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

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Unformatted text preview: 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 = . 36 mol 2 . 9 L = 0 . 124138 M POCl 3 (g) * ) POCl(g) + Cl 2 (g) ini, M . 124138-- Δ, M- x x x eq, M . 124138- x x x K c = [Cl 2 ][POCl] [POCl 3 ] = x 2 . 124138- x = 0 . 3 x 2 = 0 . 0372414- . 3 x x 2 + 0 . 3 x- . 0372414 = 0 x =- . 3 ± p (0 . 3) 2- 4(1)(- . 0372414) 2(1) = 0 . 0944205 or- . 39442 Reject- . 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- . 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) * ) 2F (g) ini, M 2 . 31 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- . 3 x Thompson, Margaret – Homework 4 – Due: Oct 3 2007, 11:00 pm – Inst: James Holcombe 2 4 x 2 + 0 . 3 x- . 693 = 0 x =- . 3 ± p . 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) * ) 2B(g) + 3C(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....
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CH 302 HW4 - Thompson Margaret – Homework 4 – Due Oct 3...

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