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# 5 050 correct version 324 exam 2 holcombe 51395 4

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Version 324 – EXAM 2 – holcombe – (51395) 4 Explanation: The expression for K c here is K c = [H 2 ] [CO] [HCHO] , so we need to know the equilibrium concen- trations of each of the species. Since each is 1 . 0 mol 2 . 0 L = 0 . 50 M, then K c = [0 . 50] [0 . 50] [0 . 50] = 0.50 012(part2of3)3.2points 2.0 moles of HCHO and 1.0 mol of CO are then added to this system. b) Which of the following statements about this reaction is now true? 1. The reaction mixture is not at equilib- rium, but will move toward equilibrium by using up more HCHO. correct 2. The reaction mixture is not at equilib- rium, but no further reaction will occur. 3. The forward rate of this reaction is the same as the reverse rate at these new concen- trations. 4. The reaction mixture is not at equilib- rium, but will move toward equilibrium by forming more HCHO. 5. The reaction mixture remains at equilib- rium. Explanation: The new concentrations for the species are [HCHO] = 3 . 0 mol 2 . 0 L = 1 . 5 M [H 2 ] = 0.50 M [CO] = 2 . 0 mol 2 . 0 L = 1.0 M Solving for Q , we have Q = [H 2 ] [CO] [HCHO] = (0 . 5) (1 . 0) 1 . 5 = 0 . 33 Since Q is less than K c , the reaction will pro- ceed forward, which means that HCHO will be consumed in order to form more products. 013(part3of3)3.2points c) What is K p for this reaction at 600 C? 1. K p cannot be determined here without the partial pressures of HCHO, H 2 , and CO. 2. 1.0 3. 0.014 4. 72 5. 0.0070 6. 0.50 7. 0.028 8. 2.0 9. 36 correct 10. 140 Explanation: K p = K c ( RT ) Δ n where Δ n = ( n gas products - n gas reactants ). Here Δ n must be 1, as there are two moles of gaseous products but only 1 mol of gaseous reactants. Since the temperature is 873 K and R is 0.08206 L · atm K · mol , K p = 0.5 [(0.08206) (873)] 1 = 36 014 3.2points Predict whether the equilibria I) N 2 O 4 (g) 2 NO 2 (g) , Δ H = +57 kJ II) CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) , Δ H = - 41 kJ will shift toward products or reactants with a temperature increase. 1. Both I and II shift toward reactants. 2. Both I and II shift toward products. 3. I shifts toward reactants and II shifts toward products. 4. Unable to determine
Version 324 – EXAM 2 – holcombe – (51395) 5 5. I shifts toward products and II shifts to- ward reactants. correct Explanation: If a reaction is exothermic, raising the tem- perature will tend to shift the reaction toward reactants; whereas if the reaction is endother- mic, a shift toward products will be observed. Reaction I is endothermic (+Δ H ); raising the temperature should favor the formation of products. Reaction II is exothermic; raising the temperature should favor the formation of reactants. 015 3.2points Nitric oxide (NO) is a toxic chemical produced in automobile engines. O 2 (g) + N 2 (g) ←→ 2 NO(g) Suppose that at 0 C, K = 0 . 5 and at 10 C, K = 4. Δ H of this reaction is (posi- tive/negative) and it would be best to run an engine as (hot/cold) as possible to reduce emissions of NO. 1. negative; hot 2. positive; hot 3. positive; cold correct 4. negative; cold Explanation: The direct proportionality between temper- ature and K demonstrates that Δ H is posi- tive and to prevent the reaction, the engine should be run as cold as possible.

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• Fall '07
• Holcombe

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