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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/The_Four_Laws_of_Thermodynamics/First_Law_of_Thermodynamics/First_Law_of_Thermodynamics

The first law of thermodynamics states that energy can be converted from one form to another, but cannot be created or destroyed. The most important and critical aspect of life revolves around the idea of energy. During the course of a single day, a person finds him or herself using energy in all sorts to live their lives. Whether driving a car or eating lunch, the consumption of some sort of energy is unavoidable. While it may seem that energy is being created for our purposes and destroyed during it, there is in fact no change in the amount of energy in the world at one time. Taking this a step farther, one may state the entirety of the energy in the universe is at a constant with energy just being converted into different forms. To obtain a better understanding of the workings of energy within the universe, it is helpful to classify it into two distinct parts. The first being the energy of a specific system, \(E_{sys}\), and the second being whatever energy was not included in the system which we label as the energy of the surroundings, E . Since these two parts are equal to the total energy of the universe, \(E_{univ}\), it can be concluded that \[E_{univ} = E_{sys} + E_{surr} \label{1}\] Now, since we stated previously that the total amount of energy within the universe does not change, one can set a change in energy of the system and surroundings to equal \[ΔE_{sys} + ΔE_{surr} = 0 \label{2}\] A simple rearrangement of Equation \ref{2} leads to the following conclusion \[ΔE_{sys} = -ΔE_{surr} \label{3}\] Equation \ref{3} represents a very important premise of energy conservation. The premise is that any change in energy of a system will result in an equal but opposite change in the surroundings. This essentially summarizes the First Law of Thermodynamics which states that energy cannot be created nor destroyed. Now that the conservation of energy has been defined, one can now study the different energies of a system. Within a system, there are three main types of energy. These three types are kinetic (the energy of motion), potential (energy stored within a system as a result of placement or configuration), and internal (energy associated with electronic and intramolecular forces). Thus, the following equation can be given \[E_{total} = KE + PE + U \label{4}\] where KE is the kinetic energy, PE is the potential energy, U is the internal energy, and E is the total energy of the system. While all forms of energy are very important, the internal energy, U, is what will receive the remainder of the focus. As stated previously, U is the energy associated with electronic and intramolecular forces. Yet, despite the abundance of forces and interactions that may be occurring within a system, it is near impossible to calculate its internal energy. Instead, the change in the U of a system, ΔU, must be measured instead. The change in ΔU of a system is affected by two distinct variables. These two variables are designated at heat, q, and work, w. Heat refers to the total amount of energy transferred to or from a system as a result of thermal contact. Work refers to the total amount of energy transferred to or from a system as a result of changes in the external parameters (volume, pressure). Applying this, the following equation can be given \[ΔU = q + w \label{5}\] If the change of ΔU is infinitesimal, then Equation \ref{5} can be altered to \[dU = dq + dw \label{6}\] Within this equation it should be noted that U is a state function and therefore independent of pathways while \(q\) and \(w\) are not.

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/4_f-Block_Elements/The_Lanthanides/Chemistry_of_Gadolinium/Gadolinium(III)_Chloride

Gadolinium trichloride, GdCl , is a colorless, hygroscopic, water-soluble solid. It is commonly encountered as hexahydrate GdCl 6H O. Gadolinium trichloride can be synthesized by the reaction of solid gadolinium with HCl at 600 °C: 2 Gd + 6 HCl 2 GdCl + 3 H Another way to prepare gadolinium trichloride is to synthesize (NH ) [GdCl ] which can be decomposed at 300°C to GdCl . 10 NH Cl + 2 Gd 2 (NH ) [GdCl ] + 6 NH + 3 H 2 (NH ) [GdCl ] {NH [Gd Cl ] + 3NH Cl} 2 GdCl + 4 NH Cl Gd ions are of special interest for magnetic resonance imaging (MRI), since the Gd species has an f electronic configuration with seven unpaired electrons. This results in a highly paramagnetic behavior. As gadolinium salts are toxic, the Gd ion has to complexed by a chelating agent (i.e. H5DTPA, diethylenetriaminepentaacetic acid) for usage in biomedical applications.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/12%3A_Thermodynamic_Processes_and_Thermochemistry/12.E%3A_Thermodynamic_Processes_(Exercises)

. A sample of \(\ce{O_2}\) gas is under an external pressure of 17 atm and contained in a cylinder with a volume of 50 L. The gas is cooled and the resulting volume is 25 L. Calculate the amount of work done the \(\ce{O_2}\) gas. \[\begin{align*} w &=-P\Delta V \nonumber \\[5pt] &= -P_{\text{ext}}(V_2-V_1) \\[5pt] &=-17atm\ (25\,L-50\,L)=425\, L\ atm \\[5pt] &=425\ atm\ L\ (101.325\dfrac{J}{atm\ L}) \\[5pt] &= 43, 063\,J \\[5pt] & =43.06\, kJ \end{align*} \nonumber \] A system containing oxygen gas is heated at a constant pressure of 40.0 atm so that its volume increases 177 L to 458 L. Express the amount of work that the system did in kilo-joules. The formula for work from the expansion of a gas at constant pressure \[w=-P_{ext} \Delta V\nonumber \] \[w=-40\; atm \times (458\; L - 177\; L) = -11240\; L\; atm \nonumber \] Convert from L atm to joules \[-11240 \; L\; atm \times 101.325\; \dfrac{J}{L\ atm} = -1138893\; J \nonumber \] Convert from J to kJ and round to get the final answer \[ -1138893\; J \times \dfrac{1\; kJ}{1000\; J}=-1140\; kJ \nonumber \] \[w=-1140\; kJ \nonumber \] The gas mixture inside one of the cylinders of an airplane expands against a constant external pressure of 5.00 atm because of the growing altitude, from an initial volume of 500 mL (at the end of the compression stroke) to the final volume of 1200 mL. Calculate the work done on the gas mixture during this process and express it in Joules. \[w = -(5.00\ \mathrm{atm}) (1200\ \mathrm{mL} - 500\ \mathrm{mL}) (\dfrac{1\ \mathrm{L}}{1000 \, \mathrm{mL}}) (101.3 \, \mathrm{\dfrac{J}{L\times atm}})= -354.55 \, \mathrm{J} \nonumber\] . A butcher for the local Chinese restaurant needs to defrost a large chunk of beef, which weights \(50\,lb\) and is currently at \(0^{\circ}C\). He wants to accomplish this by is going out onto the street and repeatedly drop it onto the ground. Suppose the potential energy of the meat completely transforms into heat each time it hits the ground, and that energy can be calculated from \[V = mg\Delta{h}\nonumber \] where \(m\) is mass of the object, \(g\) is the acceleration of gravity, and \(Δh\) is change in height. If the man is \(5\, \text{ft}\) tall and he wants to get the meat to room temperature (\(25^{\circ}\text{C}\)), how many times does he have to drop the piece of meat? Assume the environment has no effect on the meat, and that it does not lose any heat. (Specific Heat of the Meat: \(0.25\mathrm{\dfrac{J}{g \, ^{\circ}C}}\)). There are two steps to solve this equation. First, to find the amount of heat absorbed by the meat from one drop. Second, to find the total amount of heat needed to warm the meat up from \(0^{\circ}C\) to \(25^{\circ}C\). To find the total amount of heat required to bring the meat from \(0^{\circ}C\) to \(25^{\circ}C\), the equation for calculating heat has to be used: \(Q = mC_{p}\Delta{T}\) The mass of the meat can be calculated via conversion from \(lb\) to \(g\), and is as follows: \(50lb\times{453.592\dfrac{g}{lb}} = 22679.6g\) The change in temperature is just the final temperature minus the initial, so it is just \(25^{\circ}C\). Lastly, \(C_{p}\) is given. Plugging in all of these values gives: \[Q = 22679.6g\times{0.25\dfrac{J}{g^{\circ}C}}\times{25^{\circ}C} = 141747.5 J\nonumber \] Thus, \(141747.5 J\) is the total amount of heat needed to heat up the meat to \(25^{\circ}C\). Now to find the amount of heat transferred into the meat from one drop. This utilizes the potential energy equation. The question mentions how to calculate the potential energy, and all potential energy is translated into heat. Therefore. \[Q= V= mg\Delta{h}\nonumber \] The problem gives the mass, the acceleration of gravity is \(9.8\dfrac{m}{s^2}\), and height. Therefore, since the potential energy for one drop is equal to the heat for one drop, the heat can be calculated as such: \[Q= 22.6796kg\times9.8\dfrac{m}{s^2}\times{1.524m}= 338.724\dfrac{kg\;m^2}{s^2} = 338.724 J\nonumber \] From there, since no heat is lost, it carries over for each drop. Therefore, the number of times he needs to drop the meat is found by simple division: \[141747.5 J\div{338.724 J}= 418.474 drops\nonumber \] Thus, the drops needed is around 419. Calculate heat from \(Q = mC_{p}\Delta{T}\) and \(Q= V= mg\Delta{h}\). Divide the two. Suppose you have a ball \((C_p= 0.85\dfrac{J}{g^oC})\) at 25 \[\begin{align} U &=mg\Delta h \\[5pt] &=m_{ball}\cdot 86.6m\cdot 9.81\dfrac{m}{s^{2}}\\[5pt] &=850m_{ball} \end{align} \nonumber \] If we assuming all the kinetic energy at the time of collision converts to heat, then \[\begin{align} \Delta T &=\dfrac{q}{Cp\cdot m_{ball}}\\[5pt] &=\dfrac{U}{C_p\cdot m_{ball}} \\[5pt] &= \dfrac{850 m_{ball}}{85\dfrac{J}{g^oC} \cdot m_{ball}} \dfrac{1\,kg}{1,000\,g} \\[5pt] &= 1.0^{\circ}C \end{align} \nonumber \] \[\begin{align} T_{final} &=25^{\circ}C+1^{\circ}C\\[5pt] &=26^{\circ}C \end{align} \nonumber \] The shows that the molar heat capacities of most metallic elements group around a certain value “X” \(\dfrac{J}{K\ mol}\) at 25 C. By first calculating and showing the molar heat capacities of metals rhenium, silver, lead, tungsten, copper, molybdenum and hafnium (given that the respective specific heat capacities of these metals are 0.14\(\dfrac{J}{K\ g}\), 0.23\(\dfrac{J}{K\ g}\), 0.13\(\dfrac{J}{K\ g}\), 0.13\(\dfrac{J}{K\ g}\), 0.39\(\dfrac{J}{K\ g}\), 0.25\(\dfrac{J}{K\ g}\) and 0.14\(\dfrac{J}{K\ g}\) ), find the value of “X”. (hint: take the average value of the calculated molar heat capacities and round off value to nearest whole number). To first find the molar heat capacities of each metal, multiply the molar mass of each metal by their specific heat capacities: Molar heat capacity of: calculate the average molar heat capacity: Average molar heat capacity=\(\dfrac{1}{7}(26.068\dfrac{J}{K\ mol} + 24.809\dfrac{J}{K\ mol} + 26.936\dfrac{J}{K\ mol} + 23.899\dfrac{J}{K\ mol} + 24.7829\dfrac{J}{K\ mol} + 23.99\dfrac{J}{K\ mol} + 24.988\dfrac{J}{K\ mol})=25.068\dfrac{J}{K\ mol}\) \(\ce{W}= 0.13\dfrac{J}{K\ g}\times 183.84\dfrac{g}{mol}=23.899\dfrac{J}{K\ mol}\) \(\ce{Cu}= 0.39\dfrac{J}{K\ g}\times 63.546\dfrac{g}{mol}=24.7829\dfrac{J}{K\ mol}\) \(\ce{Mo}=0.25\dfrac{J}{K\ g}\times 95.96\dfrac{g}{mol}= 23.99\dfrac{J}{K\ mol}\) \(\ce{Hf}= 0.14\dfrac{J}{K\ g}\times 178.49\dfrac{g}{mol}=24.988\dfrac{J}{K\ mol}\) Multiply each of the specific heat capacities by its corresponding molar mass to obtain their molar heat capacities. Aluminum: \[0.900 \dfrac{J}{g\ K}\times 26.98 \dfrac{g}{mol} = 24.3 \dfrac{J}{mol\ K}\nonumber \] Bismuth: \[0.123 \dfrac{J}{g\ K}\times 208.98 \dfrac{g}{mol} = 25.7 \dfrac{J}{mol\ K}\nonumber \] Copper: \[0.386 \dfrac{J}{g\ K}\times 63.55 \dfrac{g}{mol} =24.5 \dfrac{J}{mol\ K}\nonumber \] Lead: \[0.128 \dfrac{J}{g\ K}\times 207.2 \dfrac{g}{mol} =26.5 \dfrac{J}{mol\ K}\nonumber \] Silver: \[0.223 \dfrac{J}{g\ K}\times 107.87 \dfrac{g}{mol} =25.1 \dfrac{J}{mol\ K}\nonumber \] An undisclosed volume of water is tightly sealed in a microwave-safe container at room temperature before it is placed in an ice bath where it is cooled by a student. \[q_{\text{sys}}=0=q_{\text{metal}}+q_{\text{water}}\nonumber \] For both sub-systems, the amount of heat gained is equal to the specific heat capacity times the mass times the temperature change: \[q_{\text{metal}}+ q_{\text{water}}= m_{\text{water}} C_{s,\text{water}} \, \Delta T_{\text{metal}} = 0\nonumber \] Solving for the specific heat capacity of the metal: \[c_{s,\text{metal}}=\dfrac{-m_{\text{water}}c_{s,\text{water}} \, \Delta T_{\text{water}}}{m_{\text{metal}} \Delta  T_{\text{metal}}}=\dfrac{-100.0\;\text{g} \times 4.18 \mathrm{\dfrac{J}{K\ g}} \times 10.15 \mathrm{^{\circ}C}} {40.0\; \text{g} \times -69.85 \mathrm{^{\circ}C}} = 1.52 \mathrm{\dfrac{J}{K\ g}} \nonumber \] Tip: Do not convert Celsius to Kelvin. The Kelvin and Celsius scales differ only in their location of their zero points. Temperature change in Celsius is the same temperature change of Kelvin. Let's say you 34.5 grams of some hot metal that is initially at 75°C and you put that metal into 64.0 grams of water that is initially at 25°C. If the the two objects reach thermal equilibrium at 39°C, what is the specific heat capacity of the metal when the specific heat capacity of water is \(4.18 \dfrac {J}{K\ g}\). The formula used to solve this question is \[m_1c_1 \Delta T_1 = -m_2c_2 \Delta T_2\nonumber \] Plug in your known values and solve for \(c_1\) \((34.15 \; g)(c_1)(36°C) = -(64 \; g)(4.18 \; \dfrac {J}{K\ g})(-14°C)\) \(c_1 = 3.02 \; \dfrac {J}{K\ g}\) A 10.00 g sample of Aluminum at 60.0 °C and a 30.0 g sample of copper at a temperature of -20.0 °C were thrown simultaneously into a 50.0 g of water at a temperature of 25.0 °C. What will be the final temperature of the system consisting of the two metal samples and the water? Assuming that this system is completely isolated from the surroundings. Use the information below for your calculations. Since our system is isolated, the thermal energy lost by one component is transferred to the other components. \[q_1=-q_2\nonumber \] which is equivalent to \[C_1 ∆T=-C_2 ∆T\nonumber \] Noting that \[C=m C_s\nonumber \] To avoid the complexity of handling three components, we can utilize the fact that temperature is a state function; we will simplify our calculations by choosing a different path to arrive to our final state. We can do this in two steps: Ignore the Aluminum sample and treat the copper and water as the only components of our system. After finding the equilibrium temperature, we add the Aluminum sample to the water and copper system, thereby reaching the same final state. Which is the same answer we got before. This should make intuitive sense because regardless of the path we take, we end up with the exact same amount of thermal energy in our system. \[m_{\ce{Cu}}×C_{s(\ce{Cu})}×(T_f-T_{i(\ce{Cu})})=-m_{\ce{H_2O}}C_{s(\ce{H_2 O})}(T_f-T_{i(\ce{H_2 O})})\nonumber \] \[30.0\,g×0.385\, J/(°C \cdot g)×(T_f+20 °C)=-50.0\,g × 4.184 \,J/(°C \cdot g)×(T_f-25.0 °C)\nonumber \] Solving for \(T_f\) for the copper and water system gives us \[T_f=22.6 °C\nonumber \] \[C_{(\ce{Cu}+\ce{H_2 O})}×(T_f-T_{i(\ce{Cu}+\ce{H_2 O})})=-m_{\ce{Al}}×C_{s(\ce{Al})}×(T_f-T_{i(\ce{Al})})\nonumber \] Noting that \[C_{(\ce{Cu}+\ce{H_2 O})}=m_{\ce{H_2O}}×C_{s(\ce{H_2 O})}+m_{\ce{Cu}}×C_{s(Cu)}\nonumber \] Combining the two equations and plugging the values gives us \[[(30.0\,g×0.385 \,J/(°C \cdot g))+(50.0\,g × 4.184 \,J/(°C \cdot g))]×(T_f-22.6 °C)= -10.0\,g×0.900\, J/(K \cdot g)×(T_f-60 °C)\nonumber \] Solving for \(T_f\): \[T_f=24.1°C\nonumber \] Which is the final temperature of the whole system. We can also choose a path where we add the Aluminum first, and then we add the copper. Again, temperature is a state function and choosing a different path will not affect the final answer. We also show the calculation for this path for the sake of completion. \[m_{\ce{Cu}}×C_{s(\ce{Al})}×(T_f-T_{i(Al)})=-m_{(\ce{H_2 O})}×C_{s(\ce{H_2 O})}×(T_f-T_{i(\ce{H_2 O})})\nonumber \] \[10.0\,g×0.900\, J/(°C \cdot g)×(T_f+20 °C)=-50.0\,g × 4.184 \,J/(°C \cdot g)×(T_f-25.0 °C)\nonumber \] Solving for \(T_f\) for the aluminum and water system gives us \[T_f=26.4 °C\nonumber \] \[[(m_{(\ce{H_2 O})}×C_{s(\ce{H_2O})})+(m_{\ce{Al}}×C_{s(\ce{Al})}) )]×(T_f-T_{i(\ce{Al}+\ce{H_2 O})})=-m_{\ce{Al}}×C_{s(\ce{Cu})}×(T_f-T_{i(\ce{Cu})})\nonumber \] Plugging the numbers, \[[(10.0\,g×0.900\, J/(°C \cdot g))+(50.0\,g × 4.184 \,J/(°C \cdot g)]×(T_f-26.4 °C)=-30.0\,g×0.385 \,J/(K \cdot g)×(T_f+20 °C)\nonumber \] Solving for \(T_f\) \[T_f=24.1°C\nonumber \] Calculated the heat required to melt 3.00 g of ice and the heat required to change the temperature of water from 0°C to 100°C. What is the proportionality of the heat necessary to melt ice compared to the heat required to change the temperature of water from 0°C to 100°C? Use values from and assume \(\Delta H_{f}\) =334 J g for calculations? Why is the heat positive instead of negative? First lets determine the amount of heat needed to melt ice: \(q = m\Delta H_{f}\) \(q = 1,002\ \mathrm{J}\) Next lets determine the heat required to raise the temperature of water 100°C: \(q = mC_s \Delta{T}\) \(q = 1,254\ \mathrm{J}\) So the proportionality was determined to be 4:5 The heat is positive because the heat is required. The heat is needed to make the states go from solid to liquid when melting and then liquid to gas when the temperature is raised from 0 to 100 degrees Celsius. An observation in the 18th century stated that the heat that raised a certain mass of water from its freezing point to boiling point is equal to four-thirds of the heat required to melt the same mass of ice. Using the theory behind the observation, estimate the heat required to melt 10 g of ice, know that the heat capacity of water is 4.18 J/ C. Let the heat required to raise the temperature of water from its freezing point to its boiling point is \(q_1\): \[q = m c \Delta T\] \[m = \text{mass}\] \[c = 4.18\dfrac{J}{g\ ^{o}C}\] \[\Delta T = \text{freezing point} - \text{boiling point} = 100 - 0 = 100^{o}C\] \[q_{1}=(10 g)(4.18\dfrac{J}{g\ ^{o}C})(100^{o}C)\] \[q_{1} = 4180 \, \text{J} = 4.18 \, \text{kJ}\] Let the heat required to melt 10 g of ice is \(q_2\): \[q_{1}=\dfrac{4}{3}q_{2}\] \[q_{2}=\dfrac{3}{4}q_{1}\] \[q_{2}=\dfrac{3}{4}(4.18 \, \text{kJ})=3.135 \, \text{kJ}\] For his birthday, his John's parents have given him a compressible oven filled with 0.250 mol argon. If he sets this compressible oven at 1.00 atm and 273 K and let it contract from a constant external pressure of 0.100 atm until the gas pressure reaches 10.00 atm and the temperature reaches 400 K, what is the work done on the gas, the internal energy change, and the heat absorbed by the gas? Using the ideal gas law \(P V = n R T\), we can see that: \[V_o = \dfrac{n R T_o}{P_o} \nonumber \] \[V_f = \dfrac{n R T_f}{P_f} \nonumber \] where \(V_o\) and \(V_f\) represent the initial and final volume of the chamber. Now, the pressure clearly changes inside the chamber, but outside the chamber, the pressure is held constant, so from the perspective of the surroundings: \[W= -P(V_f - V_o)\nonumber \] where \(P\) represents the constant external pressure. Now substituting in V and V we can see that we've solved for work. Now, for any thermodynamic process: \[\Delta{U}= \dfrac{3}{2} n R (T_f - T_o) \nonumber \] So plugging in those values allows us to solve for ∆U. Now, for heat, we simply subtract W from ∆U, by the first law of thermodynamics. Take 4 moles of ideal, monatomic gas going through expansion processes. The gas was initially put at a pressure of 5.00 atm and a temperature of 30°C. The gas first goes through isothermal expansion until the volume doubled. An isochoric process follows as the pressure is halved. \(C_v = \dfrac{3}{2}R\) Isothermal Process: \[\Delta U = 0\] \[W = Q\] \[W = nRT\ln \left( \dfrac{V_f}{V_i} \right) =(4\,\text{mol})(8.314\, \mathrm{J \cdot K^{-1} mol^{-1}})(303.15\, \text{K}) \ln(2) = 6988 \, \text{J} = Q\] Isochoric Process \[W = 0\] \[\Delta U = Q\] First find the initial pressure of this process. Use ideal law to relate pressure to volume. \(PV=nRT\), in which P is inversely proportional to V. Thus, when volume is doubled, pressure is halved so \(P_f = 2.50 \,\text{atm}\). Now, find the final temperature of this process. Once again, relate P to T using ideal gas law. T is found to be directly proportional to P. Thus, if the pressure is halved from 2.50 atm, the temperature must also be halved from 303.15K. \(\dfrac{1}{2}(303.15 \, \text{K}) = 151.58 \, \text{K}\) \[Q= nC_v\Delta T=(4.00 \, \text{mol})(\dfrac{3}{2})(8.314 \mathrm{\dfrac{J}{K\, mol}})(-151.58K)=-7561 \, \text{J} = \Delta U\] and A 150 L vessel contains 8.00 moles of neon at 270 K is compressed adiabatically, so that there is no Since Ne is a monatomic ideal gas and the volume of the vessel remains constant, the heat capacity of Ne can be expressed as: \[C_{p} = \dfrac{3}{2} \times R = \dfrac{3}{2} \times 8.314 \mathrm{\dfrac{J}{K \cdot mol}} = 12.47 \mathrm{\dfrac{J}{K \cdot mol}} \] \[\Delta U = nC_{p} \Delta T\] \[\Delta U = 8.00 \times 12.47 \mathrm{\dfrac{J}{K \cdot mol}} \times (470-2700) \; \mathrm{K} = 19952 \; \mathrm{J} = 20.0 \; \mathrm{kJ}\] Since the vessel is adiabatically compressed, no heat is added, therefore \(q = 0\) \[\Delta U = q + w \] \[20.0 \; \mathrm{kJ} = 0 + w \] \[w = 20.0 \; \mathrm{kJ}\] A gas expands at constant external pressure of 3.00 atm until its volume has increased by 9.00 to 15.00 L. During this process, it absorbs 800J of heat from the surroundings. a) \[\Delta U = q+w\] \[\Delta U= 800J + -((3.00 \, \mathrm{atm})(15.00 \mathrm{L} - 9.00 \mathrm{L}) (101.3 \mathrm{\dfrac{J}{L\ atm}})) =-1023.4 \mathrm{J}\] b) \[q=0\] \[\Delta U = w\] \[\Delta U= -((3.00 \mathrm{atm})(15.00 \mathrm{L} - 9.00 \mathrm{L}) (101.3 \mathrm{\dfrac{J}{L\ atm}}))= -1823.4 \mathrm{J}\] \(\ce{O2}\) is a linear diatomic particle, thus it has 3 translational degrees of freedom and 2 rotational degrees of freedom. The low percentage of \(C_p\) due to vibrational modes in both molecules indicates that the vibrational motion is extremely small and can be neglected. \[DOF({\ce{O2}}) = {f_t} + {f_r} = 3 + 2 = 5\nonumber \] \[DOF(\ce{CO}) = {f_t} + {f_r} = 3 + 2 = 5\nonumber \] \[U = \left( DOF\right) \cdot \left( \dfrac{1}{2}RT \right)\nonumber \] \[U \left( \ce{O_2} \right) = \left( 5 \right) \cdot \left( \dfrac{1}{2}RT \right)\nonumber \] \[U\left( \right) = U\left( \ce{CO} \right) = \dfrac{5}{2}RT\nonumber \] \[C_v= \left( \dfrac{\partial U}{\partial T} \right) = \left( \dfrac{\partial \dfrac{5}{2}RT}{\partial T} \right) = \dfrac{5}{2}R\nonumber \] \[C_p = C_v + R = \dfrac{5}{2}R + R = \dfrac{7}{2}R = \dfrac{7}{2} \times 8.314 \dfrac{J}{mol\ K} = 29.099\dfrac{J}{mol\ K}\nonumber \] The percent of the experimental value that results from vibration motion is the difference between calculated and experimental value as a percentage of the experimental value. \[\% \text{vibrational motion}(\ce{O_2}) = \dfrac{\text{experimental value} - \text{calculated value}}{ \text{experimental value}} \times 100 \% = \dfrac{ 29.36 - 29.099 }{29.099} \times 100 \% = 0.897\% \nonumber \] \[\% \text{vibrational motion} (\ce{CO}) = \dfrac{\text{experimental value} - \text{calculated value}}{ \text{experimental value}} \times 100 \% = \dfrac{ 29.14 - 29.099}{ 29.099} \times 100 \% = 0.141\% \nonumber \] \[6.11 \text{ for } \ce{F2 (g)} = 0.735R\] a) \(q=mC_s\Delta T\) \(m=69\ \mathrm{g}\) \(\Delta T=927-753=174\ \mathrm{K}\) \(C_{\text{s(zinc)}}=0.39 \mathrm{\dfrac{J}{g\ ^oC}}\) \(q=69\times 0.39\times 174=4682.34\ \mathrm{J}=4.56\ \mathrm{kJ}\) b) \(q=nC_p\Delta T\) \(n=3\) \(\Delta T=258-204=54\ \mathrm{K}\) \(C_p=132.42\) \(q=3\times 132.42\times 54=21452.04\ \mathrm{J}= 21.4504\ \mathrm{kJ}\) What is the change in enthalpy when 8.19 grams of ethane (\(\ce{C_2H_6}\)) vaporizes, assuming a normal boiling point and \(\Delta H_{\text{vap}}= 14.72\ \mathrm{\dfrac{kJ}{mol}}\). First, convert grams of ethane to moles: \[(8.19\; \cancel{g\; \ce{C­_{2}H_{6}}}) \left(\frac{1\: mol}{30.07\: \cancel{g\; \ce{C_{­2}H_{6}}}} \right) = 0.272\; mol\nonumber \] This step is to match quantity units of ethane to those in the given value of \(\ce{\Delta H_{vap}}\). Then, given the \(\Delta H_{\text{vap}}\), \[(0.272\; \cancel{mol \;\ce{C­_{2}H_{6}}} ) \left (\frac{14.72\: kJ}{1\: \cancel{mol\; \ce{C_{­3}H_{8}}}} \right ) = 4.004\; kJ\nonumber \] \(\ce{\Delta H}=4.004\; kJ\) This value is the change in enthalpy, or thermodynamic energy, when the specified amount of ethane undergoes the above reaction. The heat capacity \(C_p\) of ice is \(38 \mathrm{\dfrac{J}{K\ mol}}\) and \(C_p\) of water is \(75 \mathrm{\dfrac{J}{K\ mol}}\). A 24.0 g ice cube at -15 C is placed into 120 g of water at room temperature (25 C). What is the temperature of the water when it reaches equilibrium? \[q_{\mathrm{ice}} = -q_{\mathrm{water}}\] \[m_{\mathrm{ice}}C_{\mathrm{ice}}(T_{f}-T_{\mathrm{ice}})=-m_{\mathrm{water}}C_{\mathrm{water}}(T_{f}-T_{\mathrm{water}})\] \[\left(\dfrac{24g\ \mathrm{ice}}{18.02\dfrac{g}{mol}}\right)\times (38\dfrac{J}{K\ mol})\times (T_f - (-15^oC)) = -\left(\dfrac{120g\ \mathrm{water}}{18.02\dfrac{g}{mol}}\right)\times (75\dfrac{J}{K\ mol})\times (T_f - 25^oC)\] \[T_{f}=21.316^{o}C\] An alternative way to view this is to inspect the reactions under comparison \[\ce{2C6H6(g) + 15O2(g) -> 12CO2(g) + 6H2O(g)} \nonumber\] vs. \[\ce{2C6H6(l) + 15O2(g) -> 12CO2(g) + 6H2O(g)} \nonumber\] Recognizing that the difference between these equations is \[\ce{2C6H6(l) -> 2C6H6(g)} \nonumber\] at the \(∆H\) of the conversion of \(\ce{C6H6(g)}\) to \(\ce{C6H6(l)}\) is positive, and therefore, is an endothermic process. Consequently, ten pounds of \(\ce{C6H6(g)}\) contains more enthalpy than 10 pounds of \(\ce{C6H6(l)}\). Since these will both produce carbon dioxide and water when burned, their products are of the same final energy level. As a result, \(\ce{C6H6(g)}\) will give off more energy than \(\ce{C6H6(l)}\) when burned and \(\ce{C6H6(l)}\) will require less energy to form a solid. \[\ce{C3H6O(l) + 4 O2 (g) → 3 CO2 (g) + 3 H2O(l)} \nonumber \] \[ΔH^o_{rxn} = \sum ΔH^o_{f,\, products} – \sum ΔH^o_{f,\, reactants} \nonumber\] A 10 gram sample of pure hydrogen is burned completely with excess oxygen to generate liquid water in a constant volume calorimeter at 25 °C. The amount of heat evolved is 1,420 kJ. A consequence of the constant-volume condition is that the heat released corresponds to \(q_v\) and thus to the internal energy change \(ΔU_{sys}\) rather than to \(ΔH_{sys}\) under constant pressure conditions. \[\Delta U_{sys} = q_v\nonumber \] This comes from the definition of enthalpy \[H_{sys} = U_{sys} + PV\nonumber \] and associated change \[\Delta H_{sys} = \Delta U_{sys} + \Delta (PV)\nonumber \] or using the chain rule \[\Delta H_{sys} = \Delta U_{sys} + \cancelto{0}{P\Delta V} + V \Delta P \label{Change in Bomb}\nonumber \] which simplifies to \[\Delta H_{sys} = \Delta U_{sys} + V \Delta P\nonumber \] If we assuming ideal gas law for the gases \[PV=nRT\nonumber \] or \[\Delta P = \dfrac{\Delta n R T}{V}\nonumber \] substituting this into the equation for enthalpy change gives the enthalpy change under conditions in terms of changing number of moles \[ΔH_{sys} = \Delta U_{sys} + Δn_gRT\nonumber \] or \[ΔH_{sys} = q_v + Δn_gRT\nonumber \] where \(Δn_g\)  is the change in the number of moles of gases in the reaction. Carbon dioxide (\(\ce{CO_2}\)) is a common byproduct of the combustion of fossil fuels. Estimate the standard enthalpy of formation (\(\Delta H_f\)) of carbon dioxide at 25°C, use . First write the equation: \[\ce{C (s) + O2 (g) -> CO2 (g)}\nonumber \] The bond enthalpies in allows us to calculate \(\Delta H°\) because 2 C=O and 2 C-O bonds are formed: \(\Delta H°\) 2 mol (-192.0 kJ mol ) + 2 mol (- = -555 kJ Next we need to write the equations for "striping" the atoms from there standard state to single atoms. Each of the atomization has an enthalpy that was found in : \[\ce{C (s) -> C (g)}\nonumber \] \(\Delta H°\) = 1 mol (716.7 kJ mol ) = 716.7 kJ \[\ce{O2 (g) -> 2 O(g) }\nonumber \] \(\Delta H°\) = 2 mol (249.2 kJ mol ) = 498.4 kJ Combine the results of all three equations to calculate \(\Delta H_f\) of 1 mol of CO : -555 kJ + 716.7 kJ + 498.4 kJ = 660.1 kJ Given the table of average bond enthalpies shown below, estimate the enthalpy change \(\Delta H°\) for the following reaction: To calculate the total enthalpy change of the reaction, we use Hess's Law. Thus, the total enthalpy of formation is equal to the sum of the enthalpy of formation of the reactants and the enthalpy of formation for the products. Step 1. \(\Delta H_1\) = the bond enthalpies of \(\ce{2 H2 (g) + O2 (g)}\) = \(2(\ce{H-H}) + (\ce{O=O})\) = \(2(436) \; \mathrm{kJ/mol} + 498 \; \mathrm{kJ/mol}\) = \(+1370 \; \mathrm{kJ/mol}\) Step 2. \(\Delta H_2\) = the bond enthalpies of \(2 \ce{H2O} (l)\) = \(4(463) \; \mathrm{kJ/mol}\) = \(-1852 \; \mathrm{kJ/mol}\) Step 3. \(\Delta H°\) = \(\Delta H_1\) + \(\Delta H_2\) = \(1370 \; \mathrm{kJ/mol} - 1852 \; \mathrm{kJ/mol}\) = \(-482 \; \mathrm{kJ/mol}\) \(\Delta H_1\) \(\Delta H_2\) × \[\dfrac{P_{2}}{P_{1}} = e^{-(E_{2} - E_{1}) / k_B T}\nonumber \] , \(k_B\) i \[\dfrac{P_{2}}{P_{1}}=e^{-(0.6 \cdot 10^{-21} \, \mathrm{J})/1.38\cdot 10^{-23} \mathrm{K^{-1}} (298 \, \mathrm{K}))}\nonumber \] \[\dfrac{P_{2}}{P_{1}}=0.86\nonumber \] By utilizing the Harmonic Oscillator Model, calculate the relative population of the first energy state and the ground state, both at \(278.15 \, \text{K}\), for \(\ce{H2}\). The force constant for \(\ce{H2}\) is \(510 \mathrm{\dfrac{N}{m}}\) A few pieces of information need to be understood before going about solving this equation. First, the question asks for the “relative population” of the first energy state at a given temperature (\(278.15 \, \mathrm{K}\)) to the energy state at \(0 \, \mathrm{K}\). Recall that energy states are basically a way of saying that every chemical species (molecule, atom, etc.) can “have” a specific (discreet) amount of energy. Each of these values are a “state”, and the lowest- value state can be considered the “ground state”, and all states above are “excited states”. Also recall that the states can be named by quantum number; i.e if \(n\) were the quantum number, than the ground state would be \(n=0\) and the first state after that, an excited state, has a quantum number \(n=1\). The next step is to understand what “relative population is”. Basically, this refers to the probability that the species, in this case \(\ce{H2}\), will be found in one energy state versus another. The equation that describes the probability of a species being in energy level \(n\) is as follows: \(P(n)= Ce^{-\varepsilon_{n}/k_{b}T}\) Where C is a constant, n is quantum number (energy level), \(k_{b}\) is Boltzmann’s constant, T is temperature, and \(\varepsilon\) is the energy of the molecule, which can be determined as such: \(\varepsilon_{n} = \left(n + \dfrac{1}{2}\right)hv\) \(hv = \dfrac{h}{2\pi}\sqrt{\dfrac{k}{\mu}}\) n, as already seen, is the quantum number. \(hv\) in the equation for \(\varepsilon\) is defined as the product of planck’s constant (\(h\)) divided by \(2\pi\) and the \(\sqrt{\dfrac{k}{\mu}}\), where k is the force constant, which is given in the problem, and the \(\mu\) is the , which can be calculated as follows: \(\mu = \dfrac{m_{1}m_{2}}{m_{1} + m_{2}}\) Where \(m\) is just the mass of each of the atoms. Since the mass of both hydrogens are 1 amu (\(u\)), \(\mu = \dfrac{1\times1}{1 + 1} = \dfrac{1}{2} u\) \(\dfrac{1}{2} u = 8.3027\times10^{-28}kg\) Because the question is only interested in population, the actual number of molecules there are doesn’t really matter, since means to take the ratio between the probability of one energy state to another. Since the two energy states in comparison for this problem is \(n=1, n=0\), (remember ground state is \(n=0\)), the equation is as follows. \(\dfrac{P(1)= Ce^{-\varepsilon_{1}/k_{b} 278.15 \, \mathrm{K}}}{P(0)= Ce^{-\varepsilon_{0}/k_{b} 278.15 \, \mathrm{K}}}\) By simplification, \(= e^{\dfrac{\varepsilon_{1} - \varepsilon_{0}}{K_{b}T}}\) And, substituting in the formula for energy(\(\varepsilon\)), \(= e^{\dfrac{\left[\left(n + \dfrac{1}{2}\right)hv - \dfrac{1}{2}hv\right]}{K_{b}T}}\) \(= e^{nhv/K_{b}T}\) Here, \(n=1\) (because the 0 has already been taken into consideration) and \(hv\) is as follows: \(hv = \dfrac{h}{2\pi}\sqrt{\dfrac{k}{\mu}}\) \(hv = \dfrac{6.62607\times10^{-34}\dfrac{kgm^{2}}{s^2}}{2\pi}\sqrt{\dfrac{510\dfrac{N}{m}}{8.3027\times10^{-28}kg}}\) \(hv = \dfrac{6.62607\times10^{-34}Js^{-1}}{2\pi}\sqrt{\dfrac{510\dfrac{N}{m}}{8.3027\times10^{-28}kg}}\) \(hv= 8.2650\times 10^{-20} J\) All the variables have been found, so all there is left is to substitute and solve. \(= e^{\dfrac{-\left(1\right)\left(8.2650\times 10^{-20} J\right)}{1.3806\times10^{-23}\dfrac{m^{2}kg}{s^{2}K}\left(278.15K\right)}}\) \(= e^{\dfrac{-\left(1\right)\left(8.2650\times 10^{-20} J\right)}{1.3806\times10^{-23}\dfrac{J}{K}\left(278.15K\right)}}\) \(\approx e^{-21.5229}\) \(\approx 4.495\times10^{-10}\) Thus, the probability of the \(\ce{H2}\) molecules being in the 1st energy state when compared to the vibrational state of the lowest energy is \(\approx 4.495\times10^{-10}\). This extremely low probability makes sense, because intuition says that the compound will most likely be found at its lowest energy. : Find difference in energy, \(\varepsilon\), divide by Boltzmann's constant (\(K_{b}\)) and Temperature. Plug as \(e^{x}\) and solve. A container holds 2 L of gas under 5.00 atm and a ball floating in 10 L of NaOH. As the volume of the gas expands to 20 L, the ball is turned upside down. This turn is caused by the temperature increase of the NaOH after the gas expands. Assuming that no heat is lost, the density of NaOH is 2.13 g/cm and the specific heat is 4.184 J/g K, calculate the increase in temperature of the NaOH. The "work" completed by the ball to turn equals the negative value of the work the gas absorbs. w = -(-P (V -V )) w = -(-5 atm) (20-2) L w = 90 atm L (101.325 J/ 1 atm L) w = 9119.25 J The work carried out by the ball by the NaOH (in order to turn the ball) correlates to the increase in 10 L of NaOH temperature . 10 L of NaOH (1000 mL/1 L) = 10000 mL = 10000 cm (2.13 g/cm ) = 21300 g of NaOH q = mC = (q NaOH) / (Specific Heat NaOH * g NaOH) ΔT = (9119.25 J) / (4.184 J/K g)(21300 g) ΔT = 0.102 K gained by the NaOH to turn the ball Some gas in a piston expands against a constant pressure of 1.2 atm from a volume of 3 L to 18 L. The piston turns an egg beater submerged in 150 g of water. If the water was originally at 25°C, what is its temperature once the gas stops expanding? Assume that all of heat goes into the water and the specific heat capacity of water is 4.184 J K g . From the first law of thermodynamics \[\Delta U = q + w\nonumber \] In this case,\(\Delta U =0\), so \[q =w = P_{ext}\Delta V = P_{ext} (V_f - V_i)\nonumber \] \[ q = (1.2 \; \mathrm{atm} )(18\; \mathrm{L} - 3\; \mathrm{L}) = 14.4 \; \mathrm{L\; atm} \nonumber \] Convert From L atm to joules using the conversion factor \[14.4 \;  \mathrm{L\; atm} \times 101.325\; \mathrm{J \; L^{-1} atm^{-1}} = 1459\; \mathrm{J} \nonumber \] Now connect the heat transferred to the temperature increased via the specific heat \(c_{sp}\) via \[ q = m c_{sp} \Delta T\nonumber \] \[ \Delta T = \dfrac{q}{m c_{sp}}\nonumber \] \[\Delta T = \dfrac{1459\; \mathrm{J}}{(150\; \mathrm{g})(4.184\; \mathrm{J\; K^{-1} g^{-1}})} = 2.325 \; \mathrm{K} \nonumber \] Since initial temperatures is 25° = 298 K so \[T_{f} = 298 \; K + 2.325 \; \mathrm{K}\nonumber \] \[T_{f}=300.325\; K = 27.325 \mathrm{°C} \nonumber \]

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Sometimes, when an excited state species relaxes, giving off a photon, the wavelength of the photon is different from the one that initially led to excitation. When this happens, the photon is invariably red-shifted; its wavelength is longer than the initial one. This situation is called "fluorescence." How can that be? Isn't energy quantized? How is the molecule suddenly taking a commission out of the energy the original photon brought with it? This discrepancy is related to the Franck-Condon principle from the previous page. When an electron is promoted to an electronic excited state, it often ends up in an excited vibrational state as well. Thus, some of the energy put into electronic excitation is immediately passed into vibrational energy. Vibrational energy, however, doesn't just travel in photons. It can be gained or lost through molecular collisions and heat transfer. The electron might simply drop down again immediately; a photon would be emitted of exactly the same wavelength as the one that was previously absorbed. On the other hand, if the molecule relaxes into a lower vibrational state, some of that initial energy will have been lost as heat. When the electron relaxes, the distance back to the ground state is a little shorter. The photon that is emitted will have lower energy and longer wavelength than the initial one. Just how does a molecule undergo vibrational relaxation? Vibrational energy is the energy used to lengthen or shorten bonds, or to widen or squeeze bond angles. Given a big enough molecule, some of this vibrational energy could be transferred into bond lengths and angles further away from the electronic transition. Otherwise, if the molecule is small, it may transfer some of its energy in collisions with other molecules. There are lots of examples of energy being transferred this way in everyday life. In a game of pool, one billiard ball can transfer its energy to another, sending it toward the pocket. Barry Bonds can transfer a considerable amount of energy through his bat into a baseball, sending it out of the park, just as Serena Williams can send a whole lot of energy whizzing back at her sister. In curling, one stone can transfer its energy to another, sending it out of the house and giving Canada the gold over Sweden. In molecules, as one molecule drops to a lower vibrational state, the other will hop up to a higher vibrational state with the energy it gains. In the drawing below, the red molecule is in an electronic excited and vibrational state. In a collision, it transfers some of its vibrational energy to the blue molecule. If electrons can get to a lower energy state, and give off a little energy at a time, by hopping down to lower and lower vibrational levels, do they need to give off a giant photon at all? Maybe they can relax all the way down to the ground state via vibrational relaxation. That is certainly the case. Given lots of vibrational energy levels, and an excited state that is low enough in energy so that some of its lower vibrational levels overlap with some of the higher vibrational levels of the ground state, the electron can hop over from one state to the other, without releasing a photon. This event is called a "radiationless transition", because it occurs without release of a photon. The electron simply slides over from a low vibrational state of the excited electronic state to a high vibrational state of the electronic ground state. We will see a couple of iinds of radiationless transitions. Specifically, if the electron simply keeps dropping a vibrational level at a time back to the ground state, the process is called "internal conversion". Internal conversion has an important consequence. Because the absorption of UV and visible light can result in energy transfer into vibrational states, much of the energy that is absorbed from these sources is converted into heat. That can be a good thing if you happen to be a marine iguana trying to warm up in the sun after a plunge in the icy Pacific. It can also be a tricky thing if you are a process chemist trying to scale up a photochemical reaction for commercial production of a pharmaceutical, because you have to make sure the system has adequate cooling available. There is a very similar event, called "intersystem crossing", that leads to the electron getting caught between the excited state and the ground state. Just as, little by little, vibrational relaxation can lead the electron back onto the ground state energy surface, it can also lead the electron into states that are intermediate in energy. For example, suppose an organic molecule undergoes electronic excitation. Generally, organic molecules have no unpaired electrons. Their ground states are singlet states. According to one of our selection rules for electronic excitation, the excited state must also have no unpaired electrons. In other words, the spin on the electron that gets excited is the same after excitation as it was before excitation. However, that's not the lowest possible energy state for that electron. When we think about atomic orbital filling, there is a rule that governs the spin on the electrons in degenerate orbitals: in the lowest energy state, spin is maximized. In other words, when we draw a picture of the valence electron configuration of nitrogen, we show nitrogen's three p electrons each in its own orbital, with their spins parallel. The picture with three unpaired electrons, all with parallel spins, shows a nitrogen in the quartet spin state. Having one of those spins point the other way would result in a different spin state. One pair of electrons in the p level would be spin-paired, one up and one down, even though they are in different p orbitals. That would leave one electron without an opposite partner. The nitrogen would be in a doublet spin state. That isn't what happens. The quartet spin state is lower in energy than the doublet state. That's just one of the rules of quantum mechanics: maximize spin when orbitals are singly occupied. It's the same in a molecule. The triplet state is lower in energy than the singlet state. Why didn't the electron get excited to the triplet state in the first place? That's against the rules. But sliding down vibrationally onto the triplet state from the singlet excited state isn't, because it doesn't involve absorption of a photon. Intersystem crossing can have important consequences in reaction chemistry because it allows access to triplet states that are not normally avaiable in many molecules. Because triplet states feature unpaired electrons, their reactivity is often typified by radical processes. That means an added suite of reactions can be accessed via this process. Intersystem crossing is one way a system can end up in a triplet excited state. Even though this state is lower in energy than a singlet excited state, it can't be accessed directly via electronic excitation because that would violate the spin selection rule. That's where the electron gets stuck, though. The quick way back down to the bottom is by emitting a photon, but because that would involve a change in spin state, it isn't allowed. Realistically speaking, that means it takes a long time. By "a long time", we might mean a few seconds, several minutes, or possibly even hours. Eventually, the electron can drop back down, accompanied by the emission of a photon. This situation is called "phosphorescence". Molecules that display phosphorescence are often incorporated into toys and shirts so that they will glow in the dark. We have already seen that an excited state molecule can transfer some vibrational energy to another molecule via a collision. What about the energy of the electroic excited state? Can a molecule transfer a large quantum of energy to another -- essentially a photon's worth, but without the photon? The answer is yes. In a collision, one molecule in an electronic excited state can transfer its energy to another. In the process, the first molecule returns to the ground state and the second is excited. This process is called "photosensitization". Photosensitization can occur in a couple of different ways. Because photosensitization does not involve absorption or emission of a photon, it can also lead to formation of a triplet excited state. The significance of photosensitization is that compounds that do not have strong chromophores can still access electronic excited states if they come into contact with other molecules that do have strong chromophores. There are a number of compounds that are routinely used to induce excitation in other molecules; these photochemical enablers are referred to as photosensitizers. ,

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The entropy of gases can be experimentally measured using calorimetry (\(S^\circ_{\text{exp}}\)) or calculated using spectroscopic methods (\(S^\circ_{\text{calc}}\)). For most molecules, the experiment and calculated values are in a good agreement, however, this is not true for all molecules. The discrepancy is referred to as residual entropy: \[\bar{S}_{\text{calc}}-\bar{S}_{\text{exp}} \nonumber \] Residual entropy arises from a material that can have many different states a 0 K. The third law of thermodynamics states that at zero kelvin, a substance will have an entropy of zero. In substances, such as glass, ice, and carbon monoxide, the substance can exist in many different configurations; it is not a perfect crystal, but must still have zero entropy according to third law. The material has residual entropy.

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Metal ores are typically salts, such as oxides, carbonates or sulfides. Conversion of these ores into metals requires oxidation/reduction reactions. That's not always the case. Some early forays into metallurgy involved native gold (native meaning the metal is found in its elemental state in nature). Gold is relatively soft. It could be easily worked and shaped by heating it. Occasionally, native silver and copper can also be found. Explain, with the help of a table of standard reduction potentials, why silver and gold can sometimes be found as elements rather than salts. However, a major leap forward came when people learned to make alloys, mixing in small amounts of other metals to make harder, sturdier materials. For example, the addition of tin to copper ushered in "the bronze age". Tin itself had to be made from its ore via smelting; the earliest evidence for this process comes from what is now Turkey, where it was performed over eight thousand years ago. However, alloys were apparently not discovered until several thousand years later. In smelting, ore is heated to a high temperature in the presence of carbon sources, such as charcoal or coke. The partial combustion of the carbon source produces carbon monoxide which acts as a reducing agent. Show the half-reactions involved in the reduction of tin oxide with carbon monoxide. Use them to come up with a balanced reaction for the process, and calculate the standard potential for the reaction. Another major advance in metallurgy involved the conversion of iron ores into iron and steel. There is evidence that iron smelting in sub-Saharan Africa and Sri Lanka about three thousand years ago. Archaeological evidence in Sri Lanka shows that smelters were located on mountainsides facing the ocean, where constant winds provided ample oxygen to produce fires hot enough for smelting. In the United States, the discovery of iron ores in the states along the Great Lakes, the use of the Great Lakes as a transportation network, and the availability of anthracite coal in Pennsylvania fueled the development of an American steel industry and the rise of a major industrial power. The fact that the great lakes states are still referred to as the "rust belt" is a testament to the manufacturing prowess of the region throughout the twentieth century, which proceeded from having all the necessary features for an iron-based economy in close geographic proximity. Show the half-reactions involved in the reduction of iron oxide with carbon monoxide. Use them to come up with a balanced reaction for the process, and calculate the standard potential for the reaction. Show the half-reactions involved in the reduction of aluminum oxide with carbon monoxide. Use them to come up with a balanced reaction for the process, and calculate the standard potential for the reaction. Aluminum is a very important material in our economy. It is lightweight, strong, and forms a very hard oxide coating when exposed to the elements, rather than the rust that results from weathering steel. In contrast to the steel industry, the aluminum industry is a far-flung operation in which ore mined on one continent may be shipped to another for processing. However, aluminum metal isn't accessible via smelting. So how is it done? Just as a thermodynamically favoured redox reaction can produce a voltage in a circuit, if we already have a voltage produced by another source, we can drive an unfavourable redox reaction to completion. We can drive the reaction backwards. Quebec is a major producer of aluminum, despite being endowed with virtually no aluminum ore. Bauxite, the major aluminum-containing ore, is a mixture of minerals of formulae Al(OH) or AlO(OH) found amalgamated with other clays and minerals. It is found near the earth's surface in tropical and sub-tropical areas, left behind after of millenia of erosion and drainage of more soluble materials from underlying bedrock. The major producers of bauxite are Vietnam, Australia and Guinea, as well as a number of countries in South America. Why ship bauxite all the way to the taiga to make aluminum? Aluminum production requires a lot of electrons, and those electrons can't be provided by coal or coke. Instead, they usually come from massive hydroelectric generating stations, such as the 16,000 megawatt James Bay Project in northern Quebec. To make aluminum, you go where electricity is cheap and plentiful. The bauxite is first processed to help remove all those other materials that come mixed with the aluminum ore. It is dissolved in base, filtered and re-precipitated with acid. The residue is heated to drive off water, leaving pure alumina (Al O ). Instead of performing this redox reaction in aqueous solution, it is done in the molten state. Alumina has a melting point around 2,000 C, but that temperature drops to a much more manageable 1,000 C if a "flux" is added. To avoid contaminating the aluminum ions, cryolite has often been used as the flux, because it is also an aluminum salt. The alumina is melted in an iron vat, which conveniently functions as one of the electrodes in the redox reaction. It is the cathode, supplying electrons. Graphite anodes draw electrons out of the bath to complete a circuit. Two reactions occur: aluminum ions are reduced to aluminum at the cathode, which drops to the bottom of the vat and is drained away periodically. Oxide ions are oxidized to molecular oxygen at the anodes. However, at these temperatures, the oxygen quickly reacts with the carbon anodes to produce carbon dioxide -- that is, the anodes actually disappear as the reaction proceeds. Take a look at the redox reaction happening in the vat. Cryolite is added to get alumina to melt at a lower temperature. Unlike bauxite, it's a somewhat rare mineral found in Greenland and Quebec. Presumably, the aluminum ions in the cryolite also get reduced. Wouldn't the rare cryolite quickly get used up? Explain why this isn't a problem. ,

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The Actinide series contains elements with atomic numbers 89 to 103 and is in the sixth period and the third group of the periodic table. The series is the row below the Lanthanide series, which is located underneath the main body of the periodic table. and Actinide Series are both referred to as Rare Earth Metals. These elements all have a high diversity in oxidation numbers and all are radioactive. The most common and known element is Uranium, which is used as nuclear fuel when its converted into plutonium, through a nuclear reaction. The first actinides to be discovered were Uranium by Klaproth in 1789 and Thorium by Berezelius in 1829, but most of the Actinides were man-made products of the 20th century. Actinium and Protactinium are found in small portions in nature, as decay products of 253-Uranium and 238-Uranium. Microscopic amounts of Plutonium are made by neutron capture by Uranium, and yet occur naturally. Monazite is the principle Thorium ore. It is a phosphate ore that contains great amounts of Lanthanides in it. The main Uranium ore is U O and is known as pitchblende, because it occurs in black, pitch-like masses. An example of pitchblende is located in the picture below. All elements past Uranium are man-made. Actinides require special handling, because many of them are radioactive and/or unstable. The radiation in actinides plays a large role in the chemistry and arrangement of particles in crystals. *This is a picture of U O , a uranium, pitchblende ore, by Geomartin. Actinides have been crucial in understanding nuclear chemistry and have provided valuable usage today, such as nuclear power. These examples illustrate their importance in understanding key concepts in nuclear chemistry and related topics. Actinides are in the f-block of the periodic table. The electron configuration of uranium is [Rn] 5f 6d 7s . The reason for this arrangement unlike other conventional electron configurations such as Na with configuration of [Ne]3s . Results from difference in energy levels due to the fact that some orbitals fill in faster than others and explains why Actinides are transition metals. (see transition metals) The electron configurations of the actinides are due to the following: Knowing about the solubility and precipitations of actinides, help chemists understand their properties better. Not all Actinides have the same properties when it comes solubility and precipitation, but most of the Actinides have similar traits and characteristics. The fluorides, hydroxides and oxalates of actinides have low solubilities. The lighter trivalent, having a valence of three electrons, are unstable in aqueous solutions, such as transplutonium elements. The actinides of light weight have a higher valence; for example, oxidation states of 4, 5 and 6 where they are more stable; and they form compounds with low solubility. Trivalent Actinides can be separated from slightly acidic solutions as phosphates, mildly acidic solutions as oxalates, strongly acidic solutions as fluorides, and from basic solutions as hydroxides or hydrous oxides. Hydrogen peroxide precipitates Pu , Th , Ce , U , Np and Pa . Pu , Th , Np , UO and all trivalent actinides are precipitated by carbonate-free ammonium hydroxide. The solubility of Actinide hydroxides or hydrous oxides in strong ammonium carbonate solutions allow the separation or Uranium and Thorium from other members of the Ammonium hydroxide group, such as Fe, Ti, Al and other rare earth metals. Thorium can be precipitated from ammonia with aluminum or from aluminum as a fluoride or oxalate. Uranium is precipitated with hydrogen sulfide, unless a complexing agent is present, such as Tartaric acid \[\underbrace{\ce{Ac(OH)3 + 3HF ->[700°C] AcF3 + 3H2O}}_{Fluorides} \] \[\underbrace{\ce{Ac2O3 + 6NH4Cl ->[250°C] 2AcCl3 + 6NH3 + 3H2O}}_{Chlorides}\] \[\underbrace{\ce{Ac2O3 + 2AlBr3 ->[750°C] 2AcBr3 + Al2O3 }}_{Bromides}\] \[\underbrace{\ce{Th + 2I2->[400°C] ThI4 }}_{Iodides}\] The halides are very important binary compounds, sometimes the most important. Although they may have radioactivity that causes problems, they are useful to study to understand the trends in the Actinide series. Below are some halide characteristics of the actinide series from Actinium to Einsteinium. All Actinides form oxides with different oxidation states. The most common oxides are of the form M O , where M would be one of the elements in the Actinide series. The earliest actinides have a closer relation to the transition metals, where the oxidation state is equal to the number of electrons on the outer shell. The +4 state is more stable in the Actinide series than in the Lanthanides. The following are the different oxides of the Actinide elements: (?)-Neptunium is unknown in this oxide state, but scientists assume it does exist. The oxyhalides of Actinides are not binary but some are formed by earlier Actinides, for example, an aqueous Protactinium fluoride reacts with air to make PaO F, Thorium can form ThOX , Neptunium can make NpOF , Plutonium forms PuOF and UO F is made by Uranium (III) oxide and Hydrofluoric acid. The uranium in uranium hydride has an oxidation state of +3, which is strongly reducing, so it forms a stable hydride. Uranium hydride reacts with hydrogen gas at 250 degrees and swells up into a fine black powder, which is pyrophoric in air. It decomposes at higher temperatures to hydrogen and uranium powders. It can also react with many other compounds at certain temperatures to make halides, oxides, and other compounds, including the following: U + n → U → fission fragments + neutrons + 3.20 x10 J. 235-Uranium is bombarded with neutrons and turns into 236-Uranium. The 236-Uranium then converts into smaller pieces, 2-3 neutrons are released along with energy. These extra neutrons help create a chain reaction as more neutrons come into contact with Uranium. This uncontrolled energy eventually leads to an explosion, which is the basis of the atomic bomb. To see a lab presentation of what this reaction looks like click on the video below. The nucleus of Uranium emits radioactivity in the form of alpha particles. Alpha particles are Helium atoms with the atomic mass of 4 and an atomic number of 2. Alpha particles produce ions but have weak penetrating power that can easily be stopped by sheets of paper. Alpha particles produce large numbers of ions because of their collisions and near collisions with atoms, while traveling through matter. Their positive charge allows them to be deflected by electric and magnetic fields. By using a nuclear equation and following two rules: we can write equations, such as the following, where \(\ce{^4_2He}\) represents the alpha particle. \[\ce{ ^{238}U \rightarrow ^{234}Th + ^4_2He}\] The rules for writing a nuclear equation can apply to other radioactive decay processes, such as beta particle or gamma ray decay. Beta particles are deflected by the electric and magnetic fields in the opposite direction from alpha particles. Because they are not as big or massive as alpha particles, they are deflected more strongly than alpha particles are. Beta particles are electrons that came from the nuclei of atoms in nuclear decay processes. Although beta particles do not have actual atomic numbers, its charge is extremely close to the atomic number of -1. Many times a beta particle is small enough that its charge could be ignored during calculations. When we think of beta decay processes, we can look at the following equations and imagine a neutron within the nucleus of an atom converting to a proton or electron spontaneously. The proton remains in the nucleus and the electron is produced as a beta particle. The extra proton causes the atomic number to increase by one unit, but the mass number is unchanged. \[^{239}U \rightarrow ^{239}Np + \beta\] \[^{239}Np \rightarrow ^{239}Pu+ \beta^-\] \[^{237}U \rightarrow ^{237}Np+ \beta^-\] \[^{234}Pa \rightarrow ^{234}U + \beta^-\] \[^{234}Th \rightarrow ^{234}Pa + \beta^-\] Gamma rays (\(\gamma\)) are radiation that is emitted when the nucleus is in an excited state due to excess energy. This energy is then released in the form of gamma rays. Gamma Rays are a form of electromagnetic energy such as visible light and hence are undeflected by electric and magnetic fields. The following are some examples of gamma ray reactions: \[\ce{^{230}Th + ^1_0n \rightarrow ^{231}Th +} \gamma\] \[\ce{^{238}U + ^1_0n \rightarrow ^{239}U +} \gamma\] \[\ce{^{230}Th \rightarrow ^{230}Th +} \gamma\]

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From left to right on the periodic table, go from basic to acidic. - Ionic Bonding: no distribution of electron wave function - Ionic oxides are usually basic (element act as a base when reacting with H2O) Na2O(s) + H2O(l) --> 2NaOH(aq) --> 2Na+(aq) + 2OH-(aq) B. Oxide B. Hydroxide - Semimetal are amphoteric (elements acts as an acid and/or base when reacting depending on pH of solution) Al2O3 --> Al(OH)3 --(3H+)--> [Al(H2O)6]^(3+) (aq) --(OH-)--> [Al(OH)4]-(aq) - Covalent Bonding: almost complete distribution of electron wave function - Covalent oxides are usually acidic (elements act as an acid when reacts with H2O) SO3 + H2O(l) -> H2SO4(aq) -> H+ + HSO4- A. Oxide A Hydroxide - Ionic Bonding: no distribution of electron wave function - Bronsted Basic because they will react with proton - Lewis Basic because they can be ligands CaH2 + 2H2O -> 2H2 + Ca(OH)2 H- H+ H2 -In this case, CaH2 is basic because it reacts with water (an acid in this case) to form many hydrides by reducing a proton. - Covalent Bonding: almost complete distribution of electron wave function HF + H2O -> F- + H3O+ ....can also be written as HF(aq) <--> H+(aq) + F-(aq) H+ H+ H+ - HF is a weak acid that is bronsted acid because it will loose a proton. Therefore, HF is the weak acid, where the water acts as a silent water, and F- is the weak conjugate base.

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Nitrogen is present in almost all proteins and plays important roles in both biochemical applications and industrial applications. Nitrogen forms strong bonds because of its ability to form a triple bond with its self, and other elements. Thus, there is a lot of energy in the compounds of nitrogen. Before 100 years ago, little was known about nitrogen. Now, nitrogen is commonly used to preserve food, and as a fertilizer. Nitrogen, which makes up about 78% of our atmosphere, is a colorless, odorless, tasteless and chemically unreactive gas at room temperature. It is named from the Greek nitron + genes for soda forming. For many years during the 1500's and 1600's scientists hinted that there was another gas in the atmosphere besides carbon dioxide and oxygen. It was not until the 1700's that scientists could prove there was in fact another gas that took up mass in the atmosphere of the Earth. Discovered in 1772 by Daniel Rutherford (and independently by others such as Priestly and Cavendish) who was able to remove oxygen and carbon dioxide from a contained tube full of air. He showed that there was residual gas that did not support like oxygen or carbon dioxide. While his experiment was the one that proved that nitrogen existed, other experiments were also going in London where they called the substance "burnt" or "dephlogisticated air". Nitrogen is the fourth most abundant element in humans and it is more abundant in the known universe than carbon or silicon. Most commercially produced nitrogen gas is recovered from liquefied air. Of that amount, the majority is used to manufacture ammonia (\(NH_3\)) via the . Much is also converted to nitric acid (\(HNO_3\)). Nitrogen has two naturally occurring isotopes, and which can be separated with chemical exchanges or thermal diffusion. Nitrogen also has isotopes with 12, 13, 16, 17 masses, but they are . The two most common compounds of nitrogen are Potassium Nitrate (KNO ) and Sodium Nitrate (NaNO ). These two compounds are formed by decomposing organic matter that has potassium or sodium present and are often found in fertilizers and byproducts of industrial waste. Most nitrogen compounds have a positive Gibbs free energy (i.e., reactions are not spontaneous). The dinitrogen molecule (\(N_2\)) is an "unusually stable" compound, particularly because nitrogen forms a triple bond with itself. This triple bond is difficult hard to break. For dinitrogen to follow the octet rule, it must have a triple bond. Nitrogen has a total of 5 valence electrons, so doubling that, we would have a total of 10 valence electrons with two nitrogen atoms. The octet requires an atom to have 8 total electrons in order to have a full valence shell, therefore it needs to have a triple bond. The compound is also very inert, since it has a triple bond. Triple bonds are very hard to break, so they keep their full valence shell instead of reacting with other compounds or atoms. Think of it this way, each triple bond is like a rubber band, with three rubber bands, the nitrogen atoms are very attracted to each other. Nitrides are compounds of nitrogen with a less electronegative atom; in other words it's a compound with atoms that have a less full valence shell. These compounds form with lithium and . Nitrides usually has an oxidation state of -3. \[3Mg + N_2 \rightarrow Mg_3N_2 \label{1}\] When mixed with water, nitrogen will form ammonia and, this nitride ion acts as a very strong base. \[N + 3H_2O_{(l)} \rightarrow NH_3 + 3OH^-_{(aq)} \label{2}\] When nitrogen forms with other compounds it primarily forms covalent bonds. These are normally done with other metals and look like: MN, M N, and M N. These compounds are typically hard, inert, and have high melting points because nitrogen's ability to form triple covalent bonds. Nitrogen goes through fixation by reaction with hydrogen gas over a catalyst. This process is used to produce ammonia. As mentioned earlier, this process allows us to use nitrogen as a fertilizer because it breaks down the strong triple bond held by N The famous process for synthesis of ammonia looks like this: \[N_2 + 3H_2 \rightarrow 2NH_3 \label{3}\] Ammonia is a base and is also used in typical acid-base reactions. \[2NH_{3(aq)} + H_2SO_4 \rightarrow (NH_4)_2SO_{4(aq)} \label{4}\] Nitride ions are very strong bases, especially in aqueous solutions. Nitrides use a variety of different oxidation numbers from +1 to +5 to for oxide compounds. Almost all the oxides that form are gasses, and exist at 25 degrees Celsius. Oxides of nitrogen are acidic and easily attach protons. \[N_2O_5 + H_2O \rightarrow 2HNO_{3 (aq)} \label{5}\] The oxides play a large role in living organisms. They can be useful, yet dangerous. Hydrides of nitrogen include ammonia (NH ) and hyrdrazine (N H ). N + ___H → ___NH H N O → ? 2NH + CO → ? __Mg + N → Mg N N H + H O → ? N + 3H → 2NH (Haber process) H N O → HNO 2NH + CO → (NH ) CO + H O 2Mg + 3N → Mg N N H + H O → N + H + H O Stable forms include nitrogen-14 and nitrogen-15

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The volume ( ) occupied by moles of any gas has a pressure ( ) at temperature ( ) in Kelvin. The relationship for these variables, \[P V = n R T\] where is known as the gas constant, is called the or . Properties of the gaseous state predicted by the ideal gas law are within 5% for gases under ordinary conditions. In other words, given a set of conditions, we can predict or calculate the properties of a gas to be within 5% by applying the ideal gas law. How to apply such a law for a given set of conditions is the focus of general chemistry. At a temperature much higher than the critical temperature and at low pressures, however, the ideal gas law is a very good model for gas behavior. When dealing with gases at low temperature and at high pressure, correction has to be made in order to calculate the properties of a gas in industrial and technological applications. One of the common corrections made to the ideal gas law is the van der Waal's equation, but there are also other methods dealing with the deviation of gas from ideality. Repeated experiments show that at standard temperature (273 K) and pressure (1 atm or 101325 N/m ), one mole ( = 1) of gas occupies 22.4 L volume. Using this experimental value, you can evaluate the \(\begin{align} R &= \dfrac{P V}{n T} = \mathrm{\dfrac{1\: atm\:\: 22.4\: L}{1\: mol\:\: 273\: K}}\\ \\ &= \mathrm{0.08205\: \dfrac{L\: atm}{mol\cdot K}} \end{align}\) When SI units are desirable, = 101325 N/m (Pa for pascal) instead of 1 atm. The volume is 0.0224 m . The numerical value and units for are \(\begin{align} R &= \mathrm{\dfrac{101325\: \dfrac{N}{m^2}\: 0.0224\: m^3}{1\: mol\: 273\: K}}\\ \\ &= \mathrm{8.314\: \dfrac{J}{mol\cdot K}} \end{align}\) Note that \(\mathrm{1\: L\: atm = 0.001\: m^3 \times 101325\: \dfrac{N}{m^2} = 101.325\: J\: (or\: N\: m)}\). Since energy can be expressed in many units, other numerical values and units for are frequently in use. For your information, the gas constant can be expressed in the following values and units. \(\begin{align} R &= \mathrm{0.08205\: \dfrac{L\: atm}{mol \cdot K}} &&\textrm{Notes:} \\ &= \mathrm{8.3145\: \dfrac{L\: kPa}{mol\cdot K}} &&\mathrm{1\: atm = 101.32\: kPa} \\ &= \mathrm{8.3145\: \dfrac{J}{mol\cdot K}} &&\mathrm{1\: J = 1\: L\: kPa} \\ &= \mathrm{1.987\: \dfrac{cal}{mol\cdot K}} &&\mathrm{1\: cal = 4.182\: J} \\ &= \mathrm{62.364\: \dfrac{L\: torr}{mol\cdot K}} &&\mathrm{1\: atm = 760\: torr}\\ \end{align}\) The gas constant is such a universal constant for all gases that its values are usually listed in the "Physical Constants" of textbooks and handbooks. It is also listed in Constants of our HandbookMenu at the left bottom. Although we try to use SI units all the time, the use of atm for pressure is still common. Thus, we often use R = 8.314 J / (mol·K) or 8.3145 J / mol·K. The volume occupied by one mole, = 1, of substance is called the , \(V_{\textrm{molar}} = \dfrac{V}{n}\). Using the molar volume notation, the ideal gas law is: \(P V_{\textrm{molar}} = R T\) The ideal gas law has four parameters and a constant, , \(P V = n R T\), and it can be rearranged to give an expression for each of or . For example, \(P = \dfrac{n R T}{V}\) (Boyle's law) \(P = \left(\dfrac{n R}{V}\right) T\) (Charles's law) These equations are Boyle's law and Charles's law respectively. Similar expressions can be derived for and in terms of other variables. Thus, there are many applications. However, you must make sure that you use the proper numerical value for the gas constant according to the units you have for the parameters. Furthermore, \(\dfrac{n}{V}\) is number of moles per unit volume, and this quantity has the same units as the concentration ( ). Thus, the concentration is a function of pressure and temperature, \(C = \dfrac{P}{R T}\) At 1.0 atm pressure and room temperature of 298 K, the concentration of an ideal gas is 0.041 mol/L. Avogadro's law can be further applied to correlate gas density \(\rho\) (weight per unit volume or ) and molecular mass of a gas. The following equation is easily derived from the ideal gas law: \[P M =\dfrac{n M}{V}R T\] Thus, we have \(\begin{align} P M &= \dfrac{d R T}{M}\\ \rho &= \dfrac{n M}{V} \leftarrow \textrm{definition, and}\\ \rho &= \dfrac{P M}{R T}\\ M &= \dfrac{d R T}{P} \end{align}\) An air sample containing only nitrogen and oxygen gases has a density of 1.3393 g / L at STP. Find the weight and mole percentages of nitrogen and oxygen in the sample. From the density \(\rho\), we can evaluate an average molecular weight (also called molar mass). \(\begin{align} P M &= d R T\\ M &= 22.4 \times d\\ &= \mathrm{22.4\: L/mol \times 1.3393\: g/L}\\ &= \mathrm{30.0\: g / mol} \end{align}\) Assume that we have 1.0 mol of gas, and mol of which is nitrogen, then (1 - ) is the amount of oxygen. The average molar mass is the mole weighted average, and thus, \(\mathrm{28.0\, x + 32.0 (1 - x) = 30.0}\) \(\mathrm{- 4\, x = - 2}\) \(\mathrm{x = 0.50\: mol\: of\: N_2,\: and\: 1.0 - 0.50 = 0.50\: mol\: O_2}\) Now, to find the weight percentage, find the amounts of nitrogen and oxygen in 1.0 mol (30.0 g) of the mixture. \(\mathrm{Mass\: of\: 0.5\: mol\: nitrogen = 0.5 \times 28.0 = 14.0\: g}\) \(\mathrm{Mass\: of\: 0.5\: mol\: oxygen = 0.5 \times 32.0 = 16.0\: g}\) \(\mathrm{Percentage\: of\: nitrogen = 100 \times \dfrac{14.0}{30.0} = 46.7 \% }\) \(\mathrm{Percentage\: of\: oxygen = 100 \times \dfrac{16.0}{30.0} = 100 - 46.7 = 53.3 \%}\) We can find the density of pure nitrogen and oxygen first and evaluate the fraction from the density. \(\rho \mathrm{\: of\: N_2 = \dfrac{28.0}{22.4} = 1.2500\: g/L}\) \(\rho \mathrm{\: of\: O_2 = \dfrac{32.0}{22.4} = 1.4286\: g/L}\) \(\mathrm{1.2500\, x + 1.4286 (1 - x) = 1.3393}\) Solving for x gives \(\mathrm{x = 0.50}\) (same result as above) Now, repeat the calculations for a mixture whose density is 1.400 g/L. What is the density of acetone (\(\ce{C3H6O}\)) vapor at 1.0 atm and 400 K? The molar mass of acetone = 3*12.0 + 6*1.0 + 16.0 = 58.0. Thus, \(\begin{align} \rho &= \dfrac{P M}{R T}\\ \\ &= \mathrm{\dfrac{1.0 \times 58.0\: atm\: \dfrac{g}{mol}} {0.08205\: \dfrac{L\: atm}{mol\: K} \times 400\: K}}\\ \\ &= \mathrm{1.767\: g / L} \end{align}\) The density of acetone is 1.767 g/L; calculate its molar mass. \(P V = n R T\)? Hint: number of moles of gas in a closed system. Describe the ideal gas law. Hint: one The ideal gas equation shows the interdependence of the variables. Only one of them can be varied independently. Hint: 41.0 Evaluate molar volume at any condition. Hint: Both pressure and temperature will increase. Explain a closed system and apply ideal gas law. Hint: a Explain Charles's law. Hint: d Apply the skills acquired in math courses to chemical problem solving. Hint: b Rearrange a mathematical equation. Use conversion factors, for example: \(\mathrm{8.314\: J\times\dfrac{1\: cal}{4.184\: J}=\: ?\: cal}\) \(\ce{H2}\) Hint: 0.045 mol/L There are many methods for calculating this value. \(\ce{CO2}\) \(\mathrm{CO_2 = 44}\) Hint: 5.89 g in 3 L It contains \(n \mathrm{= \dfrac{1\: atm \times 3\: L}{0.08205\: \dfrac{L\: atm}{mol\cdot K}\times 273\: K}}\) \(\ce{N2}\) Hint: 82.1 atm Depending on the numerical value and units of you use, you will get the pressure in various units. At 1000 K, some of the \(\ce{N2}\) molecules may dissociate. If that is true, the pressure will be higher! \(\ce{N2}\) Hint: 240 K At = 240 K, ideal gas law may not apply to \(\ce{CO2}\), because this gas liquifies at a rather high temperature. The ideal gas law is still good for \(\ce{N2}\), \(\ce{H2}\), \(\ce{O2}\) etc, because these gases liquify at much lower temperatures.

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In this section, we describe several systems that illustrate the roles transition metals play in biological systems. Our goal is for you to understand why the chemical properties of these elements make them essential for life. We begin with a discussion of the strategies organisms use to extract transition metals from their environment. The section continues with a brief discussion of the use of transition metals in reactions that involve the transfer of electrons, reactions of small molecules such as O , Lewis-acid catalysis, and the generation of reactive organic radicals. There are three possible dietary levels for any essential element: deficient, optimal, and toxic, in order of increasing concentration in the diet. If the concentration of an essential element in the diet is too low, an organism must be able to extract the element from the environment and concentrate it. If the concentration of an essential element in the diet is too high, an organism must be able to limit its intake to avoid toxic effects. Moreover, organisms must be able to switch off the uptake process rapidly if dietary levels rise suddenly, and they must be able to store essential elements for future use. Three distinct steps are involved in transition metal uptake. First, the metal must be “mobilized” from the environment and brought into contact with a cell in a form that can be absorbed. Second, the metal must be transported across the cell membrane into the cell. Third, the element must be transported to its point of utilization within a cell or to other cells within the organism. In our discussion, we focus on the uptake, transport, and storage of iron, which illustrates the most important points. Because iron deficiency (anemia) is the most widespread nutritional deficiency known in humans, the uptake of iron is especially well understood. The solubility of metal ions such as Fe , which form highly insoluble hydroxides, depends on the pH and the presence of complexing agents. In an oxygen-containing atmosphere, iron exists as Fe(III) because of the positive reduction potential of Fe (Fe + e → Fe ; E° = +0.77 V). Because ferric hydroxide [Fe(OH) ] is highly insoluble (K ≈ 1 × 10 ), the equilibrium concentration of Fe (aq) at pH 7.0 is very low, about 10 M. You would have to drink 2 × 10 L of iron-saturated water per day (roughly 5 mi ) to consume the recommended daily intake of Fe for humans, which is about 1 mg/day. Animals such as humans can overcome this problem by consuming concentrated sources of iron, such as red meat, but microorganisms cannot. Consequently, most microorganisms synthesize and secrete organic molecules called siderophores to increase the total concentration of available iron in the surrounding medium. Siderophores are generally cyclic compounds that use bidentate ligands, such as the hydroxamate and catecholate groups shown here, to bind Fe in an octahedral arrangement. Typical siderophores are ferrichrome, a cyclic peptide produced by fungi, and enterobactin, a cyclic ester produced by bacteria (Figure \(\Page {1}\)). Attaching the three iron ligands to a cyclic framework greatly increases the stability of the resulting Fe complex due to the chelate effect described in Section 23.4. The formation constants for the Fe complexes of ferrichrome and enterobactin are about 10 and 10 , respectively, which are high enough to allow them to dissolve almost any Fe(III) compound. Siderophores increase the [Fe ] in solution, providing the bacterium that synthesized them (as well as any competitors) with a supply of iron. In addition, siderophores neutralize the positive charge on the metal ion and provide a hydrophobic “wrapping” that enables the Fe –siderophore complex to be recognized by a specific protein that transports it into the interior of a cell. Once it is inside a cell, the iron is reduced to Fe , which has a much lower affinity for the siderophore and spontaneously dissociates. In contrast, multicellular organisms can increase the concentration of iron in their diet by lowering the pH in the gastrointestinal tract. At pH 1.0 (the approximate pH of the stomach), most Fe(III) salts dissolve to form Fe (aq), which is absorbed by specific proteins in the intestinal wall. A protein called transferrin forms a complex with iron(III), allowing it to be transported to other cells. Proteins that bind tightly to Fe(III) can also be used as antibacterial agents because iron is absolutely essential for bacterial growth. For example, milk, tears, and egg white all contain proteins similar to transferrin, and their high affinity for Fe allows them to sequester iron, thereby preventing bacteria from growing in these nutrient-rich media. Iron is released from transferrin by reduction to Fe , and then it is either used immediately (e.g., for the synthesis of hemoglobin) or stored in a very large protein called ferritin for future use (Figure \(\Page {2}\)). Ferritin uses oxygen to oxidize Fe to Fe , which at neutral pH precipitates in the central cavity of the protein as a polymeric mixture of Fe(OH) and FePO . Because a fully loaded ferritin molecule can contain as many as 4500 Fe atoms, which corresponds to about 25% Fe by mass, ferritin is an effective way to store iron in a highly concentrated form. When iron is needed by a cell, the Fe is reduced to the much more soluble Fe by a reductant such as ascorbic acid (vitamin C). The structure of ferritin contains channels at the junctions of the subunits, which provide pathways for iron to enter and leave the interior of a molecule. A protein that contains one or more metal ions tightly bound to amino acid side chains is called a metalloprotein; some of the most common ligands provided by amino acids are shown here. A metalloprotein that catalyzes a chemical reaction is a metalloenzyme. Thus all metalloenzymes are metalloproteins, but the converse is not true. Recent estimates suggest that more than 40% of all known enzymes require at least one metal ion for activity, including almost all the enzymes responsible for the synthesis, duplication, and repair of DNA (deoxyribonucleic acid) and RNA (ribonucleic acid). Proteins whose function is to transfer electrons from one place to another are called electron-transfer proteins. Because they do not catalyze a chemical reaction, electron-transfer proteins are not enzymes; they are biochemical reductants or oxidants consumed in an enzymatic reaction. The general reaction for an electron-transfer protein is as follows: Because many transition metals can exist in more than one oxidation state, electron-transfer proteins usually contain one or more metal ions that can undergo a redox reaction. Incorporating a metal ion into a protein has three important biological consequences: Three important classes of metalloproteins transfer electrons: blue copper proteins, cytochromes, and iron–sulfur proteins, which generally transfer electrons at high (> 0.20 V), intermediate (±0 V), and low (−0.20 to −0.50 V) potentials, respectively (Table 23.12). Although these electron-transfer proteins contain different metals with different structures, they are all designed to ensure rapid electron transfer to and from the metal. Thus when the protein collides with its physiological oxidant or reductant, electron transfer can occur before the two proteins diffuse apart. For electron transfer to be rapid, the metal sites in the oxidized and reduced forms of the protein must have similar structures. Blue copper proteins were first isolated from bacteria in the 1950s and from plant tissues in the early 1960s. The intense blue color of these proteins is due to a strong absorption band at a wavelength of about 600 nm. Although simple Cu complexes, such as [Cu(H O) ] and [Cu(NH ) ] , are also blue due to an absorption band at 600 nm, the intensity of the absorption band is about 100 times less than that of a blue copper protein. Moreover, the reduction potential for the Cu /Cu couple in a blue copper protein is usually +0.3 to +0.5 V, considerably more positive than that of the aqueous Cu /Cu couple (+0.15 V). The copper center in blue copper proteins has a distorted tetrahedral structure, in which the copper is bound to four amino acid side chains (Figure \(\Page {3}\)). Although the most common structures for four-coordinate Cu and Cu complexes are square planar and tetrahedral, respectively, the structures of the oxidized (Cu ) and reduced (Cu ) forms of the protein are essentially identical. Thus the protein forces the Cu ion to adopt a higher-energy structure that is more suitable for Cu , which makes the Cu form easier to reduce and raises its reduction potential. Moreover, by forcing the oxidized and reduced forms of the metal complex to have essentially the same structure, the protein ensures that electron transfer to and from the copper site is rapid because only minimal structural reorganization of the metal center is required. Kinetics studies on simple metal complexes have shown that electron-transfer reactions tend to be slow when the structures of the oxidized and reduced forms of a metal complex are very different, and fast when they are similar. You will see that other metal centers used for biological electron-transfer reactions are also set up for minimal structural reorganization after electron transfer, which ensures the rapid transfer of electrons. The cytochromes (from the Greek cytos, meaning “cell”, and chroma, meaning “color”) were first identified in the 1920s by spectroscopic studies of cell extracts. Based on the wavelength of the maximum absorption in the visible spectrum, they were classified as cytochromes a (with the longest wavelength), cytochromes b (intermediate wavelength), and cytochromes c (shortest wavelength). It quickly became apparent that there was a correlation between their spectroscopic properties and other physical properties. For examples, cytochromes c are generally small, soluble proteins with a reduction potential of about +0.25 V, whereas cytochromes b are larger, less-soluble proteins with reduction potentials of about 0 V. All cytochromes contain iron, and the iron atom in all cytochromes is coordinated by a planar array of four nitrogen atoms provided by a cyclic tetradentate ligand called a porphyrin. The iron–porphyrin unit is called a heme group. The structures of a typical porphyrin (protoporphyrin IX) and its iron complex (protoheme) are shown here. In addition to the four nitrogen atoms of the porphyrin, the iron in a cytochrome is usually bonded to two additional ligands provided by the protein, as shown in Figure \(\Page {4}\). A cytochrome. Shown here is protoporphyrin IX and its iron complex, protoheme. In contrast to the blue copper proteins, two electron configurations are possible for both the oxidized and reduced forms of a cytochrome, and this has significant structural consequences. Thus Fe is d and can be either high spin (with four unpaired electrons) or low spin (with no unpaired electrons). Similarly, Fe is d and can also be high spin (with five unpaired electrons) or low spin (with one unpaired electron). In low-spin heme complexes, both the Fe and the Fe ions are small enough to fit into the “hole” in the center of the porphyrin; hence the iron atom lies almost exactly in the plane of the four porphyrin nitrogen atoms in both cases. Because cytochromes b and c are low spin in both their oxidized and reduced forms, the structures of the oxidized and reduced cytochromes are essentially identical. Hence minimal structural changes occur after oxidation or reduction, which makes electron transfer to or from the heme very rapid. Electron transfer reactions occur most rapidly when minimal structural changes occur during oxidation or reduction. Although all known bacteria, plants, and animals use iron–sulfur proteins to transfer electrons, the existence of these proteins was not recognized until the late 1950s. Iron–sulfur proteins transfer electrons over a wide range of reduction potentials, and their iron content can range from 1 to more than 12 Fe atoms per protein molecule. In addition, most iron–sulfur proteins contain stoichiometric amounts of sulfide (S ). These properties are due to the presence of four different kinds of iron–sulfur units, which contain one, two, three, or four iron atoms per Fe–S complex (Figure \(\Page {5}\)). In all cases, the Fe and Fe ions are coordinated to four sulfur ligands in a tetrahedral environment. Due to tetrahedral coordination by weak-field sulfur ligands, the iron is high spin in both the Fe and Fe oxidation states, which results in similar structures for the oxidized and reduced forms of the Fe–S complexes. Consequently, only small structural changes occur after oxidation or reduction of the Fe–S center, which results in rapid electron transfer. Although small molecules, such as O , N , and H , do not react with organic compounds under ambient conditions, they do react with many transition-metal complexes. Consequently, virtually all organisms use metalloproteins to bind, transport, and catalyze the reactions of these molecules. Probably the best-known example is hemoglobin, which is used to transport O in many multicellular organisms. Under ambient conditions, small molecules, such as O , N , and H , react with transition-metal complexes but not with organic compounds. Many microorganisms and most animals obtain energy by respiration, the oxidation of organic or inorganic molecules by O . At 25°C, however, the concentration of dissolved oxygen in water in contact with air is only about 0.25 mM. Because of their high surface area-to-volume ratio, aerobic microorganisms can obtain enough oxygen for respiration by passive diffusion of O through the cell membrane. As the size of an organism increases, however, its volume increases much more rapidly than its surface area, and the need for oxygen depends on its volume. Consequently, as a multicellular organism grows larger, its need for O rapidly outstrips the supply available through diffusion. Unless a transport system is available to provide an adequate supply of oxygen for the interior cells, organisms that contain more than a few cells cannot exist. In addition, O is such a powerful oxidant that the oxidation reactions used to obtain metabolic energy must be carefully controlled to avoid releasing so much heat that the water in the cell boils. Consequently, in higher-level organisms, the respiratory apparatus is located in internal compartments called mitochondria, which are the power plants of a cell. Oxygen must therefore be transported not only to a cell but also to the proper compartment within a cell. Three different chemical solutions to the problem of oxygen transport have developed independently in the course of evolution, as indicated in Table \(\Page {2}\). Mammals, birds, reptiles, fish, and some insects use a heme protein called hemoglobin to transport oxygen from the lungs to the cells, and they use a related protein called myoglobin to temporarily store oxygen in the tissues. Several classes of invertebrates, including marine worms, use an iron-containing protein called hemerythrin to transport oxygen, whereas other classes of invertebrates (arthropods and mollusks) use a copper-containing protein called hemocyanin. Despite the presence of the hem- prefix, hemerythrin and hemocyanin do not contain a metal–porphyrin complex. Myoglobin is a relatively small protein that contains 150 amino acids. The functional unit of myoglobin is an iron–porphyrin complex that is embedded in the protein (Figure 26.8.1). In myoglobin, the heme iron is five-coordinate, with only a single histidine imidazole ligand from the protein (called the proximal histidine because it is near the iron) in addition to the four nitrogen atoms of the porphyrin. A second histidine imidazole (the distal histidine because it is more distant from the iron) is located on the other side of the heme group, too far from the iron to be bonded to it. Consequently, the iron atom has a vacant coordination site, which is where O binds. In the ferrous form (deoxymyoglobin), the iron is five-coordinate and high spin. Because high-spin Fe is too large to fit into the “hole” in the center of the porphyrin, it is about 60 pm above the plane of the porphyrin. When O binds to deoxymyoglobin to form oxymyoglobin, the iron is converted from five-coordinate (high spin) to six-coordinate (low spin; Figure 26.8.2). Because low-spin Fe and Fe are smaller than high-spin Fe , the iron atom moves into the plane of the porphyrin ring to form an octahedral complex. The O pressure at which half of the molecules in a solution of myoglobin are bound to O (P ) is about 1 mm Hg (1.3 × 10 atm). A vacant coordination site at a metal center in a protein usually indicates that a small molecule will bind to the metal ion, whereas a coordinatively saturated metal center is usually involved in electron transfer. Hemoglobin consists of two subunits of 141 amino acids and two subunits of 146 amino acids, both similar to myoglobin; it is called a tetramer because of its four subunits. Because hemoglobin has very different O -binding properties, however, it is not simply a “super myoglobin” that can carry four O molecules simultaneously (one per heme group). The shape of the O -binding curve of myoglobin (Mb; Figure \(\Page {7}\)) can be described mathematically by the following equilibrium: \[MbO_2 \rightleftharpoons Mb + O_ 2 \label{26.8.1a}\] \[K_{diss}=\dfrac{[Mb,O_2]}{[MbO_2]} \label{26.8.1b}\] In contrast, the O -binding curve of hemoglobin is S shaped (Figure \(\Page {8}\)). As shown in the curves, at low oxygen pressures, the affinity of deoxyhemoglobin for O is substantially lower than that of myoglobin, whereas at high O pressures the two proteins have comparable O affinities. The physiological consequences of the unusual S-shaped O -binding curve of hemoglobin are enormous. In the lungs, where O pressure is highest, the high oxygen affinity of deoxyhemoglobin allows it to be completely loaded with O , giving four O molecules per hemoglobin. In the tissues, however, where the oxygen pressure is much lower, the decreased oxygen affinity of hemoglobin allows it to release O , resulting in a net transfer of oxygen to myoglobin. The S-shaped O -binding curve of hemoglobin is due to a phenomenon called cooperativity, in which the affinity of one heme for O depends on whether the other hemes are already bound to O . Cooperativity in hemoglobin requires an interaction between the four heme groups in the hemoglobin tetramer, even though they are more than 3000 pm apart, and depends on the change in structure of the heme group that occurs with oxygen binding. The structures of deoxyhemoglobin and oxyhemoglobin are slightly different, and as a result, deoxyhemoglobin has a much lower O affinity than myoglobin, whereas the O affinity of oxyhemoglobin is essentially identical to that of oxymyoglobin. Binding of the first two O molecules to deoxyhemoglobin causes the overall structure of the protein to change to that of oxyhemoglobin; consequently, the last two heme groups have a much higher affinity for O than the first two. Oxygen is not unique in its ability to bind to a ferrous heme complex; small molecules such as CO and NO bind to deoxymyoglobin even more tightly than does O . The interaction of the heme iron with oxygen and other diatomic molecules involves the transfer of electron density from the filled t orbitals of the low-spin d Fe ion to the empty π* orbitals of the ligand. In the case of the Fe –O interaction, the transfer of electron density is so great that the Fe–O unit can be described as containing low-spin Fe (d ) and O . We can therefore represent the binding of O to deoxyhemoglobin and its release as a reversible redox reaction: \[Fe^{2+} + O_2 \rightleftharpoons \ce{Fe^{3+}–O_2^−} \label{26.8.2}\] As shown in Figure \(\Page {9}\), the Fe–O unit is bent, with an Fe–O–O angle of about 130°. Because the π* orbitals in CO are empty and those in NO are singly occupied, these ligands interact more strongly with Fe than does O , in which the π* orbitals of the neutral ligand are doubly occupied. Although CO has a much greater affinity for a ferrous heme than does O (by a factor of about 25,000), the affinity of CO for deoxyhemoglobin is only about 200 times greater than that of O , which suggests that something in the protein is decreasing its affinity for CO by a factor of about 100. Both CO and NO bind to ferrous hemes in a linear fashion, with an Fe–C(N)–O angle of about 180°, and the difference in the preferred geometry of O and CO provides a plausible explanation for the difference in affinities. As shown in Figure \(\Page {9}\), the imidazole group of the distal histidine is located precisely where the oxygen atom of bound CO would be if the Fe–C–O unit were linear. Consequently, CO cannot bind to the heme in a linear fashion; instead, it is forced to bind in a bent mode that is similar to the preferred structure for the Fe–O unit. This results in a significant decrease in the affinity of the heme for CO, while leaving the O affinity unchanged, which is important because carbon monoxide is produced continuously in the body by degradation of the porphyrin ligand (even in nonsmokers). Under normal conditions, CO occupies approximately 1% of the heme sites in hemoglobin and myoglobin. If the affinity of hemoglobin and myoglobin for CO were 100 times greater (due to the absence of the distal histidine), essentially 100% of the heme sites would be occupied by CO, and no oxygen could be transported to the tissues. Severe carbon-monoxide poisoning, which is frequently fatal, has exactly the same effect. Thus the primary function of the distal histidine appears to be to decrease the CO affinity of hemoglobin and myoglobin to avoid self-poisoning by CO. Hemerythrin is used to transport O in a variety of marine invertebrates. It is an octamer (eight subunits), with each subunit containing two iron atoms and binding one molecule of O . Deoxyhemerythrin contains two Fe ions per subunit and is colorless, whereas oxyhemerythrin contains two Fe ions and is bright reddish violet. These invertebrates also contain a monomeric form of hemerythrin that is located in the tissues, analogous to myoglobin. The binding of oxygen to hemerythrin and its release can be described by the following reaction, where the HO ligand is the hydroperoxide anion derived by the deprotonation of hydrogen peroxide (H O ): \[\ce{2Fe^{2+} + O2 + H^{+} <=> 2Fe^{3+}–O2H} \label{23.17}\] Thus O binding is accompanied by the transfer of two electrons (one from each Fe ) and a proton to O . Hemocyanin is used for oxygen transport in many arthropods (spiders, crabs, lobsters, and centipedes) and in mollusks (shellfish, octopi, and squid); it is responsible for the bluish-green color of their blood. The protein is a polymer of subunits that each contain two copper atoms (rather than iron), with an aggregate molecular mass of greater than 1,000,000 amu. Deoxyhemocyanin contains two Cu ions per subunit and is colorless, whereas oxyhemocyanin contains two Cu ions and is bright blue. As with hemerythrin, the binding and release of O correspond to a two-electron reaction: \[\ce{2Cu^{+} + O2 <=> Cu^{2+}–O2^{2−}–Cu^{2+}} \label{23.18}\] Although hemocyanin and hemerythrin perform the same basic function as hemoglobin, these proteins are not interchangeable. In fact, hemocyanin is so foreign to humans that it is one of the major factors responsible for the common allergies to shellfish. Myoglobin, hemoglobin, hemerythrin, and hemocyanin all use a transition-metal complex to transport oxygen. Many of the enzymes involved in the biological reactions of oxygen contain metal centers with structures that are similar to those used for O transport. Many of these enzymes also contain metal centers that are used for electron transfer, which have structures similar to those of the electron-transfer proteins discussed previously. In this section, we briefly describe two of the most important examples: dioxygenases and methane monooxygenase. Dioxygenases are enzymes that insert both atoms of O into an organic molecule. In humans, dioxygenases are responsible for cross-linking collagen in connective tissue and for synthesizing complex organic molecules called prostaglandins, which trigger inflammation and immune reactions. Iron is by far the most common metal in dioxygenases; and the target of the most commonly used drug in the world, aspirin, is an iron enzyme that synthesizes a specific prostaglandin. Aspirin inhibits this enzyme by binding to the iron atom at the active site, which prevents oxygen from binding. Methane monooxygenase catalyzes the conversion of methane to methanol. The enzyme is a monooxygenase because only one atom of O is inserted into an organic molecule, while the other is reduced to water: \[\ce{CH_4 + O_2 + 2e^- + 2H^+ \rightarrow CH_3OH + H_2O} \label{23.19}\] Because methane is the major component of natural gas, there is enormous interest in using this reaction to convert methane to a liquid fuel (methanol) that is much more convenient to ship and store. Because the C–H bond in methane is one of the strongest C–H bonds known, however, an extraordinarily powerful oxidant is needed for this reaction. The active site of methane monooxygenase contains two Fe atoms that bind O , but the details of how the bound O is converted to such a potent oxidant remain unclear. Reactions catalyzed by metal ions that do not change their oxidation states during the reaction are usually group transfer reactions, in which a group such as the phosphoryl group (−PO ) is transferred. These enzymes usually use metal ions such as Zn , Mg , and Mn , and they range from true metalloenzymes, in which the metal ion is tightly bound, to metal-activated enzymes, which require the addition of metal ions for activity. Because tight binding is usually the result of specific metal–ligand interactions, metalloenzymes tend to be rather specific for a particular metal ion. In contrast, the binding of metal ions to metal-activated enzymes is largely electrostatic in nature; consequently, several different metal ions with similar charges and sizes can often be used to give an active enzyme. Metalloenzymes generally contain a specific metal ion, whereas metal-activated enzymes can use any of several metal ions of similar size and charge. A metal ion that acts as a Lewis acid can catalyze a group transfer reaction in many different ways, but we will focus on only one of these, using a zinc enzyme as an example. Carbonic anhydrase is found in red blood cells and catalyzes the reaction of CO with water to give carbonic acid. \[\ce{CO_2(g) + H_2O(l) \rightleftharpoons H^{+}(aq) + HCO^{-}3(aq)} \label{23.20}\] Although this reaction occurs spontaneously in the absence of a catalyst, it is too slow to absorb all the CO generated during respiration. Without a catalyst, tissues would explode due to the buildup of excess CO pressure. Carbonic anhydrase contains a single Zn ion per molecule, which is coordinated by three histidine imidazole ligands and a molecule of water. Because Zn is a Lewis acid, the pK of the Zn –OH unit is about 8 versus 14 for pure water. Thus at pH 7–8, a significant fraction of the enzyme molecules contain the Zn –OH group, which is much more reactive than bulk water. When carbon dioxide binds in a nonpolar site next to the Zn –OH unit, it reacts rapidly to give a coordinated bicarbonate ion that dissociates from the enzyme: \[\ce{Zn^{2+}–OH^{-} + CO_2 \rightleftharpoons Zn^{2+}–OCO_2H^- \rightleftharpoons Zn^{2+} + HCO_3^{-}} \label{23.21}\] Thus the function of zinc in carbonic anhydrase is to generate the hydroxide ion at pH 7.0, far less than the pH required in the absence of the metal ion. An organic radical is an organic species that contains one or more unpaired electrons. Chemists often consider organic radicals to be highly reactive species that produce undesirable reactions. For example, they have been implicated in some of the irreversible chemical changes that accompany aging. It is surprising, however, that organic radicals are also essential components of many important enzymes, almost all of which use a metal ion to generate the organic radical within the enzyme. These enzymes are involved in the synthesis of hemoglobin and DNA, among other important biological molecules, and they are the targets of pharmaceuticals for the treatment of diseases such as anemia, sickle-cell anemia, and cancer. In this section, we discuss one class of radical enzymes that use vitamin B . Vitamin B was discovered in the 1940s as the active agent in the cure of pernicious anemia, which does not respond to increased iron in the diet. Humans need only tiny amounts of vitamin B , and the average blood concentration in a healthy adult is only about 3.5 × 10 M. The structure of vitamin B , shown in Figure \(\Page {10}\), is similar to that of a heme, but it contains cobalt instead of iron, and its structure is much more complex. In fact, vitamin B has been called the most complex nonpolymeric biological molecule known and was the first naturally occurring organometallic compound to be isolated. When vitamin B (the form present in vitamin tablets) is ingested, the axial cyanide ligand is replaced by a complex organic group. The cobalt–carbon bond in the enzyme-bound form of vitamin B and related compounds is unusually weak, and it is particularly susceptible to homolytic cleavage: Homolytic cleavage of the Co –CH R bond produces two species, each of which has an unpaired electron: a d Co derivative and an organic radical, ·CH R, which is used by vitamin B -dependent enzymes to catalyze a wide variety of reactions. Virtually all vitamin B -catalyzed reactions are rearrangements in which an H atom and an adjacent substituent exchange positions: In the conversion of ethylene glycol to acetaldehyde, the initial product is the hydrated form of acetaldehyde, which rapidly loses water: The enzyme uses the ·CH R radical to temporarily remove a hydrogen atom from the organic substrate, which then rearranges to give a new radical. Transferring the hydrogen atom back to the rearranged radical gives the product and regenerates the ·CH R radical. The metal is not involved in the actual catalytic reaction; it provides the enzyme with a convenient mechanism for generating an organic radical, which does the actual work. Many examples of similar reactions are now known that use metals other than cobalt to generate an enzyme-bound organic radical. Nearly all vitamin B -catalyzed reactions are rearrangements that occur via a radical reaction. Three separate steps are required for organisms to obtain essential transition metals from their environment: mobilization of the metal, transport of the metal into the cell, and transfer of the metal to where it is needed within a cell or an organism. The process of iron uptake is best understood. To overcome the insolubility of Fe(OH) , many bacteria use organic ligands called siderophores, which have high affinity for Fe(III) and are secreted into the surrounding medium to increase the total concentration of dissolved iron. The iron–siderophore complex is absorbed by a cell, and the iron is released by reduction to Fe(II). Mammals use the low pH of the stomach to increase the concentration of dissolved iron. Iron is absorbed in the intestine, where it forms an Fe(III) complex with a protein called transferrin that is transferred to other cells for immediate use or storage in the form of ferritin. Proteins that contain one or more tightly bound metal ions are called metalloproteins, and metalloproteins that catalyze biochemical reactions are called metalloenzymes. Proteins that transfer electrons from one place to another are called electron-transfer proteins. Most electron-transfer proteins are metalloproteins, such as iron–sulfur proteins, cytochromes, and blue copper proteins that accept and donate electrons. The oxidized and reduced centers in all electron-transfer proteins have similar structures to ensure that electron transfer to and from the metal occurs rapidly. Metalloproteins also use the ability of transition metals to bind small molecules, such as O , N , and H , to transport or catalyze the reactions of these small molecules. For example, hemoglobin, hemerythrin, and hemocyanin, which contain heme iron, nonheme iron, and copper, respectively, are used by different kinds of organisms to bind and transfer O . Other metalloenzymes use transition-metal ions as Lewis acids to catalyze group transfer reactions. Finally, some metalloenzymes use homolytic cleavage of the cobalt–carbon bond in derivatives of vitamin B to generate an organic radical that can abstract a hydrogen atom and thus cause molecular rearrangements to occur.

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The hydrogen atom, consisting of an electron and a proton, is a two-particle system, and the internal motion of two particles around their center of mass is equivalent to the motion of a single particle with a reduced mass. This reduced particle is located at \(r\), where \(r\) is the vector specifying the position of the electron relative to the position of the proton. The length of \(r\) is the distance between the proton and the electron, and the direction of \(r\) and the direction of \(r\) is given by the orientation of the vector pointing from the proton to the electron. Since the proton is much more massive than the electron, we will assume throughout this chapter that the reduced mass equals the electron mass and the proton is located at the center of mass. Since the internal motion of any two-particle system can be represented by the motion of a single particle with a reduced mass, the description of the hydrogen atom has much in common with the description of a diatomic molecule discussed previously. The Schrödinger Equation for the hydrogen atom \[ \hat {H} (r , \theta , \varphi ) \psi (r , \theta , \varphi ) = E \psi ( r , \theta , \varphi) \label {6.1.1}\] employs the same kinetic energy operator, \(\hat {T}\), written in spherical coordinates. For the hydrogen atom, however, the distance, r, between the two particles can vary, unlike the diatomic molecule where the bond length was fixed, and the rigid rotor model was used. The hydrogen atom Hamiltonian also contains a potential energy term, \(\hat {V}\), to describe the attraction between the proton and the electron. This term is the Coulomb potential energy, \[ \hat {V} (r) = - \dfrac {e^2}{4 \pi \epsilon _0 r } \label {6.1.2}\] where \(r\) is the distance between the electron and the proton. The Coulomb potential energy depends inversely on the distance between the electron and the nucleus and does not depend on any angles. Such a potential is called a central potential. It is convenient to switch from Cartesian coordinates \(x, y, z\) to spherical coordinates in terms of a radius \(r\), as well as angles \(\phi\), which is measured from the positive x axis in the xy plane and may be between 0 and \(2\pi\), and \(\theta\), which is measured from the positive z axis towards the xy plane and may be between 0 and \(\pi\). The time-indepdent Schrödinger equation (in spherical coordinates) for a electron around a positively charged nucleus is then \[ \left \{ -\dfrac {\hbar ^2}{2 \mu r^2} \left [ \dfrac {\partial}{\partial r} \left (r^2 \dfrac {\partial}{\partial r} \right ) + \dfrac {1}{\sin \theta } \dfrac {\partial}{\partial \theta } \left ( \sin \theta \dfrac {\partial}{\partial \theta} \right ) + \dfrac {1}{\sin ^2 \theta} \dfrac {\partial ^2}{\partial \varphi ^2} \right ] - \dfrac {e^2}{4 \pi \epsilon _0 r } \right \} \psi (r , \theta , \varphi ) = E \psi (r , \theta , \varphi ) \label {6.1.4}\] Since the angular momentum operator does not involve the radial variable, \(r\), we can in Equation \(\ref{6.1.4}\) by using a . We know that the eigenfunctions of the angular momentum operator are the , \(Y (\theta , \varphi )\), so a good choice for a product function is \[ \color{red} \psi (r , \theta , \varphi ) = R (r) Y (\theta , \varphi ) \label {6.1.8}\] The Spherical Harmonic \(Y (\theta , \varphi )\) functions provide information about where the electron is around the proton, and the radial function \(R(r)\) describes how far the electron is away from the proton. A solution for both \(R(r)\) and \(Y (\theta , \varphi ) \) with \(E_n\) that depends on only one quantum number \(n\), although others are required for the proper description of the wavefunction: \[ \color{red} E_n = -\dfrac {m_e e^4}{8\epsilon_0^2 h^2 n^2}\label{6}\] with \(n=1,2,3 ...\infty\) The hydrogen atom wavefunctions, \(\psi (r, \theta , \phi )\), are called atomic orbitals. An atomic orbital is a function that describes one electron in an atom. The wavefunction with n = 1, \(l\) \(l\) = 0 is called the 1s orbital, and an electron that is described by this function is said to be “in” the ls orbital, i.e. have a 1s orbital state. The constraints on n, \(l\) \(l)\), and \(m_l\) that are imposed during the solution of the hydrogen atom Schrödinger equation explain why there is a single 1s orbital, why there are three 2p orbitals, five 3d orbitals, etc. We will see when we consider multi-electron atoms, these constraints explain the features of the Periodic Table. In other words, the Periodic Table is a manifestation of the Schrödinger model and the physical constraints imposed to obtain the solutions to the Schrödinger equation for the hydrogen atom. Schrödinger’s approach requires three quantum numbers (\(n\), \(l\), and \(m_l\)) to specify a wavefunction for the electron. The quantum numbers provide information about the spatial distribution of an electron. Although \(n\) can be any positive integer (NOT zero), only certain values of \(l\) and \(m_l\) are allowed for a given value of \(n). One of three quantum numbers that tells the average relative distance of an electron from the nucleus. indicates the energy of the electron and the average distance of an electron from the nucleus \[ n = 1,\;2,\;3,\;4,\;.\;.\;.\; \label{2.2.2} \] As\( \)n increases for a given atom, so does the average distance of an electron from the nucleus. A negatively charged electron that is, on average, closer to the positively charged nucleus is attracted to the nucleus more strongly than an electron that is farther out in space. This means that electrons with higher values of n are easier to remove from an atom. All wave functions that have the same value of n are said to constitute a principal shell. All the wave functions that have the same value of n because those electrons have similar average distances from the nucleus. because those electrons have similar average distances from the nucleus. As you will see, the principal quantum number n corresponds to the n used by Bohr to describe electron orbits and by Rydberg to describe atomic energy levels. The second quantum number is often called the azimuthal quantum number (l). One of three quantum numbers that describes the shape of the region of space occupied by an electron.. The value of l describes the shape of the region of space occupied by the electron. The allowed values of l depend on the value of n and can range from 0 to n − 1: \[ l = 0,\;1,.,2,\;3,\;.\;.\;\left (n-1 \right ) \label{2.2.3} \] For example, if n = 1, l can be only 0; if n = 2, l can be 0 or 1; and so forth. For a given atom, all wave functions that have the same values of both n and l form a subshell. A group of wave functions that have the same values of n and l. The regions of space occupied by electrons in the same subshell usually have the same shape, but they are oriented differently in space. The third quantum number is the magnetic quantum number (\(m_l\)). One of three quantum numbers that describes the orientation of the region of space occupied by an electron with respect to an applied magnetic field.. The value of \(m_l\) describes the orientation of the region in space occupied by an electron with respect to an applied magnetic field. The allowed values of \(m_l\) depend on the value of l: \(m_l\) can range from −l to l in integral steps: \[ m = -l,\;-l+1,\,.\;.\;.0,\;\,.\;.\;.l-1,l \label{2.2.4} \] For example, if \(l = 0\), ml can be only 0; if l = 1, \(m_l\) can be −1, 0, or +1; and if l = 2, \(m_l\) can be −2, −1, 0, +1, or +2. Each wave function with an allowed combination of n, l, and ml values describes an atomic orbital A wave function with an allowed combination of n, l and ml quantum numbers., a particular spatial distribution for an electron. For a given set of quantum numbers, each principal shell has a fixed number of subshells, and each subshell has a fixed number of orbitals. How many subshells and orbitals are contained within the principal shell with n = 4? : value of n Asked for: number of subshells and orbitals in the principal shell A Given n = 4, calculate the allowed values of l. From these allowed values, count the number of subshells. B For each allowed value of l, calculate the allowed values of ml. The sum of the number of orbitals in each subshell is the number of orbitals in the principal shell. A We know that l can have all integral values from 0 to n − 1. If n = 4, then l can equal 0, 1, 2, or 3. Because the shell has four values of l, it has four subshells, each of which will contain a different number of orbitals, depending on the allowed values of ml. B For l = 0, ml can be only 0, and thus the l = 0 subshell has only one orbital. For l = 1, ml can be 0 or ±1; thus the l = 1 subshell has three orbitals. For l = 2, ml can be 0, ±1, or ±2, so there are five orbitals in the l = 2 subshell. The last allowed value of l is l = 3, for which ml can be 0, ±1, ±2, or ±3, resulting in seven orbitals in the l = 3 subshell. The total number of orbitals in the n = 4 principal shell is the sum of the number of orbitals in each subshell and is equal to n2: \( \mathop 1\limits_{(l = 0)} + \mathop 3\limits_{(l = 1)} + \mathop 5\limits_{(l = 2)} + \mathop 7\limits_{(l = 3)} = 16\; {\rm{orbitals}} = {(4\; {\rm{principal\: shells}})^2} \)​ How many subshells and orbitals are in the principal shell with = 3? three subshells; nine orbitals Rather than specifying all the values of and every time we refer to a subshell or an orbital, chemists use an abbreviated system with lowercase letters to denote the value of for a particular subshell or orbital: The principal quantum number is named first, followed by the letter , , , or as appropriate. These orbital designations are derived from corresponding spectroscopic characteristics of lines involving them: harp, rinciple, iffuse, and undamental. A 1 orbital has = 1 and = 0; a 2 subshell has = 2 and = 1 (and has three 2 orbitals, corresponding to = −1, 0, and +1); a 3 subshell has = 3 and = 2 (and has five 3 orbitals, corresponding to = −2, −1, 0, +1, and +2); and so forth. We can summarize the relationships between the quantum numbers and the number of subshells and orbitals as follows (Table \(\Page {1}\)): Each principal shell has subshells, and each subshell has 2 + 1 orbitals. The first six radial functions are provided in Table \(\Page {2}\). Note that the functions in the table exhibit a dependence on \(Z\), the atomic number of the nucleus. Other one electron systems have electronic states analogous to those for the hydrogen atom, and inclusion of the charge on the nucleus allows the same wavefunctions to be used for all one-electron systems. For hydrogen, \(Z = 1\) and for helium, \(Z=2\). \(\psi_{32\pm\ 2} = \dfrac {1}{162\sqrt {\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} \rho^2 e^{-\rho/3}{\sin}^2(\theta)e^{\pm\ 2i\phi}\) Visualizing the variation of an electronic wavefunction with r,\(\theta\), and \(\phi\) is important because the absolute square of the wavefunction depicts the charge distribution (electron probability density) in an atom or molecule. The charge distribution is central to chemistry because it is related to chemical reactivity. For example, an electron deficient part of one molecule is attracted to an electron rich region of another molecule, and such interactions play a major role in chemical interactions ranging from substitution and addition reactions to protein folding and the interaction of substrates with enzymes. Methods for separately examining the radial portions of atomic orbitals provide useful information about the distribution of charge density within the orbitals. Graphs of the radial functions, R(r), for the 1s, 2s, and 2p orbitals plotted in Figure \(\Page {2}\) left). The quantity \(R(r)^* R(r)\) gives the radial probability density; i.e., the probability density for the electron to be at a point located the distance r from the proton. Radial probability densities for three types of atomic orbitals are plotted in Figure \(\Page {2}\) (right). For the hydrogen atom, the peak in the radial probability plot occurs at = 0.529 Å (52.9 pm), which is exactly the radius calculated by Bohr for the = 1 orbit. Thus the obtained from quantum mechanics is identical to the radius calculated by classical mechanics. In Bohr’s model, however, the electron was assumed to be at this distance 100% of the time, whereas in the Schrödinger model, it is at this distance only some of the time. The difference between the two models is attributable to the wavelike behavior of the electron and the Heisenberg uncertainty principle. Figure \(\Page {3}\) compares the electron probability densities for the hydrogen 1 , 2 , and 3 orbitals. Note that all three are spherically symmetrical. For the 2 and 3 orbitals, however (and for all other orbitals as well), the electron probability density does not fall off smoothly with increasing . Instead, a series of minima and maxima are observed in the radial probability plots (part (c) in Figure \(\Page {3}\)). The minima correspond to spherical nodes (regions of zero electron probability), which alternate with spherical regions of nonzero electron probability. The angular component of the wavefunction \(Y(\theta,\phi)\) in Equation \(\ref{6.1.8}\) does much to give an orbital its distinctive shape. \(Y(\theta,\phi)\) is typically normalized so the the integral of \(Y^2(\theta,\phi)\) over the unit sphere is equal to one. In this case, \(Y^2(\theta,\phi)\) serves as a probability function. The probability function can be interpreted as the probability that the electron will be found on the ray emitting from the origin that is at angles \((\theta,\phi)\) from the axes. The probability function can also be interpreted as the probability distribution of the electron being at position \((\theta,\phi)\) on a sphere of radius , given that it is distance from the nucleus. \(Y_{l,m_l}(\theta,\phi)\) are also the wavefunction solutions to Schrödinger’s equation for a rigid rotor consisting of rotating bodies, for example a diatomic molecule. These are called . Three things happen to orbitals as increases (Figure 6.6.2): Orbitals are generally drawn as three-dimensional surfaces that enclose 90% of the electron density. Although such drawings show the relative sizes of the orbitals, they do not normally show the spherical nodes in the 2 and 3 orbitals because the spherical nodes lie inside the 90% surface. Fortunately, the positions of the spherical nodes are not important for chemical bonding. Only orbitals are spherically symmetrical. As the value of increases, the number of orbitals in a given subshell increases, and the shapes of the orbitals become more complex. Because the 2 subshell has = 1, with three values of (−1, 0, and +1), there are three 2 orbitals). The electron probability distribution for one of the hydrogen 2 orbitals is shown in Figure \(\Page {4}\). Because this orbital has two lobes of electron density arranged along the axis, with an electron density of zero in the plane (i.e., the plane is a nodal plane), it is a 2 orbital. As shown in Figure \(\Page {5}\), the other two 2 orbitals have identical shapes, but they lie along the axis (2 ) and axis (2 ), respectively. Note that each orbital has just one nodal plane. In each case, the phase of the wave function for each of the 2 orbitals is positive for the lobe that points along the positive axis and negative for the lobe that points along the negative axis. It is important to emphasize that these signs correspond to the of the wave that describes the electron motion, to positive or negative charges. The surfaces shown enclose 90% of the total electron probability for the 2 , 2 , and 2 orbitals. Each orbital is oriented along the axis indicated by the subscript and a nodal plane that is perpendicular to that axis bisects each 2 orbital. The phase of the wave function is positive (orange) in the region of space where , , or is positive and negative (blue) where , , or is negative. Just as with the orbitals, the size and complexity of the orbitals for any atom increase as the principal quantum number increases. The shapes of the 90% probability surfaces of the 3 , 4 , and higher-energy orbitals are, however, essentially the same as those shown in Figure Figure \(\Page {5}\). Subshells with = 2 have five orbitals; the first principal shell to have a subshell corresponds to = 3. The five orbitals have values of −2, −1, 0, +1, and +2 (Figure \(\Page {6}\)). The hydrogen 3d orbitals have more complex shapes than the 2p orbitals. All five 3d orbitals contain two nodal surfaces, as compared to one for each p orbital and zero for each s orbital. In three of the d orbitals, the lobes of electron density are oriented between the x and y, x and z, and y and z planes; these orbitals are referred to as the \(3d_{xy}\), \)3d_{xz}\_, and \)3d_{yz}\) orbitals, respectively. A fourth d orbital has lobes lying along the x and y axes; this is the 3dx2−y2 orbital. The fifth 3d orbital, called the \(3d_{z^2}\) orbital, has a unique shape: it looks like a \(2p_z\) orbital combined with an additional doughnut of electron probability lying in the xy plane. Despite its peculiar shape, the \(3d_{z^2}\) orbital is mathematically equivalent to the other four and has the same energy. In contrast to p orbitals, the phase of the wave function for d orbitals is the same for opposite pairs of lobes. As shown in Figure \(\Page {6}\), the phase of the wave function is positive for the two lobes of the \(dz^2\) orbital that lie along the z axis, whereas the phase of the wave function is negative for the doughnut of electron density in the xy plane. Like the s and p orbitals, as n increases, the size of the d orbitals increases, but the overall shapes remain similar to those depicted in Figure 6.6.5. Principal shells with = 4 can have subshells with = 3 and values of −3, −2, −1, 0, +1, +2, and +3. These subshells consist of seven orbitals. Each orbital has three nodal surfaces, so their shapes are complex (not shown). The constraint that \(n\) be greater than or equal to \(l +1\) also turns out to quantize the energy, producing the same quantized expression for hydrogen atom energy levels that was obtained from the Bohr model of the hydrogen atom. \[E=−\dfrac{Z^2}{n^2}Rhc \label{6.6.1}\] or \[ E_n = - \dfrac {Z^2 \mu e^4}{8 \epsilon ^2_0 h^2 n^2} \] The relative energies of the atomic orbitals with ≤ 4 for a hydrogen atom are plotted in Figure \(\Page {7}\) ; note that the orbital energies depend on the principal quantum number . Consequently, the energies of the 2 and 2 orbitals of hydrogen are the same; the energies of the 3 , 3 , and 3 orbitals are the same; and so forth. The orbital energies obtained for hydrogen using quantum mechanics are exactly the same as the allowed energies calculated by Bohr. In contrast to Bohr’s model, however, which allowed only one orbit for each energy level, quantum mechanics predicts that there are orbitals with different electron density distributions in the = 2 principal shell (one 2 and three 2 orbitals), in the = 3 principal shell, and in the = 4 principal shell. For a single electron system, the energy of that electron is a function of only the principal quantum number (Equation \(\ref{6.6.1}\)). As we have just seen, however, quantum mechanics also predicts that in the hydrogen atom, all orbitals with the same value of (e.g., the three 2 orbitals) are degenerate, meaning that they have the same energy. Figure \(\Page {7}\) shows that the energy levels become closer and closer together as the value of increases, as expected because of the 1/ dependence of orbital energies. In general, both energy and radius decrease as the nuclear charge increases. Thus the most stable orbitals (those with the lowest energy) are those closest to the nucleus. For example, in the ground state of the hydrogen atom, the single electron is in the 1s orbital, whereas in the first excited state, the atom has absorbed energy and the electron has been promoted to one of the n = 2 orbitals. In ions with only a single electron, the energy of a given orbital depends on only n, and all subshells within a principal shell, such as the px, py, and pz orbitals, are degenerate. It is interesting to compare the results obtained by solving the Schrödinger equation with Bohr’s model of the hydrogen atom. There are several ways in which the Schrödinger model and Bohr model differ. The quantum numbers \(n, \ l, \ m\) are not sufficient to fully characterize the physical state of the electrons in an atom. In 1926, Otto Stern and Walther Gerlach carried out an experiment that could not be explained in terms of the three quantum numbers \(n, \ l, \ m\) and showed that there is, in fact, another quantum-mechanical degree of freedom that needs to be included in the theory. The experiment is illustrated in the Figure \(\Page {8}\). The fact that the beam splits into 2 beams suggests that the electrons in the atoms have a degree of freedom capable of coupling to the magnetic field. That is, an electron has an \(M\) arising from a degree of freedom that has no classical analog. The magnetic moment must take on only 2 values according to the Stern-Gerlach experiment. The intrinsic property that gives rise to the magnetic moment must have some analog to a , \(S\); u The implication of the Stern-Gerlach experiment is that we need to include a fourth quantum number, \(m_s\) in our description of the physical state of the electron. That is, in addition to give its principle, angular, and magnetic quantum numbers, we also need to say if it is a spin-up electron or a spin-down electron. U George Uhlenbeck (1900–1988) and Samuel Goudsmit (1902–1978), proposed that the splittings were caused by an electron spinning about its axis, much as Earth spins about its axis. When an electrically charged object spins, it produces a magnetic moment parallel to the axis of rotation, making it behave like a magnet. Although the electron cannot be viewed solely as a particle, spinning or otherwise, it is indisputable that it does have a magnetic moment. This magnetic moment is called electron spin. In an external magnetic field, the electron has two possible orientations (Figure \(\Page {9}\)). These are described by a fourth quantum number ( ), which for any electron can have only two possible values, designated +½ (up) and −½ (down) to indicate that the two orientations are opposites; the subscript is for spin. An electron behaves like a magnet that has one of two possible orientations, aligned either with the magnetic field or against it. The implications of electron spin for chemistry were recognized almost immediately by an Austrian physicist, Wolfgang Pauli (1900–1958; Nobel Prize in Physics, 1945), who determined that each orbital can contain no more than two electrons whoe developed the Pauli exclusion principle. No two electrons in an atom can have the same values of all four quantum numbers (\(n\), \(l\), \(m_l\), \(m_s\)). By giving the values of , , and , we also specify a particular orbital (e.g., 1 with = 1, = 0, = 0). Because has only two possible values (+½ or −½), two electrons, , can occupy any given orbital, one with spin up and one with spin down. With this information, we can proceed to construct the entire periodic table, which was originally based on the physical and chemical properties of the known elements. List all the allowed combinations of the four quantum numbers ( , , , ) for electrons in a 2 orbital and predict the maximum number of electrons the 2 subshell can accommodate. orbital allowed quantum numbers and maximum number of electrons in orbital For a 2 orbital, we know that = 2, = − 1 = 1, and = − , (− +1),…, ( − 1), . There are only three possible combinations of ( , , ): (2, 1, 1), (2, 1, 0), and (2, 1, −1). Because is independent of the other quantum numbers and can have values of only +½ and −½, there are six possible combinations of ( , , , ): (2, 1, 1, +½), (2, 1, 1, −½), (2, 1, 0, +½), (2, 1, 0, −½), (2, 1, −1, +½), and (2, 1, −1, −½). Hence the 2 subshell, which consists of three 2 orbitals (2 , 2 , and 2 ), can contain a total of six electrons, two in each orbital. List all the allowed combinations of the four quantum numbers ( , , , ) for a 6 orbital, and predict the total number of electrons it can contain. (6, 0, 0, +½), (6, 0, 0, −½); two electrons ")

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The freezing points of solutions are all lower than that of the pure solvent and is directly proportional to the of the solute. \[\begin{align*}\Delta{T_f} &= T_f(solvent) - T_f (solution) \\[4pt] &= K_f \times m \end{align*}\] where \(\Delta{T_f}\) is the freezing point depression, \(T_f\) (solution) is the freezing point of the solution, \(T_f\) (solvent) is the freezing point of the solvent, \(K_f\) is the freezing point depression constant, and is the molality. Nonelectrolytes are substances with no ions, only molecules. , on the other hand, are composed mostly of ionic compounds, and essentially all soluble ionic compounds form electrolytes. Therefore, if we can establish that the substance that we are working with is uniform and is not ionic, it is safe to assume that we are working with a nonelectrolyte, and we may attempt to solve this problem using our formulas. This will most likely be the case for all problems you encounter related to freezing point depression and boiling point elevation in this course, but it is a good idea to keep an eye out for ions. It is worth mentioning that these equations work for both volatile and nonvolatile solutions. This means that for the sake of determining freezing point depression or boiling point elevation, the vapor pressure does not effect the change in temperature. Also, remember that a pure solvent is a solution that has had nothing extra added to it or dissolved in it. We will be comparing the properties of that pure solvent with its new properties when added to a solution. Adding solutes to an ideal solution results in a positive \(ΔS\), an increase in entropy. Because of this, the newly altered solution's chemical and physical properties will also change. The properties that undergo changes due to the addition of solutes to a solvent are known as . These properties are dependent on the number of solutes added, not on their identity. Two examples of colligative properties are boiling point and freezing point: due to the addition of solutes, the boiling point tends to increase, and freezing point tends to decrease. The freezing point and boiling point of a pure solvent can be changed when added to a solution. When this occurs, the freezing point of the pure solvent may become lower, and the boiling point may become higher. The extent to which these changes occur can be found using the formulas: \[\Delta{T}_f = -K_f \times m\] \[\Delta{T}_b = K_b \times m\] where \(m\) is the solute and \(K\) values are proportionality constants; (\(K_f\) and \(K_b\) for freezing and boiling, respectively). Molality is defined as the number of moles of solute per kilogram . Be careful not to use the mass of the entire solution. Often, the problem will give you the change in temperature and the proportionality constant, and you must find the molality first in order to get your final answer. If solving for the proportionality constant is not the ultimate goal of the problem, these values will most likely be given. Some common values for \(K_f\) and \(K_b\) respectively, are in Table \(\Page {1}\): The solute, in order for it to exert any change on colligative properties, must fulfill two conditions. First, it must not contribute to the vapor pressure of the solution, and second, it must remain suspended in the solution even during phase changes. Because the solvent is no longer pure with the addition of solutes, we can say that the of the solvent is lower. Chemical potential is the molar Gibb's energy that one mole of solvent is able to contribute to a mixture. The higher the chemical potential of a solvent is, the more it is able to drive the reaction forward. Consequently, solvents with higher chemical potentials will also have higher vapor pressures. The boiling point is reached when the chemical potential of the pure solvent, a liquid, reaches that of the chemical potential of pure vapor. Because of the decrease in the chemical potential of mixed solvents and solutes, we observe this intersection at higher temperatures. In other words, the boiling point of the impure solvent will be at a higher temperature than that of the pure liquid solvent. Thus, occurs with a temperature increase that is quantified using \[\Delta{T_b} = K_b m\] where \(K_b\) is known as the and \(m\) is the molality of the solute. Freezing point is reached when the chemical potential of the pure liquid solvent reaches that of the pure solid solvent. Again, since we are dealing with mixtures with decreased chemical potential, we expect the freezing point to change. Unlike the boiling point, the chemical potential of the impure solvent requires a colder temperature for it to reach the chemical potential of the pure solid solvent. Therefore, a is observed. 2.00 g of some unknown compound reduces the freezing point of 75.00 g of benzene from 5.53 to 4.90 \(^{\circ}C\). What is the molar mass of the compound? First we must compute the molality of the benzene solution, which will allow us to find the number of moles of solute dissolved. \[ \begin{align*} m &= \dfrac{\Delta{T}_f}{-K_f} \\[4pt] &= \dfrac{(4.90 - 5.53)^{\circ}C}{-5.12^{\circ}C / m} \\[4pt] &= 0.123 m \end{align*}\] \[ \begin{align*} \text{Amount Solute} &= 0.07500 \; kg \; benzene \times \dfrac{0.123 \; m}{1 \; kg \; benzene} \\[4pt] &= 0.00923 \; m \; solute \end{align*}\] We can now find the molecular weight of the unknown compound: \[ \begin{align*} \text{Molecular Weight} =& \dfrac{2.00 \; g \; unknown}{0.00923 \; mol} \\[4pt] &= 216.80 \; g/mol \end{align*}\] The freezing point depression is especially vital to aquatic life. Since saltwater will freeze at colder temperatures, organisms can survive in these bodies of water. Benzophenone has a freezing point of 49.00 C. A 0.450 molal solution of urea in this solvent has a freezing point of 44.59 C. Find the freezing point depression constant for the solvent. \(9.80\,^oC/m\) Road salting takes advantage of this effect to lower the freezing point of the ice it is placed on. Lowering the freezing point allows the street ice to melt at lower temperatures. The maximum depression of the freezing point is about −18 °C (0 °F), so if the ambient temperature is lower, \(\ce{NaCl}\) will be ineffective. Under these conditions, \(\ce{CaCl_2}\) can be used since it dissolves to make three ions instead of two for \(\ce{NaCl}\).

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Two compounds that have the same formula and the same connectivity do not always have the same shape. There are two reasons why this may happen. In one case, the molecule may be flexible, so that it can twist into different shapes via rotation around individual sigma bonds. This phenomenon is called conformation, and it is covered in a different chapter. The second case occurs when two molecules appear to be connected the same way on paper, but are connected in two different ways in three dimensional space. These two, different molecules are called . One simple example of stereoisomers from inorganic chemistry is diammine platinum dichloride, (NH ) PtCl . This important compound is sometimes called "platin" for short. As the formula implies, it contains a platinum ion that is coordinated to two ammonia ligands and two chloride ligands (remember, a ligand in inorganic chemistry is an electron donor that is attached to a metal atom, donating a pair of electrons to form a bond). Platin is an example of a coordination compound. The way the different pieces of bond together is discussed in the chapter of . For reasons arising from molecular orbital interactions, platin has a square planar geometry at the platinum atom. That arrangement results in two possible ways the ligands could be connected. The two sets of like ligands could be connected on the same side of the square or on opposite corners. These two arrangements result in two different compounds; they are isomers that differ only in three-dimensional space. Although these two compounds are very similar, they have slightly different physical properties. Both are yellow compounds that decompose when heated to 270 degrees C, but trans-platin forms pale yellow crystals and is more soluble than cis-platin in water. Cis-platin has clinical importance in the treatment of ovarian and testicular cancers. The biological mechanism of the drug's action was long suspected to involve binding of the platinum by DNA. Further details were worked out by MIT chemist Steve Lippard and graduate student Amy Rosenzweig in the 1990's. Inside the cell nucleus, the two ammines in cis-platin can be replaced by nitrogen donors from a DNA strand. To donate to the Lewis acidic platinum, the DNA molecule must bend slightly. Normally that bend is detected and repaired by proteins in the cell. However, ovarian and testicular cells happen to contain a protein that is just the right shape to fit around this slightly bent DNA strand. The DNA strand becomes lodged in the protein and can't be displaced, and so it is unable to bind with other proteins used in DNA replication. The cell becomes unable to replicate, and so cancerous growth is stopped. Draw the cis and trans isomers of the following compounds: Only one isomer of (tmeda)PtCl is possible [tmeda = (CH ) NCH CH N(CH ) ; both nitrogens connect to the platinum]. Draw this isomer and explain why the other isomer is not possible. ,

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There are few reactions of aldehydes and ketones that do not in some way affect the carbonyl function. For this reason, it may be necessary to protect the carbonyl function when it is desirable to avoid reaction at this function. For example, you may plan to synthesize 4-cyclohexylidene-2-butanone by way of a Wittig reaction (Section 16-4A), which would involve the following sequence: ​​​​​​ This synthesis would in the second step because the phenyllithium would add irreversibly to the carbonyl group. To avoid this, the carbonyl group would have to be protected or blocked, and the most generally useful method of blocking is to convert the carbonyl group to a ketal, usually a cyclic ketal: With the carbonyl group suitably protected, the proposed synthesis would have a much better chance of success: Notice that the carbonyl group is regenerated by acid hydrolysis in the last step. and (1977)

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One of the most important technical reactions of alkenes is their conversion to higher-molecular-weight compounds or (Table 10-4). A polymer is defined as a . Thus polymerization of propene gives a long-chain hydrocarbon with recurring units: Most technically important polymerizations of alkenes occur by chain mechanisms and may be classed as anion, cation, or radical reactions, depending upon the character of the chain-carrying species. In each case, the key steps involve successive additions to molecules of the alkene, the differences being in the number of electrons that are supplied by the attacking agent for formation of the new carbon-carbon bond. For simplicity, these steps will be illustrated by using ethene, even though it does not polymerize very easily by any of them: Initiation of alkene polymerization by the anion-chain mechanism may be formulated as involving an attack by a nucleophilic reagent \(\ce{Y}^\ominus\) on one end of the double bond and formation of a carbanion: Attack by the carbanion on another alkene molecule would give a four-carbon carbanion, and subsequent additions to further alkene molecules would lead to a high-molecular-weight anion: The growing chain can be terminated by any reaction (such as the addition of a proton) that would destroy the carbanion on the end of the chain: Anionic polymerization of alkenes is quite difficult to achieve because few anions (or nucleophiles) are able to add readily to alkene double bonds (see ). Anionic polymerization occurs readily only with alkenes substituted with sufficiently powerful electron-attracting groups to expedite nucleophilic attack. By this reasoning, alkynes should polymerize more readily than alkenes under anionic conditions, but there appear to be no technically important alkyne polymerizations in operation by this or any other mechanism. Perhaps this is because the resultant polymer would be highly conjugated, and therefore highly reactive, and may not survive the experimental conditions: Polymerization of an alkene by acidic reagents can be formulated by a mechanism similar to the addition of hydrogen halides to alkene linkages. First, a proton from a suitable acid adds to an alkene to yield a carbocation. Then, in the absence of any other reasonably strong nucleophilic reagent, another alkene molecule donates an electron pair and forms a longer-chain cation. Continuation of this process can lead to a high-molecular-weight cation. Termination can occur by loss of a proton. The following equations represent the overall reaction sequence: Ethene does not polymerize by the cationic mechanism because it does not have sufficiently electron-donating groups to permit easy formation of the intermediate growing-chain cation. 2-Methylpropene has electron-donating alkyl groups and polymerizes much more easily than ethene by this type of mechanism. The usual catalysts for cationic polymerization of 2-methylpropene are sulfuric acid, hydrogen fluoride, or a complex of boron trifluoride and water. Under nearly anhydrous conditions a very long chain polymer called polyisobutylene is formed. Polyisobutylene fractions of particular molecular weights are very tacky and are used as adhesives for pressure-sealing tapes. In the presence of \(60\%\) sulfuric acid, 2-methylpropene is converted to a long-chain polymer, but to a mixture of eight-carbon alkenes. The mechanism is like that of the polymerization of 2-methylpropene under nearly anhydrous conditions, except that chain termination occurs after only one 2-methylpropene molecule has been added: The short chain length is due to the ; the intermediate carbocation loses a proton to water it can react with another alkene molecule. The proton can be lost in two different ways, and a mixture of alkene isomers is obtained. The alkene mixture is known as "diisobutylene" and has a number of commercial uses. Hydrogenation yields 2,2,4-trimethylpentane (often erroneously called "isooctane"), which is used as the standard "100 antiknock rating" fuel for internal-combustion gasoline engines: Ethene can be polymerized with peroxide catalysts under high pressure (\(1000 \: \text{atm}\) or more, literally in a cannon barrel) at temperatures in excess of \(100^\text{o}\). The initiation step involves formation of radicals, and chain propagation entails stepwise addition of radicals to ethene molecules. Chain termination can occur by any reaction resulting in combination or disproportionation of free radicals. The polyethene produced in this way has from 100 to 1000 ethene units in the hydrocarbon chain. The polymer possesses a number of desirable properties as a plastic and is used widely for electrical insulation, packaging films, piping, and a variety of molded articles. Propene and 2-methylpropene do not polymerize satisfactorily by radical mechanisms. A relatively low-pressure, low-temperature ethene polymerization has been achieved with an aluminum-molybdenum oxide catalyst, which requires occasional activation with hydrogen (Phillips Petroleum process). Ethene also polymerizes quite rapidly at atmospheric pressure and room temperature in an alkane solvent containing a suspension of the insoluble reaction product from triethylaluminum and titanium tetrachloride ( ). Both the Phillips and Ziegler processes produce very high-molecular-weight polyethene with exceptional physical properties. The unusual characteristics of these reactions indicate that no simple anion, cation, or radical mechanism can be involved. It is believed that the catalysts act by coordinating with the alkene molecules in somewhat the same way that hydrogenation catalysts combine with alkenes ( ). Polymerization of propene by the Ziegler process gives a very useful plastic material. It can be made into durable fibers or molded into a variety of shapes. Copolymers (polymers with more than one kind of monomer unit in the polymer chains) of ethene and propene made by the Ziegler process have highly desirable rubberlike properties and are potentially the cheapest useful elastomers (elastic polymers). A Nobel Prize was shared in 1963 by K. Ziegler and G. Natta for their work on alkene polymerization. The properties and uses of polymers are discussed in greater detail in and . The most important alkene monomers used in addition polymerizations are listed in Table 10-4 along with some names and uses of the corresponding polymers. and (1977)

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The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals. Many of the physical and chemical properties of elements can be correlated to their unique electron configurations. The valence electrons, electrons in the outermost shell, are the determining factor for the unique chemistry of the element.

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Bismuth, the heaviest non-radioactive naturally occurring element, was isolated by Basil Valentine in 1450. It is a hard, brittle metal with an unusually low melting point (271 C). Alloys of bismuth with other low-melting metals such as tin and lead have even lower melting points and are used in electrical solders, fuse elements and automatic fire sprinkler heads. The metal can be found in nature, often combined with copper or lead ores, but can also be extracted from bismuth(III) oxide by roasting with carbon. Compounds of bismuth are used in pigments for oil painting and one is in a popular pink preparation for the treatment of common stomach upset.

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Germanium, categorized as a metalloid in group 14, the Carbon family, has five naturally occurring isotopes. Germanium, abundant in the Earth's crust has been said to improve the immune system of cancer patients. It is also used in transistors, but its most important use is in fiber-optic systems and infrared optics. The metalloid was one of the elements predicted by Mendeleev in 1871 (ekasilicon) to fill out his periodic table and was discovered in 1886 by Winkler. In a mine near Freiberg, Saxony, a new mineral was found in 1885. First the mineral was called argyrodite but later when Clemens Winkler examined this mineral he discovered that it was similar to antimony. At first he wanted to name it neptunium, but because this name was already taken he named it germanium in honor of his fatherland Germany. The position of where germanium should be placed on the periodic table was under discussion during the time due to its similarities in arsenic and antimony. Due to Mendeleev's prediction of ekasilicon, germanium's place on the periodic table was confirmed because of the similar properties predicted and similar properties deduced from examining the mineral.   Like silicon, germanium is used in the manufacture of semi-conductor devices. Unlike silicon, it is rather rare (only about 1 part in 10 million parts in the earth's crust). The physical and chemical properties of germanium closely parallel those of . Germanium (Ge), has an atomic number of 32. It is grayish-white, lustrous, hard and has similar chemical properties to tin and silicon. Germanium is brittle and silvery-white under standard conditions. Germanium under this condition is known as which has a diamond cubic crystal structure. When germanium is above 120 kilobars, germanium has a different allotrope known as . Germanium is one of the few substances like water that expands when it solidifies. Germanium, a semiconductor, is the first metallic metal to become a superconductor in the presence of strong electromagnetic field. The five naturally occurring isotopes of Germanium have the atomic masses of 70, 72, 73, 74, and 76. Germanium 76 is slightly radioactive and is the least common. Germanium 74 is the most common isotope having the greatest natural abundance of the five. Under the condition of being bombarded with alpha particles, Germanium 72 generates stable Se 77. At temperature 250 °C, germanium slowly oxidizes to \(GeO_2\). Germanium dissolves slowly in concentrated sulfuric acid, and is insoluble in diluted acids and alkalis. It will react violently with molten alkalis to produce [GeO ] . The common oxidation state that Germanium occurs in is +4 and +2. Under rare conditions, Germanium also occurs in oxidation states of +3, +1, and -4. There are two forms of oxides of germanium, germanium dioxide (\(GeO_2\)) and germanium monoxide (\(GeO\)). By roasting germanium sulfide, germanium dioxide can be obtained. As for germanium monoxide, it can be obtained by the high temperature reaction of germanium dioxide and germanium metal. Germanium dioxide has the unusual property of refractive index for light but transparency to infrared light. Like silicon, germanium is used in the manufacture of semi-conductor devices. Unlike silicon, it is rather rare (only about 1 part in 10 million parts in the earth's crust). Although it forms a compound, germanium dioxide, just like silicon, it is generally extracted from the by-products of zinc refining. Germanium is used mainly for fiber-optic systems and infrared optics. It is also used for polymerization catalysts, electronics, and as phosphors, metallurgy, and chemotherapy. Because germanium dioxide have a high index of refraction and low optical dispersion it is useful for wide-angle camera lenses. As stated before, it is an important semi-conductor so it is used in transistors.

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This page discusses the trends in some atomic and physical properties of the Group 1 elements - lithium, sodium, potassium, rubidium and cesium. Sections below cover the trends in atomic radius, first ionization energy, electronegativity, melting and boiling points, and density. The chart below shows the increase in atomic radius down the group. The radius of an atom is governed by two factors: Compare the electronic configurations of lithium and sodium: In each element, the outer electron experiences a net charge of +1 from the nucleus. The positive charge on the nucleus is canceled out by the negative charges of the inner electrons. This effect is illustrated in the figure below: This is true for each of the other atoms in Group 1. The only factor affecting the size of the atom is the number of layers of inner electrons which surround the atom. More layers of electrons take up more space, due to electron-electron repulsion. Therefore, the atoms in size down the group. The first ionization energy of an atom is defined as the energy required to remove the most loosely held electron from each of one mole of gaseous atoms, producing one mole of singly charged gaseous ions; in other words, it is the energy required for 1 mole of this process: \[ X(g) \rightarrow X^+ (g) + e^-\] A graph showing the first ionization energies of the Group 1 atoms is shown above. Notice that first ionization energy decreases down the group. Ionization energy is governed by three factors: Down the group, the increase in nuclear charge is exactly offset by the increase in the number of inner electrons. As mentioned before, in each of the elements Group 1, the outermost electrons experience a net charge of +1 from the center. However, the distance between the nucleus and the outer electrons increases down the group; electrons become easier to remove, and the ionization energy falls. Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. It is usually measured on the Pauling scale, on which the most electronegative element (fluorine) is given an electronegativity of 4.0 ( ). A graph showing the electronegativities of the Group 1 elements is shown above. Each of these elements has a very low electronegativity when compared with fluorine, and the electronegativities decrease from lithium to cesium. Picture a bond between a sodium atom and a chlorine atom. The bond can be considered covalent, composed of a pair of shared electrons. The electron pair will be pulled toward the chlorine atom because the chlorine nucleus contains many more protons than the sodium nucleus. This is illustrated in the figure below: The electron pair is so close to the chlorine that an effective electron transfer from the sodium atom to the chlorine atom occurs—the atoms are ionized. This strong attraction from the chlorine nucleus explains why chlorine is much more electronegative than sodium. Now compare this with a lithium-chlorine bond. The net pull from each end of the bond is the same as before, but the lithium atom is smaller than the sodium atom. That means that the electron pair is going to be more strongly attracted to the net +1 charge on the lithium end, and thus closer to it. In some lithium compounds there is often a degree of covalent bonding that is not present in the rest of the group. Lithium iodide, for example, will dissolve in organic solvents; this is a typical property of covalent compounds. The iodine atom is so large that the pull from the iodine nucleus on the pair of electrons is relatively weak, and a fully-ionic bond is not formed. As the metal atoms increase in size, any bonding electron pair becomes farther from the metal nucleus, and so is less strongly attracted towards it. This corresponds with a decrease in electronegativity down Group 1. With the exception of some lithium compounds, the Group 1 elements each form compounds that can be considered ionic. Each is so weakly electronegative that in a Group 1-halogen bond, we assume that the electron pair on a more electronegative atom is pulled so close to that atom that ions are formed. The figure above shows melting and boiling points of the Group 1 elements. Both the melting and boiling points decrease down the group. When any of the Group 1 metals is melted, the metallic bond is weakened enough for the atoms to move more freely, and is broken completely when the boiling point is reached. The decrease in melting and boiling points reflects the decrease in the strength of each metallic bond. The atoms in a metal are held together by the attraction of the nuclei to electrons which are delocalized over the whole metal mass. As the atoms increase in size, the distance between the nuclei and these delocalized electrons increases; therefore, attractions fall. The atoms are more easily pulled apart to form a liquid, and then a gas. As previously discussed, each atom exhibits a net pull from the nuclei of +1. The increased charge on the nucleus down the group is offset by additional levels of screening electrons. As before, the trend is determined by the distance between the nucleus and the bonding electrons. The densities of the Group 1 elements increase down the group (except for a downward fluctuation at potassium). This trend is shown in the figure below: The metals in this series are relatively light—​lithium, sodium, and potassium are less dense than water (less than 1 g cm ). It is difficult to develop a simple explanation for this trend because density depends on two factors, both of which change down the group. The atoms are packed in the same way, so the two factors considered are how many atoms can be packed in a given volume, and the mass of the individual atoms. The amount packed depends on the individual atoms' volumes; these volumes, in turn, depends on their atomic radius. Atomic radius increases down a group, so the volume of the atoms also increases. Fewer sodium atoms than lithium atoms, therefore, can be packed into a given volume. However, as the atoms become larger, their masses increase. A given number of sodium atoms will weigh more than the same number of lithium atoms. Therefore, 1 cm of sodium contains fewer atoms than the same volume of lithium, but each atom weighs more. Mathematical calculations are required to determine the densities. Jim Clark ( )

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Now that we have mathematical expressions for the wavefunctions and energies for the particle-in-a-box, we can answer a number of interesting questions. The answers to these questions use quantum mechanics to predict some important and general properties for electrons, atoms, molecules, gases, liquids, and solids. What is the lowest energy for an electron? The lowest energy level is \(E_1\), and it is important to recognize that this lowest energy is not zero. This finite (meaning neither zero nor infinite) energy is called the zero-point energy, and the motion associated with this energy is called the zero-point motion. Any system that is restricted to some region of space is said to be bound. The zero-point energy and motion are manifestations of the wave properties and the Heisenberg Uncertainty Principle, and are general properties of bound quantum mechanical systems. The position and momentum of the particle cannot be determined exactly. According to the Heisenberg Uncertainty Principle, the product of the uncertainties, i.e. standard deviations in these quantities, must be greater than or equal to ħ/2. If the energy were zero, then the momentum would be exactly zero, which would violate the Heisenberg Uncertainty Principle unless the uncertainty in position were infinite. The system then would not be localized in space to any extent at all, which we will find to be true for the case of a free particle, which is not bound. The uncertainty in the position of a bound system is not infinite, so the uncertainty in the momentum cannot be zero, as it would be if the energy were zero. Use your solution to Exercise \(\Page {15}\) to write a few sentences answering this question about the location of the electron. What insight do you gain from the graphs you made for the probability distribution at very large n compared to n = 1? Use the general form of the particle-in-a-box wavefunction sin(kx) for any n to find the mathematical expression for the position expectation value \(\left \langle x \right \rangle\) for a box of length L. How does \(\left \langle x \right \rangle\) depend on n? Evaluate the integral. Calculate the probability of finding an electron at L/2 in an interval ranging from \(\dfrac {L}{2} - \dfrac {L}{200}\) to \(\dfrac {L}{2} + \dfrac {L}{200}\) for n = 1 and n = 2. Since the length of the interval, L/100, is small compared to L, you can get an approximate answer without integrating. The particle-in-a-box wavefunctions are not eigenfunctions of the momentum operator. Show that the particle-in-a-box wavefunctions are not eigenfunctions of the momentum operator. Even though the wavefunctions are not momentum eigenfunctions, we can calculate the expectation value for the momentum. Show that the expectation or average value for the momentum of an electron in the box is zero in every state. First write the expectation value integral for the momentum. Then insert the expression for the wavefunction and evaluate the integral as shown here. \[\left \langle P \right \rangle = \int \limits ^L_0 \psi ^*_n (x) \left ( -i\hbar \dfrac {d}{dx} \right ) \psi _n (x) dx\] \[ = \int \limits ^L_0 \left (\dfrac {2}{L} \right )^{1/2} \sin (\dfrac {n \pi x}{L}) \left ( -i\hbar \dfrac {d}{dx} \right ) \left (\dfrac {2}{L} \right )^{1/2} \sin (\dfrac {n \pi x }{L} ) dx \] \[ = -i \hbar \left (\dfrac {2}{L} \right ) \int \limits ^L_0 \sin (\dfrac {n \pi x}{L}) \left ( \dfrac {d}{dx} \right ) \sin (\dfrac {n \pi x}{L}) dx\] \[ = -i \hbar \left (\dfrac {2}{L} \right ) \left ( \dfrac {n \pi}{L} \right ) \int \limits ^L_0 \sin (\dfrac {n \pi x}{L}) \cos (\dfrac {n \pi x}{L}) dx\] \[= 0\] It may seem that the electron does not have any momentum, which is not correct because we know the energy is never zero. In fact, the energy that we obtained for the particle-in-a-box is entirely kinetic energy because we set the potential energy at 0. Since the kinetic energy is the momentum squared divided by twice the mass, it is easy to understand how the average momentum can be zero and the kinetic energy finite. It must be equally likely for the particle-in-a-box to have a momentum -p as +p. The average of +p and –p is zero, yet \(p^2\) and the average of \(p^2\) are not zero. The information that the particle is equally likely to have a momentum of +p or –p is contained in the wavefunction. The sine function is a representation of the two momentum eigenfunctions \(e^{ikx}\) and \(e^{-ikx}\) as shown by Exercise \(\Page {5}\). Write the particle-in-a-box wavefunction as a normalized linear combination of the momentum eigenfunctions \(e^{ikx}\) and \(e^{-ikx}\) by using Euler’s formula. Show that the momentum eigenvalues for these two functions are p = +ħk and -ħk. The interpretation of these results is physically interesting. The exponential wavefunctions in the linear combination for the sine function represent the two opposite directions in which the electron can move. One exponential term represents movement to the left and the other term represents movement to the right. The electrons are moving, they have kinetic energy and momentum, yet the average momentum is zero. Does the fact that the average momentum of an electron is zero and the average position is L/2 violate the Heisenberg Uncertainty Principle? No, of course not, because the Heisenberg Uncertainty Principle pertains to the uncertainty in the momentum and in the position, not to the average values. Quantitative values for these uncertainties can be obtained to compare with the limit set by the Heisenberg Uncertainty Principle for the product of the uncertainties in the momentum and position. First, we need a quantitative definition of uncertainty. Here, just like in experimental measurements, a good definition of uncertainty is the standard deviation or the root mean square deviation from the average. It can be shown by working Problem 6 at the end of this chapter that the standard deviation in the position of the particle-in-a-box is given by \[ \sigma _x = \dfrac {L}{2 \pi n } \sqrt{ \dfrac {\pi ^2}{3} n^2 - 2 } \label {4-34}\] and the standard deviation in the momentum by \[ \sigma _p = \dfrac {n \pi \hbar}{L} \label {4-35}\] Even for n = 1, the lowest value for n, σx is finite and proportional to L. As L increases the uncertainty in position of the electron increases. On the other hand, as L increases, σp decreases, but the product is never zero; and the uncertainty principle holds. Evaluate the product σx σp for n = 1 and for general n. Is the product greater than ħ/2 for all values of n and L as required by the Heisenberg Uncertainty Principle? Are the eigenfunctions of the particle-in-a-box Hamiltonian orthogonal? Two functions \(Ψ_A\) and \(Ψ_B\) are orthogonal if \[ \int \limits _{all space} \psi _A^* \psi _B d\tau = 0 \] In general, eigenfunctions of a quantum mechanical operator with different eigenvalues are orthogonal. Evaluate the integral \(\int \psi ^*_1 \psi _3 dx\) and as many other pairs of particle-in-a-box eigenfunctions as you wish (use symmetry arguments whenever possible) and explain what the results say about orthogonality of the functions. What happens to the energy level spacing for a particle-in-a-box when \(mL^2\) becomes much larger than \(h^2\)? What does this result imply about the relevance of quantization of energy to baseballs in a box between the pitching mound and home plate? What implications does quantum mechanics have for the game of baseball in a world where h is so large that baseballs exhibit quantum effects? The first derivative of a function is the rate of change of the function, and the second derivative is the rate of change in the rate of change, also known as the curvature. A function with a large second derivative is changing very rapidly. Since the second derivative of the wavefunction occurs in the Hamiltonian operator that is used to calculate the energy by using the Schrödinger equation, a wavefunction that has sharper curvatures than another, i.e. larger second derivatives, should correspond to a state having a higher energy. A wavefunction with more nodes than another over the same region of space must have sharper curvatures and larger second derivatives, and therefore should correspond to a higher energy state. Identify a relationship between the number of nodes in a wavefunction and its energy by examining the graphs you made in Exercise \(\Page {15}\). A node is the point where the amplitude passes through zero. What does the presence of many nodes mean about the shape of the wavefunction?

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In the presence of dioxygen, iron(II) species are readily oxidized to iron(III) species. In the presence of water, iron(III) species frequently associate into \(\mu\)-oxodiiron(III) dimers. For iron(II)-porphyrin complexes this process may take only milliseconds at room temperature. The following mechanism was proposed in 1968 for the irreversible oxidation of iron(II)-porphyrinato species; subsequent work has largely confirmed it. \[Fe^{II} + O_{2} \rightleftharpoons Fe^{III}—O_{2}^{I-} \tag{4.29a}\] \[Fe^{III}—O_{2}^{I-} + Fe^{II} \rightleftharpoons Fe^{III}—O_{2}^{II-}—Fe^{III} \tag{4.29b}\] \[Fe^{III}—O_{2}^{II-}—Fe^{III} \rightarrow 2Fe^{IV}=O \tag{4.29c}\] \[Fe^{IV}=O + Fe^{II} \rightarrow Fe^{III}—O—Fe^{III} \tag{4.29d}\] In particular, the dimerization reaction (4.29b) may be rendered less favorable by low temperatures (< -40 °C) or by sterically preventing the bimolecular contact of an Fe -O moiety with an Fe moiety. In the latter case, sterically bulky substituents on the equatorial ligand surround the coordinated O ligand and the other axial position, trans to the coordinated dioxygen ligand, is protected with a nitrogenous base, such as imidazole, or with additional bulky substituents on the equatorial ligand (Figure 4.14). The protein effectively provides such protection and thus plays a key role in preventing the bimolecular contact of two hemes. The first observation of reversible binding of dioxygen to an iron(II)-porphyrin in the absence of protein was made in 1958. In that pioneering study, a heme group was immobilized on a polymer support specially modified to contain imidazole functions. The structurally characterizable hemoglobin or myoglobin species was replaced by a noncrystalline structurally uncharacterized polymer. Why does this irreversible oxidation not occur analogously for cobalt systems? Step (4.29c) involves cleavage of the O—O bond, which in H O has a bond energy of 34.3 kcal/mol or in Na O of 48.4 kcal/mol. By way of comparison, for O the bond energy is 117.2 and for HO • it is 55.5 kcal/mol. A simple molecular orbital picture gives insight into why an Fe =O species is stabilized relative to the analogous Co =O species. From Figure 4.15 we see that for metals with electronic configuration d , where n \(\leq\) 5, no electrons occupy the antibonding orbital \(\pi\)* for Fe -O or Fe =O moieties. For Co (d ) the extra electron goes into the antibonding orbital \(\pi\)*. As predicted by the model, Mn is observed indeed to behave like Fe . A second oxidation pathway does not require the bimolecular contact of two iron(II)-porphyrins. Coordinated dioxygen may be released not as O , as in normal dioxygen transport, but, as noted in Section I.C, as a superoxide radical anion O in a process called : \[Fe^{III}—O_{2}^{I-} \rightleftharpoons Fe^{III} + O_{2} \tag{4.30}\] This process is assisted by the presence of other nucleophiles that are stronger than the superoxide anion, such as chloride, and by protons that stabilize the O anion as HO : \[Fe_{III} + Cl^{-} \rightleftharpoons Fe^{III}—Cl^{-} \tag{4.31}\] The formation of methemoglobin occurs , probably by the above mechanism, at the rate of ~ 3 percent of total hemoglobin per day. If exogenous reductants are present, then further reduction of dioxygen can occur: \[2H^{+} + Fe^{III}—O_{2}^{I-} + e^{-} \rightarrow Fe^{V}=O + H_{2}O \tag{4.32}\] Such processes are important, for example, in the cytochrome P-450 system. With suitably small reductants, oxygenase activity also has been observed for hemoglobin A. This has led to the characterization of hemoglobin as a "frustrated oxidase." Note the formal similarity between this process (Equation 4.32) and the bimolecular irreversible oxidation of iron(II) porphyrins: the second Fe(II) complex in Reaction (4.29b) functions like the electron in Reaction (4.32). The end products of the irreversible bimolecular oxidation of Fe species contain the Fe —O—Fe fragment. Given the facile formation of \(\mu\)-oxodiiron(III) species, it is not surprising that the Fe—O—Fe motif is incorporated into a variety of metalloproteins, including the oxygen-carrier hemerythrin (Figure 4.10), the hydrolase purple acid phosphatase, the oxidoreductases ribonucleotide reductase and methane monoxygenase, an iron-sulfur protein rubrerythrin, and the iron-transport protein ferritin. In ferritin higher-order oligomers are formed. This \(\mu\)-oxodiiron(III) moiety has a distinctive fingerprint that has made it easy to identify this motif in proteins. Regardless of the number (4, 5, 6, or 7), geometry (tetrahedral, square pyramidal, tetragonally distorted octahedral, or pentagonal bipyramidal), and type of ligands (halide, RO , RCOO , aliphatic N, or aromatic N) around the iron center, and of the Fe—O—Fe angle, the magnetic susceptibility at room temperature lies in the range 1.5 to 2.0 Bohr magnetons per Fe —O—Fe group, equivalent to about one unpaired electron. In other words, the high-spin (S = \(\frac{5}{2}\)) iron centers are strongly antiferromagnetically coupled. Other bridging groups, such as OH , Cl , carboxylate, alkoxide, or phenoxide, give very weak coupling. The asymmetric Fe—O stretch, (Fe—O), lies in the range 730 to 880 cm ; in multiply bridged complexes this mode is weak in the infrared region. The symmetric vibration, (Fe—O), forbidden in the infrared region for linear, symmetric Fe—O—Fe groups, occurs in the range 360 to 545 cm . The symmetric mode is usually, but not always, observed by resonance Raman techniques upon irradiating on the low-energy side of the Fe—O chargetransfer band that occurs at about 350 nm. Few dinuclear iron(II) complexes are known where the ligands approximately resemble those believed or known to occur in the family of \(\mu\)-oxodiiron(III) proteins. The dioxygen-binding process in hemerythrin has no close nonbiological analogue. Although spectroscopically similar to oxyhemerythrin, the unstable monomeric purple peroxo complex formed by the addition of hydrogen peroxide to basic aqueous Fe (EDTA) solutions remains structurally uncharacterized. Iron porphyrins, the active sites of the hemoglobin family, have a rich magnetochemistry. Iron porphyrins may be octahedral (two axial ligands), square pyramidal (one axial ligand), or square planar (no axial ligand). The metal d orbitals, now having partial porphyrin \(\pi\)* character, are split, as shown in Figure 4.16. The radius of the metal atom is much greater when it is high spin (S = 2 for Fe , S = \(\frac{5}{2}\) for Fe ) than when it is low spin (S = 0 for Fe , S = \(\frac{1}{2}\) for Fe ). This difference influences Fe—N separations, porphyrin conformation, and the displacement of the iron center with respect to the porphyrin plane. For iron(II)-porphyrins, two strong-field axial ligands, such as a pair of imidazoles or an imidazole and carbon monoxide, lead to diamagnetic complexes (S = 0) with the six 3d electrons occupying those orbitals of approximate t symmetry. In a classic experiment in 1936, Pauling and Coryell proved that oxyhemoglobin and carbonmonoxyhemoglobin are diamagnetic. * * There was a considerable flurry of interest when an Italian group, using a SQUID (Superconducting Quantum Mechanical Interference Device), reported that at room temperature oxyhemoglobin was significantly paramagnetic. Not surprisingly, several theoretical papers followed that "proved" the existence of low-lying triplet and quintet excited states Subsequently, the residual paramagnetism was doubted and shown to arise from incomplete saturation of hemoglobin by O ; in other words, small amounts of deoxy hemoglobin remained Since oxygen affinity increases with decreased temperature, the concentration of paramagnetic impurity decreased with decreasing temperature. No axial ligands at all may lead to a spin state of S = 1, with unpaired electrons in the d and d orbitals. Five-coordinate iron(II)-porphyrinato complexes are commonly high spin, S = 2, although strong \(\sigma\)-donor \(\pi\)-acceptor ligands, such as phosphines, carbon monoxide, nitric oxide, and benzyl isocyanide, enforce a low-spin state. Five-coordinate iron(II)-porphyrinato complexes with aromatic nitrogenous axial ligands, such as pyridine or 1-methylimidazole, bind a second such axial ligand 10 to 30 times more avidly than the first to give the thermodynamically and kinetically (d , S = 0) stable hemochrome species, a process that is avoided by hemoglobins. That is, the equilibrium constant for the following disproportionation reaction is greater than unity, \[Fe—N + Fe—N \rightleftharpoons N—Fe—N +Fe \tag{4.33}\] except for bulky ligand N, such as 2-methylimidazole and 1,2-dimethylimidazole, for which the five-coordinate species predominates at room temperature even with a mild excess of ligand: \(\tag{4.34}\) For iron(III)-porphyrinato complexes, strong-field ligands lead to low-spin (S = \(\frac{1}{2}\)) complexes. A pair of identical weak-field ligands, such as tetrahydrofuran, leads to intermediate-spin (S = \(\frac{3}{2}\)) species. Five-coordinate species are, with few exceptions, high-spin (S = \(\frac{5}{2}\)), with all five 3d electrons in separate orbitals. Spin equilibria S = \(\frac{1}{2} \rightleftharpoons\) S = \(\frac{5}{2}\) and S = \(\frac{3}{2} \rightleftharpoons\) S = \(\frac{5}{2}\) are not unusual. Specific examples of these spin systems are given in Table 4.4. Higher oxidation states are found in some other hemoproteins. Fe(V)-porphyrin systems actually occur as Fe(IV)-porphyrin cation radical species, and Fe(I)-porphyrin systems exist as Fe(II)-porphyrin anion radical species. Substantial structural changes occur upon the addition of ligands and upon changes in spin state. In one mechanism of cooperativity these changes are the "trigger" (metrical details are deferred until the next section). Spectral changes in the UV-visible region are observed also (Figure 4.17) and may be monitored conveniently to evaluate the kinetic and thermodynamic parameters of ligand binding to hemoglobin. a) Could be placed in Fe column b) Could be placed in Fe column with spin = 0. c) Non-linear Fe—NCS moiety.

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The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. In contrast to molecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, molecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances . The properties of liquids are intermediate between those of gases and solids but are more similar to solids. Intermolecular forces determine bulk properties such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire enough thermal energy to overcome the intermolecular forces that lock them into place in the solid. Through various experiments, Charles Augustin de Coulomb found a way to explain the interactions between charged particles, which in turn helped to explain where the stabilities and instabilities of various particles come from. While the entities that hold atoms together within a molecule can be attributed to bonds, the forces that create these bonds can be explained by Coulomb Forces. Thus, the physical basis behind the bonding of two atoms can be explained. Coulomb’s findings indicate that like charges repel each other and unlike charges attract one another. Thus electrons, which are negatively charged, repel each other but attract protons. Likewise, protons repel each other. Each atom is made up of a nucleus in the center, which consists of a number of protons and neutrons, depending upon the element in question. Surrounding the nucleus are electrons that float around the nucleus in what can be thought of as a cloud. As two atoms approach one another, the protons of one atom attract the electrons of the other atom. Similarly, the protons of the other atom attract the electrons of the first atom. As a result, the simultaneous attraction of the components from one atom to another create a bond. This interaction can be summarized mathematically and is known as : In this mathematical representation of Coulomb's observations, From Equation \ref{C}, the electrostatic force between two charges is proportional to the of the distance separating the two atoms. The interactions between ions (ion - ion interactions or charge-charge interactions) are the easiest to understand since such interactions are just a simple application of Coulombic forces (Equation \ref{C}). This specific interaction operates over relatively long distances in the gas phase and is responsible for the attraction of opposite charge ions and the repulsion of like charged ions. Coulombic forces are also involved in all forms of chemical bonding; when they act between separate charged particles they are especially strong. Thus the energy required to pull a mole of \(\ce{Na^{+}}\) and \(\ce{F^{–}}\) ions apart in the sodium fluoride crystal is greater than that needed to break the a covalent bonds of a mole of \(\ce{H2}\). The effects of ion-ion attraction are seen most directly in salts such as \(\ce{NaF}\) and \(\ce{NaCl}\) that consist of oppositely-charged ions arranged in inter-penetrating crystal lattices. According to Coulomb's Law the force between two charged particles is given by \[ \underbrace{F= \dfrac{q_1q_2}{4\pi\epsilon_0 r^2}}_{\text{ion-ion Force}} \label{7.2.1}\] Instead of using SI units, chemists often prefer to express atomic-scale distances in picometers and charges as electron charge (±1, ±2, etc.) Using these units, the proportionality constant \(1/4\pi\epsilon\) works out to \(2.31 \times 10^{16}\; J\; pm\). The sign of \(F\) determines whether the force will be attractive (–) or repulsive (+); notice that the latter is the case whenever the two 's have the same sign. Two oppositely-charged particles flying about in a vacuum will be attracted toward each other, and the force becomes stronger and stronger as they approach until eventually they will stick together and a considerable amount of energy will be required to separate them. They form an , a new particle which has a positively-charged area and a negatively-charged area. There are fairly strong interactions between these ion pairs and free ions, so that these clusters tend to grow, and they will eventually fall out of the gas phase as a liquid or solid (depending on the temperature). Equation \ref{7.2.1} is an example of an ; the force falls off as the square of the distance. A similar law governs the manner in which the illumination falls off as you move away from a point light source; recall this the next time you walk away from a street light at night, and you will have some feeling for what an inverse square law means. The stronger the attractive force acting between two particles, the greater the amount of work required to separate them. Work represents a flow of energy, so the foregoing statement is another way of saying that when two particles move in response to a force, their potential energy is lowered. This work is found by integrating the negative of the force function with respect to distance over the distance moved. Thus the energy that must be supplied in order to completely separate two oppositely-charged particles initially at a distance \(r_0\) is given by \[ w= - \int _{r_o} ^{\infty} \dfrac{q_1q_2}{4\pi\epsilon_0 r^2}dr = - \dfrac{q_1q_2}{4\pi\epsilon_0 r_o} \label{7.2.2}\] hence, the potential (\(V_{ion-ion}\)) responsible for the ion-ion force is \[ \underbrace{V_{ion-ion} = \dfrac{q_1q_2}{4\pi\epsilon_0 r} }_{\text{ion-ion potential}} \label{7.2.3}\] When sodium chloride is melted, some of the ion pairs vaporize and form neutral \(\ce{NaCl}\) dimers. How much energy would be released when one mole of \(\ce{Na^{+}}\) and \(\ce{Cl^{–}}\) ions are brought together to generate dimers in this way? The bondlength of \(\ce{NaCl}\) is 237 pm. The interactions involved in forming \(\ce{NaCl}\) dimers is the ion-ion forces with a potential energy given by Equation \ref{7.2.3}. However, this is the energy of interaction for one pair of \(\ce{Na^{+}}\) and \(\ce{Cl^{–}}\) ion and needs to be scaled by a mole. So the energy released will be \[\begin{align*}E &= N_a V(\ce{NaCl}) \\[4pt] &= N_a\dfrac{q_1q_2}{4\pi\epsilon_0 r} \end{align*}\] The \(r\) in this equation is the distance between the two ions, which is the bondlength of 237 pm (\(237 \times 10^{-12}m\)). \[\begin{align*}E &= (6.022 \times 10^{23} ) \underbrace{(8.987 \times 10^9 N m^2/C^2 )}_{1/4\pi\epsilon_o} \dfrac{(+1.6 \times 10^{-19}C) (-1.6 \times 10^{-19}C) }{ 237 \times 10^{-12} m} \\[4pt] &= –584 \;kJ/mol \end{align*}\] This is not the energy needed to separate one mole of NaCl since that is a lattice and has more than pairwise interactions and require addressing the geometric orientation of the lattice (see for more details). Intermolecular forces are electrostatic in nature; that is, they arise from the electrostatic interaction between positively and negatively charged species. Like covalent and ionic bonds, intermolecular interactions are the sum of both attractive and repulsive components. Because electrostatic interactions fall off rapidly with increasing distance between molecules, intermolecular interactions are most important for solids and liquids, where the molecules are close together. These interactions become important for gases only at very high pressures, where they are responsible for the observed deviations from the ideal gas law at high pressures (i.e., real gases). A dipole that is close to a positive or negative ion will orient itself so that the end whose partial charge is opposite to the ion charge will point toward the ion. This kind of interaction is very important in aqueous solutions of ionic substances; H O is a highly polar molecule, so that in a solution of sodium chloride, for example, the Na ions will be enveloped by a shell of water molecules with their oxygen-ends pointing toward these ions, while H O molecules surrounding the Cl ions will have their hydrogen ends directed inward. As a consequence of ion-dipole interactions, all ionic species in aqueous solution are hydrated; this is what is denoted by the suffix in formulas such as K (aq), etc. The strength of ion-dipole attraction depends on the magnitude of the dipole moment and on the charge density of the ion. This latter quantity is just the charge of the ion divided by its volume. Owing to their smaller sizes, positive ions tend to have larger charge densities than negative ions, and they should be more strongly hydrated in aqueous solution. The hydrogen ion, being nothing more than a bare proton of extremely small volume, has the highest charge density of any ion; it is for this reason that it exists entirely in its hydrated form H O in water. Since there is now both attractive and repulsive interactions and they both get weaker as the ion and dipole distance increases while also approaching each other in strength, the net ion-dipole is an inverse square relationship as shown in Equation \ref{11.2.2}. \[ \underbrace{ E\: \propto \: \dfrac{-|q_1|\mu_2}{r^2}}_{\text{ion-dipole potential}} \label{11.2.2}\] There are several differences between ion-ion potential (Equation \ref{7.2.3}) and the ion-dipole potential (Equation \ref{11.2.2}) interactions. It needs to be understood that the molecules in a solution are rotating and vibrating and actual systems are quite complicated (Figure \(\Page {4}\)). What is important to realize is that these interactions are Coulombic in nature and how the mathematical equations describe this in terms of the magnitude of the charges and their distances from each other. In this course we will not be calculating dipole moments or the magnitudes of them, but understanding how to read the equations, and developing qualitative understandings that allow us to predict trends. Polar covalent bonds behave as if the bonded atoms have localized fractional charges that are equal but opposite (i.e., the two bonded atoms generate a ). If the structure of a molecule is such that the individual bond dipoles do not cancel one another, then the molecule has a net dipole moment. Molecules with net dipole moments tend to align themselves so that the positive end of one dipole is near the negative end of another and vice versa, as shown in part (a) in Figure \(\Page {1}\). These arrangements are more stable than arrangements in which two positive or two negative ends are adjacent (Figure \(\Page {1c}\)). Hence , such as those in Figure \(\Page {5b}\), are , whereas those in Figure \(\Page {5d}\) are . Because molecules in a liquid move freely and continuously, molecules always experience both attractive and repulsive dipole–dipole interactions simultaneously, as shown in Figure \(\Page {2}\). On average, however, the attractive interactions dominate. Because each end of a dipole possesses only a fraction of the charge of an electron, dipole–dipole interactions are substantially weaker than the interactions between two ions, each of which has a charge of at least ±1, or between a dipole and an ion, in which one of the species has at least a full positive or negative charge. In addition, the attractive interaction between dipoles falls off much more rapidly with increasing distance than do the ion–ion interactions. Recall that the attractive energy between two ions is proportional to 1/ , where is the distance between the ions. Doubling the distance ( → 2 ) decreases the attractive energy by one-half. In contrast, the energy of the interaction of two dipoles is proportional to 1/ , so doubling the distance between the dipoles decreases the strength of the interaction by 2 , or 64-fold: \[V=-\dfrac{2\mu_{A}^2\mu_{B}^2}{3(4\pi\epsilon_{0})^2r^6}\dfrac{1}{k_{B}T} \label{5}\] Thus a substance such as HCl, which is partially held together by dipole–dipole interactions, is a gas at room temperature and 1 atm pressure, whereas NaCl, which is held together by interionic interactions, is a high-melting-point solid. Within a series of compounds of similar molar mass, the strength of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in Table \(\Page {1}\). Using what we learned about predicting relative bond polarities from the electronegativities of the bonded atoms, we can make educated guesses about the relative boiling points of similar molecules. The attractive energy between two ions is proportional to 1/ , whereas the attractive energy between two dipoles is proportional to 1/ . Arrange ethyl methyl ether (\(\ce{CH3OCH2CH3}\)), 2-methylpropane [isobutane, \(\ce{(CH3)2CHCH3}\)], and acetone (\(\ce{CH3COCH3}\)) in order of increasing boiling points. Their structures are as follows: compounds order of increasing boiling points Compare the molar masses and the polarities of the compounds. Compounds with higher molar masses and that are polar will have the highest boiling points. The three compounds have essentially the same molar mass (58–60 g/mol), so we must look at differences in polarity to predict the strength of the intermolecular dipole–dipole interactions and thus the boiling points of the compounds. The first compound, 2-methylpropane, contains only C–H bonds, which are not very polar because C and H have similar electronegativities. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point. Ethyl methyl ether has a structure similar to H O; it contains two polar C–O single bonds oriented at about a 109° angle to each other, in addition to relatively nonpolar C–H bonds. As a result, the C–O bond dipoles partially reinforce one another and generate a significant dipole moment that should give a moderately high boiling point. Acetone contains a polar C=O double bond oriented at about 120° to two methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should result in both a rather large dipole moment and a high boiling point. Thus we predict the following order of boiling points: 2-methylpropane < ethyl methyl ether < acetone. This result is in good agreement with the actual data: 2-methylpropane, boiling point = −11.7°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, boiling point = 7.4°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D. Arrange carbon tetrafluoride (CF ), ethyl methyl sulfide (CH SC H ), dimethyl sulfoxide [(CH ) S=O], and 2-methylbutane [isopentane, (CH ) CHCH CH ] in order of decreasing boiling points. dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (boiling point = 67°C) > 2-methylbutane (boiling point = 27.8°C) > carbon tetrafluoride (boiling point = −128°C) Thus far we have considered only interactions between polar molecules, but other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature, and others, such as iodine and naphthalene, are solids. Even the noble gases can be liquefied or solidified at low temperatures, high pressures, or both (Table \(\Page {2}\)). What kind of attractive forces can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German physicist who later worked in the United States. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could result in the formation of short-lived , which produce attractive forces called between otherwise nonpolar substances. Consider a pair of adjacent He atoms, for example. On average, the two electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in constant motion, however, their distribution in one atom is likely to be asymmetrical at any given instant, resulting in an instantaneous dipole moment. As shown in part (a) in Figure \(\Page {3}\), the instantaneous dipole moment on one atom can interact with the electrons in an adjacent atom, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The net effect is that the first atom causes the temporary formation of a dipole, called an , in the second. Interactions between these temporary dipoles cause atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing distance. London was able to show with quantum mechanics that the attractive energy between molecules due to temporary dipole–induced dipole interactions falls off as 1/ . Doubling the distance therefore decreases the attractive energy by 2 , or 64-fold. Instantaneous dipole–induced dipole interactions between nonpolar molecules can produce intermolecular attractions just as they produce interatomic attractions in monatomic substances like Xe. This effect, illustrated for two H molecules in part (b) in Figure \(\Page {3}\), tends to become more pronounced as atomic and molecular masses increase (Table \(\Page {2}\)). For example, Xe boils at −108.1°C, whereas He boils at −269°C. The reason for this trend is that the strength of London dispersion forces is related to the ease with which the electron distribution in a given atom can be perturbed. In small atoms such as He, the two 1 electrons are held close to the nucleus in a very small volume, and electron–electron repulsions are strong enough to prevent significant asymmetry in their distribution. In larger atoms such as Xe, however, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. As a result, it is relatively easy to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an atom or molecule is called its . Because the electron distribution is more easily perturbed in large, heavy species than in small, light species, we say that heavier substances tend to be much more than lighter ones. For similar substances, London dispersion forces get stronger with increasing molecular size. The polarizability of a substance also determines how it interacts with ions and species that possess permanent dipoles. Thus London dispersion forces are responsible for the general trend toward higher boiling points with increased molecular mass and greater surface area in a homologous series of compounds, such as the alkanes (part (a) in Figure \(\Page {4}\)). The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in Figure \(\Page {4}\) shows 2,2-dimethylpropane (neopentane) and -pentane, both of which have the empirical formula C H . Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas -pentane has an extended conformation that enables it to come into close contact with other -pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of -pentane (36.1°C). All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present. In general, however, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the former predominate. Arrange -butane, propane, 2-methylpropane [isobutene, (CH ) CHCH ], and -pentane in order of increasing boiling points. compounds order of increasing boiling points Determine the intermolecular forces in the compounds and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point. The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and -pentane should have the highest, with the two butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and -butane has the more extended shape. Consequently, we expect intermolecular interactions for -butane to be stronger due to its larger surface area, resulting in a higher boiling point. The overall order is thus as follows, with actual boiling points in parentheses: propane (−42.1°C) < 2-methylpropane (−11.7°C) < -butane (−0.5°C) < -pentane (36.1°C). Arrange GeH , SiCl , SiH , CH , and GeCl in order of decreasing boiling points. GeCl (87°C) > SiCl (57.6°C) > GeH (−88.5°C) > SiH (−111.8°C) > CH (−161°C) Intermolecular forces are electrostatic in nature and include van der Waals forces and hydrogen bonds. Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions. Transitions between the solid and liquid or the liquid and gas phases are due to changes in intermolecular interactions but do not affect intramolecular interactions. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as ), and hydrogen bonds. arise from the electrostatic interactions of the positive and negative ends of molecules with permanent dipole moments; their strength is proportional to the magnitude of the dipole moment and to 1/ , where is the distance between dipoles. are due to the formation of in polar or nonpolar molecules as a result of short-lived fluctuations of electron charge distribution, which in turn cause the temporary formation of an in adjacent molecules. Like dipole–dipole interactions, their energy falls off as 1/ . Larger atoms tend to be more than smaller ones because their outer electrons are less tightly bound and are therefore more easily perturbed.

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This section explores the use of symmetry to determine selection rules. Here we derive an analytical expression for the transition dipole moment integral for the particle-in-a-box model. The result that the magnitude of this integral increases as the length of the box increases explains why the absorption coefficients of the longer cyanine dye molecules are larger. We use the and the trigonometric forms of the particle-in-a-box wavefunctions to get Equation \(\ref{4-27}\) for an electron making a transition from orbital \(i\) to orbital \(f\). \[ \begin{align} \mu _T &= \dfrac {-2e}{L} \int \limits _0^L \sin \left (\dfrac {f \pi x}{L} \right ) x \sin \left ( \dfrac {i \pi x }{L} \right ) dx \\[4pt] &= \dfrac {-2e}{L} \int \limits _0^L x \sin \left (\dfrac {f \pi x}{L} \right ) \sin \left ( \dfrac {i \pi x }{L} \right ) dx \label {4-27} \end{align}\] Why is there a factor \(2/L\) in Equation \(\ref{4-27}\)? What are the units associated with the dipole moment and the transition dipole moment? Simplify the integral in Equation \(\ref{4-27}\) by substituting the \[\sin \psi \sin \theta = \dfrac {1}{2} \left[ \cos (\psi - \theta ) - \cos (\psi + \theta) \right] \label {4-28}\] and also redefine the sum and difference terms: \[ \Delta n = f - i \nonumber\] and \[n_T = f + i \nonumber \] So Equation \(\ref{4-27}\) \[ \begin{align} \mu _T &= \dfrac {-e}{L} \int \limits _0^L x \left[ \cos \left (\dfrac {\Delta n \pi x}{L} \right ) - \cos \left (\dfrac {n_T \pi x}{L} \right ) \right ] dx \\[4pt] &= \dfrac {-e}{L} \left[ \int \limits _0^L x \cos \left (\dfrac {\Delta n \pi x}{L} \right ) dx - \int \limits _0^L x \cos \left (\dfrac { n_T \pi x}{L} \right ) dx \right ] \label{step 3} \end{align}\] These two definite integrals can be directly evaluated using this relationship \[ \int \limits _0^L x \cos (ax) dx = \left[ \dfrac {1}{a^2} \cos (ax) + \dfrac {x}{a} \sin (ax) \right]^L_0 \label {4-29}\] where \(a\) is any nonzero constant. Using Equation \ref{4-29} in Equation \ref{step 3} produces \[T = \dfrac {-e}{L} {\left(\dfrac {L}{\pi}\right)}^2 \left[ \dfrac {1}{\Delta n^2} (\cos (\Delta n \pi) - 1) - \dfrac {1}{n^2_T} (\cos (n_T \pi) - 1) + \dfrac {1}{\Delta n} \sin (\Delta n \pi ) - \dfrac {1}{n_T} \sin (n_T \pi) \right] \label {4-31}\] Show that if \(Δn\) is an even integer, then \(n_T\) must be an even integer and \(μ_T = 0\). Show that if \(i\) and \(f\) are both even or both odd integers then \(Δn\) is an even integer and \(μ_T = 0\). Show that if \(Δn\) is an odd integer, then \(n_T\) must be an odd integer and \(μ_T\) is given by Equation \(\ref{4-32}\). \[ \mu _T = \dfrac {-2eL}{\pi ^2} \left(\dfrac {1}{n^2_T} - \dfrac {1}{\Delta n^2} \right) = \dfrac {8eL}{\pi^2} \left( \dfrac {f_i}{{(f^2 - i^2)}^2} \right) \label {4-32}\] Show that the two expressions for the transition moment in Equation \(\ref{4-32}\) are in fact equivalent. What is the value of the transition moment integral for transitions 1→3 and 2→4? For these two transitions, either n and f are both odd or they are both even integers. In either case, \(Δn\) and \(n_T\) are even integers. The cosine of an even integer multiple of \(π\) is +1 so the cosine terms in Equation \(\ref{4-31}\) become (1-1) = 0. The sine terms are zero because the sine of an even integer multiple of \(π\) is zero. Therefore, \(μ_T = 0\) for these transitions and they are forbidden. The same reasoning applies to any transitions that have both i and f as even or as odd integers. What is the value of the transition moment for the n = 8 to f = 10 transition? What is the value of the transition moment integral for transitions 1→2 and 2→3? For these two transitions Δn = 1 and nT = 3 and 5, respectively, all odd integers. The cosine of an odd-integer multiple of π is -1 so the cosine terms in Equation \(\ref{4-31}\) become (-1-1) = -2. The sine terms in Equation \ref{4-31} are zero because the sine of an odd integer multiple of π is zero. Therefore, \(μ_T\) has some finite value given by Equation \ref{4-32}. The same reasoning is used to evaluate the transition moment integral for any transitions that have Δn and nT as odd integers, e.g. 2→7 and 3→8. In these cases Δn = 5 and nT = 9 and 11, respectively. Again the transition moment integral for each of these transitions is finite. Explain why one of the following transitions occurs with excitation by light and the other does not: \(i = 1\) to \(f = 7\) and \(i = 3\) to \(f = 6\). From Examples \(\Page {1}\) and \(\Page {2}\), we can formulate the selection rules for the particle-in-a-box model: Transitions are forbidden if \(Δn = f - i\) is an even integer. Transitions are allowed if \(Δn = f - i \) is an odd integer. In the next section we will see that these selection rules can be understood in terms of the symmetry of the wavefunctions. Through the evaluation of the transition moment integral, we can understand why the spectra of cyanine dyes are very simple. The spectrum for each dye consists only of a single peak because other transitions have very much smaller transition dipole moments. We also see that the longer molecules have the larger absorption coefficients because the transition dipole moment increases with the length of the molecule. The lowest energy transition is from the HOMO to the LUMO, which were defined previously. Compute the value of the transition moment integral for the HOMO to LUMO transition \(E_3→E_4\) for a cyanine dye with 3 carbon atoms in the conjugated chain. What is the next lowest energy transition for a particle-in-a-box? Compute the value of the transition moment integral for the next lowest energy transition that is allowed for this dye. What are the quantum numbers for the energy levels associated with this transition? How does the probability of this transition compare in magnitude with that for 3→4?

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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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Students often wonder why many chemical reactions yield an equilibrium mixture in which a significant amount of the reactants are present, even though the products have a lower standard free energy than the reactants. One might at first think that as long as reactants are present, the free energy could be reduced if conversion of reactants to products were complete. The short answer is that by "contaminating" some of the product with reactants, the free energy of the system (including both reactants and products) can be reduced below that of the pure products alone. This additional drop in the free energy is due to the of reactants with products. Unless you are enrolled in a more advanced course, you are probably not expected to know how to calculate free energies of mixing. All you really need to know is that it is formally equivalent to the expansion of gases (or to the dilution of a solute) into a larger volume. for more details. This example illustrates how the free energies of the reaction components combine with the free energies of mixing reactants with products to minimize the Gibbs function in the equilibrium mixture. To keep things as simple as possible, we will deal with the isomerization equilibrium between the two butanes C H at 298 K: )

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Iron is an essential elements for human. , an iron-storage protein found in animals and plants, contains the largest cluster currently known, an ordered aggregate {(FeOOH) (FeO H PO )} containing up to 4,500 Fe(III) atoms. The cluster occurs within a protein cavity roughly 80 angstroms in diameter. Proteins in which heavy metal ions are bound directly to some of the side chains of histidine, cysteine, or some other amino acid are called metalloproteins. Two metalloproteins, transferrin and ceruloplasmin, occur in the globulin fractions of blood serum; they act as carriers of iron and copper, respectively. Transferrin has a molecular weight of 84,000 and consists of two identical subunits, each of which contains one ferric ion (Fe ) that seems to be bound to tyrosine. Several genetic variants of transferrin are known to occur in man. Another iron protein, , which contains 20 to 22 percent iron, is the form in which iron is stored in animals; it has been obtained in crystalline form from liver and spleen. A molecule consisting of 20 subunits, its molecular weight is approximately 480,000. The iron can be removed by reduction from the ferric (Fe3+) to the ferrous (Fe2+) state. The iron-free protein, apoferritin, is synthesized in the body before the iron is incorporated.

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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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When discussing the concept of Stability it is necessary to distinguish between thermodynamic and kinetic stability. Here B is at lower energy than A so that ΔG is negative. The reaction should therefore proceed spontaneously and B is the more thermodynamically stable species. The reaction as shown though has a barrier to the progress of the reaction called the Activation Barrier (Ea) and so the reaction may proceed very slowly. The thermodynamics describes only the starting and ending position of the reaction and not the intermediate or transition state. If the kinetics is slow, A is described as being while if it proceeds quickly then A is described as being . The conditions that distinguish them are: if the reaction takes longer than 1 minute under the conditions of concentration 0.1 M, temperature 25°C, then it is INERT, if under the same conditions the reaction time < 1 minute, then it is LABILE. In the lecture on isomerism, we depend on the samples being kinetically stable i.e. inert. In the lecture on Chelation and Stability we concentrate on thermodynamic stability and look at the changes in free energy, enthalpy and entropy during the reaction. When we consider thermodynamic stability we need to be familiar with 2 formulae: ΔG = - RT ln(K) or alternatively ΔG = - 2.303RT log (K) ---(1) ΔG = ΔH - TΔS ---(2) The first relates the free energy to the stability constant and the second shows the breakdown into the component enthalpy and entropy terms.

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Raman spectroscopy is a chemical instrumentation technique that exploits molecular vibrations. It does not require large sample sizes and is non-destructive to samples. It is capable of qualitative analysis of samples and the intensity of spectral bands produced assist in quantitative analysis as well. Raman spectroscopy is even being used in areas outside of physical science (i.e. archeology and art preservation) due to the characteristics mentioned above. Raman spectroscopy is based on scattering of radiation (Raman scattering), which is a phenomenon discovered in 1928 by physicist Sir C. V. Raman. The field of Raman spectroscopy was greatly enhanced by the advent of laser technology during the 1960s. Resonance Raman also helped to advance the field. This technique is more selective compared to non-resonance Raman spectroscopy. It works by exciting the analyte with incident radiation corresponding to the electronic absorption bands. This causes an augmentation of the emission up to a factor of 10 in comparison to non-resonance Raman. In this section readers will be introduced to the theory behind resonance and non-resonance Raman spectroscopy. Each technique has its share of advantages and challenges. Each of these aspects will be explored. Raman scattering is the basis of the two Raman techniques. A molecule must have polarizability to Raman scatter and its symmetry must be even (or gerade) for it to have polarizability . Furthermore, the more electrons a molecule has gererally increases its polarizability. Polarizability ( ) is a measure of an applied electronic field’s ( ) ability to generate a dipole moment (µ) in the molecule. In other words, it is an alteration of a molecule's electron cloud. Mathematically, this can be determined by the following equation: \[ \mu = \alpha E \label{1} \] To help provide a better visualization of how Raman spectroscopy works, a generic diagram can be seen in Figure 2. A sample is irradiated with monochromatic laser light; which is then scattered by the sample. The scattered light passes through a filter to remove any stray light that may have also been scattered by the sample. The filtered light is then dispersed by the diffraction grating and collected on the detector. This set-up works for both the non-resonance and resonance Raman techniques. scattering occurs when the radiation interacts with a molecule resulting in polarization of the molecule’s electrons. The increase in energy from the radiation excites the electrons to an unstable virtual state; therefore, the interaction is almost immediately discontinued and the radiation is emitted (scattered) at a slightly different energy than the incident radiation. scattering occurs in a similar fashion. However, the incident radiation is at a frequency near the frequency of an electronic transition of the molecule of interest. This provides enough energy to excite the electrons to a higher electronic state. Figure 1 provides a visual depiction of what non-resonance and resonance Raman scattering looks like in terms of energy levels. Instrumental techniques each have certain strengths that make them better suited for some jobs as oppose to others. Non-resonance is a good example of this notion. It is considered better suited for analyzing water containing samples due to water’s low polarizability. Non-resonance and resonance Raman each have the capability to analyze samples in the gaseous, liquid, or solid state. Their non-destructive nature makes it a great candidate for doing analysis of delicate materials. Archeologists and art historians even find resonance Raman spectroscopy useful for studying and authentication of artifacts and artwork. Monochromatic light in the ultraviolet or near-infrared regions is generally used for both resonance and non-resonance Raman spectroscopy. A tunable laser is preferred for resonance Raman and can be an advantage. That is because only one laser is necessary to do analyses of multiples samples in which each one requires a different excitation wavelength. This allows the user to switch out samples without having to switch out the lasers as well. It becomes a matter of just changing the setting on the tunable laser. If the laboratory is not equipped with a tunable laser, any laser that is available can be used to achieve the enhancement of the Raman signal. The only stipulation being that the laser available must have a frequency as near as possible to one of the analyte’s electronic transitions. Therefore, researchers conducting resonance Raman spectroscopy without a tunable laser are at the mercy of whatever laser they do have in the laboratory. Resonance Raman spectroscopy has greater sensitivity compared to its non-resonance counterpart. It is capable of analyzing samples with concentrations as low as 10 M. Non-resonance Raman can analyze samples with concentrations no lower than 0.1 M. Resonance Raman spectroscopy produces a spectrum with relatively few lines. The reason being that the technique only augments Raman signals affiliated with chromophores in the analyte. This makes the technique particularly useful for analysis of larger molecules like biomolecules. Fluorescence is a problem for both Resonance Raman techniques, particularly when using sources in the visible range. Non-resonance Raman signals are generally weak and can be easily overwhelmed by fluorescence signals. In addition, fluorescence has a longer excited state lifetime compared to Raman scattering, causing an inability to detect Raman signals. Even when the analyte is not a fluorescent molecule, the signal could be a result of the sample matrix content (i.e. solvent or contaminants). Resonance Raman is particularly at risk of inducing fluorescence because it uses sources at frequencies near to that of a molecule’s electronic transition. The radiation is more likely to absorb resulting in fluorescence as a possible mechanism for the electrons return to the ground state. Thus, highly fluorescent molecules should be avoided when using Raman spectroscopy; especially resonance Raman. Figure 3 is a general illustration of how a fluorescence signal can overwhelm Raman signals. There are techniques that spectroscopists use to avoid fluorescence interference. For instance, background subtraction could be done. Another example is to use near-infrared radiation to excite the sample as a means to overcome fluorescence. A more elaborate method was used by Matousek et al. They took advantage of the differences in excitation lifetimes for Raman and fluorescence. It required implementing shifted excitation Raman difference spectroscopy (SERDS) in conjunction with a device known as a Kerr gate to successfully obtain a resonance Raman spectrum of the rhodamine 6G dye. SERDS is a technique that uses two excitation wavelengths to produce two Raman spectra. The excitation wavelengths have a difference in value that corresponds to the bandwidth of the Raman signal. The two spectra are subtracted from each other and the difference spectra is recreated by means of mathematical processes. A Kerr gate can be used to remove fluorescence from a Raman signal based on their different lifetimes. The device consists of a couple of crossed polarizers, Kerr medium, and an additional laser to provide a gating pulse. Now consider a sample that has been irradiated resulting in fluorescence and Raman scattering. The fluorescence and Raman scatter would pass through the crossed polarizer and then through the Kerr medium (Matousek et al. used carbon disulfide as the Kerr medium). The Kerr gate is referred to as being open when a laser pulse (the gating pulse) strikes the Kerr medium as the fluorescence and Raman scattered light pass through. Furthermore, the gate remains open for a length of time corresponding to the lifetime of the Raman scatter, The interaction of the gating pulse with the Kerr medium causes the light to become anisotropic and transmit beyond the Kerr medium. The light then goes from being polarized in a linear direction to elliptical polarization. However, Raman scattered light can be selectively switched back to linear polarization by selecting the appropriate propagation length for Kerr medium transmission; or by altering the degree of anisotropy. The Raman scattered light is then allowed to pass through the second crossed polarizer and on to the spectrometer. Any fluorescence that passes through the Kerr medium is prevented from entering the spectrometer due to its inability to transmit through the second crossed polarizer on account of its new elliptical polarization. Figure 4 should assist with the visualization of such process. This module was meant to provide an introduction to the similarities and differences between non-resonance and resonance Raman spectroscopy. Notice that they each have their own advantages making both of them powerful analytical techniques. Non-resonance Raman is more advantageous when compared to IR spectroscopy. However, resonance Raman appears to have the upper hand when compared to its non-resonance counterpart. The important thing is to choose the technique that is most appropriate for the work to be done.

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Prior to studying this section, you may wish to revise your solution calculations and your molar calculations. Aquatic life is dependent upon gases dissolved in the water that they live in. Gases, such as carbon dioxide (CO ) will be required by aquatic plants for photosynthesis, whereas most aquatic plants and animals require oxygen for anaerobic respiration. Microorganisms will also require oxygen as they go about decomposing organic matter. One important indicator of water quality is the dissolved oxygen content of the water. At a pressure of 1atm and a temperature of 20 C, the maximum solubility of oxygen is about 9ppm, which equates to 0.009 g dm . Oxygen, being a non-polar molecule, has a low solubility in water, which is a polar solvent. The polar water molecule induces a dipole moment on the oxygen molecule and the two molecules are now weakly attracted. The diagram below shows the polar water molecule on the left inducing a dipole moment on the non-polar oxygen molecule leading to a weak force of attraction; To ensure a balanced and diverse aquatic community, the oxygen content should not fall below 6ppm, although some species of fish can survive in environments with oxygen contents as low as 3ppm. Bacteria are able to survive in water with even lower levels of oxygen. When organic matter decomposes aerobically in water, the bacteria responsible for this process use up some of the dissolved oxygen present in the water. The amount of oxygen required by the bacteria to decompose this organic matter is defined as biological oxygen demand (BOD). This is often measured in a fixed volume of water over a fixed period of time such as 5 days. If water has a high BOD without the means of replenishing the used oxygen, then very soon it will not be able to support aquatic life. There is a high risk of this happening in bodies of water that are still and do not have much mechanical mixing, e.g. lakes and ponds, whereas fast flowing rivers are able to replenish this oxygen via the mechanical action of its flow. So what factors could cause an increase in BOD? If excessive biodegradable materials find their way into water, this will lead to an increase in decomposing bacteria and hence an increase in BOD. Possible sources of this material include sewerage and industrial wastes from food processing or paper mills. BOD of water can also increase due to the addition of nutrients such as nitrates and phosphates that can be found in fertilizers or laundry detergents. This causes an increase in algae growth, which eventually dies, and decays. As plant growth becomes excessive, the volume of dead and decaying organic material increases rapidly. This decay requires O , leading to the depletion of O in the water. We can use BOD values as an indicator of water quality. The BOD of a sample of water can be tested using a redox titration called the Winkler method. The principle of the Winkler method is as follows; oxygen in a water sample is made to oxidise iodide ions into iodine. The amount of iodine produced is determined by titrating with a standard thiosulphate solution. The amount of oxygen present in the original sample of water can be determined from the titer. The reactions are summarized as follows: 1) 2 Mn + 4 OH + O → 2 MnO + 2 H O 2) MnO + 4 H + 2 I → Mn + I + 2 H O 3) I + 2 S O → S O + 2 I   Using the steps outlined above, please calculate the following. A 500 cm sample of water was saturated with oxygen and left for 5 days. The final oxygen content was measured using the sequence of reactions highlighted above. It was found that 5.00cm of a 0.0500 mol dm solution of Na S O was required to react with the iodine produced. a) Calculate how many moles of Na S O reacted with the iodine in reaction (3) b) Deduce how many moles of iodine had been produced in reaction (2). c) Deduce how many moles of MnO had been produced in reaction (1). d) Deduce how many moles of O were present in the water. e) Calculate the solubility of oxygen in the water in g dm . f) Assume the maximum solubility of the water is 0.009 g dm and deduce the BOD of the water sample. Solution a) Amount of Na S O = 5.00 x 0.0500/1000 = 2.5 x 10 moles. b) Amount of I = ½ (2.50 x 10 moles) = 1.25 x 10 moles. c) Amount of MnO = 1.25 x 10 moles. d) Amount of O = ½ (1.25 x 10 ) moles = 6.25 x 10 moles. e) Amount of O2(g) in 1dm3 = 1.25 x 10-4 moles Mass in 1dm3 = 0.004 g dm-3 f) Oxygen used by bacteria (BOD) = 0.009 g dm – 0.004 g dm = 0.005 g dm

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Consideration of the quantum mechanical description of the particle-in-a-box exposed two important properties of quantum mechanical systems. We saw that the eigenfunctions of the Hamiltonian operator are orthogonal, and we also saw that the position and momentum of the particle could not be determined exactly. We now examine the generality of these insights by stating and proving some fundamental theorems. These theorems use the Hermitian property of quantum mechanical operators, which is described first. Since the eigenvalues of a quantum mechanical operator correspond to measurable quantities, the eigenvalues must be real, and consequently a quantum mechanical operator must be Hermitian. We start with the premises that ψ and φ are functions, \(\int d\tau\) represents integration over all coordinates, and the operator  is Hermitian by definition if \[ \int \psi ^* \hat {A} \psi d\tau = \int (\hat {A} ^* \psi ^* ) \psi d\tau \label {4-37}\] This equation means that the complex conjugate of  can operate on ψ* to produce the same result after integration as  operating on φ, followed by integration. To prove that a quantum mechanical operator  is Hermitian, consider the eigenvalue equation and its complex conjugate. \[\hat {A} \psi = a \psi \label {4-38}\] \[\hat {A}^* \psi ^* = a^* \psi ^* = a \psi ^* \label {4-39}\] Note that a* = a because the eigenvalue is real. Multiply Equations \ref{4-38} and \ref{4-39} from the left by ψ* and ψ, respectively, and integrate over all the coordinates. Note that ψ is normalized. The results are \[ \int \psi ^* \hat {A} \psi d\tau = a \int \psi ^* \psi d\tau = a \label {4-40}\] \[ \int \psi \hat {A}^* \psi ^* d \tau = a \int \psi \psi ^* d\tau = a \label {4-41}\] Since both integrals equal a, they must be equivalent. \[ \int \psi ^* \hat {A} \psi d\tau = \int \psi \hat {A}^* \psi ^* d\tau \label {4-42}\] The operator acting on the function, \(\hat {A}^* \int \psi ^* \hat {A} \psi d\tau = \int \psi \hat {A} ^* \psi ^* d\tau_* \), produces a new function. Since functions commute, Equation \ref{4-42} can be rewritten as \[ \int \psi ^* \hat {A} \psi d\tau = \int (\hat {A}^*\psi ^*) \psi d\tau \label{4-43}\] This equality means that  is Hermitian. Eigenfunctions of a Hermitian operator are orthogonal if they have different eigenvalues. Because of this theorem, we can identify orthogonal functions easily without having to integrate or conduct an analysis based on symmetry or other considerations. ψ and φ are two eigenfunctions of the operator  with real eigenvalues \(a_1\) and \(a_2\), respectively. Since the eigenvalues are real, \(a_1^* = a_1\) and \(a_2^* = a_2\). \[\hat {A} \psi = a_1 \psi \] \[\hat {A}^* \psi ^* = a_2 \psi ^* \label {4-44}\] Multiply the first equation by φ* and the second by ψ and integrate. \[\int \psi ^* \hat {A} \psi d\tau = a_1 \int \psi ^* \psi d\tau \] \[\int \psi \hat {A}^* \psi ^* d\tau = a_2 \int \psi \psi ^* d\tau \label {4-45}\] Subtract the two equations in (4-45)to obtain \[\int \psi ^*\hat {A} \psi d\tau - \int \psi \hat {A} ^* \psi ^* d\tau = (a_1 - a_2) \int \psi ^* \psi d\tau \label {4-46}\] The left-hand side of (4-46) is zero because  is Hermitian yielding \[ 0 = (a_1 - a_2 ) \int \psi ^* \psi d\tau \label {4-47}\] If a1 and a2 in (4-47) are not equal, then the integral must be zero. This result proves that nondegenerate eigenfunctions of the same operator are orthogonal. Draw graphs and use them to show that the particle-in-a-box wavefunctions for n = 2 and n = 3 are orthogonal to each other. If the eigenvalues of two eigenfunctions are the same, then the functions are said to be degenerate, and linear combinations of the degenerate functions can be formed that will be orthogonal to each other. Since the two eigenfunctions have the same eigenvalues, the linear combination also will be an eigenfunction with the same eigenvalue. Degenerate eigenfunctions are not automatically orthogonal but can be made so mathematically. The proof of this theorem shows us one way to produce orthogonal degenerate functions. If ψ and φ are degenerate but not orthogonal, define \(Φ = φ - Sψ\) where \(S\) is the overlap integral \(\int \psi ^* \psi d\tau \), then ψ and Φ will be orthogonal. \[\int \psi ^* \phi d\tau = \int \psi ^* (\varphi - S\psi ) d\tau = \int \psi ^* \psi d\tau - S \int \psi ^*\psi d\tau \label {4-48}\] \[= S - S = 0\] Find \(N\) that normalizes Φ if \(Φ = N(φ − Sψ)\) where ψ and φ are normalized and S is their overlap integral. If two operators commute, then they can have the same set of eigenfunctions. By definition, two operators \(\hat {A}\) and \(\hat {B}\)commute if the effect of applying \(\hat {A}\) then \(\hat {B}\) is the same as applying \(\hat {B}\) then \(\hat {A}\), i.e. \(\hat {A}\hat {B} = \hat {B} \hat {A}\). For example, the operations brushing-your-teeth and combing-your-hair commute, while the operations getting-dressed and taking-a-shower do not. This theorem is very important. If two operators commute and consequently have the same set of eigenfunctions, then the corresponding physical quantities can be evaluated or measured exactly simultaneously with no limit on the uncertainty. As mentioned previously, the eigenvalues of the operators correspond to the measured values. If \(\hat {A}\) and \(\hat {B}\) commute and \(ψ\) is an eigenfunction of \(\hat {A}\) with eigenvalue \(b\), then \[\hat {B} \hat {A} \psi = \hat {A} \hat {B} \psi = \hat {A} b \psi = b \hat {A} \psi \label {4-49}\] Equation (4-49) says that \(\hat {A} \psi \) is an eigenfunction of \(\hat {B}\) with eigenvalue b, which means that when \(\hat {A}\) operates on ψ, it cannot change ψ. At most, \(\hat {A}\) operating on \(ψ\) can produce a constant times ψ. \[\hat {A} \psi = a \psi \label {4-50}\] \[\hat {B} (\hat {A} \psi ) = \hat {B} (a \psi ) = a \hat {B} \psi = ab\psi = b (a \psi ) \label {4-51}\] Equation \ref{4-51} shows that Equation \ref{4-50} is consistent with Equation \ref{4-49}. Consequently ψ also is an eigenfunction of \(\hat {A}\) with eigenvalue \(a\). Write definitions of the terms orthogonal and commutation. Show that the operators for momentum in the x-direction and momentum in the y-direction commute, but operators for momentum and position along the x-axis do not commute. Since differential operators are involved, you need to show whether \[\hat {P} _x \hat {P} _y f (x,y) = \hat {P} _y \hat {P} _x f (x, y)\] \[\hat {P} _x \hat {x} f(x) = \hat {x} \hat {P} _x f(x) \] where f is an arbitrary function, or you could try a specific form for f, e.g. f = 6xy. lthough it will not be proven here, there is a general statement of the uncertainty principle in terms of the commutation property of operators. If two operators \(\hat {A}\) and \(\hat {B}\) do not commute, then the uncertainties (standard deviations σ) in the physical quantities associated with these operators must satisfy \[\sigma _A \sigma _B \ge | \int \psi ^* [ \hat {A} \hat {B} - \hat {B} \hat {A} ] \psi d\tau \label {4-52}\] where the integral inside the square brackets is called the commutator, and ││signifies the modulus or absolute value. If \(\hat {A}\) and \(\hat {B}\) commute, then the right-hand-side of equation (4-52) is zero, so either or both σA and σ B could be zero, and there is no restriction on the uncertainties in the measurements of the eigenvalues a and b. If \(\hat {A}\) and \(\hat {B}\) do not commute, then the right-hand-side of equation (4-52) will not be zero, and neither σA nor σB can be zero unless the other is infinite. Consequently, both a and b cannot be eigenvalues of the same wavefunctions and cannot be measured simultaneously to arbitrary precision. Show that the commutator for position and momentum in one dimension equals –i ħ and that the right-hand-side of Equation (4-52) therefore equals ħ/2 giving \(\sigma _x \sigma _{px} \ge \frac {\hbar}{2}\) In a later chapter you will learn that the operators for the three components of angular momentum along the three directions in space (x, y, z) do not commute. What is the relevance of this mathematical property to measurements of angular momentum in atoms and molecules? Write the definition of a Hermitian operator and statements of the Orthogonality Theorem, the Schmidt Orthogonalization Theorem, and the Commuting Operator Theorem. Reconstruct proofs for the Orthogonality Theorem, the Schmidt Orthogonalization Theorem, and the Commuting Operator Theorem. Write a paragraph summarizing the connection between the commutation property of operators and the uncertainty principle.

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Although you have been introduced to some of the interactions that hold molecules together in a liquid, we have not yet discussed the consequences of those interactions for the bulk properties of liquids. We now turn our attention to four unique properties of liquids that intimately depend on the nature of intermolecular interactions: If liquids tend to adopt the shapes of their containers, then why do small amounts of water on a freshly waxed car form raised droplets instead of forming a thin, continuous film? The answer lies in a property called , which depends on intermolecular forces. Surface tension is the energy required to increase the surface area of a liquid by a unit amount and varies greatly from liquid to liquid based on the nature of the intermolecular forces, e.g., water with hydrogen bonds has a surface tension of \(7.29 \times 10^{-2} J/m^2\) (at 20°C), while mercury with metallic (electrostatic) bonds has as surface tension that is 6.5-times greater: \(4.86 \times 10^{-1} J/m^2\) (at 20°C). Figure \(\Page {1}\) presents a microscopic view of a liquid droplet. A typical molecule in the of the droplet is surrounded by other molecules that exert attractive forces from all directions. Consequently, there is no force on the molecule that would cause it to move in a particular direction. In contrast, a molecule on the experiences a net attraction toward the drop because there are no molecules on the outside to balance the forces exerted by adjacent molecules in the interior. Because a sphere has the smallest possible surface area for a given volume, intermolecular attractive interactions between water molecules cause the droplet to adopt a spherical shape. This maximizes the number of attractive interactions and minimizes the number of water molecules at the surface. Hence raindrops are almost spherical, and drops of water on a waxed (nonpolar) surface, which does not interact strongly with water, form round beads. A dirty car is covered with a mixture of substances, some of which are polar. Attractive interactions between the polar substances and water cause the water to spread out into a thin film instead of forming beads. The same phenomenon holds molecules together at the surface of a bulk sample of water, almost as if they formed a skin. When filling a glass with water, the glass can be overfilled so that the level of the liquid actually extends the rim. Similarly, a sewing needle or a paper clip can be placed on the surface of a glass of water where it “floats,” even though steel is much denser than water. Many insects take advantage of this property to walk on the surface of puddles or ponds without sinking. This is better demonstrated in the zero-gravity conditions of space (Figure \(\Page {2}\)). Such phenomena are manifestations of , which is defined as the energy required to increase the surface area of a liquid by a specific amount. Surface tension is therefore measured as energy per unit area, such as joules per square meter (J/m ). The values of the surface tension of some representative liquids are listed in Table \(\Page {2}\). Note the correlation between the surface tension of a liquid and the strength of the intermolecular forces: the stronger the intermolecular forces, the higher the surface tension. For example, water, with its strong intermolecular hydrogen bonding, has one of the highest surface tension values of any liquid, whereas low-boiling-point organic molecules, which have relatively weak intermolecular forces, have much lower surface tensions. Mercury is an apparent anomaly, but its very high surface tension is due to the presence of strong metallic bonding. Adding soaps and detergents that disrupt the intermolecular attractions between adjacent water molecules can reduce the surface tension of water. Because they affect the surface properties of a liquid, soaps and detergents are called surface-active agents, or . In the 1960s, US Navy researchers developed a method of fighting fires aboard aircraft carriers using “foams,” which are aqueous solutions of fluorinated surfactants. The surfactants reduce the surface tension of water below that of fuel, so the fluorinated solution is able to spread across the burning surface and extinguish the fire. Such foams are now used universally to fight large-scale fires of organic liquids. Any material - solid, liquid or (non-ideal) gas - wants to bond to itself. This is why condensed phase materials "sticks" together in the first place. However, a surface disrupts this bonding, and so incurs an energy penalty. This is why liquids in zero gravity ball up into spherical drops ( the sphere is the shape with the lowest surface area for a fixed volume. We can describe this with dimensions of energy per unit area, which is the amount of extra energy needed to create new surface or extend a surface (e.g., cracking a solid or parting a liquid). Hence, surface tension is typically given in J/m units (Table \(\Page {2}\)). Intermolecular forces also cause a phenomenon called capillary action, which is the tendency of a polar liquid to rise against gravity into a small-diameter tube (a ), as shown in Figure \(\Page {3}\). When a glass capillary is put into a dish of water, water is drawn up into the tube. The height to which the water rises depends on the diameter of the tube and the temperature of the water but on the angle at which the tube enters the water. The smaller the diameter, the higher the liquid rises. The height of the water does depend on the angle at which the capillary is tilted. Capillary action is the net result of two opposing sets of forces: , which are the intermolecular forces that hold a liquid together, and , which are the attractive forces between a liquid and the substance that composes the capillary. Water has both strong adhesion to glass, which contains polar \(\ce{SiOH}\) groups, and strong intermolecular cohesion. When a glass capillary is put into water, the surface tension due to cohesive forces constricts the surface area of water within the tube, while adhesion between the water and the glass creates an upward force that maximizes the amount of glass surface in contact with the water. If the adhesive forces are stronger than the cohesive forces, as is the case for water, then the liquid in the capillary rises to the level where the downward force of gravity exactly balances this upward force. If, however, the cohesive forces are stronger than the adhesive forces, as is the case for mercury and glass, the liquid pulls itself down into the capillary below the surface of the bulk liquid to minimize contact with the glass (part (a) in Figure \(\Page {4}\)). The upper surface of a liquid in a tube is called the , and the shape of the meniscus depends on the relative strengths of the cohesive and adhesive forces. In liquids such as water, the meniscus is concave; in liquids such as mercury, however, which have very strong cohesive forces and weak adhesion to glass, the meniscus is convex (Figure \(\Page {4b}\)). Fluids and nutrients are transported up the stems of plants or the trunks of trees by capillary action. Plants contain tiny rigid tubes composed of cellulose, to which water has strong adhesion. Because of the strong adhesive forces, nutrients can be transported from the roots to the tops of trees that are more than 50 m tall. Cotton towels are also made of cellulose; they absorb water because the tiny tubes act like capillaries and “wick” the water away from your skin. The moisture is absorbed by the entire fabric, not just the layer in contact with your body. Polar substances are drawn up a glass capillary and generally have concave meniscuses and nonpolar substances general avoid the capillary and exhibit convex meniscuses. is the resistance of a liquid to flow. Some liquids, such as gasoline, ethanol, and water, flow very readily and hence have a . Others, such as motor oil, molasses, and maple syrup, flow very slowly and have a . The two most common methods for evaluating the viscosity of a liquid are The higher the viscosity, the slower the liquid flows through the tube and the steel balls fall. Viscosity is expressed in units of the poise (mPa•s); the higher the number, the higher the viscosity. The viscosities of some representative liquids are listed in Table \(\Page {1}\) and show a correlation between viscosity and intermolecular forces. Because a liquid can flow only if the molecules can move past one another with minimal resistance, strong intermolecular attractive forces make it more difficult for molecules to move with respect to one another. The addition of a second hydroxyl group to ethanol, for example, which produces ethylene glycol (HOCH CH OH), increases the viscosity 15-fold. This effect is due to the increased number of hydrogen bonds that can form between hydroxyl groups in adjacent molecules, resulting in dramatically stronger intermolecular attractive forces. There is also a correlation between viscosity and molecular shape. Liquids consisting of long, flexible molecules tend to have higher viscosities than those composed of more spherical or shorter-chain molecules. The longer the molecules, the easier it is for them to become “tangled” with one another, making it more difficult for them to move past one another. London dispersion forces also increase with chain length. Due to a combination of these two effects, long-chain hydrocarbons (such as motor oils) are highly viscous. Viscosity increases as intermolecular interactions or molecular size increases. Motor oils and other lubricants demonstrate the practical importance of controlling viscosity. The oil in an automobile engine must effectively lubricate under a wide range of conditions, from subzero starting temperatures to the 200°C that oil can reach in an engine in the heat of the Mojave Desert in August. Viscosity decreases rapidly with increasing temperatures because the kinetic energy of the molecules increases, and higher kinetic energy enables the molecules to overcome the attractive forces that prevent the liquid from flowing (Table \(\Page {3}\)). As a result, an oil that is thin enough to be a good lubricant in a cold engine will become too “thin” (have too low a viscosity) to be effective at high temperatures. The viscosity of motor oils is described by an SAE (Society of Automotive Engineers) rating ranging from SAE 5 to SAE 50 for engine oils: the lower the number, the lower the viscosity. So-called can cause major problems. If they are viscous enough to work at high operating temperatures (SAE 50, for example), then at low temperatures, they can be so viscous that a car is difficult to start or an engine is not properly lubricated. Consequently, most modern oils are , with designations such as SAE 20W/50 (a grade used in high-performance sports cars), in which case the oil has the viscosity of an SAE 20 oil at subzero temperatures (hence the W for winter) and the viscosity of an SAE 50 oil at high temperatures. These properties are achieved by a careful blend of additives that modulate the intermolecular interactions in the oil, thereby controlling the temperature dependence of the viscosity. Many of the commercially available oil additives “for improved engine performance” are highly viscous materials that increase the viscosity and effective SAE rating of the oil, but overusing these additives can cause the same problems experienced with highly viscous single-grade oils. Based on the nature and strength of the intermolecular cohesive forces and the probable nature of the liquid–glass adhesive forces, predict what will happen when a glass capillary is put into a beaker of SAE 20 motor oil. Will the oil be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)? (Hint: the surface of glass is lined with Si–OH groups.) substance and composition of the glass surface behavior of oil and the shape of meniscus Motor oil is a nonpolar liquid consisting largely of hydrocarbon chains. The cohesive forces responsible for its high boiling point are almost solely London dispersion forces between the hydrocarbon chains. Such a liquid cannot form strong interactions with the polar Si–OH groups of glass, so the surface of the oil inside the capillary will be lower than the level of the liquid in the beaker. The oil will have a convex meniscus similar to that of mercury. Predict what will happen when a glass capillary is put into a beaker of ethylene glycol. Will the ethylene glycol be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)? Capillary action will pull the ethylene glycol up into the capillary. The meniscus will be concave. Superfluid helium-4 is the superfluid form of helium-4, an isotope of the element helium. A superfluid is a state of matter in which the matter behaves like a fluid with zero viscosity. The substance, which looks like a normal liquid, flows without friction past any surface, which allows it to continue to circulate over obstructions and through pores in containers which hold it, subject only to its own inertia. Many ordinary fluids, like alcohol or petroleum, creep up solid walls, driven by their surface tension. However, in the case of superfluid helium-4, the flow of the liquid in the layer is not restricted by its viscosity but by a critical velocity which is about 20 cm/s. This is a fairly high velocity so superfluid helium can flow relatively easily up the wall of containers, over the top, and down to the same level as the surface of the liquid inside the container. In a container, lifted above the liquid level, it forms visible droplets as seen above. The of a liquid is defined as the pressure exerted by a vapor in equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system (discussed in more detail in next Sections of Chapter). As the temperature of a liquid increases, the vapor pressure of the liquid increases until it equals the external pressure, or the atmospheric pressure in the case of an open container. Bubbles of vapor begin to form throughout the liquid, and the liquid begins to boil. The temperature at which a liquid boils at exactly 1 atm pressure is the of the liquid. For water, the normal boiling point is exactly 100°C. The normal boiling points of the other liquids in Figure \(\Page {8}\) are represented by the points at which the vapor pressure curves cross the line corresponding to a pressure of 1 atm. Although we usually cite the normal boiling point of a liquid, the boiling point depends on the pressure. At a pressure greater than 1 atm, water boils at a temperature greater than 100°C because the increased pressure forces vapor molecules above the surface to condense. Hence the molecules must have greater kinetic energy to escape from the surface. Conversely, at pressures less than 1 atm, water boils below 100°C. Typical variations in atmospheric pressure at sea level are relatively small, causing only minor changes in the boiling point of water. For example, the highest recorded atmospheric pressure at sea level is 813 mmHg, recorded during a Siberian winter; the lowest sea-level pressure ever measured was 658 mmHg in a Pacific typhoon. At these pressures, the boiling point of water changes minimally, to 102°C and 96°C, respectively. At high altitudes, on the other hand, the dependence of the boiling point of water on pressure becomes significant. Table \(\Page {4}\) lists the boiling points of water at several locations with different altitudes. At an elevation of only 5000 ft, for example, the boiling point of water is already lower than the lowest ever recorded at sea level. The lower boiling point of water has major consequences for cooking everything from soft-boiled eggs (a “three-minute egg” may well take four or more minutes in the Rockies and even longer in the Himalayas) to cakes (cake mixes are often sold with separate high-altitude instructions). Conversely, pressure cookers, which have a seal that allows the pressure inside them to exceed 1 atm, are used to cook food more rapidly by raising the boiling point of water and thus the temperature at which the food is being cooked. As pressure , the boiling point of a liquid and vice versa. Use Figure \(\Page {8}\) to estimate the following. Mercury boils at 356 °C at room pressure. To see video go to www.youtube.com/watch?v=0iizsbXWYoo data in Figure \(\Page {8}\), pressure, and boiling point corresponding boiling point and pressure Ethylene glycol is an organic compound primarily used as a raw material in the manufacture of polyester fibers and fabric industry, and polyethylene terephthalate resins (PET) used in bottling. Use the data in Figure \(\Page {8}\) to estimate the following. 200°C 450 mmHg ​​​​​​ Surface tension, capillary action, boiling points, and viscosity are unique properties of liquids that depend on the nature of intermolecular interactions. is the energy required to increase the surface area of a liquid by a given amount. The stronger the intermolecular interactions, the greater the surface tension. are molecules, such as soaps and detergents, that reduce the surface tension of polar liquids like water. is the phenomenon in which liquids rise up into a narrow tube called a capillary. It results when , the intermolecular forces in the liquid, are weaker than , the attraction between a liquid and the surface of the capillary. The shape of the , the upper surface of a liquid in a tube, also reflects the balance between adhesive and cohesive forces. The of a liquid is its resistance to flow. Liquids that have strong intermolecular forces tend to have high viscosities. The of a liquid is the temperature when the of the liquid equals the external pressure, or the atmospheric pressure in the case of an open container. ( )

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There are numerous reactions in organic chemistry that proceed through cyclic transition states. They may be classified generally as reactions. An important and familiar example is the Diels-Alder reaction, in which a conjugated diene cycloadds to an alkene or alkyne: This reaction has been described previously ( ) and is an example of a [4 + 2] cycloaddition. Such reactions occur thermally (by simply heating the reactants) and appear to be entirely concerted. By this we mean that the reactants are converted to products , without involving the formation of reaction intermediates. The principal evidence for the concertedness of [4 + 2] cycloadditions is the fact that they are highly stereospecific and involve suprafacial addition of both components. The configuration of substituents in the diene and the dienophile is retained in the adduct: In contrast to the [4 + 2] cycloaddition, thermal [2 + 2] cycloadditions seldom are observed, and when they are observed, they are not stereospecific and evidently are stepwise reactions (see ): Why are the [4 + 2] and [2 + 2] cycloadditions different? Simple molecular orbital theory provides an elegant explanation of this difference based on the \(4n + 2\) rule described in . To understand this, we need to look in more detail at how the \(p\) orbitals of the double bonds interact in concerted addition mechanisms by suprafacial overlap, as in \(36\) and \(37\): There is a way around the \(4n + 2\) rule that is not very important for substances analogous to benzene, but is quite important for cycloaddition reactions. Let us see how this works for a cyclic conjugated polyene. From the molecular-orbital diagrams of Figures 21-5, 21-7, 21-9, and 21-14, you will see that the -energy \(\pi\) molecular orbital has nodes (changes of phase). A model of such an orbital, which usually is called a , can be constructed by joining the ends of a ribbon or strip of parallel \(p\) orbitals, as represented on the left side of Figure 21-15. However, one could join the orbitals by making in the strip, which then would give a lowest-energy orbital with one node, as on the right side of Figure 21-15. A strip with one such twist is called a \(^8\) and has the topological property of having only one side. If we now calculate the orbital energies for the Möbius orbitals, as was down for the normal Hückel \(\pi\) orbitals in Figure 21-13, we get the results shown in Figure 21-16. From this, we see that the \(4n\) situation now is favored and \(4n + 2\) is unfavorable. Whereas the energies of the \(\pi\) molecular orbitals in the Hückel arrangement can be obtained by inscribing a polygon in a circle with a ( ), in the Möbius arrangement the orbital energies are obtained from the polygon inscribed with a . If you compare the orbital energies of the Hückel and Möbius cyclic \(\pi\) systems (Figures 21-13 and 21-16), you will see that the Hückel systems have only lowest-energy MO, whereas the Möbius systems have . Hückel systems have an odd number of bonding orbitals (which, when full, accommodate 2, 6, 10, 14, or \(4n + 2\) electrons) and the Möbius systems have an even number of bonding orbitals (which, when full, accommodate 4, 8, 12, or \(4n\) electrons). The Hückel molecular orbitals have or an of nodes (see, for example, the benzene MOs, Figure 21-5); the Möbius molecular orbitals are not shown, but they have or an of nodes. The relevance of all this may seem tenuous, especially because no example of a simple cyclic polyene with a Möbius \(\pi\) system is known. However, the Möbius arrangement is relevant to cycloaddition because we can conceive of alkenes, alkadienes, and so on approaching each other to produce Möbius transition states when \(4n\) electrons are involved. For example, consider two molecules of ethene, which we showed previously would violate the \(4n + 2\) rule by undergoing cycloaddition through a transition state represented by \(37\). There is an alternative transition state, \(38\), in which the four \(p\) orbitals come together in the Möbius arrangement (with one node for minimum energy). To achieve this arrangement the ethene molecules approach each other in roughly perpendicular planes so that the \(p\) orbitals overlap suprafacially in one ethene and antarafacially in the other, as shown in \(38\): This pathway is electronically favorable, but the steric interference between the groups attached to the double bond is likely to be severe. Such repulsions can be relieved if there are no groups sticking out sidewise at one end of the double bond, as with the central carbon of 1,2-propadiene, \(\ce{CH_2=C=CH_2}\), and ketene, \(\ce{CH_2=C=O}\). These substances often undergo [2 + 2] cycloadditions rather readily ( ), and it is likely that these are concerted additions occurring by the Möbius route. A much less strained Möbius [4 + 4] transition state can be formed from two molecules of 1,3-butadiene. When 1,3-butadiene is heated by itself, a few percent of 1,5-cyclooctadiene is formed, but it is not known for sure whether the mechanism is that shown: The principal reaction is a Diels-Alder [4 + 2] cycloaddition, with butadiene acting both as a diene and as a dienophile: Much of what we have said about the electronic factors controlling whether a cycloaddition reaction can be concerted or not originally was formulated by the American chemists R. B. Woodward and R. Hoffmann several years ago, in terms of what came to be called the principles, or the . Orbital symmetry arguments are too complicated for this book, and we shall, instead, use the \(4n + 2\) electron rule for normal Hückel arrangements of \(\pi\) systems and the \(4n\) electron rule for Möbius arrangements. This is a particularly simple approach among several available to account for the phenomena to which Woodward and Hoffmann drew special attention and explained by what they call "conservation of orbital symmetry". The cycloaddition reactions that we have discussed so far in this chapter ([2 + 2], [4 + 2], etc.) have involved ring formation by bringing two unsaturated molecules together. Thus [4 + 2] addition is represented by the Diels-Alder reaction of ethene and 1,3-butadiene: We can conceive of similar cyclizations involving only single molecules, that is, . Such reactions are called . Two examples follow to show cyclization of a diene and a triene: Cyclization of 1,3,5-hexatriene occurs only when the central double bond has the cis configuration. The reaction is reversible at elevated temperatures because of the gain in entropy on ring opening (see ). The cyclobutene-1,3-butadiene interconversion proceeds much less readily, even in the thermodynamically favorable direction of ring opening. However, substituted dienes and cyclobutenes often react more rapidly. A related group of reactions involves shifts of substituent groups from one atom to another; for example, with \(\ce{H}\), alkyl, or aryl groups as \(\ce{R}\): These reactions are called and, in general, they are subject to the \(4n + 2\) rule and the Möbius orbital modification of it. Potential sigmatropic rearrangements can be recognized by the fact that the single bond to the migrating group \(\left( \ce{R} \right)\) is "conjugated" with the \(\pi\) bonds, and the group moves from a saturated \(sp^3\) to an \(sp^2\) carbon at a different part of the \(\pi\) system. A striking feature of thermal electrocyclic reactions that proceed by concerted mechanisms is their high degree of stereospecificity. Thus when -3,4-dimethylcyclobutene is heated, it affords only one of the three possible cis-trans isomers of 2,4-hexadiene, namely, , -2,4-hexadiene: We can see how this can occur if, as the ring opens, the ends of the diene twist in the direction (\(\curvearrowright \curvearrowright\) or \(\curvearrowleft \curvearrowleft\), ) as indicated in the equation. You will notice that with this particular case, if conrotation occurs to the left, rather than the right, the same final product results: The conrotatory movement of groups is typical of thermal ring openings of cyclobutenes and other rings involving \(4n\) electrons. When a cyclobutene is so constituted that conrotation cannot occur for steric reasons, then the concerted reaction cannot occur easily. Substances that otherwise might be predicted to be highly unstable often turn out to be relatively stable. An example is bicyclo[2.1.0]-2-pentene, which at first sight might seem incapable of isolation because of the possibility of immediate arrangement to 1,3-cyclopentadiene. This rearrangement does occur, but not so fast as to preclude isolation of the substance: How can we explain the fact that this substance can be isolated? The explanation is that, if the reaction has to be conrotatory, then the product will not be ordinary 1,3-cyclopentadiene, but , -1,3-cyclopentadiene - surely a very highly strained substance. (Try to make a ball-and-stick model of it!) This means that the concerted mechanism is not favorable: It is of great interest and importance that, with systems of \(4n + 2\) electrons, the groups move in directions (\(\curvearrowleft \curvearrowright\) or \(\curvearrowright \curvearrowleft\), ). For example, In this case, the disrotation of the groups one another would lead to the cis,cis,cis product. Because this product is not formed, it seems likely that rotation of the methyl groups toward each other must be sterically unfavorable: How can we account for the stereoselectivity of thermal electrocyclic reactions? Our problem is to understand why it is that concerted \(4n\) electrocyclic rearrangements are conrotatory, whereas the corresponding \(4n + 2\) processes are disrotatory. From what has been said previously, we can expect that the conrotatory processes are related to the Möbius molecular orbitals and the disrotatory processes are related to Hückel molecular orbitals. Let us see why this is so. Consider the electrocyclic interconversion of a 1,3-diene and a cyclobutene. In this case, the Hückel transition state ( ) is formed by , but is unfavorable with four (that is, \(4n\)) electrons: In contrast, the Möbius transition state ( ) is formed by and is favorable with four \(\left( 4n \right)\) electrons: You will notice that the ring closure of a 1,3-diene through the favorable Möbius transition state may appear to be able to form only an arrangement of the overlapping \(\sigma\) orbitals, which would correspond to a high-energy cyclobutene. In fact, the normal cyclobutene would be formed, because on the way down from the transition state, the phases of the orbitals that will become the \(\sigma\) bond change to give the arrangement of the \(\sigma\) orbitals expected for the ground state. The reverse occurs in ring opening so that this reaction also can go through the favorable Möbius transition state. The same reasoning can be extended to electrocyclic reactions of 1,3,5-trienes and 1,3-cyclohexadienes, which involve \(4n + 2\) electrons and consequently favor Hückel transition states attained by . The three principal types of pericyclic reactions are , , and : The factors that control if and how these cyclization and rearrangement reactions occur in a concerted manner can be understood from the aromaticity or lack of aromaticity achieved in their transition states. For a concerted pericyclic reaction to be thermally favorable, the transition state must involve \(4n + 2\) participating electrons if it is a Hückel orbital system, or \(4n\) electrons if it is a Möbius orbital system. A Hückel transition state is one in which the cyclic array of participating orbitals has no nodes (or an even number) and a Möbius transition state has an odd number of nodes. We summarize here a procedure to predict the feasibility and the stereochemistry of reactions involving . The 1,2 rearrangement of carbocations will be used to illustrate the approach. This is a very important reaction of carbocations which we have discussed in other chapters. We use it here as an example to illustrate how qualitative MO theory can give insight into how and why reactions occur: The first step of the procedure is to draw the orbitals as they are expected to be involved in the transition state. There may be several possible arrangements. There are two such arrangements, \(41\) and \(42\), for the rearrangement of carbocations; the dotted lines show the regions of bond-making and bond-breaking (i.e., orbital overlap): The second step is to determine whether the transition states are Hückel or Möbius from the number of nodes. This is readily done by assigning signs to the lobes of the orbitals corresponding to their phases and counting the number of nodes that develop in the circle of overlapping orbitals. An odd number denotes a Möbius transition state, whereas an even number, including zero, denotes a Hückel transition state. There are alternative ways of node-counting for transition states \(41\) and \(42\). Diagrams \(43abc\) and \(44abc\) represent molecular orbitals of different energies - those with more nodes having the higher energies (cf. ).\(^9\) We show these diagrams with more than one node for the sake of completeness. It is not necessary to draw more than one such diagram to determine whether the transition state is Möbius or Hückel. Finally, we evaluate the transition states according to the \(4n\) or \(4n + 2\) rule. In the example here, because only two electrons occupy the molecular orbitals, the Hückel transition state (\(43a\)) is the favorable one. A bonus coming from these formulations is that the stereochemistry of the reaction can be predicted when we have predicted which transition state is the favored one. Thus the migrating group in 1,2-carbocation rearrangements should move with of configuration by a Hückel transition state - and this has been verified experimentally. The alternative Möbius transition state predicts of the configuration of the migrating group: You can use the procedures just outlined to determine whether any thermal reaction with a cyclic transition state is likely to be favorable. A good place to start is the Diels-Alder [4 + 2] cycloaddition, which proceeds thermally by a suprafacial (Hückel) transition state. We suggest that you apply the procedure to the Diels-Alder reaction of 1,3-butadiene and ethene, and following that, show the electrocyclic ring opening of a cyclobutene ring to be thermally favorable only by a conrotatory opening of the \(\ce{C-C}\) bond. Many pericyclic reactions take place photochemically, that is, by irradiation with ultraviolet light. One example is the conversion of norbornadiene to quadricyclene, described in . This reaction would have an unfavorable [2 + 2] mechanism if it were attempted by simple heating. Furthermore, the thermodynamics favor ring opening rather than ring closure. However, quadricyclene can be isolated, even if it is highly strained, because to reopen the ring thermally involves the reverse of some unfavorable [2 + 2] cycloaddition mechanism. Photochemical activation can be used to achieve forward or reverse cycloadditions and electrocyclic reactions that are thermodynamically unfavorable or have unfavorable concerted thermal mechanisms. Thus the thermodynamically unstable disrotatory [2 + 2] product can be obtained from 1,3-cyclopentadiene by irradiation with ultraviolet light: The stereochemical results of electrocyclic and cycloaddition reactions carried out photochemically often are opposite to what is observed for corresponding thermal reactions. However, exceptions are known and the degree of stereospecificity is not always as high as in the thermal reactions. Further examples of photochemical pericyclic reactions are given in . \(^8\)Named after the mathematician A. F. Möbius. \(^9\)The assignment of orbital phases must take appropriate account of molecular symmetry, and although this is easy for open-chain systems, it is much less straightforward for cyclic ones. You usually will be able to avoid this problem by always trying to set up the orbitals so that the transition state will have no nodes, or just one node at a point where a bond is being made or broken. and (1977)

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In this chapter, we have described various galvanic cells in which a spontaneous chemical reaction is used to generate electrical energy. In an electrolytic cell, however, the opposite process, called , occurs: an external voltage is applied to drive a nonspontaneous reaction. In this section, we look at how electrolytic cells are constructed and explore some of their many commercial applications. If we construct an electrochemical cell in which one electrode is copper metal immersed in a 1 M Cu solution and the other electrode is cadmium metal immersed in a \(\,1\; M\, Cd^{2+}\) solution and then close the circuit, the potential difference between the two compartments will be 0.74 V. The cadmium electrode will begin to dissolve (Cd is oxidized to Cd ) and is the anode, while metallic copper will be deposited on the copper electrode (Cu is reduced to Cu), which is the cathode (Figure \(\Page {1a}\)). The overall reaction is as follows: \[ \ce{Cd (s) + Cu^{2+} (aq) \rightarrow Cd^{2+} (aq) + Cu (s)} \nonumber \] with \(E°_{cell} = 0.74\; V\) This reaction is thermodynamically spontaneous as written (\(ΔG^o < 0\)): \[ \begin{align*} \Delta G^\circ &=-nFE^\circ_\textrm{cell} \\[4pt] &=-(\textrm{2 mol e}^-)[\mathrm{96,485\;J/(V\cdot mol)}](\mathrm{0.74\;V}) \\[4pt] &=-\textrm{140 kJ (per mole Cd)} \end{align*} \nonumber \] In this direction, the system is acting as a galvanic cell. In an electrolytic cell, an external voltage is applied to drive a reaction. The reverse reaction, the reduction of Cd by Cu, is thermodynamically nonspontaneous and will occur only with an input of 140 kJ. We can force the reaction to proceed in the reverse direction by applying an electrical potential greater than 0.74 V from an external power supply. The applied voltage forces electrons through the circuit in the reverse direction, converting a galvanic cell to an electrolytic cell. Thus the copper electrode is now the anode (Cu is oxidized), and the cadmium electrode is now the cathode (Cd is reduced) (Figure \(\Page {1b}\)). The signs of the cathode and the anode have switched to reflect the flow of electrons in the circuit. The half-reactions that occur at the cathode and the anode are as follows: \[\ce{Cd^{2+}(aq) + 2e^{−} \rightarrow Cd(s)}\label{20.9.3} \] with \(E^°_{cathode} = −0.40 \, V\) \[\ce{Cu(s) \rightarrow Cu^{2+}(aq) + 2e^{−}} \label{20.9.4} \] with \(E^°_{anode} = 0.34 \, V \) \[\ce{Cd^{2+}(aq) + Cu(s) \rightarrow Cd(s) + Cu^{2+}(aq) } \label{20.9.5} \] with \(E^°_{cell} = −0.74 \: V\) Because \(E^°_{cell} < 0\), the overall reaction—the reduction of \(Cd^{2+}\) by \(Cu\)—clearly occur spontaneously and proceeds only when sufficient electrical energy is applied. The differences between galvanic and electrolytic cells are summarized in Table \(\Page {1}\). At sufficiently high temperatures, ionic solids melt to form liquids that conduct electricity extremely well due to the high concentrations of ions. If two inert electrodes are inserted into molten \(\ce{NaCl}\), for example, and an electrical potential is applied, \(\ce{Cl^{-}}\) is oxidized at the anode, and \(\ce{Na^{+}}\) is reduced at the cathode. The overall reaction is as follows: \[\ce{ 2NaCl (l) \rightarrow 2Na(l) + Cl2(g)} \label{20.9.6} \] This is the reverse of the formation of \(\ce{NaCl}\) from its elements. The product of the reduction reaction is liquid sodium because the melting point of sodium metal is 97.8°C, well below that of \(\ce{NaCl}\) (801°C). Approximately 20,000 tons of sodium metal are produced commercially in the United States each year by the electrolysis of molten \(\ce{NaCl}\) in a Downs cell (Figure \(\Page {2}\)). In this specialized cell, \(\ce{CaCl2}\) (melting point = 772°C) is first added to the \(\ce{NaCl}\) to lower the melting point of the mixture to about 600°C, thereby lowering operating costs. Similarly, in the Hall–Heroult process used to produce aluminum commercially, a molten mixture of about 5% aluminum oxide (Al O ; melting point = 2054°C) and 95% cryolite (Na AlF ; melting point = 1012°C) is electrolyzed at about 1000°C, producing molten aluminum at the cathode and CO gas at the carbon anode. The overall reaction is as follows: \[\ce{2Al2O3(l) + 3C(s) -> 4Al(l) + 3CO2(g)} \label{20.9.7} \] Oxide ions react with oxidized carbon at the anode, producing CO (g). There are two important points to make about these two commercial processes and about the electrolysis of molten salts in general. In the Hall–Heroult process, C is oxidized instead of O or F because oxygen and fluorine are more electronegative than carbon, which means that C is a weaker oxidant than either O or F . Similarly, in the Downs cell, we might expect electrolysis of a NaCl/CaCl mixture to produce calcium rather than sodium because Na is slightly less electronegative than Ca (χ = 0.93 versus 1.00, respectively), making Na easier to oxidize and, conversely, Na more difficult to reduce. In fact, the reduction of Na to Na is the observed reaction. In cases where the electronegativities of two species are similar, other factors, such as the formation of complex ions, become important and may determine the outcome. If a molten mixture of MgCl and KBr is electrolyzed, what products will form at the cathode and the anode, respectively? identity of salts electrolysis products The possible reduction products are Mg and K, and the possible oxidation products are Cl and Br . Because Mg is more electronegative than K (χ = 1.31 versus 0.82), it is likely that Mg will be reduced rather than K. Because Cl is more electronegative than Br (3.16 versus 2.96), Cl is a stronger oxidant than Br . Electrolysis will therefore produce Br at the anode and Mg at the cathode. Predict the products if a molten mixture of AlBr and LiF is electrolyzed. Br and Al Electrolysis can also be used to drive the thermodynamically nonspontaneous decomposition of water into its constituent elements: H and O . However, because pure water is a very poor electrical conductor, a small amount of an ionic solute (such as H SO or Na SO ) must first be added to increase its electrical conductivity. Inserting inert electrodes into the solution and applying a voltage between them will result in the rapid evolution of bubbles of H and O (Figure \(\Page {3}\)). The reactions that occur are as follows: For a system that contains an electrolyte such as Na SO , which has a negligible effect on the ionization equilibrium of liquid water, the pH of the solution will be 7.00 and [H ] = [OH ] = 1.0 × 10 . Assuming that \(P_\mathrm{O_2}\) = \(P_\mathrm{H_2}\) = 1 atm, we can use the standard potentials to calculate E for the overall reaction: \[\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log(P_\mathrm{O_2}P^2_\mathrm{H_2}) \\ &=-\textrm{1.23 V}-\left(\dfrac{\textrm{0.0591 V}}{4}\right)\log(1)=-\textrm{1.23 V}\end{align} \label{20.9.11} \] Thus E is −1.23 V, which is the value of E° if the reaction is carried out in the presence of 1 M H rather than at pH 7.0. In practice, a voltage about 0.4–0.6 V greater than the calculated value is needed to electrolyze water. This added voltage, called an , represents the additional driving force required to overcome barriers such as the large activation energy for the formation of a gas at a metal surface. Overvoltages are needed in all electrolytic processes, which explain why, for example, approximately 14 V must be applied to recharge the 12 V battery in your car. In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. The p-block metals and most of the transition metals are in this category, but metals in high oxidation states, which form oxoanions, cannot be reduced to the metal by simple electrolysis. Active metals, such as aluminum and those of groups 1 and 2, react so readily with water that they can be prepared only by the electrolysis of molten salts. Similarly, any nonmetallic element that does not readily oxidize water to O can be prepared by the electrolytic oxidation of an aqueous solution that contains an appropriate anion. In practice, among the nonmetals, only F cannot be prepared using this method. Oxoanions of nonmetals in their highest oxidation states, such as NO , SO , PO , are usually difficult to reduce electrochemically and usually behave like spectator ions that remain in solution during electrolysis. In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. In a process called , a layer of a second metal is deposited on the metal electrode that acts as the cathode during electrolysis. Electroplating is used to enhance the appearance of metal objects and protect them from corrosion. Examples of electroplating include the chromium layer found on many bathroom fixtures or (in earlier days) on the bumpers and hubcaps of cars, as well as the thin layer of precious metal that coats silver-plated dinnerware or jewelry. In all cases, the basic concept is the same. A schematic view of an apparatus for electroplating silverware and a photograph of a commercial electroplating cell are shown in Figure \(\Page {4}\). The half-reactions in electroplating a fork, for example, with silver are as follows: The overall reaction is the transfer of silver metal from one electrode (a silver bar acting as the anode) to another (a fork acting as the cathode). Because \(E^o_{cell} = 0\, V\), it takes only a small applied voltage to drive the electroplating process. In practice, various other substances may be added to the plating solution to control its electrical conductivity and regulate the concentration of free metal ions, thus ensuring a smooth, even coating. If we know the stoichiometry of an electrolysis reaction, the amount of current passed, and the length of time, we can calculate the amount of material consumed or produced in a reaction. Conversely, we can use stoichiometry to determine the combination of current and time needed to produce a given amount of material. The quantity of material that is oxidized or reduced at an electrode during an electrochemical reaction is determined by the stoichiometry of the reaction and the amount of charge that is transferred. For example, in the reaction \[\ce{Ag^{+}(aq) + e^{−} → Ag(s)} \nonumber \] 1 mol of electrons reduces 1 mol of \(\ce{Ag^{+}}\) to \(\ce{Ag}\) metal. In contrast, in the reaction \[\ce{Cu^{2+}(aq) + 2e^{−} → Cu(s)} \nonumber \] 1 mol of electrons reduces only 0.5 mol of \(\ce{Cu^{2+}}\) to \(\ce{Cu}\) metal. Recall that the charge on 1 mol of electrons is 1 faraday (1 F), which is equal to 96,485 C. We can therefore calculate the number of moles of electrons transferred when a known current is passed through a cell for a given period of time. The total charge (\(q\) in coulombs) transferred is the product of the current (\(I\) in amperes) and the time (\(t\), in seconds): \[ q = I \times t \label{20.9.14} \] The stoichiometry of the reaction and the total charge transferred enable us to calculate the amount of product formed during an electrolysis reaction or the amount of metal deposited in an electroplating process. For example, if a current of 0.60 A passes through an aqueous solution of \(\ce{CuSO4}\) for 6.0 min, the total number of coulombs of charge that passes through the cell is as follows: \[\begin{align*} q &= \textrm{(0.60 A)(6.0 min)(60 s/min)} \\[4pt] &=\mathrm{220\;A\cdot s} \\[4pt] &=\textrm{220 C} \end{align*} \nonumber \] The number of moles of electrons transferred to \(\ce{Cu^{2+}}\) is therefore \[\begin{align*} \textrm{moles e}^- &=\dfrac{\textrm{220 C}}{\textrm{96,485 C/mol}} \\[4pt] &=2.3\times10^{-3}\textrm{ mol e}^- \end{align*} \nonumber \] Because two electrons are required to reduce a single Cu ion, the total number of moles of Cu produced is half the number of moles of electrons transferred, or 1.2 × 10 mol. This corresponds to 76 mg of Cu. In commercial electrorefining processes, much higher currents (greater than or equal to 50,000 A) are used, corresponding to approximately 0.5 F/s, and reaction times are on the order of 3–4 weeks. A silver-plated spoon typically contains about 2.00 g of Ag. If 12.0 h are required to achieve the desired thickness of the Ag coating, what is the average current per spoon that must flow during the electroplating process, assuming an efficiency of 100%? mass of metal, time, and efficiency current required We must first determine the number of moles of Ag corresponding to 2.00 g of Ag: \(\textrm{moles Ag}=\dfrac{\textrm{2.00 g}}{\textrm{107.868 g/mol}}=1.85\times10^{-2}\textrm{ mol Ag}\) The reduction reaction is Ag (aq) + e → Ag(s), so 1 mol of electrons produces 1 mol of silver. Using the definition of the faraday, The current in amperes needed to deliver this amount of charge in 12.0 h is therefore \[\begin{align*}\textrm{amperes} &=\dfrac{1.78\times10^3\textrm{ C}}{(\textrm{12.0 h})(\textrm{60 min/h})(\textrm{60 s/min})}\\ & =4.12\times10^{-2}\textrm{ C/s}=4.12\times10^{-2}\textrm{ A}\end{align*} \nonumber \] Because the electroplating process is usually much less than 100% efficient (typical values are closer to 30%), the actual current necessary is greater than 0.1 A. A typical aluminum soft-drink can weighs about 29 g. How much time is needed to produce this amount of Al(s) in the Hall–Heroult process, using a current of 15 A to reduce a molten Al O /Na AlF mixture? 5.8 h Electroplating: In electrolysis, an external voltage is applied to drive a reaction. The quantity of material oxidized or reduced can be calculated from the stoichiometry of the reaction and the amount of charge transferred. Relationship of charge, current and time: \[ q = I \times t \nonumber \] In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. Electrolysis can also be used to produce H and O from water. In practice, an additional voltage, called an overvoltage, must be applied to overcome factors such as a large activation energy and a junction potential. Electroplating is the process by which a second metal is deposited on a metal surface, thereby enhancing an object’s appearance or providing protection from corrosion. The amount of material consumed or produced in a reaction can be calculated from the stoichiometry of an electrolysis reaction, the amount of current passed, and the duration of the electrolytic reaction.

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So far, we have come across one big rule of photon absorbance. In order to be absorbed, a photon's energy has to match an energy difference within the compound that is absorbing it. In the case of visible or ultraviolet light, the energy of a photon is roughly in the region that would be appropriate to promote an electron to a higher energy level. Different wavelengths would be able to promote different electrons, depending on the energy difference between an occupied electronic energy level and an unoccupied one. Other types of electromagnetic radiation would not be able to promote an electron, but they would be coupled to other events. For example, absorption of infrared light is tied to vibrational energy levels. Microwave radiation is tied to rotational energy levels in molecules. Thus, one reason a photon may or may not be absorbed has to do with whether its energy corresponds to the available energy differences within the molecule or ion that it encounters. Photons face other limitations. One of these is a moderate variation on our main rule. It is called the Frank Condon Principle. According to this idea, when an electron is excited from its normal position, the ground state, to a higher energy level, the optimal positions of atoms in the molecule may need to shift. Because electronic motion is much faster than nuclear motion, however, any shifting of atoms needed to optimize positions as they should be in the excited state will have to wait until after the electron gets excited. In that case, when the electron lands and the atoms aren't yet in their lowest energy positions for the excited state, the molecule will find itself in an excited vibrational state as well as an excited electronic state. That means the required energy for excitation doesn't just correspond to the difference in electronic energy levels; it is fine-tuned to reach a vibrational energy level, which is quantized as well. There are other restrictions on electronic excitation. Symmetry selection rules, for instance, state that the donor orbital (from which the electron comes) and the acceptor orbital (to which the electron is promoted) must have different symmetry. The reasons for this rule are based in the mathematics of quantum mechanics. What constitutes the same symmetry vs. different symmetry is a little more complicated than we will get into here. Briefly, let's just look at one "symmetry element" and compare how two orbitals might differ with respect to that element. If an orbital is centrosymmetric, one can imagine each point on the orbital reflecting through the very centre of the orbital to a point on the other side. At the end of the operation, the orbital appears unchanged. That means the orbital is symmetric with respect to a centre of inversion.. If we do the same thing with a sigma antibonding orbital, things turn out differently. In the drawing, the locations of the atoms are labelled A and B, but the symmetry of the orbital itself doesn't depend on that. If we imagine sending each point on this orbital through the very centre to the other side, we arrive at a picture that looks exactly the opposite of what we started with. These two orbitals have different symmetry. A transition from one to the other is allowed by symmetry. Decide whether each of the following orbitals is centrosymmetric. a) an s orbital b) a p orbital c) a d orbital d) a π orbital e) a π* orbital Decide whether each of the following transitions would be allowed by symmetry. a) π → π* b) p → π* c) p → σ* d) d → d Symmetry selection rules are in reality more like "strong suggestions." They depend on the symmetry of the molecule remaining strictly static, but all kinds of distortions occur through molecular vibrations. Nevertheless, these rules influence the likelihood of a given transition. The likelihood of a transition, similarly, has an influence upon the extinction coefficient, ε. Let's take a quick look at one last rule about electronic emissions. This rule concerns the spin of the excited electron, or more correctly, the "spin state" of the excited species. The spin state describes the number of unpaired electrons in the molecule or ion. The rule says that in an electronic transition, the spin state of the molecule must be preserved. That means if there are no unpaired electrons before the transition, then the excited species must also have no unpaired electrons. If there are two unpaired electrons before the transition, the excited state must also have two unpaired electrons. ,

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in a gas is the random motion of particles involved in the net movement of a substance from an area of high concentration to an area of low concentration. Each particle in a given gas continues to collide with other particles. In regions of the gas where the particle density is the highest, the particles bounce off each other and the boundary of their container at a greater rate than particles in less-dense regions. For a gas, the rate at which diffusion occurs is proportional to the square root of the density of the gas. The density of a gas is equal to the mass of the gas divided by the volume of the gas. If the volume is held constant one gas is compared with another with another, \(\dfrac{R_2}{R_1} = \sqrt{\dfrac{M_1}{M_2}}\) where R is the rate of diffusion in mol/s and M is the molar mass in g/mol. This is known as For a volume of solution that does not change: Two different particles colliding may be represented as a : \(A + B \rightarrow AB\) \[AB = K_d[A,B] \nonumber \] Notice the use of K to denote the . If circumstances change and either of the particles is able to diffuse out of the s , then the following 1st order reaction \(AB \rightarrow A + B\) is possible, then: \[AB = K_d'[AB] \nonumber \] There now exists a reaction for the formation of the AB complex as well as the breakdown of the AB complex into products. \[k = \dfrac{K_aK_d}{K_a + K_d'} \nonumber \] \(V = K[A,B]\) \[A + B \rightarrow AB \rightarrow Products \nonumber \] The net rate of formation for AB can now be determined: \(\dfrac{d[AB]}{dt} = (A+B \rightarrow AB) - (AB \rightarrow A + B) - (AB \rightarrow Products)\) \[\dfrac{d[AB]}{dt} = K_d[A,B] - K_d'[AB] - K_a[AB] \nonumber \] Assuming steady state conditions: \[\dfrac{d[AB]}{dt} = K_d[A,B] - K_d'[AB] - K_a[AB] = 0 \nonumber \] \([AB] = \dfrac{K_d[A,B]}{K_a + K_d'}\) The final rate of product formation taking into account both and : \(V = \dfrac{K_d[A,B]}{K_a + K_d'}\) If the rate at which particle A encounters particle B is much slower than the rate at which AB dissociates, then Kd' is essentially zero. \(V = \dfrac{K_d[A,B]}{K_a}\) The rate at which particle A encounters and reacts with particle B may exceed the rate at which the AB complex breaks apart into a product by a significant quantity. If the rate of at which AB decomposes is slow enough that K in the denominator may be ignored, the following results: \(V = \dfrac{K_d[A,B]}{K_d'}\) Viscosity and rate of diffusion may be related by the following formula: \(K_d = \dfrac{8RT}{3n}\) where n is the viscosity of the solution. 1. Using Graham's law of diffusion: (Rate /Rate ) = (Mass /Mass ) (RateF /RateCl ) = (70.9g/32g) = 2. Using Graham's law of diffusion: (Rate /Rate ) = (Mass /Mass ) (Rate /Rate ) = (Mass /Mass ) 0.75 = (32g/Mass ) 0.75 =(32g/Mass ) Mass = (32g/0.5625) 3. Using Fick's first law: J = -D(dc/dx) Where: J = unknown flux D = 0.29×10 cm /s dc = (C - C ) = 50mg/L - 290mg/L = -240mg/L, which is equivalent to -240mg/1000cm =-0.24mg/cm dx = 0.5cm J = (0.29×10 cm /s)[(-0.24mg/cm )/(0.5cm)] = 1.39×10 mg/s×cm 4. Using Fick's second law: T = x /2D Where: T = our unknown (time) x = 0.5 cm D = 1×10 cm /s T = (0.5cm) /[2(1×10 cm /s)] 5. Using Fick's second law: First, rearrange the equation T = x /2D to solve for D --> D = x /2T Where: D = our unknown (diffusivity constant) x = 0.01 cm (distance from the outside to the center of the cell) T = 5s D = (0.01cm) /[2(5s)]

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The boron family contains the semi-metal (B) and metals (Al), (Ga), (In), and (Tl). The boron family adopts oxidation states +3 or +1. The +3 oxidation states are favorable except for the heavier elements, such as Tl, which prefer the +1 oxidation state due to its stability; this is known as the . The elements generally follow except for certain Tl deviations: tends to forms hydrides, the simplest of which is diborane, \(B_2H_6\). Boron hydrides are used to synthesize organic compounds. One of the main compounds used to form other boron compounds is boric acid, which is a weak acid and is formed in the following two-step reaction: \[B_2O_{3 \;(s)} + 3 H_2O _{(l)} \rightarrow 2 B(OH)_{3 (aq)}\] \[B(OH)_{3 \;(aq)} + 2 H_2O_{(l)} \rightarrow H_3O^+_{(aq)} + B(OH)^-_{4\; (aq)}\] Boron can be crystallized from a solution of hydrogen peroxide and borax to produce sodium perborate, a bleach alternative. The bleaching ability of perborate is due to the two peroxo groups bound to the boron atoms. is an active metal with the electron configuration [Ne] 2s 2p , and usually adopts a +3 oxidation state. This element is the most abundant metal in the Earth's crust (7.5-8.4%). Even though it is very abundant, before 1886 aluminum was considered a semiprecious metal; it was difficult to isolate due to its high melting point. Aluminum is very expensive to produce, because the electrolysis of one mole of aluminum requires three moles of electrons: \[Al^{3+} + 3e^- \rightarrow Al(l)\] Aluminum can dissolve in both acids and bases—it is amphoteric. In an aqueous OH solution it produce Al(OH) , and in an aqueous H O solution it produce [Al(H O) ] . Another important feature of aluminum is that it is a good reducing agent due to its +3 oxidation state. It can therefore react with acids to reduce H (aq) to H (g). For example: \[2Al (s) + 6H^+(aq) \rightarrow 2Al^{3+}(aq) + 3H_2(g)\] Aluminum can also extract oxygen from any metal oxide. The following reaction, which is known as the thermite reaction, is very exothermic: \[Fe_2O_3(s) + 2 Al(s) \rightarrow Al_2O_3(s) +2 Fe(l)\] has the chemical symbol Ga and the atomic number 31. It has the electron configuration [Ar] 2s 2p and a +3 oxidation state. Gallium is industrially important because it forms gallium arsenide (GaAs), which converts light directly into electricity. Gallium is also used in conjunction with aluminum to generate hydrogen. In a process similar to the thermite reaction, aluminum extracts oxygen from water and releases hydrogen gas. However, as mentioned above, aluminum forms a protective coat in the presence of water. Combining gallium and aluminum prevents the formation of this protective layer, allowing aluminum to reduce water to hydrogen. has the electron configuration [Kr] 2s 2p1 and may adopt the +1 or +3 oxidation state; however, the +3 state is more common. Indium is soluble in acids, but does not react with oxygen at room temperature. It is obtained by separation from zinc ores. Indium is mainly used to make alloys, and only a small amount is required to enhance the metal strength. Thallium has the electron configuration [Xe] 2s 2p and has a +3 or +1 oxidation state. Because thallium is heavy, it has a greater stability in the +1 oxidation state ( ). Therefore, it is found more commonly in its +1 oxidation state. Thallium is soft and malleable. It is poisonous, but used in high-temperature superconductors. Both Be and Al are hydrated to produce [Be(H O) ] and Al(H O) , respectively. When reacted with water, both compounds produce hydronium ions, making them slightly acidic. Another similarity between aluminum and beryllium is that they are amphoteric, and their hydroxides are very basic. Both metals also react with oxygen to produce oxide coatings capable of protecting other metals from corrosion. Both metals also react with halides that can act as Lewis acids.

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Phosphorus (P) is an essential part of life as we know it. Without the phosphates in biological molecules such as ATP, ADP and DNA, we would not be alive. Phosphorus compounds can also be found in the minerals in our bones and teeth. It is a necessary part of our diet. In fact, we consume it in nearly all of the foods we eat. Phosphorus is quite reactive. This quality of the element makes it an ideal ingredient for matches because it is so flammable. Phosphorus is a vital element for plants and that is why we put phosphates in our fertilizer to help them maximize their growth. Phosphorus plays a big role in our existence but it can also be dangerous. When fertilizers containing phosphorus enter the water, it produces rapid algae growth. This can lead to eutrophication of lakes and rivers; i.e., the ecosystem has an increase of chemical nutrients and this can led to negative environmental effects. With all the excess phosphorus, plants grow rapidly then die, causing a lack of in the water and an overall reduction of water quality. It is thus necessary to remove excess phosphorus from our wastewater. The process of removing the phosphorus is done chemically by reacting the phosphorus with compounds such as ferric chloride, ferric sulfate, and aluminum sulfate or aluminum chlorohydrate. Phosphorus, when combined with aluminum or iron, becomes an insoluble salt. The of \(FePO_4\) and \(AlPO_4\) are 1.3x10 and 5.8x10 , respectively. With solubility's this low, the resulting precipitates can then be filtered out. Another example of the dangers of phosphorus is in the production of matches. The flammable nature and cheap manufacturing of white phosphorus made it possible to easily make matches around the turn of the 20th century. However, white phosphorus is highly toxic. Many workers in match factories developed brain damage and a disease called "phosphorus necrosis of the jaw" from exposure to toxic phosphorus vapors. Excess phosphorus accumulation caused their bone tissue to die and rot away. For this reason, we now use red phosphorus or phosphorus sesquisulfide in "safety" matches. Named from the Greek word phosphoros ("bringer of light"), elemental Phosphorus is not found in its elemental form because this form is quite reactive. Because of this factor it took a long period of time for it to be "discovered". The first recorded isolation of phosphorus was by alchemist Hennig Brand in 1669 involving about 60 pails of urine. After letting a large amount of urine putrefy for a long time, Brand distilled the liquid to a paste, heated the paste, discarded the salt formed, and put the remaining substance under cold water to form solid white phosphorus. Brand's process was not very efficient; the salt he discarded actually contained most of the phosphorus. Nevertheless, he obtained some pure, elemental phosphorus for his efforts. Others of the time improved the efficiency of the process by adding sand, but still continued to discard the salt. Later, phosphorus was manufactured from bone ash. Currently, the process for manufacturing phosphorus does not involve large amounts of putrefied urine or bone ash. Instead, manufacturers use calcium phosphate and coke (Emsley). Phosphorus is a nonmetal, solid at room temperature, and a poor conductor of heat and electricity. Phosphorus occurs in at least 10 allotropic forms, the most common (and reactive) of which is so-called white (or yellow) phosphorus which looks like a waxy solid or plastic. It is very reactive and will spontaneously inflame in air so it is stored under water. The other common form of phosphorus is red phosphorus which is much less reactive and is one of the components on the striking surface of a match book. Red phosphorus can be converted to white phosphorus by careful heating. White phosphorus consists of \(\ce{P4}\) molecules, whereas the crystal structure of red phosphorus has a complicated network of bonding. White phosphorus has to be stored in water to prevent natural combustion, but red phosphorus is stable in air. When burned, red phosphorus also forms the same oxides as those obtained in the burning of white phosphosrus, \(\ce{P4O6}\) when air supply is limited, and \(\ce{P4O10}\) when sufficient air is present. Diphosphorus (\(P_2\)) is the gaseous form of phosphorus that is thermodynamically stable above 1200 °C and until 2000 °C. It can be generated by heating white phosphorus (see below) to 1100 K and is very reactive with a (117 kcal/mol or 490 kJ/mol) half that of dinitrogen (\(N_2\)). White phosphorus (P ) has a tetrahedral structure. It is soft and waxy, but insoluble in water. Its glow occurs as a result of its vapors slowly being oxidized by the air. It is so thermodynamically unstable that it combusts in air. It was once used in fireworks and the U.S. military still uses it in incendiary bombs. This link shows various experiments with white phosphorus, which help show the physical and chemical properties of it. It also shows white phosphorus combusting with air. Red Phosphorus has more atoms linked together in a network than white phosphorus does, which makes it much more stable. It is not quite as flammable, but given enough energy it still reacts with air. For this reason, we now use red phosphorus in matches. Violet phosphorus is obtained from heating and crystallizing red phosphorus in a certain way. The phosphorus forms pentagonal "tubes". Black phosphorus is the most stable form; the atoms are linked together in puckered sheets, like . Because of these structural similarities black phosphorus is also flaky like graphite and possesses other similar properties. There are many isotopes of phosphorus, only one of which is stable ( P). The rest of the isotopes are radioactive with generally very short half-lives, which vary between a few nanoseconds to a few seconds. Two of the radioactive phosphorus isotopes have longer half-lives. P has a half-life of 14 days and P has a half-life of 25 days. These half-lives are long enough to be useful for analysis and for this reason the isotopes can be used to mark DNA. P played an important role in the 1952 Hershey-Chase Experiment. In this experiment, Alfred Hershey and Martha Chase used radioactive isotopes of phosphorus and to determine that DNA was genetic material and not proteins. Sulfur can be found in proteins but not DNA, and phosphorus can be found in DNA but not proteins. This made Phosphorus and Sulfur effective markers of DNA and protein, respectively. The experiment was set up as follows: Hershey and Chase grew one sample of a virus in the presence of radioactive S and another sample of a virus in the presence of P. Then, they allowed both samples to infect bacteria. They blended the S and the P samples separately and centrifuged the two samples. Centrifuging separated the genetic material from the non-genetic material. The genetic material penetrated the solid that contained the bacterial cells at the bottom of the tube while the non-genetic material remained in the liquid. By analyzing their radioactive markers, Hershey and Chase found that the P remained with the bacteria, and the S remained in the supernatant liquid. These results were confirmed by further tests involving the radioactive Phosphorus.. We get most elements from nature in the form of minerals. In nature, phosphorus exists in the form of phosphates. Rocks containing phosphate are fluoroapatite (\(\ce{3Ca3(PO4)2.CaF2}\)), chloroapaptite, (\(\ce{3Ca3(PO4)2. CaCl2}\)), and hydroxyapatite (\(\ce{3Ca3(PO4)2. Ca(OH)2}\)). These minerals are very similar to the bones and teeth. The arrangements of atoms and ions of bones and teeth are similar to those of the phosphate containing rocks. In fact, when the \(\ce{OH-}\) ions of the teeth are replaced by \(\ce{F-}\), the teeth resist decay. This discovery led to a series of social and economical issues. Nitrogen, phosphorus and potassium are key ingredients for plants, and their contents are key in all forms of fertilizers. From an industrial and economical view point, phosphorus-containing compounds are important commodities. Thus, chemistry of phosphorus has academic, commercial and industrial interests. As a member of the , on the Periodic Table, Phosphorus has 5 valence shell electrons available for bonding. Its valence shell configuration is 3s 3p . Phosphorus forms mostly covalent bonds. Any phosphorus rock can be used for the production of elemental phosphorus. Crushed phosphate rocks and sand (\(\ce{SiO2}\)) react at 1700 K to give phosphorus oxide, \(\ce{P4O10}\): \[\ce{2 Ca3(PO4)2 + 6 SiO2 \rightarrow P4O10 + 6 CaSiO3} \label{1}\] \(\ce{P4O10}\) can be reduced by carbon: \[\ce{P4O10 + 10 C \rightarrow P4 + 10 CO}. \label{2}\] Waxy solids of white phosphorus are molecular crystals consisting of \(\ce{P4}\) molecules. They have an interesting property in that they undergo spontaneous combustion in air: \[\ce{P4 + 5 O2 \rightarrow P4O10} \label{3}\] The structure of \(\ce{P4}\) can be understood by thinking of electronic configuration ( ) of \(\ce{P}\) in bond formation. Sharing three electrons with other \(\ce{P}\) atoms gives rise to the 6 \(\ce{P-P}\) bonds, leaving a lone pair occupying the 4th position in a distorted tetrahedron. When burned with insufficient oxygen, \(\ce{P4O6}\) is formed: \[\ce{P4 + 3 O2 \rightarrow P4O6} \label{4}\] To each of the \(\ce{P-P}\) bonds, an \(\ce{O}\) atom is inserted. Burning phosphorus with excess oxygen results in the formation of \(\ce{P4O10}\). An additional \(\ce{O}\) atom is attached to the \(\ce{P}\) directly: \[\ce{P4 + 5 O2 \rightarrow P4O10} \label{5}\] Thus, the oxides \(\ce{P4O6}\) and \(\ce{P4O10}\) share interesting features. Oxides of phosphorus, \(\ce{P4O10}\), dissolve in water to give phosphoric acid, \[\ce{P4O10 + 6 H2O \rightarrow 4 H3PO4} \label{6}\] Phosphoric acid is a polyprotic acid, and it ionizes at three stages: \[\ce{H3PO4 \rightleftharpoons H+ + H2PO4-} \label{7a}\] \[\ce{H2PO4- \rightleftharpoons H+ + HPO4^2-} \label{7b}\] \[\ce{HPO4^2- \rightarrow H+ + PO4^3-} \label{7c}\] Phosphoric Acid is a , which makes it an ideal buffer. It gets harder and harder to separate the hydrogen from the phosphate making the pK values increase in basicity: 2.12, 7.21, and 12.67. The conjugate bases H PO , HPO , and PO can be mixed to form buffer solutions. Commercially, phosphorus compounds are used in the manufacture of phosphoric acid (\(H_3PO_4\)) (found in soft drinks and used in fertilizer compounding). Other compounds find applications in fireworks and, of course, phosphorescent compounds which glow in the dark. Phosphorus compounds are currently used in foods, toothpaste, baking soda, matches, pesticides, nerve gases, and fertilizers. Phosphoric acid is not only used in buffer solutions it is also a key ingredient of Coca Cola and other sodas! Phosphorus compounds were once used in detergents as a water softener until they raised concerns about pollution and eutrophication. Pure phosphorus was once prescribed as a medicine and an aphrodisiac until doctors realized it was poisonous (Emsley). The middle number, (for example, 6-5-8) specifies the percentage of phosphorus compound in a fertilizer. Phosphorus is an important element for plant life. \[\ce{P4O10 + 6 H2O \rightarrow 4 H3PO4}\] \[\ce{PCl3 + 3 H2O \rightarrow H3PO3 + 3 HCl}\] This is a weaker acid than \(\ce{H3PO4}\).

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The stopped-flow technique allows for the evaluation of solution-based kinetics on a milliseconds timescale with a very small volume of reactants used. Imagine you wanted to compute the initial rate of a reaction, one that, when the reactants combined changed color or fluoresced, however the reaction was far too fast to detect via the human eye alone. How would you go about determining the reaction rate? Could you use the ? But what if you only had a limited volume of the reactants at your disposal? The continuous flow technique requires that you have enough of the reactants to constantly flow through the spectrophotometer at a continuous rate. In cases like this, the Stopped Flow technique is probably more appropriate. Stopped-Flow allows for the rate of a solution-based reaction to be determined in milliseconds, and with a very small volume of reactants. This technique involves two reactants held in separate reservoirs that are prevented from freely flowing by syringe pumps. The reaction is initiated by depressing the reactant syringes, and thus releasing the reactants into the connecting "mixing chamber" where the solutions are mixed. The reaction is monitored by observing the change in absorbance of the reaction solution as a function of time. As the reaction progresses it fills the “stop syringe” which then expands until it hits a block at the point when the reaction has reached a continuous flow rate, thereby stopping the flow and the reaction, and thus allowing the researcher to calculate the exact initial rate of reaction. See Figure 1 below: The Stopped Flow technique works because within milliseconds of combining the two reactants the absorbance can be read. In addition, the stop syringe assures for a steady rate of flow pas the spectophotometer so that reactants are being added to solution and forming products at a consistant rate. Imagine a reaction that proceeds as follows: \[A + B \rightarrow C + D \nonumber \] Substance C is the product and Substance D is known to fluoresce. The optical path is 1 cm and the e value is 1 L/mol cm . Based on the following data gathered by the Stopped Flow method, determine the rate constant and the order of the reaction. Time (seconds) Absorbance .001 .2 .002 .4 .003 .6 .004 .8 .005 1 .006 1.2 .007 1.4 .008 1.6 .009 1.8 (Hint: Make a graph and use that to find the order and the rate. If you need more help, these pages might help: and ) This is a zero order reaction with a rate constant of 200 M/S. We know that the absorbance is proportional to the change of the concentration of D over time because D fluoresces. However, from the reaction above, the concentration of C=D because the stoichiometry is 1:1. Now lets start with the relationship between absorbance and concentration. From Beer's Law we know that absorbance is directly proportional to concentration, therefore we know, in this case, absorbance is equal to concentration because we divide by 1 and 1 ( ). Now for the reaction rate and the reaction order. You can figure the order out by graphing the either concentration by time, ln(concentration) by time, or 1/concentration by time depending on the order of the reaction. If concentration by time is linear, then the reaction is zero order. If ln(concentration) by time is linear, then the reaction is first order. Finally, if 1/concentration by time is linear, then the reaction is 2nd order. Once you have determined that this reaction is zero order, you can find the slope of the line to solve for \(k\), and the rate of the reaction.

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The Unit Cell refers to a part of a simple crystal lattice, a repetitive unit of solid, brick-like structures with opposite faces, and equivalent edge points. In 1850, Auguste Bravais proved that crystals could be split into fourteen unit cells. Although there are several types of unit cells found in cubic lattices, we will be discussing the basic ones: Simple Cubic, Body-centered Cubic, and Face-centered Cubic. If any atom recrystalizes, it will eventually become the original lattice. Crystallization refers the purification processes of molecular or structures;. The Unit Cell contains seven crystal systems and fourteen crystal lattices. These unit cells are given types and titles of symmetries, but we will be focusing on . One of the most commonly known unit cells is NaCl (Sodium Chloride), an octahedral geometric unit cell. The whole lattice can be reproduced when the unit cell is duplicated in a three dimensional structure. These unit cells are imperative for quite a few metals and ionic solids crystallize into these cubic structures. Calculating with unit cells is a simple task because edge-lengths of the cell are equal along with all 90⁰ angles. Simple Cubic unit cells indicate when lattice points are only at the corners. They are the simplest (hence the title) repetitive unit cell. The lattice points at the corners make it easier for metals, ions, or molecules to be found within the crystalline structure. This phenomena is rare due to the low packing of density, but the closed packed directions give the cube shape. Since the edges of each unit cell are equidistant, each unit cell is identical. In order to be labeled as a "Simple Cubic" unit cell, each eight cornered same particle must at each of the eight corners. This unit cell only contains one atom. Its packing efficiency is about 52%. To packing efficiency, we multiply eight corners by one-eighth (for only one-eighth of the atom is part of each unit cell), giving us one atom. To calculate edge length in terms of the equation is as follows: An example of a Simple Cubic unit cell is Polonium. Body-centered Cubic (BCC) unit cells indicate where the lattice points appear not only at the corners but in the center of the unit cell as well. The atoms touch one another along the cube's diagonal crossing, but the atoms don't touch the edge of the cube. All atoms are identical. This type of unit cell is more common than that of the Simple Cubic unit cell due to tightly packed atoms. Its packing efficiency is about 68% compared to the Simple Cubic unit cell's 52%. This unit cells contains two atoms. To determine this, we multiply the previous eight corners by one-eighth and one for the additional lattice point in the center. To calculate edge length in terms of the equation is as follows: \[\dfrac{4r}{\sqrt{3}}\] Some examples of BCCs are Iron, Chromium, and Potassium. It is a common mistake for CsCl to be considered bcc, but it is not. Instead, it is non-closed packed. Face-centered Cubic (FCC) unit cells indicate where the lattice points are at both corners and on each face of the cell. This is a more common type of unit cell since the atoms are more tightly packed than that of a Simple Cubic unit cell. Like the BCC, the atoms don't touch the edge of the cube, but rather the atoms touch diagonal to each face. Its packing efficiency is the highest with a percentage of 74%. Atoms touch one another along the face diagonals. All atoms are identical. This unit cell contains four atoms. To determine this, we take the equation from the aforementioned Simple Cubic unit cell and to the parenthesized six faces of the unit cell multiplied by one-half (due to the lattice points on each face of the cubic cell). To calculate edge length in terms of the equation is as follows: \[ 2 \sqrt{2}r \] We end up with 1.79 x 10 g/atom. Next we find the mass of the unit cell by multiplying the number of atoms in the unit cell by the mass of each atom (1.79 x 10 g/atom)(4) = 7.167 x 10 grams. Next we find the edge length by: \[{2}\sqrt{2}*{160pm}\] Which equals 4.525 x 10 meters. Now we find the volume which equals the edge length to the third power. We convert meters into centimeters by dividing the edge length by 1 cm/10 m to the third power. (4.525 x 10 m x 1cm/10 m = 9.265 x 10 cubic centimeters.) Finally, we find the density by mass divided by volume. So, 7.167 x 10 grams/9.265 x 10 cubic centimeters = 7.74 g/cm

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Given the knowledge that a particular reaction will proceed at a suitable rate, a host of practical considerations are necessary for satisfactory operation. These considerations include interference by possible side reactions that give products other than those desired, the ease of separation of the desired product from the reaction mixture, and costs of materials, apparatus, and labor. We shall consider these problems in connection with the important synthetic reactions discussed in this book. The chlorination of saturated hydrocarbons can be induced by light, but also can be carried out at temperatures of about \(300^\text{o}\) in the dark. Under such circumstances the mechanism is similar to that of light-induced chlorination, except that the chlorine atoms are formed by thermal dissociation of chlorine molecules. Solid carbon surfaces catalyze thermal chlorination, possibly by aiding in the cleavage of the chlorine molecules. Direct monohalogenation of saturated hydrocarbons works satisfactorily only with chlorine and bromine. For the general reaction the calculated \(\Delta H^\text{0}\) value is negative and very large for fluorine, negative and moderate for chlorine and bromine, and positive for iodine (see Table 4-7). With fluorine, the reaction evolves so much heat that it may be difficult to control, and products from cleavage of carbon-carbon as well as of carbon-hydrogen bonds may be obtained. The only successful, direct fluorination procedure for hydrocarbons involves diffusion of minute amounts of fluorine mixed with helium into liquid or solid hydrocarbons at low temperatures, typically \(-78^\text{o}\) (Dry Ice temperature). As fluorination proceeds, the concentration of fluorine can be increased. The process is best suited for preparation of completely fluorinated compounds, and it has been possible to obtain in this way amounts of \(\left( CF_3 \right)_4C\) and \(\left( CF_3 \right)_3 C-C \left( CF_3 \right)_3\) from 2,2-dimethylpropane and 2,2,3,3-tetramethylbutane corresponding to \(10\)-\(15\%\) yields based on the fluorine used. Bromine generally is much less reactive toward hydrocarbons than chlorine is, both at high temperatures and with activation by light. Nonetheless, it usually is possible to brominate saturated hydrocarbons successfully. Iodine is unreactive. The chlorination of methane does not have to stop with the formation of chloromethane (methyl chloride). It is usual when chlorinating methane to obtain some of the higher chlorination products: dichloromethane (methylene chloride), trichloromethane (chloroform), and tetrachloromethane (carbon tetrachloride): In practice, one can control the degree of substitution to a considerable extent by controlling the methane-chlorine ratio. For example, for monochlorination to predominate, a high methane-chlorine ratio is necessary such that the chlorine atoms react with \(CH_4\) and not with \(CH_3Cl\). For propane and higher hydrocarbons for which more than one monosubstitution product is generally possible, difficult separation problems bay arise when a particular product is desired. For example, the chlorination of 2-methylbutane \(3\) at \(300^\text{o}\) gives all four possible monosubstitution products, \(4\), \(5\), \(6\), and \(7\): On a purely statistical basis, we may expect the ratio of products from \(3\) to correlate with the number of available hydrogens at the various positions of substitution. That is, \(4\), \(5\), \(6\), and \(7\) would be formed in the ratio 6:3:2:1 (\(50\%\):\(25\%\):\(17\%\):\(8\%\)). However, as can be seen from Table 4-6, the strengths of hydrogen bonds to primary, secondary, and tertiary carbons are not the same and, from the argument given in we would expect the weaker \(C-H\) bonds to be preferentially attacked by \(Cl \cdot\). The proportion of \(7\) formed is about three times that expected on a statistical basis which is in accord with our expectation that the tertiary \(C-H\) bond of 2-methylbutane should be the weakest of the \(C-H\) bonds. (See Table 4-6.) The factors governing selectivity in halogenation of alkanes follow: 1. The rates at which the various \(C-H\) bonds of 2-methylbutane are broken by attack of chlorine atoms approach 1:1:1 as the temperature is raised above \(300^\text{o}\). At higher temperatures both chlorine atoms and hydrocarbons become more reactive because of increases in their thermal energies. Ultimately, temperatures are attained where a chlorine atom essentially removes the first hydrogen with which it collides regardless of position on the hydrocarbon chain. In such circumstances, the composition of monochlorination products will correspond to that expected from simple statistics. 2. Bromine atoms are far more selective than chlorine atoms. This is not unexpected because is endothermic, whereas corresponding reactions with a chlorine atoms usually are exothermic (data from Table 4-6). Bromine removes only those hydrogens that are relatively weakly bonded to a carbon atom. As predicted, attack of \(Br \cdot\) on 2-methylbutane leads mostly to 2-bromo-2-methylbutane, some secondary bromide, and essentially no primary bromides: 3. The selectivity of chlorination reactions carried on in is increased markedly in the presence of benzene or alkyl-substituted benzenes because benzene and other arenes form loose complexes with chlorine atoms. This substantially cuts down chlorine-atom reactivity, thereby making the chlorine atoms behave more like bromine atoms. It is possible to achieve chlorination of alkanes using sulfuryl chloride (\(SO_2Cl_2\), bp \(69^\text{o}\)) in place of chlorine: The reaction has a radical-chain mechanism and the chains can be initiated by light or by chemicals, usually peroxides, \(ROOR\). Chemical initiation requires an with a weak bond that dissociates at temperatures between \(40\)-\(80^\text{o}\). Peroxides are good examples. The \(O-O\) bond is very weak (\(30\)-\(50 \: \text{kcal}\)) and on heating dissociates to alkoxyl radicals, \(RO \cdot\), which are reactive enough to generate the chain-propagating radicals from the reactants. The exact sequence of chemical initiation is not always known, but a plausible route in the present case would have \(RO \cdot\) abstract hydrogen from the alkane: The propagation steps that would follow are Chlorination with sulfuryl chloride of alkanes with more than one kind of hydrogen gives a mixture of alkyl chlorides resembling that obtained with chlorine itself. However, in some circumstances the mixture of chlorides is not the same mixture obtained with chlorine itself and when this is true, the hydrogen-abstraction step probably involves \(\cdot SO_2Cl\) rather than \(Cl \cdot\). The alternative propagation steps then are Different product ratios are expected from \(Cl \cdot\) and \(ClSO_2 \cdot\) for the same reason that \(Cl \cdot\) and \(Br \cdot\) lead to different product ratios ( ). Other reagents that sometimes are useful halogenating agents in radical-chain reactions include and (1977)

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Sulfenyl chloride additions are initiated by the attack of an electrophilic sulfur species on the pi-electrons of the double bond. The resulting cationic intermediate may be stabilized by the non-bonding valence shell electrons on the sulfur in exactly the same way the halogens exerted their influence. Indeed, a cyclic sulfonium ion intermediate analogous to the bromonium ion is believed to best represent this intermediate (see drawing below). Figure 1: cyclic sulfonium ion intermediate Two advantages of the oxymercuration method of adding water to a double bond are its high anti-stereoselectivity and the lack of rearrangement in sensitive cases. These characteristics are attributed to a mercurinium ion intermediate, analogous to the bromonium ion discussed above. In this case it must be d-orbital electrons that are involved in bonding to carbon. A drawing of this intermediate is shown below. Figur mercurinium The hydroboration reaction is among the few simple addition reactions that proceed cleanly in a syn fashion. , this is a single-step reaction. Since the bonding of the double bond carbons to boron and hydrogen is concerted, it follows that . Furthermore, rearrangements are unlikely inasmuch as a discrete carbocation intermediate is never formed. These features are illustrated for the hydroboration of α-pinene in the following equation. Since the hydroboration procedure is most commonly used to hydrate alkenes in an anti-Markovnikov fashion, we also need to know the stereoselectivity of the second oxidation reaction, which substitutes a hydroxyl group for the boron atom. Independent study has shown this reaction takes place with so the overall addition of water is also syn. The hydroboration of α-pinene also provides a nice example of steric hindrance control in a chemical reaction. In the less complex alkenes used in earlier examples the plane of the double bond was often a plane of symmetry, and addition reagents could approach with equal ease from either side. In this case, one of the methyl groups bonded to C-6 (colored blue in the equation) covers one face of the double bond, blocking any approach from that side. All reagents that add to this double bond must therefore approach from the side opposite this methyl.

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where \(P^o_{solvent}\) is the vapor pressure of the pure solvent and \(\chi_{solvent}\) is the mole fraction of the solvent. Since this is a two-component system (solvent and solute), then \[\chi_{solvent} + \chi_{solute} = 1\] where \(\chi_{solute}\) is the mole fraction of the solvent or solute. The change in vapor pressure (\(\Delta P\)) can be expressed \[\Delta P = P_{solution} - P^o_{solvent} = \chi_{solvent}P^o_{solvent} - P^o_{solvent}\] \[\Delta P = ( \chi_{solvent} - 1) P^o_{solvent} = \chi_{solute} P^o_{solvent} \label{DP}\] Calculate the vapor pressure of a solution made by dissolving 50.0 g glucose, \(C_6H_{12}O_6\), in 500 g of water. The vapor pressure of pure water is 47.1 torr at 37°C To use Raoult's Law (Equation \(\ref{RLaw}\)), we need to calculate the mole fraction of water (the solvent) in this sugar-water solution. \[ \chi_{solvent} = \dfrac{ \text{moles of water}}{\text{moles of solute} + \text{moles of solvent}} \nonumber\] \[ \chi_{solvent} = \dfrac{ n_{water}}{ n_{glucose} + n_{water} } \nonumber\] The molar mass of glucose is 180.2 g/mol and of water is 18 g/mol. So \[n_{water} = \dfrac{500\,g}{18\,g /mol} = 27.7 \,mol \nonumber\] and \[n_{glucose} = \dfrac{50\,g}{180.2\,g /mol} = 0.277 \,mol \nonumber\] and \[ \chi_{solvent} = \dfrac{ 27.7 \,mol}{ 0.277 \,mol + 27.7 \,mol } = 0.99 \nonumber\] Note that this still relatively dilute. The pressure of the solution is then calculated via Raoult's Law (Equation \(\ref{RLaw}\)): \[ P_{solution} = 0.99 \times 47.1 = 46.63 \, torr \nonumber\] Calculate the vapor pressure of a solution made by dissolving 50.0 g CaCl , \(C_6H_{12}O_6\), in 500 g of water. The vapor pressure of pure water is 47.1 torr at 37°C To use Raoult's Law (Equation \(\ref{RLaw}\)), we need to calculate the mole fraction of water (the solvent) in this salt-water solution. \[ \chi_{solvent} = \dfrac{ \text{moles of water}}{\text{moles of solute} + \text{moles of solvent}} \nonumber\] \[ \chi_{solvent} = \dfrac{ n_{water}}{ n_{solutes} + n_{water} } \nonumber\] The molar mass of \(CaCl_2\) is 111 g/mol and of water is 18 g/mol. So \[n_{water} = \dfrac{500\,g}{18\,g /mol} = 27.7 \,mol \nonumber\] and \[n_{solutes} = \dfrac{50\,g}{111 \,g /mol} = 0.45 \,mol \nonumber\] but this is really: and \[ \chi_{solvent} = \dfrac{ 27.7 \,mol}{ 0.45 \,mol + 0.9 \,mol + 27.7 \,mol } = 0.953 \nonumber\] Note that this still relatively dilute. The pressure of the solution is then calculated via Raoult's Law (Equation \(\ref{RLaw}\)): \[ P_{solution} = 0.953 \times 47.1 = 44.88\, torr \nonumber \] Example \(\Page {1A}\) At 25 C the vapor pressure of pure benzene is 93.9 torr. When a non-volatile solvent is dissolved in benzene, the vapor pressure of benzene is lowered to 91.5 torr. What is the concentration of the solute and the solvent, expressed in mole fraction? Vapor pressure lowering \(\Delta P= 2.4\; torr\) with \(\chi_{solute} = 0.026\).

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Substitution reactions involve the replacement of one atom or group \(\left( \ce{X} \right)\) by another \(\left( \ce{Y} \right)\): \[\ce{RX} + \ce{Y} \rightarrow \ce{RY} + \ce{X}\] We already have described one very important type of substitution reaction, the halogenation of alkanes (Section 4-4), in which a hydrogen atom is replaced by a halogen atom (\(\ce{X} = \ce{H}\), \(\ce{Y} =\) halogen). The chlorination of 2,2-dimethylpropane is an example: Reactions of this type proceed by radical-chain mechanisms in which the bonds are broken and formed by atoms or radicals as reactive intermediates. This mode of bond-breaking, in which one electron goes with \(\ce{R}\) and the other with \(\ce{X}\), is called bond cleavage: There are a large number of reactions, usually occurring in solution, that do involve atoms or radicals but rather involve ions. They occur by cleavage as opposed to homolytic cleavage of electron-pair bonds. In heterolytic bond cleavage, the electron pair can be considered to go with one or the other of the groups \(\ce{R}\) and \(\ce{X}\) when the bond is broken. As one example, \(\ce{Y}\) is a group such that it has an unshared electron pair and also is a negative ion. A heterolytic substitution reaction in which the \(\ce{R} \colon \ce{X}\) bonding pair goes with \(\ce{X}\) would lead to \(\ce{RY}\) and \(\colon \ce{X}^\ominus\), A specific substitution reaction of this type is that of chloromethane with hydroxide ion to form methanol: In this chapter, we shall discuss substitution reactions that proceed by ionic or polar mechanisms in which the bonds cleave heterolytically. We also will discuss the mechanistically related reactions that result in the formation of carbon-carbon multiple bonds: These reactions often are influenced profoundly by seemingly minor variations in the structure of the reactants, in the solvent, or in the temperature. It is our purpose to show how these reactions can be understood and how they can be used to prepare other useful organic compounds. But first it will be helpful to introduce the concepts of and reagents, and to consider the \(\Delta H\) values for heterolytic bond breaking. and (1977)

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Iron and copper are two common metals in biology, and they are both involved in electron relays in which electrons are passed from one metal to another to carry out transformations on substrates in cells. But which one passes the electron to which? Or can it go either way? How easily one metal can pass an electron to another, or how easily one metal can reduce another, is a pretty well-studied question. There are some preferences, some of which can be easily understood. In order to look at this question, electrochemists typically measure the voltage produced when a circuit is set up that includes an electron made of the metal in question and an electrode made of a "standard hydrogen electrode". In a standard hydrogen electrode, hydrogen gas (H ) reacts on a platinum surface to produce two electrons and two protons. The electrons travel along a wire to the other electrode. The other electrode sits in an aqueous solution containing a salt of the metal in question. How easily is the metal ion reduced by accepting the electrons from the standard hydrogen electrode? The more downhill energetically this process is, the more positive is the voltage measured in the circuit. For example, we might put some copper(II) salts, such as CuSO , into the solution together with a copper electrode. Then we would see whether the copper in solution is spontaneously reduced to copper metal. That would happen, essentially, if the copper ion is more easily reduced than a proton. Thus, an electron released by hydrogen flows from the platinum electrode to the copper electrode and is picked up by copper ions waiting in solution. Otherwise, what would happen if the reaction were not spontaneous? If no reaction occurred at all, maybe there would be no voltage. However, if the opposite reaction were to occur -- if copper were able to provide an electron to convert protons into molecular hydrogen -- then current would flow through the circuit in the opposite direction. A voltage would register, but it would be negative. As it happens, the reaction is spontaneous, the hydrogen does send electrons through the wire, turning into protons in the process, and at the other end of the wire, the copper ion is converted into copper metal, putting another layer on the surface of the electrode. The voltage is about 0.34V. This is a comparative, rather than absolute, measurement. We are measuring the intrinsic potential of an electron to be transferred from one species, a hydrogen molecule, to another, a Cu(II) ion. It is also an intensive, rather than extensive, property. It does not matter how much copper we have, or how much hydrogen; the electron still has the same natural tendency to flow from the hydrogen to the copper. The results of many such studies, carefully measured under specific conditions for maximum reproducibility, are gathered in a table of reduction potentials. The reactions are referred to as "half-reactions" because they each provide only half the picture of what is going on. The electron in each reaction doesn't come from nowhere; every reaction in the table would involve transfer of an electron from elemental hydrogen to form a proton. Much more extensive tables of reduction potentials can be found; for example, see the following . A positive reduction potential indicates a spontaneous reaction. That makes sense, for instance, in the reaction of fluorine to give fluoride ion. For that reaction, E = 2.87 V. Of course, fluorine is a very electronegative element, and it will spontaneously accept an electron to obtain a noble gas configuration. A negative reduction potential, on the other hand, indicates a reaction that would not occur spontaneously. For example, we would not expect lithium cation to accept an electron. We are used to thinking about alkali metals easily giving up their electrons to become cations. The reduction of lithium ion has a reduction potential E = -3.04 V. This reaction would only occur if it were driven by an expenditure of energy. The opposite reaction, on the other hand, would be the oxidation of lithium metal to give a lithium cation. That reaction would occur spontaneously, and would have a spontaneous "oxidation potential". In fact, that value is + 3.04 V. The oxidation potential is always the same magnitude of the reduction potential for the reverse reaction, but with the opposite sign. These signs may seem counter-intuitive if you are used to thinking of free energy changes. A negative free energy change means energy is lost in a reaction. A positive free energy change means energy must be put into a reaction to drive it forward. In fact, reduction potential and free energy are closely linked by the following expression: \[ \Delta G = - nFE^0 \] in which n = number of electrons transferred in the reaction; F = Faraday's constant, 96 500 Coulombs/mol. So, a positive reduction potential translates into a negative free energy change. Note that reduction potentials are pretty sensitive to changes in the environment. Factors that may stabilize one particular metal ion may not have the exact same effect on another, and so the preference for one state versus another will be altered slightly under different conditions. For example, permanganate ion (MnO ) has a more positive reduction potential under "acidic conditions" (with excess protons in solution) compared to "basic conditions" (with a paucity of protons in solution and instead an excess of hydroxide ion). The reduction potential under acidic conditions is +1.23V, compared to +0.59 V under basic conditions. Tables of reduction potentials are also useful in assessing the opposite reaction. For example, lithium metal spontaneously reduces protons to produce hydrogen gas, becoming lithium ion in turn. The potential for that reaction is simply the opposite of the reduction potential of lithium ion; this is called the oxidation potential of lithium metal. The more positive a metal's oxidation potential, the more easily it is oxidized. However, we don't need a separate table of those values; they are just the opposite of the reduction potentials. Reactions with negative reduction potentials easily go backwards, reducing the proton to hydrogen gas by taking an electron from the reducing agent. However, the most important use of standard reduction potentials is combining them to find out the potential of new reactions. For example, when it says in the table that \[ Cu^+ + e^- \rightarrow Cu (s) E^0 = 0.53 V \] It really means that is the potential produced for a specific reaction involving electron transfer between hydrogen and copper ion: \[ H_2 (g) + 2 Cu^+ \rightarrow 2 Cu (s) + 2 H^+ E^0 = 0.53 V \] And when it says that \[ Fe ^{2+} + 2 e^- \rightarrow Fe (s) E^0 = - 0.44 V \] It really means that \[ H_2 (g) + Fe^{2+} \rightarrow Fe (s) + 2 H^+ E^0 = -0.44 V \] But now we know that reaction is endergonic, with a negative reduction potential and a positive free energy change. However, the reverse reaction \[ Fe (s) + 2 H^+ \rightarrow H_2 (g) + Fe^{2+} E^0 = 0.44 V \] has a positive reduction potential and would proceed easily. Now, if we combine those previous reactions, simply by adding them together \[ H_2 (g) + 2 Cu^+ + Fe (s) + 2 H^+ \rightarrow H_2 (g) + Fe^{2+} + 2 Cu (s) + 2 H^+ E^0 = 0.44 + 0.53 V \] and simplifying \[ 2 Cu^+ + Fe (s) \rightarrow Fe^{2+} + 2 Cu (s) E^0 = 0.97 V \] That means the hydrogen reaction doesn't need to be involved at all. It's just a common reference point for all the other reactions. If we know how far uphill (or downhill) any two reactions are compared to that one, then we know how they compare to each other, too. We should note that electrochemistry is a business that demands great care. There are a number of factors that can cause variations in the potential that is measured, and so we need to be very careful to control for those factors. For example, if there is a buildup of charge in one solution or another (because we are taking cations out of solution in one case and putting them into solution in the other), the ability to remove more electrons at one electrode and deliver them at another may be hindered. For that reason, a "salt bridge" is incorporated into the design of the system; this bridge allows ions to diffuse from one cell to the other in order to keep charge balanced. Also, the solutions are maintained at a standard concentration to make sure measurements are always made in comparable circumstances. Finally, non-reactive electrolytes (salts) are added to solution to aid in conductivity and maintain a constant ionic strength. Calculate oxidation states to confirm that the manganese ion is being reduced in the following reaction: MnO + 2 H O + 3e → MnO (s) + 4 OH Balance the following half reactions by adding the right number of electrons to one side or the other, based on oxidation state. Then add water molecules and protons to help balance oxygens and overall charge. Why is the reduction potential of Li so negative? Why is the reduction potential of F so positive? Rank the following metals from most easily oxidized to least easily oxidized: gold, iron, aluminum, copper, lithium. Calculate reduction potentials for the following reactions: In general, if one reaction is combined with the reverse of a reaction above it in the table, will the overall reaction be spontaneous? What about if a reaction is combined with the reverse of a reaction below it? ,

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The colligative properties of a solution depend on the relative numbers (concentration) of solute and solvent particles, they do not depend on the nature of the particles. Colligative properties change in proportion to the concentration of the solute particles. We distinguish between four colligative properties: vapor pressure lowering, freezing point depression, boiling point elevation, and osmotic pressure. All four colligative properties fit the relationship property = solute concentration x constant The determination of colligative properties allows us to determine the concentration of a solution and calculate molar masses of solutes The boiling points of solutions are all higher than that of the pure solvent. Difference between the boiling points of the pure solvent and the solution is proportional to the concentration of the solute particles: \[\Delta{T_b} = T_b (solution) - T_b (solvent) = K_b \times m\] where \(\Delta{T_b}\) is the boiling point elevation, \(K_b\) is the boiling point elevation constant, and is the molality (mol/kg solvent) of the solute. A solution is prepared when 1.20 g of a compound is dissolved in 20.0 g of benzene. The boiling point of the solution is 80.94 C. Answer

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\(3.6 \times 10^{-19}\;J\)? \[E= hv = \dfrac{hc}{λ} \nonumber \] \[3.6 \times 10^{-19}\; J = \dfrac{(6.626 \times 10^{-34}\; J\;s) ( 3 \times 10^8\; ms^{-1})}{\lambda}\nonumber \] \[λ = 0.55 \times 10^{-7}\; m = 550\; nm\nonumber \] is \(6.5 \times 10^{-14}\; s^{-1}\) ? \[E= hv\nonumber \] \[E=( 6.626 \times 10^{-34}J\;s) ( 6.5 \times 10^{-14} s^{-1})\nonumber \] \[E= 4.3 \times 10^{-19}\;J\nonumber \] Bohr model: \[\dfrac{1}{λ} = R_H \left(\dfrac{1}{n_a^2} -\dfrac{1}{n_b^2} \right)\nonumber \] \[\dfrac{1}{λ} = ( 1.09678 \times 10^{-2} nm^{-1} ) \left(\dfrac{1}{9}- \dfrac{1}{25}\right) \nonumber \] \[λ = 1,282\; nm\nonumber \] Calculate the different energy between \(n=3\) and \(n=4\) of atomic hydrogen based on Bohr orbits \[\Delta E=-hcR_H \left(\dfrac{1}{3^2}\ - \dfrac{1}{4^2} \right) \nonumber \] \[\Delta E=-6.626 \times 10^{-34} J.s)(2.9979 \times 10^8\; m/s)(109,737\; cm^{-1}) \left(\dfrac{1}{3^2} - \dfrac{1}{4^2}\right))\nonumber \] \[\Delta E= 1.8 \times 10^{-21}\; J\nonumber \] Calculate the wavenumber of the wavelength of the light emitted from the \(n=8\) to \(n=6\) transition. This is a simple application of the Rydberg equation \[ \tilde{\nu} R_H \left (\dfrac{1}{n_{f}^{2}} -\dfrac{1}{n_{i}^{2}} \right)\nonumber \] with \(n_i = 6\) and \(n_f = 8\). \[ \tilde{\nu}=109,737cm^{-1}\left (\dfrac{1}{6^{2}}-\dfrac{1}{8^{2}} \right)=1333.61 \; cm^{-1}\nonumber \] Calculate the wavelength associated with a 42 g baseball with speed of 80 m/s. \[ \lambda =\dfrac{h}{p} = \dfrac{h}{mv}\nonumber \] \[ \lambda = \dfrac{6.626 \times 10^{34} J s}{(0.042 \; kg)( 80 m/s)} = 1.97 \times 10^{-34} \, m\nonumber \] This is a very small wavelength as expected since a ball behaves rather classically (i.e., non-quantum). Calculate the energy of a 530-nm photon. \[p =\dfrac{E}{c}\nonumber \] \[\lambda = \dfrac{h}{p}\nonumber \] \[\lambda = \dfrac{hc}{E}\nonumber \] \[ E= \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 1-^{-34})( 3 \times 10^{8})}{530 \times 10^{-9}} = 3.75 \times 10^{-9} \; J\nonumber \] Describe Planck's three experimental observations that explain the photoelectric effect. "When light of a certain frequency shines on a clean metal surface, electrons are ejectedd from the metal. Experimentally, it is found that: What is the difference between classical and quantum mechanics? Provide the equation relating the energy of emitted radiation to frequency. Classical mechanics predicts that the radiant energy produced by oscillating objects is continuous. Quantum mechanics predicts that their energy can be viewed as existing in discrete levels. The equation is \[E = h\nu\nonumber \]. Energy is emitted in discrete multiples of \(h\nu\), where h = Planck's constant. Do you expect the ionization of an atom to be an endothermic process or an exothermic process? It will be endothermic because energy is needed to ionize the atom. Calculate the wavelength of light with an energy of \(5.22 \times 10^{-19}\; J\). Rearrange the equation \[E = \dfrac{hc}{\lambda }\nonumber \] so that the unknown value is the wavelength of light \[\lambda = \dfrac{hc}{E}\nonumber \] \[\lambda =\dfrac{(6.626\times 10^{-34}Js)(3.0\times 10^{8}m/s)}{5.22\times 10^{-19}J} = 3.81\times 10^{-7}m = 381 nm\nonumber \] You realize after class that you forgot to write down the final wavelength of your proton. You do, however, remember that the potential difference was 114 V. From this information, figure out the final wavelength: \(\frac{1}{2}mv^{2}=eV\) given the uncertainty for position of an electron circling an atom is 0.5A . Find uncertainty in its velocity? Δp ≥ h/ 4πΔx = 6.626 x 10 Js/ 4π(0.5 x 10 m) = 1.311.05 x 10 kg m s Δ v= Δp/ m ≥ 1.3 x 10 kg m s / 9.109 x 10 kg = 1 x 10 m s If the uncertainty of measuring the position of an electron is 2.0 Å, what is the uncertainty of simultaneously measuring its velocity? A typical mass for a horse is 510 kg, and a typical galloping speed is 22 kilometers per hour. Use these values to answer the following questions. (a) 1. Use the relationship between velocity and momentum to find the momentum. \[ p = m v \nonumber \] \[ p = 510\ kg \times 22\ \dfrac{km}{hr} \times \dfrac{1000\ m}{km} \times \dfrac{hr}{3600\ s} \nonumber \] \[ p = 3.1 \times 10^3\ \dfrac{kg\ m}{s} \nonumber \] 2. Find the de Broglie wavelength. \[ \lambda = \dfrac{h}{p} \nonumber \] \[ \lambda = \dfrac{6.626 \times 10^{-34}\ J\ s}{3.1 \times 10^3\ \dfrac{kg\ m}{s}} \times \dfrac{\dfrac{kg\ m^2}{s^2}}{J} \nonumber \] \[ \lambda = 2.1 \times 10^{-37}\ m \nonumber \] (b) 1. Find the uncertainty of momentum from the uncertainty of velocity. \[ \Delta p = m \Delta v \nonumber \] \[ \Delta v = 0.01 \times v \nonumber \] \[ \Delta p = 510\ kg \times 0.01 \times 22\ \dfrac{km}{hr} \times \dfrac{1000\ m}{km} \times \dfrac{hr}{3600\ s} \nonumber \] \[ \Delta p = 31 \dfrac{kg\ m}{s} \nonumber \] 2. Use Heisenberg's uncertainty principle to find the uncertainty of position. \[ \Delta x \Delta p \geq \dfrac{h}{4 \pi} \nonumber \] \[ \Delta x \geq \dfrac{h}{4 \pi \Delta p} \nonumber \] \[ \Delta x \geq \dfrac{6.626 \times 10^{-34}\ J\ s}{4 \pi \times 3.1 \times 10^3 \dfrac{kg\ m}{s}} \times \dfrac{\dfrac{kg\ m^2}{s^2}}{J} \nonumber \] \[ \Delta x \geq 1.7 \times 10^{-36}\ m \nonumber \] (c) The calculations that involve Planck's constant (h) will change. This much larger value will make the wave-like properties of the horse more important; its wavelength and uncertainty of position will increase dramatically: \[ \lambda ' = \dfrac{h'}{p} = \dfrac{0.010\ J\ s}{3.1 \times 10^3\ \dfrac{kg\ m}{s} \times \dfrac{\dfrac{kg\ m^2}{s^2}}{J}} \nonumber \] \[ \lambda ' = 3.2 \times 10^{-6}\ m = 3200\ nm \nonumber \] \[ \Delta x' \geq \dfrac{h'}{4 \pi \Delta p} = \dfrac{0.010\ J\ s}{4 \pi \times 3.1 \times 10^3 \dfrac{kg\ m}{s}} \times \dfrac{\dfrac{kg\ m^2}{s^2}}{J} \nonumber \] \[ \Delta x' \geq 2.6 \times 10^{-5} m = 26\ \mu m \nonumber \] The momentum would not change. how fast must a person weighing 42 kg move through a door 0.5 m wide in order to be diffracted? p= h/ λ = 6.626 x 10 J s/ 0.5 = 3.313 x 10 kg m s V= p/m = 3.313 x 10 / 42 kg = 7.88 x 10 ms When a particle passes through a slit, diffraction occurs if the particle's wavelength is on the same order of magnitude as the width of the slit. At approximately what velocity would a 1670 kg car have to move through the Lincoln Tunnel (6.6 m width) for diffraction to occur? 1. Use de Broglie's equation to relate velocity to wavelength; solve for velocity. \[ \lambda = \dfrac{h}{m v} \nonumber \] \[ v = \dfrac{h}{m \lambda} \nonumber \] 2. Set the wavelength equal to the tunnel width and solve. \[ v = \dfrac{6.626 \times 10^{-34}\ J\ s}{1670\ kg \times 6.6\ m} \times \dfrac{\dfrac{kg\ m^2}{s^2}}{J} \nonumber \] Answer: the car would have to be driving slowly. \[ v = 6.0 \times 10^{-38}\ m/s \nonumber \] The Paschen emission spectrum is the collection of spectral lines emitted by H atoms, where the final state is n = 3. What are the shortest and longest wavelengths (in nm) of Paschen spectral lines? Spectral lines of Lyman and Balmer series do not overlap. Verify this statement. A photon has a specific energy according to the energy between the excited and ground state. E = (1/nf2 - 1/ni2), A = hv The lyman series is when nf is the ground state or nf =1 and Balmer series is when nf =2 nf = 1 the value , ni from 2 to infinity with the values between 1 and ½. For the balmer series ni is from 3 to infinity the range between 1/4 and 5/36. Thus, these intervals don not overlap --> the energies do not overlap --> the lines in the series cant overlap Assume the Rydberg constant for He is 9.72x10 nm, calculate the wavelength of He ions from n=3 to n=2. \[\dfrac{1}{\lambda}=R_{H} \left | \dfrac{1}{n^{2}_{i}}-\dfrac{1}{n^{2}_{f}} \right | \nonumber \] \[\dfrac{1}{\lambda}=9.72\times10^{-18} \left | \dfrac{1}{3^{2}}-\dfrac{1}{2^{2}} \right | \nonumber \] \[\lambda=7.41\times10^{17}nm\nonumber \] Derive from the following equation to solve for wavelength. \(\Delta E=hcR_{H}(\dfrac{1}{n_{i}^{2}}-\dfrac{1}{n_{f}^{2}})\) \[\Delta E=R_{H} \left | \dfrac{1}{n^{2}_{i}}-\dfrac{1}{n^{2}_{f}} \right | \nonumber \] \[\dfrac{v}{c}=\Delta E=R_{H} \left | \dfrac{1}{n^{2}_{i}}-\dfrac{1}{n^{2}_{f}} \right | \nonumber \] \[\dfrac{1}{\lambda}=\dfrac{v}{c}=\Delta E=R_{H} \left | \dfrac{1}{n^{2}_{i}}-\dfrac{1}{n^{2}_{f}} \right | \nonumber \] Photosynthesis occurs when light excites the electrons in the chloroplasts of leaves. If you flash a light of 500 nm at the plant and assuming your flashlight emits 4.5x10 numbers of photons, how much energy in joules was that one flash? What is an antiatom? How is it different from a regular hydrogen? What will happen if these two collided with one another? An antiatom is a hydrogen atom with reversed electrical charges compared to that of a hydrogen atom. The proton is called an antiproton and it bears a negative charge rather than a positive charge. The electron, or antielectron, has a positive charge rather than a negative charge. As a result, an antiatom and regular hydrogen would obliterate one another should they collide. Subsequent energy would then be released. A scientist determined the kinetic energy released electrons of cesium metal through a photoelectric experiment. Determine h and the \work function for cesium graphically from the following results: 0.37 First, convert wavelength to frequency and volt to energy in J. Then graph the values and determine the linear equation. The linear equation is \[y=4.7 \times 10^{-16} f - 1.89\nonumber \] Use the equation below: hf- =E (max kinetic energy) h=slope = y-intercept h= 4.7 X 10 = 1.89 Calculate the de Broglie wavelength of a Cl molecule at 300 K. \[ v_{rms} = \sqrt{\dfrac{3RT}{M}}\nonumber \] =((3)(8.314)(300)/(70.8X10 kg) = 325 m/s \[ \lambda = \dfrac{h}{mv_{rms}}\nonumber \] =(6.626X10 J*s)/[(70.8 amu)(1.66X10 kg/amu)(325 m/s)] = 1.73 X 10 m Consider a balloon with a diameter of \(2.5 \times 10^{-5}\; m\). What is the uncertainty of the velocity of an oxygen molecule that is trapped inside. Use the direct application of the uncertainty principle: \[ \Delta x \Delta p \ge \dfrac{h}{4 \pi}\nonumber \] Let's consider the molecule has an uncertainly that is \(\pm\, radius\) of the balloon. \[\Delta x = 1.3 \times 10^{-5}\; m\nonumber \] The uncertainty of the momentum of the molecule can be estimated via the uncertainty principle. \[ \Delta p =\dfrac{h}{4 \pi \Delta x} = \dfrac{(6.626 \times 10^{-34} J*s)}{(4\pi) (1.3 \times 10^{-5}\; m)} = 4.05 \times 10^{-30} kg\,m\,s^{-1}\nonumber \] This can be converted to uncertainty in velocity via \[p=mv\nonumber \] or \[ \Delta p =m_e \Delta v\nonumber \] with the electron mass \(m_e\) equal to \(9.109 \times 10^{-31}\; kg\). since the mass of the molecule is uncertain. \[ \Delta v = \dfrac{ \Delta p }{ m_e } = \dfrac{4.05 \times 10^{-30}}{9.109 \times 10^{-31}\; kg} = 4.5\; m/s\nonumber \] What conditions must be satisfied for the Schrodinger's equation for \(\psi\) to be an acceptable wavefunction. What are some examples of unacceptable wave functions? The equation for calculating the energies of the electron in a hydrogen atom or a hydrogenlike ion is given by En = -(2.18 x10^-18 J)Z^2(1/n^2), where Z is the atomic number of the element. One way to modify this equation for many-electron atoms is to replace Z with (Z- [sigma]), where [sigma] is a positive dimensionless quantity called the shielding constant. Consider the lithium atom as an example. The physical significance of sigma that it represents the extent of shielding that the two 1s electrons exert on each other. Thus the quantity (Z - [sigma]) is appropriated called the "effective nuclear change." Use the first ionization energy of lithium to calculate the value of [sigma]. Calculate the wavelength and frequency of an emitted gamma particle that has energy of 3.11 x 10^12 J mol^-1.

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The de Broglie wavelength is the wavelength, \(\lambda\), associated with a object and is related to its momentum and mass. In 1923, Louis de Broglie, a French physicist, proposed a hypothesis to explain the theory of the atomic structure. By using a series of substitution de Broglie hypothesizes particles to hold properties of waves. Within a few years, de Broglie's hypothesis was tested by scientists shooting electrons and rays of lights through slits. What scientists discovered was the electron stream acted the same was as light proving de Broglie correct. De Broglie derived his equation using well established theories through the following series of substitutions: De Broglie first used Einstein's famous equation relating : \[ E = mc^2 \label{0}\] with Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation: \[ E= h \nu \label{1}\] with Since de Broglie believed particles and wave have the same traits, he hypothesized that the two energies would be equal: \[ mc^2 = h\nu \label{2}\] Because real particles do not travel at the speed of light, De Broglie submitted velocity (\(v\)) for the speed of light (\(c\)). \[ mv^2 = h\nu \label{3}\] Through the equation \(\lambda\), de Broglie substituted \( v/\lambda\) for \(\nu\) and arrived at the final expression that relates wavelength and particle with speed. \[ mv^2 = \dfrac{hv}{\lambda} \label{4}\] Hence \[ \lambda = \dfrac{hv}{mv^2} = \dfrac{h}{mv} \label{5} \] A majority of problems are simple plug and chug via Equation \ref{5} with some variation of canceling out units Find the de Broglie wavelength for an electron moving at the speed of \(5.0 \times 10^6\; m/s\) (mass of an electron is \(9.1 \times 10^{-31}\; kg\)). \[ \lambda = \dfrac{h}{p}= \dfrac{h}{mv} =\dfrac{6.63 \times 10^{-34}\; J \cdot s}{(9.1 \times 10^{-31} \; kg)(5.0 \times 10^6\, m/s)}= 1.46 \times 10^{-10}\;m\] Although de Broglie was credited for his hypothesis, he had no actual experimental evidence for his conjecture. In 1927, Clinton J. Davisson and Lester H. Germer shot electron particles onto onto a nickel crystal. What they saw was the diffraction of the electron similar to waves diffraction against crystals (x-rays). In the same year, an English physicist, George P. Thomson fired electrons towards thin metal foil providing him with the same results as Davisson and Germer.

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This page provides an overview of how an isotope can affect the frequencies of the vibrational modes of a molecule. Isotopic substitution is a useful technique A diatomic molecule, as seen in Figure \(\Page {1}\), contains two atoms, which can either be composed of the same or different elements. It is easier to focus on these types of molecules when analyzing and calculating vibrational frequencies because they are simpler systems than polyatomic molecules. Whether or not the diatomic consists of the same or different elements, a diatomic molecule will have only one vibrational frequency. This singular normal mode is because of the diatomic's linear symmetry, so the only vibration possible occurs along the bond connecting the two atoms. Normal modes describe the possible movements/vibrations of each of the atoms in a system. There are many different types of that molecules can undergo, like stretching, bending, wagging, rocking, and twisting, and these types can either be out of plane, asymmetric, symmetric, or degenerate. Molecules have 3n possible movements due to their 3-dimensionality, where n is equal to the number of atoms in the molecule. Three movements are subtracted from the total because they are saved for the displacement of the center of mass, which keeps the distance and angles between the atoms constant. Another 3 movements are subtracted from the total because they are for the rotations about the 3 principle axes. This means that for nonlinear molecules, there are 3n – 6 normal modes possible. Linear molecules, however, will have 3n – 5 normal modes because it is not possible for internuclear axis rotation, meaning there is one less possible rotation for the molecule. This explain why diatomic molecules only have 1 vibrational frequency, because 3(2) – 5 = 1. Molecular vibrations are often thought of as masses attached by a spring (Figure \(\Page {2}\)), and Hook’s law can be applied \[ F=-kx \] where \[ k=\left (\dfrac{\partial^2V(r)}{\partial r^2} \right)_{r_{eq}}\] in which \( V(r)=\dfrac{1}{2}k(r-r_{eq}) \), which comes from incorporating Hook’s law to the . The diatomic molecule is thought of as two masses (m and m ) on a spring, they will have a reduced mass, µ, so their vibrations can be mathematically analyzed. \[ \mu=\dfrac{m_{1}m_{2}}{m_{1}+m_{2}}\] When an atom in a molecule is changed to an isotope, the mass number will be changed, so \(µ\) will be affected, but \(k\) will not (mostly). This change in reduced mass will affect the vibrational modes of the molecule, which will affect the vibrational spectrum. Vibrational energy levels, \(\nu_{e}\), are affected by both k and µ, and is given by \[ \nu_e=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{\mu}} \] These vibrational energy levels correspond to the peaks which can be observed in IR and Raman spectra. IR spectra observe the asymmetric stretches of the molecule, while Raman spectra observe the symmetric stretches. When an atom is replaced by an isotope of larger mass, µ increases, leading to a smaller \(\nu_{e}\) and a downshift (smaller wavenumber) in the spectrum of the molecule. Taking the diatomic molecule HCl, if the hydrogen is replaced by its isotope deuterium, µ is doubled and therefore \(\nu_{e}\) will be decreased by \(\sqrt{2}\). Deuterium substitution leads to an isotopic ratio of 1.35-1.41 for the frequencies corresponding to the hydrogen/deuterium vibrations. There will also be a decrease by \(\sqrt{2}\) in the band width and integrated band width for the vibrational spectra of the substituted molecule. Isotopic substitution will affect the entire molecule (to a certain extent), so it is not only the vibrational modes for the substituted atom that will change, but rather the vibrational modes of all the atoms of the molecule. The change in frequency for the atoms not directly invovled in the substitution will not display as large a change, but a downshift can still occur. When polyaniline (Figure \(\Page {4}\)) is fully deuterated, the vibrational peaks will downshift slightly. The following data was summarized from Quillard Changing hydrogen to deuterium leads to the largest effect in a vibrational spectrum since the mass is doubled. Other isotopic substitutions will also lead to a shift in the vibrational energy level, but because the mass change is not as significant, µ will not change by much, leading to a smaller change in \(\nu_{e}\). This smaller change in vibrational frequency is seen in the sulfur substitution of sulfur hexafluoride (Figure \(\Page {5}\):), from S to S. The frequencies as reported by These two examples show the consistency of downshifted vibrational frequencies for atoms substituted with an isotope of higher mass. Substituting atoms with isotopes has been shown to be very useful in determining normal mode vibrations of organic molecules. When analyzing the spectrum of a molecule, isotopic substitution can help determine the vibrational modes specific atoms contribute to. Those normal modes can be assigned to the peaks observed in the spectrum of the molecule. There are specific CH rocks and torsions, as well as CH bends that can be identified in the spectrum upon deuterium substitution. Other torsion bands from hydroxyl and amine groups can also be assigned when hydrogen is replaced with deuterium. Experimental data has also shown that using deuterium substitution can help with symmetry assignments and the identification of metal hydrides. Isotopic substitution can also be used to determine the force constants of the molecule. Calculations can be done using the frequencies of the normal modes in determining these values, based on both calculated frequencies and experimental frequencies. Researchers have also attempted to contribute peak shape changes and splits in peaks of vibrational spectra to naturally occurring isotopes in molecules. It has been shown, however, that the shape of a peak is not related to the size of the atom, so substitution to an atom of larger mass will not affect the peak shape in the molecule's spectrum. As previously stated, isotopic substitution of atoms of higher mass will not have a significant enough effect on the shifts in frequencies for the corresponding vibrations, so analyzing the frequency shifts of smaller mass isotopes, like deuterium and C is necessary. As depicted in the rough representation of the vibrational spectra of the molecule tetrachlorinated dibenzodioxin (TCDD), the C substituted TCDD spectrum is slightly downshifted compared to the unsubstituted TCDD spectrum. Although the shifts and split peaks do occur in the spectra of isotopically substituted molecules, not all observed peaks can be attributed to the isotope. This is because the intensities of the peaks shown are not large enough to relate to the natural abundance of the C isotope, and not all peaks can be accounted for by the substitution.

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In the ideal case, we have seen that the thermodynamic potential for species \(i\) can be written as: \[μ_i^{sln}=μ^o_i + RT \ln \,x_i=μ^o_2 + RT \ln \left(\dfrac{P_i}{P^o}\right) \nonumber \] One approach to non-ideality is to simply the problem and say: \[μ_i^{sln} \equiv μ^o_i + RT \ln\, a_i \nonumber \] ≡ indicates this is actually a definition. The newly defined variable a is known as the . Alternatively we can define it as: \[[a_i] \equiv \dfrac{P_i}{P^o} \nonumber \] As at high enough values of the mole fraction we know that we can still apply Raoult law. So \(a_i\) must approach \(x_i\) in this limit, but for other concentrations this will no longer hold. Often this is expressed in terms of an \(\gamma\): \[[a_i] = \gamma_i x_i \nonumber \] For high values of \(x_i\), \(\gamma_i\) will approach unity. If we model the non-ideality with a Margules function we see that: \[ P_i = x_iP^of_{Mar} \] \[ [a_i] = \left[\frac{P_i}{P^0}\right] = \left[\frac{x_iP^0f_{Mar}}{P^0}\right] = [x_if_{Mar}] \] The activity coefficient and the Margules function are the same thing in this description. A special and simplest case of a Margules function is the case where all but one Margules parameters (\(a\)) can be neglected. Such a system is called a In this case, we can write: \[ a_1 = x_1e^{ax_2^2} \] We can use Gibbs-Duhem to show that this : \[ a_2 = x_2e^{ax_1^2} \] Consider the change in Gibbs free energy when we mix two components to form a regular solution: \[ \Delta G_{mix} = n_1\mu_1^{sln}+n_2\mu_2^{sln} - (n_1\mu_1^* + n_2\mu_2^*) \] Dividing by the total number of moles, we get: \[ \Delta_{mix}G = x_1\mu_1^{sln}+x_2\mu_2^{sln} - (x_1\mu_1^* + x_2\mu_2^*) \] Using: \[ \mu_i^{sln} \equiv \mu_i^* + RT\ln{a_i} \] and: We get: \[ \frac{\Delta_{mix} G}{RT} = x_1\ln{x_1} + x_2\ln{x_2} + x_1\ln{\gamma_1} + x_2\ln{\gamma_2} \] For a regular solution: \[ \ln{\gamma_1} = \ln{f_{Mar}} = ax_2^2 \] \[ \ln{\gamma_2} = \ln{f_{Mar}} = ax_1^2 \] This gives: \[\begin{split} \frac{\Delta_{mix} G}{RT} &= x_1\ln{x_1} + x_2\ln{x_2} + x_1ax_2^2 + x_2ax_1^2 \\ &= x_1\ln{x_1} + x_2\ln{x_2} + a[x_1+x_2]x_1x_2 \\ &= x_1\ln{x_1} + x_2\ln{x_2} + ax_1x_2 \end{split}\] Since: \[ x_1 + x_2 = 1 \] In this expression we see that we have an additional term to the entropy of mixing term we had seen before. Its coefficient \(a\) is dimensionless but represents the fact that the (strong!) interactions between the molecules are different depending on who is the neighbor. In general. \(a\) can be written as \(W/RT\), where \(W\) represents an energy (enthalpy) that brings the difference in interaction energies into account. \(W\) does not depend strongly on temperature. We could look at \(W\) as the difference in average interaction energies: \[ W = 2U_{12} - U_{11} - U_{22} \] Rearranging we get: \[ \frac{\Delta_{mix} G}{W} = \frac{RT}{W} [x_1\ln{X_1}+x_2\ln{x_2}]+x_1x_2 \] The two terms will as a function of temperature. The mixing entropy will be more important at high temperatures, the interaction enthalpy at low temperatures. The entropy term has a at x =0.5, the enthalpy term a if W is positive. So, one tends to favor mixing, the other segregation and we will get a compromise between the two. Depending on the value of \(RT/W\) (read: temperature), we can either get one or two minima. This means that at low temperatures there will be a solubility limit of 1 into 2 and . At higher temperatures the two components can mix completely. At the transition between these two regimes we will have Notice that even though we used the vapor pressures of the gas to develop our theory, they are conspicuously absent from the final result. The same thing we said about melting points hold true here. Because we are dealing with the miscibility behavior of two condensed phases, the outcome should not depend very strongly on the total pressure of our experiment. Although in regular solutions the consolute point is predicted to be a maximum in temperature, we can find them as minima as well in practice. The nicotine-water system even has two consolute points, an upper and a lower one. When heating up a mixture of these we first observe mixing, then segregation and then mixing again. Obviously this behavior is far more complicated than we can describe with just one Margules parameter. What we said above about volumes simply being additive in the ideal case is no longer true here. \[ \left(\frac{\partial\Delta G_{mix}}{\partial P}\right)_T = \Delta V_\text{regular} \] \[ \left(\frac{\partial\Delta H_{mix}+RT(n_1\ln{[x_1]}+n_2\ln{[x_2]})}{\partial P}\right)_T = \Delta V_\text{mix} \] \[ \left(\frac{\partial\Delta H_{mix}}{\partial P}\right)_T = \Delta V_\text{mix} \] In general the enthalpy of mixing does depend on pressure as it is related to the interactions between the molecules in solution. (\(W\) depends on the distance between them). This means that partial molar volumes now become a function of composition and volume is no longer simply additive. Notice that the curves are around \(x=0.5\). This implies that it is as easy (or not) dissolving A into B as vice versa. In many cases this is not realistic. Many systems diverge more seriously from ideal behavior that the regular one. Up to a point we can model that by adding more terms to the Margules function. For example, adding a β-term undoes the symmetry (see example 24-7). However, many systems are so non-ideal that the Margules expression become unwieldy with too many parameters. For ideal solutions we have seen that there is a lense shaped two-phase region between the gas and the liquid phase. For non-ideal systems the two-phase region can attain different shapes. In many cases there is either a minimum or a maximum. As such a point the phase gap closes to a point that is known as an . It represents a composition of the liquid that boils . That means that the vapor and the liquid have the same composition for a change. Azeotropes impose an important limitation unto distillation: they represent the end point of a distillation beyond which we can not purify by this method. Another point to be made is that in the diagram with the consolute point we are assuming the pressure to be constant. If we lower the pressure this would affect the boiling points strongly: the whole gas-liquid gap would come down in temperature (see animation). The mixing behavior is only weakly affected. (The reason is that one involves the volume term of the gas, the other only of the liquid(s)). At lower pressures it is possible therefore that the consolute point is the gas-liquid gap. In other words: the mixtures will boil before they get a chance to mix. The boiling points will be lower there than for the pure compounds. There will be a composition for which the boiling point is at a minimum and where the mixture boils congruently (i.e. to a vapor with the same (overall) composition). The mutual solubility limits increase as temperature increases, just as happens in the critical mixing case, but that due to the competition from the vapor phase this process comes to an end at the eutectic temperature. At this temperature one liquid boils always completely, the other one in part. At the eutectic composition they both boil away simultaneously.

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The Eyring Equation, developed by Henry Eyring in 1935, is based on transition state theory and is used to describe the relationship between reaction rate and temperature. It is similar to the , which also describes the temperature dependence of reaction rates. However, whereas Arrhenius Equation can be applied only to gas-phase kinetics, the Eyring Equation is useful in the study of gas, condensed, and mixed-phase reactions that have no relevance to the collision model. The Eyring Equation gives a more accurate calculation of rate constants and provides insight into how a reaction progresses at the molecular level. The Equation is given below: \[ k = \dfrac{k_BT}{h}e^{-\left ( \frac{\bigtriangleup H^\ddagger}{RT} \right )}e^{ \left ( \frac{\bigtriangleup S^\ddagger}{R} \right )} \label{1} \] Consider a bimolecular reaction: \[A~+B~\rightarrow~C \label{2} \] \[K = \dfrac{[C]}{[A,B]} \label{3} \] where \(K\) is the equilibrium constant. In the transition state model, the activated complex AB is formed: \[A~+~B~\rightleftharpoons ~AB^\ddagger~\rightarrow ~C \label{4} \] \[K^\ddagger=\dfrac{[AB]^\ddagger}{[A,B]} \label{5} \] There is an energy barrier, called activation energy, in the reaction pathway. A certain amount of energy is required for the reaction to occur. The transition state, \(AB^\ddagger\), is formed at maximum energy. This high-energy complex represents an unstable intermediate. Once the energy barrier is overcome, the reaction is able to proceed and product formation occurs. The rate of a reaction is equal to the number of activated complexes decomposing to form products. Hence, it is the concentration of the high-energy complex multiplied by the frequency of it surmounting the barrier. \[\begin{eqnarray} rate~&=&~v[AB^\ddagger] \label{6} \\ &=&~v[A,B]K^\ddagger \label{7} \end{eqnarray} \] The rate can be rewritten: \[rate~=~k[A,B] \label{8} \] Combining Equations \(\ref{8}\) and \(\ref{7}\) gives: \[ \begin{eqnarray} k[A,B]~&=&~v[A,B]K^\ddagger \label{9} \\ k~&=&~vK^\ddagger \label{10} \end{eqnarray} \] where The frequency of vibration is given by: \[v~=~\dfrac{k_BT}{h} \label{11} \] where Substituting Equation \(\ref{11}\) into Equation \(\ref{10}\) : \[k~=~\dfrac{k_BT}{h}K^\ddagger \label{12} \] Equation \({ref}\) is often tagged with another term \((M^{1-m})\) that makes the units equal with \(M\) is the molarity and \(m\) is the molecularly of the reaction. \[k~=~\dfrac{k_BT}{h}K^\ddagger (M^{1-m}) \label{E12} \] The following thermodynamic equations further describe the equilibrium constant: \[ \begin{eqnarray} \Delta G^\ddagger~&=&~-RT\ln{K^\ddagger}\label{13} \\ \Delta G^\ddagger~&=&~\Delta H^\ddagger~-~T\Delta S^\ddagger \label{14} \end{eqnarray} \] where \(\Delta G^\ddagger\) is the Gibbs energy of activation, \(\Delta H^\ddagger\) is the and \(\Delta S^\ddagger\) is the . Combining Equations \(\ref{10}\) and \(\ref{11}\) to solve for \(\ln K ^\ddagger \) gives: \[\ln{K}^\ddagger~=~-\dfrac{\Delta H^\ddagger}{RT}~+~\dfrac{\Delta S^\ddagger}{R} \label{15} \] The Eyring Equation is finally given by substituting Equation \(\ref{15}\) into Equation \(\ref{12}\): \[ k~=~\dfrac{k_BT}{h}e^{-\frac{\Delta H^\ddagger}{RT}}e^\frac{\Delta S^\ddagger}{R} \label{16} \] The linear form of the Eyring Equation is given below: \[\ln{\dfrac{k}{T}}~=~\dfrac{-\Delta H^\dagger}{R}\dfrac{1}{T}~+~\ln{\dfrac{k_B}{h}}~+~\dfrac{\Delta S^\ddagger}{R} \label{17} \] The values for \(\Delta H^\ddagger\) and \(\Delta S^\ddagger\) can be determined from kinetic data obtained from a \(\ln{\dfrac{k}{T}}\) vs. \(\dfrac{1}{T}\) plot. The Equation is a straight line with negative slope, \(\dfrac{-\Delta H^\ddagger}{R}\), and a y-intercept, \(\dfrac{\Delta S^\ddagger}{R}+\ln{\dfrac{k_B}{h}}\).

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Thermodynamics puts constraints on the behavior of macroscopic systems without referencing the underlying microscopic properties. In particular, it does not provide a quantitative connection to the origin of its fundamental quantities \(U\) and \(S\). For \(U\), this is less of a problem because we know from mechanics that \[ U=\dfrac{1}{2} \sum m_i v_i^2 + V(x_i),\] and the macroscopic formula arises by integrating over most coordinates and velocities. Somehow the thermal motions end up as \(TS\), and the mechanical and electrical motions end up as terms such as \(–PV+\mu n\). Statistical mechanics makes the macro-micro connection and provides a quantitative description of U and S is terms of microscopic quantities. For large systems (except near the critical point), its results are in agreement with thermodynamics: one can derive thermodynamic postulates 0 – 3 from statistical mechanics. For systems undergoing large fluctuations (small systems or those systems near a critical point), its prediction are different and more accurate. In addition as the ‘mechanics’ implies, statistical mechanics can deal with time-varying systems and systems out of equilibrium. Averages over x(t) and p(t)=mv(t) of the microscopic particles are done, but not in such a way that all time-dependent information is lost, as in thermodynamics. Unlike mechanics, statistical mechanics is not intended to discuss the time-dependence of an isolated particle. Rather, the time-dependent (e.g. diffusion coefficient) and time independent properties of whole systems of particles, and the averaged properties of whole ensembles of such systems, are of interest. We begin with an introduction to important facts from mechanics and statistics, then proceed to the postulates of statistical mechanics, consider in detail equilibrium systems, and finally non-equilibrium systems. Hamiltonian dynamics applies to the density operator \(\hat{\rho}_{i}\) of any finite closed system; fully specified by its extensive constrainnt parameters and the Hamiltonian. The principle of equal probabilities holds in its ensemble (weak) form and is assumed in it strong (time) form i) weak form: all W microscopic realizations of a system satisfying I have equal probability. The ensemble density matrix is therefore given by \(\hat{\rho}=\frac{1}{W} \sum_{i=1}^{W} \hat{\rho}_{i}\). The ensemble of these \(W\) systems is the microcanonical ensemble. ii) strong form: for any ensemble satisfying i) at equilibrium, \(\left\langle\hat{\rho}_{i}\right\rangle_{t}=\left\langle\hat{\rho}_{i}\right\rangle_{e}=\hat{\rho}\) (ergodic principle). This states that averaging over time is equivalent to averaging over the ensemble of \(\mathrm{W}\) microstates. The entropy of a n ensemble of systems satisfying postuilates I and II.i) is given by \[S=-\operatorname{Tr}\{\hat{\rho} \ln \hat{\rho}\}\] Before we use them, these postulates require some explanation. I) This is a strong statement; the system i usually has a \(>10^{20}\)-dimensional phase space, and we assume that the dynamics are the same as for a few degrees of freedom! Classically, \(\hat{\rho}_{i}\) corresponds to a specific trajectory; quantum mechanically, to a specific initial condition of the system. Among the extensive variables fixed in a closed finite system: U (always by postulate I). Other constrained variables: \(\mathrm{V}\) (or \(\mathrm{L}, \mathrm{A}, \mathrm{N}_{\mathrm{i}}=\) particle number ... depending on the system). Note that if \(\hat{H}\) is independent of time, the system is closed, and \(U\) is therefore constant (as it needs to be for 'full' specification of the system), P1 of thermodynamics is automatically satisfied. II) This is the postulate that lets us perform macroscopic averages over the individual density matrices, so we can derive properties for the energyconserving (microcanonical) ensemble. i. Classically, this says that as long as a trajectory stratifies the constraints in I (has specific energy U), we can combine it with equal weight with all other such trajectories to obtain \(\rho\left(q_{i}, p_{i}, t\right)\), the classical density function. Quantum-mechanically, this means that all the linearly independent pure density matrices \(\hat{\rho}_{i}\) characterizing a system with the same extensive parameters (i.e. all the members of the microcanonical ensemble) can be averaged with equal weights to obtain the ensemble density matrix. consider a state of energy \(U\) that can be realized in \(W\) ways (W-fold degenerate or W microstates). One set of initial conditions \(\hat{\rho}_{i}\) would be \[\rho_{1}=\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right), \rho_{2}=\left(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right), \ldots\] These are pure states. All of these are equally likely because they have the same energy (and volume, etc.), so \[\hat{\rho}=\frac{1}{W} \sum_{i=1}^{W} \hat{\rho}_{i}=\left(\begin{array}{ccc} 1 / W & & 0 \\ & \ddots & \\ 0 & & 1 / W \end{array}\right)\] This is a 'mixed' sate of constant energy U. Note that there is a potentially embarrassing problem with this: a finite quantum system (e.g. particle in a box) for which all extensive parameters (e.g. \(U\), or \(L\) for particles in a 1-D box) have been specified has as discrete energy spectrum given by \(H\left|\varphi_{i}\right\rangle=E_{i}\left|\varphi_{i}\right\rangle .\) For a large system, the level spacing may be very narrow, but it is nonetheless discrete. Thus, at some every U we pick, there is likely to be no state, so we have nothing to average! In practice, this is resolved by having an energy window \(\delta U\), and by considering all W levels within it. As discussed in more detail in III below, as the number of degrees of freedom \(N=6 n\) of the system approaches infinity, the size of \(\delta U\) rigorously has no effect on the result. ii. This says we could take a single trajectory, or a single initial condition \(\hat{\rho}_{i}(0)\), propagate it in time, and all the possible microscopic states will also be visited in turn to yield again \(\rho\left(q_{i}, p_{i}, t\right)\) (classically) or \(\hat{\rho}(t)\) (quantum mechanically). This is a much stronger statement than i): the full ensemble of W microstates by definition includes all realizations i of the macroscopic system compatible with \(\mathrm{H}\) and the constraints; on the other hand ii) says a single microstate will, in time, evolve to explore all the others, or at least come arbitrarily close to them. This property is know as 'ergodicity.' In practice, ergodicity cannot really be satisfied, but we can use ii) for 'all practical purposes.' We will use a discrete system to illustrate. Consider a box with \(M=\frac{V}{V_{0}}\) cells, filled with \(N \ll M\) particles of volume \(V_{0}\). The dynamics is that the particles hop randomly to unoccupied neighboring cells at each time step \(\Delta t\). This model is crlled a hatice ideal gas. The number of arrangementsfor \(N\) identical particles \[W=\frac{M!}{(M-N) !} \cdot \frac{1}{N!}\] Large factorial n! (or gamma fuactious \(\Gamma (n-1)-N!)\) can be approximated by \[W=\frac{M!}{(M-N) !} \cdot \frac{1}{M!} \cdot M^{M}(M -N)^{N-M} M^{M}=\left(\frac{M}{N}\right)^{N}\] Let us plug realistic numbers into this \(V_{0}=10 \hat{A}_{o}\), V=1 cm , \( \rightarrow M=\frac{V}{V_{0}}=10^{23}\). For N ~ 10 gas molecules (~1 atm) \( \rightarrow\) M/N =10 \( v_{gas}=\left ( \frac{U_{o}}{m} \right )= 300 m/s \) (O at room temperature) \( \Delta t= \frac{L_{o}}{v} = \frac{V_{o}^{1/3}}{v} =10^{-12}s = 1\;ps \) \( W_{possible}= \frac{10^{12}s}{10^{18}s}=10^{30} \) Lifetime of universe: \( \leqslant 10^{11} a \ldots 10^{18} s\) \(W_{\text {actual }}\) = \(\left(10^{4}\right)^{10^{19}}=\) 1 googol \( \geqslant 10^{30}\) The possible \(W\) that can te visited during the lifetime of the universe is a mere 10 , negligible compared to the actual rnumber of microsta W at constant energy Clearly, not even a warm gas, a system about as random as conceivable, even touches the true microcanonical degeneracy W. Although the a priori probability of microstates (classically: of trajectories) may be the same (i), they simply cannot all be sampled in finite time as seen in (iii), this provides a practical solution to the quantum dilemma outlined in (i) Why assume (ii) at all? In real life \(\hat{\rho}_{i}(t) \) is always observed, but it is difficult to compute. W or \(\hat{\rho}(t) \) are often much easier to compute. Although (ii) fails by a factor, surprisingly it still works in most situations: most microstates in the ensemble of W microstates are indistinguishable(e.g. the gas atoms in the room right now vs 10 seconds from now), so leaving many of them out of the average still yields the same average; sampling only one in 10 still gives the same result as true ensemble averaging. There are cases where this reasoning fails: in glasses, members of the ensemble can be so slowly interconnecting and so different from one another, that \(\hat{\rho}\) is not at all like \(\left\langle\rho_{i}\right\rangle_{t}\) unless very special care is taken. III) This definition of the entropy was made plausible in our mathematical review, on grounds of information content: a system with many microstates has a greater potential for disorder than a system of a few microstates. But instead of measuring disorder multiplicatively, we want an additive (extensive) quantity. This postulate proves the microscopic definition for thermodynamic entropy \(\left(\hat{\rho}=\hat{\rho}_{e q} \& N \rightarrow \infty\right)\), just as energy is microscopically defined as \[U=\langle H\rangle_{e} \text { where } H=\frac{1}{2} \sum_{j} \frac{p_{j}^{2}}{m_{j}}+V\left(x_{i}\right)\] so \[S_{e q} \equiv S=-k_{B} T_{r}\left\{\hat{\rho}_{e q} \ln \hat{\rho}_{e q}\right\}\] gives the thermodynamic entropy \(S\) in terms of the equilibrium density matrix. We must have \(\operatorname{Tr}\left\{\hat{\rho}_{e q}\right\}=1,\left[\hat{\rho}_{e q}, H\right]=0\), and by postulate II.i), all elements of \(\hat{\rho}_{e q}\) must be of equal size if we are in the microcanonical (constant energy U) ensemble. This is satisfied only by \[\hat{\rho}_{e q}=\left(\begin{array}{ccc}\frac{1}{W} & & 0 \\0 & & \frac{1}{W}\end{array}\right)\] Where \(\hat{\rho}\) is a diagonal \(\mathrm{W} \times \mathrm{W}\) matrix. Inserting into \(\mathrm{S}\) and evaluating the trace in the eigenfunction basis of \(\hat{H}\) (and \(\hat{\rho}\) ), which we can call \(|j\rangle:\) \[\begin{aligned}&S=-k_{B} \sum_{j=1}^{W}\left\langle j\left|\hat{\rho}_{e q} \ln \hat{\rho}_{e q}\right| j\right\rangle=-k_{B} \sum_{j=1}^{W} \frac{1}{W} \ln \frac{1}{W} \\ &\Rightarrow S=k_{B} \ln W \quad \text { where } k_{B} \approx 1.38 \cdot 10^{-23} J / K \text { is Boltzmann's constant. } \end{aligned}\] This is Boltzmann's famous formula for the entropy. Postulate III is more general, but at equilibrium Boltzmann's formula holds. It secures for \(\mathrm{S}\) all the properties in postulates 2 and 3 of thermodynamics, and provides a microscopic interpretation for \(\mathrm{S}\) : W specifies disorder in a system: the more possible microstates correspond to the same macrostate, the more disorder a system has. For two independent systems, \(W_{\text {tot }}=W_{\mathrm{i}} \cdot W_{2} .\) However, thermodynamic entropy has the property of additivity: \(S_{\text {tot }}=S_{1}+S_{2}\). The function that uniquely effects the transformation from multiplication to addition is the log function (within a constant factor) \(\Rightarrow S_{i}=\ln W_{i}\) must be true so that both relations at the beginning of this paragraph are satisfied. The constant factor \(k_{\mathrm{B}}\) is provided to match the energy and temperature scales, which were independently defined in the early \(19^{\text {th }}\) century when the equivalence of temperature and average energy was not understood. Consider a system divided into subsystems by constraints, with \(W=W_{0} .\) When the constraints are removed at \(t=0\), then by II.ii) the system now explores additional ensemble members as time goes on. Thus \(\mathrm{W}(\mathrm{t}>0)=\mathrm{W}_{1}>\mathrm{W}_{0}\). If macroscopic equilibrium is reached \(W_{e q}>W_{1}>W_{0} \Rightarrow S_{e q}>S_{0}\). Thus S in stat mech postulate III satisfies all requirements of postulate P2 of thermodynamics, which is a simple consequence of the fact that microscopic degrees of freedom tend to explore all available states (= all the available phase space in classical mechanics). Also, because \(W\) is monotonic in \(U\) (at higher energy \(U\), there are always more quantum states in a multidimensional system) and because \(S\) is monotonic in \(W\) (property of the ln function), \(S\) is monotonic in \(U\). finally, we shall see in detail later that when \(\left(\frac{\partial U}{\partial S}\right)_{V, N}=T\), only the ground state is populated, so \(W \rightarrow 1 \Rightarrow \lim _{T \rightarrow 0} S=k_{B} \ln (1)=0\). Thus, the third postulate is also satisfied as long as the ground state is singly degenerate and the system can get to it during the experiment. (Glasses again would be a problem here!) The error of thermodynamics: it identifies the most probable value of a quantity with its average, by assuming the spread is negligible. We will derive examples of this spread later on. Thermodynamic limit: \(\mathrm{N}\) goes to infinity but \(\mathrm{N} / \mathrm{V}\) or any other ratio of extensive quantities remains constant. To conclude this chapter, we turn to the problem of computing \(W\) in the quantum case. A closed finite quantum system has a discrete spectrum \(\mathrm{E}_{i}\). The figure below shows the number of states below energy \(U\) as a function of \(U\). Because of quantum mechanics, the density of states \[\Omega(U)=\frac{\partial \tilde{\Omega}}{\partial U}\] is discontinuous, and the integrated density of states (= total number of states up to energy \(U\) ) has steps in it: \[\tilde{\Omega}=\sum_{j} \operatorname{Step}\left(U-E_{j}\right) \Rightarrow \Omega=\sum_{j} \delta\left(U-E_{j}\right)\] At any randomly picked \(U, \Omega\) is mostly likely zero, so \(W=0\) also! However, because \(\Omega\) increases so enormously rapidly with energy, the states are very (understatement!) closely spaced in energy for any system with even just a few particles. If a system is observed for a finite time \(\delta t\), the states are broadened by the uncertainty principle: \[\delta E \sim \frac{\hbar}{2 \delta t} \Rightarrow \Omega=\sum_{i} L_{i}\left(U-E_{i}, \delta t\right)\] where \(L\) indicates a broadened profile of finite width that replaces the delta function. L still counts a single state, so \(\int_{0}^{\infty} d U L_{i}\left(U-E_{i}, \delta t\right)=1 ;\) often \(L_{i}\) is taken as a Lorentzian \(L_{i}=\frac{1}{\pi} \frac{\delta U}{\left(U-E_{i}\right)^{2}+\delta U^{2}} .\) Thus, \(\Omega\) can be taken as a smooth function and its value \(\Omega(E, \delta t)=W(U)\) tells us how many states contribute to the degeneracy at energy U. It is clear from the above that if \(\delta U \gg\left|E_{j}-E_{i}\right|\), (the broadening is greater than the spacing of adjacent levels), then \(\Omega(U)\) is indeed independent of the choice of \(\delta U\) or \(\delta t\). This is guaranteed by the astronomical number of states for a macroscopic system (see example in II.ii)). Because \(\tilde{\Omega}\) in the above figure grows so fast, \(\Omega(U) \delta U \approx \tilde{\Omega}(U)\), as illustrated in the bottom right panel of the figure Another way to look at it is in state space (classically: action space). It has \(\mathrm{N}\) coordinates for \(\mathrm{N}\) degrees of freedom. \(\tilde{\Omega}\) is the number of states under the surface \(\mathrm{U}\) \(=\) constant. If \(U \gg U_{0}\) (where \(U_{0}\) is the average characteristic energy step for one degree of freedom) then \[\tilde{\Omega}_{1} \sim\left(\frac{U}{U_{0}}\right)^{N}\] Letting \(\delta U\) now be an uncertainty in U instead of in individual energy levels, the number of states in the interval \((U-\delta U, U)\) is \[ \tilde{\Omega}(U)-\tilde{\Omega}(U-\delta U) \sim\left(\frac{U}{U_{0}}\right)^{N}-\left(\frac{U-\delta U}{U_{0}}\right)^{N} \approx\left(\frac{U}{U_{0}}\right)^{N}\left\{1-\left[1-\frac{\delta U}{U}\right]^{N}\right\} \approx\left(\frac{U}{U_{0}}\right)^{N} \] Because \(\mathrm{N} \sim 10^{20}\), as long as \(\delta U<U\) (even if only a small amount!), the number of states in any width shell \(\delta U\) is the same as the total number of states up to U: Thus \(\tilde{\Omega}(U) \approx \Omega(U, \delta U) \approx W(U)\) to extreme precision. This topic will be taken up once more in the examples of microcanonical calculations given in the next chapter.

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Gallium is the chemical element with the atomic number 31 and symbol Ga on the periodic table. It is in the Boron family (group 13) and in period 4. Gallium was discovered in 1875 by Paul Emile Lecoq de Boisbaudran. Boisbaudran named his newly discovered element after himself, deriving from the Latin word, “Gallia,” which means “Gaul.” Elemental Gallium does not exist in nature but gallium (III) salt can be extracted in small amounts from bauxite and zinc ores. Also, it is known for liquefying at temperatures just above room temperature. Gallium is one of the elements originally predicted by Mendeleev in 1871 when he published the first form of the periodic table. He dubbed it ekaaluminum, indicating that it should have chemical properties similar to aluminum. The actual metal was isolated and named (from the Latin Gallia, for France) by Paul-Emile Lecoq de Boisbaudran in 1875. The detective work behind the isolation of gallium depended on the recognition of unexpected lines in the emission spectrum of a zinc mineral, sphalerite. Eventual extraction and characterization followed. Today, most gallium is still extracted from this zinc mineral. Although once considered fairly obscure, gallium became an important commercial item in the '70s with the advent of gallium arsenide LEDs and laser diodes. At room temperature gallium is as soft as lead and can be cut with a knife. Its melting point is abnormally low and it will begin to melt in the palm of a warm hand. Gallium is one of a small number of metals that expands when freezing. Gallium has a few notable characteristics which are summarized below: Gallium usually cannot be found in nature. It exists in the earth's crust, where its abundance is about 16.9 ppm. It is extracted from bauxite and sometimes sphalerite. Gallium can also be found in coal, diaspore and germanite. : While Gallium can be found in the human body in very small amounts, there is no evidence for it harming the body. In fact, Gallium (III) salt is used in many pharmaceuticals, used as treatment for hypercalcemia, which can lead to growth of tumors on bones. Further, it has even been suggested that it can be used to treat cancer, infectious disease, and inflammatory disease. However, exposure to large amounts of Gallium can cause irritation in the throat, chest pains, and the fume it produces can lead to very serious conditions. : Roughly 90-95% of gallium consumption is in the electronics industry. In the United States, Gallium arsenide (GaAs) and gallium nitride (GaN) represent approximately 98% of the gallium consumption. Gallium arsenide (GaAs) can convert light directly into electricity. Further, gallium arsenide is also used in LEDs and transistors. Gallium has the property to wet porcelain and even glass surfaces. As a result, gallium can be used to create dazzling mirrors. Scientists employ an alloy with Gallium for the plutonium pits of nuclear weapons to stabilize the alloptropes of plutonium. As a result, some have issue with the element.

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Phase transition is when a substance changes from a solid, liquid, or gas state to a different state. Every element and substance can transition from one phase to another at a specific combination of temperature and pressure. Each substance has three phases it can change into; solid, liquid, or gas . Every substance is in one of these three phases at certain temperatures. The temperature and pressure at which the substance will change is very dependent on the intermolecular forces that are acting on the molecules and atoms of the substance . There can be two phases coexisting in a single container at the same time. This typically happens when the substance is transitioning from one phase to another. This is called a two-phase state . In the example of ice melting, while the ice is melting, there is both solid water and liquid water in the cup. There are six ways a substance can change between these three phases; melting, freezing, evaporating, condensing, sublimination, and deposition . These processes are reversible and each transfers between phases differently: There are two variables to consider when looking at phase transition, pressure (P) and temperature (T). For the gas state, The relationship between temperature and pressure is defined by the equations below: Ideal Gas Law: \[ PV=nRT\] van der Waals Equation of State: \[ \left(P+a*\frac{n^2}{V^2}\right)\left(V-nb\right)=nRT\] Where V is volume, R is the gas constant, and n is the number of moles of gas. The ideal gas law assumes that no intermolecular forces are affecting the gas in any way, while the van der Waals equation includes two constants, a and b, that account for any intermolecular forces acting on the molecules of the gas. Temperature can change the phase of a substance. One common example is putting water in a freezer to change it into ice. In the picture above, we have a solid substance in a container. When we put it on a heat source, like a burner, heat is transferred to the substance increasing the kinetic energy of the molecules in the substance. The temperature increases until the substance reaches its melting point . As more and more heat is transferred beyond the melting point, the substance begins to melt and become a liquid . This type of phase change is called an isobaric process because the pressure of the system stays at a constant level. Each substance has a melting point. The melting point is the temperature that a solid will become a liquid. At different pressures, different temperatures are required to melt a substance. Each pure element on the periodic table has a normal melting point, the temperature that the element will become liquid when the pressure is 1 atmosphere . Each substance also has a boiling point. The boiling point is the temperature that a liquid will evaporate into a gas. The boiling point will change based on the temperature and pressure. Just like the melting point, each pure element has a normal boiling point at 1 atmosphere . Pressure can also be used to change the phase of the substance. In the picture above, we have a container fitted with a piston that seals in a gas. As the piston compresses the gas, the pressure increases. Once the boiling point has been reached, the gas will condense into a liquid. As the piston continues to compress the liquid, the pressure will increase until the melting point has been reached. The liquid will then freeze into a solid. This example is for an isothermal process where the temperature is constant and only the pressure is changing. Phase transition can be represented with a phase diagram. A phase diagram is a visual representation of how a substance changes phases. This is an example of a phase diagram. Often, when you are asked about a phase transition, you will need to refer to a phase diagram to answer it. These diagrams usually have the normal boiling point and normal melting point marked on them, and have the pressures on the y-axis and temperatures on the x-axis. The bottom curve marks the temperature and pressure combinations in which the substance will subliminate . The left left marks the temperature and pressure combinations in which the substance will melt . Finally, the right line marks the conditions under which the substance will evaporate . 1. Using the phase diagram for carbon dioxide below, explain what phase carbon dioxide is normally in at standard temperature and pressure, 1 atm and 273.15 K. Phase diagram for CO2.from . 2: Looking at the same diagram, we see that carbon dioxide does not have a normal melting point or a normal boiling point. Explain what kind of a change carbon dioxide makes at 1 atm and estimate the temperature of this point. 1: Before we can completely answer the question, we need to convert the given information to match the units in the diagram. First we convert 25 degrees Kelvin into Celsius: \(K=273.15+C\) \[ 298.15-273.25C\] Now we can look at the diagram and determine its phase. At 25 degrees Celsius and 1 atm carbon dioxide is in the gas phase. 2: Carbon dioxide sublimes at 1 atm because it transitions from the solid phase directly to the gas phase. The temperature of sublimation at 1 atm is about -80 degrees Celsius.

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There are numerous reactions in organic chemistry that proceed through cyclic transition states. They may be classified generally as reactions. An important and familiar example is the Diels-Alder reaction, in which a conjugated diene cycloadds to an alkene or alkyne: This reaction has been described previously ( ) and is an example of a [4 + 2] cycloaddition. Such reactions occur thermally (by simply heating the reactants) and appear to be entirely concerted. By this we mean that the reactants are converted to products , without involving the formation of reaction intermediates. The principal evidence for the concertedness of [4 + 2] cycloadditions is the fact that they are highly stereospecific and involve suprafacial addition of both components. The configuration of substituents in the diene and the dienophile is retained in the adduct: In contrast to the [4 + 2] cycloaddition, thermal [2 + 2] cycloadditions seldom are observed, and when they are observed, they are not stereospecific and evidently are stepwise reactions (see ): Why are the [4 + 2] and [2 + 2] cycloadditions different? Simple molecular orbital theory provides an elegant explanation of this difference based on the \(4n + 2\) rule described in . To understand this, we need to look in more detail at how the \(p\) orbitals of the double bonds interact in concerted addition mechanisms by suprafacial overlap, as in \(36\) and \(37\): There is a way around the \(4n + 2\) rule that is not very important for substances analogous to benzene, but is quite important for cycloaddition reactions. Let us see how this works for a cyclic conjugated polyene. From the molecular-orbital diagrams of Figures 21-5, 21-7, 21-9, and 21-14, you will see that the -energy \(\pi\) molecular orbital has nodes (changes of phase). A model of such an orbital, which usually is called a , can be constructed by joining the ends of a ribbon or strip of parallel \(p\) orbitals, as represented on the left side of Figure 21-15. However, one could join the orbitals by making in the strip, which then would give a lowest-energy orbital with one node, as on the right side of Figure 21-15. A strip with one such twist is called a \(^8\) and has the topological property of having only one side. If we now calculate the orbital energies for the Möbius orbitals, as was down for the normal Hückel \(\pi\) orbitals in Figure 21-13, we get the results shown in Figure 21-16. From this, we see that the \(4n\) situation now is favored and \(4n + 2\) is unfavorable. Whereas the energies of the \(\pi\) molecular orbitals in the Hückel arrangement can be obtained by inscribing a polygon in a circle with a ( ), in the Möbius arrangement the orbital energies are obtained from the polygon inscribed with a . If you compare the orbital energies of the Hückel and Möbius cyclic \(\pi\) systems (Figures 21-13 and 21-16), you will see that the Hückel systems have only lowest-energy MO, whereas the Möbius systems have . Hückel systems have an odd number of bonding orbitals (which, when full, accommodate 2, 6, 10, 14, or \(4n + 2\) electrons) and the Möbius systems have an even number of bonding orbitals (which, when full, accommodate 4, 8, 12, or \(4n\) electrons). The Hückel molecular orbitals have or an of nodes (see, for example, the benzene MOs, Figure 21-5); the Möbius molecular orbitals are not shown, but they have or an of nodes. The relevance of all this may seem tenuous, especially because no example of a simple cyclic polyene with a Möbius \(\pi\) system is known. However, the Möbius arrangement is relevant to cycloaddition because we can conceive of alkenes, alkadienes, and so on approaching each other to produce Möbius transition states when \(4n\) electrons are involved. For example, consider two molecules of ethene, which we showed previously would violate the \(4n + 2\) rule by undergoing cycloaddition through a transition state represented by \(37\). There is an alternative transition state, \(38\), in which the four \(p\) orbitals come together in the Möbius arrangement (with one node for minimum energy). To achieve this arrangement the ethene molecules approach each other in roughly perpendicular planes so that the \(p\) orbitals overlap suprafacially in one ethene and antarafacially in the other, as shown in \(38\): This pathway is electronically favorable, but the steric interference between the groups attached to the double bond is likely to be severe. Such repulsions can be relieved if there are no groups sticking out sidewise at one end of the double bond, as with the central carbon of 1,2-propadiene, \(\ce{CH_2=C=CH_2}\), and ketene, \(\ce{CH_2=C=O}\). These substances often undergo [2 + 2] cycloadditions rather readily ( ), and it is likely that these are concerted additions occurring by the Möbius route. A much less strained Möbius [4 + 4] transition state can be formed from two molecules of 1,3-butadiene. When 1,3-butadiene is heated by itself, a few percent of 1,5-cyclooctadiene is formed, but it is not known for sure whether the mechanism is that shown: The principal reaction is a Diels-Alder [4 + 2] cycloaddition, with butadiene acting both as a diene and as a dienophile: Much of what we have said about the electronic factors controlling whether a cycloaddition reaction can be concerted or not originally was formulated by the American chemists R. B. Woodward and R. Hoffmann several years ago, in terms of what came to be called the principles, or the . Orbital symmetry arguments are too complicated for this book, and we shall, instead, use the \(4n + 2\) electron rule for normal Hückel arrangements of \(\pi\) systems and the \(4n\) electron rule for Möbius arrangements. This is a particularly simple approach among several available to account for the phenomena to which Woodward and Hoffmann drew special attention and explained by what they call "conservation of orbital symmetry". The cycloaddition reactions that we have discussed so far in this chapter ([2 + 2], [4 + 2], etc.) have involved ring formation by bringing two unsaturated molecules together. Thus [4 + 2] addition is represented by the Diels-Alder reaction of ethene and 1,3-butadiene: We can conceive of similar cyclizations involving only single molecules, that is, . Such reactions are called . Two examples follow to show cyclization of a diene and a triene: Cyclization of 1,3,5-hexatriene occurs only when the central double bond has the cis configuration. The reaction is reversible at elevated temperatures because of the gain in entropy on ring opening (see ). The cyclobutene-1,3-butadiene interconversion proceeds much less readily, even in the thermodynamically favorable direction of ring opening. However, substituted dienes and cyclobutenes often react more rapidly. A related group of reactions involves shifts of substituent groups from one atom to another; for example, with \(\ce{H}\), alkyl, or aryl groups as \(\ce{R}\): These reactions are called and, in general, they are subject to the \(4n + 2\) rule and the Möbius orbital modification of it. Potential sigmatropic rearrangements can be recognized by the fact that the single bond to the migrating group \(\left( \ce{R} \right)\) is "conjugated" with the \(\pi\) bonds, and the group moves from a saturated \(sp^3\) to an \(sp^2\) carbon at a different part of the \(\pi\) system. A striking feature of thermal electrocyclic reactions that proceed by concerted mechanisms is their high degree of stereospecificity. Thus when -3,4-dimethylcyclobutene is heated, it affords only one of the three possible cis-trans isomers of 2,4-hexadiene, namely, , -2,4-hexadiene: We can see how this can occur if, as the ring opens, the ends of the diene twist in the direction (\(\curvearrowright \curvearrowright\) or \(\curvearrowleft \curvearrowleft\), ) as indicated in the equation. You will notice that with this particular case, if conrotation occurs to the left, rather than the right, the same final product results: The conrotatory movement of groups is typical of thermal ring openings of cyclobutenes and other rings involving \(4n\) electrons. When a cyclobutene is so constituted that conrotation cannot occur for steric reasons, then the concerted reaction cannot occur easily. Substances that otherwise might be predicted to be highly unstable often turn out to be relatively stable. An example is bicyclo[2.1.0]-2-pentene, which at first sight might seem incapable of isolation because of the possibility of immediate arrangement to 1,3-cyclopentadiene. This rearrangement does occur, but not so fast as to preclude isolation of the substance: How can we explain the fact that this substance can be isolated? The explanation is that, if the reaction has to be conrotatory, then the product will not be ordinary 1,3-cyclopentadiene, but , -1,3-cyclopentadiene - surely a very highly strained substance. (Try to make a ball-and-stick model of it!) This means that the concerted mechanism is not favorable: It is of great interest and importance that, with systems of \(4n + 2\) electrons, the groups move in directions (\(\curvearrowleft \curvearrowright\) or \(\curvearrowright \curvearrowleft\), ). For example, In this case, the disrotation of the groups one another would lead to the cis,cis,cis product. Because this product is not formed, it seems likely that rotation of the methyl groups toward each other must be sterically unfavorable: How can we account for the stereoselectivity of thermal electrocyclic reactions? Our problem is to understand why it is that concerted \(4n\) electrocyclic rearrangements are conrotatory, whereas the corresponding \(4n + 2\) processes are disrotatory. From what has been said previously, we can expect that the conrotatory processes are related to the Möbius molecular orbitals and the disrotatory processes are related to Hückel molecular orbitals. Let us see why this is so. Consider the electrocyclic interconversion of a 1,3-diene and a cyclobutene. In this case, the Hückel transition state ( ) is formed by , but is unfavorable with four (that is, \(4n\)) electrons: In contrast, the Möbius transition state ( ) is formed by and is favorable with four \(\left( 4n \right)\) electrons: You will notice that the ring closure of a 1,3-diene through the favorable Möbius transition state may appear to be able to form only an arrangement of the overlapping \(\sigma\) orbitals, which would correspond to a high-energy cyclobutene. In fact, the normal cyclobutene would be formed, because on the way down from the transition state, the phases of the orbitals that will become the \(\sigma\) bond change to give the arrangement of the \(\sigma\) orbitals expected for the ground state. The reverse occurs in ring opening so that this reaction also can go through the favorable Möbius transition state. The same reasoning can be extended to electrocyclic reactions of 1,3,5-trienes and 1,3-cyclohexadienes, which involve \(4n + 2\) electrons and consequently favor Hückel transition states attained by . The three principal types of pericyclic reactions are , , and : The factors that control if and how these cyclization and rearrangement reactions occur in a concerted manner can be understood from the aromaticity or lack of aromaticity achieved in their transition states. For a concerted pericyclic reaction to be thermally favorable, the transition state must involve \(4n + 2\) participating electrons if it is a Hückel orbital system, or \(4n\) electrons if it is a Möbius orbital system. A Hückel transition state is one in which the cyclic array of participating orbitals has no nodes (or an even number) and a Möbius transition state has an odd number of nodes. We summarize here a procedure to predict the feasibility and the stereochemistry of reactions involving . The 1,2 rearrangement of carbocations will be used to illustrate the approach. This is a very important reaction of carbocations which we have discussed in other chapters. We use it here as an example to illustrate how qualitative MO theory can give insight into how and why reactions occur: The first step of the procedure is to draw the orbitals as they are expected to be involved in the transition state. There may be several possible arrangements. There are two such arrangements, \(41\) and \(42\), for the rearrangement of carbocations; the dotted lines show the regions of bond-making and bond-breaking (i.e., orbital overlap): The second step is to determine whether the transition states are Hückel or Möbius from the number of nodes. This is readily done by assigning signs to the lobes of the orbitals corresponding to their phases and counting the number of nodes that develop in the circle of overlapping orbitals. An odd number denotes a Möbius transition state, whereas an even number, including zero, denotes a Hückel transition state. There are alternative ways of node-counting for transition states \(41\) and \(42\). Diagrams \(43abc\) and \(44abc\) represent molecular orbitals of different energies - those with more nodes having the higher energies (cf. ).\(^9\) We show these diagrams with more than one node for the sake of completeness. It is not necessary to draw more than one such diagram to determine whether the transition state is Möbius or Hückel. Finally, we evaluate the transition states according to the \(4n\) or \(4n + 2\) rule. In the example here, because only two electrons occupy the molecular orbitals, the Hückel transition state (\(43a\)) is the favorable one. A bonus coming from these formulations is that the stereochemistry of the reaction can be predicted when we have predicted which transition state is the favored one. Thus the migrating group in 1,2-carbocation rearrangements should move with of configuration by a Hückel transition state - and this has been verified experimentally. The alternative Möbius transition state predicts of the configuration of the migrating group: You can use the procedures just outlined to determine whether any thermal reaction with a cyclic transition state is likely to be favorable. A good place to start is the Diels-Alder [4 + 2] cycloaddition, which proceeds thermally by a suprafacial (Hückel) transition state. We suggest that you apply the procedure to the Diels-Alder reaction of 1,3-butadiene and ethene, and following that, show the electrocyclic ring opening of a cyclobutene ring to be thermally favorable only by a conrotatory opening of the \(\ce{C-C}\) bond. Many pericyclic reactions take place photochemically, that is, by irradiation with ultraviolet light. One example is the conversion of norbornadiene to quadricyclene, described in . This reaction would have an unfavorable [2 + 2] mechanism if it were attempted by simple heating. Furthermore, the thermodynamics favor ring opening rather than ring closure. However, quadricyclene can be isolated, even if it is highly strained, because to reopen the ring thermally involves the reverse of some unfavorable [2 + 2] cycloaddition mechanism. Photochemical activation can be used to achieve forward or reverse cycloadditions and electrocyclic reactions that are thermodynamically unfavorable or have unfavorable concerted thermal mechanisms. Thus the thermodynamically unstable disrotatory [2 + 2] product can be obtained from 1,3-cyclopentadiene by irradiation with ultraviolet light: The stereochemical results of electrocyclic and cycloaddition reactions carried out photochemically often are opposite to what is observed for corresponding thermal reactions. However, exceptions are known and the degree of stereospecificity is not always as high as in the thermal reactions. Further examples of photochemical pericyclic reactions are given in . \(^8\)Named after the mathematician A. F. Möbius. \(^9\)The assignment of orbital phases must take appropriate account of molecular symmetry, and although this is easy for open-chain systems, it is much less straightforward for cyclic ones. You usually will be able to avoid this problem by always trying to set up the orbitals so that the transition state will have no nodes, or just one node at a point where a bond is being made or broken. and (1977)

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The universe is made up of bits of matter, and the bits of matter have mass and energy. Neither mass nor energy may be created or destroyed, but in some special cases interconversions between mass and energy may occur. The things that happen in nature occur because matter is trying to gather to itself as much mass as it can (gravity) and to release as much of its energy as it can (the Sun, for example). This gathering of mass and loss of energy requires a price to be paid - the price of freedom. It seems that mass, energy, and freedom are the coins of the universe. Mass and energy must be conserved: if one part of the universe gathers more "wealth" as mass or energy, those must come from somewhere else in the universe, so there is always competition in the trading of these commodities. Freedom, on the other hand, is not conserved. However, there are some strange restraints on freedom. The freedom of the universe can take many forms, and appears to be able to increase without limit. However, the total freedom of the universe is not allowed to decrease. The energy or the mass of a part of the universe may increase or decrease, but only if there is a corresponding decrease or increase somewhere else in the universe. The freedom in that part of the universe may increase with no change in the freedom of the rest of the universe. There might be decreases in freedom in the rest of the universe, but the sum of the increase and decrease must result in a net increase. There can be a decrease in the freedom in one part of the universe, but ONLY if there is an equal or greater increase in the rest of the universe. There can be a decrease in the freedom in one part of the universe, but if there is an equal or greater increase in the rest of the universe Most of us have a general idea of what mass and energy are, and we may have a fair understanding of how we can quantify them, or to say how much of them we have. Freedom is a more complicated concept. The freedom within a part of the universe may take two major forms: the freedom of the mass and the freedom of the energy. The amount of freedom is related to the number of different ways the mass or the energy in that part of the universe may be arranged while not gaining or losing any mass or energy. We will concentrate on a specific part of the universe, perhaps within a closed container. If the mass within the container is distributed into a lot of tiny little balls (atoms) flying blindly about, running into each other and anything else (like walls) that may be in their way, there is a huge number of different ways the atoms could be arranged at any one time. Each atom could at different times occupy any place within the container that was not already occupied by another atom, but on average the atoms will be uniformly distributed throughout the container. If we can mathematically estimate the number of different ways the atoms may be arranged, we can quantify the freedom of the mass. If somehow we increase the size of the container, each atom can move around in a greater amount of space, and the number of ways the mass may be arranged will increase. Now let us turn to the freedom of the energy within the container. If the mass is in the form of atoms flying around, energy is only in the form of the kinetic energy of these atoms, and the energy of the electrons in the atoms moving around their nucleus (and sometimes we have to consider the neutrons, protons, and other stuff moving around in the nucleus). An atom's kinetic energy is related to its mass and its velocity. The energy in the container is the sum of the kinetic energies of all of the atoms, but the velocity is not the same for each atom, and the atoms are continually exchanging this energy through collisions with each other and through collisions with the walls. In the same way that the mass may have freedom in the number of ways the atoms may be arranged in space, the energy may have freedom in the number of ways that the velocities and directions of the atoms may be arranged. The velocities of the molecules are closely related to the temperature. The energy and freedom of gaseous atoms appear only in velocities and directions, and is called translational energy and translational freedom. In the case of molecules (a molecule is a group of two or more atoms held together by chemical bonds) there are additional freedoms and additional forms of energy. Bonds between atoms act as springs allowing the atoms to vibrate within the molecule, so that the molecules may contain different levels of vibrational energy and vibrational freedom. Additionally, the entire molecule may rotate on different axes, allowing different levels of rotational energy and rotational freedom. While the freedom of mass is related to the volume in which the mass is distributed, the freedom of energy is related to the temperature. An increase in the temperature of a gas leads directly to an increase in energy, and this can only occur if there is a decrease in energy somewhere else in the universe. We say that energy is transferred to the gas and the container (we will call that the "system")) from somewhere in the remainder of the universe (we will call that the "surroundings"). The increase in energy is accompanied by an increase in the energetic freedom of the system. The thermodynamic term for quantifying freedom is entropy, and it is given the symbol \(S\). Like freedom, the entropy of a system increases with the temperature and with volume. The effect of volume is more easily seem in terms of concentration, especially in the case of mixtures. For a certain number of atoms or molecules, an increase in volume results in a decrease in concentration. Therefore, the entropy of a system increases as the concentrations of the components decrease. The part of entropy which is determined by energetic freedom is called , and the part that is determined by concentration is called entropy. The units of entropy are the same as those of heat capacity and of the gas law constant. The product of entropy (or a change in entropy) and the absolute temperature has the same units as energy (or a change in energy). If the temperature of a gas is increased while the volume remains constant, the energy of the gas is increased and there is an increase in energetic freedom or thermal entropy. There is no change in concentration, and no change in configurational entropy. If the volume of the gas is increased while the temperature remains constant, the energy of the gas does not change and there is no change in thermal entropy. The increase in volume lowers the concentration and there is an increase in configurational entropy. → The opening paragraph stated that matter tends to draw more matter to itself and tries to reduce its energy. There is also the tendency to increase its freedom and thus its entropy. There is a natural conflict between these tendencies to lower energy and increase entropy, since a reduction in energy is usually accompanied by a reduction in freedom and entropy. For changes in which the initial and final temperatures are the same, these are combined into a net tendency for a system to change, in which the symbol \(U\) is used for energy and \(T\) is the absolute temperature. \[ΔU - TΔS \label{eq10}\] in which \(ΔU\) and \(ΔS\) represent the changes in energy and entropy that would be measured for the change IF it occurred. If this net quantity is positive, the change cannot occur without some additional help. If this quantity is negative, the change might occur but there is no guarantee that it will occur. However, when the quantity is negative, the reverse change cannot occur without additional help. Changes that actually occur when this quantity is less than zero are said to be spontaneous or irreversible. If the quantity in Equation \ref{eq10} is to zero, the change will not occur, but it could be pushed either forward or backward with very little additional help. This condition is described as , and the change is said to be . : A crystalline solid has very little movement and it is in a very low energy state. Movement of the atoms or molecules is limited to vibrations around a fixed point, so there is very little thermal entropy. The atoms/molecules are very close together (high concentration) and they are arranged in a very specific configuration (the crystal structure) which is repeated over and over within the crystal. There is very little freedom in the ways that the mass can be arranged, so the crystalline state has very little configurational entropy. : When a solid melts, the atoms/molecules begin to move around, and perhaps also rotating and vibrating. The liquid has considerably more energy than the solid and thus has more thermal entropy. The volume does not change appreciably when a solid melts. Normally there is a small increase in volume on melting (the solid sinks in the liquid), but a few materials (water is one of them) show a decrease in volume on melting (the solid floats in the liquid). Normally there is a small increase in configurational entropy, but for materials like water there is a small decrease. Overall, there is an increase in both energy and entropy when a solid melts. : When a liquid vaporizes, the atoms/molecules receive a huge increase in energy - so large that it seems like they are no longer subject to gravity. There is a large increase in volume and the concentration becomes very small. There is a correspondingly large increase in freedom, so that both the thermal entropy and the configurational entropy are greatly increased. Both energy and entropy increase on melting, so ΔU and ΔS are positive for fusion. At low temperatures (below the melting point) the positive ΔU contributes more than \(TΔS\) so the quantity \[ΔU - TΔS\] is positive, and melting cannot occur. As the temperature is raised however, both ΔU and ΔS increase but TΔS increases much more rapidly than ΔU. The quantity above will eventually become equal to zero at some temperature (the melting point) and the solid will spontaneously melt at any higher temperature. The opposite of melting is freezing. : Both \(ΔU\) and \(ΔS\) are large and positive for vaporization. The configurational entropy of the gas is related to the concentration of the gas molecules, so ΔS is greater for smaller concentrations of molecules in the vapor. This allows a balance between \(ΔU\) and \(TΔS\) at low temperatures, providing the concentration of gas molecules (and the vapor pressure) is sufficiently low. As the temperature is raised, this balance is maintained by an increase in the concentration of the gas molecules (and an increase in vapor pressure). This results in a wide range of temperatures for a liquid and its vapor to be in equilibrium, with the vapor pressure increasing as the temperature is raised. The temperature at which the vapor pressure is exactly 1 atmosphere is defined as the normal boiling point of the liquid. The opposite of vaporization is condensation. In the same way that \(ΔU\) and \(TΔS\) can balance for a liquid with its vapor at very low temperature, a similar balance can occur for a solid and its vapor below the melting point. For most solids, the vapor pressure is so low that we aren't concerned about it. However, we can see the effect in snow disappearing from a rooftop on a cold day without ever melting. This process is called sublimation. For a few solids (carbon dioxide or "dry ice" is one) the vapor pressure reaches 1 atmosphere at a temperature below the melting point, so that we never see the liquid at atmospheric pressure. The liquid state can be observed, however, at higher pressures and pressurized cylinders of carbon dioxide usually contain a mixture of the liquid and the gas. The opposite of sublimation is deposition. When pure materials in some physical state are mixed to form a solution in the same physical state at the same temperature and pressure, the process is called mixing. The energy change on mixing may be either positive or negative, but the entropy change is always positive. The entropy change is due mainly to the decrease in the concentrations of the individual components in the change from a limited volume in the pure state to a much larger volume in the mixed state. Small Change in Energy Small Change in Thermal Entropy --> Small Change in Volume Decrease in Concentration of Red Molecules Decrease in Concentration of Blue Molecules Increase in Configurational Entropy When the mixed components are roughly equal in concentration, they are usually just called components (component A and component B, or components 1 & 2). For these solutions, concentrations are usually expressed in mole fractions, and occasionally in mass fractions or volume fractions. When the concentration of one component in a liquid solution is much greater than the others, it is usually called the solvent, and the less-concentrated components are called solutes. For these solutions, the concentration of the solute may be expressed as mole fraction, molality (moles of solute per kilogram of solvent), or molarity (moles of solute per liter of solution). If needed, the concentration of solvent is usually expressed as mole fraction. Before mixing, the pure solute may have been a gas, a liquid, or a solid. The first bit of solute that dissolves will have an extremely low concentration, so the change in configurational entropy for that dissolution will be very large if the pure solute is a liquid or a solid. If the pure solute is a gas, the change in configurational entropy will depend on the concentration of the gas. In all cases, the change in thermal entropy may be positive or negative, but the configurational entropy usually dominates at very low concentrations. We can get a "feel" for the energy change as something dissolves. If the solution becomes colder as the solute dissolves, the energy change is positive since energy will have to be transferred to the solution in order to bring it back to the original temperature. If the solution becomes hotter as the solute dissolves, the energy change is negative since energy will have to be removed from the solution in order to return to the initial temperature. If the energy change is positive and unfavorable the positive entropy change may allow a small amount of the solute to dissolve. However, as more of the solute dissolves the concentration increases and the entropy change decreases until the energy change is exactly balanced by \(TΔS\) and no more of the solute will dissolve. The concentration at this point determines the solubility of the solute in the solvent at this temperature. When there is a positive energy change for dissolution, an increase in temperature increases the effect of the positive entropy change and the solubility of the solute increases. In the case of gases dissolving in liquids, the solubility of the gas depends on the concentration of the gas molecules. As the concentration in the gas phase increases, the concentration in the liquid phase increases. At low concentrations there is a direct proportionality between the concentrations in the two phases. This relationship is generally known as Henry's Law. Its most common form is for the concentration in the vapor phase to be represented by the partial pressure of the solute and for the concentration in the liquid phase to be represented by the mole fraction of the solute. Henry's Law is normally associated with the solute in dilute solutions, but it also applies to the solvent in these dilute solutions. The application to the solvent is a special case of Henry's Law, called Raoult's Law. Raoult's Law states that the vapor pressure (or partial pressure) of the solvent becomes equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent when the solutes become very dilute. Colligative properties of solutions are properties which depend only on the concentration of the solvent, and are independent of what the solute might be. Raoult's Law is the most basic example of a colligative property, because the vapor pressure (or partial pressure) of the solvent is determined only by the mole fraction of the solvent in a sufficiently dilute solution. This is usually stated as vapor pressure lowering by a non-volatile solute: The difference between the vapor pressure of the solution and that of the pure solvent is proportional to the mole fraction of solute. All colligative properties are based on the fact that the concentration of the solvent is greatest when the solvent is pure, and the concentration of the solvent decreases as solute is added. When applied to freezing and melting, the entropy change is greater for the frozen pure solvent to melt when a solute is present than when the liquid phase is pure, and the solvent can melt at a lower temperature. This freezing point depression depends only on the concentration of the solvent and not on the nature of the solute, and is therefore a colligative property. In a similar fashion, when a solution containing a solvent and a non-volatile solute is heated to near its boiling point at 1 atmosphere pressure, the entropy of the solvent in the solution is greater than in the pure liquid. This makes the entropy change for vaporization from the solution less than from the pure liquid, and the solvent will not boil until the temperature is higher than the normal boiling point. This boiling point elevation is also a colligative property. In some cases, a solvent may be separated from a solution containing a solute by a semi-permeable membrane. The solvent can move through the pores in this membrane, but the solute is unable to move through the pores because of its size or perhaps its charge or polarity. This movement of the solvent is called osmosis. Since the concentration of the solvent is lower in the solution than in the pure liquid, the solvent molecules have greater entropy in the solution, and there is a tendency for the molecules to move through the membrane. As the solvent molecules move, the liquid level of the solution becomes higher than that of the pure liquid, and eventually the pressure becomes sufficient to prevent any more solvent molecules from crossing the membrane. This pressure difference between the solution and the pure solvent is called the osmotic pressure. Osmotic pressure is a colligative property. This discussion has carefully avoided associating entropy with order and disorder. Instead, the focus has been on different types of freedom. Many textbooks use the order/disorder interpretation, and refer to examples such as the entropy of a shuffled deck of cards, or the entropy of a messy desk (or room) in comparison to a neat one. While these are strong images, the concept is basically incorrect. The example with a deck of cards assumes that there is some order of the cards which represents perfect order. This simply happens to be the order given to a sealed deck by the manufacturer, namely each suit increasing from the Ace sequentially to the King, with alternating colors of the suits. That is no more orderly than starting with the 2 and increasing through the King to the Ace, or the reverse order. The deck could be arranged with all four Aces in some order of suits, followed by the four 2's in the same order of suits, etc. - that is also ordered. A new manufacturer may decide to package the decks in some order that has significance only to that manufacturer. Does that arrangement then become the perfect order for his product? Perhaps we should also be concerned with whether the cards are all facing the same direction. The point is that there are many possible arrangements of the cards in a deck. When the cards are randomly and thoroughly shuffled, each of these arrangements has equal probability of occurring. When an arrangement is established, there is no freedom for the deck to acquire any other arrangement unless someone shuffles it or rearranges it in some manner. Since there is no freedom for that arrangement, there is no entropy associated with a single arrangement. On the other hand, if a very large number of decks of cards are thoroughly and randomly shuffled, there are many possibilities for the arrangements of the cards within the different decks. When this idea of different arrangements of the cards is applied to atoms and molecules in solids and liquids, the number of possible arrangements create freedom of arrangement, a type of configurational entropy - not for any specific deck, but for the group of decks - or the ensemble (technically, the ensemble is the huge imaginary group of all of these decks arranged in all of the possible arrangements) of decks. The number of possible arrangements of the 52 cards in a deck is 52! (fifty-two factorial) which is equal to 52 x 51 x 50 x 49 x...x 3 x 2 x 1, an astronomical number. If these cards were atoms or molecules, the entropy due to this incredible amount of freedom would be related to the number of randomized decks multiplied by the natural logarithm of 52!. We can calculate the entropy per deck by dividing the entropy of the group of decks by the number of decks, but it is important to differentiate between the entropy per deck (which is a property of the group) and the entropy of a single deck (which really has no meaning in chemical systems). , Professor of Chemistry, University of Missouri-Rolla

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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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The size and shape of molecules are as much a part of molecular structure as is the order in which the component atoms are bonded. Contrary to the impression you may get from structural formulas, complex molecules are not flat and formless, but have well-defined spatial arrangements that are determined by the lengths and directional character of their chemical bonds. It is not easy to visualize the possible arrangements of the bonds in space and it is very helpful to have some kind of mechanical model that reflects the molecular geometry, including at least an approximation to the relative lengths of the bonds. "Ball-and-stick" models such as the ones used by Paterno ( ) fill this purpose admirably. It is well established that the normal carbon atom forms its four single bonds in compounds of the type \(CX_4\) so that the four attached atoms lie at the corners of a regular tetrahedron. The bond angles \(X-C-X\) are \(109.5^\text{o}\) and this value is the "normal" valence angle of carbon. For many purposes, ball-and-stick models of organic compounds give useful information about the spatial relationships of the atoms, and for \(CX_4\) the angles between sticks are set at \(109.5^\text{o}\) (Figure 2-1). Organic molecules strongly resist deformation forces that alter their valence angles from normal values. Therefore ball-and-stick models correspond better to the behavior of actual molecules if the connectors representing single bonds are made to be rather stiff. Whereas methane, \(CH_4\), is tetrahedral, ethene, \(C_2H_4\), is not. According to the best available physical measurements, all six atoms of ethene lie in a single plane and the \(H-C-H\) bond angles are \(117.3^\text{o}\). Methanal (formaldehyde) also is a planar molecule with an \(H-C-H\) bond angle of \(118^\text{o}\). Ethyne, \(C_2H_2\), has been established experimentally to be a linear molecule; that is, the \(H-C-C\) bond angle is \(180^\text{o}\): Structural units that have \(C-C-C\) valence angles substantially less than the tetrahedral value include double and triple bonds, and small rings such as cyclopropane. Several bent bonds are required to construct models of compounds containing these units. Interestingly, such compounds are much less stable and more reactive than otherwise similar molecules for which models can be constructed with straight sticks at tetrahedral angles. The length of a chemical bond is the average distance between the nuclei of two bonded atoms, regardless of where the bonding electrons happen to be. The customary unit of length is the angstrom\(^1\) (\(\text{A} = 10^{-10} \: \text{m}\)), and measurements often can be made with an accuracy of \(0.001 \: \text{A}\) by using the techniques of molecular spectroscopy, x-ray diffraction (for crystalline solids), and electron diffraction (for volatile compounds). Bond lengths vary considerably with structure and depend on the identity of both atoms, the type of bonding (single, double, or triple), and the nature of other atoms or groups bonded to the two atoms in question. These effects are apparent in the data of Table 2-1, which lists the bond lengths in several simple organic compounds. Multiple bonds, double or triple, clearly are shorter than single bonds, and it can be stated as a general observation that the more bonding electrons in a given bond, the shorter (and stronger) the bond. The lengths of single \(C-C\) bonds also vary significantly depending on what other atoms or groups are attached to the carbons. Thus Table 2-1 shows that single \(C-C\) bonds become progressively shorter as the number of multiple bonds or electronegative atoms attached to the carbons increases. Although molecular models cannot represent the subtle variations in bond lengths and bond angles that actual molecules exhibit, most kinds of commercially available molecular models do attempt to reproduce relative bond lengths with some degree of reality. In the ball-and-stick type, the sticks usually come in various lengths to simulate different kinds of bonds; \(C-H\) bonds typically are scaled to represent \(1.1 \: \text{A}\), \(C-C\) bonds to be \(1.54 \: \text{A}\), and \(C=C\) and \(C \equiv C\) to be correspondingly shorter. In some model sets the bonds can be cut to any desired length. While the ball-and-stick models of molecules are very useful for visualizing the relative positions of the atoms in space, they are unsatisfactory whenever we also want to show how large the atoms are. Actually, atomic radii are so large relative to the lengths of chemical bonds that when a model of a molecule such as methyl chloride is constructed with atomic radii and bond lengths, both to scale, the bonds connecting the atoms are not clearly evident. None- As we shall see, such crowding has many chemical consequences. Ideally, a model should reflect not only the size and shape of the molecule it represents but also the flexibility of the molecule. By this we mean that it should simulate the type of motions available to the molecule, particularly bond rotation. For example, it is known that rotation normally occurs about single bonds in open-chain compounds but is restricted about double bonds. Motions of this kind are demonstrated easily with ball-and-stick models, but are not at all obvious with the space-filling type. For this reason, ball-and-stick models or their equivalent are more generally useful than the space-filling models for visualizing structures and the positions of the atoms relative to one another. \(^1\)The angstrom unit likely will be replaced eventually by the nanometer (\(1 \: \text{nm} = 10^{-9} \: \text{m} = 10 \: \text{A}\)). and (1977)

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Sir William Crookes discovered thallium in 1861, positively identifying it by a bright green line in its spectrum (hence the name, which is from the Greek, thallos, for "green twig"). Although in appearance thallium resembles lead, it does not have the corrosion resistance of lead and so has few commercial applications. Thallium has the chemical symbol Tl and atomic number 81. It has the electron configuration [Xe] 6s 6p and has a +3 or +1 oxidation state. As stated above, because thallium is heavy, it has a greater stability in the +1 oxidation state (inert pair effect). Therefore, it is found more commonly in its +1 oxidation state. Thallium is soft and malleable. Thallium compounds are quite toxic and some have been used as rat poisons. A few compounds are used in glasses for special infra-red lenses. Because of its toxicity, thallium was widely used in insecticide and rat poison until this usage was prohibited in 1975 in the U.S.

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Before beginning this section, you should know and understand: Chemical reactions between gaseous materials are quite similar to reactions between solids and liquids, except the Ideal Gas Law (\(PV=nRT\)) can now be included in the calculations. If a chemical reaction is reversible (such as the decay and formation of dinitrogen tetraoxide), then Dalton's Law of Partial Pressure may be used to determine the moles of reactants and products at which the reaction ceases (and subsequently, the temperature, pressure and volume of each gas can be determined as well). A non-reversible reaction uses the reactants to form the products. The reaction goes in one direction; that is, using the product to recreate the reactants has greatly different requirements. One of the most common forms of non-reversible reactions is combustion (once an organic molecule has been converted to water and hydrogen gas, it is extremely difficult to reform). Other non-reversible reactions produce a state change, such as Hydrogen Peroxide (the gaseous material produces water, a liquid). To understand how the Ideal Gas Law applies to reactions, we shall use a nonreversible reaction as an example. If 4.000 grams of hydrogen peroxide is placed within a sealed 250 milliliter container at 500 K. What is the pressure of the oxygen gas produced in atmospheres? \[\ce{2H_2O_2 \rightarrow 2H_2O + O_2} \nonumber\] First, we need to determine the moles of \(\ce{O_2}\) produced, just like any other stoichiometric problem. \[(4g\; \cancel{\ce{H_2O_2}}) \times \left(\dfrac{1\; mol\; \cancel{\ce{H_2O_2}}}{34.016\;g\; \cancel{ \ce{H_2O_2}}} \right) \left(\dfrac{1\; mol\; \ce{O_2}}{2\;mol\; \cancel{\ce{H_2O_2}}} \right) = 0.0588 \;mol\; \ce{O_2} \nonumber\] With the moles of oxygen determined, we can now use the Ideal Gas Law to determine the pressure. \[PV=nRT \nonumber\] The volume (250 mL = 0.25 L) and temperature (500 K) are already given to us, and R (0.0820574 Latm mol K ) is a constant. \[ \begin{align*} P &=\dfrac{nRT}{V} \\[4pt] &= \dfrac{(0.0588\; mol\; O_2) \times (0.0820\; L \;atm \;mol^{-1}\;K^{-1}) \times (500 \;K)}{0.25\;L} \\[4pt] &= 9.65\; atm \end{align*}\] By using the Ideal Gas Law for unit conversions, properties such as the pressure, volume, moles, and temperature of a gas involved in a reaction can be determined. However, a different approach is needed to solve reversible reactions. For further clarification, when solving equations with gases, we must remember that gases behave differently under different conditions. For example, if we have a certain temperature or pressure, this can change the number of moles produced or the volume. This is unlike regular solids where we only had to account for the mass of the solids and solve for the mass of the product by stoichiometry. In order to solve for the temperature, pressure, or volume of a gas using chemical reactions, we often need to have information on two out of three of these variables. So we need either the temperature and volume, temperature and pressure, or pressure and volume. The mass we can find using stoichiometric conversions we have learned before. The reason why gases require additional information is because gases behave as ideal gases and ideal gases behave differently under different conditions. To account for these conditions, we use the ideal gas equation PV=nRT where P is the pressure measured in atmosphere(atm), V is the volume measured in liters (L), n is the number of moles, R is the gas constant with a value of .08206 L atm mol K , and T is the temperature measured in kelvin (K). Suppose we have the following combustion reaction (below). If we are given 2 moles of ethane at STP, how many liters of CO are produced? \[\ce{2C2H6(s) + 7O2(g) -> 6H2O(l) + 4CO2(g)}\] First use stoichiometry to solve for the number of moles of CO produced. \[(2\, mol\, \ce{C2H6} )\left(\dfrac{4 \,mol \ce{CO2}}{2\, mol\, \ce{ C2H6}}\right) = 4 mol \, \ce{CO2} \nonumber\] So 4 moles of Carbon Dioxide are produced if we react 2 moles of ethane gas. Now we simply need to manipulate the ideal gas equation to solve for the variable of interest. In this case we are solving for the number of liters. Since we are told ethane is at STP, we know that the temperature is 273 K and the pressure is 1 atm. So the variables we have are: Isolating the variable of interest from \(PV=nRT\), we get \[\begin{align*} V &=\dfrac{nRT}{P} \\[4pt] &= \dfrac{4\, mol \times 0.08206 \,L \,atm \,mol^{-1} K^{-1} \times 273\,K}{1\, atm} \\[4pt] &= 89.61\, L \end{align*}\] So we have a volume of 89.61 liters. A reversible reaction is a chemical reaction in which reactants produces a product, which then decays back to the reactants. This continues until the products and reactants are in equilibrium. In other words, the final state of the gas includes both the reactants and the products. For example, Reactant A combines with Reactant B to form Product AB, which then breaks apart into A and B, until an equilibrium of the three is reached. In a reaction between gases, determining gas properties such as partial pressure and moles can be quite difficult. For this example, we consider the theral decomposition of Dinitrogen Tetraoxide into Nitrogen Dioxide. For this example, we shall use Dinitrogen Tetraoxide, which decomposes to form Nitrogen Dioxide. \[\ce{N_2O_4 <=> 2NO_2} \nonumber \] 2 atm of dinitrogen tetraoxide is added to a 500 mL container at 273 K. After several minutes, the total pressure of N O and 2NO at equilibrium is found to be 3.2 atm. Find the partial pressures of both gases. The simplest way of solving this problem is to begin with an . \(\ce{N2O4}\) With this data, a simple equation can be derived to determine the value of X. \[P_{total} = (2-X) + 2X = 3.2\;atm\] \[X = 1.2\; atm\] \[P_{NO_2}= 2x = 2.4\; atm\] \[P_{N_20_4} = 2-x = 0.8\; atm\] This law of combining volumes was first discovered by the famous scientist Gay-Lussac who noticed this relationship. He determined that if certain gases that are products and reactions in a chemical reaction are measured at the same conditions, temperature and pressure, then the volume of gas consumed/produced is equal to the ratio between the gases or the ratio of the coefficients. If ozone, hydrogen, and oxygen were all measured at 35 C and at 753 mmHg, then how many liters of ozone was consumed if you had 5 liters of oxygen gas? \[\ce{O3(g) + H2O(l) -> H2(g) + 2O2(g)} \nonumber\] Identify what we are looking for and if any relationships can be spotted. In this case, we can see that there are three gases all at the same temperature and pressure, which follows Gay-Lussac’s Law of Combining Volumes. We can now proceed to use his law. We simply change the coefficients to volumetric ratios. So for every 1 Liter of Ozone gas we have, we produce 1 Liter of H gas and 2 Liter of \(O_2\) gas. We are given 5 liters of Oxygen gas and want to solve for the amount of liters of ozone consumed. We simply use the 2:1 stoichiometry of the reaction. \[5 L O_2 \left(\dfrac{1\; L\; O_3}{2\; L\; O_2}\right) = 2.5\; L\; O_3\]

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Cyclic voltammetry is a commonly used method of measuring the reduction potential of a species in solution. The species may be a coordination complex or a redox-active organic compound, for example. Cyclic voltammetry provides additional data that can be interpreted to make conclusions about the reduction / oxidation reaction and the stability of the species resulting from the electron transfer. In cyclic voltammetry, rather than measuring the voltage produced by a reaction as we discussed before, a voltage is instead applied to the solution. The voltage is changed over time and current through a circuit is monitored. When the voltage reaches a point at which a reduction/oxidation is induced, current begins to flow. A cyclic voltammogram is a plot of current versus applied voltage. In the experiment, the species of interest is dissolved along with some electrolyte, which promotes conductivity in the solution. Three electrodes are inserted into the solution. The working electrode, where the reduction / oxidation reaction takes place, is mostly covered with an insulator, but has a small disc of electrode exposed so the reaction can take place in a carefully controlled area. The counter electrode completes the circuit. A reference electrode with a known potential is also used in order to measure the potential applied to the cell. Unusually, the solution must not be stirred. To begin the experiment, a potential is applied that is much more positive than the potential of the reference electrode. This step ensures that the species of interest is completely oxidized to begin with. The voltage is then swept in the negative direction at a constant rate. That is, the potential at the working electrode gets lower and lower, possibly until it becomes negative compared to the reference electrode. At some point, the voltage sweep is reversed, and it becomes more and more positive until it returns to the initial setting. The (simulated) results of such an experiment are shown below. At point A, the potential is very positive but is then swept to lower and lower values. At point B, current begins to flow as the voltage reaches a point that allows reduction to occur. Current keeps increasing until, at point C, all of the species in the vicinity of the working electrode has already been reduced. This point is called "cathodic peak potential". Current then begins to decrease, although some still keeps flowing as more of the species slowly diffuses over to the working electrode (point D). The reasons for two of the features of the experimental design are now apparent. The reason for the very small exposed surface of the working electrode (usuallyan exposed disc about 1 mm wide) is to limit the area in which reaction takes place, so that we can observe when a controlled population of species has been reduced. The reason for not stirring the solution is similar; if we stirred the solution, more species would be continually and quickly fed to the working electrode and we would never observe a point at which the reaction was "finished". At some point, the potential is increased again (point E). Current keeps decreasing; that trend is reversed as the previously reduced species is again oxidized. This time, current flows in the opposite direction, and a negative peak is observed. At point F, all of the species in the vicinity of the working electrrode has been oxidized, and current begins to "drop" again. This point is called "anodic peak potential". The "formal potential" is the mean of anodic and cathodic peak potential. What is different about the following cyclic voltammogram compared to the previous one? Explain what is happening in this sample. What is different about the following cyclic voltammogram? Explain what is happening in this sample. ,

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We have seen that min{\(U(S,\bar{X})\)} and min{\(U(S,\hat{X})\)} imply one another. Under certain conditions, these principles are very convenient. For example, \[ dS = \dfrac{1}{T} dU - \dfrac{P}{T} dV + \sum \dfrac{\mu_i}{T} dn_i \Rightarrow max \left\{ \sum \dfrac{\mu_i}{T} dn_i \right\}\] maximizes \(S\) at constant \(U\) and \(V\). But what do we do if we are working at constant ( , ) or ( , )? We then have to deal with, where energy must flow (to keep \(T\) constant), or the volume must change (to keep constant), etc. This problem arises for many laboratory reactions in chemistry. As it turns out, if thermodynamic variables other than and are held constant, thermodynamic potentials other than entropy or energy of the open system are extremized. (Of course, \(S\) is still maximized for a closed system containing our open system of interest.) Very conveniently, these potentials can be computed just from the properties of the open system of interest alone (e.g. one where \(T\) is constant, and therefore energy must be allowed to flow in and out), and the added assumption that the corresponding variable (e.g. \(T\)) is always constant in the environment. The environment is thus treated as a “bath” or “reservoir” for the intensive variable of interest. That is, we assume the closed system containing our open system of interest is so vast that a change in extensive variable (e.g. \(U\)) of the small open system does not affect the conjugate intensive variable in the environment (e.g. (T\); the energy deposited or taken from the environments by the open system is too small to change in the environment). Our goal: we want potentials where our of intensive variable is in the dependent variables. Let’s begin by discussing an apparently obvious way of doing this, which does not work. Let’s say we have the fundamental relation for \(U\), but we want to hold \(T\) constant, not \(S\): \[ U = U(S,V,n_i) \Rightarrow T=\left( \dfrac{\partial U}{\partial S} \right)_{V,n_i} = T(S,V,n_i)\] solving for \(S=S(T,V,n_i)\), and insert in \(U\) we seem to be able to get \[ U =U(T,V,n_i).\] Now we can hold \(T\) constant. The problem with this: \(T\) is the slope of \(U(S)\), and expressing a function in terms of its own slope leaves the intercept indeterminate: \(U\) is no longer completely defined, and we do not have a fundamental relation to which the laws of thermodynamics can be applied. We do however want an equation in terms of the slope. The solution is to express the slope \(m\) of \(y(x)\) in terms of the intercept \(\phi\) (or vice-versa). The intercept as a function of slope does contain all the original information about the function \(y(x)\). The process of transforming a function \(y(x)\) into \(\phi(m)\) is called .

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Dissolution of a salt in water is a chemical process that is governed by the same laws of chemical equilibrium that apply to any other reaction. There are, however, a number of special aspects of of these equilibria that set them somewhat apart from the more general ones that are covered in the lesson set devoted specifically to chemical equilibrium. These include such topics as the common ion effect, the influence of pH on solubility, supersaturation, and some special characteristics of particularly important solubility systems... all explained in what follows. Drop some ordinary table salt into a glass of water, and watch it "disappear". We refer to this as , and we explain it as a process in which the sodium and chlorine units break away from the crystal surface, get surrounded by H O molecules, and become . \[\ce{NaCl_(s) \rightarrow Na^{+}(aq) + Cl^{–}(aq)} \nonumber\] The designation means "aqueous" and comes from , the Latin word for water. It is used whenever we want to emphasize that the ions are hydrated — that H O molecules are attached to them. Remember that solubility equilibrium and the calculations that relate to it are only meaningful when sides (solids and dissolved ions) are simultaneously present. But if you keep adding salt, there will come a point at which it no longer seems to dissolve. If this condition persists, we say that the salt has reached its , and the solution is in NaCl. The situation is now described by \[\ce{NaCl_(s) <=>Na^{+}(aq) + Cl^{–}(aq)} \nonumber\] in which the solid and its ions are . Salt solutions that have reached or exceeded their solubility limits (usually 36-39 g per 100 mL of water) are responsible for prominent features of the earth's geochemistry. They typically form when NaCl leaches from soils into waters that flow into salt lakes in arid regions that have no natural outlets; subsequent evaporation of these brines force the above equilibrium to the left, forming natural salt deposits. These are often admixed with other salts, but in some cases are almost pure NaCl. Many parts of the world contain buried deposits of NaCl (known as halite) that formed from the evaporation of ancient seas, and which are now mined. Solubilities are most fundamentally expressed in molar (mol L of solution) or molal (mol kg of water) units. But for practical use in preparing stock solutions, chemistry handbooks usually express solubilities in terms of grams-per-100 ml of water at a given temperature, frequently noting the latter in a superscript. Thus 6.9 means 6.9 g of solute will dissolve in 100 mL of water at 20° C. When quantitative data are lacking, the designations "soluble", "insoluble", "slightly soluble", and "highly soluble" are used. There is no agreed-on standard for these classifications, but a useful guideline might be that shown below. The solubilities of salts in water span a remarkably large range of values, from almost completely insoluble to highly soluble. Moreover, there is no simple way of predicting these values, or even of explaining the trends that are observed for the solubilities of different anions within a given group of the periodic table. Ultimately, the driving force for dissolution (and for chemical processes) is determined by the Gibbs free energy change. Dissolution of a salt is conceptually understood as a sequence of the two processes depicted above: The first step consumes a large quantity of energy, something that by itself would strongly discourage solubility. But the second step a large amount of energy and thus has the opposite effect. Thus the net energy change depends on the sum of two large energy terms (often approaching 1000 kJ/mol) having opposite signs. Each of these terms will to some extent be influenced by the size, charge, and polarizability of the particular ions involved, and on the lattice structure of the solid. This large number of variables makes it impossible to predict the solubility of a given salt. Nevertheless, there are some clear trends for how the solubilities of a series of salts of a given anion (such as hydroxides, sulfates, etc.) change with a periodic table group. And of course, there are a number of general . This is a condition for solubility equilibrium, but it is not by itself . True chemical equilibrium can only occur when all components are simultaneously present. A solubility system can be in equilibrium only when some of the solid is in contact with a saturated solution of its ions. Failure to appreciate this is a very common cause of errors in solving solubility problems. If the ion product is smaller than the solubility product, the system is not in equilibrium and no solid can be present. Such a solution is said to be . A solution is one in which the ion product exceeds the solubility product. A supersaturated solution is not at equilibrium, and no solid can ordinarily be present in such a solution. If some of the solid is added, the excess ions precipitate out and until solubility equilibrium is achieved. Solubility usually increases with temperature - but not always. This is very apparent from the solubility-vs.-temperature plots shown in Figure \(\Page {1}\). (Some of the plots are colored differently in order to make it easier to distinguish them where they crowd together.) The temperature dependence of any process depends on its entropy change — that is, on the degree to which thermal kinetic energy can spread throughout the system. When a solid dissolves, its component molecules or ions diffuse into the much greater volume of the solution, carrying their thermal energy along with them. So we would normally expect the entropy to increase — something that makes any process take place to a greater extent at a higher temperature. So why does the solubility of cerium sulfate (green plot) diminish with temperature? Dispersal of the Ce and SO ions themselves is still associated with an entropy increase, but in this case the entropy of the decreases even more owing to the ordering of the H O molecules that attach to the Ce ions as they become hydrated. It's difficult to predict these effects, or explain why they occur in individual cases — but they do happen. )

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explains why different reactions occur at different rates, and suggests ways to change the rate of a reaction. Collision theory states that for a chemical reaction to occur, the reacting particles must collide with one another. The rate of the reaction depends on the frequency of collisions. The theory also tells us that reacting particles often collide reacting. For collisions to be successful, reacting particles must (1) collide with (2) sufficient energy, and (3) with the proper orientation. This rule is fundamental to any analysis of an ordinary reaction mechanism. It explains why termolecular processes are so uncommon. The kinetic theory of gases states that for every 1000 binary collisions, there will be only one event in which three molecules simultaneously come together. Four-way collisions are so improbable that this process has never been demonstrated in an elementary reaction. If the two molecules A and B are to react, they must approach closely enough to disrupt some of their existing bonds and to permit the creation of any new ones that are needed in the products. Such an encounter is called a . The frequency of collisions between A and B in a gas is proportional to the concentration of each; if [A] is doubled, the frequency of A-B collisions will double, and doubling [B] will have the same effect. If all collisions lead to products, then the rate of a bimolecular process is first-order in A and in B, or second-order overall: \[rate = k[A,B] \nonumber \] For a gas at room temperature and normal atmospheric pressure, there are about 10 collisions in each cubic centimeter of space every second. If every collision between two reactant molecules yielded products, all reactions would be complete in a fraction of a second. For example, when two billiard balls collide, they simply bounce off of each other. This is the most likely outcome if the reaction between A and B requires a significant disruption or rearrangement of the bonds between their atoms. In order to effectively initiate a reaction, (or have sufficient kinetic energy) to bring about this bond disruption. This is further discussed below. There is often one additional requirement. In many reactions, especially those involving more complex molecules, the reacting species must be oriented in a manner that is appropriate for the particular process. For example, in the gas-phase reaction of dinitrogen oxide with nitric oxide, the oxygen end of N O must hit the nitrogen end of NO; altering the orientation of either molecule prevents the reaction. Owing to the extensive randomization of molecular motions in a gas or liquid, there are always enough correctly-oriented molecules for some of the molecules to react. However, the more critical this orientational requirement is, the fewer collisions will be effective. Energetic collisions between molecules cause interatomic bonds to stretch and bend, temporarily weakening them so that they become more susceptible to cleavage. Distortion of the bonds can expose their associated electron clouds to interactions with other reactants that might lead to the formation of new bonds. Chemical bonds have some of the properties of mechanical springs: their potential energies depend on the extent to which they are stretched or compressed. Each atom-to-atom bond can be described by a potential energy diagram that shows how its energy changes with its length. When the bond absorbs energy (either from heating or through a collision), it is elevated to a higher quantized vibrational state (indicated by the horizontal lines) that weakens the bond as its length oscillates between the extended limits corresponding to the curve. A particular collision will typically excite a number of bonds in this way. Within about 10 seconds, this excitation is distributed among the other bonds in the molecule in complex and unpredictable ways that can concentrate the added energy at a particularly vulnerable point. The affected bond can stretch and bend farther, making it more susceptible to cleavage. Even if the bond does not break by pure stretching, it can become distorted or twisted so as to expose nearby electron clouds to interactions with other reactants that might encourage a reaction. Until about 1921, chemists did not understand the role of collisions in unimolecular processes. It turns out that the mechanisms of such reactions are actually quite complicated, and that at very low pressures they do follow second-order kinetics. Such reactions are more properly described as . The cyclopropane isomerization described in Example 1 is typical of many decomposition reactions found to follow first-order kinetics, implying that the process is unimolecular. Consider, for example, the isomerization of cyclopropane to propene, which takes place at fairly high temperatures in the gas phase. The collision-to-product sequence can be conceptualized in the following [grossly oversimplified] way: Note that Here are some simple little "experiments" to help illustrate this concepts. These are "thought experiments," to be reasoned out rather than actually performed. In an empty classroom, blindfold a group of 10 or so students and instruct them to walk slowly around the room. Occasionally, a pair of students will bump into one another. If they are moving slowly enough, nothing much will happen (that is, no pain upon contact). But if these blindfolded students begin running around the room (more speed means more energy), then a collision is likely to be successful (painfully so). If some students move about quickly, while others stroll about at a more leisurely pace, successful reactions will still occur, but not as often as if all students are running. Next, have the students move at a brisk pace, but without running. This time, however, consider that collisions that occur shoulder-to-shoulder are not successful—shoulders act as sufficient bumpers and collisions are not painful. But if one student steps on another's toes, then a successful collision occurs. Let the movement begin. Of the frequent collisions that result, only a few will involve one student stepping on another's toes. This experiment illustrates that collisions must occur with the proper orientation. Again, collisions between students will occur, but only some will be successful. Factors that increase the rate of a reaction must influence at least one of the following: Two more topics must be examined before these can be discussed in depth: reaction mechanisms and the concept of threshold energy.

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is an obsolete name for what is now known to be a mixture of two distinct minerals: zinc carbonate (ZnCO or smithsonite) and zinc silicate (Zn Si O (OH) ·H O, or hemimorphite). the name "Calamine" is now used only for calomine lotion, which is a suspension of ZnO and Fe O . Smithsonite and hemimorphite may be similar in appearance to one another, but their appearance may be quite variable depending on location, so two samples of smithsonite (or hemimorphite) may look quite different as shown in the figures:   Hemimorphite from Mapimi, Durango, Mexico Hemimorphite Smithsonite from Tsumeb, Namibia Smithsonite from Tsumeb, Namibia The two minerals can only be reliably distinguished through chemical analysis. Carbonate minerals like calcite or smithsonite react with acids to efforvesce (fizz) while dissolving and producing CO (see equation (1) below). This test can be done with 1 M HCl, or household vinegar (crushing the sample will help if vinegar is used). While calcite (CaCO ) bubbles strongly in cold dilute acid, dolomite CaMg(CO ) ) and rhodochrosite (MnCO ) bubble weakly. Smithsonite (along with Siderite, FeCO and Magnesite, MgCO ) require heating to react. Silicates, like hemimorphite, don't generally react with cold, dilute acids at all. So we could tell if a sample contained just smithsonite, because it would ocmpletely dissolve in acid. hemimorphite would not react, and a mixture of the two would partially dissolve. It would be necessary to add an excess of HCl to the sample, otherwise it might not all dissolve because there isn't enough HCl, not because it's partially hemimorphite. If we have a 100 g sample that may contain smithsonite, hemimorphite, or both, we need to add enough acid to react with the sample, assuming it's all smithsonite, just to make sure. Example 4 from Equations and Mass Relationships also illustrates the idea that one reactant in a chemical equation may be completely consumed without using up all of another. Here we want the smithsonite to be completely consumed. When it is gone, the other excess reactants have nothing to react with and they cannot be converted to products. The substance which is used up first (the smithsonite) is the . When 100.0 g of smithsonite is reacted with 100.0 g of HCl to form carbon dioxide gas, which is the limiting reagent? What mass of product will be formed? (Note: HCl is provided as a solution with a concentration of 1-5% HCl for this purpose. The mass of HCl solution would be much (20-100 times) greater than the mass of HCl given here). The balanced equation ZnCO + 2 HCl → ZnCl + CO + H O (1) tells us that according to the atomic theory, 2 mol HCl are required for each mole of ZnCO . That is, the stoichiometric ratio S(HCl/ZnCO ) = 2 mol HCl/ 1 mol ZnCO . Let us see how many moles of each we actually have \(\begin{align} & n_{\text{HCl}}=\text{100}\text{.0 g}\times \frac{\text{1 mol HCl}}{\text{36.5 g}}=\text{2}\text{.74 mol HCl} \\ & n_{\text{ZnCO}_{\text{3}}}=\text{100}\text{.0 g}\times \frac{\text{1 mol ZnCO}_{\text{3}}}{\text{125}\text{.4 g}}=\text{0}\text{.798 mol ZnCO}_{\text{3}} \\ \end{align}\) Calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. We can calculate (hypothetically) how much of each reactant would be required if the other were completely consumed to demonstrate which is in excess, and which is limiting. For example, if all the HCl were to react, it would require \(\frac{\text{1 mol ZnCO}_3}{\text{2 mol HCl}}~~ x ~~ \text{2.74 mol HCl} = \text{1.37 mol ZnCO}_3\) Since there is not this much ZnCO present, this is impossible. HCl is in excess, and ZnCO is the limiting reactant. In the table, we've crossed out this calculation, and proceeded to calculate how much HCl would be required if all the ZnCO reacts (which is what happens). We use the amount of limiting reagent to calculate the amount of product formed. \(\frac{\text{1 mol CO}_{2}}{\text{1 mol ZnCO}_{3}}~~ x ~~ \text{0.798 mol ZnCO}_{3} = \text{0.798 mol CO}_2\) \(\text{0.798 mol CO}_2 ~~ x ~~ \frac{\text{44.0 g CO}_{2}}{\text{1 mol CO}_{2}} = \text{35.1 g CO}_2 \) When the reaction ends, 1.60 mol HCl will have reacted with 0.798 mol ZnCO and there will be (2.74 – 1.60) mol HCl = 1.14 mol HCl left over. ZnCO is therefore the limiting reagent. The left over HCl will ensure that if any material remains in a mineral test of a 100 g sample, that it can't be a carbonate. From this 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. In Example 1 this ratio of initial amounts \(\frac{n_{\text{ZnCO}_{3}}\text{(initial)}}{n_{\text{HCl}}\text{(initial)}}=\frac{\text{0.798 mol ZnCO}_{3}}{\text{2.74 mol HCl}}=\frac{\text{0}\text{.291 mol ZnCO}_{3}}{\text{1 mol HCl}}\) was less than the stoichiometric ratio \(\text{S}\left( \frac{\text{ZnCO}_{3}}{\text{HCl}} \right)=\frac{\text{1 mol ZnCO}_{3}}{\text{2 mol HCl}}~=~\frac{\text{0.5 mol ZnCO}_{3}}{\text{1 mol HCl}}\) This indicated that there was not enough Hg to react with all the bromine and mercury was the limiting reagent. 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.). This general rule for determining the limiting reagent is applied in the next example. Iron can be obtained by reacting the ore hematite (Fe O ) with coke (C). The latter is converted to CO . As manager of a blast furnace you are told that you have 20.5 Mg (megagrams) of Fe O and 2.84 Mg of coke on hand. (a) Which should you order first—another shipment of iron ore or one of coke? (b) How many megagrams of iron can you make with the materials you have? a) Write a balanced equation 2Fe O + 3C → 3CO + 4Fe The stoichiometric ratio connecting C and Fe O is \(\text{S}\left( \frac{\text{C}}{\text{Fe}_{\text{2}}\text{O}_{\text{3}}} \right)=\frac{\text{3 mol C}}{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}=\frac{\text{1}\text{.5 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}\) The initial amounts of C and Fe O are calculated using appropriate molar masses \(\begin{align} & \text{ }n_{\text{C}}\text{(initial)}=\text{2}\text{.84}\times \text{10}^{\text{6}}\text{g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{2}\text{.36}\times \text{10}^{\text{5}}\text{mol C} \\ & \\ & n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{(initial)}=\text{20}\text{.5}\times \text{10}^{\text{6}}\text{g}\times \frac{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}{\text{159}\text{.69 g}}=\text{1}\text{.28}\times \text{10}^{\text{5}}\text{mol Fe}_{\text{2}}\text{O}_{\text{3}} \\ \end{align}\) Their ratio is \(\frac{n_{\text{C}}\text{(initial)}}{n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{(initial)}}=\frac{\text{2}\text{.36}\times \text{10}^{\text{5}}\text{mol C}}{\text{1}\text{.28}\times \text{10}^{\text{5}}\text{mol Fe}_{\text{2}}\text{O}_{\text{3}}}=\frac{\text{1}\text{.84 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}\) Since this ratio is larger than the stoichiometric ratio, you have more than enough C to react with all the Fe O . Fe O is the limiting reagent, and you will want to order more of it first since it will be consumed first. b) The amount of 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 C will be left over, but all the initial amount of Fe O will be consumed. Therefore we use (initial) to calculate how much Fe can be obtained \(n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{ }\xrightarrow{S\text{(Fe/Fe}_{\text{2}}\text{O}_{\text{3}}\text{)}}\text{ }n_{\text{Fe}}\xrightarrow{M_{\text{Fe}}}\text{ }m_{\text{Fe}}\) \(m_{\text{Fe}}=\text{1}\text{.28 }\times \text{ 10}^{\text{5}}\text{ mol Fe}_{\text{2}}\text{O}_{\text{3}}\text{ }\times \text{ }\frac{\text{4 mol Fe}}{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}\text{ }\times \text{ }\frac{\text{55}\text{.85 g}}{\text{mol Fe}}=\text{1}\text{.43 }\times \text{ 10}^{\text{7}}\text{ g Fe}\) This is 1.43 × 10 g, or 14.3 Mg, Fe. 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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Indium is the 49th element, abbreviated as In. Discovered in 1863, indium rarely is found as an isolated element. Alloys of indium have low melting points and are good semiconductors; it's use in LCD displays has recently increased the demand for Indium. The element indium (named from the Latin indicum, for the color indigo) was discovered in 1863 by Reich and Richter. It is a rare metal, with an abundance similar to that of silver. It is generally found in deposits with zinc and refineries which produce this more common metal often sell indium as well. The pure metal is so soft that you can "wipe" it onto other materials in much the same way as lead (or even pencil graphite). It is corrosion resistant. As with gallium, identification of indium involved the recognition of new emission spectrum lines (its name was chosen because of indigo lines in its spectrum). Curiously enough, Reich who did the initial chemical isolation work was color blind and had to turn over his experiment to an assistant (Richter) who was the first to observe the characteristic lines. Like pure tin, pure indium emits a squealing sound when bent. Indium has the chemical symbol In and the atomic number 49. It has the electron configuration [Kr] 2s 2p1 and may adopt the +1 or +3 oxidation state; however, the +3 state is more common. It is a soft, malleable metal that is similar to gallium. Indium forms InAs, which is found in photoconductors in optical instruments. The physical properties of indium include its silver-white color and the "tin cry" it makes when bent. Indium is soluble in acids, but does not react with oxygen at room temperature. It is obtained by separation from zinc ores. Indium is mainly used to make alloys, and only a small amount is required to enhance the metal strength. For example, indium is added to gold or platinum to make the metals more useful industrial tools. Almost all of Indium is produced as a byproduct of zinc production. A small amount is produced from tin production. Indium is produced from the slag of zinc production (the waste matter separated from metals during the smelting or refining of ore) using a leaching process. Leaching is an technique used to extract metals which converts them to an aqueous state. Indium is leached using \(HCl\) or \(H_2SO_4\). There are many different processes to extract Indium from the zinc slag, and they vary from processor to processor. Indium recovered from this process is metal of a low-grade. It is further refined to be a high purity metal in refineries. Indium is produced in a lot of different forms, such as foil, ribbon, ingot, plates, powder, shot and pellets, and wire.

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Carbon is the fourth most abundant element in the known universe but not nearly as common on the earth, despite the fact that living organisms contain significant amounts of the element. Common carbon compounds in the environment include the gases carbon dioxide (\(CO_2\)) and methane (\(CH_4\)). Inorganic carbon is carbon extracted from ores and minerals, as opposed to organic carbon found in nature through plants and living things. Some examples of inorganic carbon are carbon oxides such as carbon monoxide and carbon dioxide; polyatomic ions, cyanide, cyanate, thiocyanate, carbonate and carbide in carbon. Carbon is an element that is unique to itself. Carbon forms strong single, double and triple bonds, therefore it would take more energy to break these bonds than if carbon were to bond to another element. For carbon monoxide the reaction is as follows: \[2C_{(s)} + O_2 \rightarrow 2CO_{(g)} \;\;\;\; \Delta H = -110.52\; kJ/mol \;CO\] \[C_{(s)} + O_2 \rightarrow CO_{2(g)} \;\;\;\; \Delta H = -393.51\; kJ/mol\; CO_2\] CO and CO are both gases. CO has no odor or taste and can be fatal to living organisms if exposed at even very small amounts (about a thousandth of a gram). This is because CO will bind to the hemoglobin that carries oxygen in the blood. CO will not become fatal unless living organisms are exposed to larger amounts of it, about 15%. CO influences the atmosphere and effects the temperature through the greenhouse gas effect. As heat is trapped in the atmosphere by CO gases, the Earth's temperature increases. The main source for CO in our atmosphere, amongst many is volcanoes. Carbon exists in several forms called allotropes. Diamond is one form with a very strong crystal lattice, known as a precious gem from the most ancient records. Graphite is another allotrope in which the carbon atoms are arranged in planes which are loosely attracted to one another (hence its use as a lubricant). The recently discovered fullerenes are yet another form of carbon. Carbon has a very high melting and boiling point and rapidly combines with oxygen at elevated temperatures. In small amounts it is an excellent hardener for iron, yielding the various steel alloys upon which so much of modern construction depends. An important (but rare) radioactive isotope of carbon, C-14, is used to date ancient objects of organic origin. It has a half-life of 5730 years but there is only 1 atom of C-14 for every 1012 atoms of C-12 (the usual isotope of carbon).

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Quantum mechanics incorporates the idea of wave mechanics that demonstrates the idea of wave-particle duality.This notion suggests that matter can display simultaneously both particle and wave-like properties. This new approach came from Louis de Broglie who built upon Einstein's conception that light possessed particle-like properties in his attempt to explain the photoelectric effect. Albert Einstein showed that the dependence on frequency could not be justified by the classical wave theory alone, so he provided a particle perspective. In 1905 he declared that photons (named by G.N. Lewis), were "particles of light" that had similar energy to that of Planck's equation. This equation states that the frequency and energy of a quantum of electromagnetic radiation are proportional. Einstein's idea was revolutionary because he brought a new perspective at looking at light not only as a wave, but as a particle. E=hv h=6.626x10 Js The photoelectric effect phenomenon that electrons are emitted when light strikes the surface of metals was discovered by Heinrich Hertz in 1888. This process holds true when the incident light has a higher frequency than a certain threshold value. The amount of electrons ejected is determined by the intensity of the incident light, however, the frequency of the light effects the kinetic energies of the emitted electrons. In other words, the intensity can be described as the brightness from a source of light. So by increasing brightness, intensity increases, and so does energy released. The energy output will be greater and when this happens the amplitude of the light wave increases. But it doesn't matter how much energy is increased or how much you increase the amplitude when it comes to trying to emit electrons from a metallic surface. To do so, frequency must increase. Increase brightness (maintains frequency and energy)-->Increases Intensity (increases#of phot ons)-->Increases # of electrons emitted Increase frequency-->Increases kinetic energy of electrons Einstein explanation was that light had the characteristic of a particle (photon) with the photon energy of E=hv. He concluded that if the threshold frequency of the metal was greater than the frequency of the photon, then the photon will have no effect when it bombards the metal surface. However, if the photon reached the threshold frequency it could cause one electron to be emitted. In order to emit more electrons, the light source must be brightened to increase intensity, which still maintains frequency of light and same energy, but increases the number of photons. V = (eV )/h = work function/Planck's constant The photoelectric effect can occur even with the lowest frequency light called the threshold frequency. Photelectrons are released when photon energy (hv) is greater than the work function. Energy in excess is released as kinetic energy in the ejected photoelectron and is proportional to frequency of light. The diagram above illustrates an electron being striked by a photon of energy, which allows it to overcome the work function binding it to the metal surface. As a result, a photoelectron is emitted with kinetic energy. By applying the law of conservation of energy we get the equation: hv =eV +(1/2)mv To summarize, regardless of the intensity of light, no electrons will be emitted if the frequency of light is below the threshold frequency (V ) of the metal surface. In 1924, Louis de Broglie used Einstein's equation E=mc2 and incorporated it with Planck's equation. This brought together the relationship of the mass of a photon and speed of light with the photon energy. E=mc m=mass of photon c=speed of light : E=hv E=mc & E=hv --> mc =hv --> mc=(hv)/c=p where p is momentum Since wavelegth is speed of light/frequency as shown by λ=c/f momentum is equivalent to p=(hv)/c=h/λ In order for this equation to apply for the purpose of a material particle (electron), de Broglie changed the equation to its equivalent So p=h/λ changed to λ=h/p=h/(mv) where p is the product of mass of particle and velocity demonstrates that particles have wavelike properties as shown by the integration of wavength (λ), Planck's constant (h), and momentum of particle (p=mv). This equation is also known as the "de Broglie wavelength." He reasoned that if things that acted like waves had particle characteristics then it should work vice versa. Therefore, things that behave like particles should also have wave characteristics. He coined the term "matter waves" to describe waves that involved material particles, such as electrons. Pattern is made up of emitted particles Pattern is similar to wave interference When waves travel in the same medium, interference can occur. Interference is the net effect caused by two different waves. In the example below, the superposition principle is in effect. The diagram shows that when two waves are traveling with the same frequency through the same medium, the individual waves pass one another without disturbance. There are two types of interferences known as Constructive and Destructive Interference. 1. When waves are "in phase" this means that their crests and troughs coincide. The net result produces a combination of the two waves that produces high crests and deep troughs causing amplification. The waves overlap and create an even larger wave. 2. This occurs when waves cancel one another because the crest of one wave occurs at the trough of another and is considered "out of phase." The wave-particle duality limits our understanding in determining the position and momentum of subatomic particles. In the 1920's, Werner Heisenberg and Niels Bohr tried different experiments to precisely measure the behavior of particles. However, they were unable to simultaneously measure position and momentum at the same time. This meant that even if they knew where a particle was, they couldn't find where it was going nor where it came from, after all the wave is spread out. They concluded that it was impossible to accurately measure and determine an electron's momentum and position simultaneously, hence the uncertainty principle is also referred to as the principle of indeterminacy. 1. From this diagram determine the number of crests and troughs. Remember crests are the high peaks and troughs form the deep ends. Therefore, there are and . 2. If a radiation has a wavelength of 302.5nm, what is the energy of one photon? To do this, we must look at Planck's equation and find the frequency of the radiation. Frequency is v= c/λ and c=speed of light (2.998x10 m/s) --> v= (2.998x10 m/s)/302.5x10 m = 9.911x10 s Planck's equation E=hv = (6.626x10 Js/photon)(9.911x10 s )= 3. A photon has a frequency of 5.5x10 Hz. Convert this to wavelength. c=λv so λ=c/v= (2.998x10 m/s)/5.5x10 Hz=5.5x10 m Remember Hz is equivalent to s 5.5x10 m (1nm/10 m)= Remember wavelengths are expressed in nanometers (nm) 4. What is the de Broglie wavelength of a 1.5g bouncy ball traveling at 20m/s? λ=h/mv Convert g to kg so 1.5g(1kg/10 g)=,0015kg Joule=Nm Newton=kgm/s λ=h/mv = 6.626x10 Js/(.0015kgx20m/s)=2.209x10 m Stoichiometry=Js/[kg(m/s)]=(Ns /kg)=(kgms /kgs )=m 2.209x10 m (1nm/10 m)= 5. If a metal had a threshold frequency of 7.38x10 Hz, would it display the photelectric effect with infrared light? The maximum frequency of infrared light is about 3x10 Hz. Since this is above the threshold frequency for the metal, it will display the photoelectric effect. However, if the threshold frequency was greater than the infrared light frequency, no photoelectric effect would occur. 1. How many crests and troughs are shown here? 2. The higher the frequency, the __________(smaller/greater) the wavelength. 3. Intensity of the incident light effects the amount of ______ emitted, whereas the frequency of incident light effects the _____ energy. 4. Using Planck's equation determine the frequency of radiation if it has an energy of 3.21x10 J/photon. 5. What is the energy (in joules) of a photon with a threshold frequency radiation of 8.67X10 Hz? 6. What is the de Broglie wavelenth of a 8.5g tennis bal traveling at 12.6m/s?

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A common paradigm in solving for pHs in weak acids and bases is that the equilibria of solutions containing one weak acid or one weak base. In most cases, the amount of \(\ce{H+}\) from the autoionization of water is negligible. For very dilute solutions, the amount of \(\ce{H+}\) ions from the autoionization of water must also be taken into account. Thus, a strategy is given here to deal with these systems. When two or more acids are present in a solution, the concentration of \(\ce{H+}\) (or pH) of the solution depends on the concentrations of the acids and their acidic constants . The hydrogen ion is produced by the ionization of all acids, but the ionizations of the acids are governed by their equilibrium constants, 's. Similarly, the concentration of \(\ce{OH-}\) ions in a solution containing two or more weak bases depends on the concentrations and values of the bases. For simplicity, we consider two acids in this module, but the strategies used to discuss equilibria of two acids apply equally well to that of two bases. If the pH is between 6 and 8, the contribution due to autoionization of water to \(\ce{[H+]}\) should also be considered. When autoionization of water is considered, the method is called the or the . This method is illustrated below. When the contribution of pH due to self-ionization of water cannot be neglected, there are two coupled equilibria to consider: \[\ce{HA \rightleftharpoons H+ + A-} \] and \[\ce{H2O \rightleftharpoons H+ + OH-}\] Thus, \[\begin{align} \ce{[H+]} &= ({\color{Red} x+y})\\ \ce{[A- ]} &= x\\ \ce{[OH- ]} &= y \end{align}\] and the two equilibrium constants are \[K_{\large\textrm{a}} = \dfrac{ ({\color{Red} x + y})\, x}{C - x} \label{1}\] and \[K_{\large\textrm{w}} = ({\color{Red} x + y})\, y \label{2}\] Although you may use the method of successive approximation, the formula to calculate the pH can be derived directly from Equations \(\ref{1}\) and \(\ref{2}\). Solving for \(\color{ref} x\) from Equation \(\ref{2}\) gives \[x = \dfrac{K_{\large\textrm{w}}}{y} - y\] and substituting this expression into \(\ref{1}\) results in \[K_{\large\textrm{a}} = \dfrac{({\color{Red} x+y}) \left(\dfrac{K_{\large\textrm{w}}}{y} - y\right)}{C - \dfrac{K_{\large\textrm{w}}}{y} + y}\] Rearrange this equation to give: \[\begin{align} \ce{[H+]} &= ({\color{Red} x+y})\\ &= \dfrac{C - \dfrac{K_{\large\textrm{w}}}{y} + y}{\dfrac{K_{\large\textrm{w}}}{y} - y}\, K_{\large\textrm{a}} \end{align}\] Note that \[\dfrac{K_{\large\textrm w}}{y} = \ce{[H+]}\] so \[y = \dfrac{K_{\large\textrm w}}{\ce{[H+]}}.\] Thus, we get: \[\ce{[H+]} = \dfrac{ C - \ce{[H+]} + \dfrac{K_{\large\textrm{w}}}{\ce{[H+]}}}{\ce{[H+]} - \dfrac{K_{\large\textrm{w}}}{\ce{[H+]}}}\, K_{\large\textrm{a}} \label{Exact}\] As written, Equation \(\ref{Exact}\) is complicated, but can be put into a polynomial form \[\ce{[H+]^3} + K_{\large\textrm{a}} \ce{[H+]^2} - \left( K_{\large\textrm{w}} + C K_{\large\textrm{a}} \right) \ce{[H+]} - K_{\large\textrm{w}} K_{\large\textrm{a}} =0 \label{Exact2}\] Solving for the exact hydronium concentration requires solving a third-order polynomial. While this is , it is an awkward equation to handle. Instead, we often consider two approximations to Equation \(\ref{Exact2}\) that can made under limiting conditions. If \([H^+] > 1 \times 10^{-6}\), then \[\dfrac{K_w}{[H^+]} < 1 \times 10^{-8}.\] This is small indeed compared to \([H^+]\) and \(C\) in Equation \(\ref{Exact}\). Thus, \[[H^+] \approx \dfrac{C - [H^+]}{[H^+]} K_{\large\textrm{a}}\] \[[H^+]^2 + K_{\large\textrm{a}} [H^+] - C K_{\large\textrm{a}} \approx 0 \label{Quad}\] Equation \(\ref{Quad}\) is a quadratic equation with two solutions. However, only one will be positive and real: \[[H^+] \approx \dfrac{-K_{\large\textrm{a}} + \sqrt{K_{\large\textrm{a}}^2 + 4 C K_{\large\textrm{a}}}}{2}\] If \([H^+] \ll C\), then \[C - [H^+] \approx C\] Equation \(\ref{Exact}\) can be simplified \(\begin{align} [H^+] &\approx \dfrac{C}{[H^+]}\: K_{\large\textrm{a}}\\ [H^+] &\approx \sqrt{C K_{\large\textrm{a}}} \end{align}\) The treatment presented in deriving Equation \(\ref{Exact}\) is more general, and may be applied to problems involving two or more weak acids in one solution. Calculate the \(\ce{[H+]}\), \(\ce{[Ac- ]}\), and \(\ce{[Cc- ]}\) when the solution contains 0.200 M \(\ce{HAc}\) (\(K_a = 1.8 \times 10^{-5}\)), and 0.100 M \(\ce{HCc}\) (the acidity constant \(K_c = 1.4 \times 10^{-3}\)). (\(\ce{HAc}\) is acetic acid whereas \(\ce{HCc}\) is chloroacetic acid). Assume x and y to be the concentrations of \(\ce{Ac-}\) and \(\ce{Cc-}\), respectively, and write the concentrations below the equations: \(\begin{array}{ccccc} \ce{HAc &\rightleftharpoons &H+ &+ &Ac-}\\ 0.200-x &&x &&x\\ \\ \ce{HCc &\rightleftharpoons &H+ &+ &Cc-}\\ 0.100-y &&y &&y \end{array}\) \[\ce{[H+]} = (x + y) \nonumber\] Thus, you have \[\dfrac{(x + y)\, x}{0.200 - x} = 1.8 \times 10^{-5} \label{Ex1.1}\] \[\dfrac{(x + y)\, y}{0.100 - y} = 1.4\times 10^{-3} \label{Ex1.2}\] Solving for x and y from Equations \(\ref{Ex1.1}\) and \(\ref{Ex1.2}\) may seem difficult, but you can often make some assumptions to simplify the solution procedure. Since \(\ce{HAc}\) is a weaker acid than is \(\ce{HCc}\), you expect x << y. Further, y << 0.100. Therefore, \(x + y \approx y\) and 0.100 - y => 0.100. Equation \(\ref{Ex1.2}\) becomes: \[\dfrac{ ( y)\, y}{0.100} = 1.4 \times 10^{-3} \label{Ex1.2a}\] which leads to \[\begin{align*} y &= (1.4 \times 10^{-3} \times 0.100)^{1/2}\\ &= 0.012 \end{align*}\] Substituting \(y\) in Equation \(\ref{Ex1.1}\) results in \[\dfrac{(x + 0.012)\, x}{0.200 - x} = 1.8 \times 10^{-5} \label{1'}\] This equation is easily solved, but you may further assume that \(0.200 - x \approx 0.200\), since \(x << 0.200\). Thus, \[\begin{align*} x &= \dfrac{-0.012 + (1.44\times 10^{-4} + 1.44\times 10^{-5})^{1/2}}{2}\\ &= 2.9\times 10^{-4}\:\: \longleftarrow \textrm{Small indeed compared to 0.200} \end{align*}\] You had a value of 0.012 for y by neglecting the value of x in Equation \(\ref{Ex1.2}\). You can now recalculate the value for y by substituting values for x and y in Equation \(\ref{Ex1.2}\). \[\dfrac{(2.9\times 10^{-4} + y)\, y}{0.100 - 0.012} = 1.4\times 10^{-3} \label{2"}\] Solving for y in the above equation gives \[y = 0.011 \nonumber\] You have improved the y value from 0.012 to 0.011. Substituting the new value for y in a successive approximation to recalculate the value for x improves its value from \(2.9 \times 10^{-4}\) to a new value of \(3.2 \times 10^{-4}\). Use your calculator to obtain these values. Further refinement does not lead to any significant changes for x or y. You should write down these calculations on your note pad, since reading alone does not lead to thorough understanding. A weak acid \(\ce{HA}\) has a \(K_a\) value of \(4.0 \times 10^{-11}\). What are the pH and the equilibrium concentration of \(\ce{A-}\) in a solution of 0.0010 M \(\ce{HA}\)? For the solution of this problem, two methods are given here. If you like the x and y representation, you may use method (a). The two equilibrium equations are: \[\begin{array}{ccccc} \ce{HA &\rightleftharpoons &H+ &+ &A-};\\ 0.0010-x &&x &&x\\ \\ \ce{H2O &\rightleftharpoons &H+ &+ &OH-}\\ &&y &&y \end{array}\] \[\ce{[H+]} = (x+y)\] \[\begin{align} \dfrac{(x+y)\, x}{0.0010-x} &= 4.0\times 10^{-}11 \label{3}\\ \\ (x+y)\, y &= 1\times 10^{-}14 \label{4} \end{align}\] Assume y << x, and x << 0.0010, then you have \[\begin{align} \dfrac{(x )\, x}{0.0010} &= 4.0\times 10^{-11} \label{3'} \\ x &= (0.0010 \times 4.0e^{-11})^{1/2}\\ &= 2.0\times 10^{-7} \end{align}\] Substituting \(2.0 \times ^{-7}\) for x in 4 and solving the quadratic equation for y gives, \[(2.0\times 10^{-}7+y)\, y = 1\times 10^{-14} \nonumber\] \[y = 4.1\times 10^{-8} \nonumber\] Substituting \(4.1 \times 10^{-8}\) in Equation \(\ref{3}\), but still approximating 0.0010-x by 0.0010: \[\dfrac{(x+4.1\times 10^{-8})\, x}{0.0010} = 4.0\times 10^{-11} \label{3''}\] Solving this quadratic equation for a positive root results in \[x = 1.8 \times 10^{-7} \;\text{M} \longleftarrow \textrm{Recall }x = \ce{[A- ]} \nonumber\] \[\begin{align*} \ce{[H+]} &= x + y\\ &= (1.8 + 0.41)\,1\times 10^{-7}\\ &= 2.2\times 10^{-7}\\ \ce{pH} &= 6.65 \end{align*}\] The next method uses the formula derived earlier. Using the formula from the exact treatment, and using \(2 \times 10^{-7}\) for all the \(\ce{[H+]}\) values on the right hand side, you obtain a new value of \(\ce{[H+]}\) on the left hand side, \[\begin{align*} \ce{[H+]} &= \dfrac{C - \ce{[H+]} + \dfrac{K_{\large\textrm w}}{\ce{[H+]}}}{\ce{[H+]} - \dfrac{K_{\large\textrm w}}{\ce{[H+]}}} K_{\large\textrm a}\\ &= 2.24\times 10^{-7}\\ \ce{pH} &= 6.65 \end{align*}\] The new \(\ce{[H+]}\) enables you to recalculate \(\ce{[A- ]}\) from the formula: \[\begin{align*} (2.24\times 10^{-7}) \ce{[A- ]} &= C K_{\large\textrm a} \nonumber\\ \ce{[A- ]} &= \dfrac{(0.0010) (4.0\times 10^{-11})}{2.24\times 10^{-7}} \nonumber\\ &= 1.8\times 10^{-7} \nonumber \end{align*}\] You may have attempted to use the approximation method: \[\begin{align*} x &= (C K_{\large\textrm a})^{1/2} \nonumber\\ &= 2.0\times 10^{-7}\: \mathrm{M\: A^-,\: or\: H^+;\: pH = 6.70} \nonumber \end{align*}\] and obtained a pH of 6.70, which is greater than 6.65 by less than 1%. However, when an approximation is made, you have no confidence in the calculated pH of 6.70. Water is both an acid and a base due to the autoionization, \[\ce{H2O \rightleftharpoons H+ + OH-} \nonumber\] However, the amount of \(\ce{H+}\) ions from water may be very small compared to the amount from an acid if the concentration of the acid is high. When calculating \(\ce{[H+]}\) in an acidic solution, approximation method or using the quadratic formula has been discussed in the modules on weak acids.

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Cohesive and adhesive forces are associated with bulk (or macroscopic) properties, and hence the terms are not applicable to the discussion of . When a liquid comes into contact with a surface (such as the walls of a graduated cylinder or a tabletop), both cohesive and adhesive forces will act on it. These forces govern the shape which the liquid takes on. Due to the effects of adhesive forces, liquid on a surface can spread out to form a thin, relatively uniform film over the surface, a process known as . Alternatively, in the presence of strong cohesive forces, the liquid can divide into a number of small, roughly spherical beads that stand on the surface, maintaining minimal contact with the surface. The term "cohesive forces" is a generic term for the collective (e.g., and ) responsible for the bulk property of liquids resisting separation. Specifically, these attractive forces exist between molecules of the substance. For instance, rain falls in droplets, rather than a fine mist, because water has strong cohesion, which pulls its molecules tightly together, forming droplets. This force tends to unite molecules of a liquid, gathering them into relatively large clusters due to the molecules' dislike for its surroundings. Similarly, the term "adhesive forces" refers to the attractive forces between unlike substance, such as mechanical forces (sticking together) and electrostatic forces (attraction due to opposing charges). In the case of a liquid wetting agent, adhesion causes the liquid to cling to the surface on which it rests. When water is poured on clean glass, it tends to spread, forming a thin, uniform film over the glasses surface. This is because the adhesive forces between water and glass are strong enough to pull the water molecules out of their spherical formation and hold them against the surface of the glass, thus avoiding the repulsion between like molecules. When a liquid is placed on a smooth surface, the relative strengths of the cohesive and adhesive forces acting on that liquid determine the shape it will take (and whether or not it will wet the surface). If the adhesive forces between a liquid and a surface are stronger, they will pull the liquid down, causing it to wet the surface. However, if the cohesive forces among the liquid itself are stronger, they will resist such adhesion and cause the liquid to retain a spherical shape and bead the surface. The meniscus is the curvature of a liquid's surface within a container, such as a graduated cylinder. However, before we explain why some liquid have a concave up meniscus while others share a concave down meniscus, we have to understand the adhesive forces at work of . Water, for example, is a polar molecule that consists of a partial positive charge on the hydrogens and a partial negative charge on the oxygen. Thus, within liquid water, each molecule's partial positive charge is attracted to its neighbor's partial negative charge. This is the origin of the cohesive forces within the water. Water molecules buried inside the liquid is then being pulled and pushed evenly in every direction, producing no net pull. Meanwhile, the molecules on the surface of the liquid, lacking pulling forces in the upward direction, thus encompass a net downward pull. How does this cohesive force create both a concave up and concave down surface then? The answer is in its relationship to the adhesive forces between the water molecules and the container's surface. When the cohesive force of the liquid is stronger than the adhesive force of the liquid to the wall, the liquid concaves down in order to reduce contact with the surface of the wall. When the adhesive force of the liquid to the wall is stronger than the cohesive force of the liquid, the liquid is more attracted to the wall than its neighbors, causing the upward concavity. In agitated glasses of wine, droplets of wine seemingly "float" above the meniscus of the liquid and form "tears." This age-old phenomenon is the result of surface tension and cohesive and adhesive forces. Alcohol is more volatile than water. As a result, "evaporation of alcohol produces a surface tension gradient driving a thin film up along the side of a wine glass" (Adamson). This process is called the "solutal Marangoni effect." Due to adhesive forces, some waters clings to the walls of the glass. The "tears" form from the cohesive forces within the water holding it together. It is important to note that the surface tension gradient is "the driving force for the of the liquid" (Gugliotti), but the actual formation of the tears is a result of cohesive and adhesive forces.

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Gibbs free energy, denoted \(G\), combines and into a single value. The change in free energy, \(\Delta G\), is equal to the sum of the enthalpy plus the product of the temperature and entropy of the system. \( \Delta G\) can predict the direction of the chemical reaction under two conditions: If \(ΔG\) is positive, then the reaction is nonspontaneous (i.e., an the input of external energy is necessary for the reaction to occur) and if it is negative, then it is spontaneous (occurs without external energy input). This quantity is the energy associated with a chemical reaction that can be used to do work, and is the sum of its enthalpy (H) and the product of the temperature and the entropy (S) of the system. This quantity is defined as follows: \[ G= H-TS \label{1.1} \] or more completely as \[ G= U+PV-TS \label{1.2} \] where Spontaneous - is a reaction that is consider to be natural because it is a reaction that occurs by itself without any external action towards it. Non spontaneous - needs constant external energy applied to it in order for the process to continue and once you stop the external action the process will cease. When solving for the equation, if change of G is negative, then it's spontaneous. If change of G if positive, then it's non spontaneous. The symbol that is commonly used for FREE ENERGY is G. can be more properly consider as "standard free energy change" In chemical reactions involving the changes in thermodynamic quantities, a variation on this equation is often encountered: \[ \underset{\text {change in free energy} }{\Delta G } = \underset{ \text {change in enthalpy}}{ \Delta H } - \underset{\text {(temperature) change in entropy}}{T \Delta S} \label{1.3} \] Calculate ∆G at 290 K for the following reaction: \[\ce{2NO(g) + O2(g) \rightarrow 2NO2(g)} \nonumber \] Given now all you have to do is plug in all the given numbers into Equation 3 above. Remember to divide \(\Delta S\) by 1000 \(J/kJ\) so that after you multiply by temperature, \(T\), it will have the same units, \(kJ\), as \(\Delta H\). \[\Delta S = -150 \cancel{J}/K \left( \dfrac{1\; kJ}{1000\;\cancel{J}} \right) = -0.15\; kJ/K \nonumber \] and substituting into Equation 3: \[\begin{align*} ∆G &= -120\; kJ - (290 \;\cancel{K})(-0.150\; kJ/\cancel{K}) \\[4pt] &= -120 \;kJ + 43 \;kJ \\[4pt] &= -77\; kJ \end{align*} \] What is the \(\Delta G\) for this formation of ammonia from nitrogen and hydrogen gas. \[\ce{N_2 + 3H_2 \rightleftharpoons 2NH_3} \nonumber \] The Standard free energy formations: NH =-16.45 H =0 N =0 \[\Delta G=-32.90\;kJ \;mol^{-1} \nonumber \] Since the changes of entropy of chemical reaction are not measured readily, thus, entropy is not typically used as a criterion. To obviate this difficulty, we can use \(G\). The sign of ΔG indicates the direction of a chemical reaction and determine if a reaction is spontaneous or not. The factors affect of a reaction (assume and are independent of temperature): Note: The standard Gibbs energy change (at which reactants are converted to products at 1 bar) for: \[ aA + bB \rightarrow cC + dD \label{1.4} \] \[ \Delta r G^o = c \Delta _fG^o (C) + d \Delta _fG^o (D) - a \Delta _fG^o (A) - b \Delta _fG^o (B) \label{1.5} \] \[\Delta _fG^0 = \sum v \Delta _f G^0 (\text {products}) - \sum v \Delta _f G^0 (\text {reactants}) \label{1.6} \] The standard-state free energy of reaction ( \(\Delta G^o\)) is defined as the free energy of reaction at standard state conditions: \[ \Delta G^o = \Delta H^o - T \Delta S^o \label{1.7} \] The standard-state free energy of formation is the change in free energy that occurs when a compound is formed from its elements in their most thermodynamically stable states at standard-state conditions. In other words, it is the difference between the free energy of a substance and the free energies of its constituent elements at standard-state conditions: \[ \Delta G^o = \sum \Delta G^o_{f_{products}} - \sum \Delta G^o_{f_{reactants}} \label{1.8} \] Used the below information to determine if \(NH_4NO_{3(s)}\) will dissolve in water at room temperature. This question is essentially asking if the following reaction is spontaneous at room temperature. \[\ce{NH4NO3(s) \overset{H_2O} \longrightarrow NH4(aq)^{+} + NO3(aq)^{-}} \nonumber \] This would normally only require calculating \(\Delta{G^o}\) and evaluating its sign. However, the \(\Delta{G^o}\) values are not tabulated, so they must be calculated manually from calculated \(\Delta{H^o}\) and \(\Delta{S^o}\) values for the reaction. \[ \Delta H^o = \sum n\Delta H^o_{f_{products}} - \sum m\Delta H^o_{f_{reactants}} \nonumber \] \[ \Delta H^o= \left[ \left( 1\; mol\; NH_3\right)\left(-132.51\;\dfrac{kJ}{mol} \right) + \left( 1\; mol\; NO_3^- \right) \left(-205.0\;\dfrac{kJ}{mol}\right) \right] \nonumber \] \[- \left[ \left(1\; mol\; NH_4NO_3 \right)\left(-365.56 \;\dfrac{kJ}{mol}\right) \right] \nonumber \] \[ \Delta H^o = -337.51 \;kJ + 365.56 \; kJ= 28.05 \;kJ \nonumber \] \[ \Delta S^o = \sum n\Delta S^o_{f_{products}} - \sum S\Delta H^o_{f_{reactants}} \nonumber \] \[ \Delta S^o= \left[ \left( 1\; mol\; NH_3\right)\left(113.4 \;\dfrac{J}{mol\;K} \right) + \left( 1\; mol\; NO_3^- \right) \left(146.6\;\dfrac{J}{mol\;K}\right) \right] \nonumber \] \[- \left[ \left(1\; mol\; NH_4NO_3 \right)\left(151.08 \;\dfrac{J}{mol\;K}\right) \right] \nonumber \] \[ \Delta S^o = 259.8 \;J/K - 151.08 \; J/K= 108.7 \;J/K \nonumber \] These values can be substituted into the free energy equation \[T_K = 25\;^oC + 273.15K = 298.15\;K \nonumber \] \[\Delta{S^o} = 108.7\; \cancel{J}/K \left(\dfrac{1\; kJ}{1000\;\cancel{J}} \right) = 0.1087 \; kJ/K \nonumber \] \[\Delta{H^o} = 28.05\;kJ \nonumber \] Plug in \(\Delta H^o\), \(\Delta S^o\) and \(T\) into Equation 1.7 \[\Delta G^o = \Delta H^o - T \Delta S^o \nonumber \] \[\Delta G^o = 28.05\;kJ - (298.15\; \cancel{K})(0.1087\;kJ/ \cancel{K}) \nonumber \] \[\Delta G^o= 28.05\;kJ - 32.41\; kJ \nonumber \] \[\Delta G^o = -4.4 \;kJ \nonumber \] This reaction is spontaneous at room temperature since \(\Delta G^o\) is negative. Therefore \(NH_4NO_{3(s)}\) will dissolve in water at room temperature. Calculate \(\Delta{G}\) for the following reaction at \(25\; ^oC\). Will the reaction occur spontaneously? \[NH_{3(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)} \nonumber \] given for the reaction calculate \(\Delta{G}\) from the formula \[\Delta{G} = \Delta{H} - T\Delta{S} \nonumber \] but first we need to convert the units for \(\Delta{S}\) into kJ/K (or convert \(\Delta{H}\) into J) and temperature into Kelvin The definition of Gibbs energy can then be used directly \[\Delta{G} = \Delta{H} - T\Delta{S} \nonumber \] \[\Delta{G} = -176.0 \;kJ - (298 \cancel{K}) (-0.284.8\; kJ/\cancel{K}) \nonumber \] \[\Delta{G} = -176.0 \;kJ - (-84.9\; kJ) \nonumber \] \[\Delta{G} = -91.1 \;kJ \nonumber \] Yes, this reaction is spontaneous at room temperature since \(\Delta{G}\) is negative. Let's consider the following reversible reaction: \[ A + B \leftrightharpoons C + D \label{1.9} \] The following equation relates the standard-state free energy of reaction with the free energy at any point in a given reaction (not necessarily at standard-state conditions): \[ \Delta G = \Delta G^o + RT \ln Q \label{1.10} \] At equilibrium, ΔG = 0 and Q=K. Thus the equation can be arranged into: \[\Delta{G} = \Delta{G}^o + RT \ln \dfrac{[C,D]}{[A,B]} \label{1.11} \] with The Gibbs free energy depends primarily on the reactants' nature and concentrations (expressed in the term and the logarithmic term of Equation 1.11, respectively). At equilibrium, \(\Delta{G} = 0\): no driving force remains The equilibrium constant is defined as \[K_{eq} = \dfrac{[C,D]}{[A,B]} \label{1.14} \] When \(K_{eq}\) is large, almost all reactants are converted to products. Substituting \(K_{eq}\) into Equation 1.14, we have: Rearrange, \[K_{eq} = 10^{-\Delta{G}^{o}/(2.303RT)} \label{1.17} \] This equation is particularly interesting as it relates the free energy difference under standard conditions to the properties of a system at equilibrium (which is rarely at standard conditions). What is for isomerization of dihydroxyacetone phosphate to glyceraldehyde 3-phosphate? If at equilibrium, we have \(K_{eq} = 0.0475\) at 298 K and pH 7. We can calculate: \[\Delta{G}^{o} = -2.303\;RT log_{10} K_{eq}= (-2.303) * (1.98 * 10^{-3}) * 298 * (log_{10} 0.0475) = 1.8 \;kcal/mol \nonumber \] Given: From equation 2: = 1.8 kcal/mol + 2.303 RT log (3*10 M/2*10 M) = -0.7 kcal/mol Under non-standard conditions (which is essential all reactions), the spontaneity of reaction is determined by , not . The relates the standard-state cell potential with the cell potential of the cell at any moment in time: \[ E = E^o - \dfrac {RT}{nF} \ln Q \label{1.18} \] with By rearranging this equation we obtain: \[ E = E^o - \dfrac {RT}{nF} \ln Q \label{1.19} \] multiply the entire equation by \(nF\) \[ nFE = nFE^o - RT \ln Q \label{1.20} \] which is similar to: \[ \Delta G = \Delta G^o + RT \ln Q \label{1.21} \] By juxtaposing these two equations: \[ nFE = nFE^o - RT \ln Q \label{1.22} \] \[ \Delta G = \Delta G^o + RT \ln Q \label{1.23} \] it can be concluded that: \[ \Delta G = -nFE \label{1.24} \] Therefore, \[ \Delta G^o = -nFE^o \label{1.25} \]

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The law of Conservation of Energy refers to an isolated system in which there is no net change in energy and where energy is neither created nor destroyed. Although there is no change in energy, energy can change forms, for example from potential to kinetic energy. In other words, potential energy (V) and kinetic energy (T) sum to a constant total energy (E) for a specific isolated system. \(E = T + V\) Another way that energy can change forms is heat (q) and work (w). As heat is applied to a closed system, the system does work by increasing its volume. \(w = P_{ext}\Delta{V}\) where P is the external pressure, and delta V is the change in volume. A classic example of this is a piston. As heat is added to the cylinder, the pressure inside the cylinder increases. The piston then rises to relieve the pressure difference between the pressure inside the cylinder and the external pressure. By increasing the volume in the cylinder, the piston has just done work. Reference the picture below. The sum of heat and work is the change in internal energy, \(\Delta{U}\). In an isolated system, \(q = -w\). Therefore, \(\Delta{U} = 0\). In quantum mechanics, the equation \(\hat{H}\psi_n = E_n\psi_n\) Where, The equation is analogous to the equation: \(E = T + V\)

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Because the elements of group 4 have a high affinity for oxygen, all three metals occur naturally as oxide ores that contain the metal in the +4 oxidation state resulting from losing all four ns (n − 1)d valence electrons. They are isolated by initial conversion to the tetrachlorides, as shown for Ti: \[\ce{2FeTiO3(s) + 6C(s) + 7Cl2(g) \rightarrow 2TiCl4(g) + 2FeCl3(g) + 6CO(g) }\label{1.1.1}\] followed by reduction of the tetrachlorides with an active metal such as Mg. The chemistry of the group 4 metals is dominated by the +4 oxidation state. Only Ti has an extensive chemistry in lower oxidation states. In contrast to the elements of group 3, the group 4 elements have important applications. Titanium (melting point = 1668°C) is often used as a replacement for aluminum (melting point = 660°C) in applications that require high temperatures or corrosion resistance. For example, friction with the air heats the skin of supersonic aircraft operating above Mach 2.2 to temperatures near the melting point of aluminum; consequently, titanium is used instead of aluminum in many aerospace applications. The corrosion resistance of titanium is increasingly exploited in architectural applications, as shown in the chapter-opening photo. Metallic zirconium is used in UO -containing fuel rods in nuclear reactors, while hafnium is used in the control rods that modulate the output of high-power nuclear reactors, such as those in nuclear submarines. Consistent with the periodic trends, the group 4 metals become denser, higher melting, and more electropositive down the column (Table \(\Page {1}\)). Unexpectedly, however, the atomic radius of Hf is slightly smaller than that of Zr due to the lanthanide contraction. Because of their ns (n − 1)d valence electron configurations, the +4 oxidation state is by far the most important for all three metals. Only titanium exhibits a significant chemistry in the +2 and +3 oxidation states, although compounds of Ti are usually powerful reductants. In fact, the Ti (aq) ion is such a strong reductant that it rapidly reduces water to form hydrogen gas. Reaction of the group 4 metals with excess halogen forms the corresponding tetrahalides (MX ), although titanium, the lightest element in the group, also forms dihalides and trihalides (X is not F). The covalent character of the titanium halides increases as the oxidation state of the metal increases because of increasing polarization of the anions by the cation as its charge-to-radius ratio increases. Thus TiCl is an ionic salt, whereas TiCl is a volatile liquid that contains tetrahedral molecules. All three metals react with excess oxygen or the heavier chalcogens (Y) to form the corresponding dioxides (MO ) and dichalcogenides (MY ). Industrially, TiO , which is used as a white pigment in paints, is prepared by reacting TiCl with oxygen at high temperatures: \[\ce{TiCl4(g) + O2(g) \rightarrow TiO2(s) + 2Cl2(g)}\label{1.1.2}\] The group 4 dichalcogenides have unusual layered structures with no M–Y bonds holding adjacent sheets together, which makes them similar in some ways to graphite (Figure \(\Page {1}\)). The group 4 metals also react with hydrogen, nitrogen, carbon, and boron to form hydrides (such as TiH ), nitrides (such as TiN), carbides (such as TiC), and borides (such as TiB ), all of which are hard, high-melting solids. Many of these binary compounds are nonstoichiometric and exhibit metallic conductivity.

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The first person to realize that white light was made up of the colors of the rainbow was Isaac Newton, who in 1666 passed sunlight through a narrow slit, then a prism, to project the colored spectrum on to a wall. This effect had been noticed previously, of course, not least in the sky, but previous attempts to explain it, by Descartes and others, had suggested that the white light became colored when it was refracted, the color depending on the angle of refraction. Newton clarified the situation by using a second prism to reconstitute the white light, making much more plausible the idea that the white light was composed of the separate colors. He then took a monochromatic component from the spectrum generated by one prism and passed it through a second prism, establishing that no further colors were generated. That is, light of a single color did not change color on refraction. He concluded that white light was made up of all the colors of the rainbow, and that on passing through a prism, these different colors were refracted through slightly different angles, thus separating them into the observed spectrum. The spectrum of , which turned out to be crucial in providing the first insight into atomic structure over half a century later, was first observed by Anders Angstrom in Uppsala, Sweden, in 1853. His communication was translated into English in 1855. Angstrom, the son of a country minister, was a reserved person, not interested in the social life that centered around the court. Consequently, it was many years before his achievements were recognized, at home or abroad (most of his results were published in Swedish). Most of what is known about atomic (and molecular) structure and mechanics has been deduced from spectroscopy. Figure 1.4.1 shows two different types of spectra. A continuous spectrum can be produced by an incandescent solid or gas at high pressure , for example, is a continuum). An emission spectrum can be produced by a gas at low pressure excited by heat or by collisions with electrons. An absorption spectrum results when light from a continuous source passes through a cooler gas, consisting of a series of dark lines characteristic of the composition of the gas. In 1802, William Wollaston in England had discovered (perhaps by using a thinner slit or a better prism) that in fact the solar spectrum itself had tiny gaps - there were many thin dark lines in the rainbow of colors. These were investigated much more systematically by Joseph von Fraunhofer, beginning in 1814. He increased the dispersion by using more than one prism. He found an "almost countless number" of lines. He labeled the strongest dark lines A, B, C, D, etc. Frauenhofer between 1814 and 1823 discovered nearly 600 dark lines in the solar spectrum viewed at high resolution and designated the principal features with the letters A through K, and weaker lines with other letters (Table \(\Page {1}\)). Modern observations of sunlight can detect many thousands of lines. It is now understood that these lines are caused by absorption by the outer layers of the Sun. The Fraunhofer lines are typical spectral absorption lines. These dark lines are produced whenever a cold gas is between a broad spectrum photon source and the detector. In this case, a decrease in the intensity of light in the frequency of the incident photon is seen as the photons are absorbed, then re-emitted in random directions, which are mostly in directions different from the original one. This results in an , since the narrow frequency band of light initially traveling toward the detector, has been turned into heat or re-emitted in other directions. By contrast, if the detector sees photons emitted directly from a glowing gas, then the detector often sees photons emitted in a narrow frequency range by quantum emission processes in atoms in the hot gas, resulting in an . In the Sun, Fraunhofer lines are seen from gas in the outer regions of the Sun, which are too cold to directly produce emission lines of the elements they represent. Gases heated to incandescence were found by Bunsen, Kirkhoff and others to emit light with a series of sharp wavelengths. The emitted light analyzed by a spectrometer (or even a simple prism) appears as a multitude of narrow bands of color. These so called are characteristic of the atomic composition of the gas. The line spectra of several elements are shown in Figure \(\Page {3}\). Obviously, if any pattern could be discerned in the spectral lines for a specifc atom (in contract to the mixture that Fraunhofer lines represent), that might be a clue as to the internal structure of the atom. One might be able to build a model. A great deal of effort went into analyzing the spectral data from the 1860's on. The big breakthrough was made by Johann Balmer, a math and Latin teacher at a girls' school in Basel, Switzerland. Balmer had done no physics before, and made his great discovery when he was almost sixty. Balmer decided that the most likely atom to show simple spectral patterns was the lightest atom, hydrogen. Angstrom had measured the four visible spectral lines to have wavelengths 656.21, 486.07, 434.01 and 410.12 nm (Figure \(\Page {4}\)). Balmer concentrated on just these four numbers, and found they were represented by the formula: \[\lambda = b \left( \dfrac{n_2^2}{n_2^2 -4} \right) \label{1.4.1}\] where The first four wavelengths of Equation \(\ref{1.4.1}\) (with n_2=3,4,5,6) were in excellent agreement with the experimental lines from Angstrom (Table \(\Page {2}\)). Balmer predicted that other lines exist in the infrared that correspond to = 7, 8, etc., and in fact some of them had already been observed, unbeknown to Balmer. The \(n_2\) integer in the Balmer series extends theoretically to infinity and the series resents a monotonically increasing energy (and frequency) of the absorption lines with increasing \(n_2\) values. Moreover, the energy difference between successive lines decreased as \(n_2\) increases (Figure 1.4.4). This behavior converges to a highest possible energy as Example \(\Page {1}\) demonstrates. If the lines are plot according to their \(\lambda\) on a linear scale, you will get the appearance of the spectrum in Figure \(\Page {4}\); these lines are called the . Balmer's general formula (Equation \(\ref{1.4.1}\)) can be rewritten in terms of the inverse wavelength typically called the (\(\widetilde{\nu}\)). \[ \widetilde{\nu}= \dfrac{1}{ \lambda} =R_H \left( \dfrac{1}{4} -\dfrac{1}{n_2^2}\right) \label{1.4.2} \] where He further conjectured that the 4 could be replaced by 9, 16, 25, … and this also turned out to be true - but these lines, further into the infrared, were not detected until the early twentieth century, along with the ultraviolet lines. The relation between wavelength and frequency for electromagnetic radiation is \[\lambda \nu= c\] In the SI system of units the wavelength, \lambda\) is measured in meters (m) and since wavelengths are usually very small one often uses the nanometer (nm) which is \(10^{-9}\; m\). The frequency (\(\nu\)) in the SI system is measured in reciprocal seconds 1/s − which is called a Hertz (after the discover of the ) and is represented by Hz. It is common to use the reciprocal of the wavelength in centimeters as a measure of the frequency of radiation. This unit is called a wavenumber and is represented by (\(\widetilde{\nu}\)) and is defined by \[ \widetilde{\nu}= \dfrac{1}{ \lambda} = \dfrac{\nu}{c}\] \[E = \dfrac{hc}{\lambda}\] \[E = hc \times \dfrac{1}{\lambda}\] \[E = hc\widetilde{\nu}\] \[E \propto \widetilde{\nu}\] From the behavior of the Balmer equation (Equation \(\ref{1.4.1}\) and Table \(\Page {2}\)), the value of \(n_2\) that gives the longest (i.e., greatest) wavelength (\(\lambda\)) is the smallest value possible of \(n_2\), which is \(n_2=3\) for this series. This results in \[\lambda_{longest} = (364.56 \;nm) \left( \dfrac{9}{9 -4} \right)\] \[\lambda_{longest} = (364.56 \;nm) \left( 1.8 \right) = 656.2\; nm\] This is also known as the \(H_{\alpha}\) line of atomic hydrogen and is bight red (Figure 1.4.4). For the shortest wavelength, it should be recognized that the shortest wavelength (greatest energy) is obtained at the limit of greatest (n_2\): \[ \lambda_{shortest} = \lim_{n_2 \rightarrow \infty} (364.56 \;nm) \left( \dfrac{n_2^2}{n_2^2 -4} \right) \] This can be solved via , or alternatively the limit can be expressed via the equally useful energy expression (Equation 1.4.2) and simply solved: \[ \widetilde{\nu}_{greatest}= \lim_{n_2 \rightarrow \infty} R_H \left( \dfrac{1}{4} -\dfrac{1}{n_2^2}\right) \] \[ \widetilde{\nu}_{greatest}= \lim_{n_2 \rightarrow \infty} R_H \left( \dfrac{1}{4}\right) = 27,434 \;cm^{-1} \] Since \( \dfrac{1}{\widetilde{\nu}}= \lambda\) in units of cm, this converts to 364 nm as the shortest wavelength possible for the Balmer series. The Balmer series is particularly useful in astronomy because the Balmer lines appear in numerous stellar objects due to the abundance of hydrogen in the universe, and therefore are commonly seen and relatively strong compared to lines from other elements. In an amazing demonstration of mathematical insight, in 1885 Balmer came up with a simple formula for predicting the wavelength of any of the lines in atomic Hydrogen in what we now know as the Balmer series (Equation \(\ref{1.4.2}\)). Three years later, Rydberg generalized this so that it was possible to determine the wavelengths of any of the lines in the hydrogen emission spectrum. Rydberg suggested that all atomic spectra formed families with this pattern (he was unaware of Balmer's work). It turns out that there are families of spectra following Rydberg's pattern, notably in the alkali metals, sodium, potassium, etc., but not with the precision the hydrogen atom lines fit the Balmer formula, and low values of \(n_2\) predicted wavelengths that deviate considerably. Rydberg's general equation is as follows: \[ \color{ref} \widetilde{\nu}= \dfrac{1}{ \lambda} =R_H \left( \dfrac{1}{n_1^2} -\dfrac{1}{n_2^2}\right) \label{1.5.1} \] where For the Balmer lines, \(n_1 =2\) and \(n_2\) can be any whole number between 3 and infinity. The various combinations of numbers that can be substituted into this formula allow the calculation the wavelength of any of the lines in the hydrogen emission spectrum; there is close agreement between the wavelengths generated by this formula and those observed in a real spectrum. The results given by Balmer and Rydberg for the spectrum in the visible region of the electromagnetic radiation start with \(n_2 = 3\), and \(n_1^2=2\). Is there a different series with the following formula (e.g., \(n_1=1\)? \[\dfrac{1}{\lambda} = R_{\textrm H} \left(\dfrac{1}{1^2} - \dfrac{1}{n^2} \right ) \label{1.5.2}\] The values for \(n_2\) and wavenumber \(\widetilde{\nu}\) for this series would be: These lines are in the UV region, and they are not visible, but they are detected by instruments; these lines form a . The existences of the Lyman series and Balmer's series suggest the existence of more series. For example, the series with \(n_2^2 = 3\) and \(n_1^2\) = 4, 5, 6, 7, ... is called . The spectral lines are grouped into series according to \(n_1\) values. Lines are named sequentially starting from the longest wavelength/lowest frequency of the series, using Greek letters within each series. For example, the (\(n_1=1/n_2=2\)) line is called "Lyman-alpha" (Ly-α), while the (\(n_1=3/n_2=7\)) line is called "Paschen-delta" (Pa-δ). The first six series have specific names: The so-called Lyman series of lines in the emission spectrum of hydrogen corresponds to transitions from various excited states to the = 1 orbit. Calculate the wavelength of the lowest-energy line in the Lyman series to three significant figures. In what region of the electromagnetic spectrum does it occur? lowest-energy orbit in the Lyman series wavelength of the lowest-energy Lyman line and corresponding region of the spectrum We can use the Rydberg equation to calculate the wavelength: \[ \dfrac{1}{\lambda }=\Re \left ( \dfrac{1}{n_{1}^{2}} - \dfrac{1}{n_{2}^{2}}\right ) \] For the Lyman series, \(n_1 = 1\). \[ \dfrac{1}{\lambda }=\Re \left ( \dfrac{1}{n_{1}^{2}} - \dfrac{1}{n_{2}^{2}}\right )=1.097 \times m^{-1}\left ( \dfrac{1}{1}-\dfrac{1}{4} \right )= 8.228 \times 10^{6}\; m^{-1} \] Spectroscopists often talk about energy and frequency as equivalent. The cm unit is particularly convenient. The infrared range is roughly 200 - 5,000 cm , the visible from 11,000 to 25.000 cm and the UV between 25,000 and 100,000 cm . The units of cm are called wavenumbers, although people often verbalize it as inverse centimeters. We can convert the answer in part A to cm . \[ \widetilde{\nu} =\dfrac{1}{\lambda }=8.228\times 10^{6}\cancel{m^{-1}}\left (\dfrac{\cancel{m}}{100\;cm} \right )=82,280\: cm^{-1} \] and \[\lambda = 1.215 \times 10^{−7}\; m = 122\; nm \] This emission line is called Lyman alpha. It is the strongest atomic emission line from the sun and drives the chemistry of the upper atmosphere of all the planets producing ions by stripping electrons from atoms and molecules. It is completely absorbed by oxygen in the upper stratosphere, dissociating O molecules to O atoms which react with other O molecules to form stratospheric ozone This wavelength is in the ultraviolet region of the spectrum. The Pfund series of lines in the emission spectrum of hydrogen corresponds to transitions from higher excited states to the \( = 5\) orbit. Calculate the wavelength of the line in the Pfund series to three significant figures. In which region of the spectrum does it lie? 4.65 × 10 nm; infrared The above discussion presents only a phenomenological description of hydrogen emission lines and fails to provide a probe of the nature of the atom itself. Clearly a continuum model based on classical mechanics is not applicable.  (Beams Professor,  , 

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Under certain conditions, the 2nd order kinetics can be approximated as first-order kinetics. These greatly simplify quantifying the reaction dynamics. A can be challenging to follow mostly because the two reactants involved must be measured simultaneously. There can be additional complications because certain amounts of each reactant are required to determine the reaction rate, for example, which can make the cost of one's experiment high if one or both of the needed reactants are expensive. To avoid more complicated, expensive experiments and calculations, we can use the pseudo-1 -order reaction, which involves treating a 2 order reaction like a . In with two reactant species, \[\ce{A + B -> products}\nonumber \] the rate of disappearance of \(A\) is \[\dfrac{d[A]}{dt}= -k[A,B]\nonumber \] as discussed previously ( ), the integrated rate equation under the condition that [A] and [B] are not equal is \[ \dfrac{1}{[B]_o - [A]_o}\ln \dfrac{[B,A]_o}{[A,B]_o} = kt \label{2nd} \] However, when \([B]_0 \gg [A]_0\), then \([B]_0 \approx [B]\) and Equation \(\ref{2nd}\) becomes \[\dfrac{1}{[B]_0-[A]_0}\ln \dfrac{[B,A]_0}{[B]_0[A]} \approx \dfrac{1}{[B]}\ln \dfrac{[A]_0}{[A]}=kt \nonumber \] or \[[A] = [A]_0 e^{-[B]kt}\nonumber \] This functional form of the decay kinetics is similar ot the first order kinetics and the system is said to operate under pseudo-first order kinetics. To reach a pseudo-1 -order reaction, we can manipulate the initial concentrations of the reactants. One of the reactants, \(\ce{B}\), for example, would have a significantly higher concentration, while the other reactant, \(A\), would have a significantly lower concentration. We can then assume that the concentration of reactant \(B\) effectively remains constant during the reaction because its consumption is so small that the change in concentration becomes negligible. Because of this assumption, we can multiply the reaction rate, \(k \), with the reactant with assumed constant concentration, \(B\), to create a new rate constant (\(k'=k[B]\)) that will be used in the new rate equation, \[\text{Rate}=k'[A]\nonumber \] as the new rate constant so we can treat the 2 order reaction as a 1 order reaction. One can also do the same by overloading the initial concentration of \(A\) so that it effectively remains constant during the course of the reaction For example, if one were to dump a liter of 5 M \(\ce{HCl}\) into a 55 M ocean, the concentration of the mixture would be closer or equal to that of the ocean because there is so much water physically compared to the \(\ce{HCl}\). Even if the amount of water was one liter this would still be the case because 55 M is relatively large compared to 5 M. If we have an instance where there are more than two reactants involved in a reaction, all we would have to do to make the reaction pseudo-1 -order is to make the concentrations of all but one of the reactants very large. If there were three reactants, for example, we would make two of the three reactants be in excess (whether in amount or in concentration) and then monitor the dependency of the third reactant. We can write the pseudo -order reaction equation as: \[ [A] = [A]_0 e^{-[B]_{0}kt}\label{eq1} \] or \[[A] = [A]_0 e^{-k^{'}t} \nonumber \] where By using natural log to both sides of the pseudo-1 -order equation we get: \[ \ln \left ( \dfrac{A}{A_0} \right ) =-k[B]_0 t\nonumber \] or \[\ln \left ( \dfrac{A}{A_0} \right ) = -k^{'}t \nonumber \] If a 2 order reaction has the rate equation \(\text{Rate} = k[A,B]\), and the rate constant, \(k\), is \(3.67\, M^{-1}s^{-1}\), \([A]\) is \(4.5\, M\) and \([B]\) is \(99\, M\), what is the rate constant of its pseudo-1st-order reaction? Because \([B]\) is in excess we multiply \(99\, M\) with \(3.67\, M^{-1}s^{-1}\) \[(99\,M)(3.67\, M^{-1}s^{-1}) = 363.33\,s^{-1}\nonumber \] If [A] = 55 M at 39 s, [A] = 99 M, and [B] = 1000 M, what is the 2 order reaction rate constant? Use the Equation \ref{eq1} (55 M) = (99 M)e k' = 1.507 x 10 M s What is the concentration of A at time 45 s if [A] = 1M, [B] = 45 M, and 2 order rate constant is 0.6 M s ? Use the Equation \ref{eq1} [A] = (1)e [A] = 1.88 x 10 M What is the rate of a reaction if [A] = 560 M, [B] = 0.2M, and 2 order rate constant is 0.1 M s ? Use the equation R = k'[A,B] R = (0.1M s )(560M)(0.2M) Rate = 11.2Ms Half-life refers to the time required to decrease the concentration of a reactant by half, so we must solve for \(t\). Here, \([B]\) will be the reactant in excess, and its concentration will stay constant. \([A]_o\) is the initial concentration of \(A\); thus the half-life concentration of \(A\) is \(0.5[A]_o\). The pseudo-1 -order reaction equation can be wr \[ [A] = [A]_o e^{-[B]kt} \;\;or\;\; \dfrac{[A]} {[A]_o}= e^{-k^{'}t}\nonumber \] \[ \ln \left ( \dfrac{[A]} {[A]_o} \right )= k^{'}t \nonumber \] \[ \ln \left ( \dfrac{1/2[A]_o} {[A]_o} \right ) = \ln \left ( \dfrac {1}{2} \right )=- k^{'}t_{1/2} \nonumber \] Recalling that \( k^{'}=k [B] \), \( [B] \approx [B]_{0} \) and that \(-\ln (1/2) = \ln 2\): \[\ln(2)=k[B]_{0}t_{1/2}\nonumber \] or \[ t_{1/2}=\dfrac{\ln 2}{k[B]_{0}} \nonumber \] What is the half-life of a reaction with [A] = 109 M, [B] = 1 M, k' = 45 M s ? Because [A] is in excess we can multiply the k' with [A] to find k (109M)(45M s ) = 4905s t = (ln0.5 / -k) t = 1.41 X 10 s

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The linear [Ag(NH ) ] ion Although [Ag(en)]ClO involves a normally bidentate ligand, in this case the structure is polymeric and the silver ion still retains a CN=2 with the N atoms (from different ligands) at ~180 degrees to each other. [Cu(CN) ] [Cu(PPh ) Br] To help view more easily, the H atoms are turned off. [Rh(PPh ) ] To help view more easily, the H atoms are turned off. Copyr Cl cisplatin - cis-Pt(NH ) Cl The isomer is a powerful anti-cancer drug whereas the is inactive. [Ni(CN) ] tris(cis-1,2-diphenylethene-1,2-dithiolato)rhenium The ReS6 geometry is perfectly trigonal prismatic. Hexol The first 'truly' inorganic complex to be resolved into its optical isomers. [Co(en) ]Cl The classic example of optical isomerism in octahedral coordination complexes (H atoms not shown). [Co(NH ) CO ] K [NbOF ] [V(III)(Hedta)(H O)]H O bis-(tert-butylacac) (DMSO)di-oxoUranium The UO7 geometry fits a pentagonal bipyramid. Zr(acac) (NO ) [Zr(C O ) ] is reported to have this shape as well. U(acac) UO (OAc) Hydrated salts of the lanthanide elements eg Eu(H O) ] Tetrakis(nitrato-O,O')-bis(triphenylphosphine oxide) cerium(IV) Another example is [Ce(NO ) ] This is not a common stereochemistry. In aqua-(12-crown-4)-tris(nitrato-O,O')-cerium(III) (12-crown-4) solvate and (15-crown-5)-tris(nitrato-O,O')-cerium(III) the Cerium ion is 11 coordinate. Ceric ammonium nitrate -(NH ) Ce(NO )

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Reaction rates depend on the energy required (the activation barrier) and the energy available. They also may involve collisions between molecules. If two molecules need to collide in order for a reaction to take place, then factors that influence the ease of collisions will be important. The more energy there is available to the molecules, the faster they will move around, and the more likely they are to bump into each other. Higher temperatures ought to lead to more collisions and a greater frequency of reactions between molecules. In the drawing below, the cold, sluggish molecules on the left are not likely to collide, but the energetic molecules on the right are due to collide at any time. These principles are summarized as follows: Phase also has a pronounced effect on the mobility of molecules. Molecules in the gas phase are free to move around, and they do so quickly. On the other hand, they are pretty well spread out. Nevertheless, collisions in the gas phase happen quite easily, which make gas-phase reactions happen more readily. At the opposite extreme, molecules in the solid phase are not very mobile at all (reactions may involve atoms or ions, rather than molecules, but the same arguments apply). Not many collisions happen. As a result, reactions often happen extremely slowly in the solid state. Reactions are mostly limited to the grain boundaries: the surfaces of the grains, where they are in contact with each other. Nothing happens in the middle of a lump of solid, which remains unreacted. If a solid is heated, the molecules can move around a little more. They may even leave their crystal lattice (if the solid is crystalline) and diffuse very slowly through the solid. Many solid state reactions are run at elevated temperature. There are also many solid state reactions that are conducted in combination with gas-phase reactants. The solid reactants are often heated in a furnace while gas-phase reactants flow over them. Many reactions are performed in the liquid state, either because the reactants are already liquid of because the solid reactants are heated past their melting point. In the liquid phase, molecules are much more mobile and collisions are much more frequent than in the solid phase. It is also very common to run reactions in solution. In solution, a compound that is meant to undergo reaction is dissolved in a solvent. The solvent must be a liquid at the temperature at which the reaction will be run, so that molecules will be very mobile, but will still be close together, so collisions are favored. There are many advantages to running reactions in solution. The reactant molecules are very mobile and pretty close together, so that collisions are facilitated. If the reaction is exothermic and gives of a lot of heat, the excess heat can be absorbed by the solvent molecules and carried away. That can be important in controlling reactions and avoiding decomposition. Also, we will see that the rate of collisions can be controlled by adding more solvent or less, in order to slow the reaction down or speed it up. In this way, the reaction rate can be controlled to some extent. There are many liquids that are commonly used as solvents. Dichloromethane, toluene, dimethylformamide, tetrahydrofuran and acetonitrile are some common "organic" solvents, so called because they are based on carbon, which forms the basis of molecules in organisms. These different solvents offer a range of polarities, offering an ability to dissolve a variety of different reactants. Water may be the most common solvent on the planet, and it is non-toxic, so it is very appealing for use in large-scale, industrial reactions. However, it is not very good at dissolving non-polar reactants. The reactant compound could be a liquid or a solid. It only has to have strong enough intermolecular attractions with the solvent molecules so that it can be dissolved. Individual molecules of the reactant become lost among the solvent molecules move freely. The disadvantage of using a solvent is that the solvent must be removed at the end of the reaction so that the desired product can be isolated and used. That means the solvent may eventually be thrown away as waste. That practice is less efficient and less environmentally friendly, although the solvent could possibly be recycled. One method of dissolving reactants that is potentially greener is the use of supercritical fluids. In this approach, gases such as carbon dioxide are pressurized until they turn into liquids. In this form, carbon dioxide is a pretty good solvent, and reactions can be run when reactants are dissolved in it. At the end of the reaction, a valve is opened, releasing the pressure, and the carbon dioxide turns back into a gas. It can be stored and re-pressurized for another reaction. ,

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Born forces are one type of force that acts upon atoms in an ionic lattice. In simplest terms, because ions have some finite size, electron-electron and nucleus-nucleus interactions occur and give rise to repulsion forces and electrostatic potential, both called Born forces. Lattice energy is the energy released when gaseous cations and anions bond to form a solid ionic compound. With the Born-Lande equation one can calculate the lattice energy of a crystalline ionic compound. Born and Lande theorized around the turn of the century, that lattice energy of a crystalline ionic compound could be found by calculating terms of electrostatic potential and a repulsive potential. \[\Delta U = \dfrac{-LA|Z_+|\,|Z_-| e^2}{4\pi\epsilon_o r} \tag{1}\] with The first potential is the force of attraction. It is a negative value because it pulls the two atoms closer together, and the forming of a bond is energetically favorable. The negative value of Avogadro’s number (6.022x10 ) times a Madelung Constant (varies) times the absolute value of the charge of the cation times the absolute value of the charge of the anion, times the charge of an electron (1.6022x10 C) all over four times pi times the emissivity of space times the ion radius. \[\Delta U = \dfrac{-LB}{r^n} \tag{2}\] with The second equation is the repulsive force. It found by multiplying Avogadro’s number (6.022x10 ) by a repulsion coefficient, and dividing that by the ionic radius raised to the power of a Born exponent (some number between 5 and 12). Attractive forces are affected by the charge of the ions and their radii. Ions with large charges (like Mg or O ) have greater attractive potential than those with smaller charges (like Na or F ). Smaller ions (like Li or Cl ) also have greater electrostatic potential than larger ions (like I or Cs ). The Madelung constant is dependent on the crystal structure type. This value is found in tables online or in a text, but in general can be thought of as large with larger cation-cation distances and anion-anion distances. A structure with fluorite geometry has a relatively large A value, whereas rock salt crystals have a much lower A value. Repulsive forces are mainly determined by the born exponent. The Born exponent is dictated by the electronic configuration of the noble gas in the row above it on the periodic table (a closed shell). Max Born was a German physicist and mathematician who was instrumental in the development of quantum mechanics. He also made contributions to solid-state physics and optics and supervised the work of a number of notable physicists in the 1920s and 30s. Born won the 1954 Nobel Prize in Physics for his "fundamental research in Quantum Mechanics, especially in the statistical interpretation of the wave function". Plug in and cancel SI units to the Born-Lande equation to find the units of lattice energy, electrostatic potential, and repulsion forces. All energies are in units of kJ/mol

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Just as there were two mechanisms for nucleophilic substitution, there are two elimination mechanisms. The E1 mechanism is nearly identical to the S 1 mechanism, differing only in the course of reaction taken by the carbocation intermediate. As shown by the following equations, a carbocation bearing beta-hydrogens may function either as a Lewis acid (electrophile), as it does in the S 1 reaction, or a Brønsted acid, as in the E1 reaction. Thus, hydrolysis of tert-butyl chloride in a mixed solvent of water and acetonitrile gives a mixture of 2-methyl-2-propanol (60%) and 2-methylpropene (40%) at a rate independent of the water concentration. The alcohol is the product of an S 1 reaction and the alkene is the product of the E1 reaction. The characteristics of these two reaction mechanisms are similar, as expected. They both show first order kinetics; neither is much influenced by a change in the nucleophile/base; and both are relatively non-stereospecific. (CH ) – + H → [ (CH ) ] + + H → (CH ) – H + (CH ) =CH + H + H To summarize, when carbocation intermediates are formed one can expect them to react further by one or more of the following modes: Since the S 1 and E1 reactions proceed via the same carbocation intermediate, the product ratios are difficult to control and both substitution and elimination usually take place. Having discussed the many factors that influence nucleophilic substitution and elimination reactions of alkyl halides, we must now consider the practical problem of predicting the most likely outcome when a given alkyl halide is reacted with a given nucleophile. As we noted earlier, several variables must be considered, . The nature of the halogen substituent on the alkyl halide is usually not very significant if it is Cl, Br or I. In cases where both S 2 and E2 reactions compete, chlorides generally give more elimination than do iodides, since the greater electronegativity of chlorine increases the acidity of beta-hydrogens. Indeed, although alkyl fluorides are relatively unreactive, when reactions with basic nucleophiles are forced, elimination occurs (note the high electronegativity of fluorine).

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We have not yet considered the factors that influence elimination reactions, such as example in the group presented at the of this section. (CH ) - + CN → (CH ) =CH + + HCN We know that t-butyl bromide is not expected to react by an S 2 mechanism. Furthermore, the ethanol solvent is not sufficiently polar to facilitate an S 1 reaction. The other reactant, cyanide anion, is a good nucleophile; and it is also a decent base, being about ten times weaker than bicarbonate. Consequently, a base-induced elimination seems to be the only plausible reaction remaining for this combination of reactants. To get a clearer picture of the interplay of these factors consider the reaction of a 2º-alkyl halide, isopropyl bromide, with two different nucleophiles. In the methanol solvent used here, methanethiolate has greater nucleophilicity than methoxide by a factor of 100. Methoxide, on the other hand is roughly 10 times more basic than methanethiolate. As a result, we see a clear-cut difference in the reaction products, which reflects (bonding to an electrophilic carbon) versus (bonding to a proton). Kinetic studies of these reactions show that they are both second order (first order in R–Br and first order in Nu: ), suggesting a bimolecular mechanism for each. The substitution reaction is clearly . The corresponding designation for the elimination reaction is . An energy diagram for the single-step bimolecular E2 mechanism is shown below. We should be aware that the E2 transition state is less well defined than is that of S 2 reactions. More bonds are being broken and formed, with the possibility of a continuum of states in which the extent of C–H and C–X bond-breaking and C=C bond-making varies. For example, if the R–groups on the beta-carbon enhance the acidity of that hydrogen, then substantial breaking of C–H may occur before the other bonds begin to be affected. Similarly, groups that favor ionization of the halogen may generate a transition state with substantial positive charge on the alpha-carbon and only a small degree of C–H breaking. For most simple alkyl halides, however, it is proper to envision a balanced transition state, in which there has been an equal and synchronous change in all the bonds. Such a model helps to explain an important regioselectivity displayed by these elimination reactions. If two or more structurally distinct groups of beta-hydrogens are present in a given reactant, then several constitutionally isomeric alkenes may be formed by an E2 elimination. This situation is illustrated by the 2-bromobutane and 2-bromo-2,3-dimethylbutane elimination examples given below. By using the strongly basic hydroxide nucleophile, we direct these reactions toward elimination. In both cases there are two different sets of beta-hydrogens available to the elimination reaction (these are colored red and magenta and the alpha carbon is blue). If the rate of each possible elimination was the same, we might expect the amounts of the isomeric elimination products to reflect the number of hydrogens that could participate in that reaction. For example, since there are three 1º-hydrogens (red) and two 2º-hydrogens (magenta) on beta-carbons in 2-bromobutane, statistics would suggest a 3:2 ratio of 1-butene and 2-butene in the products. This is not observed, and the latter predominates by 4:1. This departure from statistical expectation is even more pronounced in the second example, where there are six 1º-beta-hydrogens compared with one 3º-hydrogen. These results point to a strong regioselectivity favoring the more highly substituted product double bond, an empirical statement generally called the . The main factor contributing to Zaitsev Rule behavior is the stability of the alkene. We noted earlier that carbon-carbon double bonds are stabilized (thermodynamically) by alkyl substituents, and that this stabilization could be evaluated by appropriate measurements. Since the E2 transition state has significant carbon-carbon double bond character, alkene stability differences will be reflected in the transition states of elimination reactions, and therefore in the activation energy of the rate-determining steps. From this consideration we anticipate that if two or more alkenes may be generated by an E2 elimination, the more stable alkene will be formed more rapidly and will therefore be the predominant product. This is illustrated for 2-bromobutane by the energy diagram below. The propensity of E2 eliminations to give the more stable alkene product also influences the distribution of product stereoisomers. In the elimination of 2-bromobutane, for example, we find that trans-2-butene is produced in a 6:1 ratio with its cis-isomer. The Zaitsev Rule is a good predictor for simple elimination reactions of alkyl chlorides, bromides and iodides as long as relatively small strong bases are used. Thus hydroxide, methoxide and ethoxide bases give comparable results. Bulky bases such as tert-butoxide tend to give higher yields of the less substituted double bond isomers, a characteristic that has been attributed to steric hindrance. In the case of 2-bromo-2,3-dimethylbutane, described above, tert-butoxide gave a 4:1 ratio of 2,3-dimethyl-1-butene to 2,3-dimethyl-2-butene ( essentially the opposite result to that obtained with hydroxide or methoxide). This point will be discussed further once we know more about the the structure of the E2 transition state.

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In practice, the Joule-Thomson experiment is done by allowing gas from a pressure vessel to pass through an insulated tube. The tube contains a throttling valve or a porous plug through which gas flows slowly enough so that the gas upstream from the plug is at a uniform pressure \(P_{\mathrm{1}}\), and the gas downstream is at a uniform pressure \(P_{\mathrm{2}}\). In general, the temperature of the downstream gas is different from that of the upstream gas. Depending on the initial temperature and pressure, the pressure drop, and the gas, the temperature of the gas can either decrease or increase as it passes through the plug. (We see below that it must be constant if the gas is ideal.) The temperature change is called the Joule-Thomson effect. The enthalpy of the gas remains constant. If the measured temperature and pressure changes are \(\mathrm{\Delta }T\) and \(\mathrm{\Delta }P\), their ratio is called the Joule-Thomson coefficient, \({\mu }_{JT}\). We define \[\mathrm{\ }{\mu }_{JT} = {\left(\frac{\partial T}{\partial P}\right)}_H\mathrm{\approx }\frac{\mathrm{\Delta }T}{\mathrm{\Delta }P}\] To see that the enthalpy of the gas is the same on both sides of the plug, we consider an idealized version of the experiment, in which the flow of gas through the plug is controlled by the coordinated movement of two pistons. (See Figure 3.) We suppose that the gas is pushed through the plug in such a way that the upstream pressure remains constant at \(P_1\) and the downstream pressure remains constant at \(P_2\). Let us consider the changes that result when one mole of gas passes through the plug under these conditions. Initially, there are \(n_1+1\) moles of gas on the upstream side at a pressure \(P_1\), occupying a volume \(\left(n_1+1\right){\overline{V}}_1\), at a temperature \(T_1\), and having an energy per mole of \({\overline{E}}_1\). On the downstream side, there are \(n_2\) moles of gas at a pressure \(P_2\), occupying a volume \(n_2{\overline{V}}_2\), but having a temperature \(T_2\) and an energy per mole of \({\overline{E}}_2\). When the process is complete, there are \(n_1\) moles of gas on the upstream side, still at a pressure \(P_1\) and temperature \(T_1\), but occupying a volume \(n_1{\overline{V}}_1\). On the downstream side, there are \(n_2+1\) moles of gas at pressure \(P_2\), occupying volume \(\left(n_2+1\right){\overline{V}}_2\) at a temperature \(T_2\) and with an energy per mole of \({\overline{E}}_2\). On the upstream side, \(\Delta E_1=-\overline{E}_1\) and \(w_1=-P_1\left[n_1 \overline{V}_1-\left(n_1+1\right) \overline{V}_1 \right]=P_1 \overline{V}_1\) On the downstream side, \(\Delta E_2={\overline{E}}_2\), and \[w_2=-P_2\left[\left(n_2+1\right){\overline{V}}_1-n_2{\overline{V}}_1\right]=-P_2{\overline{V}}_2\] Since the process is adiabatic, any heat taken up by the upstream gas must be surrendered by the downstream gas, so that \(q_1+q_2=0\). For the process of moving the mole of gas across the plug, \[\Delta E=\Delta E_1+\Delta E_2=-\overline{E}_1+\overline{E}_2=q_1+q_2+w_1+w_2=P_1\overline{V}_1-P_2\overline{V}_2\] from which \[{\overline{E}}_1+P_1{\overline{V}}_1={\overline{E}}_2+P_2{\overline{V}}_2\] or \[{\overline{H}}_1={\overline{H}}_2\] so that we have \(\Delta H=0\) for the expansion. In practice, it is convenient to measure downstream pressures and temperatures, \(P_2\) and \(T_2\), in a series of experiments in which the upstream pressure and temperature, \(P_1\) and \(T_1\), are constant. The enthalpy of the gas is the same at each of these pressure-temperature points. A graph of these points is an isenthalpic (constant enthalpy) curve. At any given pressure and temperature, the Joule-Thomson coefficient, \({\mu }_{JT}\), is the slope of this curve. We can also express \(\mathrm{\ }{\mu }_{JT}\) as a function of the heat capacity, \(C_P\), and the coefficient of thermal expansion, \(\alpha\), where \(\alpha =V^{-1}{\left({\partial V}/{\partial T}\right)}_P\). We begin by expressing \(d\overline{H}\) as a function of temperature and pressure: \[d\overline{H}={\left(\frac{\partial \overline{H}}{\partial T}\right)}_PdT+{\left(\frac{\partial \overline{H}}{\partial P}\right)}_TdP\] If we divide through by \(dP\) and hold \(\overline{H}\) constant, we obtain \[0={\left(\frac{\partial \overline{H}}{\partial T}\right)}_P{\left(\frac{\partial T}{\partial P}\right)}_{\overline{H}}+{\left(\frac{\partial \overline{H}}{\partial P}\right)}_T\] so that \[{\mu }_{JT}={\left(\frac{\partial T}{\partial P}\right)}_{\overline{H}}=-{{\left(\frac{\partial \overline{H}}{\partial P}\right)}_T}/{{\left(\frac{\partial \overline{H}}{\partial T}\right)}_P}=-\frac{1}{C_P}{\left(\frac{\partial \overline{H}}{\partial P}\right)}_T\] If we substitute the coefficient of thermal expansion into the expression for \({\left({\partial \overline{H}}/{\partial P}\right)}_T\) that we develop in , we have \[{\left(\frac{\partial \overline{H}}{\partial P}\right)}_T=\overline{V}-T{\left(\frac{\partial \overline{V}}{\partial T}\right)}_P=\overline{V}-\alpha \overline{V}T=\overline{V}\left(1-\alpha T\right)\] For an ideal gas, \({\left({\partial \overline{V}}/{\partial T}\right)}_P={\overline{V}}/{T}\), so that both \({\left({\partial \overline{H}}/{\partial P}\right)}_T\) and \({\mu }_{JT}\) are zero. For real gases, we substitute into the expression for \({\mu }_{JT}\) to find \[{\mu }_{JT}={\left(\frac{\partial T}{\partial P}\right)}_{\overline{H}}=-\frac{1}{C_P}{\left(\frac{\partial \overline{H}}{\partial P}\right)}_T=-\frac{\overline{V}}{C_P}\left(1-\alpha T\right)\] Given \(\overline{V}\) and any two of \({\mu }_{JT}\), \(C_P\), or \(\alpha\), we can find the third from this relationship. Making the same substitutions using the partial derivatives we found above for a van der Waals gas, we find \[{\mu }_{JT}=-\frac{1}{C_P}\left(\overline{V}-\frac{RT}{\gamma \left(P,\overline{V}\right)}\right)\] Given that the van der Waals equation oversimplifies the effects of intermolecular forces, we can anticipate that calculation of the Joule-Thomson coefficient from the van der Waals parameters is likely to be qualitatively correct, but in poor quantitative agreement with experimental results. Figure 4 compares calculated and experimental curves for the Joule-Thomson coefficient of nitrogen gas at 0 C from 1 to 200 bar. (Calculated values take \(a=0.137\mathrm{\ Pa}\ {\mathrm{m}}^{\mathrm{6}}\mathrm{\ }{\mathrm{mol}}^{--\mathrm{2}}\) and \(b=3.81\times {10}^{-5}\ {\mathrm{m}}^{\mathrm{3}}\mathrm{\ }{\mathrm{mol}}^{--\mathrm{1}}\). The experimental data are from reference 1.) We anticipate that the Joule-Thomson coefficient becomes zero at pressures and temperatures where the effects of intermolecular attractions and repulsions exactly offset one another. For interactions between molecules, attractive forces have the dominant effect at long distances, while repulsive forces dominate at short distances. The lower the pressure, the greater the average distance between gas molecules. Therefore, at any given temperature and a sufficiently low pressure, the effects of intermolecular attractive forces are more important than those of intermolecular repulsive forces. At low pressures, the Joule-Thomson coefficient should be positive. As the pressure increases, the effects of both attractive and repulsive forces must both increase, but at a sufficiently high pressure, the average intermolecular distance becomes so small that the effects of intermolecular repulsive forces become dominant. Therefore, we anticipate that the Joule-Thomson coefficient decreases as the pressure increases, eventually becoming negative. Experiments confirm these expectations. A temperature and pressure at which the Joule-Thomson coefficient becomes zero is called a Joule-Thomson inversion point. The experimentally determined curve for nitrogen gas is graphed in Figure 5. The van der Waals model also exhibits this effect. The inversion curve can be found from the expression for \({\mu }_{JT}\) developed above for a van der Waals gas. The inversion curve for nitrogen that is found in this way is also graphed in Figure 5. Qualitatively, the agreement is a satisfying confirmation of the basic interpretation that we have given for the role of intermolecular forces. Quantitatively, the agreement is poor, as we expect given the overly simple character of the van der Waals model. The Joule-Thomson coefficient for an ideal gas is zero, and we normally expect the properties of real gases to approach those of an ideal gas as the pressure falls to zero. However, both experiment and the van der Waals model indicate that the Joule-Thomson coefficient converges to a finite value as the pressure decreases to zero at a fixed temperature. A statistical thermodynamic model also predicts this outcome. This model calculates the coefficients in the virial equation of state. In it, the second virial coefficient reflects the net effect of attractive and repulsive forces between a pair of molecules, and it is the second virial coefficient and its temperature derivative determine that the value of \({\left({\partial \overline{H}}/{\partial P}\right)}_T\). (Higher-order virial coefficients reflect interactions among larger numbers of molecules.)

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Using the rule "like dissolves like" with the formation of ionic solutions, we must assess first assess two things: 1) the strength of the ion-dipole forces of attraction between water and the ionic compound and 2) the strength of the interionic bond of the ionic compound. For an ionic compound to form a solution, the ion-dipole forces between water and ionic compound must be greater than the interionic bonds. Therefore, to form a compound: ion-dipole forces > interionic bonds When the ionic compound is surrounded by water, the water dipoles surround the crystal's clustered structure. The water's negative ends of the dipole will be attracted to the positive dipoles of the ion and the positive ends of the water's dipole will be attracted to the negative dipoles of the ion. If the force of this attraction is stronger than the interionic bonds, the crystal's interionic bonds will be broken, then surrounded by the water molecules or hydrated . There is a 3-step process that we can use to approach the energy involved in ionic solution formation. 1) Breaking apart the ionic compound is endothermic and requires energy. 2) Hydrating cation is exothermic and therefore releases energy. 3) Hydrating the anion is exothermic and also releases energy. The sum of these 3 steps will then give us the enthalpy of the solution. 1) CaCl (s) -> Ca (g) + Cl (g) energy > 0 2) Ca (g) > Ca (aq) energy < 0 3) Cl (g) > Cl (aq) energy < 0 CaCl (s) > Ca (aq) + Cl (aq) energy > 0 The dissolution is endothermic because in the formation of ionic solutions, you must take into account entropy in addition to the enthalpy of the solution to determine whether it will occur spontaneously.

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Recent reports about increased risk for diabetes resulting from diets high in sugars and carbohydrates have led to suggestions for healthier diets, including the low glycemic index diet. The (GI) is a measure of the rate at which foods increase blood glucose . Foods which increase the blood glucose concentration rapidly seem to increase the risk of diabetes more than other foods which may have high sugar, but lead to moderated increases in blood glucose concentration. The glycemic load (GL) is an estimate of the probable impact of standard servings of food in typical meals on glucose concentrations. The surprises in this table may be honey, sucrose, chocolate, and ice cream. Fats (in honey and ice cream) may slow the uptake of sugar in the intestine, while fructose (in honey, hydrolyzed sucrose, and high fructose corn syrup) has a low impact on blood glucose. There are many sources of glycemic index values. The glycemic index is determined by plotting blood glucose concentrations vs. time for two hours after ingesting 50 g of glucose or another food, and comparing the graphs which are typically like the one below: Note that the blood glucose concentration rises much faster when glucose is ingested (with a GI = 100) than when white bread is ingested (GI = 70). The glucose concentration is measured in , the abbreviation for "milliMolar". The (M) concentration of a solution is a measure of how much solute (the component of a solution present in lesser amount, often a solid) there is in a given volume of . \(\text{Concentration of solute in molar units}=\frac{\text{amount of solute, mol}}{\text{volume of solution, L}}\) \(c_{\text{M}}~=~\frac{n_{\text{solute, mol}}}{V_{\text{solution,L}}}\text{ (1)}\) or equivalently, \(c_{\text{M}}~=~\frac{n_{\text{solute, mmol}}}{V_{\text{solution,mL}}}\) So we can define mM as \(c_{\text{mM}}~=~\frac{n_{\text{solute, mmol}}}{V_{\text{solution,L}}}\text{ (1)}\) The units moles per liter (mol liter ) or moles per cubic decimeter (mol dm ) are used to express molar concentration. They are equivalent (since 1 dm = 1 liter). It's important to note that we are referring to the volume of , not the volume of . The solvent is the component of a solution present in the largest amount, and is often a liquid. The solution will generally have a different volume than the solvent from which it was made. Solutes and solvents may be solids, liquids or gases (air is a solutioin of mostly oxygen gas in the solvent nitrogen gas; brass is a solution of solid zinc in solid copper; and gin is a solution of liquid ethanol in liquid water; etc.) There are many other concentration units. Glucose blood concentration is sometimes given in mg/dL (milligrams per cubic deciliter), alcoholic beverages in percent by volume (C = V(solute)/V(solvent) x 100%), and physical chemists sometimes use "molal" units (where C = n(solute, mol)/m(solvent, kg)). Macroscopically a is defined as a homogeneous mixture of two or more substances, that is, a mixture which appears to be uniform throughout. On the microscopic scale a solution involves the random arrangement of one kind of atom or molecule with respect to another. Blood is heterogeneous, because it contains red blood cells which can be separated by centrifugation, leaving the , so blood is a solution. The plasma, however, is a solution which contains the solute glucose among many other substances. We often use "plasma glucose concentration" and "blood glucose concentration" interchangeably, although the latter is not precisely correct. Solutions are common in the body, because atoms or molecules of solids dissolved in a liquid are close together but still able to move past one another. They contact each other more frequently than if two solids were placed next to each other. This “intimacy” in liquid solutions often facilitates chemical reactions. Blood glucose concentration measurements are usually made with an instrument ("glucometer") that needs to be calibrated by measuring solutions of known concentrations. If a pure substance like glucose is soluble in water, it is easy to prepare a solution of known concentration. A container with a sample of the substance is weighed accurately, and an appropriate mass of sample is poured through a funnel into a volumetric flask, as shown in the figure. The container is then reweighed. Any solid adhering to the funnel is rinsed into the flask, and water is added until the flask is about three-quarters full. After swirling the flask to dissolve the solid, water is added carefully until the bottom of the meniscus coincides with the calibration mark on the neck of the flash. This process is shown in detail in Figure 1: The following movie shows a difficult step in this process: adding water to dilute the solution to the proper concentration. In this movie, water is added to the volumetric flask until it is 2cm from the mark. A wash bottle is then used to bring the solution level to within a few millimeters of the mark. Finally, a dropper is used to fill to the mark and ensure the calibration mark is not overshot. A standard solution of glucose was prepared as described above. The initial mass of the container plus glucose was 43.2874 g, and the final mass after pouring was 42.9724 g. The volume of the flask was 250.00 ml. What is the concentration of the solution? The concentration can be calculated by dividing the amount of solute by the volume of solution [Eq. (1)]: We obtain from the mass of glucose added to the flask: = 43.2874 - 42.9724 g = 0.3150 g = 0.3150 g × \(\frac{\text{1 mol}}{\text{180}\text{.157 g}}\) = 1.748 × 10 mol The volume of solution is 250.00 ml, or = 250.00 cm × \(\frac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}}\) = 2.5000 × 10 dm Thus \(c_{\text{glucose}}=\frac{n_{\text{glucose}}}{V_{\text{solution}}}=\frac{\text{1}\text{.748}\times \text{10}^{\text{-3}}\text{ mol }}{\text{2}\text{.500 }\times \text{10}^{\text{-1}}\text{ dm}^{\text{3}}}=\text{6}\text{.994 }\times 10^{^{\text{-3}}}\text{mol dm}^{\text{-3}}\) This could be expressed in millimolar: \(\text{6}\text{.994 }\times 10^{^{\text{-3}}}\text{mol dm}^{\text{-3}}~\times~\frac{\text{1 mM}}{10^{-3}\text{M}}~=~\text{6.99 mM}\) Note that the definition of concentration is entirely analogous to the definitions of density, molar mass, and stoichiometric ratio that we have previously encountered. Concentration will serve as a conversion factor relating the volume of solution to the amount of dissolved solute. \(\text{Volume of solution}~~\overset{concentration}{\longleftrightarrow}\text{amount of solute}\) \(V~~\overset{c}{\longleftrightarrow}~~n\) Since solutions offer a convenient medium for carrying out chemical reactions, it is often necessary to know how much of one solution will react with a given quantity of another. Examples in other sections have shown that the amount of substance is the quantity which determines how much of one material will react with another. The ease with which solution volumes may be measured suggests that it would be very convenient to know the amount of substance dissolved per unit volume of solution. Then by measuring a certain volume of solution, we would also be measuring a certain amount of substance. Because the volume of a liquid can be measured quickly and easily, concentration is a much- used quantity. The next two examples show how this conversion factor may be applied to commonly encountered solutions in which water is the solvent ( solutions). Karo Lite corn syrup is an aqueous (water) solution of nearly 100% glucose [represented or written glucose( )] whose concentration we'll explore below. It is made by hydrolyzing cornstarch, but it contains no "high fructose corn syrup", which is corn syrup that has been enzymatically converted from 100% glucose to about 55% fructose and 45% glucose to increase its sweetness. A sample of Karo syrup has a glucose( )] concentration of 1.30 mol dm . a. If 24.71 cm (24.71 ml) of this solution is delivered from a buret to prepare a sample for a GI test, what amount of glucose has been delivered? b. What is the mass of glucose in the sample, and the glucose concentration of Karo syrup in g/mL? a. Using concentration as a conversion factor, we have \(n_{\text{glucose}}=\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{1}\text{.30 mol}}{\text{1 dm}^{\text{3}}}\) The volume units will cancel if we supply a unity factor to convert cubic centimeters to cubic decimeters: \(n_{\text{glucose}}=\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{1}\text{.30 mol}}{\text{1 dm}^{\text{3}}}\times \left( \frac{\text{1 dm}}{\text{10 cm}} \right)^{\text{3}}\) \(=\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{1}\text{.30 mol}}{\text{1 dm}^{\text{3}}}\times \frac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}}\) b. m = n x M = 0.0321 mol x 180.157 g/mol = 5.78 g Thus Karo syrup has a concentration of 5.78 g / 24.74 mL or 0.234 g/mL, consistent with the nutritional label which states that a 2 Tbsp (30 mL) serving provides 30 ml x 0.234 g/ml = 7 g of glucose. The concentration units of moles per cubic decimeter are often abbreviated , pronounced . That is, a 0.1- (one-tenth molar) solution contains 0.1 mol solute per cubic decimeter of solution. This abbreviation is very convenient for labeling laboratory bottles and for writing textbook problems; however, when doing calculations, it is difficult to see that 1 dm × 1 = 1 mol Therefore we recommend that you when doing any calculations involving solution concentrations. It is sometimes easier to use the unit liter, which is equivalent to cubic deciliters: \(\text{1 dm}^{\text{3}}\times \text{1 }\frac{\text{mol}}{\text{dm}^{\text{3}}}=\text{1mol}\) \(\text{1 L}\times \text{1 }\frac{\text{mol}}{\text{L}}=\text{1mol}\) Problems such as Example 2 are easier for some persons to solve if the solution concentration is expressed in millimoles per cubic centimeter (mmol cm ) or millimoles per ml (1 ml = 1 cm ) instead of moles per cubic decimeter. Since the SI prefix m means 10 , 1 mmol = 10 mol, and \(\text{1 M}~=~\frac{\text{1 mol}}{\text{1 dm}^{\text{3}}} ~\times~ \frac{\text{1 dm}^{\text{3}}}{\text{1 L}} ~=~\frac{\text{1 mol}}{\text{L}}\) \(\text{1 M} ~=~\frac{\text{1 mol}}{\text{L}} ~\times~\frac{\text{10}^{\text{-3}}\text{ L}}{\text{1 ml}} ~\times~\frac{\text{1 mmol}}{\text{10}^{\text{-3}}\text{ mol}} ~=~\frac{\text{1 mmol}}{\text{1 ml}}\) \(\text{1 M} ~=~\frac{\text{1 mol}}{\text{1 dm}^{\text{3}}} \times \frac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}} \times \frac{\text{1 mmol}}{\text{10}^{\text{-3}}\text{ mol}} ~=~ \frac{\text{1 mmol}}{\text{1 cm}^{\text{3}}} \) Thus a concentration of 1.30 mol dm (1.30 ) can also be expressed as 1.30 mmol cm , 1.30 mol/L or 1.30 mmol/mL. Expressing the concentration this way is very convenient when dealing with laboratory glassware calibrated in milliliters or cubic centimeters. Diabetes is diagnosed when the fasting blood plasma glucose level is ≥ 7.0 mmol/L (126 mg/dL) (greater or equal to 7.0 mmol/L), or when plasma glucose ≥ 11.1 mmol/L (200 mg/dL) two hours after a 75 g oral glucose load as in a glucose tolerance test. Suppose the 24.71 cm (24.71 ml) sample of Karo syrup with a glucose( ) concentration of 1.30 mol dm in Example 2 were ingested. In Example 2 we found that it contains 0.0321 mol of glucose. What blood concentration would result in a person with the typical blood volume of 4.7L, assuming that all the glucose showed up in the blood at the time of the test, half an hour later? Is this a reasonable assumption, comparing it to the graph above where 50 g of glucose was ingested? \(c_{\text{glucose}}=\frac{n_{\text{glucose}}}{V_{\text{solution}}}=\frac{\text{0}\text{.0321 mol}}{\text{4}\text{.7 dm}^{\text{3}}}=\text{6}\text{.84 }\times 10^{\text{-3}}\text{mol dm}^{\text{-3}}\) This could be expressed in millimolar: \(\text{6}\text{.84 }\times 10^{^{\text{-3}}}\text{mol dm}^{\text{-3}}~\times~\frac{\text{1 mM}}{10^{-3}\text{M}}~=~\text{6.84 mM}\) Here only 5.78 g sample (see Example 2) led to a calculated plasma glucose level of 6.84 mM. In the glycemic index test, a 50 g dose (nearly 10x as much) leads to a concentration of only 9 mM in a normal person, so a lot of glucose regulation is done by the body in a short time. Note: The symbols and refer to the amount and mass of the glucose, respectively. They do refer to the solution. If we wanted to specify the mass of aqueous glucose solution, the symbol could be used.

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Created in the early 17th century, the gas laws have been around to assist scientists in finding volumes, amount, pressures and temperature when coming to matters of gas. The gas laws consist of three primary laws: Charles' Law, Boyle's Law and Avogadro's Law (all of which will later combine into the General Gas Equation and Ideal Gas Law). The three fundamental gas laws discover the relationship of pressure, temperature, volume and amount of gas. Boyle's Law tells us that the volume of gas increases as the pressure decreases. Charles' Law tells us that the volume of gas increases as the temperature increases. And Avogadro's Law tell us that the volume of gas increases as the amount of gas increases. The ideal gas law is the combination of the three simple gas laws. Ideal gas, or perfect gas, is the theoretical substance that helps establish the relationship of four gas variables, , , the and . It has characters described as follow: Real gas, in contrast, has real volume and the collision of the particles is not elastic, because there are attractive forces between particles. As a result, the volume of real gas is much larger than of the ideal gas, and the pressure of real gas is lower than of ideal gas. All real gases tend to perform ideal gas behavior at low pressure and relatively high temperature. The tells us how much the real gases differ from ideal gas behavior. \[ Z = \dfrac{PV}{nRT} \] For ideal gases, \( Z = 1 \). For real gases, \( Z\neq 1 \). In 1662, Robert Boyle discovered the correlation between and (assuming and remain constant): \[ P\propto \dfrac{1}{V} \rightarrow PV=x \] where x is a constant depending on amount of gas at a given temperature. Another form of the equation (assuming there are 2 sets of conditions, and setting both constants to eachother) that might help solve problems is: \[ P_1V_1 = x = P_2V_2 \] A 17.50mL sample of gas is at 4.500 atm. What will be the volume if the pressure becomes 1.500 atm, with a fixed amount of gas and temperature? \[ V_2= \dfrac {P_1 \centerdot V_1}{P_2} \] \[ =\dfrac{4.500 atm \centerdot 17.50mL}{1.500 atm} \] \[ = 52.50mL \] In 1787, French physicists Jacques Charles, discovered the correlation between and (assuming and remain constant): \[ V \propto T \rightarrow V=yT \] where y is a constant depending on amount of gas and pressure. Volume is directly proportional to Temperature Another form of the equation (assuming there are 2 sets of conditions, and setting both constants to eachother) that might help solve problems is: \[ \dfrac{V_1}{T_1} = y = \dfrac{V_2}{T_2} \] A sample of Carbon dioxide in a pump has volume of 20.5 mL and it is at 40.0 C. When the amount of gas and pressure remain constant, find the new volume of Carbon dioxide in the pump if temperature is increased to 65.0 C. \[ V_2=\dfrac{V_1 \centerdot T_2}{T_1}\] \[ =\dfrac{20.5mL \centerdot (60+273.15K)}{40+273.15K}\] \[ = 22.1mL \] In 1811, Amedeo Avogadro fixed Gay-Lussac's issue in finding the correlation between the and (assuming and remain constant): \[ V \propto n \rightarrow V = zn\] where z is a constant depending on Pressure and Temperature. Another form of the equation (assuming there are 2 sets of conditions, and setting both constants to eachother) that might help solve problems is: \[ \dfrac{P_1}{n_1} = z= \dfrac{P_2}{n_2}\] A 3.80 g of oxygen gas in a pump has volume of 150 mL. constant temperature and pressure. If 1.20g of oxygen gas is added into the pump. What will be the new volume of oxygen gas in the pump if temperature and pressure held constant? V =150 mL \[ n_1= \dfrac{m_1}{M_oxygen gas} \] \[ n_2= \dfrac{m_2}{M_oxygen gas} \] \[ V_2=\dfrac{V_1 \centerdot n_2}{n_1}\] \[ = \dfrac{150mL\centerdot \dfrac{5.00g}{32.0g \centerdot mol^-1} \dfrac{3.80g}{32.0g\centerdot mol^-1} \] \[ = 197ml\] The ideal gas law is the combination of the three simple gas laws. By setting all three laws directly or inversely proportional to Volume, you get: \[ V \propto \dfrac{nT}{P}\] Next replacing the directly proportional to sign with a constant(R) you get: \[ V = \dfrac{RnT}{P}\] And finally get the equation: \[ PV = nRT \] where P= the absolute pressure of ideal gas Here, R is the called the gas constant. The value of R is determined by experimental results. Its numerical value changes with units. At 655mm Hg and 25.0 C, a sample of Chlorine gas has volume of 750mL. How many moles of Chlorine gas at this condition? n=? \[ n=\frac{PV}{RT} \] \[ =\frac{655mm Hg \centerdot \frac{1 atm}{760mm Hg} \centerdot 0.75L}{0.082057L \centerdot atm \centerdot mol^-1 \centerdot K^-1 \centerdot (25+273.15K) }\] \[ =0.026 mol\] You can get the numerical value of gas constant, R, from the ideal gas equation, PV=nRT. At standard temperature and pressure, where temperature is 0 C, or 273.15 K, pressure is at 1 atm, and with a volume of 22.4140L, \[ R= \frac{PV}{RT} \] \[ \frac{1 atm \centerdot 22.4140L}{1 mol \centerdot 273.15K} \] \[ =0.082057 \; L \centerdot atm \centerdot mol^{-1} K^{-1} \] \[ R= \frac{PV}{RT} \] \[ = \frac{1 atm \centerdot 2.24140 \centerdot 10^{-2}m^3}{1 mol \centerdot 273.15K} \] \[ = 8.3145\; m^3\; Pa \centerdot mol^{-1} \centerdot K^{-1} \] In an Ideal Gas situation, \( \frac{PV}{nRT} = 1 \) (assuming all gases are "ideal" or perfect). In cases where \( \frac{PV}{nRT} \neq 1 \) or if there are multiple sets of conditions (Pressure(P), Volume(V), number of gas(n), and Temperature(T)), use the General Gas Equation: Assuming 2 set of conditions: Initial Case: Final Case: \[ P_iV_i = n_iRT_i \; \; \; \; \; \; P_fV_f = n_fRT_f \] Setting both sides to R (which is a constant with the same value in each case), one gets: \[ R= \dfrac{P_iV_i}{n_iT_i} \; \; \; \; \; \; R= \dfrac{P_fV_f}{n_fT_f} \] If one substitutes one R for the other, one will get the final equation and the General Gas Equation: \[ \dfrac{P_iV_i}{n_iT_i} = \dfrac{P_fV_f}{n_fT_f} \] If in any of the laws, a variable is not give, assume that it is given. For constant temperature, pressure and amount: 2. Pressure: 1 Atmosphere (760 mmHg) 3. Amount: 1 mol = 22.4 Liter of gas 4. In the Ideal Gas Law, the gas constant R = 8.3145 Joules · mol · K = 0.082057 L · atm·K · mol Dutch physicist Johannes Van Der Waals developed an equation for describing the deviation of real gases from the ideal gas. There are two correction terms added into the ideal gas equation. They are \( 1 +a\frac{n^2}{V^2}\), and \( 1/(V-nb) \). Since the attractive forces between molecules do exist in real gases, the pressure of real gases is actually lower than of the ideal gas equation. This condition is considered in the van der waals equation. Therefore, the correction term \( 1 +a\frac{n^2}{V^2} \) corrects the pressure of real gas for the effect of attractive forces between gas molecules. Similarly, because gas molecules have volume, the volume of real gas is much larger than of the ideal gas, the correction term \(1 -nb \) is used for correcting the volume filled by gas molecules. To solve this question you need to use Boyle's Law: \[ P_1V_1 = P_2V_2 \] Keeping the key variables in mind, temperature and the amount of gas is constant and therefore can be put aside, the only ones necessary are: Plugging these values into the equation you get: V =(1.43atm x 4 L)/(2.39atm) = 2.38 L To solve this question you need to use Charles's Law: Once again keep the key variables in mind. The pressure remained constant and since the amount of gas is not mentioned, we assume it remains constant. Otherwise the key variables are: Since we need to solve for the final temperature you can rearrange Charles's: Once you plug in the numbers, you get: T =(308.15 K x .75 L)/(1.25 L) = 184.89 K Using Avogadro's Law to solve this problem, you can switch the equation into \( V_2=\frac{n_1\centerdot V_2}{n_2} \). However, you need to convert grams of Helium gas into moles. \[ n_1 = \frac{4.00g}{4.00g/mol} = \text{1 mol} \] Similarily, n =2 mol \[ V_2=\frac{n_2 \centerdot V_2}{n_1}\] \[ =\frac{2 mol \centerdot 500mL}{1 mol}\] \[ = \text{1000 mL or 1L } \]

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With 70% of our earth being ocean water and 65% of our bodies being water, it is hard to not be aware of how important it is in our lives. There are 3 different forms of water, or H O: (ice), (water), and (steam). Because water seems so ubiquitous, many people are unaware of the unusual and unique properties of water, including: If you look at the periodic table and locate tellurium (atomic number: 52), you find that the boiling points of hydrides decrease as molecule size decreases. So the hydride for tellurium: has a boiling point of . Moving up, the next hydride would be with a boiling point of . One more up and you find that has a boiling point at . The next hydride would be . And we all know that the boiling point of water is . So despite its molecular weight, water has an incredibly boiling point. This is because water requires more energy to break its hydrogen bonds before it can then begin to boil. The same concept is applied to freezing point as well, as seen in the table below. The boiling and freezing points of water enable the molecules to be very slow to boil or freeze, this is important to the ecosystems living in water. If water was very easy to freeze or boil, drastic changes in the environment and so in oceans or lakes would cause all the organisms living in water to die. This is also why sweat is able to cool our bodies. Besides mercury, water has the highest for all liquids. Water's high surface tension is due to the hydrogen bonding in water molecules. Water also has an exceptionally high . Vaporization occurs when a liquid changes to a gas, which makes it an endothermic reaction. Water's heat of vaporization is 41 kJ/mol. is inversely related to intermolecular forces, so those with stronger intermolecular forces have a lower vapor pressure. Water has very strong intermolecular forces, hence the low vapor pressure, but it's even lower compared to larger molecules with low vapor pressures. All substances, including water, become less dense when they are heated and more dense when they are cooled. So if water is cooled, it becomes more dense and forms ice. Water is one of the few substances whose solid state can float on its liquid state! Why? Water continues to become more dense until it reaches 4°C. After it reaches 4°C, it becomes dense. When freezing, molecules within water begin to move around more slowly, making it easier for them to form hydrogen bonds and eventually arrange themselves into an open crystalline, hexagonal structure. Because of this open structure as the water molecules are being held further apart, the volume of water about 9%. So molecules are more tightly packed in water's liquid state than its solid state. This is why a can of soda can explode in the freezer. It is very rare to find a compound that lacks carbon to be a liquid at standard temperatures and pressures. So it is unusual for water to be a liquid at room temperature! Water is liquid at room temperature so it's able to move around quicker than it is as solid, enabling the molecules to form fewer hydrogen bonds resulting in the molecules being packed more closely together. Each water molecule links to four others creating a tetrahedral arrangement, however they are able to move freely and slide past each other, while ice forms a solid, larger hexagonal structure. As water boils, its hydrogen bonds are broken. Steam particles move very far apart and fast, so barely any hydrogen bonds have the time to form. So, less and less hydrogen bonds are present as the particles reach the critical point above steam. The lack of hydrogen bonds explains why steam causes much worse burns that water. Steam contains all the energy used to break the hydrogen bonds in water, so when steam hits your face you first absorb the energy the steam has taken up from breaking the hydrogen bonds it its liquid state. Then, in an exothermic reaction, steam is converted into liquid water and heat is released. This heat adds to the heat of boiling water as the steam condenses on your skin. Because of water's polarity, it is able to dissolve or dissociate many particles. Oxygen has a slightly negative charge, while the two hydrogens have a slightly positive charge. The slightly negative particles of a compound will be attracted to water's hydrogen atoms, while the slightly positive particles will be attracted to water's oxygen molecule; this causes the compound to dissociate. Besides the explanations above, we can look to some attributes of a water molecule to provide some more reasons of water's uniqueness: The properties of water make it suitable for organisms to survive in during differing weather conditions. Ice freezes as it expands, which explains why ice is able to float on liquid water. During the winter when lakes begin to freeze, the surface of the water freezes and then moves down toward deeper water; this explains why people can ice skate on or fall through a frozen lake. If ice was not able to float, the lake would freeze from the bottom up killing all ecosystems living in the lake. However ice floats, so the fish are able to survive under the surface of the ice during the winter. The surface of ice above a lake also shields lakes from the cold temperature outside and insulates the water beneath it, allowing the lake under the frozen ice to stay liquid and maintain a temperature adequate for the ecosystems living in the lake to survive.

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The pH of an aqueous solution is the measure of how acidic or basic it is. The pH of an aqueous solution can be determined and calculated by using the concentration of concentration in the solution. The pH of an aqueous solution is based on the pH scale which typically ranges from 0 to 14 in water (although as discussed below this is not an a formal rule). A pH of 7 is considered to be neutral. A pH of less than 7 is considered acidic. A pH of greater than 7 is then considered basic. Acidic solutions have high hydronium concentrations and lower hydroxide concentrations. Basic solutions have high hydroxide concentrations and lower hydronium concentrations. In the self-ionization of water, the amphiprotic ability of water to act as a proton donor and acceptor allows the formation of hydronium (\(H_3O^+\)) and hydroxide ions (\(OH^-\)). In pure water, the concentration of hydronium ions equals that of hydroxide ions. At 25 C, the concentrations of both hydronium and hydroxide ions equal \(1.0 \times 10^{-7}\). The ion product of water, \(K_w\), is the equilibrium condition for the self-ionization of water and is express as follows: \(K_w = [H_{3}O^+,OH^-] = 1.0 \times 10^{-14}\). Another equation can be used that relates the concentrations of hydronium and hydroxide concentrations. This equation is derived from the equilibrium condition for the self-ionization of water, \K_w\). It brings the three equations for pH, pOH, and \K_w\) together to show that they are all related to each other and either one can be found if the other two are known. The following equation is expressed by taking the negative \logarithm of the \K_w\) expression for the self-ionization of water at room temperature: \(K_w = [H_{3}O^+,OH^-] = 1.0 \times 10^{-14}\) \(pK_w = pH + pOH = 14\). The ionization of strong acids and strong bases in dilute aqueous solutions essentially go to completion. In aqueous solutions of strong acids and strong bases, the self-ionization of water only occurs to a small extent. Since it only occurs to a small extent, the self-ionization of water is an insignificant source of hydronium and hydroxide ions. Knowing this, we can say in calculating hydronium concentration in an aqueous solution of a strong acid that the strong acid is the main source of hydronium ions. We can also say that in calculating hydroxide concentration in an aqueous solution of a strong base that the strong base is the main source of hydroxide ions. This is usually true unless the solutions of strong acids and strong bases are extremely dilute. Weak acids only partially dissociate in aqueous solutions and reach a condition of equilibrium, therefore how much they dissociate is given by the equilibrium equation for that acid in solution: \[K_a = \dfrac{[H_{3}O^+,A^-]}{[HA]}\] with Weak bases also only partially dissociate in aqueous solutions and reach a condition of equilibrium. The equation for the partial dissociation of a base is then the equilibrium equation for that base in solution: \[K_b = \dfrac{[OH^-,B+]}{[B]}\] \[ [OH^-] = Hydroxide\;Concentration\] \[[B^+] = Ion\] \[ [B] = Weak\;Base\] 1. Use the pH equation which is: \(pH = -\log[H_{3}O^+]\). 0.055 M HBr, HBr is a strong acid [H O ] = 5.5 X 10 M pH = -\log(5.5 X 10 ) = 2. Use the pH equation \(pH = -\log[H_{3}O^+]\) and pK equation \(pK_w = pH + pOH = 14\). 0.00025 M HCl, HCl is a strong acid [H O ] = 2.5 X 10 M pH = -\log(2.5 X 10 ) = Then solve for the pOH: pH + pOH = 14 pOH = 14 - pH pOH = 14 - 3.6 = 3. Use the pOH equation \(pH = -\log[OH^-]\) and pK equation \(pK_w = pH + pOH = 14\). 0.0035 M LiOH, LiOH is a strong base [OH ] = 3.5 X 10 pOH = -\log(3.5 X 10 ) = Now solve for pH: pH = 14 - pOH pH = 14 - 2.46 = 4. 0.0045 M hydrofluoric acid, hydrofluoric acid is a weak acid. Use K equation \(K_a = \dfrac{[H_{3}O^+,A^-]}{[HA]}\) and . 5. 0.0085 M ammonia, ammonia is a weak base. Use K equation \(K_b = \dfrac{[OH^-,B+]}{[B]}\) and .

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One of the major problems in organic synthesis is the suppression of unwanted side reactions. Frequently the desired reaction is accompanied by reaction at other parts of the molecule, especially when more than one functional group is present. Functional groups usually are the most reactive sites in the molecule, and it may be difficult or even impossible to insulate one functional group from a reaction occurring at another. Therefore any proposed synthesis must be evaluated at each step for possible side reactions that may degrade or otherwise modify the structure in an undesired way. To do this will require an understanding of how variations in structure affect chemical reactivity. Such understanding is acquired through experience and knowledge of reaction mechanism and reaction stereochemistry. To illustrate the purpose and practice of protecting groups in organic synthesis, let us suppose that the synthesis of -2-octene, which we outlined in , has to be adapted for the synthesis of 5-octyn-1-ol. We could write the following: However, the synthesis as written would fail because the alkyne is a weaker acid than the alcohol ( ), and the alkynide anion would react much more rapidly with the acidic proton of the alcohol than it would displace bromide ion from carbon: The hydroxyl group of 4-bromo-1-butanol therefore must be protected it is allowed to react with the alkynide salt. There are a number of ways to protect hydroxyl groups, but one method, which is simple and effective, relies on the fact that unsaturated ethers of the type are very reactive in electrophilic addition reactions (Section 10-4). An alcohol readily adds to the double bond of such an ether in the presence of an acid catalyst: The protected compound is a much weaker acid than the alkyne, and the displacement reaction can be carried out with the alkynide salt without difficulty. To obtain the final product, the protecting group must be removed, and this can be done in dilute aqueous acid solution by an \(S_\text{N}1\) type of substitution (Sections 8-7D and 8-7E): Hydroxyl \(\left( \ce{OH} \right)\) protecting groups in Organic Synthesis Protection of alcohols: Acetyl \(\left( \ce{Ac} \right)\) – Removed by acid or base. Benzoyl \(\left( \ce{Bz} \right)\) – Removed by acid or base, more stable than \(\ce{Ac}\) group. Benzyl (\(\ce{Bn}\), \(\ce{Bnl}\)) – Removed by hydrogenolysis. \(\ce{Bn}\) group is widely used in sugar and nucleoside chemistry. \(\beta\)-Methoxyethoxymethyl ether (MEM) – Removed by acid. Dimethoxytrityl, [bis-(4-methoxyphenyl)phenylmethyl] (DMT) – Removed by weak acid. DMT group is widely used for protection of 5′-hydroxy group in nucleosides, particularly in oligonucleotide synthesis. Methoxymethyl ether (MOM) – Removed by acid. Methoxytrityl [(4-methoxyphenyl)diphenylmethyl, MMT) – Removed by acid and hydrogenolysis. p-Methoxybenzyl ether (PMB) – Removed by acid, hydrogenolysis, or oxidation. Methylthiomethyl ether – Removed by acid. Pivaloyl \(\left( \ce{Piv} \right)\) – Removed by acid, base or reductant agents. It is substantially more stable than other acyl protecting groups. Tetrahydropyranyl (THP) – Removed by acid. Trityl (triphenylmethyl, \(\ce{Tr}\)) – Removed by acid and hydrogenolysis. Silyl ether (most popular ones include trimethylsilyl (TMS), tert-butyldimethylsilyl (TBDMS), tri-iso-propylsilyloxymethyl (TOM), and triisopropylsilyl (TIPS) ethers) – Removed by acid or fluoride ion. (such as \(\ce{NaF}\), TBAF (Tetra-n-butylammonium fluoride, HF-Py, or HF-NEt3)). TBDMS and TOM groups are used for protection of 2′-hydroxy function in nucleosides, particularly in oligonucleotide synthesis. Methyl Ethers – Cleavage is by TMSI in DCM or MeCN or Chloroform. An alternative method to cleave methyl ethers is BBr3 in DCM Ethoxyethyl ethers (EE) – Cleavage more trivial than simple ethers e.g. 1N Hydrochloric acid Amine protecting groups in Organic Synthesis Protection of amines: Carbobenzyloxy (Cbz) group – Removed by hydrogenolysis p-Methoxybenzyl carbonyl (Moz or MeOZ) group – Removed by hydrogenolysis, more labile than Cbz tert-Butyloxycarbonyl (BOC) group (Common in solid phase peptide synthesis) – Removed by concentrated, strong acid. (such as HCl or CF3COOH) 9-Fluorenylmethyloxycarbonyl (FMOC) group (Common in solid phase peptide synthesis) – Removed by base, such as piperidine Acetyl (Ac) group is common in oligonucleotide synthesis for protection of N4 in cytosine and N6 in adenine nucleic bases and is removed by treatment with a base, most often, with aqueous or gaseous ammonia or methylamine. Ac is too stable to be readily removed from aliphatic amides. Benzoyl (Bz) group is common in oligonucleotide synthesis for protection of N4 in cytosine and N6 in adenine nucleic bases and is removed by treatment with a base, most often with aqueous or gaseous ammonia or methylamine. Bz is too stable to be readily removed from aliphatic amides. Benzyl (Bn) group – Removed by hydrogenolysis Carbamate group – Removed by acid and mild heating. p-Methoxybenzyl (PMB) – Removed by hydrogenolysis, more labile than Benzyl 3,4-Dimethoxybenzyl (DMPM) – Removed by hydrogenolysis, more labile than p-methoxybenzyl p-methoxyphenyl (PMP) group – Removed by Ammonium cerium(IV) nitrate (CAN) Tosyl (Ts) group – Removed by concentrated acid (HBr, H2SO4) & strong reducing agents (sodium in liquid ammonia or sodium naphthalenide) Other Sulfonamides (Nosyl & Nps) groups – Removed by samarium iodide, tributyltin hydride Carbonyl protecting groups in Organic Synthesis Protection of carbonyl groups: Acetals and Ketals – Removed by acid. Normally, the cleavage of acyclic acetals is easier than of cyclic acetals. Acylals – Removed by Lewis acids. Dithianes – Removed by metal salts or oxidizing agents. Carboxylic acid protecting groups in Organic Synthesis Protection of carboxylic acids: Methyl esters – Removed by acid or base. Benzyl esters – Removed by hydrogenolysis. tert-Butyl esters – Removed by acid, base and some reductants. Silyl esters – Removed by acid, base and organometallic reagents. Orthoesters – Removed by mild aqueous acid to form ester, which is removed according to ester properties. Oxazoline – Removed by strong hot acid (pH < 1, T > 100 °C) or alkali (pH > 12, T > 100 °C), but not e.g. LiAlH4, organolithium reagents or Grignard organomagnesium) reagents Phosphate protecting groups in Organic Synthesis 2-cyanoethyl – removed by mild base. The group is widely used in oligonucleotide synthesis. Methyl (Me) – removed by strong nucleophiles e.c. thiophenole/TEA. Terminal alkyne protecting groups in Organic Synthesis propargyl alcohols in the Favorskii reaction, silyl groups, especially in protection of the acetylene itself Orthogonal protection in Organic Synthesis Orthogonal protection is a strategy allowing the deprotection of multiple protective groups one at a time each with a dedicated set of reaction conditions without affecting the other. It was introduced in the field of peptide synthesis by Robert Bruce Merrifield in 1977. As a proof of concept orthogonal deprotection is demonstrated in a photochemical transesterification by trimethylsilyldiazomethane utilizing the kinetic isotope effect: Due to this effect the quantum yield for deprotection of the right-side ester group is reduced and it stays intact. Significantly by placing the deuterium atoms next to the left-side ester group or by changing the wavelength to 254 nm the other monoarene is obtained. and (1977)

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Upon binding a second axial ligand, the iron center together with the axial base move toward the plane of the porphyrin, initiating a change in spin state from high-spin to low-spin when the sixth ligand is O or CO or any other strong ligand with an even number of valence electrons. Given these general features, what are the structural differences between systems that bind O with high affinity and those that bind O with low affinity? The answers to this question are relevant to understanding at the molecular level the mechanism of cooperativity, where a low-affinity conformation, the T state, and a high-affinity conformation, the R state, are in dynamic equilibrium in one tetrameric molecule. In looking at crystallographic data one sees a particular conformation frozen in the crystal, usually the one of lowest free energy among many in equilibrium in the solution state. The R \(\rightleftharpoons\) T equilibrium for hemoglobin is moderately rapid, at 4 x 10 s ; hemocyanin also switches quaternary conformations with a similar rate constant. Human hemoglobins are a heterogeneous group. Many mutants are known, and several have been structurally characterized. A structural alteration that affects the equilibrium between R and T states has a marked effect on ligand affinity and cooperativity in hemoglobin. If a specific amino-acid substitution destabilizes the T state, then the transition to the R state will occur earlier in the ligation process, and the hemoglobin will have an increased oxygen affinity. Hemoglobin Kempsey is an example. In this mutant an aspartic acid on the \(\beta\) chain is replaced by asparagine. Conversely, if the R state is destabilized, then the hemoglobin will have a lowered oxygen affinity. Hemoglobin Kansas is an example. Here an asparagine on the \(\beta\) chain has been replaced by threonine. It was proposed earlier that the molecule Fe(PF)(1-MeIm)(O ) in the solid or solution state was a fair approximation to the reference gas-phase molecule. The axial base, although not oriented for minimization of contacts with the porphyrin (i.e., \(\phi\) = 45°), is well-removed from an eclipsing orientation where \(\phi\) = 0 ± 10°; the Fe atom is centered in the plane of a highly planar porphyrin; the O ligand is oriented for minimization of contacts with the porphyrin, and its geometry is largely unconstrained by distal groups (the pickets); no groups are hydrogen-bonded to the axial base. The major difference from the reference state is that there is a significant attractive interaction between the electronegative dioxygen moiety and the amide groups on the pickets, and a smaller repulsive interaction with the picket t-butyl groups. For the CO adduct, contacts with the pickets are all at ideal van der Waals' separations and the Fe—CO moiety is free to assume its normal linear geometry. For CO binding the reference molecule is again the carbonyl adduct of the iron picket-fence porphyrinato molecule. In contrast to O binding, there are no specific distal effects, such as hydrogen bonding, by which CO affinity may be increased; there remain many ways, as with O binding, by which CO affinities may be reduced. Thus, the CO binder with highest affinity is the iron picket-fence porphyrin. The O affinities of myoglobin, R-state hemoglobin, and the Fe(PF)(1-MeIm) system are similar. However, the means by which this is achieved are different, and this difference is reflected most clearly in the kinetics of binding and release of O , which for Mb are much slower. The similarities and differences are summarized in Table 4.9, which is culled from Tables 4.2, 4.4, and 4.5. Fe(PF-Im) Fe(PF)(1-MeIm) Fe(Poc-PF) (1-MeIm) His (H O) polar, protic H-N(amide) polar, aprotic H-N (amide) phenyl NH 4.06(5) CH 2.67(6) NH 5.0 CH 3.3 Fe-N (deoxy), Å For the cobalt-dioxygen derivative, the putative hydrogen bonding between the dioxygen and the amide groups of the pickets assumes greater importance because the coordinated dioxygen is substantially more negative. Again the picket-fence porphyrin, being structurally characterized, is the reference system. Although no Co picket-fence porphyrin structures have been determined, the structures may be predicted with confidence from the iron analogues together with related structures of Co and Co tetraphenylporphyrinato systems.* * For CoHbO , single-crystal EPR spectra have been interpreted in terms of a nearly triangularly coordinated O , although a crystal structure of CoMbO shows a bent CoOO group. There is no precedent for this triangular arrangement in any Co -superoxo (O ) system, whereas there are many for angularly coordinated O in electronically not dissimilar square-planar Schiff-base systems. Regardless of geometry, the picket amide • • • O contacts do not change substantially. The 2-methyl substituent on 2-methylimidazole is not sterically active in the five-coordinate structures Fe(PF)(2-MeIm) and Fe(TPP)(2-MeIm), since the iron atom is displaced from the plane of the porphyrin by the expected amount and the Fe—N bond is unstretched and similar to that in deoxyhemoglobin (low O affinity) and deoxyMb (higher O affinity). Moreover, resonance Raman measurements also indicate little strain in this bond. In other words, there is no "tension at the heme," a key concept in early discussions of cooperativity before structures on model systems and high-resolution, refined protein structures became available. On moving into the plane of the porphyrin upon oxygenation, the 2-methyl substituent prevents the Fe-imidazole group from achieving its optimum geometry with the iron at the center of the porphyrin hole, as seen in the structure of Fe(PF)(1-MeIm)(O ). Thus, the sterically active 2-methyl substituent leads to lowered O (and CO) affinity relative to the 1-methyl analogue. In metrical terms the lowered affinity is reflected in an increase in the sum of the axial bond lengths from 1.75 + 2.07 = 3.82 Å to 1.90 + 2.11 = 4.01 Å. In the crystal structure of Fe(C Cap)(1-MeIm)(CO) the cap is about 5.6 Å from the porphryin plane. Hence, in the crystal structures of the free base H2(C Cap) and FeCI(C Cap) species, in which the cap is screwed down to approximately 4.0 Å from the porphyrin plane, considerable conformational rearrangement of the cap and the four chains attaching it to the porphyrin is needed to provide room for a small ligand such as CO. This is even more pronounced in a Co(C Cap) complex where the cap is only 3.49 Å from the mean porphyrin plane. Thus not only is affinity for CO lowered, but some additional discrimination against it is induced, since a linear, perpendicular coordination creates considerable strain energy elsewhere in the molecule. For the pocket porphyrin (Figure 4.23), structural data are available on the carbonyl adduct. The CO ligand is unable to achieve the linear perpendicular geometry seen in the high-affinity picket-fence porphyrin derivative, Fe(PF)(1-MeIm)(CO), and distortion of the porphyrin core is greater. In the pocket-porphyrin system, O affinity is unaffected, but CO affinity is lowered. The crystal structure of partially oxygenated hemoglobin, [\(\alpha\)-FeO ] [\(\beta\)-Fe] , reveals that the quaternary structure, except in the immediate vicinity of the \(\alpha\) hemes, which have O coordinated, resembles that of T-state deoxyhemoglobin rather than R-state liganded hemoglobin. In accord with the low affinity of T-state hemoglobin, the Fe—N bonds for the six-coordinate \(\alpha\)-hemes at 2.37 Å are significantly longer than those in fully oxygenated R-state oxyhemoglobin, [\(\alpha\)-FeO ] [\(\beta\)-FeO ] in the notation above, (1.94 (\(\alpha\)-hemes) and 2.06 Å (\(\beta\) hemes)) and that found in oxymyoglobin (2.07 Å). In contrast to the R-state structure and oxyMb, the a-hemes are folded as seen in the deoxy parent, leaving the Fe still substantially displaced (0.2 Å) from the plane of the four pyrrole nitrogen atoms. The deoxyhemoglobin T-state quaternary structure also has been observed in two other partially liganded hybrid hemoglobins, [\(\alpha\)-FeCO] [\(\beta\)-Mn(II)] and [\(\alpha\)-Ni] [\(\beta\)-FeCO] . Again, structural changes upon coordination do not propagate beyond the immediate vicinity of the liganded heme to the critical \(\alpha_{1} \beta_{2}\) interfaces. Note that although the crystal structure of hemoglobin A reveals that access to the binding site for the \(\beta\) chains is blocked by groups at the entrance to the cavity above the iron center, this does not prevent facile access to the binding site; the rate of O binding is slowed by a factor of only five. A similar situation occurs also for vertebrate myoglobins. The large structural differences that exist between deoxy (T) and oxy (R) hemoglobin and the much smaller differences between deoxy (T) and partially liganded (T) hybrid hemoglobin are shown in Figure 4.32. Because of the steric hindrance afforded by the distal histidine, all biological systems have low affinity for CO relative to the picket-fence porphyrins, with the exception of mutants where the distal histidine has been replaced by glycine. Thus low affinity to CO is associated primarily with the inability of the Fe-CO group to achieve its preferred linear geometry perpendicular to the porphyrin. Low-affinity O binding in the hemoglobins appears to be associated with the inability of the Fe-proximal histidine unit to move into the plane of the porphyrin and less so to distal effects, such as a cavity too small to accommodate the coordinated ligand. The blocked access to the site affects the kinetics but not necessarily the thermodynamics of ligand binding, as evidenced by the structure of T-state [\(\alpha\)-FeNi] [\(\beta\)-FeCO] Some similarities between the structures and properties of partially oxygenated (T-state) [\(\alpha\)-FeO ] [\(\beta\)-Fe] hemoglobin and Fe(PF)(2-MeIm)(O ) are provided in Table 4.10. In the synthetic systems low O affinity can be induced by 2-methyl substituents—a restraint on the movement of the Fe-imidazole moiety analogous to that provided by the protein chain. A second means is by distal effects, such as caps and straps. HbA [\(\alpha\)-FeO ] [\(\beta\)-Fe] Fe(PF)(2-MeIm) Fe(PF)(1,22-Me Im) histidine polar H-N(amide) polar, aprotic NH 3.88 CH 2.77(3) NH 4.9 CH 3.5 Fe-N (deoxy), Å \(\dagger\) Shortly (10 -10 s) after a ligand dissociates, a large difference in (Fe—N ) between R and T structures is observed, prior to relaxation to the equilibrium R and T conformations. Few structural data are available for high-affinity oxygen carriers. The crystal structures of two leghemoglobin derivatives, a monomeric myoglobin-like oxygen carrier found in the nitrogen-fixing nodules of legumes, are known at 2.0 and 3.3 Å. The binding pocket appears more open, perhaps allowing H O to enter and partake in stronger hydrogen bonding than that offered by the distal imidazole. Consistent with this notion is the more rapid rate of autoxidation observed for oxyleghemoglobin. oxymyoglobin, which lacks a distal histidine, also autoxidizes rapidly, although a distal arginine further along the helix E, E10Arg, fulfills the role of the distal histidine by hydrogen bonding to the sixth ligand, at least in the fluoride derivative, met-MbF. Although no structural data are available, a tenfold increase in O affinity was observed between an ester-strapped porphyrin, offering no hydrogen-bonding possibilities, and its conformationally very similar amide analogue. O • • • amide hydrogen bonding was demonstrated by means of NMR shift data (Zn and Fe—CO complexes vs. the Fe—O complex) and from infrared spectroscopy, which showed shifted amide N—H absorptions. The structural features that lead to the extraordinarily high affinity for O and low affinity for CO in hemoglobin remain unidentified. This high affinity is due to an slow dissociation rate of O of only 0.1s ; in most hemoglobins the rate is about 10 to 2,500 s (Table 4.2). Dioxygen binding is thus close to irreversible. Figure 4.27 shows that hydrogen bonding to the coordinated dioxygen ligand, unrestrained motion of the Fe-proximal histidine group into the plane of the porphyrin, hydrogen bonding to the proximal histidine, and, in the deoxy form, compression of the Fe—N bond and decrease in the out-of-plane displacement of the Fe atom will all increase O affinity over that of a system where these effects are absent. When hydrogen bonding is impossible, as in various synthetic systems (Table 4.5) as well as hemoglobin and Mb(E7His → Gly), O affinity is much lower than when hydrogen bonding can occur (see Table 4.6), especially for the cobalt analogues. But caution is needed in the absence of complete structural information: the lowered affinity of hemoglobin had been attributed to the lack of a distal histidine and its attendant hydrogen-bonding capabilities. However, the crystal structure reveals that an arginine residue, normally directed out into the solution, is capable of folding back into the ligand-binding pocket and of hydrogen bonding to ligands at the sixth site. In oxyhemerythrin the hydrogen bonding of the coordinated hydroperoxy group to the oxo bridge linking the two iron atoms (Figure 4.10B), described in Section II.F.1, may not only increase the stability of oxyhemerythrin, but also facilitate electron transfer that occurs in dioxygen binding.

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Collisional Frequency is the average rate in which two reactants collide for a given system and is used to express the average number of collisions per unit of time in a defined system. To fully understand how the collisional frequency equation is derived, consider a simple system (a jar full of helium) and add each new concept in a step-by-step fashion. Before continuing with this topic, it is suggested that the articles on and are reviewed, as these topics are essential to understanding collisional frequency. In determining the collisional frequency for a single molecule, \(Z_i\), picture a jar filled with helium atoms. These atoms collide by hitting other helium atoms within the jar. If every atom except one is frozen and the number of collisions in one minute is counted, the collisional frequency (per minute) of a single atom of helium within the container could be determined. This is the basis for the equation. \[Z_i = \dfrac{(\text{Volume of Collisional Cylinder}) (\text{Density})}{\text{Time}} \nonumber \] While the helium atom is moving through space, it sweeps out a collisional cylinder, as shown above. If the center of another helium atom is present within the cylinder, a collision occurs. The length of the cylinder is the helium atom's mean relative speed, \(\sqrt{2}\langle c \rangle\), multiplied by change in time, \(\Delta{t}\). The mean relative speed is used instead of average speed because, in reality, the other atoms are moving and this factor accounts for some of that. The area of the cylinder is the helium atom's Although \[\text{Volume of Collisional Cylinder} = \sqrt{2}\pi{d^2}\langle c \rangle\Delta{t} \nonumber \] Next, account must be taken of the other atoms that are moving that helium can hit; which is simply the density \(\rho\) of helium within the system. The density component can be expanded in terms of \(N\) and \(V\). \(N\) is the number of atoms in the system, and \(V\) is the volume of the system. Alternatively, the density in terms of pressure (relating pressure to volume using the perfect gas law equation, \(PV = nRT\): \[\rho = \left(\dfrac{N}{V}\right) = \left(\dfrac{\rho{N_A}}{V}\right) = \left(\dfrac{\rho{N_A}}{RT}\right) = \left(\dfrac{\rho}{kT}\right) \nonumber \] When you substitute in the values for \(Z_i\), the following equation results: \[{Z_{i} = \dfrac{\sqrt{2}\pi d^{2} \left \langle c \right \rangle\Delta{t}\left(\dfrac{N}{V}\right)}{\Delta{t}}} \nonumber \] Cancel Δt: \[Z_{i} = \sqrt{2}\pi d^{2} \left \langle c \right \rangle\left(\dfrac{N}{V}\right) \nonumber \] Now imagine that all of the helium atoms in the jar are moving again. When all of the collisions for every atom of helium moving within the jar in a minute are counted, \(Z_{ii}\) results. The relation is thus: \[Z_{ii} = \dfrac{1}{2}Z_{i}\left(\dfrac{N}{V}\right) \nonumber \] This expands to: \[Z_{ii} = \dfrac{\sqrt{2}}{2}\pi d^{2}\left \langle c \right \rangle\left(\dfrac{N}{V}\right)^2 \nonumber \] Consider a system of hydrogen in a jar: \[\ce{H_A + H_{BC} <=> H_{AB} + H_{C}} \nonumber \] In considering hydrogen in a jar instead of helium, there are several problems. First, the H ions have a smaller radius than the H molecules. This is easily solved by accounting for the different radii which changes \(d^{2}\) to \(\left(r_A + r_B\right)^2\). The second problem is that the number of H ions could be much different than the number of H molecules. So we expand \(\dfrac{\sqrt{2}}{2}\left(\dfrac{N}{V}\right)^2\) to account for the number of both reacting molecules to get \(N_AN_B\). Because two reactants are considered, Z becomes Z , and the two changes are combined to give the following equation: \[Z_{AB} = N_{A}N_{B}\pi\left(r_{A} + r_{B}\right)^2 \left \langle c \right \rangle \nonumber \] , \( \left \langle c \right \rangle \), can be expanded: \[ \left \langle c \right \rangle = \sqrt{\dfrac{8k_BT}{\pi m}} \nonumber \] This leads to the final change to the collisional frequency equation. Because two different molecules must be taken into account, the equation must accommodate molecules of different masses (\(m\)). So, mass (\(m\)) must be converted to reduced mass, \( \mu_{AB} \), converting a two bodied system to a one bodied system. Now we substitute \( \left \langle c \right \rangle \) in the Z equation to obtain: \[Z_{AB} = N_{A}N_{B}\left(r_{A} + r_{B}\right)^2\pi\sqrt{ \dfrac{8k_{B}T}{\pi\mu_{AB}}} \nonumber \] Cancel \(\pi\): \[Z_{AB} = N_{A}N_{B}\left(r_{A} + r_{B}\right)^2\sqrt{\dfrac{8\pi{k_{B}T}}{\mu_{AB}}} \nonumber \] with The following assumptions are used when deriving and calculating the collisional frequency: If the temperature of the system was increased, how would the collisional frequency be affected? If the masses of both the reactants were increased, how would the collisional frequency be affected? 0.4 moles of N gas (molecular diameter= 3.8x10 m and mass= 28 g/mol) occupies a 1-liter (0.001m ) container at 1 atm of pressure and at room temperature (298 K)

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One of the major goals of chemical thermodynamics is to establish criteria for predicting whether a particular reaction or process will occur spontaneously. We have developed one such criterion, the change in entropy of the universe: if \(ΔS_{univ} > 0\) for a process or a reaction, then the process will occur spontaneously as written. Conversely, if \(ΔS_{univ} < 0\), a process cannot occur spontaneously; if \(ΔS_{univ} = 0\), the system is at equilibrium. The sign of \(ΔS_{univ}\) is a universally applicable and infallible indicator of the spontaneity of a reaction. Unfortunately, using \(ΔS_{univ}\) requires that we calculate \(ΔS\) for . This is not particularly useful for two reasons: we are normally much more interested in the system than in the surroundings, and it is difficult to make quantitative measurements of the surroundings (i.e., the rest of the universe). A criterion of spontaneity that is based solely on the state functions of a system would be much more convenient and is provided by a new state function: the Gibbs free energy. The Gibbs energy (also known as the ) is defined as \[G_{sys} = H_{sys} – T S_{sys} \label{23.4.1}\] 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, we can expand Equation \ref{23.4.1} to \[ΔG_{sys} = ΔH_{sys} – T ΔS_{sys} \label{23.4.2}\] How does this simple equation relate to the entropy change of the universe \(ΔS_{univ}\) that we know is the sole criterion for spontaneous change from the second law of thermodynamics? Starting with the definition \[ΔS_{univ} = ΔS_{surr} + ΔS_{sys} \label{23.4.3}\] 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 heat \(q_p\) across the system boundary (we are considering constant pressure processes). The enthalpy change of the reaction \(ΔH_{sys}\) is the "flow" of heat into the system from the surroundings under constant pressure, so the heat "withdrawn" from the surroundings will be \(–q_p\). From the thermodynamic definition of entropy, the change of the entropy of the surroundings will be \[ΔS_{surr} = –\dfrac{q_p}{T} = –\dfrac{ΔH_{sys}}{T}.\] We can therefore rewrite Equation \(\ref{23.4.3}\) as \[ΔS_{univ} = \dfrac{– ΔH_{sys}}{T} + ΔS_{sys} \label{23.4.4}\] Multiplying through by \(–T\) , we obtain \[–T ΔS_{univ} = ΔH_{sys} – T ΔS_{sys} \label{23.4.5}\] which expresses the entropy change of the universe in terms of thermodynamic properties of the exclusively. \(–TΔS_{univ}\) is The criterion for predicting spontaneity is based on (\(ΔG\)), the change in \(G\), at constant temperature and pressure. Although very few chemical reactions actually occur under conditions of constant temperature and pressure, most systems can be brought back to the initial temperature and pressure without significantly affecting the value of thermodynamic state functions such as \(G\). At constant temperature and pressure, \[ ΔG = ΔH − TΔS \label{Eq2}\] where all thermodynamic quantities are those of the system. Recall that at constant pressure, \(ΔH = q\), whether a process is reversible or irreversible, and \[TΔS = q_{rev}.\] Using these expressions, we can reduce Equation \(\ref{Eq2}\) to \[ΔG = q − q_{rev}.\] Thus \(ΔG\) is the difference between the heat released during a process (via a reversible or an irreversible path) and the heat released for the same process occurring in a reversible manner. Under the special condition in which a process occurs reversibly, \(q = q_{rev}\) and \(ΔG = 0\). As we shall soon see, if \(ΔG\) is zero, the system is at equilibrium, and there will be no net change. What about processes for which \(ΔG ≠ 0\)? To understand how the sign of \(ΔG\) for a system determines the direction in which change is spontaneous, we can rewrite the relationship between \(\Delta S \) and \(q_{rev}\), discussed earlier. \[ \Delta S= \dfrac{q_{rev}}{T} \nonumber\] with the definition of \(\Delta H\) in terms of \(q_{rev}\) \[ q_{rev} = ΔH \nonumber \] to obtain \[\Delta S_{\textrm{surr}}=-\dfrac{\Delta H_{\textrm{sys}}}{T} \label{Eq3}\] Thus the entropy change of the surroundings is related to the enthalpy change of the system. We have stated that for a spontaneous reaction, \(ΔS_{univ} > 0\), so substituting we obtain \[\begin{align*} \Delta S_{\textrm{univ}}&=\Delta S_{\textrm{sys}}+\Delta S_{\textrm{surr}}>0 \\[4pt] &=\Delta S_{\textrm{sys}}-\dfrac{\Delta H_{\textrm{sys}}}{T}>0\end{align*}\] Multiplying both sides of the inequality by \(−T\) reverses the sign of the inequality; rearranging, \[ΔH_{sys}−TΔS_{sys}<0\] which is equal to \(ΔG\) (Equation \(\ref{Eq2}\)). We can therefore see that for a spontaneous process, \(ΔG_{sys} < 0\). The relationship between the entropy change of the surroundings and the heat gained or lost by the system provides the key connection between the thermodynamic properties of the system and the change in entropy of the universe. The relationship shown in Equation \(\ref{Eq2}\) allows us to predict spontaneity by focusing exclusively on the thermodynamic properties and temperature of the system. We predict that highly exothermic processes (\(ΔH \ll 0\)) that increase the entropy of a system (\(ΔS_{sys} \gg 0\)) would therefore occur spontaneously. An example of such a process is the decomposition of ammonium nitrate fertilizer. Ammonium nitrate was also used to destroy the Murrah Federal Building in Oklahoma City, Oklahoma, in 1995. For a system at constant temperature and pressure, we can summarize the following results: To further understand how the various components of \(ΔG\) dictate whether a process occurs spontaneously, we now look at a simple and familiar physical change: the conversion of liquid water to water vapor. If the conversion of liquid water to water vapor is carried out at 1 atm and the normal boiling point of 100.00 °C (373.15 K), we can calculate \(ΔG\) from the experimentally measured value of \(ΔH_{vap}\) (40.657 kJ/mol). For vaporizing 1 mol of water, \(ΔH = 40,657; J\), so the process is highly endothermic. From the definition of \(ΔS\) (Equation \(\ref{Eq3}\)), we know that for 1 mol of water, \[ \begin{align*} \Delta S_{\textrm{vap}} &= \dfrac{\Delta H_{\textrm{vap}}}{T_\textrm b} \\[4pt] &=\dfrac{\textrm{40,657 J}}{\textrm{373.15 K}} \\[4pt] &=\textrm{108.96 J/K} \end{align*}\] Hence there is an increase in the disorder of the system. At the normal boiling point of water, \[ \begin{align*} \Delta G_{100^\circ\textrm C} &=\Delta H_{100^\circ\textrm C}-T\Delta S_{100^\circ\textrm C} \\[4pt] &=\textrm{40,657 J}-[(\textrm{373.15 K})(\textrm{108.96 J/K})] \\[4pt] &=\textrm{0 J} \end{align*}\] The energy required for vaporization offsets the increase in entropy of the system. Thus \(ΔG = 0\), and the liquid and vapor are in equilibrium, as is true of any liquid at its boiling point under standard conditions. Now suppose we were to superheat 1 mol of liquid water to 110°C. The value of \(ΔG\) for the vaporization of 1 mol of water at 110°C, assuming that \(ΔH\) and \(ΔS\) do not change significantly with temperature, becomes \[ \begin{align*} \Delta G_{110^\circ\textrm C} &=\Delta H-T\Delta S \\[4pt] &=\textrm{40,657 J}-[(\textrm{383.15 K})(\textrm{108.96 J/K})] \\[4pt] &=-\textrm{1,091 J} \end{align*}\] Since \(ΔG < 0\), the vaporization of water is predicted to occur spontaneously and irreversibly at 110 °C. We can also calculate \(ΔG\) for the vaporization of 1 mol of water at a temperature below its normal boiling point—for example, 90°C—making the same assumptions: \[ \begin{align*}\Delta G_{90^\circ\textrm C} &= \Delta H-T\Delta S \\[4pt] &=\textrm{40,657 J}-[(\textrm{363.15 K})(\textrm{108.96 J/K})]\\[4pt] &=\textrm{1,088 J} \end{align*}\] Since \(ΔG > 0\), water does not spontaneously convert to water vapor at 90 °C. When using all the digits in the calculator display in carrying out our calculations, ΔG = 1090 J = −ΔG , as we would predict. We can also calculate the temperature at which liquid water is in equilibrium with water vapor. Inserting the values of \(ΔH\) and \(ΔS\) into the definition of \(ΔG\) (Equation \(\ref{Eq2}\)), setting \(ΔG = 0\), and solving for \(T\), \[\begin{align*} 0 &=40,657\, J−T(108.96\, J/K) \\[4pt] T&=373.15 \,K \end{align*}\] Thus \(ΔG = 0\) at \(T = 373.15\, K\) and \(1\, atm\), which indicates that liquid water and water vapor are in equilibrium; this temperature is called the normal boiling point of water. At temperatures greater than 373.15 K, ΔG is negative, and water evaporates spontaneously and irreversibly. Below 373.15 K, \(ΔG\) is positive, and water does not evaporate spontaneously. Instead, water vapor at a temperature less than 373.15 K and 1 atm will spontaneously and irreversibly condense to liquid water. Figure \(\Page {1}\) shows how the \(ΔH\) and \(TΔS\) terms vary with temperature for the vaporization of water. When the two lines cross, \(ΔG = 0\), and \(ΔH = TΔS.\) A similar situation arises in the conversion of liquid egg white to a solid when an egg is boiled. The major component of egg white is a protein called albumin, which is held in a compact, ordered structure by a large number of hydrogen bonds. Breaking them requires an input of energy (\(ΔH > 0\)), which converts the albumin to a highly disordered structure in which the molecules aggregate as a disorganized solid (\(ΔS > 0\)). At temperatures greater than 373 K, the \(TΔS\) term dominates, and \(ΔG < 0\), so the conversion of a raw egg to a hard-boiled egg is an irreversible and spontaneous process above 373 K. 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. Consider the following possible states for two different types of molecules with some attractive force: There would appear to be greater entropy on the left (state 1) than on the right (state 2). Thus the entropic change for the reaction as written (i.e. going to the right) would be (-) in magnitude, and the energetic contribution to the free energy change would be (+) (i.e. unfavorable) for the reaction as written. In going to the right, there is an attractive force and the molecules adjacent to each other is a lower energy state (heat energy, \(q\), is liberated). To go to the left, we have to overcome this attractive force (input heat energy) and the left direction is unfavorable with regard to heat energy q. The change in enthalpy is (-) in going to the right (q released), and this enthalpy change is negative (-) in going to the right (and (+) in going to the left). This reaction as written, is therefore, and . Hence, It is . From Table \(\Page {1}\), it would appear that we might be able to get the reaction to go to the right at low temperatures (lower temperature would minimize the energetic contribution of the entropic change). Looking at the same process from an opposite direction: This reaction as written, is , and it is . From Table \(\Page {1}\), it would appear that we might be able to get the reaction to go to the right at high temperatures (high temperature would increase the energetic contribution of the entropic change). In the previous subsection, we learned that the value of \(ΔG\) allows us to predict the spontaneity of a physical or a chemical change. In addition, the magnitude of \(ΔG\) for a process provides other important information. The change in Gibbs energy is equal to the maximum amount of work that a system can perform on the surroundings while undergoing a spontaneous change (at constant temperature and pressure): \[ΔG = w_{max}.\] To see why this is true, let’s look again at the relationships among free energy, enthalpy, and entropy expressed in Equation \(\ref{Eq2}\). We can rearrange this equation as follows: \[ ΔH = ΔG + TΔS \label{Eq4}\] This equation tells us that when energy is released during an exothermic process (\(ΔH < 0\)), such as during the combustion of a fuel, some of that energy can be used to do work (\(ΔG < 0\)), while some is used to increase the entropy of the universe (\(TΔS > 0\)). Only if the process occurs infinitely slowly in a perfectly reversible manner will the entropy of the universe be unchanged. Because no real system is perfectly reversible, the entropy of the universe increases during all processes that produce energy. As a result, no process that uses stored energy can ever be 100% efficient; that is, \(ΔH\) will never equal \(ΔG\) because \(ΔS\) has a positive value. One of the major challenges facing engineers is to maximize the efficiency of converting stored energy to useful work or converting one form of energy to another. As indicated in Table \(\Page {2}\), the efficiencies of various energy-converting devices vary widely. For example, an internal combustion engine typically uses only 25%–30% of the energy stored in the hydrocarbon fuel to perform work; the rest of the stored energy is released in an unusable form as heat. In contrast, gas–electric hybrid engines, now used in several models of automobiles, deliver approximately 50% greater fuel efficiency. A large electrical generator is highly efficient (approximately 99%) in converting mechanical to electrical energy, but a typical incandescent light bulb is one of the least efficient devices known (only approximately 5% of the electrical energy is converted to light). In contrast, a mammalian liver cell is a relatively efficient machine and can use fuels such as glucose with an efficiency of 30%–50%. We have seen that there is no way to measure absolute enthalpies, although we can measure changes in enthalpy (ΔH) during a chemical reaction. Because enthalpy is one of the components of Gibbs free energy, we are consequently unable to measure absolute free energies; we can measure only changes in free energy. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free-energy change can be calculated from the definition of free energy, if the standard enthalpy and entropy changes are known, using Equation \(\ref{Eq5}\): \[ΔG° = ΔH° − TΔS° \label{Eq5}\] If ΔS° and ΔH° for a reaction have the same sign, then the sign of \(ΔG^o\) depends on the relative magnitudes of the ΔH° and TΔS° terms. It is important to recognize that a positive value of \(ΔG^o\) for a reaction does not mean that no products will form if the reactants in their standard states are mixed; it means only that at equilibrium the concentrations of the products will be less than the concentrations of the reactants. A positive \(ΔG^o\) means that the equilibrium constant is less than 1. Calculate the standard free-energy change (ΔG°) at 25°C for the reaction \[\ce{ H2(g) + O2(g) <=>  H2O2(l)} \nonumber\] At 25°C, the standard enthalpy change (ΔH°) is −187.78 kJ/mol, and the absolute entropies of the products and reactants are: Is the reaction spontaneous as written? : balanced chemical equation, ΔH° and S° for reactants and products spontaneity of reaction as written : A To calculate \(ΔG^o\) for the reaction, we need to know ΔH°, ΔS°, and T. We are given ΔH°, and we know that T = 298.15 K. We can calculate ΔS° from the absolute molar entropy values provided using the “products minus reactants” rule: \[\begin{align*}\Delta S^\circ &=S^\circ(\mathrm{H_2O_2})-[S^\circ(\mathrm{O_2})+S^\circ(\mathrm{H_2})] \\[4pt] &=[\mathrm{1\;mol\;H_2O_2}\times109.6\;\mathrm{J/(mol\cdot K})] \\[4pt] &-\left \{ [\textrm{1 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]+[\textrm{1 mol O}_2\times 205.2\;\mathrm{J/(mol\cdot K)}] \right \} \\[4pt] &=-226.3\textrm{ J/K }(\textrm{per mole of }\mathrm{H_2O_2}) \end{align*}\] As we might expect for a reaction in which 2 mol of gas is converted to 1 mol of a much more ordered liquid, ΔS° is very negative for this reaction. B Substituting the appropriate quantities into Equation \(\ref{Eq5}\), \[\begin{align*}\Delta G^\circ &=\Delta H^\circ -T\Delta S^\circ \\[4pt] &=-187.78\textrm{ kJ/mol}-(\textrm{298.15 K}) [-226.3\;\mathrm{J/(mol\cdot K)}\times\textrm{1 kJ/1000 J}] \\[4pt] &=-187.78\textrm{ kJ/mol}+67.47\textrm{ kJ/mol} \\[4pt] &=-120.31\textrm{ kJ/mol} \nonumber \end{align*}\] The negative value of \(ΔG^o\) indicates that the reaction is spontaneous as written. Because ΔS° and ΔH° for this reaction have the same sign, the sign of \(ΔG^o\) depends on the relative magnitudes of the ΔH° and TΔS° terms. In this particular case, the enthalpy term dominates, indicating that the strength of the bonds formed in the product more than compensates for the unfavorable ΔS° term and for the energy needed to break bonds in the reactants. Calculate the standard free-energy change (ΔG°) at 25°C for the reaction \[\ce{ 2H2(g) + N2(g) <=>  N2H4(l)}. \nonumber \] At 25°C, the standard enthalpy change (ΔH°) is 50.6 kJ/mol, and the absolute entropies of the products and reactants are S°(N H ) = 121.2 J/(mol•K), S°(N ) = 191.6 J/(mol•K), and S°(H ) = 130.7 J/(mol•K). Is the reaction spontaneous as written? 149.5 kJ/mol; no Tabulated values of standard free energies of formation allow chemists to calculate the values of \(ΔG^o\) for a wide variety of chemical reactions rather than having to measure them in the laboratory. The standard free energy of formation (\(ΔG^∘_f\))of a compound is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. By definition, the standard free energy of formation of an element in its standard state is zero at 298.15 K. One mole of Cl gas at 298.15 K, for example, has \(ΔG^∘_f = 0\). The standard free energy of formation of a compound can be calculated from the standard enthalpy of formation (ΔH ) and the standard entropy of formation (ΔS ) using the definition of free energy: \[ΔG^o_{f} =ΔH^o_{f} −TΔS^o_{f} \label{Eq6}\] Using standard free energies of formation to calculate the standard free energy of a reaction is analogous to calculating standard enthalpy changes from standard enthalpies of formation using the familiar “products minus reactants” rule: \[ΔG^o_{rxn}=\sum mΔG^o_{f} (products)− \sum n ΔG^o_{f} (reactants) \label{Eq7a}\] where m and n are the stoichiometric coefficients of each product and reactant in the balanced chemical equation. A very large negative \(ΔG^o\) indicates a strong tendency for products to form spontaneously from reactants; it does not, however, necessarily indicate that the reaction will occur rapidly. To make this determination, we need to evaluate the kinetics of the reaction. The \(ΔG^o\) of a reaction can be calculated from tabulated \(ΔG^o_f\) values ( ) using the “products minus reactants” rule. Calculate \(ΔG^o\) for the reaction of isooctane with oxygen gas to give carbon dioxide and water. Use the following data: Is the reaction spontaneous as written? : balanced chemical equation and values of ΔG° for isooctane, CO , and H O : spontaneity of reaction as written : Use the “products minus reactants” rule to obtain ΔG , remembering that ΔG° for an element in its standard state is zero. From the calculated value, determine whether the reaction is spontaneous as written. The balanced chemical equation for the reaction is as follows: \[\ce{ C8H18(l) + 25/2 O2(g) -> 8CO2(g) + 9H2O(l)} \nonumber\] We are given ΔG values for all the products and reactants except O (g). Because oxygen gas is an element in its standard state, ΔG (O ) is zero. Using the “products minus reactants” rule, \[\begin{align*} \Delta G^\circ &=[8\Delta G^\circ_\textrm f(\mathrm{CO_2})+9\Delta G^\circ_\textrm f(\mathrm{H_2O})]-\left[1\Delta G^\circ_\textrm f(\mathrm{C_8H_{18}})+\dfrac{25}{2}\Delta G^\circ_\textrm f(\mathrm{O_2})\right] \\[4pt] &=[(\textrm{8 mol})(-394.4\textrm{ kJ/mol})+(\textrm{9 mol})(-237.1\textrm{ kJ/mol})] \\[4pt] &-\left [(\textrm{1 mol})(-353.2\textrm{ kJ/mol})+\left(\dfrac{25}{2}\;\textrm{mol}\right)(0 \textrm{ kJ/mol}) \right ] \\[4pt]  &=-4935.9\textrm{ kJ }(\textrm{per mol of }\mathrm{C_8H_{18}}) \end{align*}\] Because \(ΔG^o\) is a large negative number, there is a strong tendency for the spontaneous formation of products from reactants (though not necessarily at a rapid rate). Also notice that the magnitude of \(ΔG^o\) is largely determined by the ΔG of the stable products: water and carbon dioxide. Calculate \(ΔG^o\) for the reaction of benzene with hydrogen gas to give cyclohexane using the following data Is the reaction spontaneous as written? 92.8 kJ; no Calculated values of \(ΔG^o\) are extremely useful in predicting whether a reaction will occur spontaneously if the reactants and products are mixed under standard conditions. We should note, however, that very few reactions are actually carried out under standard conditions, and calculated values of \(ΔG^o\) may not tell us whether a given reaction will occur spontaneously under nonstandard conditions. What determines whether a reaction will occur spontaneously is the free-energy change (ΔG) under the actual experimental conditions, which are usually different from ΔG°. If the ΔH and TΔS terms for a reaction have the same sign, for example, then it may be possible to reverse the sign of ΔG by changing the temperature, thereby converting a reaction that is not thermodynamically spontaneous, having K < 1, to one that is, having a K > 1, or vice versa. Because ΔH and ΔS usually do not vary greatly with temperature in the absence of a phase change, we can use tabulated values of ΔH° and ΔS° to calculate \(ΔG^o\) at various temperatures, as long as no phase change occurs over the temperature range being considered. In the absence of a phase change, neither \(ΔH\) nor \(ΔS\) vary greatly with temperature. Calculate (a) \(ΔG^o\) and (b) ΔG for the reaction N (g)+3H (g)⇌2NH (g), assuming that ΔH and ΔS do not change between 25°C and 300°C. Use these data: : balanced chemical equation, temperatures, S° values, and ΔH for NH : \(ΔG^o\) and ΔG at 300°C : A To calculate \(ΔG^o\) for the reaction using Equation \(\ref{Eq5}\), we must know the temperature as well as the values of ΔS° and ΔH°. At standard conditions, the temperature is 25°C, or 298 K. We can calculate ΔS° for the reaction from the absolute molar entropy values given for the reactants and the products using the “products minus reactants” rule: \[\begin{align*}\Delta S^\circ_{\textrm{rxn}}&=2S^\circ(\mathrm{NH_3})-[S^\circ(\mathrm{N_2})+3S^\circ(\mathrm{H_2})] \\[4pt] &=[\textrm{2 mol NH}_3\times192.8\;\mathrm{J/(mol\cdot K)}] \\[4pt] &-\left \{[\textrm{1 mol N}_2\times191.6\;\mathrm{J/(mol\cdot K)}]+[\textrm{3 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]\right \} \\[4pt] &=-198.1\textrm{ J/K (per mole of N}_2)\end{align*} \] We can also calculate ΔH° for the reaction using the “products minus reactants” rule. The value of ΔH (NH ) is given, and ΔH is zero for both N and H : \[\begin{align*}\Delta H^\circ_{\textrm{rxn}}&=2\Delta H^\circ_\textrm f(\mathrm{NH_3})-[\Delta H^\circ_\textrm f(\mathrm{N_2})+3\Delta H^\circ_\textrm f(\mathrm{H_2})] \\[4pt] &=[2\times(-45.9\textrm{ kJ/mol})]-[(1\times0\textrm{ kJ/mol})+(3\times0 \textrm{ kJ/mol})] \\[4pt] &=-91.8\textrm{ kJ(per mole of N}_2) \end{align*} \] B Inserting the appropriate values into Equation \(\ref{Eq5}\) \[\begin{align*} \Delta G^\circ_{\textrm{rxn}} &=\Delta H^\circ-T\Delta S^\circ \\[4pt] &=(-\textrm{91.8 kJ})-(\textrm{298 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J}) \\[4pt] &=-\textrm{32.7 kJ (per mole of N}_2) \end{align*}\] C To calculate \(ΔG\) for this reaction at 300°C, we assume that ΔH and ΔS are independent of temperature (i.e., ΔH = H° and ΔS = ΔS°) and insert the appropriate temperature (573 K) into Equation \(\ref{Eq2}\): \[\begin{align*}\Delta G_{300^\circ\textrm C}&=\Delta H_{300^\circ\textrm C}-(\textrm{573 K})(\Delta S_{300^\circ\textrm C}) \\[4pt] &=\Delta H^\circ -(\textrm{573 K})\Delta S^\circ \\[4pt] &=(-\textrm{91.8 kJ})-(\textrm{573 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J}) \\[4pt] &=21.7\textrm{ kJ (per mole of N}_2) \end{align*} \] In this example, changing the temperature has a major effect on the thermodynamic spontaneity of the reaction. Under standard conditions, the reaction of nitrogen and hydrogen gas to produce ammonia is thermodynamically spontaneous, but in practice, it is too slow to be useful industrially. Increasing the temperature in an attempt to make this reaction occur more rapidly also changes the thermodynamics by causing the −TΔS° term to dominate, and the reaction is no longer spontaneous at high temperatures; that is, its K is less than one. This is a classic example of the conflict encountered in real systems between thermodynamics and kinetics, which is often unavoidable. Calculate for the following reaction \[\ce{2NO(g) + O2 (g) <=>  2NO2 (g)} \nonumber\] which is important in the formation of urban smog. Assume that \(ΔH\) and \(ΔS\) do not change between 25.0°C and 750°C and use these data: −72.5 kJ/mol of \(O_2\) 33.8 kJ/mol of \(O_2\) The effect of temperature on the spontaneity of a reaction, which is an important factor in the design of an experiment or an industrial process, depends on the sign and magnitude of both ΔH° and ΔS°. The temperature at which a given reaction is at equilibrium can be calculated by setting \(ΔG^o\) = 0 in Equation \(\ref{Eq5}\), as illustrated in Example \(\Page {4}\). As you saw in Example \(\Page {3}\), the reaction of nitrogen and hydrogen gas to produce ammonia is one in which ΔH° and ΔS° are both negative. Such reactions are predicted to be thermodynamically spontaneous at low temperatures but nonspontaneous at high temperatures. Use the data in Example \(\Page {3}\) to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous, assuming that ΔH° and ΔS° are independent of temperature. : ΔH° and ΔS° : temperature at which reaction changes from spontaneous to nonspontaneous : Set \(ΔG^o\) equal to zero in Equation \(\ref{Eq5}\) and solve for T, the temperature at which the reaction becomes nonspontaneous. In Example \(\Page {3}\), we calculated that ΔH° is −91.8 kJ/mol of N and ΔS° is −198.1 J/K per mole of N , corresponding to \(ΔG^o\) = −32.7 kJ/mol of N at 25°C. Thus the reaction is indeed spontaneous at low temperatures, as expected based on the signs of ΔH° and ΔS°. The temperature at which the reaction becomes nonspontaneous is found by setting \(ΔG^o\) equal to zero and rearranging Equation \(\ref{Eq5}\) to solve for T: \[\begin{align*}\Delta G^\circ &=\Delta H^\circ - T\Delta S^\circ=0 \\[4pt] \Delta H^\circ &=T\Delta S^\circ \\[4pt] T=\dfrac{\Delta H^\circ}{\Delta S^\circ}&=\dfrac{(-\textrm{91.8 kJ})(\textrm{1000 J/kJ})}{-\textrm{198.1 J/K}}=\textrm{463 K}\end{align*}\] This is a case in which a chemical engineer is severely limited by thermodynamics. Any attempt to increase the rate of reaction of nitrogen with hydrogen by increasing the temperature will cause reactants to be favored over products above 463 K. As you found in the exercise in Example \(\Page {3}\), ΔH° and ΔS° are both negative for the reaction of nitric oxide and oxygen to form nitrogen dioxide. Use those data to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous. 792.6 K \[ΔG = ΔH − TΔS \nonumber\] \[ΔG° = ΔH° − TΔS° \nonumber\] We can predict whether a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written but is spontaneous in the reverse direction, ΔG > 0. At constant temperature and pressure, ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free energy of formation (ΔG ), is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. Tabulated values of standard free energies of formation are used to calculate \(ΔG^o\) for a reaction. ) ( )

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Fourier transform is a mathematical technique that can be used to transform a function from one real variable to another. It is a unique powerful tool for spectroscopists because a variety of spectroscopic studies are dealing with electromagnetic waves covering a wide range of frequency. In Fourier transform term \(\ e^{ - 2\pi ixy} \ \), when x represents frequency, the corresponding y is time. This provides an alternate way to process signal in time domain instead of the conventional frequency domain. To realize this idea, Fourier transform from time domain to frequency domain is the essential process that enable us to translate raw data to readable spectra. Recent prosperity of Fourier transform in spectroscopy should also attribute to the development of efficient Fast Fourier Transform algorithm. The nature of trigonometric function enables Fourier transform to convert a function from the domain of one variable to another and reconstruct it later on. This is a robust mathematical tool to process data in different domains under different circumstances. Taking this principal idea and applying it in spectroscopy showed many impressive results in the early stage, which in other ways are very difficult to resolve. These benefits triggered a wide exploration of Fourier transform based methodology in a variety of spectroscopic techniques. At the same time, Fourier transform spectroscopic instruments are developed with great efforts by physicists and engineers. All these factors give rise to the wide use of Fourier transform spectroscopy. In the following topics, the relevant mathematical background, the implementation of Fourier transform in spectroscopy and a brief overview of various Fourier transform Spectrometers will be addressed in sequence. The motivation of Fourier transform arises from Fourier series, which was proposed by French mathematician and physicist Joseph Fourier when he tried to analyze the flow and the distribution of energy in solid bodies at the turn of the 19th century. He claimed that the temperature distribution could be described as an infinite series of sines and cosines of the form shown in equation (1): \[f(x)= \dfrac{a_{0}}{2} + \displaystyle \sum_{n=1}^{\infty} \left(a_{n}\cos\dfrac{n\pi x}{L}+b_{n} \sin \frac{n \pi x}{L}\right) \tag{1}\] It turns out that this combination of sines and cosines series can be used to express any periodical function. As \(n\) increases, the series will approach to the original function more closely. If we use Euler's Identity (equation 2), as well as the exponential representations of sine (equation 3 and cosine (equation 4 \[ e^{i\theta}=\cos{\theta}+ i\sin{\theta} \tag{2}\] \[\cos{\theta}=\dfrac{1}{2} (e^{i\theta} + e^{-i\theta}) \tag{3}\] \[\sin{\theta}=\dfrac{1}{2i} (e^{i\theta} - e^{-i\theta}) \tag{4}\] \[C_n =\dfrac{1}{2}(C_n-ib_n) \tag{5}\] \[f(x) = \sum_{n=-\infty}^{\infty} C_n e^{i\theta} \tag{6a}\] \[\theta = \dfrac{2\pi n x}{L} \tag{6b}\] Writing the Fourier series in this exponential form helps to simplify many formulas and expressions involved in the transformation. Then we can consider an extreme case, when L in equation 1, the summation becomes an integral as shown in equation (4) \[ f(x) = \sum\limits_{-\infty} ^\infty c_n e^{n\frac{i\pi x}{L}} = \int_{ -\infty}^\infty c_n e^{n\frac{i\pi x}{L}} dn = \int_{-\infty}^\infty c_n e^{\frac{i\pi xn}{L}} dn \tag{7}\] This naturally gives the Fourier transform pair of f(x) and F(y). The relationships are shown below : \[F(y) = \int_{ -\infty }^{ +\infty } {f(x)e^{ - i2\pi yx} dx} \tag{8a}\] \[f(x) = \int_{ -\infty }^{ +\infty } {F(y)e^{ i2\pi xy} dx} \tag{8b}\] Another important formula widely used to benefit In other cases, it is used to simplify the integral in the Fourier transform based on the symmetry of the function. But so far, all these are just about mathematics. Its story with spectroscopy should start from the mathematical description of electromagnetic waves. Maxwell–Faraday equation and Ampère's circuital law give us electromagnetic wave equations to describe the characteristics of an electromagnetic wave. Using the linearity of Maxwell's equations in a vacuum, the solutions of the equation can be decomposed into a superposition of sinusoids as shown below : \[E(r,t) = \overrightarrow {E_0 } \cos (2\pi ft - \overrightarrow k \cdot r + \phi _0 ) \tag{9a}\] \[B(r,t) = \overrightarrow {B_0 } \cos (2\pi ft - \overrightarrow k \cdot r + \phi _0 ) \tag{9b}\] Where t is time, f is the frequency, k=(k ,k ,k ) is the wave vector and \(\phi\) is the phase angle. This indicates that electromagnetic wave can be written as the sum of trigonometric functions with specific frequencies. Scientists already discovered the fact that frequency and time is a classic Fourier transform pair in Fourier transform relationship. All the Fourier transform pairs are connected by the Fourier transform term \(e^{ - i2\pi yx}\). Regarding this case, we can use the term to transform between two variables in this pair, namely time and frequency. In this way, we can measure the properties of the electromagnetic wave in both conventional frequency domain and somehow more robust time domain. Fourier transform are widely involved in spectroscopy in all research areas that require high accuracy, sensitivity, and resolution. All these spectroscopic techniques using Fourier transform are considered Fourier transform spectroscopy. By definition, Fourier transform spectroscopy is a spectroscopic technique where interferograms are collected by measurements of the coherence of an electromagnetic radiation source in the time-domain or space-domain, and translated into frequency domain through Fourier transform. How to introduce a time-domain or space-domain variable in the spectrometer is the primary question that needed to be addressed when we consider constructing a Fourier transform spectrometer. In the experimental set-up, a Michelson interferometer is commonly used to solve this problem. Different from the classical Michelson interferometer with two fixed mirrors ( ), the interferometer used in Fourier transform spectrometer has a moving mirror at one arm ( ). Scheme for Michelson interferometer [components: coherent light source; half-silvered beam-splitting mirror; two highly polished reflective mirrors; detector] Stationary version [two fixed mirrors] Movable version [One movable mirror and one movable mirror] As shown in Figure 1.b, when a parallel beam of coherent light hits a half-silvered mirror, it is divided into two beams of equal intensities by partial reflection and transmission. After being reflected back, the two beams meet at the half-silvered mirror and recombine to produce an interference pattern, which is later detected by the detector. Manipulating the difference between these two paths of light is the core of Michelson interferometer. If these two paths differ by a whole number of wavelengths, the resulting constructive interference will give a strong signal at the detector. If they differ by a whole number and a half of wavelengths, destructive interference will cancel the intensity of the signal. With a Fourier transform spectrometer equipped with an interferometer, we can easily vary the parameter in time domain or spatial domain by changing the position of the movable mirror. But how data are collected by a Fourier transform spectrometer? A quick comparison between a conventional spectrometer and a Fourier transform Spectrometer may help to find the answer. Monochromator is commonly used. It can block off all other wavelengths except for a certain wavelength of interest. Then measuring the intensity of a monochromic light with that particular wavelength becomes practical. To collect the full spectrum over a wide wavelength range, monochromator needs to vary the wavelength setting every time. Rather than allowing only one wavelength to pass through the sample at a time, an interferometer can let through a beam with the whole wavelength range at once, and measure the intensity of the total beam at that optical path difference. Then by changing the position of the moving mirror, a different optical path difference is modified and the detector can measure another intensity of the total beam as the second data point. If the beam is modified for each new data point by scanning the moving mirror along the axis of the moving arm, a series of intensity versus each optical path length difference are collected. So instead of obtaining a scan spectrum directly, raw data recorded by the detector in a Fourier transform spectrometer is less intuitive to reveal the property of the sample. The raw data is actually the intensity of the interfering wave versus the optical path difference (also called Interferogram). The spectrum of the sample is actually encoded into this interferogram. Based on the previous discussion, it is predictable that, without further translation, the raw data collected on a Fourier transform spectrometer will be quite difficult to read. A Fourier transform needs to be performed to decode interferogram and extract actual spectrum I(\(\overline v\)) from it. The following shows how to conduct a Fourier transform to decode: The intensity collected by the detector is a function of the path length differences in the interferometer p and wavenumber \(\overline v\) : \[I(p,\overline v ) = I(\overline v )[1 + \cos (2\pi \overline v p)] \tag{9}\] Thus, the total intensity measured at a certain optical path length difference (for each data point at a certain optical pathlength difference p) is: \[I(p) = \int_0^\infty {I(p,\overline v ) = I(\overline v )[1 + \cos (2\pi \overline v p)]} \cdot d\overline v \tag{10}\] It shows that they have a cosine Fourier transform relationship. So by computing an inverse Fourier transform, we can resolve the desired spectrum in terms of the measured raw data I(p) (10): \[I(\overline v ) = 4\int_0^\infty {[I(p) - \frac{1} {2}I(p = 0)]} \cos (2\pi \overline v p) \cdot dp \tag{11}\] An example to illustrate the raw data and the resolved spectrum is also shown in . Fourier transform between interferogram and actual spectrum Fast Fourier Transform (FFT) is a very efficient algorithm to compute Fourier transform. It applies to Discrete Fourier Transform (DFT) and its inverse transform. DFT is a method that decomposes a sequence of signals into a series of components with different frequency or time intervals. This operation is useful in many fields, but in most cases computing it directly from definition is too slow to be practical. Fast Fourier Transform algorithm can help to reduce DFT computation time by several orders of magnitude without losing the accuracy of the result. This benefit becomes more significant when the number of the components is very large. FFT is considered a huge improvement to make many DFT-based algorithms practical. In Fourier transform spectrometer, signals are often collected by a series of optical or digital channels at the detector. Then FFT is of great importance to quickly achieve the following signal processing and data extraction based on DFT method. Combining all these steps together, we can take a look at how the data from the sample are processed. The diagram is shown in . (Figure from ThermoNiolet at ) Continuous Fourier transform spectroscopy refers to the scanning form of FTS, in which by step moving one mirror, the whole range of optical path difference is measured. This is the most widely used mode in FTS, like most absorption spectra and emission spectra obtained by FTS. In some Fourier transform spectrometers, depending on the feature of the involved spectroscopic technique and purpose of measurement, a pulsed Fourier transform technique may be applied instead of the scanning mode. Pulsed FTS is different from conventional continuous FTS. It is not based on the transmittance technique, which is widely used in the absorption spectra, like FTIR. Instead, in pulse FTS, the idea is that the sample is first exposed to an energizing event, and this pulse induces a periodic response. The frequency of this response relative to the field strength is determined by the properties of the sample. Using Fourier transform to resolve the frequency will tell the information about the targeted analyte. Pulse FTS is a relatively new improvement of FTS. Some examples are from pulse-Fourier Transform-Nuclear Magnetic Resonnance (FT-NMR), pulse-Fourier Transform-Electron Paramagnetic Resonnance (FFT-EPR) and Fourier Transform-Mass Spectrometry (FT-MS). Please refer to the following topics for more details about how they work. In addition to the continuous/scanning mode of FTS, a number of stationary Fourier transform spectrometers are also available to meet special needs. The principle of the interferometer and the analysis of its output signal is similar to the typical scanning FTS. But the signal is collected at certain optical path length differences rather than scanning over the whole range of the path difference. Fourier transform spectroscopy can be applied to a variety of regions of spectroscopy and it continues to grow in application and utilization including optical spectroscopy, infrared spectroscopy (IR), nuclear magnetic resonance, electron paramagnetic resonance spectroscopy, mass spectrometry, and magnetic resonance spectroscopic imaging (MRSI). Among them, Fourier Transform Infrared Spectroscopy ( ) has been most intensively developed, which uses scanning Fourier transform to measure the mid-IR absorption spectra. . (NMR) and Spectroscopy (EPR) are two magnetic techniques that use pulse Fourier transform mode. A Radio Frequency Pulse (RF Pulse) in a strong ambient magnetic field background is used as the energizing event. This RF Pulse directs the magnetic particles at an angle to the ambient strong magnetic field, causing gyration of the particle. Then the resulting gyrating spin induces a periodic current in the detector coil. This periodic current is recorded as the signal. Each gyrating spin has a characteristic frequency relative to the strength of the ambient magnetic field, which is also governed by the properties of the sample. Fourier Transform Mass Spectrometry (MS) is also operated at pulse Fourier transform mode. Different from NMR and EPR, the injection of the charged sample into the strong electromagnetic field of a cyclotron acts as the energizing event in MS. The injected charged particles travel in circles under the strong electromagnetic field. The circular pathway will thus induce a current in a fixed coil at one point in their circle. Each traveling particle exhibits a characteristic cyclotron frequency relative to the field strength, which is determined by the masses in the sample. 1. Firstly, Fourier transform spectrometers have a multiplex advantage (Fellgett advantage) over dispersive spectral detection techniques for signal, but a multiplex disadvantage for noise; Moreover, measurement of a single spectrum is faster(in the FTIR technique) because the information at all frequencies is collected simultaneously. This allows multiple samples to be collected and averaged together also resulting in an improvement in sensitivity; In addition, FT spectrometers are cheaper than conventional spectrometers because building of interferometers is easier than the fabrication of a monochromator (in the FTIR technique). So most commercial IR spectrometers are built based on FTIR techniques. practical frequency regions limited (FT UV-vis is not quite practical) 2. See the following figure for the solution: 3. Interferometer vs. Monochromator Interometer: a. Collect signal in time or spatial domain; b. Measure all frequencies in the incident beam at one time; c. Determined by the interferometer, raw data from FT spectrometer is an interogram, which needs to be Fourier Transform back to get spectrum. Monochromator: a. Collect signal in frequency domain; b. Scan each wavelength and measure the intensity for each single wavelength at a time; c. Determined by the feasure of monochromator, spectrum can be directly collected from the spectrometer; 4. Interferometer and Fast Fourier Transform Data Analyzer Interferometer: to generate continuous optical path length difference and enable the idea to collect data in the time or spatial domain; Fast Fourier Transform Data Analyzer: to quickly transform the raw data (interferogram) to spectrum by using fast Fourier transform algorithm; 5. FTIR instrumentation: A FTIR Spectrometer Layout (Figure from ThermoNiolet at ) FT-NMR instrumentation: A modern high resolution liquid FT-NMR instrumentation is shown in : A schematic diagram of liquid FT-NMR

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Upon binding a second axial ligand, the iron center together with the axial base move toward the plane of the porphyrin, initiating a change in spin state from high-spin to low-spin when the sixth ligand is O or CO or any other strong ligand with an even number of valence electrons. Given these general features, what are the structural differences between systems that bind O with high affinity and those that bind O with low affinity? The answers to this question are relevant to understanding at the molecular level the mechanism of cooperativity, where a low-affinity conformation, the T state, and a high-affinity conformation, the R state, are in dynamic equilibrium in one tetrameric molecule. In looking at crystallographic data one sees a particular conformation frozen in the crystal, usually the one of lowest free energy among many in equilibrium in the solution state. The R \(\rightleftharpoons\) T equilibrium for hemoglobin is moderately rapid, at 4 x 10 s ; hemocyanin also switches quaternary conformations with a similar rate constant. Human hemoglobins are a heterogeneous group. Many mutants are known, and several have been structurally characterized. A structural alteration that affects the equilibrium between R and T states has a marked effect on ligand affinity and cooperativity in hemoglobin. If a specific amino-acid substitution destabilizes the T state, then the transition to the R state will occur earlier in the ligation process, and the hemoglobin will have an increased oxygen affinity. Hemoglobin Kempsey is an example. In this mutant an aspartic acid on the \(\beta\) chain is replaced by asparagine. Conversely, if the R state is destabilized, then the hemoglobin will have a lowered oxygen affinity. Hemoglobin Kansas is an example. Here an asparagine on the \(\beta\) chain has been replaced by threonine. It was proposed earlier that the molecule Fe(PF)(1-MeIm)(O ) in the solid or solution state was a fair approximation to the reference gas-phase molecule. The axial base, although not oriented for minimization of contacts with the porphyrin (i.e., \(\phi\) = 45°), is well-removed from an eclipsing orientation where \(\phi\) = 0 ± 10°; the Fe atom is centered in the plane of a highly planar porphyrin; the O ligand is oriented for minimization of contacts with the porphyrin, and its geometry is largely unconstrained by distal groups (the pickets); no groups are hydrogen-bonded to the axial base. The major difference from the reference state is that there is a significant attractive interaction between the electronegative dioxygen moiety and the amide groups on the pickets, and a smaller repulsive interaction with the picket t-butyl groups. For the CO adduct, contacts with the pickets are all at ideal van der Waals' separations and the Fe—CO moiety is free to assume its normal linear geometry. For CO binding the reference molecule is again the carbonyl adduct of the iron picket-fence porphyrinato molecule. In contrast to O binding, there are no specific distal effects, such as hydrogen bonding, by which CO affinity may be increased; there remain many ways, as with O binding, by which CO affinities may be reduced. Thus, the CO binder with highest affinity is the iron picket-fence porphyrin. The O affinities of myoglobin, R-state hemoglobin, and the Fe(PF)(1-MeIm) system are similar. However, the means by which this is achieved are different, and this difference is reflected most clearly in the kinetics of binding and release of O , which for Mb are much slower. The similarities and differences are summarized in Table 4.9, which is culled from Tables 4.2, 4.4, and 4.5. Fe(PF-Im) Fe(PF)(1-MeIm) Fe(Poc-PF) (1-MeIm) His (H O) polar, protic H-N(amide) polar, aprotic H-N (amide) phenyl NH 4.06(5) CH 2.67(6) NH 5.0 CH 3.3 Fe-N (deoxy), Å For the cobalt-dioxygen derivative, the putative hydrogen bonding between the dioxygen and the amide groups of the pickets assumes greater importance because the coordinated dioxygen is substantially more negative. Again the picket-fence porphyrin, being structurally characterized, is the reference system. Although no Co picket-fence porphyrin structures have been determined, the structures may be predicted with confidence from the iron analogues together with related structures of Co and Co tetraphenylporphyrinato systems.* * For CoHbO , single-crystal EPR spectra have been interpreted in terms of a nearly triangularly coordinated O , although a crystal structure of CoMbO shows a bent CoOO group. There is no precedent for this triangular arrangement in any Co -superoxo (O ) system, whereas there are many for angularly coordinated O in electronically not dissimilar square-planar Schiff-base systems. Regardless of geometry, the picket amide • • • O contacts do not change substantially. The 2-methyl substituent on 2-methylimidazole is not sterically active in the five-coordinate structures Fe(PF)(2-MeIm) and Fe(TPP)(2-MeIm), since the iron atom is displaced from the plane of the porphyrin by the expected amount and the Fe—N bond is unstretched and similar to that in deoxyhemoglobin (low O affinity) and deoxyMb (higher O affinity). Moreover, resonance Raman measurements also indicate little strain in this bond. In other words, there is no "tension at the heme," a key concept in early discussions of cooperativity before structures on model systems and high-resolution, refined protein structures became available. On moving into the plane of the porphyrin upon oxygenation, the 2-methyl substituent prevents the Fe-imidazole group from achieving its optimum geometry with the iron at the center of the porphyrin hole, as seen in the structure of Fe(PF)(1-MeIm)(O ). Thus, the sterically active 2-methyl substituent leads to lowered O (and CO) affinity relative to the 1-methyl analogue. In metrical terms the lowered affinity is reflected in an increase in the sum of the axial bond lengths from 1.75 + 2.07 = 3.82 Å to 1.90 + 2.11 = 4.01 Å. In the crystal structure of Fe(C Cap)(1-MeIm)(CO) the cap is about 5.6 Å from the porphryin plane. Hence, in the crystal structures of the free base H2(C Cap) and FeCI(C Cap) species, in which the cap is screwed down to approximately 4.0 Å from the porphyrin plane, considerable conformational rearrangement of the cap and the four chains attaching it to the porphyrin is needed to provide room for a small ligand such as CO. This is even more pronounced in a Co(C Cap) complex where the cap is only 3.49 Å from the mean porphyrin plane. Thus not only is affinity for CO lowered, but some additional discrimination against it is induced, since a linear, perpendicular coordination creates considerable strain energy elsewhere in the molecule. For the pocket porphyrin (Figure 4.23), structural data are available on the carbonyl adduct. The CO ligand is unable to achieve the linear perpendicular geometry seen in the high-affinity picket-fence porphyrin derivative, Fe(PF)(1-MeIm)(CO), and distortion of the porphyrin core is greater. In the pocket-porphyrin system, O affinity is unaffected, but CO affinity is lowered. The crystal structure of partially oxygenated hemoglobin, [\(\alpha\)-FeO ] [\(\beta\)-Fe] , reveals that the quaternary structure, except in the immediate vicinity of the \(\alpha\) hemes, which have O coordinated, resembles that of T-state deoxyhemoglobin rather than R-state liganded hemoglobin. In accord with the low affinity of T-state hemoglobin, the Fe—N bonds for the six-coordinate \(\alpha\)-hemes at 2.37 Å are significantly longer than those in fully oxygenated R-state oxyhemoglobin, [\(\alpha\)-FeO ] [\(\beta\)-FeO ] in the notation above, (1.94 (\(\alpha\)-hemes) and 2.06 Å (\(\beta\) hemes)) and that found in oxymyoglobin (2.07 Å). In contrast to the R-state structure and oxyMb, the a-hemes are folded as seen in the deoxy parent, leaving the Fe still substantially displaced (0.2 Å) from the plane of the four pyrrole nitrogen atoms. The deoxyhemoglobin T-state quaternary structure also has been observed in two other partially liganded hybrid hemoglobins, [\(\alpha\)-FeCO] [\(\beta\)-Mn(II)] and [\(\alpha\)-Ni] [\(\beta\)-FeCO] . Again, structural changes upon coordination do not propagate beyond the immediate vicinity of the liganded heme to the critical \(\alpha_{1} \beta_{2}\) interfaces. Note that although the crystal structure of hemoglobin A reveals that access to the binding site for the \(\beta\) chains is blocked by groups at the entrance to the cavity above the iron center, this does not prevent facile access to the binding site; the rate of O binding is slowed by a factor of only five. A similar situation occurs also for vertebrate myoglobins. The large structural differences that exist between deoxy (T) and oxy (R) hemoglobin and the much smaller differences between deoxy (T) and partially liganded (T) hybrid hemoglobin are shown in Figure 4.32. Because of the steric hindrance afforded by the distal histidine, all biological systems have low affinity for CO relative to the picket-fence porphyrins, with the exception of mutants where the distal histidine has been replaced by glycine. Thus low affinity to CO is associated primarily with the inability of the Fe-CO group to achieve its preferred linear geometry perpendicular to the porphyrin. Low-affinity O binding in the hemoglobins appears to be associated with the inability of the Fe-proximal histidine unit to move into the plane of the porphyrin and less so to distal effects, such as a cavity too small to accommodate the coordinated ligand. The blocked access to the site affects the kinetics but not necessarily the thermodynamics of ligand binding, as evidenced by the structure of T-state [\(\alpha\)-FeNi] [\(\beta\)-FeCO] Some similarities between the structures and properties of partially oxygenated (T-state) [\(\alpha\)-FeO ] [\(\beta\)-Fe] hemoglobin and Fe(PF)(2-MeIm)(O ) are provided in Table 4.10. In the synthetic systems low O affinity can be induced by 2-methyl substituents—a restraint on the movement of the Fe-imidazole moiety analogous to that provided by the protein chain. A second means is by distal effects, such as caps and straps. HbA [\(\alpha\)-FeO ] [\(\beta\)-Fe] Fe(PF)(2-MeIm) Fe(PF)(1,22-Me Im) histidine polar H-N(amide) polar, aprotic NH 3.88 CH 2.77(3) NH 4.9 CH 3.5 Fe-N (deoxy), Å \(\dagger\) Shortly (10 -10 s) after a ligand dissociates, a large difference in (Fe—N ) between R and T structures is observed, prior to relaxation to the equilibrium R and T conformations. Few structural data are available for high-affinity oxygen carriers. The crystal structures of two leghemoglobin derivatives, a monomeric myoglobin-like oxygen carrier found in the nitrogen-fixing nodules of legumes, are known at 2.0 and 3.3 Å. The binding pocket appears more open, perhaps allowing H O to enter and partake in stronger hydrogen bonding than that offered by the distal imidazole. Consistent with this notion is the more rapid rate of autoxidation observed for oxyleghemoglobin. oxymyoglobin, which lacks a distal histidine, also autoxidizes rapidly, although a distal arginine further along the helix E, E10Arg, fulfills the role of the distal histidine by hydrogen bonding to the sixth ligand, at least in the fluoride derivative, met-MbF. Although no structural data are available, a tenfold increase in O affinity was observed between an ester-strapped porphyrin, offering no hydrogen-bonding possibilities, and its conformationally very similar amide analogue. O • • • amide hydrogen bonding was demonstrated by means of NMR shift data (Zn and Fe—CO complexes vs. the Fe—O complex) and from infrared spectroscopy, which showed shifted amide N—H absorptions. The structural features that lead to the extraordinarily high affinity for O and low affinity for CO in hemoglobin remain unidentified. This high affinity is due to an slow dissociation rate of O of only 0.1s ; in most hemoglobins the rate is about 10 to 2,500 s (Table 4.2). Dioxygen binding is thus close to irreversible. Figure 4.27 shows that hydrogen bonding to the coordinated dioxygen ligand, unrestrained motion of the Fe-proximal histidine group into the plane of the porphyrin, hydrogen bonding to the proximal histidine, and, in the deoxy form, compression of the Fe—N bond and decrease in the out-of-plane displacement of the Fe atom will all increase O affinity over that of a system where these effects are absent. When hydrogen bonding is impossible, as in various synthetic systems (Table 4.5) as well as hemoglobin and Mb(E7His → Gly), O affinity is much lower than when hydrogen bonding can occur (see Table 4.6), especially for the cobalt analogues. But caution is needed in the absence of complete structural information: the lowered affinity of hemoglobin had been attributed to the lack of a distal histidine and its attendant hydrogen-bonding capabilities. However, the crystal structure reveals that an arginine residue, normally directed out into the solution, is capable of folding back into the ligand-binding pocket and of hydrogen bonding to ligands at the sixth site. In oxyhemerythrin the hydrogen bonding of the coordinated hydroperoxy group to the oxo bridge linking the two iron atoms (Figure 4.10B), described in Section II.F.1, may not only increase the stability of oxyhemerythrin, but also facilitate electron transfer that occurs in dioxygen binding.

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For a monatomic ideal gas we have seen that energy \(\langle E \rangle\) observed as \(U= 3/2 nRT\). This means that energy is only dependent on temperature and if a gas is compressed , then the internal energy does not change: \[ΔU_{isothermal-ideal gas} = 0 \nonumber \] This means that the reversible work must the reversible heat: \[ΔU_{rev} = w_{rev} + q_{rev} = 0 \nonumber \] Therefore \[w_{rev} = - q_{rev} \nonumber \] so from the expression of the reversible work for expansion in the last section \[ q_{rev}= nRT\ln \dfrac{V_2}{V_1} \nonumber \] If \(V_2>V_1\) (expansion), then you (or the environment) put heat into the system because this is a positive number. Now suppose you make sure that no heat can enter the cylinder. (Put it in styrofoam adiabatic This bat- part comes from a Greek verb (baino) that means walking, compare acro , someone who goes high places (acro-). The (dia) part means 'through' (cf. diagram, diorama, diagonal etc.) and the prefix (a-) denies it all (compare versus ). So the styrofoam prevents the heat from walking through the wall. When expanding the gas from V to V it still does reversible work but where does that come from? It can only come from the internal energy itself. So in this case any energy change should consist of work (adiabatic means: \(δq=0\)). \[dU = δw_{rev} \nonumber \] This implies that the temperature must drop, because if \(U\) changes, then \(T\) must change. The change of energy with temperature at constant volume is known as the heat capacity (at constant volume) \(C_v\) \[ C_v =\left( \dfrac{\partial U}{\partial T} \right)_V \nonumber \] For an ideal gas \(U\) only changes with temperature, so that or: We can now compare two paths to go from state \(P_1,V_1,T_1\) to state \(P_2,V_2,T_1\): Notice that the temperature remains \(T_1\) for path A (isotherm!), but that it drops to \(T_2\) on the adiabat B, so that the cylinder has to be isochorically warmed up, C, to regain the same temperature. \(\Delta U_{tot}\) should be the same for both path A and the combined path B+C, because the end points are the same (\(U\) is a state function!). As the and points are at the same temperature and \(U\) only depends on \(T\): \[\Delta U_{tot}=0 \nonumber \] Along adiabat B: \[q_{rev}=0 \nonumber \] Along isochoric heating C, there is no volume work because the volume is kept constant, so that: This is the only reversible heat involved in path B+C. However, we know that \(\Delta U_{tot}\) for path A is zero (isothermal!). This means that the volume work along B must cancel the heat along C: The book keeping looks as follows, all paths are reversible: \[\Delta U_{B+C} = \Delta U_A = 0 = q_B + w_B + q_C + w_C \nonumber \] We know that \(q_B=0\) since it is an adiabat and \(w_C=0\) since it is an isochore: \[\Delta U_{B+C} = \Delta U_A = 0 = 0 + w_B + q_C + 0 \nonumber \] Therefore: \[w_B=-q_C \nonumber \] We had already seen before that along the isotherm A: \[w_A = - q_A = - nRT \ln \dfrac{V_1}{V_2} \nonumber \] As you can see \w_A\) and \(w_B\) are the same. Work is a path function, even if reversible. As we are working with an ideal gas we can be more precise about \(w_B\) and \(q_c\) as well. The term \(w_B\) along the adiabat is reversible volume work. Since there is no heat along B we can write a straight \(d\) instead of \(\delta\) for the work contributions (It is the only contribution and must be identical to the state function \(dU\)):

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From section , we learned that the entropy at constant pressure changes with temperature by: \[\Delta S=\int_{T_1}^{T_2}{\frac{C_P(T)}{T}dT} \nonumber \] From section , we learned that the entropy of a phase transition is: \[\Delta_{trs} S=\frac{\Delta_{trs}H}{T_{trs}} \nonumber \] Both the heat capacity and enthalpy of transition can be experimentally determined using calorimetry. Using experimental values with the two above expressions and the convention that the entropy at absolute zero (0 K) is zero, we can calculate the practical absolute entropy of a substance for any temperature. For example, the entropy of CO gas at 300 K can be calculated by: \[S(T)=\int_{0 K}^{T_{sub}}{\frac{C_P^s(T)}{T}dT}+\frac{\Delta_{sub}H}{T_{sub}}+\int_{T_{sub}}^{300\text{ K}}{\frac{C_P^g(T)}{T}dT} \nonumber \] Where the temperature of sublimation (\(T_{sub}\)) is 194.7 K.

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As noted previously, the Diels Alder reaction is a classic example of a pericyclic reaction. Figure PR3.1. The Diels Alder reaction. Unlike the Cope and Claisen rearrangements, this reaction often occur intermolecularly (between two molecules). It always occurs between an alkene on one molecule and a conjugated diene on the other molecule. The alkene is referred to as a "dienophile"; it reacts with the conjugated pair of double bonds. Figure PR3.2. The diene and dienophile in the Diels Alder reaction. Draw curved arrows to keep track of electrons in the Diels Alder reaction. Draw the aromatic transition state of the Diels Alder reaction. Many pericyclic reactions are reversible. The reversible process is usually named the same way as the forward reaction, but with the prefix "retro". For example, a retro-Diels Alder reaction is shown below. Once again, the reaction can be thought of in terms of a reorganization of electrons between these two molecules. In the Diels Alder reaction, we can think of an interaction between the LUMO on one molecule and the HOMO on the other. As it happens, the LUMO on one molecule has the correct symmetry such that it can overlap and form a bonding interaction with the HOMO on the other molecule. Figure PR3.3. Qualitative molecular orbital picture of the Diels Alder reaction. Pay attention to the p orbital drawings on the carbons that will bond to each other to form the six-membered ring. It is important that those orbitals are able to overlap with each other to form an in-phase interaction. In that way, these carbon atoms at the ends of the diene and dienophile are able to bond with each other. A Diels Alder reaction is sometimes called a [2+4] addition reaction. A 2-carbon unit on one molecule interacts with a 4-carbon unit on another molecule. In contrast, the addition of one regular alkene to another regular alkene would be called a [2+2] addition reaction. If this reaction occurred, two alkenes would come together to form a four-membered ring. Figure PR3.4. A [2+2] addition reaction. However, [2+2] addition reactions don't occur without special circumstances. There are a couple of reasons why, and you may be able to suggest some at this point. You might say that the four-membered ring would be much more strained than the six-membered ring formed by the Diels Alder reaction. That is true, but it may not be reason enough to prevent the reaction from happening. Four-membered rings do occur in nature despite their strain energy. You might also say that the benzene-like transition state that stabilizes the pathway through a Cope or Diels Alder reaction isn't possible in a [2+2] addition. In fact, the transition state would be more like antiaromatic cyclobutadiene. The transition state would be very high in energy. Another problem shows up if we look at the orbital interactions in a [2+2] addition reaction. The HOMO on one alkene and LUMO on the other alkene do not overlap so that bonds can form between the two ends. If the p orbitals on one end are in phase, the p orbitals on the other end must be out of phase. The concerted reorganization of bonding possible for the Diels Alder reaction can't happen here. Figure PR3.5. Qualitative molecular orbital picture of [2+2] addition reaction. In fact, there is a way around that problem. Irradiating an alkene with UV light leads to promotion of an electron from the LUMO to the HOMO. The alkene is now in an "excited state". Figure PR3.6. Excitation of an electron in an alkene. This does not happen with 100% efficiency, so only some of the alkenes will become excited. In the excited state alkene, the HOMO now resembles the LUMO of the ground state alkene. Because of the matching symmetry between these orbitals, the addition reaction can proceed. Figure PR3.7. HOMO-LUMO interaction between a ground-state alkene and an excited-state alkene. A [4+2] reaction is sometimes referred to as "thermally-allowed", whereas a [2+2] addition is sometimes referred to as "photolytically-allowed." This distinction refers to the need for electronic excitation to accomplish the latter type of reaction. ,

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Mentioned in the Hebrew scriptures, tin is of ancient origins. Tin is an element in Group 14 (the carbon family) and has mainly metallic properties. Tin has atomic number 50 and an atomic mass of 118.710 atomic mass units. Mentioned in the Hebrew scriptures, tin is of ancient origins. Early metal smiths were quick to learn that mixing copper with tin created a more durable metal (bronze) and it is principally for its alloys that tin is valued today. Named after the Etruscan god Tinia, the chemical symbol for tin is taken from the Latin . The metal is silvery white and very soft when pure. It has the look of freshly cut aluminum, but the feel of lead. Polished tin is slightly bluish. It has been used for many years in the coating of steel cans for food because it is more resistant to corrosion than iron. It forms a number of useful low-melting alloys (solders) which are used to connect electrical circuits. Bending a bar of tin produces a characteristic squealing sound called "tin cry". Tin shares chemical similarities with and . Tin mining began in Australia in 1872 and today Tin is used extensively in industry and commerce. 5.77g/cm When heated in it, tin produces stannic oxide \[Sn_{(s)} + O_{2(g)} \rightarrow SnO_{2(s)}\] \[Sn_{(s)} + 2H_2O_{(g)} \rightarrow SnO_{2(s)} + 2H_{2(g)}\] There are 10 known stable isotopes of Tin, the most of any elements on the periodic table. This high number of stable isotopes could be attributed to the fact that the atomic number of \(\ce{^{50}Sn}\) is a ' ' in nuclear physics. Tin has 3 allotropes: alpha, beta and gamma tin. Alpha tin is the most unstable form of tin. Beta tin is the most commonly found allotrope of tin and gamma tin only exists at very high temperatures. Tin, although it is found in Group 14 of the periodic table, is consistent with the trend found in Group 13 where the lower oxidation state is favored farther down a group. Tin can exist in two oxidation states, +2 and +4, but Tin displays a tendency to exist in the +4 . Tin forms two main oxides, SnO and SnO Tin has a ground state electron configuration of and can form covalent tin (II) compounds with its two unpaired p-electrons. In the three dimensional figure below, the first and most inner electron shell is represented by blue electrons, the second electron shell made up of eight electrons is represented by red electrons, the third shell containing eighteen electrons is represented with green electrons, and the next outer electron again contains eighteen electrons and represented in purple. Early metal smiths were quick to learn that mixing copper with tin created a more durable metal (bronze) and it is principally for its alloys that tin is valued today. Nearly half of the tin metal produced is used in solders, which are low melting point alloys used to join wires. Solders are important in electrician work and plumbing. Tin is also used as a coating for lead, zinc, and steel to prevent corrosion. Tin cans are widely used for storing foods; the first tin can was used in London in 1812. Find the oxidation state of tin in the following compounds: a. SnCl^2 answer:2 b. SnO^2 answer:4 Write an equation for the reaction of tin with water. Under what conditions does this reaction take place? answer: Sn(s) + 2H O(g) → SnO (s) + 2H (g) Reaction takes place if water is heated to a high temperature to form steam. Which of these reactions take place. a. tin with oxygen Answer: YES b. tin with hydrogen Answer: NO c. tin with argon Answer: NO d. tin with chlorine Answer: YES Arrange the following in order of increasing atomic radius: Sn, K, Ag, C, Pb Answer: C<Sn<Pb<Ag<K Arrange the following in order of decreasing ionization energy: Sn, Si, Pb, I, In. Answer: Si> I > Sn > In > Pb

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The states that, in an atom or molecule, no two electrons can have the four electronic quantum numbers. As an orbital can contain a maximum of only two electrons, the two electrons must have opposing spins. This means if one electron is assigned as a spin up (+1/2) electron, the other electron must be spin-down (-1/2) electron. Electrons in the same orbital have the same first three quantum numbers, e.g., \(n=1\), \(l=0\), \(m_l=0\) for the 1 subshell. Only two electrons can have these numbers, so that their spin moments must be either \(m_s = -1/2\) or \(m_s = +1/2\). If the 1 orbital contains only one electron, we have one \(m_s\) value and the electron configuration is written as 1 (corresponding to hydrogen). If it is fully occupied, we have two \(m_s\) values, and the electron configuration is 1 (corresponding to helium). Visually these two cases can be represented as As you can see, the 1 and 2s subshells for beryllium atoms can hold only two electrons and when filled, the electrons must have opposite spins. Otherwise they will have the same four quantum numbers, in violation of the Pauli Exclusion Principle.

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Astatine formerly known as alabamine. . Astatine At 85 (210.0) amu 302.0 °C (575.15 K, 575.6 °F) 337.0 °C (610.15 K, 638.6 °F) 85 125 Halogen Unknown Unknown Unknown 1940 D.R. Corson From the Greek word (unstable) No uses known Man-made -1, +5 Astatine is the last of the known halogens and was synthesized in 1940 by Corson and others at the University of California. It is radioactive and its name, from the Greek astatos, means "unstable". The element can be produced by bombarding targets made of bismuth-209 with high energy alpha particles (helium nuclei). Astatine 211 is the product and has a half-life of 7.2 hours. The most stable isotope of astatine is 210 which has a half-life of 8.1 hours. Not much is known about the chemical properties of astatine but it is expected to react like the other halogens, although much less vigorously, and it should be more metallic than iodine. There should be tiny quantities of astatine in the earth's crust as products of other radioactive decays, but their existence would be short-lived.