Homework 04 Solutions

# Homework 04 Solutions - Gadhia, Tejas Homework 4 Due: Oct 3...

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Unformatted text preview: Gadhia, Tejas 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. 1.9 M 2. 1.7 M 3. 0.20 M 4. 0.10 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 . 38 moles of POCl 3 are placed in a 2 . 2 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 . 100227 M 2. final concentration = 0 . 0501137 M cor- rect 3. final concentration = 0 . 391727 M 4. final concentration = 0 . 219 M 5. final concentration = 0 . 123 M Explanation: K c = 0.30 V container = 2 . 2 L [POCl 3 ] initial = . 38 mol 2 . 2 L = 0 . 172727 M POCl 3 (g) * ) POCl(g) + Cl 2 (g) ini, M . 172727-- , M- x x x eq, M . 172727- x x x K c = [Cl 2 ][POCl] [POCl 3 ] = x 2 . 172727- x = 0 . 3 x 2 = 0 . 0518182- . 3 x x 2 + 0 . 3 x- . 0518182 = 0 x =- . 3 p (0 . 3) 2- 4(1)(- . 0518182) 2(1) = 0 . 122614 or- . 422614 Reject- . 422614 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 . 122614 M and [POCl 3 ] = 0 . 172727 M- . 122614 M = 0 . 0501137 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 . 18 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 . 262344 mol. Explanation: n F 2 = 1 . 18 mol V container = 1.0 L [F 2 ] = 1 . 18 mol 1 L = 1 . 18 M F 2 (g) * ) 2F (g) ini, M 1 . 18 , M- x 2 x eq, M 1 . 18- x 2 x K = [F] 2 [F 2 ] = 0 . 3 (2 x ) 2 1 . 18- x = 0 . 3 4 x 2 = 0 . 354- . 3 x Gadhia, Tejas Homework 4 Due: Oct 3 2007, 11:00 pm Inst: James Holcombe 2 4 x 2 + 0 . 3 x- . 354 = 0 x =- . 3 p . 09 + 4(4)(0 . 354) 8 = 0 . 262344 mol / L n F 2 dissociated = 1 . 00 L (0 . 262344 mol / L) = 0 . 262344 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 . 7 atm and the system is allowed to come to equilibrium, the final pressure is found to be 15 . 28 atm. What is G r for this reaction? Correct answer:- 7 . 68565 kJ / mol....
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## This note was uploaded on 03/19/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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Homework 04 Solutions - Gadhia, Tejas Homework 4 Due: Oct 3...

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