Chem2 hw4s

# Chem2 hw4s - Yoon Seong Meen – Homework 4 – Due Oct 3...

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Unformatted text preview: Yoon, Seong Meen – 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. 0.20 M 2. 0.10 M 3. 0.32 M correct 4. 1.9 M 5. 1.7 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 . 35 moles of POCl 3 are placed in a 2 L container with initial concen- trations 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 . 102277 M 2. final concentration = 0 . 124 M 3. final concentration = 0 . 382 M 4. final concentration = 0 . 207 M 5. final concentration = 0 . 0511387 M cor- rect Explanation: K c = 0.30 V container = 2 L [POCl 3 ] initial = . 35 mol 2 L = 0 . 175 M POCl 3 (g) * ) POCl(g) + Cl 2 (g) ini, M . 175-- Δ, M- x x x eq, M . 175- x x x K c = [Cl 2 ][POCl] [POCl 3 ] = x 2 . 175- x = 0 . 3 x 2 = 0 . 0525- . 3 x x 2 + 0 . 3 x- . 0525 = 0 x =- . 3 ± p (0 . 3) 2- 4(1)(- . 0525) 2(1) = 0 . 123861 or- . 423861 Reject- . 423861 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 . 123861 M and [POCl 3 ] = 0 . 175 M- . 123861 M = 0 . 0511387 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 . 2 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 . 370429 mol. Explanation: n F 2 = 2 . 2 mol V container = 1.0 L [F 2 ] = 2 . 2 mol 1 L = 2 . 2 M F 2 (g) * ) 2F (g) ini, M 2 . 2 Δ, M- x 2 x eq, M 2 . 2- x 2 x K = [F] 2 [F 2 ] = 0 . 3 (2 x ) 2 2 . 2- x = 0 . 3 4 x 2 = 0 . 66- . 3 x Yoon, Seong Meen – Homework 4 – Due: Oct 3 2007, 11:00 pm – Inst: James Holcombe 2 4 x 2 + 0 . 3 x- . 66 = 0 x =- . 3 ± p . 09 + 4(4)(0 . 66) 8 = 0 . 370429 mol / L n F 2 dissociated = 1 . 00 L × (0 . 370429 mol / L) = 0 . 370429 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 . 4 atm and the system is allowed to come to equilibrium, the final pressure is found to be 15 . 05 atm. What is Δ G ◦ r for this reaction? Correct answer:- 7 . 95728 kJ / mol. Explanation: Considering the pressures (in atm), A * ) 2B + 3C ini 10 . 4 Δ- x +2 x +3 x fin 10 . 4- x +2 x +3 x The total pressure is P total = 10 . 4- x + 2 x + 3 x 15 . 05 atm = 10 . 4 + 4 x x = 1 . 1625 atm , so P A = 10 . 4- x = 9 . 2375 atm P B = 2 x = 2 . 325 atm P C = 3 x = 3 . 4875 atm The equilibrium expression is K = P 2 B ·...
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## This note was uploaded on 05/14/2009 for the course CH 53575 taught by Professor Hackert during the Spring '09 term at University of Texas.

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Chem2 hw4s - Yoon Seong Meen – Homework 4 – Due Oct 3...

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