( 1+ I K I. λ = v P σ 2. z 11. = N 1 V σ 2 v Z 11. = 1 2 ( N 1 V ) 2 σ 2 v. J = D ( x ) D = 1 3 λ v D = r 2. 2(dimension)t. where K m,app.

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Physical Chemistry I Final Exam Name: KEY CHEM 4641 Fall 2013 median = average = 18 /30 high = 30 /30 7 problems worth a total of 30 points. Show your work. Gas constant: R = J mol -1 K -1 =

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Physical Chemistry I Final Exam Name: KEY CHEM 4641 Fall 2013 median = average = 18 /30 high = 30 /30 7 problems worth a total of 30 points. Show your work. Gas constant: R = J mol -1 K -1 = L bar mol -1 K -1. Boltzmann constant k = 1.381X10-23 J/K = cm -1 /K Units: 1 bar = 10 5 Pa. 1 GPa = 10 9 Pa 1 Joule = 1 Pa m 3. 1 L bar = 100 J. 1L=10-3 m 3. Formulae: z 11 = N 1 V σ 2 v Z 11 = 1 2 ( N 1 V ) 2 σ 2 v λ = v z 11 = k T P σ 2 J = D ( Ñ x ) D = 1 3 λ v D = r 2 2(dimension)t competitve inhibition rate= R max S 0 S 0 +K m, app where K m,app = K m ( 1+ I K I ) Score: / (ch 5, 5 points) Consider the reaction CaCO 3 (s) + SO 2 (g) CaSO 4 (s) + C(s). These thermodynamic data will be helpful. 298K CaCO 3 (s) SO 2( g) CaSO 4 (s) C(s) S m o (J/mol K) C p,m (J/mol K) The reaction enthalpy at 298K, ΔH o 298 = kj/mol. The tabulated heat capacities should be considered constant. a. (1 point) Calculate the reaction entropy, ΔS 0, at 298K. ΔS = = J/mol K b. (1 point) Calculate the entropy change of the surroundings and of the universe, assuming that the system and surroundings are at 298K. Δ S surr = q J/mol = = Δ H T 298 K 298 K = 12540J/mol K Δ S universe = Δ S + Δ S surr = == J/mol K c. (1 point) Based on ΔS universe, is the reaction at 298K spontaneous or not? ΔS univ is negative, so the reaction is not spontaneous d. (2 points) Calculate the entropy change for the chemical reaction, ΔS 0, at T=596 K. ΔC P = = 5.4 J/mol K ΔS 596 = ΔS ΔC P /T dt = ΔS ΔC P ln(596/298) = ln(2) = J/mol K 2. (ch 9, 5 points) At o C, the vapor pressure of pure ethyl bromide is 1330 Pa and that of pure ethyl chloride is 5330 Pa. Consider a mixture of ethyl bromide and ethyl chloride at o C. The solution is ideal. (Raoult's law P = x P* applies.) The total vapor pressure is 1790 Pa. The mole fraction of ethyl chloride in the vapor is a. (1 point) What is the vapor pressure of ethyl chloride above the mixture? P(chloride)=0.340 X 1790 = 609 Pa b. (1 point) What is the mole fraction ethyl chloride in the liquid? Raoult's law is that P(chloride) = x(chloride) P(pure chloride) so x(chloride) = P(chloride) / P(pure chloride) = 609 / 5330 = c. (2 points) If there are 8.0 mol liquid and 2.0 mol vapor present at the same pressure as in part a, what is the overall mole fraction ethyl chloride in the system? n(cl liquid) = X 8.0 = mol n(cl vapor) = X 2.0 = mol n(chloride total) = mol overall mole fraction chloride = / 10 = d. (1 point) Now consider a slight non-ideality of the mixture. In particular, intermolecular chloride-chloride and bromide-bromide attractions are slightly weaker than chlroide-bromide attractions. Is the effect of that nonideality to lower the vapor pressure, relative to that of an ideal solution, or to raise the vapor pressure? Stronger chloride-bromide attractions tend to keep molecules in the liquid phase, lowering the vapor pressure. compared to the pressure of an ideal solution. 3. (ch 15, 5 points) T = K. Consider the following energy levels and associated degeneracies for atomic Fe: level(n) energy(cm -1 ) degeneracy a. (2 points) Calculate the atomic electronic partition function, q elec. 2 q elec = g n e E n /(k T ) n=0 q elec = 9e e 416 cm 1 /(0.695 K 1 cm K ) + 5e 704 cm 1 /(0.695 K 1 cm K ) q elec = 9 + 7e e 1.69 = = 12.5 b. (3 points) Calculate the average electronic internal energy of atomic Fe. Give the result in cm -1. E = 1 2 q E n g n e E n /(k T ) elec n=0 E = 1 [ ] E = cm 1 = 12.5 = 139cm 1 4. (ch 16, 4 points) Consider a gas of pure carbon dioxide at K and bar. mass of CO 2 = amu = 7.31X10-26 kg. collision cross section σ = 5.2X10-19 m 2. a. (3 points) Calculate the single-molecule collision frequency, z 11. Give z 11 in s -1. z 11 = N 1 V N 1 V = P k T = σ 2 v 10 7 Pa J/K 400 K v = ( 8k T π m ) 1/2 = (192292m 2 /s 2 ) 1/2 = 439 m/s z 11 = s 1 = m 3 b. (1 point) The vibrational period in carbon dioxide is 13 fs. (1 fs = s) Is that greater or smaller than the typical time between collisions? Δt ~ 1/z 11 = 1.7X10-12 s = 1700 fs, so the vibration time is much less than the collision time. 5. (ch 17, 3 points) Philipp Nauer, et al., measured diffusion of methane through soil. (Environmental Science and Technology, 2013, 47, 11122) A graph of their measured methane concentration versus depth into soil is at right. The concentration unit is (mg methane) per (m 3 of soil). They measured the diffusion coefficient to be D = 0.13 m 2 /day. a. (1 point) Based on the graph, is the direction of net methane diffusion to the left or to the right? Explain briefly how you know. Net diffusion is to the right, because diffusion occurs down the concentration gradient. b. (1 point) At the top of the soil (i.e., at 0 depth) what is the concentration gradient? Answer in (mg CH 4 )/m 4. c. (1 point) At the top of the soil, calculate the flux of methane in mg/m 2 /day. Flux = - D (concentration gradient) = 0.13 * 10 = 1.3 mg CH 4 /(m 2 day). (downward) 6. (ch 18, 4 points) Consider the first-order reversible reaction Suppose that concentrations of A and B are measured in molarity, moles per liter, and that time is measured in seconds. a. Write the differential rate expression for the product. d [ B] d t = k A [ A] k B [ B] d ([ A]+[B ]) b. For the given mechanism, = 0 at all times, t. Either explain in words why this d t is true, or derive it mathematically. Answers: (i) The mechanism neither creates nor consumes molecules, it only converts one type to the other. That is why the total concentration does not change. (ii) d [B ] = k [ A ] k [B] and d[ A ] = k [B] k [ A ] d t A B d t B A Sum the two differential rate expressions: d ([ A ]+[B]) = 0 which means that ([A]+[B]) is constant. d t c. Write an equation that relates the equilibrium constant to the rate coefficients. K = k A k B d. What are the units of k A? s -1 7. (ch 19, 4 points) Sesame cake, the remains of sesame seeds after extracting oil, contains much potentially edible protein. Hydrolysis of sesame cake protein by the enzyme alcalase was studied by Demirhan, et al., Journal of Food Science, 2011, 76(1), Michaelis-Menten treatment worked. At right is a double-reciprocal (Lineweaver-Burk) plot of rate (milliequivalents per liter per minute) and substrate concentration (g/l). a. (2 points) From the graph, calculate R max and K m. R max = 1/(0.1) = 10 mequiv/(l minute) K m = 1/0.025 L/g = 40 g/l (Demirhan wrote 9.2 and 41.) b. (1 point) Michaelis-Menten analysis of enzyme kinetics applies the steady-state approximation to which one of the following? (Circle the answer.) enzyme substrate enzyme-substrate complex product c. (1 point) Demirhan, et al., tested the effect of an inhibitor. Their Lineweaver-Burk plot with inhibitor is shown at right. The lowest line shows data (the same as above) with no inhibitor. Inhibitor concentrations increase in the direction shown by the arrow. The parallel lines in the plot are characteristic of a type of inhibition called uncompetitive, in which the inhibitor binds to the enzyme-substrate complex. Engel and Reid's book describes competitive inhibition. How would the inhibition plot look in the case of competitive inhibition? Sketch the expected appearance on the blank graph below. R max is the same at all inhibitor concentrations.
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