HW 4 Solutions - Wright Jacqueline Homework 4 Due Oct 3...

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Wright, Jacqueline – Homework 4 – Due: Oct 3 2007, 11:00 pm – Inst: James Holcombe 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points K c = 9 . 7 at 900 K for the reaction NH 3 (g) + H 2 S(g) NH 4 HS(s) . If the initial concentrations of NH 3 (g) and H 2 S(g) are 2.0 M, what is the equilibrium concentration of NH 3 (g)? 1. 0.10 M 2. 0.20 M 3. 1.9 M 4. 1.7 M 5. 0.32 M correct Explanation: 002 (part 1 of 1) 10 points For the reaction POCl 3 (g) * ) POCl(g) + Cl 2 (g) K c = 0 . 30. An initial 0 . 3 moles of POCl 3 are placed in a 3 . 7 L container with initial concentrations of POCl and Cl 2 equal to zero. What is the final concentration of POCl 3 ? ( Note: You must solve a quadratic equation.) 1. final concentration = 0 . 0146918 M cor- rect 2. final concentration = 0 . 185 M 3. final concentration = 0 . 066 M 4. final concentration = 0 . 0293836 M 5. final concentration = 0 . 266081 M Explanation: K c = 0.30 V container = 3 . 7 L [POCl 3 ] initial = 0 . 3 mol 3 . 7 L = 0 . 0810811 M POCl 3 (g) * ) POCl(g) + Cl 2 (g) ini, M 0 . 0810811 - - Δ, M - x x x eq, M 0 . 0810811 - x x x K c = [Cl 2 ] [POCl] [POCl 3 ] = x 2 0 . 0810811 - x = 0 . 3 x 2 = 0 . 0243243 - 0 . 3 x x 2 + 0 . 3 x - 0 . 0243243 = 0 x = - 0 . 3 ± p (0 . 3) 2 - 4 (1) ( - 0 . 0243243) 2 (1) = 0 . 0663893 or - 0 . 366389 Reject - 0 . 366389 as x because it leads to negative concentrations for POCl and Cl 2 and a concentration larger that the orig- inal concentration for POCl 3 . Therefore x = 0 . 0663893 M and [POCl 3 ] = 0 . 0810811 M - 0 . 0663893 M = 0 . 0146918 M 003 (part 1 of 1) 10 points The equilibrium constant for thermal dissoci- ation of F 2 into atoms is 0.300. If 1 . 77 moles of F 2 are placed in a 1.00 liter container, how many moles of F 2 have dissociated at equilib- rium? Correct answer: 0 . 328773 mol. Explanation: n F 2 = 1 . 77 mol V container = 1.0 L [F 2 ] = 1 . 77 mol 1 L = 1 . 77 M F 2 (g) * ) 2 F (g) ini, M 1 . 77 0 Δ, M - x 2 x eq, M 1 . 77 - x 2 x K = [F] 2 [F 2 ] = 0 . 3 (2 x ) 2 1 . 77 - x = 0 . 3 4 x 2 = 0 . 531 - 0 . 3 x
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Wright, Jacqueline – Homework 4 – Due: Oct 3 2007, 11:00 pm – Inst: James Holcombe 2 4 x 2 + 0 . 3 x - 0 . 531 = 0 x = - 0 . 3 ± p 0 . 09 + 4(4)(0 . 531) 8 = 0 . 328773 mol / L n F 2 dissociated = 1 . 00 L × (0 . 328773 mol / L) = 0 . 328773 mol 004 (part 1 of 1) 10 points Consider the equilibrium A(g) * ) 2 B(g) + 3 C(g) at 25 C. When A is loaded into a cylinder at 10 atm and the system is allowed to come to equilibrium, the final pressure is found to be 16 . 69 atm. What is Δ G r for this reaction? Correct answer: - 12 . 7203 kJ / mol. Explanation: Considering the pressures (in atm), A * ) 2 B + 3 C ini 10 0 0 Δ - x +2 x +3 x fin 10 - x +2 x +3 x The total pressure is P total = 10 - x + 2 x + 3 x 16 . 69 atm = 10 + 4 x x = 1 . 6725 atm , so P A = 10 - x = 8 . 3275 atm P B = 2 x = 3 . 345 atm P C = 3 x = 5 . 0175 atm The equilibrium expression is K = P 2 B · P 3 C P A = (3 . 345) 2 (5 . 0175) 3 8 . 3275 = 169 . 723 and the energy is Δ G = - R T ln K = - 8 . 314 J K · mol (298 K) × ln (169 . 723) = - 12720 . 3 J / mol = - 12 . 7203 kJ / mol . 005 (part 1 of 1) 10 points When 0 . 0172 mol HI is heated to 500 K in a 2 L sealed container, the resulting equilibrium mixture contains 1 . 9 g of HI. Calculate K c for the decomposition reaction 2 HI(g) * ) H 2 (g) + I 2 (g) .
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