HW_ANSWERS_CH13 - ANSWERS TO CHAPTER 13 13.28 (a) (1) and...

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ANSWERS TO CHAPTER 13 13.28 (a) (1) and (3) because the number of A and B’s are the same in the third and fourth Box. (b) Kc = [A]/[B] = 6/4 (c) Because the same number of molecules appear on both sides of the equation, the volume terms in K c all cancel. Therefore, we can calculate K c , without including the volume. 13.30 (a) Only reaction (3), Kc = 2, is at equilibrium (b) Qc = 15 for reaction (1). Because Qc > Kc the reaction will go in the reverse Reaction to reach equilibrium. Qc = 1/3 for reaction (2). Because Qc < Kc, the reaction will go in the forward direction to reach equilibrium.
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13.39 (a) (b) (c) 13.47 The two reactions are the reverse of each other. Kp(reverse) = 1/Kp(forward) = 1.99 x 10 -2 13.52 (a) (b) See Worked Example 13.9 and 13.10 on page 545 of textbook (c) x = 0.65 mol Kc = (0.65) 2 /(.35) 2 = 3.4
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13.59 (a) Kc = [H 2 O] 3 /[H 2 ] 3 Kp = (P H2O ) 3 /(P H2 ) 3 (b) Kc = 1/[Ag+][Cl - ] (c) Kc = [HCl]/[H 2 O] 3 Kp = (P HCl ) 6 /(P H2O ) 3 (d) Kc = [CO 2 ] Kp = P
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HW_ANSWERS_CH13 - ANSWERS TO CHAPTER 13 13.28 (a) (1) and...

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