Physical Chemistry I CHEM 4641 Final Exam 13 questions, 30 points Name: ___________KEY____________ Gas constant: R = 8.314 J·mol-1·K-1 = 0.008314 kJ·mol-1·K-1. h = 6.626×10-34 J·s Boltzmann constant k = 1.381×10-23 J/K = 0.6950 cm-1/K c = 2.998×108 m/s Units: 1 bar = 105 Pa. 1 GPa = 109Pa 1 Joule = 1 Pa·m3 . 1 L·bar = 100 J. 1L=10-3m3. Formulae: A = - kT ln Q Q = qN or qN/N! Λ= h / √ 2 π m k B T B = h2 /(8 π2 μ R2 ) μindist = - kT ln(q/N) μdist = - kT ln(q) 1 H 2 He 1.01 4.00 3 Li 4 Be 6.94 9.01 11 Na 12 Mg 5 B 20 Ca 7 N 8 O 9 F 10 Ne 10.81 12.01 14.01 16.00 19.00 20.18 13 Al 22.99 24.31 19 K 6 C 14 Si 15 P 16 S 17 Cl 18 Ar 26.98 28.09 30.97 32.06 35.45 39.95 21 Sc 22 Ti 23 V 24 Cr 25 Mn 26 Fe 27 Co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr 39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.70 63.55 65.38 69.72 72.59 74.92 78.96 79.90 83.80 37 Rb 38 Sr 39 Y 40 Zr 41 Nb 42 Mo 85.47 87.62 88.91 91.22 92.91 95.94 55 Cs 56 Ba 57 La 72 Hf 73 Ta 74 W 43 Tc 98 75 Re 44 Ru 45 Rh 46 Pd 47 Ag 48 Cd 49 In 50 Sn 51 Sb 223 88 Ra 89 Ac 226.0 227.0 58 Ce 54 Xe 76 Os 77 Ir 78 Pt 79 Au 80 Hg 81 Tl 82 Pb 83 Bi 84 Po 85 At 86 Rn 209 210 222 104 Rf 105 Db 106 Sg 107 Bh 108 Hs 109 Mt 110 Ds 111 Rg 112 Cn 113 Nh 114 Fl 115 Mc 116 Lv 117 Ts 118 Og 267 268 269 270 269 278 281 280 285 286 289 289 293 294 294 59 Pr 60 Nd 61 Pm 62 Sm 63 Eu 64 Gd 65 Tb 66 Dy 67 Ho 68 Er 69 Tm 70 Yb 71 Lu 140.1 140.9 144.2 90 Th 53 I 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 132.9 137.3 138.9 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 87 Fr 52 Te 91 Pa 92 U 232.0 231.0 238.0 145 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0 93 Np 94 Pu 95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr 237 244 243 247 247 251 252 257 258 259 262 1. (2 points) The word "reversible" has related but different meanings in thermodynamics and in kinetics. Explain what "reversible" means in the context of the expansion of a gas, and what "reversible" means in a reaction mechanism. A few sentences can suffice. Expansion: An ionfinitessimal change in the driving force (e.g., pressure) changes expansion to compression. Mechanism: Reaction goes both ways: reactants to products and products to reactants. 2. (1 point) When graphite begins to melt at 4800K, does the graphite sink in the liquid or does it float on top of the liquid? Explain how you know. Answer: The left-sloping coexistence line, dP/dT < 0, implies that ΔV<0; the molar volume of graphite is larger than the molar volume of liquid carbon. That is, graphite is less dense than liquid carbon. Graphite floats. Another argument, giving the same conclusion, is that at a fixed temperature liquid is the high-pressure phase, so it must be the higher-density phase. 3. (2 points) The vibrational modes of the water molecule have frequencies 1595, 3657 and 3756 cm-1. Which mode contributes most to the zero-point energy? Answer: 3756 Which mode contributes most to the heat capacity at 1000 K? Answer: 1595 4. (3 points) Consider one mole of ideal gas. CP=38.314 J/(mol K). A three-step cycle is graphed. I. Isotherm at 302.0 K from 10.0 L to 20.0 L. II. Isobar from the end of step I, back to 10.0 L. III. Isochore from the end of step II to the beginning of step I, all at 10.0 L. Calculate ΔU, w and ΔS for step I, the reversible isotherm. Answer: ΔU = 0 for an ideal gas at constant temperature. w=−∫ P dV = −nRT ∫ ΔS =∫ d q rev T OR Δ S = dV 20 = −RT ln = −1740 J V 10 ( ) = −∫ dw n R T /V dV 20 =∫ dV = n R∫ = R ln = 5.76 J/K T T V 10 qrev −w 20 = R ln = 5.76 J/K T 10 T = ( ) ( ) 5. (4 points) Consider the gas-phase equilibrium I2 ⇌ 2 I . Both I2 and I are gases. Assume ideal-gas behavior. gas I I2 CP at 298 K ΔfGo298 ΔfHo298 So298 (kJ/mol) (kJ/mol) (J/mol·K) (J/mol·K) 70.2 106.8 180.8 20.8 19.3 62.4 260.7 36.9 a) Calculate the equilibrium constant K at 298K. b) Given that the pressure of I2 at 298 K is 1.0X10-4 bar, calculate the pressure of I. c) At 350 K, is K larger or smaller than at 298 K? You need not calculate K at 350 K, but explain how you know. d) Explain why the heat capacity of I2 is larger than the heat capacity of I. Answers: a) ΔrxnGo298 = 2×70.2 - 19.3 = 121.1 kJ/mol. Or, ΔrxnGo298 = ΔrxnHo298 - TΔSo = 151.2 - 298×10-3(2×180.8 -260.7) = 151.2 - 0.298 × 100.9 = 151.2 - 30.1 = 121.1 kJ/mol ln K = -121.1/(0.008314×298) =-48.9 . K = 5.9×10-22. b) K = (PI/Po)2 / (PI2/Po) so PI= Po [ K × 1.0×10-4]1/2 = 1 bar [ 2.4×10-13] = 2.4×10-13 bar. c) K is larger at larger T because the reaction is endothermic. A test taker might take that as obvious, this being a dissociation reaction, or might calculate ΔrxnHo298 = 2×106.8 - 62.4 = 151.2 kJ/mol. d) The heat capacity of I2 is larger because of rotation (a contribution of R) and of vibration ( about 0.9R). 6. (1 point) Consider the four-state energy level diagram that is sketched at right. Let T=288K, so kBT=200 cm-1. Calculate the canonical partition function, q. 4 −E i /(k B T ) Answer: q = ∑ e = 1 + e−200 /200 + e−400 /200 + e−600 /200 i=0 q = 1+ e−1 +e−2 +e −3 = 1+0.368 +0.135+ 0.050=1.55 7. (2 points) Consider a nitrogen molecule, N2, in a volume of 10-27 m3. Its mass is 4.65×10-26kg. At what temperature does the translational partition function, qT, equal 1.00×105? Answer: V (2 π m k B )3 / 2 T 3/ 2 V qT = 3 = Λ h3 (2 π m k B )3/ 2=(2 π×4.65×10−26 kg×1.381×10−23 J/K )3 / 2 = 8.10×10−72 ( kg⋅m/s)3 (2 π m k B )3/ 2 / h3 = 2.78×10 28 m−3 K −3/ 2 V (2 π m k B )3 / 2 / h3 = 10−27 m 3 ×2.78×10 28 m−3 K −3/ 2 = 27.8 K −3 / 2 q T = 27.86 K −3 / 2 T 3/ 2 so T = (1.00×105/ 27.86 K −3 / 2)2 /3 = (3590)2 / 3 = 234 K 8. Let E be the translational kinetic energy of one gas molecule. The probability distribution is f (E ) = 2 π E 1/2 (π k B T )3/ 2 e −E /(k B T ) ; 0≤E a) (2 points) What is the most probable kinetic energy? (Derive a formula.) Answer: df 2π 1 E 1/2 −E /(k T ) = − e d E (π k T )3/2 2 E 1/ 2 kT df 1 = 0 for E = k B T 2 dE ( ) b) (1 point) For comparison, what is the translational kinetic energy of a molecule that has the average speed? (Again, the answer is a formula, not a number.) Answer: average speed = √ 8 kB T πm 4 so E = 1 m v 2 = π k B T 2 9. (2 points) Rare earth elements produced in a nuclear reactor may enter liquid sodium coolant. Simulated diffusion of Ce atoms in liquid Na at 1000 K is graphed at right. (Samin, et al., J. Appl. Phys., 118, 234902, 2015). Recall that 1 ps = 10-12 s, and 1 Å = 10-10 m. What is the diffusion coefficient, in m2/s? Answer: < r2 > = 6Dt so 6D = slope = 3.34 A2/ps D = (3.34 /6) (10-10m)2/(10-12 s) D = 5.57 X 10-9 m2/s 10. (2 points) Marteza Waskasi, et al., (J. Am. Chem. Soc., 2016, 138, 9251-9257) measured the rate of electron transfer between a porphyrin cation and a C60 anion. The two were covalently bound. Data are below. a) At what temperature is the activation energy zero? b) At that temperature, what is the half-life (t1/2) of P+-C-60 ? Give t1/2 in seconds. Answers: a) Ea=0 where d ln(kR)/d(1/T) =0, so at 1000/T=6.1. T = 1000/6.1 = 164 K. b) ln(kR)=20.65 so kR=9.3×108 s-1. t1/2=ln(2)/kR = 7.5×10-10 s. 11. Consider the second-order reaction A + B → products. The rate law is, rate = k [A] [B]. Initial concentrations are [A]0 = 0.12 M, [B]0 = 0.24 M. The reaction is 50.0% complete in 50.0 s. [ A ]0 [ B] = k ([ B ]0− [ A ]0) t . As you may recall, the integrated rate law is ln [ A ][ B]0 ( ) a) (2 points) Calculate the rate coefficient, k. b) (1 point) After how many seconds is the reaction 75.0% complete? Answers: [B]0-[A]0 = 0.24 - 0.12 = 0.12 M. a) At 50% completion, [A]=0.06M and [B] = 0.24 -0.06 = 0.18 M. ln k= (0.12×0.18 0.06×0.24 ) ln (1.5) 0.4055 = = = 0.0676 M 0.12 M×50 s 6.0 M s 6M s −1 −1 s b) At 75% completion, [A]=0.03 M and [B] = 0.24 - 0.09 = 0.15 M. ln t= (0.12×0.15 0.03×0.24 ) 0.0676 M −1 −1 s ×0.12 M = ln (2.5) 0.008112 s −1 = 0.916 = 113 s 0.008112 s−1 12. (3 points) Consider the net reaction I- + OCl- ⇌ OI- + OCl-, which occurs in water. Here is a possible mechanism: k1 OCl- + H2O HOCl + OHk-1 k2 I- + HOCl HOI + Clk3 HOI + OHH2O + OIa) Based on the initial-rate data, what is the reaction order with respect to OCl- ? b) Write the differential rate law for HOCl: d [HOCl ] =⋯ dt c) What is the steady-state expression for [HOCl] in terms of OCl-, H2O, OH- and rate coefficients? [I-]0 (mM) [OCl-]0 (mM) initial rate (mM/s) 2.0 1.5 0.18 4.0 1.5 0.36 4.0 3.0 0.72 Answers: a) Doubling [OCl-] increases the rate by a factor of two, so the order with respect to OCl- is 1. d [ HOCl ] = k 1[ OCl - ]−k −1 [ HOCl ][OH - ]−k 2 [ HOCl ][ I - ] b) d t k 1 [OCl - ] d [ HOCl ] = 0 gives [HOCl ]ss = c) Setting d t k−1 [OH - ]+ k 2 [ I - ] 13. (2 points) Guanyi Chen, et al., (Biochemical Engineering Journal, 113, 2016, 86–92) studied biodiesel production catalyzed by bacteria. bacteria triglyceride biodiesel The reaction rate is in mg/min: milligrams of biodiesel produced per minute. The concentration of substrate (triglyceride) is in mg/mL and appears as "Cs" on the graph. Methanol was added as an inhibitor. The concentration of methanol is given as percent by volume of the reaction mixture. methanol a) In terms of Michaelis-Menten kinetics, what is the value of Rmax? b) What sort of inhibitor is methanol: competitive or non-competitive? How do you know? Answers: a) Rmax = 1/intercept = 1/(0.45 min/mg) = 2.2 mg/min. b) Inhibition is competitive because Rmax is unchanged (constant intercept) while KM (slope × Rmax) changes. Chen wrote, "higher methanol concentration gives higher slope of lines, but an invariant rmax , indicating that methanol was a competitive inhibitor of enzyme activity."

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