# ch13 - 13 13.1(a Kc = Chemical Equilibrium[SO3]2[SO2]2[O2...

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13 Chemical Equilibrium 13.1 (a) K c = ] O [ ] SO [ ] SO [ 2 2 2 2 3 (b) K c = ] SO [ ] O [ ] SO [ 2 3 2 2 2 13.2 (a) K c = ) 10 x (3.5 ) 10 x (3.0 ) 10 x (5.0 = ] O [ ] SO [ ] SO [ 3 _ 2 3 _ 2 2 _ 2 2 2 2 3 = 7.9 x 10 4 (b) K c = ) 10 x (5.0 ) 10 x (3.5 ) 10 x (3.0 = ] SO [ ] O [ ] SO [ 2 2 _ 3 _ 2 3 _ 2 3 2 2 2 = 1.3 x 10 5 13.3 (a) ] O H C [ ] O H C ][ H [ = K 3 6 3 _ 3 5 3 + c (b) = 0365)] (0.100)(0. _ [0.100 ] .0365) [(0.100)(0 = K 2 c 1.38 x 10 4 13.4 From (1), = (1)(2) (1)(2) = ] B [A][ [AB][B] = K 2 c 1 For a mixture to be at equilibrium, ] B [A][ [AB][B] 2 must be equal to 1. For (2), = (2)(1) (2)(1) = ] B [A][ [AB][B] 2 1. This mixture is at equilibrium. For (3), = (4)(2) (1)(1) = ] B [A][ [AB][B] 2 0.125. This mixture is not at equilibrium. For (4), 2 2 [AB][B] (2)( ) = = [A][ ] (4)(1) B 1.0 This mixture is at equilibrium. 13.5 K p = 0) (1.31)(10. 3) (6.12)(20. = ) P )( P ( ) P )( P ( O H CO H CO 2 2 2 = 9.48 13.6 2 NO(g) + O 2 _ 2 NO 2 (g);Δn = 2 3 = 1 K p = K c (RT) Δn , K c = K p (1/RT) Δn at 500 K: K p = (6.9 x 10 5 )[(0.082 06)(500)] 1 = 1.7 x 10 4 at 1000 K: K c = (1.3 x 10 2 ) 06)(1000) (0.082 1 1 _ = 1.1 13.7 (a) K c = ] O H [ ] H [ 3 2 3 2 , K p = ) P ( ) P ( 3 O H 3 H 2 2 , Δn = (3) (3) = 0 and K p = K c 75

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(b) K c = [H 2 ] 2 [O 2 ], K p = ) P ( ) P ( O 2 H 2 2 , Δn = (3) (0) = 3 and K p = K c (RT) 3 (c) K c = ] H ][ SiCl [ ] [HCl 2 2 4 4 , K p = ) P )( P ( ) P ( 2 H SiCl 4 HCl 2 4 , Δn = (4) (3) = 1 and K p = K c (RT) (d) K c = ] Cl ][ Hg [ 1 2 _ + 2 2 13.8 K c = 1.2 x 10 42 . Because K c is very small, the equilibrium mixture contains mostly H 2 molecules. H is in periodic group 1A. A very small value of K c is consistent with strong bonding between 2 H atoms, each with one valence electron. 13.9 The container volume of 5.0 L must be included to calculate molar concentrations. (a) Q c = L) 5.0 mol/ (1.0 ) L 5.0 mol/ (0.060 ) L 5.0 mol/ (0.80 = ] O [ ] [NO ] NO [ 2 2 t 2 2 t 2 t 2 = 890 Because Q c < K c , the reaction is not at equilibrium. The reaction will proceed to the right to reach equilibrium. (b) Q c = L) 5.0 mol/ (0.20 ) L 5.0 mol/ 10 x (5.0 ) L 5.0 mol/ (4.0 = ] O [ ] [NO ] NO [ 2 3 _ 2 t 2 2 t 2 t 2 = 1.6 x 10 7 Because Q c > K c , the reaction is not at equilibrium. The reaction will proceed to the left to reach equilibrium. 13.10 4 = ] B ][ A [ ] [AB = K 2 2 2 c ; For a mixture to be at equilibrium, ] B ][ A [ ] [AB 2 2 2 must be equal to 4. For (1), (1)(1) ) (6 = ] B ][ A [ ] [AB = Q 2 2 2 2 c = 36, Q c > K c For (2), (2)(2) ) (4 = ] B ][ A [ ] [AB = Q 2 2 2 2 c = 4, Q c = K c For (3), (3)(3) ) (2 = ] B ][ A [ ] [AB = Q 2 2 2 2 c = 0.44, Q c < K c (a) (2) (b) (1), reverse; (3), forward 13.11 K c = ] H [ ] [H 2 2 = 1.2 x 10 42 (a) [H] = )(0.10) 10 x (1.2 = ] H [ K 42 _ 2 c = 3.5 x 10 22 M (b) H atoms = (3.5 x 10 22 mol/L)(1.0 L)(6.022 x 10 23 atoms/mol) = 210 H atoms H 2 molecules = (0.10 mol/L)(1.0 L)(6.022 x 10 23 molecules/mol) = 6.0 x 10 22 H 2 molecules 13.12 CO(g)
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ch13 - 13 13.1(a Kc = Chemical Equilibrium[SO3]2[SO2]2[O2...

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