HW 4 Solutions - Wright, Jacqueline – Homework 4 – Due:...

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Unformatted text preview: 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 = . 3 mol 3 . 7 L = 0 . 0810811 M POCl 3 (g) * ) POCl(g) + Cl 2 (g) ini, M . 0810811-- Δ, M- x x x eq, M . 0810811- x x x K c = [Cl 2 ][POCl] [POCl 3 ] = x 2 . 0810811- x = 0 . 3 x 2 = 0 . 0243243- . 3 x x 2 + 0 . 3 x- . 0243243 = 0 x =- . 3 ± p (0 . 3) 2- 4(1)(- . 0243243) 2(1) = 0 . 0663893 or- . 366389 Reject- . 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- . 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) * ) 2F (g) ini, M 1 . 77 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- . 3 x Wright, Jacqueline – Homework 4 – Due: Oct 3 2007, 11:00 pm – Inst: James Holcombe 2 4 x 2 + 0 . 3 x- . 531 = 0 x =- . 3 ± p . 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) * ) 2B(g) + 3C(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....
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This note was uploaded on 12/08/2008 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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HW 4 Solutions - Wright, Jacqueline – Homework 4 – Due:...

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