Physical Chemistry I Final Exam Name: KEY CHEM 4641 Fall 017 15 questions worth a total of 3 points. Show your work, except on multiple-choice questions. 1 V 1 V α= κt = V T P V P T Gas constant R = 8.314 J mol-1 K-1 E ln q Boltzmann constant kb = 1.381 10-3 J/K = 0.695 cm-1/k S = + k B T ln Q E =N k B T T T N,V Planck's constant h = 6.63 10-34 J s d ln P ΔH Clausius-Clapeyron: = Avogadro's number NA = 6.0 103 mol-1 R d (1/T ) 8 Speed of light c = 3.00 10 m/s dp ΔH Clapeyron: = 5 9-7 dt T ΔV Units: 1 bar = 10 Pa. 1 GPa = 10 Pa 1 amu = 1.6605 10 kg ( ) ( ) ( 1 Joule = 1 Pa m3. ) 1L=10-3m3. 1 L bar = 100 J. Stirling's approximatipon: ln ( N! ) N ln ( N ) N ( q vib = 1 e ωe /( kt ) 1 ) e a x d x = Integrals: 0 x n e a x qrot = kt σb q trans = 0 Λ3 ; Λ=h / π m k T 1 a dx= V xn e a x d x = 0 1 3 5 ( n+ 1) π a n+1 an x n+1 e a x 1 a n+1 dx= 0 n! a n+1 1 H He 1.01 4.00 3 Li 4 Be 6.94 9.01 11 Na 1 Mg 5 B 0 Ca 7 N 8 O 9 F 10 Ne 10.81 1.01 14.01 16.00 19.00 0.18 13 Al.99 4.31 19 K 6 C 14 Si 15 P 16 S 17 Cl 18 Ar 6.98 8.09 30.97 3.06 35.45 39.95 1 Sc Ti 3 V 4 Cr 5 Mn 6 Fe 7 Co 8 Ni 9 Cu 30 Zn 31 Ga 3 Ge 33 As 34 Se 35 Br 36 Kr 39.10 40.08 44.96 47.88 50.94 5.00 54.94 55.85 58.93 58.70 63.55 65.38 69.7 7.59 74.9 78.96 79.90 83.80 37 Rb 38 Sr 39 Y 40 Zr 41 Nb 4 Mo 85.47 87.6 88.91 91. 9.91 95.94 55 Cs 56 Ba 57 La 7 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 3 88 Ra 89 Ac 6.0 7.0 58 Ce 54 Xe 76 Os 77 Ir 78 Pt 79 Au 80 Hg 81 Tl 8 Pb 83 Bi 84 Po 85 At 86 Rn 09 10 104 Rf 105 Db 106 Sg 107 Bh 108 Hs 109 Mt 110 Ds 111 Rg 11 Cn 113 Nh 114 Fl 115 Mc 116 Lv 117 Ts 118 Og 67 68 69 70 69 78 81 80 85 86 89 89 93 94 94 59 Pr 60 Nd 61 Pm 6 Sm 63 Eu 64 Gd 65 Tb 66 Dy 67 Ho 68 Er 69 Tm 70 Yb 71 Lu 140.1 140.9 144. 90 Th 53 I 101.1 10.9 106.4 107.9 11.4 114.8 118.7 11.8 17.6 16.9 131.3 13.9 137.3 138.9 178.5 180.9 183.8 186. 190. 19. 195.1 197.0 00.6 04.4 07. 09.0 87 Fr 5 Te 91 Pa 9 U 145 93 Np 150.4 15.0 157.3 158.9 16.5 164.9 167.3 168.9 173.0 175.0 94 Pu 95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm
3.0 31.0 38.0 37 44 43 47 47 51 5 57
Score: /3 1. Consider gas-phase 79Br. The rotation constant, B, is 0.08 cm-1. The vibration frequency is 35 cm-1. The molecule's mass is 157.8 amu. Electronically, only the ground state is populated. Suppose 98 K and 1 bar. Which is the largest partition function, q? [Note: Calculating the partition functions is not recommended.] (a) electronic (b) rotational (c) translational (d) vibrational Answer: (c) Translational q is by far the largest. Just to check the number: qtrans = 8 107, qrot=1300 and qvib=1.3.. Consider a gaseous mixture of oxygen and hydrogen molecules, O and H, at T = 500.0 K. The gas is confined to a cylinder of constant volume. Which one of the following is smallest? (a) average speed of H (b) average speed of O (c) average relative speed of H and O (d) average relative speed of O and O Answer: (b) largest mass or reduced mass gives smallest speed H mass = 4 amu, O mass = 3 amu, H-O reduced mass = 3.6, O-O reduced mass = 16 3. Supposing that the equipartition principle applies to translation and rotation, and that vibration and electronic excitation are inactive, what is the average energy of an HS molecule in the gas phase? Choose the best answer. (a) kbt (b) (3/) kbt (c) (5/) kbt (d) 3 kbt Answer: (d) 3/ translational plus 3/ rotational 4. At constant volume and number of moles, how does S depend on E? Assume a pure substance in a single phase. Choose the best graph.
Answer: (c). Recall that ( S / E )V =1/T. 5. Consider reversible adiabatic heating of an ideal gas. Which graph best describes S and E? Answer: (d). S is constant and E increases. 6. Which of the following equals the Helmholtz free energy? (a) E - T S (b) E + P V - T S (c) G + P V (d) H - P V Answer: (a) 7. Consider a reservoir and a system in close contact, surrounded by an impenetrable, insulated and rigid container. A change occurs inside the container. Which one of the following is assured by the second law of thermodynamics? (a) ΔSSreservoir<0 (b) ΔSSsystem 0 (c) ΔSSsystem 0 (d) ΔSStot 0 Answer: (d). 8. One mole of an ideal gas is heated reversibly from 73 K to 310 K at a constant pressure of 1.00 bar. CP = 3.4 J/(mol K). What is ΔSH? Choose the best of the answers listed. (a) 4.1 J/mol (b) 117 J/mol (c) 890 J/mol (d) 100 J/mol 310 Answer: d. Δ H = C P dt =C P ( 310 73)=100 J/mol 73
9. At right is a graph of ln(k/t), where k is a rate coefficient. Which of the following kinetic parameters is proportional to the slope of the graph? (a) activation enthalpy (b) activation entropy (c) activation free energy (d) activation energy Answer: (a) activation enthalpy equals -R times the slope of the Eyring plot. 10. Values of energy (E) and degeneracy (g) for a two-level system are tabulated at right. The temperature is 1464 K. a. ( points) Calculate the partition function, q. b. ( points) Calculate the average energy, <E>. E (cm-1) g 0.0 4 1745.0 10 Answers: a. q = 4 + 10 e-1745/(0.69503 1464) = 4 + 10 e-1.715 = 4 + 10 0.180 = 4 + 1.80 = 5.80 b. <E> = (4 0 + 1.80 1745 )/q = 3141 / q = 54 cm-1
11. Consider melting of solid sulfur hexafluoride. The melting pressure (in Pascals) and melting Tm c temperature (in Kelvin) are related by this equation: P m= P t + a 1, where Tt [( ) ] Tt is the triple temperature: 3.555 K Pt is the triple pressure: 3149 Pa a = 3.7 106 Pa c = 1.555 Pm is the pressure at which SF6 melts when the temperature is Tm. (Source: Allan Harvey, Journal of Physical and Chemical Reference Data, 46, 04310, 017.) For parts a and b, let T = 5.0 K. At that temperature, ΔSV = Vliquid-Vsolid = 15.06 10-6 m3/mol. a. ( points) Calculate the Clapeyron slope, dpm/dtm. Give dpm/dtm in Pa/K. b. (1 point) Calculate the molar enthalpy of fusion, ΔSfusH. Give ΔSfusH in kj/mol. c. (1 point) Why is the Clausius-Clapeyron equation not applicable in this problem? Answers: a) dp T cm 1 m = ac c dtm T t dp m 5.0 0.555 = 3.7 106 Pa 1.555 = 1.56 10 6 Pa / K 1.555 dtm 3.555 d Pm b) Δ fus H = T Δ V dt m = 1.56 10 6 Pa 1 kj 5.0 K 15.06 10 6 m3 /mol 3 = 5.3 kj/mol K 10 Pa m3 c) The Clausius-Clapeyron applies only when one of the phases is gas.
1. Consider this gas-phase reaction: Br + Cl BrCl. a. (1 point) Calculate ΔrxnHo at 98 K. Cl BrCl ΔfHo (kj/mol) 30.91 0.00 14.64 ΔfGo (kj/mol) 3.11 0.00-0.98 Cp,m (J/mol K) 36 34 35 Data are at 98K. b. (1 point) Calculate ΔrxnHo at 318 K. c. (1 point) Calculate ΔrxnGo at 98 K and calculate the equilibrium constant, K, at 98K. Br d. ( points) Assume that the reaction begins with equal moles of Br and Cl and zero moles of BrCl. No other gases are present. Assume activities are proportional to moles. a BrCl Calculate the equilibrium ratio of activities: a. Br Answers: a. 14.64-30.91 = -1.63 kj/mol b. ΔSHo(318) = ΔSHo(98) + ΔSCp ΔST = ΔSHo(98) + 0 = -1.63 kj/mol c. ΔrxnGo = (-0.98) - 3.11 = -5.07 kj/mol K = exp( - ΔrxnGo /(0.0083145 kj/mol/k 98 K) ) = exp(.05) = 7.74 d. abrcl K= a Br a Cl Moles of Br equal moles of Cl. a K = BrCl a Br ( ) a BrCl K = 7.74 =.78 = a Br
13. In the gas phase, nitric oxide reacts with carbon atoms to form either CO or CN. NO + C CN + O ; rate coefficient k1 = 1.6 10-11 cm3s-1 at 95 K. NO + C CO + N; rate coefficient k = 1.1 10-11 cm3s-1 at 95 K. Assume the initial concentrations of CN and CO are zero. a. (1 point) Write the differential rate law for [CN] and the differential rate law for [CO]. Do not integrate the rate laws, just write them. d [CN ] = k 1 [C ][NO ] Answer: d t d [CO ] = k [C ][ NO ] dt b. (1 point) Calculate the ratio of [CN] to [CO] after the reaction runs. Answer: [CN ] k 1 1.6 = = = 1.45 [CO ] k 1.1 c. (1 point) It has been observed that the product ratio, [CN]/[CO], has no temperature dependence. What does that imply about the activation energies Ea1 and Ea associated with k1 and k? Answer: Ea1 = Ea d. (1 point) A potential energy surface for one of the reactions is shown at right. On the diagram, mark two features: (i) the transition state; and (ii) the exit channel for products. (Source: S. Andersson, et al., PCCP,, 613, 000, DOI: 10.1039/A908183F)
14. Hydrogen peroxide oxidizes a sulfoxide to a sulfone. E.g., (CH3)SO+HO (CH3)SO+HO. M. D. G. de Luna, et al., studied conversion of sulfoxide to sulfone in liquid fuel. (Energy & Fuels, 017). The reaction is sulfoxide + HO sulfone + HO, first-order in sulfoxide and first-order in HO. Let C stand for concentration of sulfoxide. C0 is the initial concentration of sulfoxide. a. (1 point) Because there is great excess of hydrogen peroxide, [HO] is approximately constant. Write the differential rate law for dc/dt. Explain why [HO] being constant makes this a pseudo-first-order reaction. dc = k [ H O ]C = k ' C where k ' k [ H O ]. dt The reaction will appear to be first order in sulfoxide so long as [HO] is approximately constant. The apparent first-order rate coefficient k' is actually a product of the second-order k and [HO]. b. (1 point) From the graph of c/c0 at right, what is the half-life of sulfoxide at 50oC? Accept 11-15 min. From the Arrhenius plot, ln k = -.8, k=0.061, t1/=11 min. c. (1 point) From the graph of ln(c/c0) determine the first-order rate coefficient k at 30oC. Another way to answer this question is to read k30 from the Arrhenius plot below. That gives k30=e-3.9 = 0.00 min-1. One point: at 60 min, 1.5=60 k, k=0.01. d. (1 point) From the Arrhenius plot, calculate the activation energy, Ea. Ea = - 8.3145 J/(mol K) (-15 K) = 18 kj/mol
15. Here is a mechanism for enzyme catalysis. a. (1 point) Write the differential rate law for [EP]. Do not integrate the equation, just write it. Answer: d [EP ] = k [ ES ] + k 3 [ E ][P ] (k + k 3 )[EP ] dt b. (1 point) Using the fast-equilibrium approximation, write the equation for [EP] in terms of [ES]. Answer: [ EP ] k [ES ] k k [ ES ] + k 3 [E ][P ] Half credit for steady-state [EP ]ss = k + k 3