CH 302 HW4

# CH 302 HW4 - Thompson Margaret Homework 4 Due Oct 3 2007...

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Thompson, Margaret – 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. 0.32 M correct 3. 0.20 M 4. 1.7 M 5. 1.9 M 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 . 36 moles of POCl 3 are placed in a 2 . 9 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 . 211 M 2. final concentration = 0 . 0297174 M cor- rect 3. final concentration = 0 . 335138 M 4. final concentration = 0 . 094 M 5. final concentration = 0 . 0594349 M Explanation: K c = 0.30 V container = 2 . 9 L [POCl 3 ] initial = 0 . 36 mol 2 . 9 L = 0 . 124138 M POCl 3 (g) * ) POCl(g) + Cl 2 (g) ini, M 0 . 124138 - - Δ, M - x x x eq, M 0 . 124138 - x x x K c = [Cl 2 ] [POCl] [POCl 3 ] = x 2 0 . 124138 - x = 0 . 3 x 2 = 0 . 0372414 - 0 . 3 x x 2 + 0 . 3 x - 0 . 0372414 = 0 x = - 0 . 3 ± p (0 . 3) 2 - 4 (1) ( - 0 . 0372414) 2 (1) = 0 . 0944205 or - 0 . 39442 Reject - 0 . 39442 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 . 0944205 M and [POCl 3 ] = 0 . 124138 M - 0 . 0944205 M = 0 . 0297174 M 003 (part 1 of 1) 10 points The equilibrium constant for thermal dissoci- ation of F 2 into atoms is 0.300. If 2 . 31 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 . 380419 mol. Explanation: n F 2 = 2 . 31 mol V container = 1.0 L [F 2 ] = 2 . 31 mol 1 L = 2 . 31 M F 2 (g) * ) 2 F (g) ini, M 2 . 31 0 Δ, M - x 2 x eq, M 2 . 31 - x 2 x K = [F] 2 [F 2 ] = 0 . 3 (2 x ) 2 2 . 31 - x = 0 . 3 4 x 2 = 0 . 693 - 0 . 3 x

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Thompson, Margaret – Homework 4 – Due: Oct 3 2007, 11:00 pm – Inst: James Holcombe 2 4 x 2 + 0 . 3 x - 0 . 693 = 0 x = - 0 . 3 ± p 0 . 09 + 4(4)(0 . 693) 8 = 0 . 380419 mol / L n F 2 dissociated = 1 . 00 L × (0 . 380419 mol / L) = 0 . 380419 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 . 6 atm and the system is allowed to come to equilibrium, the final pressure is found to be 15 . 87 atm. What is Δ G r for this reaction? Correct answer: - 9 . 49574 kJ / mol. Explanation: Considering the pressures (in atm), A * ) 2 B + 3 C ini 10 . 6 0 0 Δ - x +2 x +3 x fin 10 . 6 - x +2 x +3 x The total pressure is P total = 10 . 6 - x + 2 x + 3 x 15 . 87 atm = 10 . 6 + 4 x x = 1 . 3175 atm , so P A = 10 . 6 - x = 9 . 2825 atm P B = 2 x = 2 . 635 atm P C = 3 x = 3 . 9525 atm The equilibrium expression is K = P 2 B · P 3 C P A = (2 . 635) 2 (3 . 9525) 3 9 . 2825 = 46 . 1862 and the energy is Δ G = - R T ln K = - 8 . 314 J K · mol (298 K) × ln (46 . 1862) = - 9495 . 74 J / mol = - 9 . 49574 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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