several commonly used solvents. For relatively dilute solutions, the magnitude of both properties is proportional to the solute concentration. The concentration of the solute is typically expressed as molality rather than mole fraction or molarity for two reasons. First, because the density of a solution changes with temperature, the value of molarity also varies with temperature. If the boiling point depends on the solute concentration, then by definition the system is not maintained at a constant temperature. Second, molality and mole fraction are proportional for relatively dilute solutions, but molality has a larger numerical value (a mole fraction can be only between zero and one). Using molality allows us to eliminate nonsignificant zeros. According to Table \(\Page {1}\), the molal boiling point elevation constant for water is 0.51°C/m. Thus a 1.00 m aqueous solution of a nonvolatile molecular solute such as glucose or sucrose will have an increase in boiling point of 0.51°C, to give a boiling point of 100.51°C at 1.00 atm. The increase in the boiling point of a 1.00 m aqueous \(\ce{NaCl}\) solution will be approximately twice as large as that of the glucose or sucrose solution because 1 mol of \(\ce{NaCl}\) produces 2 mol of dissolved ions. Hence a 1.00 m \(\ce{NaCl}\) solution will have a boiling point of about 101.02°C. In Example \(\Page {1}\), we calculated that the vapor pressure of a 30.2% aqueous solution of ethylene glycol at 100°C is 85.1 mmHg less than the vapor pressure of pure water. We stated (without offering proof) that this should result in a higher boiling point for the solution compared with pure water. Now that we have seen why this assertion is correct, calculate the boiling point of the aqueous ethylene glycol solution. : composition of solution : boiling point : Calculate the molality of ethylene glycol in the 30.2% solution. Then use Equation \ref{eq2} to calculate the increase in boiling point. : From Example \(\Page {1}\), we know that a 30.2% solution of ethylene glycol in water contains 302 g of ethylene glycol (4.87 mol) per 698 g of water. The molality of the solution is thus \[\text{molality of ethylene glycol}= \left(\dfrac{4.87 \;mol}{698 \; \cancel{g} \;H_2O} \right) \left(\dfrac{1000\; \cancel{g}}{1 \;kg} \right)=6.98 m\] From Equation \ref{eq2}, the increase in boiling point is therefore \[ΔT_b=mK_b=(6.98 \cancel{m})(0.51°C/\cancel{m})=3.6°C\] The boiling point of the solution is thus predicted to be 104°C. With a solute concentration of almost 7 m, however, the assumption of a dilute solution used to obtain Equation \ref{eq2} may not be valid. Assume that a tablespoon (5.00 g) of \(\ce{NaCl}\) is added to 2.00 L of water at 20.0°C, which is then brought to a boil to cook spaghetti. At what temperature will the water boil? 100.04°C, or 100°C to three significant figures. (Recall that 1 mol of \(\ce{NaCl}\) produces 2 mol of dissolved particles. The small increase in temperature means that adding salt to the water used to cook pasta has essentially no effect on the cooking time.) The phase diagram in Figure \(\Page {1}\) shows that dissolving a nonvolatile solute in water not only raises the boiling point of the water but also lowers its freezing point. The solid–liquid curve for the solution crosses the line corresponding to P = 1 atm at a lower temperature than the curve for pure water. This phenomenon is exploited in “de-icing” schemes that use salt (Figure \(\Page {3}\)), calcium chloride, or urea to melt ice on roads and sidewalks, and in the use of ethylene glycol as an “antifreeze” in automobile radiators. Seawater freezes at a lower temperature than fresh water, and so the Arctic and Antarctic oceans remain unfrozen even at temperatures below 0 °C (as do the body fluids of fish and other cold-blooded sea animals that live in these oceans). We can understand this result by imagining that we have a sample of water at the normal freezing point temperature, where there is a dynamic equilibrium between solid and liquid. Water molecules are continuously colliding with the ice surface and entering the solid phase at the same rate that water molecules are leaving the surface of the ice and entering the liquid phase. If we dissolve a nonvolatile solute such as glucose in the liquid, the dissolved glucose molecules will reduce the number of collisions per unit time between water molecules and the ice surface because some of the molecules colliding with the ice will be glucose. Glucose, though, has a very different structure than water, and it cannot fit into the ice lattice. Consequently, the presence of glucose molecules in the solution can only decrease the rate at which water molecules in the liquid collide with the ice surface and solidify. Meanwhile, the rate at which the water molecules leave the surface of the ice and enter the liquid phase is unchanged. The net effect is to cause the ice to melt. The only way to reestablish a dynamic equilibrium between solid and liquid water is to lower the temperature of the system, which decreases the rate at which water molecules leave the surface of the ice crystals until it equals the rate at which water molecules in the solution collide with the ice. By analogy to our treatment of boiling point elevation,the freezing point depression (\(ΔT_f\)) is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution: \[ ΔT_f=T^0_f−T_f \label{eq3}\] where The order of the terms is reversed compared with Equation \ref{eq1} to express the freezing point depression as a positive number. The relationship between \(ΔT_f\) and the solute concentration is given by an equation analogous to Equation \ref{eq2}: \[ΔT_f = mK_f \label{eq4}\] where Like \(K_b\), each solvent has a characteristic value of \(K_f\) (Table \(\Page {1}\)). Freezing point depression depends on the total number of dissolved nonvolatile solute particles, just as with boiling point elevation. Thus an aqueous \(\ce{NaCl}\) solution has twice as large a freezing point depression as a glucose solution of the same molality. People who live in cold climates use freezing point depression to their advantage in many ways. For example, ethylene glycol is added to engine coolant water to prevent an automobile engine from being destroyed, and methanol is added to windshield washer fluid to prevent the fluid from freezing. Heated glycols are often sprayed onto the surface of airplanes prior to takeoff in inclement weather in the winter to remove ice that has already formed and prevent the formation of more ice, which would be particularly dangerous if formed on the control surfaces of the aircraft ( \(\Page {1}\)). The decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus a pure liquid all depend on the total number of dissolved nonvolatile solute particles. In colder regions of the United States, \(\ce{NaCl}\) or \(\ce{CaCl_2}\) is often sprinkled on icy roads in winter to melt the ice and make driving safer. Use the data in Figure 13.9 to estimate the concentrations of two saturated solutions at 0°C, one of \(\ce{NaCl}\) and one of \(\ce{CaCl_2}\), and calculate the freezing points of both solutions to see which salt is likely to be more effective at melting ice. : solubilities of two compounds : concentrations and freezing points : : From Figure 13.9, we can estimate the solubilities of \(\ce{NaCl}\) and \(\ce{CaCl_2}\) to be about 36 g and 60 g, respectively, per 100 g of water at 0°C. The corresponding concentrations in molality are \[m_{\ce{NaCl}}=\left(\dfrac{36 \; \cancel{g \;NaCl}}{100 \;\cancel{g} \;H_2O}\right)\left(\dfrac{1\; mol\; NaCl}{58.44\; \cancel{ g\; NaCl}}\right)\left(\dfrac{1000\; \cancel{g}}{1\; kg}\right)=6.2\; m\] \[m_{\ce{CaCl_2}}=\left(\dfrac{60\; \cancel{g\; CaCl_2}}{100\;\cancel{g}\; H_2O}\right)\left(\dfrac{1\; mol\; CaCl_2}{110.98\; \cancel{g\; CaCl_2}}\right)\left(\dfrac{1000 \;\cancel{g}}{1 kg}\right)=5.4\; m\] The lower formula mass of \(\ce{NaCl}\) more than compensates for its lower solubility, resulting in a saturated solution that has a slightly higher concentration than \(\ce{CaCl_2}\). B Because these salts are ionic compounds that dissociate in water to yield two and three ions per formula unit of \(\ce{NaCl}\) and \(\ce{CaCl_2}\), respectively, the actual concentrations of the dissolved species in the two saturated solutions are 2 × 6.2 m = 12 m for \(\ce{NaCl}\) and 3 × 5.4 m = 16 m for \(\ce{CaCl_2}\). The resulting freezing point depressions can be calculated using Equation \(\Page {4}\): \[\ce{NaCl}: ΔT_f=mK_f=(12\; \cancel{m})(1.86°C/\cancel{m})=22°C\] \[\ce{CaCl2}: ΔT_f=mK_f=(16\;\cancel{m})(1.86°C/\cancel{m})=30°C\] Because the freezing point of pure water is 0°C, the actual freezing points of the solutions are −22°C and −30°C, respectively. Note that \(\ce{CaCl_2}\) is substantially more effective at lowering the freezing point of water because its solutions contain three ions per formula unit. In fact, \(\ce{CaCl_2}\) is the salt usually sold for home use, and it is also often used on highways. Because the solubilities of both salts decrease with decreasing temperature, the freezing point can be depressed by only a certain amount, regardless of how much salt is spread on an icy road. If the temperature is significantly below the minimum temperature at which one of these salts will cause ice to melt (say −35°C), there is no point in using salt until it gets warmer Calculate the freezing point of the 30.2% solution of ethylene glycol in water whose vapor pressure and boiling point we calculated in Example \(\Page {6}\).8 and Example \(\Page {6}\).10. −13.0°C Arrange these aqueous solutions in order of decreasing freezing points: 0.1 m \(KCl\), 0.1 m glucose, 0.1 m SrCl2, 0.1 m ethylene glycol, 0.1 m benzoic acid, and 0.1 m HCl. : molalities of six solutions relative freezing points : : Because the molal concentrations of all six solutions are the same, we must focus on which of the substances are strong electrolytes, which are weak electrolytes, and which are nonelectrolytes to determine the actual numbers of particles in solution. \(KCl\), \(SrCl_2\), and \(HCl\) are , producing two, three, and two ions per formula unit, respectively. Benzoic acid is a weak electrolyte (approximately one particle per molecule), and glucose and ethylene glycol are both nonelectrolytes (one particle per molecule). The molalities of the solutions in terms of the total particles of solute are: \(KCl\) and \(HCl\), 0.2 m; \(SrCl_2\), 0.3 m; glucose and ethylene glycol, 0.1 m; and benzoic acid, 0.1–0.2 m. Because the magnitude of the decrease in freezing point is proportional to the concentration of dissolved particles, the order of freezing points of the solutions is: glucose and ethylene glycol (highest freezing point, smallest freezing point depression) > benzoic acid > \(HCl\) = \(KCl\) > \(SrCl_2\). Arrange these aqueous solutions in order of increasing freezing points: 0.2 m \(NaCl\), 0.3 m acetic acid, 0.1 m \(\ce{CaCl_2}\), and 0.2 m sucrose. 0.2 m \(\ce{NaCl}\) (lowest freezing point) < 0.3 m acetic acid ≈ 0.1 m \(\ce{CaCl_2}\) < 0.2 m sucrose (highest freezing point) Boiling Point Elevation and Freezing Point Depression: Osmotic pressure and changes in freezing point, boiling point, and vapor pressure are directly proportional to the concentration of solute present. Consequently, we can use a measurement of one of these properties to determine the molar mass of the solute from the measurements. A solution of 4.00 g of a nonelectrolyte dissolved in 55.0 g of benzene is found to freeze at 2.32 °C. What is the molar mass of this compound? We can solve this problem using the following steps. A solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of 64.5 °C. What is the molar mass of this compound? 1.8 × 10 g/mol A 0.500 L sample of an aqueous solution containing 10.0 g of hemoglobin has an osmotic pressure of 5.9 torr at 22 °C. What is the molar mass of hemoglobin? Here is one set of steps that can be used to solve the problem: What is the molar mass of a protein if a solution of 0.02 g of the protein in 25.0 mL of solution has an osmotic pressure of 0.56 torr at 25 °C? 2.7 × 10 g/mol Finding the Molecular Weight of an Unknown using Colligative Properties: The colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as . Solutions that obey Raoult’s law are called ideal solutions. Most real solutions exhibit positive or negative deviations from Raoult’s law. The boiling point elevation (\(ΔT_b\)) and freezing point depression (\(ΔT_f\)) of a solution are defined as the differences between the boiling and freezing points, respectively, of the solution and the pure solvent. Both are proportional to the molality of the solute.
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https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.03%3A_The_Limiting_Reagent/3.3.01%3A_Cultural_Connections-_Anthropology_and_Protein_Stoichiometry
We have examined the idea of amino acid , where different foods in the diet supply the correct balance of amino acids to construct Human protein, in another section. Human proteins require stoichiometric amounts of about ten "essential" amino acids, so if only one is missing, the protein cannot be synthesized, and protein malnourishment results. This illustrates the concept of a , which is a reactant present in less than the stoichiometric amount relative to other reactants. A diverse group of surviving cultures worldwide have adopted diets, different as they may be, that supply the correct amino acid balance . Anthropologists believe that fortuitous adoption of these diets provided survival value for the cultures. In areas where food is scarce, cultures that do not adopt diets with correct amino acid balance may not survive. Here are some examples of diets that supply approximately the correct stoichiometric amounts of amino acids: In Latin America, diets combining corn tortillas or other corn products with beans, like the tostada shown here, are common. Tostada with corn tortilla, bean topping, and vegetables Beans eaten alone provide limiting amounts of sulfur containing amino acids like methionine and cysteine, so these amino acids limit the amount of Human protein that can be synthesized. Beans have large amounts of the amino acids lysine and tryptophan, so these are when used to synthesize Human protein, and will be degraded to urea and wasted. If wheat, rice, or corn are eaten alone, they typically provide amounts of lysine and tryptophan which limit the amount of Human protein that can be synthesized. But if beans are eaten with grains or corn, the of the beans the limiting reactants of the grain (and vice versa). Human protein can then be synthesized efficiently, and very small amounts of amino acids are simply excreted as urea. The origins of this diet are ancient. The Tehuacan Archaeological-Botanical Project (1961–1964) investigated the beginnings of agriculture in the New World and the concomitant rise of autochthonous (indigenous) civilization . The project is described in six volumes, the first of which explores subsistence and the environment . The domestication of maize after 8000 B.C.E. and beans (3000–4000 B.C.E.) in the Tehuacan Valley (the “Valley of Mexico”) of south central Mexico, gave inhabitants a nutritional advantage in the face of progressively deteriorating resources. Maize cob size increased rapidly under domestication from only a centimeter or two to current size, while Phaseolus vulgaris (kidney and pinto beans) similar to modern ones probably existed in preagricultural times. “Wherever these two food crops met, an immediate adaptive combination favored by human selection was formed.” . In the Middle East, a meal may consist of falafel and hummus made from chickpeas eaten with pita bread made from wheat. Pita with humus and felafel balls Dal (lintil soup) Bhat (grain or maize) In India, rice or chapatis are eaten with (lentils) to provide amino acid balance. We'll invesigate experiments dealing with corn, and the protein malnutrition that can result if it's eaten alone. It's well known that corn is a poor protein food by itself, being low in lysine and tryptophan , and even more so when it is nixtamalized. These deficiencies motivated researchers to develop the QPM (Quality Protein Maze) to increase concentrations of these essential amino acids in its protein. Experiments on rats, whose amino acid requirements are similar to Humans', provide a basis for some stoichiometric calculations . As the table below shows, there was little weight gain improvement when LYS or TRP were added separately to a "Base" diet. That means that neither is a limiting amino acid, preventing by itself the synthesis of rat protein. But when both were added, a significant increase was measured, along with a decrease in serum urea. That means that both were limiting, and when both were added, much less amino acid was wasted. The amino acids went into protein synthesis, rather than being simply metabolized and excreted as urea. While the addition of isoleucine in diet 5 made little difference (so it must be supplied adequately by corn), the addition of threonine (THR) alone to Diet 4 increased body weight gain, so it must be limiting in a diet of corn + LYS + TRP. Addition of ILE, methionine (MET), histidine (HIS) and valine (VAL) finally constitutes a nearly balanced diet, as evidenced by the large weight gains and low serum urea in Diet 8. The base diet consisted of 920.2 g corn, 30.0 g corn oil, 35.0 g mineral mix, 10 g vitamin mix, 2.5 g limestone, and 2.3 g choline. This large increase is due to the doubling of total dietary protein, rather than protein inefficiency. The same paper provides the sequence, from most limiting to least, of the amino acids in corn, and the requirement of the rat (similar to Human) published by the NRC . We've added the amino acid amounts in lentils to represent beans Amino acids are joined by a very simple "condensation" reaction where two molecules join by eliminating water, as shown in the figure below. The portion of the molecule labeled "R" stands for a variable part of the molecule that distinguishes different amino acids; the rest of the molecule is common to all amino acids.   Formation of a peptide bond This process can continue with different amino acids adding to either end of the protein chain. The H-N of an amino acid adds to the C-OH end of the protein, eliminating water (H-OH) and forming a C-N bond where the water was eliminated (or the C-OH end of an amino acid adds to the N-H end of the protein). This process continues until the protein has hundreds of different amino acids. a. Correct, or Stoichiometric Amino Acid Ratios: From data in the Table II above, calculate the optimal molar ratio of lysine (LYS, C H N O , Molar mass 146.19 g/mol) to Arginine (ARG, C H N O , Molar mass 174.2 g/mol) in the diet. b. Limiting Amino Acid: From data in the Table above, calculate which, LYS or ARG, is the in a corn diet? c. How much of the will be wasted in 1 kg of corn diet? a. The balanced equation will have coefficients that we need to determine: x LYS + y ARG + other amino acids → Protein + water According to the atomic theory, x mol LYS is required for each y mole of ARG to make useful protein. If the diet is optimal when 7.0 g of LYS is eaten with every 6.0 g of ARG (this ratio might be provided by eggs, for example), we can calculate the optimal stoichiometric ratio: \(\text{n}_{\text{LYS}}=\text{7.0 g}\times \frac{\text{1 mol LYS}}{\text{146}\text{.19 g}}=\text{0}\text{.048 mol LYS}\) \(\text{n}_{\text{ARG}}=\text{6.0 g}\times \frac{\text{1 mol ARG}}{\text{174.2 g}}=\text{0}\text{.034 mol ARG}\) The ratio is \(\frac{\text{LYS}}{ARG}~=~\frac{0.048}{0.034}~=~ \frac{1.4}{1}\) And since 1.4 is about 1-2/5 or 7/5, \(\frac{\text{7/5}}{1}~=~ \frac{7}{5}\) This exact amino acid ratio may not be found in any particular rat protein, but it is the average ratio for all the proteins in the rat. The (partial) chemical equation is That is, the stoichiometric ratio S(LYS/ARG) = 7 mol LYS / 5 mol ARG. \(\text{n}_{\text{LYS}}=\text{2.4 g}\times \frac{\text{1 mol LYS}}{\text{146}\text{.19 g}}=\text{0}\text{.016 mol LYS}\) \(\text{n}_{\text{ARG}}=\text{5.0 g}\times \frac{\text{1 mol ARG}}{\text{174.2 g}}=\text{0}\text{.029 mol ARG}\) The ratio is \(\frac{LYS}{ARG}~=~\frac{0.016}{0.029}~=~ \frac{0.55}{1}\) From the corn diet example you can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting. We must compare the stoichiometric ratio S(X/Y) with the actual ratio of amounts of X and Y which were initially mixed together. The corresponding general rule, for any reagents X and Y, is \(\begin{align} & \text{If}~\frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is less than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then X is limiting}\text{.} \\ & \\ & \text{If}~\frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is greater than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then Y is limiting}\text{.} \\ \end{align}\) Of course, when the amounts of X and Y are in exactly the stoichiometric ratio, both reagents will be completely consumed at the same time, and neither is in excess. We can verify that LYS is limiting by calculating the amount of LYS that would be required if all the ARG reacted, using a stoichiometric ratio: \(n_{\text{ARG}}\text{ }\xrightarrow{S\text{(LYS/ARG}\text{)}}\text{ }n_{\text{LYS}}\) \(\text{0.029 mol ARG}~\times~\frac{\text{7 mol LYS}}{\text{5 mol ARG}}~=~\text{0.041 mol LYS}\) Similarly, if all the LYS reacts, the required amount of ARG is \(\text{0.016 mol LYS}~\times~\frac{\text{5 mol ARG}}{\text{7 mol LYS}}~=~\text{0.018 mol ARG}\) This is much less than the amount present, so ARG is the . Let's set up a table: When the reaction ends, 3.14 g of the original 5.0 g (63%) remains, and will be metabolized to urea and excreted. Of the entire diet of 2.4 g LYS + 5.0 g ARG, we see that 3.14 g or 42% is wasted. What a waste of food and the resources to produce it! Notice that the corn serving will also have to be larger than the serving of optimal diet for the same amount of protein, because corn has only 5 g/kg ARG, while the optimal diet has 6.0 g/kg. The optimal diet might be provided by meat or eggs, but they are environmentally demanding and present health concerns. A subsistence bean diet may also lead to protein malnutrition, as inspection of Table II above shows. Although the Lysine content is much greater than in beans, the methionine (MET) and Cystine (CYS) levels are low, as are the Phenylalanine (PHE) and Tyrosine (TYR). But suppose beans and corn are eaten in the same day, in a diet that mixed equal masses of corn and lentils. Now recalculate the limiting amino acid in a diet of 2.4 + 6.3 = 8.7 g of LYS and 5.0 + 6.97 = 11.97 g of ARG. b. In this case, how much of the excess amino acid is wasted? \(\text{n}_{\text{LYS}}~=~\text{8.7 g}~\times~ \frac{\text{7 mol LYS}}{\text{146.19 g}}=\text{0}\text{.060 mol LYS}\) \(\text{n}_{\text{ARG}}~=~\text{11.97 g}~\times~ \frac{\text{1 mol ARG}}{\text{174.2 g}}=\text{0}\text{.069 mol ARG}\) Now we see that the ratio of amounts present is 0.060 mol LYS / 0.069 mol ARG = 0.869, which is still less than the stoichiometric ratio, 7 mol LYS / 5 mol ARG = 1.4 So LYS is once again the limiting amino acid. Recalculating the values in a table as above, we get In this case, 4.61 g of the excess ARG remains from a 11.97 g portion, or 38% of the ARG. The total diet is 8.7 g LYS + 11.97 g ARG = 20.7 g, and only 22% of that is wasted, a big improvement over the corn alone case above. In order to design a proper diet, complementary protein foods would have to be chosen from tables of amino acid contents of foods. In Central America, field corn or "maize" is traditionally boiled in lime (CaO) to loosen the hulls and soften the kernels in the preparation of tortilla flour. The process is shown beautifully in a Culinary Institute of America film. The Southern US breakfast "grits" are also made this way . But this process converts some of the already limiting tryptophan to 2-aminoacetophenone, exacerbating the poor nutritional, and survival value, of the corn. But the nixtamalization also makes niacin in the kernels more bioavailable, reducing the incidence of Pellagra, and more than offsetting the loss of amino acids if plenty of beans are also included in the diet. The beans & corn diet, even with nixtamalization, has survival value for cultures. To detect an amino acid (even in a fingerprint in forensic chemistry), the test is often used. Ninhydrin In the ninhydrin test, two ninhydrin molecules (C H O , shown to the left) become linked by the N attached to the first carbon of the amino acid chain, producing the blue/purple ion shown at right. Blue/Purple Product The balanced chemical equation is: 2 C H O + C H N O → (C H O )-N=(C H O ) + C H NO + CO + 3 H O If 2.00 mg of ninhydrin (Nin) is used to detect 2 mg of TRP, has enough ninhydrin been added to react with all the TRP? Which is the limiting reagent, and what mass of H O will be formed? The stoichiometric ratio connecting Nin and TRP is \(\text{S}\left( \frac{\text{Nin}}{\text{TRP}} \right)=\frac{\text{2 mol Nin}}{\text{1 mol TRP}}\) The initial amounts of Nin and TRP are calculated using appropriate molar masses(160.13 g mol for Nin and 204.23 g mol for TRP: \(\begin{align} & \text{ }n_{\text{Nin}}\text{(initial)}=\text{2}\text{.00}\times \text{10}^{\text{-3}}\text{g}\times \frac{\text{1 mol Nin}}{\text{178}\text{.14 g}}=\text{1}\text{.12}\times \text{10}^{\text{-5}}\text{mol Nin} \\ & \\ & n_{\text{TRP}}\text{(initial)}=\text{2}\text{.00}\times \text{10}^{\text{-3}}\text{g}\times \frac{\text{1 mol TRP}}{\text{204}\text{.23 g}}=\text{9}\text{.79}\times \text{10}^{\text{-6}}\text{mol TRP} \\ \end{align}\) Their ratio is \(\frac{n_{\text{Nin}}\text{(initial)}}{n_{\text{TRP}}\text{(initial)}}=\frac{\text{1}\text{.123}\times \text{10}^{\text{-5}}\text{mol Nin}}{\text{9}\text{.79}\times \text{10}^{\text{-6}}\text{mol TRP}}=\frac{\text{1}\text{.15 mol Nin}}{\text{1 mol TRP}}\) Since this ratio is smaller than the 2:1 stoichiometric ratio, you have insufficient Nin to react with all the TRP, so Nin is the limiting reagent. To ensure detection, in would be better to add excess Ninhydrin. b) The amount of water product formed in a reaction may be calculated via an appropriate stoichiometric ratio from the amount of a reactant which was . Some of the excess reactant TRP will be left over, but all the initial amount of Nin will be consumed. Therefore we use (initial) to calculate how much H O can be obtained \(n_{\text{Nin}}\text{ }\xrightarrow{S\text{(water/Nin}\text{)}}\text{ }n_{\text{water}}\xrightarrow{M_{\text{water}}}\text{ }m_{\text{water}}\) \(m_{\text{water}}=\text{1}\text{.12 }\times \text{ 10}^{\text{-5}}\text{ mol Nin }\text{ }\times \text{ }\) \(\frac{\text{3 mol water}}{\text{2 mol Nin}}\text{ }\times \text{ }\frac{\text{18}\text{.016 g}}{\text{mol water}}=\text{3}\text{.02 }\times \text{ 10}^{\text{-4}}\text{ g water}\) This is 0.302 mg water. These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. Calculations are shown for each possible case, assuming that one reactant is completely consumed and determining if enough of the other reactants is present to consume it. If not, that scenario is discarded. As you can see from the example, in a case where there is a limiting reagent, . Using the initial amount of a reagent present in excess would be incorrect, because such a reagent is not entirely consumed. The concept of a limiting reagent was used by the nineteenth century German chemist Justus von Liebig (1807 to 1873) to derive an important biological and ecological law. states that the essential substance available in the smallest amount relative to some critical minimum will control growth and reproduction of any species of plant or animal life. When a group of organisms runs out of that essential limiting reagent, the chemical reactions needed for growth and reproduction must stop. Vitamins, protein, and other nutrients are essential for growth of the human body and of human populations. Similarly, the growth of algae in natural bodies of water such as Lake Erie can be inhibited by reducing the supply of nutrients such as phosphorus in the form of phosphates. It is for this reason that many states have regulated or banned the use of phosphates in detergents and are constructing treatment plants which can remove phosphates from municipal sewage before they enter lakes or streams.
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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Laws/Avogadro's_Law
Discovering that the volume of a gas was directly proportional to the number of particles it contained was crucial in establishing the formulas for 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 \(H_2\), \(O_2\), and \(Cl_2\) was not recognized until the results of experiments involving gas volumes was interpreted. Early chemists mistakenly assumed that the formula for water was \(HO\), leading them to miscalculate the molecular weight of oxygen as 8 instead of 16. However, when chemists found that an assumed reaction of \[H+Cl→HCl\] yielded twice the volume of \(HCl\), they realized hydrogen and chlorine were diatomic molecules. The chemists revised their reaction equation to be \[H_2+Cl_2→2HCl.\] When chemists revisited their water experiment and their hypothesis that \[HO \rightarrow H + O\] they discovered that the volume of hydrogen gas consumed was twice that of oxygen. By Avogadro's Law, this meant that hydrogen and oxygen were combining in a 2:1 ratio. This discovery led to the correct molecular formula for water (\(H_2O\)) and the correct reaction \[2H_2O→2H_2+O_2.\] Boundless (www.boundless.com)
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Independently discovered in 1735 by Antonio de Ulloa and in 1741 by Charles Wood, platinum is named from the Spanish platina or "silver". The metal is classified as precious owing to its scarcity and commercial demand. It is very heavy and silvery white and is used in laboratory instruments, jewelry, medical and dental items, and electrical contacts. Because it is immune to air oxidation the metal is often found in native form in nature. It is sometimes found in a rare, naturally occurring alloy, platiniridium. Naturally occurring platinum is a mixture of six non-radioactive isotopes. Platinum is part of the the Platinum Group Metals (PGM) whic is located in the 5th and 6th rows of the transition metal section of the periodic table and includes , , , , , and Platinum. Common characteristics include resistance to wear, oxidation, and corrosion, high melting points, and oxidation states of +2 to +4. They are generally non-toxic.
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