NAME: Section Number: 10 CHEMISTRY 443, Fall, 2014 (14F) Examination 2, November 5, 2014 Answer each question in the space provided; use back of page if extra space is needed. Answer questions so the grader can READILY understand your work; only work on the exam sheet will be considered. Clearly indicate your answer and all indications of your logic in arriving at your answer. Please answer the question asked and refrain from providing irrelevant comments or information. Write answers, where appropriate, with reasonable numbers of significant figures. You may use only the "Student Handbook," a calculator, and a straight edge. DO NOT WRITE THIS SPACE IN p. 1_______/25 p. 2_______/15 p. 3_______/20 p. 4_______/20 p. 5_______/20 ============= p. 6 _______/5 (Extra credit) ============= TOTAL PTS /100 1 Problem 1 (25 Points) Lattice Models, again………… ! …. ☺ Condensed phases (liquid, solids) have historically been treated as lattice models. In this case, consider a bulk liquid as a lattice site model. There are N particles, and the particles interact with a nearest-neighbor pair-wise interaction that leads to an energy contribution of ‘w’ to the total internal energy, U, of the system from a pair of particles interacting. In this lattice model, the number of nearest neighbors for any particle in the bulk (neglect the edges) is ‘z’. For an N-particle system, the total internal energy from interactions is U interaction = 12 Nzw . The per-particle interaction contribution to U U is thus interaction = 12 zw . It is important to note that this type of interaction implicitly builds in a N reference state where the particles do not interact (i.e. ‘large’ separation, or ‘dilute’ conditions); in such a reference state, Uinteraction = 0 (much like in the case of an ideal gas). In the lattice, dynamic effects are taken into account€since the lattice is an ‘average’ representation of the positions of particles. All lattice sites are filled, and the particles are indistinguishable. € A. (5 Points). Using the definition of the Helmholtz Free Energy, please provide an appropriate equation for this thermodynamic quantity for this lattice model liquid based on the given information and in terms of the variables presented in the above discussion and associated figure. A(T,V,N) ≡ U − TS = U interaction − TS = U interaction − T ( k B ln(W )) ( ( = U interaction − T k B ln N! (N!)(0!) )) = U interaction − T ( k B ln(1)) = U interaction A(T,V,N) = 12 Nzw € B. (5 Points). Using the definition of the chemical potential, please provide an appropriate equation for this thermodynamic quantity. µliquid = ( ) ∂A ∂N T ,V = ( ∂ ( 12 Nzw) ∂N ) T ,V = 12 zw 2 C (5 Points). Using the results from Parts A and B and your knowledge of the chemical potential of a pure ideal gas, determine a relation between the vapor pressure of the liquid and the interaction strength, ‘w’. You are free to define the reference state ideal gas chemical potential to be 0. Since this question asks about vapor pressure, we think of vapor - liquid equilibrium. At conditions of vapor - liquid equilibrium, we know that vapor and liquid chemical potentials are equal. µ liquid = µ vapor We have computed the liquid chemical potential (for lattice model liquid) in Part B. For ideal gas, we know the behavior of chemical potential as a function of pressure : µ ideal,vapor = µ ref + RT ln ( ) p p ref Equating chemical potentials gives : 1 2 zw = µ ref + RT ln ( ) p p ref Feeling free to set the reference state ideal gas chemical potential to 0 : ⎛ p ⎞ zw = RT ln⎜ ref ⎟ ⎝ p ⎠ Rearranging : 1 2 p= p ref zw 2RT e € 3 D1. (5 Points). The vapor pressure of water is 22 mmHg at T = 300K and 760 mmHg at T = 373K. Estimate the vaporization enthalpy using appropriate analytics. This is a Clausius-Clapeyron analysis. Assume vaporization enthalpy is constant over the temperature and pressure ranges probed, and liquid is transferred to an ideal vapor phase: ln ( ) p2 p1 = −Δh vap ⎛ 1 1 ⎞ ⎜ − ⎟ R ⎝ T2 T1 ⎠ Δh vap = −R ln ( ⎛ 1 1 ⎞ −1 ⎜ − ⎟ ⎝ T2 T1 ⎠ ( ) p2 p1 = − 8.314 J mol −1 K −1 ) −1 ⎛ 760mmHg ⎞⎛ 1 1 ⎞ ln⎜ − ⎟ ⎟⎜ ⎝ 22mmHg ⎠⎝ 373K 300K ⎠ Δh vap = 45.14kJ mol −1 € D2. (5 Points). What is an estimate of the interaction strength ‘w’ considering water to be modeled as a lattice fluid? Consider z = 4 for a lattice model of water. Refer to your analyses of Parts A through E for inspiration. S1: In this approach, we compare the differential forms of the vapor pressure expressions we have derived and have at our disposal: From Clausius - Clapeyron Analysis ⎛ Δh vap ⎞⎛ 1 ⎞ d(ln P) = ⎜ ⎟⎜ 2 ⎟dT ⎝ R ⎠⎝ T ⎠ From Analysis of Part 1C : ⎛ −zw ⎞⎛ 1 ⎞ d(ln P) = ⎜ ⎟⎜ ⎟dT ⎝ 2R ⎠⎝ T 2 ⎠ Comparing the expressions, we have : ⎛ −zw ⎞ ⎛ Δh vap ⎞ ⎜ ⎟ = ⎜ ⎟ ⎝ 2R ⎠ ⎝ R ⎠ ⎛ −2Δh vap ⎞ w = ⎜ ⎟ z ⎝ ⎠ ⎛ −2(45.14kJ mol −1 ) ⎞ −1 −1 = ⎜ ⎟ = −22.57kJ mol = −5.4kcal mol 4 ⎝ ⎠ € 4 S2. If we think of vaporization as a molecule moving from the condensed phase (liquid) to the ideal vapor (where there are no interactions between particles), then the vaporization enthalpy corresponds to the loss of interaction per particle on a mole basis. This is simply the interaction contribution of a single particle to the potential energy of the liquid, and so: ⎛ −zw ⎞ ⎟ 2 ⎠ (Δhvap ) = U final −U init = 0 −U init = ⎜⎝ ⎛ −2Δh vap ⎞ w = ⎜ ⎟ z ⎝ ⎠ ⎛ −2(45.14kJ mol −1 ) ⎞ −1 −1 = ⎜ ⎟ = −22.57kJ mol = −5.4kcal mol 4 ⎝ ⎠ € 5 Problem 2 (15 Points) Multiple Choice? Why yes……...... For the following, match the statement in the left-hand side column with the most appropriate answer(s) from the right-hand side column. I. Under the Ehrenfest classification scheme, a first-order phase transition is one in which the Gibbs Free Energy changes continuously (smoothly) and properties related to first derivatives of G change discontinuously. The pure substance liquid-vapor phase transition is first-order, true or false? __________B______ II. How many intensive degrees of freedom are available for a system comprised of a binary liquid solution in equilibrium with its vapor (both species present in both phases)? ___P_____. A. ∞ (diverges) B, true C. false D. greater than 0 E. equal to 0 III. Negative absolute temperatures arise under particular conditions, such as: _____________N,Q___________ F. 1 G. ∑ν µ j j =0 j IV. What is the theoretical value of Cp(T) at the liquid-vapor phase transition temperature? _______A______ V. The Legendre Transform of U(S,V,N) to a function of T,V,N results in what thermodynamic potential? ______S____________ VI. The maximum non-expansion work related to a transformation (under conditions of equilibrium along the process) of a chemical system between two equilibrium thermodynamic states is equal to ___T_____. Consider the process to occur with external temperature and pressure constraints (values are constant). VII. The Gibbs Free Energy of mixing of two ideal gases, A and B, is identically zero (due to ideality of the vapor phase species). True or False? _C________ VIII. For a binary mixture of ideal gases at constant temperature and pressure, what mole fraction ratio would maximize the entropy of the system? ________F________ IX. What constraint arises when there is a reversible (equilibrium) chemical reaction? ______G______. X. In a two-state system (i.e., 2 energy levels, ‘high’ and ‘low’) with 10 noninteracting particles what value would represent the ratio of occupancies of the two states (or equivalently, what value would represent the ratio of ‘high’ to ‘low’ labeled particles?) _____F_______ 6 H. entropy I. isenthalpic J. 0 J/K K. 3 L. 4 M. internal energy N. relatively few quantum states available/accessible at a particular temperature O. T dQ P. 2 Q. population inversion of quantum states R. q is isentropic S. Helmholtz Free Energy T. Gibbs Free Energy Problem 3 (20 Points) Put me to sleep already……….. A common anesthetic drug molecule is halothane (2-bromo-2-chloro-1,1,1-trifluoroethane). Its mode of action is presumed to involve partitioning from water (state A) into lipid bilayer membranes (state B). The values of the equilibrium constant representing this reversible partitioning, determined at two different pressures, are the following: P (atm) ln K T (Kelvin) 0 7.84 300 280 7.6 300 What is the corresponding change in molar volume of the bilayer/water/drug system as a result of the pressure change from p1 = 0 atm to p2 = 280 atm? Solution: ΔGrxn = −RT ln K At constant T (data given at T = 300K = constant) d ln K = −d(ΔGrxn )T −1 = (ΔVrxn dP − ΔSrxn dT ) RT RT −1 (ΔVrxn dP ) RT ⎛∂ ln K ⎞ = −RT ⎜ ⎟ ⎝ ∂P ⎠T d ln K = ΔVrxn Assume change in V is constant; use a constant slope approximation from data given : ⎛ Δ ln K ⎞ ΔVrxn ≈ −(8.314 JmolK )(300K )⎜ ⎟ ⎝ ΔP ⎠T ⎛ (7.6 − 7.84) ⎞ = −(0.0821 L atm mol −1 K −1 )(300K )⎜ ⎟ ⎝ (280atm − 0atm) ⎠T = 0.021L / mol = 21cm 3 / mol € 7 Problem 4 (20 Points) What are Legendre Transforms Good For Anyway....... Here we will consider the utility of Legendre Transforms in determining useful thermodynamic potentials that reach extrema under conditions of chemical equilibrium with constant pressure, temperature, and chemical potential. That is, in order to use, as a control ‘knob’, a species chemical potential, we need to determine what thermodynamic potential depends on the chemical potential of interest. This is not an irrelevant question to ask, since biochemical reactions often occur under conditions of constant pH, where the concentration of proton is controlled (specified). Knowing that chemical potentials, in appropriate limits can be expressed in terms of concentrations, constant pH scenario is a case of constant proton chemical potential. Let’s consider a simple equilibrium (reversible) reaction: A+B ⇔C A. (1 Points) Because of chemical equilibrium, what constraint arises involving species chemical potentials (recall the discussion of ideal gas reaction equilibrium, though in this exercise, we are considering a general situation). € ∑ν µ j j =0 j B. (1 Points) Without taking into account your constraint, write down the total differential of the extensive Gibbs Free Energy (knowing that Gibbs Free Energy is a function of T, P, and the amounts of species, A, B, C). Be careful and provide the complete expression. Use the nomenclature: ni=moles of species ‘i’ and ‘µi’ is chemical potential of species ‘i’. dG(T,P,{n j }) = VdP − SdT + µ A dnA + µ B dnB + µ C dnC C. (5 Points) Using your constraint from Part A, eliminate the chemical potential of species C from your results of Part B. You can make the substitution: € n ,A = n A + nC ; dn ,A = dn A + dnC ; n ,B = n B + nC ; dn ,B = dn B + dnC . These substitutions are another way to say that A is ‘distributed’ between free A and ‘bound’ A that has transformed to species C; similarly for species B. € Chemical reaction equilibrium constraint : µ A + µ B = µC Thus : dG(T,P,{n j }) = VdP − SdT + µ A dnA + µ B dnB + µ C dnC = VdP − SdT + µ A dnA + µ B dnB + (µ A + µ B )dnC = VdP − SdT + µ A (dnA + dnC ) + µ B (dnB + dnC ) = VdP − SdT + µ A dnA' + µ B dnB' D. (3 Points) Based on your results of Part C, what are the variables that G really depends on? Excluding temperature and pressure, there are 2 (two). € 8 dG = VdP − SdT + µ A dnA' + µ B dnB' Thus : dnA' and dnB' € E. (7 Points) Now, we want to control the chemical potential of species B in our reaction----that is, we want to specify the value of µB. Provide a Legendre transform of the Gibbs Free Energy (the thermodynamic potential you have been working with so far in Parts A- D) to another thermodynamic potential, G’, that would leave you with µB as one of the independent variables. Show that indeed the new potential you generated is in part dependent on µB. ⎛ ∂G ⎞ ' G = G − ⎜ ' ⎟nB ⎝∂nB ⎠ ' new potential defined by Legendre transform To show that the new potential is indeed dependent on µB: ⎛ ∂G ⎞ ' G = G − ⎜ ' ⎟nB = G − µnB' new potential defined by Legendre transform ⎝∂nB ⎠ ' € dG'= dG − d(µnB' ) = VdP − SdT + µ A dnA' + µ B dnB' - µ B dnB' − nB' dµ B dG'= VdP − SdT + µ A dnA' − nB' dµ B The last expression shows that G' is dependent on µ B F. (3 Points) At constant temperature, pressure, and µB, what does the total differential of G’ become in terms of µA and dna’? That is, what is ( dG')T ,P,µ ? What is the integrated form of G’ and what B € variables does G’ depend on? dG'= µ A dnA' Integrated form : € G'= µ A nA' € 9 Problem 5 (20 Points) It’s bubbalicious…….. For the binary system acetonitrile(1)/nitromethane(2), liquid-vapor equilibrium is adequately represented through Raoult’s expression. Antoine equations for saturation pressures as a function of temperature (in Celsius) are: ln ( P1sat ) = 14.2724 − 2945.47 2972.64 ; ln ( P2sat ) = 14.2043− t + 224.0 t + 209.0 What is the bubble temperature at P = 70 kPa and initial composition of 0.6 mole fraction of acetonitrile at initial temperature t = 90 Celsius. This is best done via iterating over the total pressure. The liquid phase composition at the bubble temperature is x1=0.6 and x2=0.4. We plot in the last column the relative error in our predicted total Pressure. The last temperature gives a fairly low relative error, and we state that value as our solution. P total = x1 P1sat + x 2 P2sat relative error = total Ppredicted P total P t (x1)(P1sat) (x2)(P2sat) 70 100 106.7636578 39.16623773 70 90 79.92563735 28.39092264 70 70 70 70 70 80 70 75 75.5 76 58.70520086 42.22316204 49.92411448 50.75194804 51.59068036 20.12690971 13.92082071 16.79308198 17.10485805 17.42129796 70 70 70 76.5 76.4 76.42 52.44041316 52.26958166 52.30371247 17.7424528 17.67784218 17.69074907 € 10 diff/P 1.084712794 0.547379428 0.126173008 0.197943104 0.046897193 0.030617056 0.014114595 0.002612371 0.000751088 7.91208E-05 Problem 6. Extra Credit (5 Points) Time to jam………. In this figure, the y-axis represents the vapor pressure of the solution of an ethanol-water. Straight lines are the vapor phase partial pressures based on Raoult’s equation. Treating the vapor as ideal and still considering the liquid solution as non-ideal, what thermodynamic quantity related to the species in the liquid is also being plotted on the y-axis. Please show your work in order to receive credit. Solution: On the y-axis, we are plotting partial pressures of the species in the ideal gas vapor mixture. If we consider the fugacities of the species in vapor and liquid, we come to the following conclusion: fˆi vapor = fˆ liquid fˆi vapor = φˆivapor pi = (1)yi P total fugacity coefficiet = 1 for ideal vapor Thus fˆ liquid = pi The y-axis plots the liquid phase fugacity of species ‘i’. € 11 Potentially Useful Information Stirling’s Approximation: ln(N!) = (N ln N) − N N →∞ NA R = kB = Boltzmann Constant € Number of ways to place N indistinguishable objects into M bins: W= M! N!(M − N )! S = kB ln (W) (isolated system, statistical mechanical form of entropy for lattice model) µ i = fi dµ = RTdp fˆi vapor = φˆivapor pi = yi P total ref µ (T,P{x}) = µ still + RT ln(γci ) dµ = RTdp ref du = PRdT + µ still +∇• fˆ vapor = fˆ liquid (at mixture vapor - liquid equilibrium) i ref µ (T,P{x}) = µ still + RT ln(φci ) € 12 NAME: Section Number: 10 CHEMISTRY 443, Fall, 2014 (14F) Examination 1, October 1, 2014 Answer each question in the space provided; use back of page if extra space is needed. Answer questions so the grader can READILY understand your work; only work on the exam sheet will be considered. Clearly indicate your answer and all indications of your logic in arriving at your answer. Please answer the question asked and refrain from providing irrelevant comments or information. Write answers, where appropriate, with reasonable numbers of significant figures. You may use only the "Student Handbook," a calculator, and a straight edge. DO NOT WRITE THIS SPACE IN p. 1_______/25 p. 2_______/15 p. 3_______/20 p. 4_______/20 p. 5_______/20 ============= p. 6 _______/5 (Extra credit) ============= TOTAL PTS /100 13

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