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# hw6solution_pdf - casey(rmc2555 Homework 6 holcombe(51395...

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casey (rmc2555) – Homework 6 – holcombe – (51395) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 3.3points All of the following may shift the position of a reaction at equilibrium EXCEPT 1. a temperature decrease. 2. a concentration change. 3. a pressure change. 4. a homogeneous catalyst. correct 5. a temperature increase. Explanation: 002 3.3points For the following reaction HgO ( s ) Hg ( l ) + O 2 ( g ) A sample of solid HgO is heated to a tem- perature at which it is in equilibrium with liquid Hg and O 2 gas. At this temperature Δ G = 0 kJ mol 1 . What do you know about the O 2 gas at equilibrium? 1. the partial pressure of O 2 is 1 Torr 2. there is no way to know without the initial mass of the HgO solid 3. the partial pressure of O 2 is 1 atm cor- rect 4. the concentration of O 2 is 1 M 5. the concentration of O 2 is 1 ppm Explanation: If Δ G = 0 kJ mol 1 , then K =1. For this reaction K = P O 2 as measured relative to the standard pressure of 1 atm. Thus K = 1 yields P O 2 =1 atm 003 3.3points Knowing only the standard Δ H rxn and Δ S values of each of the different reactions I) Δ H rxn = 10 J/mol, Δ S = 1000 J/K/mol II) Δ H rxn = 10 J/mol, Δ S = 1000 J/K/mol III) Δ H rxn = 20 J/mol, Δ S = 1000 J/K/mol IV) Δ H rxn = 40 J/mol, Δ S = 2000 J/K/mol which will have the highest K eq value? As- sume all reactions are run at 293 K. 1. III only correct 2. I only 3. II only 4. I, II and III have equal highest values of K . 5. All four have equal values of K . 6. IV only 7. I and II have equal highest values of K . Explanation: 004 3.3points An endothermic reaction is most likely to have a small equilibrium constant if Δ S r is small. 1. True correct 2. False Explanation: 005 3.3points The equilibrium constant K for the reaction 2 HgO(s) 2 Hg( ) + O 2 (g) is 1 . 2 × 10 30 . Calculate K for the reaction 1 2 O 2 (g) + Hg( ) HgO(s) . 1. 4 . 2 × 10 29 2. 1 . 1 × 10 15
casey (rmc2555) – Homework 6 – holcombe – (51395) 2 3. 9 . 1 × 10 14 correct 4. 1 . 1 × 10 15 5. 8 . 3 × 10 29 Explanation: 006 3.3points The figure represents a reaction at 298 K. G A B C D E rxn progress Which statement is true? 1. The reactants possess less free energy than the products. 2. At point B, Q is less than K. correct 3. The Δ G of reaction is zero at point C. 4. K is less than 1. Explanation: Q is less than K at point B because more products are needed to get to equilibrium point C. K is greater than 1 because Δ G is negative (point E is lower free energy than point A). Point C identifies where Δ G is zero, whereas the difference between points E and A identify Δ G . Because point E is below point A, we know the products have lower free energy than the reactants. 007 3.3points For a pure solid or liquid, the molar free en- ergy always has its standard value.

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• Fall '07
• Holcombe

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