from_F02_FED_key

From_F02_FED_key - KEY Chemistry 111 Page 1 out of 8 Final Exam D Fall 2002 1 For the reaction system CoO(s H2(g CO(s H2O(g at 125C its K = 250 The

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KEY Chemistry 111 Page 1 out of 8 Final Exam D / Fall 2002 1. For the reaction system, CoO (s) + H 2 (g) CO (s) + H 2 O (g), at 125°C, its K = 250. The equilibrium constant expression for Kc is: Of course, there is a typo here – the CO (s) on the right should be CO. But since it's a solid, it doesn't matter! Recall that the activities of pure solids and liquids are 1, and so they don't appear in the equilibrium constant … K C = [H 2 O] / [H 2 ] 2. The following reaction has a Kc = 1.20 × 10 +8 . 2 A (g) + B(g) + C (s) D 2 (l) + F(g) Exactly one mol each of A, B, C, and D 2 are placed in a 1.00 L flask. What is the equilibrium concentration of gas A? Hmmm … this is one where we need to start with the initial conditions, figure out what Q is, and then go to our equilibrium position … K C = 2 [F] [A] [B] = 1.20 x 10 +8 Our initial conditions give us Q = 0, and so the reaction will head towards the right. But the K is very, very large, so we want to do our trick of going completely to the right and then come back to equilibrium. 0 1 infinity K C
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KEY Chemistry 111 Page 2 out of 8 Final Exam D / Fall 2002 A is the limiting reactant. 2 A (g) + B (g) + C(s) D 2 (l) + F (g) initially 1.0 M 1.0 M 1.0 M 1.0 M 0 M 100% to R 0 1.0 – 0.5 = 0.5 who cares? who cares? 0 + 0.5 = 0.5 at equil. 0 + 2x 0.5 + x --- --- 0.5 – x The reason we don't care about C and D 2 is the same as for problem 1 – their activities are 1.0 and so they don't appear in the equilibrium constant. K C = 2 [F] [A] [B] = 22 2 (0.5 - x) (0.5) 1 (2x) (0.5+ x) (2x) (0.5) (2x) ≈= = 1.20 x 10 +8 x = 4.56 x 10 -5 [A] = 2x = 9.13 x 10 -5 For questions 3-6 , consider the following reaction, initially at equilibrium, which has a given value of Δ H° = + 33.3 kJ/mol. 2 A (g) + B (s) C (g) + 3 D (g) 3. What would happen if the volume of the container were to be increased? a) No change b) Shift to the left. c) Shift to the right. d) Can’t tell. e) None of the above. Increasing the volume will cause the equilibrium to shift towards the side with the most number of moles of gases – in this case, to the right. In terms of Q, the partial pressures of the gases on the product side will decrease more than those on the right, so Q < K P and the reaction will shift to the right.
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This note was uploaded on 02/08/2010 for the course CHEM 111 taught by Professor Kenney during the Spring '08 term at Case Western.

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From_F02_FED_key - KEY Chemistry 111 Page 1 out of 8 Final Exam D Fall 2002 1 For the reaction system CoO(s H2(g CO(s H2O(g at 125C its K = 250 The

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