Name ANSWER KEY Chemistry 25 (Spring term 2016) Midterm Examination 1 Some like it hot 1a (5 pts) The Large Hadron Collider is designed to reach energies of 7 TeV (= 7 x 10 12 ev, with 1 ev = 1.602 x 10-19 J) per proton. What would be the corresponding temperature of an ideal gas with the same average energy? E = 3 2 k BT per molecule (3/2 RT per mole) ( ) = 3 2 ( 7 10 12 )( 1.6 10 19 ) ( 1.38 10 23 ) 7 10 12 1.6 10 19 T = 2 3 = 5 10 16 K ( 1.38 10 23)T 1b (5 pts) The total energy in the LHC beam is 360 MJ = 360 x 10 6 J. How many kg of copper could be melted if this amount of energy were completely converted into thermal energy in the metal? Assume the copper is initially at 2 K (the temperature of the superconducting magnets), which is 1356 K below the melting temperature of copper. The relevant heat capacity of solid copper is 385 J kg -1 K -1, assumed to be temperature independent for this problem, and the heat of fusion for copper = 205 kj kg -1. E = C T + H fusion = m( 385 i1356 + 205000) 360 10 6 m = 385 i1356 + 205000 = 500 kg 1
Problem 1, extra space if needed 2
2 Thermodynamics of expansion 2a (5 pts) Calculate the multiplicities of the two systems A and B depicted below. A and B each contain 1 particle ( X ). X cannot move across the partition (the dark vertical bar) in A, but it can move between the grid boxes on the left side of the partition. The partition has been completely removed in system B, so X can be located in any of the grid boxes. W A = 3! 1!2! 1= 3 W B = 9! 1!8! 1= 9 2b (3 pts) Using the statistical definition of entropy, calculate the change in entropy for the transition from A to B illustrated in 2a. Express your answer in units of k B. S = k B ln W B = k B ln 3 ( 1.1k B ) W A 3
2c (2 pts) Using the volume dependence of S calculated in class for an ideal gas (text eq. 7.69), calculate the change in entropy for the isothermal expansion of a single molecule of an ideal gas going from an initial volume V to a final volume 3V. Express you answer in terms of k B. How does this compare to your answer in 2b? S = k B ln V 2 = k B ln 3 ( 1.1k B ) V 1 2d (5 pts) An ideal gas undergoes an isothermal reversible expansion. Which of the following remain constant: q, w, U, H, S, A and G? Explain briefly. all that is required is the following U and H are constant (the energy and enthalpy of an ideal gas are only functions of T (assuming n is fixed as would be in this problem) [for additional information, here are the changes in these quantitites] U = 0 H = 0! # U = 3 " 2 nrt $ & %! # H = 5 " 2 nrt $ & % q = +nrt ln V 2 V 1 w = nrt ln V 2 V 1 S = +nrln V 2 V 1 A = G = nrt ln V 2 V 1 4
3 Having a blast 0.1 mole of nitroglycerin (C 3 H 5 N 3 O 9 ) is exploded in a confined space of volume 10 cm 3. The reaction equation is approximately C 3 H 5 N 3 O 9 ( l)! 3CO 2 ( g)+ 5 2 H O g 2 ( )+ 3 2 N g 2 ( )+ 1 4 O g 2 The change in energy, U, for this reaction is -1400 kj mol -1 nitroglycerin. 3a (10 pts) What is the expected temperature (in K) reached by the reaction? 3b (5 pts) What is the expected pressure (in Pa and atm) reached by the reaction? Assume constant molar heat capacities C V and the following values (in J mol -1 K -1 ): CO 2, 29; H 2 O, 29; N 2, 21; and O 2, 21. You may neglect the contributions to this calculation from the volumes of nitroglycerin and air originally present in the chamber. You can further assume that all the heat produced is used to increase the temperature of the system inside the chamber. 3a for a constant volume process, q = q V = U U = C T U = 0.1 mol x 1.4x10 6 J mol -1 = 1.4x10 5 J C = 0.1 mol x (3x29 + (5/2)x29 + (3/2)x21 + (1/4)x21 J mol -1 K -1 ) = 19.6 J K -1 T = U/C = 7130 K T = 273+ T ~ 7400 C ( ) 3b P = P 2 P 1 P 2 = nrt 2 /V P 1 = nrt 1 /V P = (nr/v) (T 2 -T 1 ) = (nr/v) T n = 0.1 * (3 + (5/2) + (3/2) + (1/4)) = 0.725 mol P = (0.725 mol)(8.3144 J mol -1 K -1 )/(10 cm 3 * 10-6 m 3 /cm 3 ) (7130 K) ~ 4.3 x 10 9 Pa ~ 42 000 atm 5
Problem 3, extra space if needed 6
4 Spring thaw A thermally insulated bottle at 1 atm contains 100.0 gm of liquid H 2 O at 25.0 C to which are added 10.0 gm of snow at -10.0 C. The system is allowed to reach thermal equilibrium. 4a (12 pts) What is the final temperature (in C) of the system? 4b (8 pts) What is S (in J K -1 ) for this process? The enthalpy of fusion of ice is 6.0 kj mol -1, and the specific heat capacities, C p, for ice and liquid water may be taken as 2.1 and 4.2 J gm -1 K -1, respectively. Hint 1: all the snow is melted at equilibrium. Hint 2: for those of you unfamiliar with snow, it is an ornate form of ice. 4a By the conservation of energy, the heat given up by cooling liquid water from 25 C to the final T must equal the energy required to heat ice from -10 C to 0 C; melt the ice and then heat the resulting liquid water from 0 to T. (with T in C) heat lost from liquid water = 100 gm x 4.2 J gm -1 K -1 x (T-25 C) = 420 (T-25 C) J heat required to warm up ice = 10 x 2.1 J gm -1 K -1 x 10 C = 210 J heat required to melt ice = 6000 J mol -1 x (10/18) mol = 3333 J heat required to warm water = 10 gm x 4.2 J gm -1 K -1 x (T- 0 C) = 42 (T) J 420 (T-25 C) + 210 + 3333 + 42 T = 0 462 T = 6957 T = 15 C 4b S = contributions from cooling liquid water from 298 to 288 + heating ice from 263 to 273 + melting ice at 273 + heating resulting water from 273 to 288 420 J K -1 = heat capacity of 100 gm liquid water 21 J K -1 = heat capacity of 10 gm ice 3333 J = heat of fusion of 10 gm ice 42 J K -1 = heat capacity of 10 gm water S = 420ln 288 273 + 21ln 298 263 + 3333 288 + 42ln 273 273 = 0.90 J K 1 7
Problem 4, extra space if needed 8
5 Acid rain N 2 can be fixed in other reactions besides reduction to NH 3. One example is N 2 +O 2! 2NO that is the origin of NO produced by internal combustion engines. The enthalpies and free energies of formation at 298 K and 1 atm for N 2, O 2 and NO, respectively, are H f = 0, 0, +90.3 kj mol -1 G f = 0, 0, +86.6 kj mol -1 5a (3 pts) What is G for the reaction as written at 298 K, in kj mol -1? 5b (3 pts) What is H for the reaction as written at 298 K, in kj mol -1? Is the reaction endothermic or exothermic? 5c (4 pts) What is K eq for the reaction as written at 298 K? Does K eq increase or decrease with T? Explain briefly. 5d (2 pts) Will the equilibrium composition for this reaction be as sensitive to pressure as the Haber Bosch reaction? Explain briefly. 5e (8 pts) At 2500 K, the temperature dependence of log 10 K eq may be approximated as log 10 K eq = 2.69 9451 T From this empirical expression and the van t Hoff equation, derive H for the reaction at 2500 K in kj mol -1. How does this value compare to the result of 5b for H at 298 K? What does this indicate about the heat capacities for the product and reactants (ie, is the difference in heat capacities between the products and the reactants either (i) large and negative, (ii) large and positive, or (iii) close to zero)? It may be helpful to recall that ln x = ln10 i log 10 x. 5a G = 2(86.6) 0 0 = +173.2 kj mol -1 5b H = 2(90.3) 0 0 = +180.6 kj mol -1, endothermic 5c K eq = exp(- G /RT) = 4.67 x 10-31, increases with T since H > 0 5d since there are equal moles of products and reactants, K eq is much less sensitive to P than for the Haber-Bosch reaction with 2 moles of products and 4 moles of reactants. 5e ln K T ln K T = H RT 2 log K = ln10 = 2.303 T T # % $ 2.69 9451 T & ( = ' 2.303 9451 = H T 2 RT 2 H = 2.303 9451R =181 kj mol -1 which is close to the results at 298 K, so the difference in heat capacities is close to 0. 9
Problem 5, extra space if needed 10
Problem 5, extra space if needed 11
6 It s only a phase (5 pts) The free energies of formation, G f, of benzene in the liquid and vapor phases at 298 K and 1 atm are 124.5 and 129.7 kj mol -1, respectively. Assuming the vapor behaves as an ideal gas, what is the partial pressure (in Pa, atm and mm Hg) of benzene in the vapor phase in equilibrium with liquid benzene at 298 K? µ g! + RT ln P P o = µ l! µ g = µ l at equilibrium RT ln P P o = µ l! µ g! P = exp P # o " ( ) RT µ!! l µ g $ % = exp"# ( 124.5 129.7) RT $ % = 0.123 when P o = 101325 Pa, P = 12500 Pa P o = 1 atm, P = 0.123 atm P o = 760 mm Hg, P = 93 mm Hg experimentally, P = 95 mm Hg 12
7 Can global warming mitigate the effects of acid rain? The equilibrium constant, K w, for the ionization of water H 2 O! H + ( aq)+oh ( aq) = 1.0 x 10-14 at 298 K. Since H for this reaction is non-zero, the value of the equilibrium constant is temperature dependent. 7a (3 pts) Calculate H (in kj mol -1 ) from the enthalpies of formation of H 2 O, H + (aq) and OH - (aq) (= -285.9, 0 and -230.0 kj mol -1 respectively). 7b (3 pts) Assuming these enthalpies are temperature independent, estimate the value of K w at 100 C. 7c (2 pts) What is the corresponding ph of pure water at 100 C? 7a H = -230.0 0 + 285.9 = +55.9 kj mol-1 7b from the 4/26/2016 lecture, we can conveniently use an integrated form of the van t Hoff equation ( K 2 = K 1 exp H " 1 1 % + * $ '- ) R # T 1 T 2 &, H = 55.9 kj mol 1 T 1 = 298K T 1 = 373K K 1 =1 10 14 K 2 = 9.3 10 13 7c ph = log K 2 1 2 6.0 (this was not asked for, but global warming would accentuate the effects of acid rain, at least by this simplistic analysis) 13
Problem 7, extra space if needed 14
8 Stretching things The elastic force constant for a single molecule of DNA with a total length of 1 µ is measured to be 1 x 10-7 N m -1 at 298 K. Treating the elastic properties of this molecule as a 1-dimensional random walk, 8a (4 pts) calculate N and l, the effective number of steps in 1 µ and the step length size, respectively, from this data. Express l in m. 8b (3 pts) How many base pairs are in l? (recall the base pair repeat distance of B- form DNA is 3.4 Å). 8a force constant = k B T Nl 2 =1 10 7 Nm 1 k B T = 4.11 10 21 J Nl 2 = 4.11 10 21 J 1 10 7 Nm 1 = 4 10 14 m 2 with 1J =1Nm ( ) Nl =1µ =10 6 m l = 4 10 14 m 2 10 6 m = 4 10 8 m N = 10 6 m 4 10 8 m = 25 8b l = 4 10 8 m 3.4 10 10 m / bp =120bp l is also known as the persistence length, and DNA is sufficiently stiff that it takes about 12 full helical turns (120 bp) before the directions of the two ends of a piece of DNA are uncorrelated over that distance. 15
Extra credit (+i pts) What is the enthalpy of formation of succinic acid, HOOCCH 2 CH 2 COOH, given that the enthalpy of combustion at 298K and 1 atm is -1490 kj mol -1? For reference, the enthalpies of formation of CO 2, H 2 O and NH 3 under these conditions are -393.5, -285.8 and -46.1 kj mol -1, respectively. balanced equation HOOCCH 2 CH 2 COOH + (3/2)O 2! 4CO 2 + 3H 2 O H combustion = -1490 kj mol -1 overall H = sum of the enthalpies of formation of the products minus the reactants 4(-393.5) + 3(-285.8) - H f (succinic acid) 0 = -1490 kj mol -1 H f (succinic acid) = +1490 + 4(-393.5) + 3(-285.8) = -941 kj mol -1 Note: if the balanced equation is doubled, then twice the enthalpy must be used 2 HOOCCH 2 CH 2 COOH + 3O 2! 8CO 2 + 6H 2 O H = 2 H combustion = -2980 kj mol -1 16