CHM 3400 Fundamentals of Physical Chemistry Second Hour Exam March 8, 2017 There are five problems on the exam. Do all of the problems. Show your work R = 0.08206 L atm/mole K N A = 6.022 x 10 23 R = 0.08314 L bar/mole K 1 L atm = 101.3 J R = 8.314 J/mole K 1 atm = 1.013 bar = 1.013 x 10 5 N/m 2 1 atm = 760 torr 1. (20 points) Thermodynamic data for several pure chemical substances are given below (at T =298.0 K), and may be of use in doing the following problem. Substance H f (kj/mol) G f (kj/mol) S (J/mol K) C p,m (J/mol K) CuO(s) - 157.3-129.7 42.63 42.30 Cu 2 O(s) - 168.6-146.0 93.14 63.64 O 2 (g) 0.0 0.0 205.14 29.36 a) What are the values for G rxn, H rxn, and S rxn (at T = 298.0 K) for the following reaction? 2 Cu 2 O(s) + O 2 (g) 4 CuO(s) b) What is the value of S for CuO(s) at T = 373. K? 2. (16 points) Data for the vapor pressure of a pure liquid at several temperatures are given below. A plot of ln(p) vs 1/T is also given, along with the formula for the line that best fits the data in the plot. T (K) p (torr) T (K) p (torr) 300.0 75. 330.0 260. 315.0 148. 340.0 372. Plot of ln(p) vs 1/T 6.2 ln(p) 5.2 4.2 0.0028 0.0029 0.003 0.0031 0.0032 0.0033 0.0034 0.0035 1/T (K-1) ln(p) = ( - 4071. K) + 17.90 T Based on the above information, answer the following questions. a) What is H vap, the enthalpy of vaporization for the liquid? Give your final answer in units of kj/mol. b) What is T vap, the normal boiling point for the liquid? Give your final answer in units of C.
3. (18 points) An insulated container initially holds 25.0 g of liquid water (H 2 O, M = 18.02 g/mol) at a temperature T = 20.0 C. An additional 40.0 g of liquid water at a temperature T = 50.0 C is added to the container, and the water in the container is allowed to come to equilibrium. Note that the constant pressure molar heat capacity of H 2 O( ) is C p,m = 75.3 J/mol K, and is constant for the temperature range of this problem. Find the following. a) The final temperature of the water inside the container. Give your final answer in C. b) The value for S for the process. Give your final answer in units of J/K. 4. (20 points) The vapor pressure of pure water (H 2 O, M = 18.02 g/mol) at T = 30.0 C is 31.824 torr. a) When 1.084 g of a nonvolatile and nonionizing solute is dissolved in 200.0 g of water, the vapor pressure of the solution, again measured at 30.0 C, is 0.037 torr lower than the vapor pressure of pure water at the same temperature. What is M, the molecular mass of the solute? b) What would the concentration of a nonvolative solute in water need to be for it to generate an osmotic pressure of 31.824 torr at T = 30.0 C? 5. (26 points) A phase diagram for two volatile liquids A and B is given below, and may be used to answer the following questions. Note that in this problem p = 1.00atm. a) What are the values for T A (normal boiling point of A) and T B (normal boiling point of B). b) Does the above solution form any azeotropes? If your answer is yes, give the location (temperature and mole fraction) of each azeotrope that forms. c) A closed system contains a solution of A and B at low temperature, with n A = 3.00 moles and n B = 2.00 moles (and so Z A = 0.600). The solution is slowly heated. At what temperature (in C) will the solution first begin to boil. d) We continue heating the solution until we reach a temperature T = 70.0 C. At this point, X A = 0.308 and Y A = 0.724. How many moles of A are in the vapor phase at this temperature?
Solutions. 1) a) G rxn = [ 4 G f (CuO(s)) ] [ 2 G f (Cu 2 O(s)) + G f (O 2 (g)) ] = [4 (- 129.7) ] [ 2 (- 146.0) + 1 (0.0) ] = - 226.8 kj/mol H rxn = [ 4 H f (CuO(s)) ] [ 2 H f (Cu 2 O(s)) + H f(o 2 (g)) ] = [ 4 (- 157.3) ] [ 2 (- 168.6) + 1 (0.0) ] = - 292.0 kj/mol S rxn = [ 4 S (CuO(s)) ] [ 2 S (Cu 2 O(s)) + S (O 2 (g)) ] = [ 4 (42.63) ] [ 2 (93.14) + 1 (205.14) ] = - 220.90 J/mol K b) S (CuO(s, 373. K)) = S (CuO(s, 298 K)) + 298 373 (C p,m (CuO(s))/T) dt If we assume C p,m is constant over this temperature range, then S (CuO(s, 373. K)) = S (CuO(s, 298 K)) + C p,m (CuO(s)) ln(373/298) = 42.63 J/mol K + (42.30 J/mol K) ln(373/298) = 42.63 J/mol K + 9.50 J/mol K = 52.13 J/mol K 2) a) H vap = - (slope) R = - ( - 4071 K) (8.3145 J/mol K) = 33.85 kj/mol b) ln(p) = ( - 4071. K) + 17.90 T T = ( - 4071. K) (ln(p) 17.90 At the normal boiling point p = 760. Torr, and so T = ( - 4071. K) = 361.3 K = 88.2 C ln(760) 17.90 3) a) We can think of the process occurring as follows Initial Final 25.0 g H 2 O, T = 20.0 C 65.0 g H 2 O, T = T f 40.0 g H 2 O, T = 50.0 C The container is insulated, and so q = 0. That means the heat gained by the 25.0 g of H 2 O at 20.0 C is equal in magnitude to the heat lost from the 40.0 g of H 2 O at T = 50.0 C. Since the amounts of water are given in grams, it is convenient (but not necessary) to use the specific heat of water (s, units of J/g K) rather than the molar heat capacity. s = (75.3 J/mol K) (1. mol/18.02 g) = 4.18 J/g K Finally, since the size of a degree K and a degree C are the same, we can also say s = 4.18 J/g K = 4.18 J/g C
We now proceed to do the problem. q = 0 = (25.0) s (T f - 20.0) + (40.0) s (T f - 50.0) (we have left out the units for convenience) We can divide both sides of this equation by s, to get 0 = (25.0) (T f - 20.0) + (40.0) (T f - 50.0) 0 = 25.0 T f - 500.0 + 40.0 T f - 2000.0 = 65.0 T f - 2500.0 T f = 2500.0 = 38.46 C 65.0 b) Divide the process into two steps step 1 - heat the 25.0 g of water from 20.0 C to 38.46 C step 2 - cool the 40.0 g of water from 50.0 C to 38.46 C We can use the result from problem problem 1b, which also involved a constant pressure change of temperature for a substance with a constant value for heat capacity. So S 1 = (25.0 g) (4.18 J/g K) ln(311.61 K/293.15 K) = 6.382 J/K S 2 = (40.0 g) (4.18 J/g K) ln(311.61 K/323.15 K) = - 6.079 J/K S = S 1 + S 2 = 6.382 J/K + (- 6.079 J/K) = 0.30 J/K 4) a) M = m/n. Since we know m, we need to find n (the number of moles of solute). For vapor pressure lowering we may say p = X B p A * where X B is the mole fraction of solute particles and p A * is the vapor pressure of the pure solvent. X B = p = 0.037 torr = 0.00116 p A * 31.824 torr But X B = n B n B (n A + n B ) n A So But n B = X B n A n A = 200.0 g 1 mol = 11.10 mol 18.02 g n B = (0.00116)(11.10 mol) = 0.0129 mol M B = 1.084 g = 84. g/mol 0.0129 mol
b) For osmotic pressure, = [B]RT So [B] = = (31.824 torr)(1 atm/760 torr) = 1.68 x 10-3 mol/l RT (0.08206 L atm/mol K)(303.2 K) 5) a) T A * = 60.0 C T B * = 81.5 C b) There are no azeotropes present c) For Z A = 0.600, boiling begins at T = 64.0 C d) From the lever rule n = Y A - Z A = (0.724-0.600) = 0.425 n g Z A - X A (0.600-0.308) And so n = 0.425 n g But n + n g = (0.425 n g ) + n g = 1.425 n g = n total = 5.00 mol n g = 5.00 mol = 3.51 mol 1.425 And so the number of moles of A in the vapor phase is n A,g = Y A n g = (0.724)(3.51 mol) = 2.54 mol