I affirm that I have never given nor received aid on this examination. I understand that cheating in the exam will result in a grade F for the class.

Chem340 Physical Chemistry for Biochemists Exam Mar 16, 011 Your Name _ I affirm that I have never given nor received aid on this examination. I understand that cheating in the exam will result in a grade F for the class. Signature. Sec 1 Multile Choice [ /4] Sec Calculation/MC [ /3] Sec 3 Long Question [ /45] Sec 4 Derivation [ /1] Total [ /100] Total oints are 131 oints. If you score more than 100, you will receive 100 oints. 1

Multile Choice Questions (Some questions may be similar to those one in quiz, but not the same) Each 3 oints [ /4] Q1. One mole of an ideal gas with C P = 5R/ undergoes the transformations from an initial state described by T i = 300. K and P i = 1.50 bar. The gas undergoes a reversible isothermal comression at 300 K until P becomes 1.0 bar. Choose a formula that best reresents S. (a) S = (5R/)ln(1.5/1) (b) S = (5R/)ln(1/1.5) (c) S = Rln(1/1.5) (d) S = Rln(1.5/1) Q. You have containers of ure H and He at 98 K and 1 atm ressure. Calculate G mixing relative to the unmixed gases for a mixture of 1.5 mol of H and 0.5 mol of He. Choose the most aroriate formula for G mixing (a) RT{3/4ln(3/4) + 1/4ln(1/4)} (b) RT{3/4ln(3/4) + 1/4ln(1/4)} (c) 0 (d) RT{1/4ln(3/4) + 3/4ln(1/4)} (e) RT{1/4ln(3/4) + 3/4ln(1/4)} Q3. One mole of N at 600 K and one mole of N at 300K were thermally contacted in an adiabatic container of a fixed volume. At the end, the system reached the equilibrium at 450K. Choose two correct answers below. (a) The entroy change is C v {ln(450/600) + ln(450/300)} (b) The entroy change is -C v {ln(450/600) + ln(450/300)} (c) The entroy change is 0. (d) The change in the internal energy is ositive (e) The change in the internal energy is negative (f) The change in the internal energy is zero. Q4. Consider the formation of glucose from carbon dioxide and water, that is, the of the following hotosynthetic rocess: CO (g) + H O(l) (1/6)C 6 H 1 O 6 (s) + O (g). Choose a formula that best reresents the entroy for this chemical system at T = 348 K. The following table of information will be useful in working this roblem: (a) 50x (-13.8-70 + (1/6) x 09. + 05.) + (-6x 37.1-75.3 + (1/6) x 19. + 9.4) Jmol K (b) (-6x 13.8-6x 70 + 09. + 6x 05.) + 50x (-6x 37.1-6x 75.3 + 19. + 6x 9.4) Jmol K (c) (-13.8-70 + (1/6)x 09. + 05.)ln(348/98) + (-37.1-75.3 + (1/6)x 19. + 9.4) Jmol K (d) (-6x 13.8-6x 70 + 09. + 6x 05.) + (-6x 37.1-6x 75.3 + 19. + 6x 9.4)ln(348/98) Jmol K (e) None of them Q5. Which formula best reresents the enthaly at 98 K for the in Q4 in kjmol units? (a) (-6x 393.5-6x 85.8-173.1) (b) (393.5 + 85.8 - (1/6)x173.1) (c) (-6x 393.5-6x 85.8 + 173.1) (d) (-393.5-85.8 + (1/6)x173.1

Q6. Which relationshi is correct for an irreversible rocess? Choose all correct answer(s). (a) Dqirreversible 0 (b) ds 0 (c) Dqirreversib le T ds T (d) Dqirreversible ds (e) ds 0 (f) Dqirreversib le T ds T Q7. We calculated G for two exansion rocesses of 10.00 mol of an ideal gas at 300 K from an initial ressure of 1.0 bar at 1.0 L to a final volume of 10.0 L. In rocess X, we obtained G X for the isothermal reversible exansion. In rocess Y, we obtained G Y for the irreversible isothermal exansion at the external ressure of P ext = 0.05 bar? C P, m = (5/)R. What is G Y /G X? (a) 7/5 (b) 5/3 (c) 3/ (d) 1 (e) /3 (f) 3/5 (g) 5/7 Q8. Choose the most aroriate for which a standard formation enthaly for NH 3 can be obtained as a enthaly. a. NH 3 (g) N(g) + 3H(g) b. 1/N (g) + 3/H (g) NH 3 (g) c. 1/3NH 3 (g) 1/3N(g) + H(g) d. N (g) + 3H (g) NH 3 (g) Q9. For the following, ΔU = ΔH + X. What is X? TiCl 4 (l) + H O(l) TiO (s) + 4HCl(g) (a) 3RT (b) 4RT (c) -3RT (d) -4RT Q11. In the case of this figure, melting oint T m and boiling oint T b are changed as follows when P is increased from P 0 to P 0 + P. Choose one correct answer. a) T m and T b are both decreased. b) T m is increased while T b is decreased. c) T m is decreased while T b is increased. d) None of all Q1 Choose ALL correct equations. (a) U = TdS - PdV (b) H = TdS + VdP (c) G = -SdT - PdV (d) A = -SdT + VdP (e) (G/P) T = V (f) (A/T) V = -S Q1. A choice of (e), (f) is also correct since d was missing from du & dh in (a) & (b) by our mistake. 3

Q1. Choose ALL correct statements on the right diagram. (a) V m (sl) > 0 (b) V m (sl) < 0 (c) S m (sl) > 0 (d) S m (sl) < 0 (e) Line b denotes a hase transition from liquid to solid through the melting oint. (f) Sublimation (s g) does not take lace for this system when the temerature is above that for trile oint. (g) This system undergoes a hase transition as g l s when the ressure is increased at constant T which is a little below the trile oint temerature. Q13 The figure right shows a DSC scan of a solution of a T4 lysozyme mutant. Which of the following of [A-D] reresents (i) melting temerature T m and (ii) intrinsic excess heat caacity? Choose the most suitable combination. (i, ii) = (a) (A, B) (b) (D, A) (c) (D, B) (d) (D, C) (e) (A, C) (f) (A, B) Q14. Figure in the right shows Gibbs energy (G mixture ) for a mixture of mol of N O 4 and (1-) mol of NO in a of NO (g) N O 4 (g) (*). How much is the molar Gibbs energy of ure N O 4. Choose the closest. (a) -81 kjmol (b) -79.5 kjmol (c) -77 kjmol (d) -38 kjmol (e) -4 kjmol (f) 0 kjmol 4

. Calculation/Multile Choice Question each 4 oint [ /3 oints].1 One mole of N at 600K and 6.00 bar undergoes a transformation to the state described by 300K and 3.00 bar. Calculate S if C,m /(Jmol K ) = a + b(t/k) + c(t/k). Choose the most aroriate formula that reresents S. (a) Rln + a(t f T i )/K + b{(t f /K) (T i /K) }/ + c{(t f /K) 3 (T i /K) 3 }/3 (b) - Rln + a(t f T i )/K + b{(t f /K) (T i /K) }/ + c{(t f /K) 3 (T i /K) 3 }/3 (c) Rln + a ln(t f /T i ) + b{(t f /K) (T i /K)} + c{(t f /K) (T i /K) }/ (d) -Rln + a ln(t f /T i ) + b{(t f /K) (T i /K)} + c{(t f /K) (T i /K) }/ (e) Rln + C,m [a ln(t f /T i ) + b{(t f /K) (T i /K)} + c{(t f /K) (T i /K) }/] (f) -Rln + C,m [a ln(t f /T i ) + b{(t f /K) (T i /K)} + c{(t f /K) (T i /K) }/] (g) None of them...3 Consider the reversible Carnot cycle (a b c d a) with 1 mol of an ideal gas with C V = 3/R as the working substance. The initial isothermal reversible exansion occurs at the hot reservoir temerature of T hot = 600 K (T a, T b ) from an initial volume of 5.0 L (V a ) to a volume of 10.0 L (V b ). The system then undergoes an adiabatic reversible exansion until the temerature falls to T cold = 300 K (T c, T d ). The system then undergoes an isothermal reversible comression until V= V d. Finally, the system is subject to a subsequent adiabatic reversible comression until the initial state described by T a = 600 K and V a = 5.0 L is reached. (from P5., 5.3). How much is the sum of the work (w bc + w da ) in the adiabatic rocess? Choose the closest. (a) 1.0 kj (b) -5.5 kj (c) 0 kj (d) 5.5 kj (e) 11 kj.3 How much is ΔH da? Choose the closest. (a) -9.5 kj (b) -3.6 kj (c) 0 kj (d) 3.6 kj (e) 9.5 kj.4 Calculate A and G for the isothermal irreversible comression of 1.00 mol of an ideal gas at 98 K from an initial volume of 35.0 L to a final volume of 1.0 L. Choose the closest value for each of A and G from (a-e) and (f-j), resectively. ( oints if only one is correct) A G (a) 5.5 kj (b).5 kj (c) 0kJ (d) -.5 kj (e) -5.5 kj (f) 5.5 kj (g).5 kj (h) 0kJ (i) -.5 kj (j) -5.5 kj.5 From the following data at 5 C, calculate the standard enthaly of formation of Fe O 3 (s) and choose the closest value. 5 H Fe O 3 (s) + 3C(grahite) Fe(s) + 3CO(g) 49.6 FeO(s) + C(grahite) Fe(s) + CO(g) 155.8 C(grahite) + O (g) CO (g) 393.51 CO(g) + 1/ O (g) CO (g) 8.98 (a) -900 kjmol (b) -60 kjmol (c) -400kJmol (d) 50 kjmol (e) 0 kjmol (f) 150 kjmol (f) 400 kjmol (g) 60 kjmol (h) 90 kjmol kj mol 1

.6. Calculate mixture O 98.15 K, 1 bar for oxygen in air diluted by helium, assuming that the molar fraction of O in air is 0.100 and O behave as an ideal gas. The molar entroy of O, S 0 m is 05. Jmol K. Choose the closest. (a) -66.9 kjmol (b) -64.9 kjmol (c) -6.9 kjmol (d) 6.9 kjmol (e) 64.9 kjmol (f) 66.9 kjmol -.7. 1 mole of steam (H O(g)) at 500K was gradually cooled to 300K, while forming water, H O(l) at the boiling oint T b. Assume that H va denotes a molar vaorization enthaly of water, and heat caacities for steam and water, C (steam) and C(water) are constant over the temerature. Choose the most aroriate exression that corresonds to S during the rocess. (a) H va / Tb C ( steam) 500K Tb C ( water) 300K Tb (b) H va / Tb C ( steam) T b 500K C ( water) 300K Tb (c) H va / Tb C ( steam) ln500k / Tb C ( water) ln300k / Tb (d) H / T C ( steam) lnt / 500K C ( water) ln300k / T va b b.8. One mole of a van der Waals gas at T (300K) is exanded isothermally and reversibly from an initial volume of V i to a final volume of V f. For the van der Waals gas, U / V a / V T m. Assuming that b is 0, choose the right formula that reresents S. Use van der Waals gas equation P = RT/V m a/v m, where b=0. (a) Vf R ln Vi (e) V f R ln V i 1 1 a /T Vf Vi 1 1 a /T Vf Vi (b) Vf R ln V i a Vf b 1 (c) Vf R ln (d) Vf R ln V V 1 /T Vi i i 1 a Vf 1 /T Vi 6

Calculation Question. Choose 3 questions among the following 4 questions. If you solve 4 without choosing, we will grade for 3 questions with the lowest scores. You will gain considerable oints by indicating the CORRECT formula. [ /45] Write answers in the secified lace. You should show calculations for full credit and artial credit. Please include units in calculations. 3.1 [Circle if you choose this] (Original Question P4.18) [ /15] P4.18) Given the data in the following information, calculate the single bond enthalies and energies for C F as follows: (a) List all the s needed for calculations with their formation enthalies. (6 oints) (b) By combining the s in (a) using Hess s law, obtain the from which you can elucidate the single bond enthaly for C-F as a enthaly, and calculate the single bond enthaly. (6 oints) (c) Calculate the single bond energy for C-F (3 oints). Assume that the standard temerature is 98 K (rather than 98.15K). Show equations for a major credit. Substance F (g) F(g) CF 4 (g) NF 3 (g) C(g) N(g) 0 79 95 15 717 47 H f kj mol 1 ΔH 95 kj mol ΔH 79 kj mol ΔH 716 kj mol oint each for showing a with its formation enthaly (total 6 oints) (a) Cgrahite F g CF4 g 1/F g F g Cgrahite Cg (b, c) CF4 g C grahite F g F g 4 F g Cgrahite C g ΔH 95 kj mol 4 79 kj mol ΔH 717 kj mol ΔH Additional oints (1 each) for balancing as shown above for CF 4 (g) & F(g) to roduce the below CF 4 g C g 4 F g The average C-F bond enthaly is then: H bond avrg 1959 kj mol 4 490 kj mol ΔH 1959 kj mol (equation 1 oint; calculation 1 oint) ( oints) (c) And the average C-F bond free energy: ΔU ΔH 1949 kj mol calculation 1 oint) n R T 1949.0 kj mol 1959.0 kj mol 4 8.314 J K mol 98.15 K bond Uavrg 487 kj mol (1 oint for equation) 4 (eq. 1 oint; 7

3. [Circle if you choose this] (Original Question P5.13) [ /15] One mole of an ideal gas with C P,m = 5/ R undergoes the transformations described in the following list from an initial state described by T i = 300. K and P i = 4.00 bar. The gas undergoes an adiabatic exansion against a constant external ressure of P ext until the final ressure P f ==1.00 bar. (a) Calculate q, w, and S when P ext = 0.00 (6 oints). (b) Obtain the temerature after the exansion (T f ) and S when P ext = 1.00 bar. (6 oints) Show a formula first and then calculate the value. (c) Calculate S when the gas undergoes an adiabatic reversible exansion (in lace of a constant external ressure P ext ) until the final ressure P f ==1.00 bar. What is the final temerature T f? (3 oints) T f Ti (a) U n C V,m Tf Ti n R external w f i Since P external =0, w = 0. (1 oint) From this, U= 0 using the equation above. q = U w = 0. (1 oint) (Alternatively, q = 0 since this is adiabatic rocess. Thus, U= 0 T f =T i ) Because U= 0, T f = T i. S = -nrln(p f /P i )+nc P ln(t f /T i ) (1 oint) = -nrln(1.00/4.00) ( 1 oint; for the second term is zero) = nrln3. = (8.315 Jmol K )ln(4.00). = 11.57 JK mol (1 oint) (b) T f Ti U n C V,m Tf Ti n R external w (1 oints) f i (n ext xt CV, m nr ) Tf n R Ti e ncv, m (1 oints) f i 1 3 (3/nR nr) Tf n R n R Ti T f = 7T i /10 = 10K. ( oints) 4 S = -nrln(p f /P i ) + nc P,m ln(t f /T i ) (eq. 1 oints) = -nrln(1.00/4.00) + 5nR/ ln(7/10) = 4.11 JK mol (1 oint) (c) Since this is a reversible adiabatic exansion = S = 0 (1 oint) Tf Ti 1 i f Tf Ti i f 1 (1 oint) Tf T i 4.00 bar 1.000 bar 5 1 3 5 3 4.00 5 0. 574 T f = T i * 0.574 = 17 K (1 oint) 8

3.3 [Circle if you choose this] Original Question P6.8) [ /15] (a) Calculate G for the CH 4 (g) + O (g) CO (g) + H O(l) at T =T 0 = 98 K from the combustion enthaly of methane (-890.3 kj mol ) and the entroies of the reactants and roducts. S 0 (CO, g) = 13.8 Jmol, S 0 (H O, l) = 70.0 Jmol, S 0 (CH 4, g) = 186.3 Jmol, S 0 (O, g) = 05. Jmol. (6 oint) (b) Give an equation to calculate A from G you obtained in (a) at T=T 0. Then, calculate the values. (3 oint) (c) Define K x for this and give an exression that yields K x of the at 98K at a constant total ressure P. Just show an equation (as K X = ) using G obtained in (b), P, T 0, and P 0. (4 oints) (d) Give an equation to obtain K at T =1000K from K (98K) assuming ΔH 0 is constant. ( oints) (a) All reactants and roducts are treated as ideal gases Gcombustion Hcombustion TScombustion Scombustion S CO, gs HO, ls CH 4, gs O, g 1 1 1 1 1 1 1 1 13.8 J mol K + 70.0 J mol K 186.3 J mol K 05. J mol K 1 1 4.9 J mol K 3 G 890.310 J mol 98.15 K 4.7 J mol K 817.910 J mol combustion 1 1 1 3 1 (1 oint for equation of G 0 combustion, oint for equation for S 0 combustion; oint for the correct answer) (b) G = A + (PV) A = G - (PV) = G - RTn ( oint) = G - RT(-) = -817.9 kjmol - 8.315 JK mol x 98 K x (-) = -81.9 kjmol (1 oint) (c) K X = (x CO )/{(x CH4 (x O ) } (1 oint) K (98K, 1 bar) = ex(-g 0 /RT 0 ) (1 oint) K = K x (P/P 0 ) v K x = ex(-g 0 /RT 0 ) (P/P 0 ) -v (1 oint), where v =. (1 oint) 0 H 1 1 (d) ln{ K ( T )} ln{ K ( T1 )} ( oints) R T T1 9

3.4 [Circle if you choose this] [ /15] mole of NO (g) ( mole is the initial number of mole) roduced N O 4 (g) from the following : NO (g) N O 4 (g) (*). A resultant mixture of NO (g) and N O 4 (g) is in its equilibrium between ( - ξ) mole of NO and ξ mole of N O 4 (g). (a) Assume that the takes lace at a constant ressure and temerature at 1 bar and 98 K. Exress formulae of G ure and G mixture using ξ and molar Gibbs energy for NO, G 0 m (NO, g) and N O 4, G 0 m (N O 4, g). (6 oints). (b) Exress ΔG for (*) using ξ, G 0 m (NO, g), and G 0 m (N O 4, g). You can divide ΔG into two terms as ΔG 0 + ΔG mixing (5 oints) (c) How do you find ξ in the equilibrium state (under constant P and T)? Reort two methods. You do not have to calculate the value, but show the rincile. (4 oints) (a) G ure m m 4 o o ( - ) G ( NO, g) G ( N O, g) (3 oints) (b) G - G o o G ure ( - ) G m( NO, g) G m( NO4, g) G mixture G ure ΔGmixing G ure nrt ( xno ln x NO x G ure ( - ) RT( ln ln ) (4 oints; Only ΔG mix is correct 3 oints) o m G 0 ΔG mixing o ( NO, g) G m( NO4, g) RT ( ln ln x NO ln x N (5 oints; 3 Points if G 0 is written) O 4 ) N O 4 ln x N O 4 ) (c) Find ξ that minimizes G. Find ξ that satisfies G 0 ( oints each) mixture 10

4. Derivation Questions Q1-6 oints each [ /1] Fill a formula or equation in [Q1-Q4] and answer Q5. This section has some hint for other questions. 4.1 Now, we would like to derive ΔH(308 K) for the B + C X at the standard temerature using C A (A = B, C, X) and ΔH 0 (98 K) for this. For this urose, we use dh A = CAdT C, Aμ J-T dp + ( H/ n) T, P dna (A = B, C, or X) assuming that μ J-T and C,A (A = B, C, or X) are constant over the ressure range. (Also, assume that the J-T coefficients are the same for B, C, X). First, we consider a rocess 1: B(308 K) + C(308K) B(98K) + C(98K). Since dp = 0 and dn A = 0, dha = CA dt. Thus, ΔH for this rocess (ΔH 1 ) is given by [Q1 ]. Next, we consider a rocess : B(98K) + C(98K) X(98K). ΔH for this (ΔH ) is ΔH 0 (98K). Lastly, we consider a rocess 3: X(98K) X(308K). As shown for a rocess 1, ΔH for this rocess (ΔH 3 ) can be obtained by using C,X. ΔH(308K), which can be obtained by ΔH 1 +ΔH +ΔH 3, is exressed as [Q ] using μ J-T, C,A (A = B, C, X) and ΔH 0 (1.0 bar). Q1: [ (C, B + C, C )(98K 308K) ] (check the sign is correct.) Q: [ (C, X - C, B - C, C )(308K 98K) + ΔH 0 (98K) ] 4. We rove that A / T U as follows. Fill aroriate formula in [Q3] and [Q4]. First, (1/ T ) V A / T 1 A A S A A TS [ Q3] / T ( T ) T T V T T T T V Then, A / T A / T T A / T [ Q4] U. (1/ T ) T V (1/ T ) T V V V Q3: [ ΔU ] Q4: [ -T ] 11

4.3 We examine the hase transition between solid, liquid, and gas states. From d(t, P) = -S m dt + V m dp. (1) For a constant P, d(t, P) = -S m dt. () Assuming S m is constant, (T, P) = (0, P) -S m T. (3) μ gas (T, P) μ solid (T, P) μ liquid (T, P) The figure in the right shows chemical otential for a different temerature. 3 lines reresent solid, liquid, gas. From (3), the sloes of these lines are given by molar entroies of solid (S m, solid ), liquid (S m, liquid ), and gas (S m, gas ). Among S m, solid, S m, liquid, S m, gas, S m is the greatest for [Q5 ] from the figure. At the boiling oint T b, liquid (T b, P) = gas (T b, P). (4) Now, we define the change in the chemical from liquid to solid as l g(t, P) = gas (T, P) - liquid (T, P). (5) From (4) and (5) l g(t b, P) = 0. (6) Thus, L and G are in the equilibrium at T = T b. When the temerature is increased (ΔT >0), l g(t b + ΔT, P) = [Q6 ] ΔT, (7) where we used (T b + ΔT, P) = (T b, P) - S m ΔT for liquid and solid and eq. (4). Thus, l g(t b + ΔT, P) < 0. The gas state is more stable than the suerheated liquid state. [Q5: S m, gas (or gas) ] [Q6: - (S m, gas - S m, liquid ) ] 1

dln(x)/dx = 1/x dcos(x)/dx = -Sin(x) dsin(x)/dx = Cos(x) de -x /dx = -e -x Constants L = dm 3 = 10-3 m 3 13

Calculation sheet (You can detach this sheet, if you like) 14