1178564629Xuj - Anderson, Kiel Homework 4 Due: Feb 23 2007,...

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Anderson, Kiel – Homework 4 – Due: Feb 23 2007, midnight – Inst: McCord 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 1) 10 points A mixture o± PCl 5 (g) and Cl 2 (g) is placed into a closed container. At equilibrium it is ±ound that [PCl 5 ] = 0 . 75 M, [Cl 2 ] = 0 . 41 M and [PCl 3 ] = 0 . 09 M. PCl 5 * ) PCl 3 + Cl 2 What is the value o± K c ±or the reaction? 1. 52 2. 0 . 0492 correct 3. 0 . 1476 4. 0 . 0984 5. 0 . 0246 Explanation: [PCl 5 ] = 0 . 75 M [Cl 2 ] = 0 . 41 M [PCl 3 ] = 0 . 09 M K c = [Cl 2 ] [PCl 3 ] [PCl 5 ] = (0 . 41 M)(0 . 09 M) 0 . 75 M = 0 . 0492 M 002 (part 1 o± 1) 10 points At 1000 K the equilibrium pressure o± the three gases in one mixture 2 SO 2 (g) + O 2 (g) * ) 2 SO 3 (g) were ±ound to be 0.562 atm SO 2 , 0.101 atm O 2 , and 0.332 atm SO 3 . Calculate the value K p ±or the reaction as written. 1. 2.64 atm 2. 5.83 atm 3. 0.289 atm 4. 0.171 atm 5. 3.46 atm correct Explanation: P SO 3 = 0 . 332 atm P SO 2 = 0 . 562 atm P O 2 = 0 . 101 atm K p = P 2 SO 3 P 2 SO 2 · P O 2 = (0 . 332) 2 (0 . 562) 2 (0 . 101) = 3 . 46 003 (part 1 o± 1) 10 points The system H 2 (g) + I 2 (g) * ) 2 HI(g) is at equilibrium at a fxed temperature with a partial pressure o± H 2 o± 0.200 atm, a par- tial pressure o± I 2 o± 0.200 atm, and a partial pressure o± HI o± 0.100 atm. An additional 0 . 23 atm pressure o± HI is admitted to the container, and it is allowed to come to equi- librium again. What is the new partial pres- sure o± HI? Correct answer: 0 . 146 atm. Explanation: P I 2 = P H 2 = 0 . 2 atm P HI = 0 . 1 atm H 2 (g) + I 2 (g) * ) 2 HI(g) K p = ( P HI ) 2 P H 2 · P I 2 = (0 . 1 atm) 2 (0 . 2 atm) (0 . 2 atm) = 0 . 25 new P HI = (0 . 1 + 0 . 23) atm = 0 . 33 atm Adding the products shi±ts the equilibrium to the le±t. H 2 (g) + I 2 (g) * ) 2 HI(g) ini, atm 0.200 0.200 0 . 33 Δ, atm + x + x - 2 x eq, atm 0 . 200 + x 0 . 200 + x 0 . 33 - 2 x (0 . 33 - 2 x ) 2 (0 . 2 + x ) 2 = 0 . 25 0 . 33 - 2 x 0 . 2 + x = 0 . 25 0 . 33 - 2 x = (0 . 5)(0 . 2) + 0 . 5 x 2 . 5 x = 0 . 23 x = 0 . 092
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Anderson, Kiel – Homework 4 – Due: Feb 23 2007, midnight – Inst: McCord 2 P HI = 0 . 33 atm - (2) (0 . 092 atm) = 0 . 146 atm 004 (part 1 of 1) 10 points A 2.000 liter vessel is ±lled with 4.000 moles of SO 3 and 6.000 moles of O 2 . When the reaction 2 SO 3 (gas) * ) 2 SO 2 (gas) + O 2 (gas) comes to equilibrium a measurement shows that only 1.000 mole of SO 3 remains. How many moles of O 2 are in the vessel at equilib- rium? 1. 7.000 mol 2. 12.000 mol 3. None of these is correct. 4. 7.500 mol correct 5. 3.750 mol Explanation: Initially, [SO 3 ] = 4 mol 2 L = 2 M [O 2 ] = 6 mol 2 L = 3 M 2 SO 3 (g) * ) 2 SO 2 (g) + O 2 (g) ini, M 2 0 3 Δ, M - 2 x 2 x x eq, M 2 - 2 x 2 x 3 + x At equilibrium, [SO 3 ] eq = 1 mol 2 L = 0 . 5 M, so 2 - 2 x = 0 . 5 - 2 x = - 1 . 5 x = 0 . 75 Thus [O 2 ] = 3 + x = 3 . 75 M mol O 2 = (3 . 75 mol / L) (2 L) = 7 . 5 mol . 005
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This note was uploaded on 03/20/2008 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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1178564629Xuj - Anderson, Kiel Homework 4 Due: Feb 23 2007,...

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