SQSM-05 - Chapter 5 Chemical Equilibrium 1 What are the...

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Chapter 5 Chemical Equilibrium 5-1 1. What are the units of K p and K c for each of the following: a) 2H 2 S(g) U 2H 2 (g) + S 2 (g) × == = 22 2 2 2 HS p 2 P P atm atm K a tm Pa t m × = c 2 [H ] [S ] M M M [H S] M b) 4NH 3 (g) + 3O 2 (g) U 2N 2 (g) + 6H 2 O(g) 3 × = × 32 26 NH O p 43 4 NH O PP atm atm a P P atm atm × = × 2 6 c 4 3 [N ] [H O] M M M [NH ] [O ] M M 3. Determine K c values for the following at 298 K: a) N 2 O 4 (g) U 2NO 2 (g) K p = 0.15 atm Δ × -n g cp K=K (RT ) & Δ n g = 2 - 1 = 1, so ×× c Kmo l 1 K =0.15 atm = 0.0061 M 0.0821 L atm 298 K b) NO 2 (g) + NO(g) U N 2 O 3 (g) K p = 0.86 atm -1 Δ n g = -1, so -1 -1 c 0.0821 L atm K =0.86 atm 298 K = 21 M mol K 5. Determine the value of K p for each of the following: Use Eq. 5.2, K c = K p (RT) Δ n g a) CO(g) + H 2 O(g) U CO 2 (g) + H 2 (g) K c = 23.2 at 600 K Δ n g = 0 for the reaction, so K p = K c = 23.2 b) 2 H 2 S(g) U 2 H 2 (g) + S 2 (g) K c = 2.3 x 10 -4 M at 1405 K Δ n g = 1, so ( ) × 1 -4 -2 mol L atm p Lm o l K K =(2.3 10 ) 0.0821 1450 K =2.7 10 atm 7. a) Determine K for: HI(g) U 1 / 2 H 2 (g) + 1 / 2 I 2 (s) given H 2 (g) + I 2 (s) U 2 HI(g) K= 8.6 We are asked to find the value of K for a reaction given the value for the reaction that is 1 / 2 of the reverse of the original reaction. Consequently, K 2 = (K 1 ) -0.5 = 8.6 -0.5 = 0.34 b) Determine K c for: 2 SO 2 (g) + O 2 (g) U 2 SO 3 (g) given SO 2 (g) + 1 / 2 O 2 (g) U SO 3 (g) K c = 3.61 M -1/2 This reaction is simply twice the original reaction: () === 2 2 21 K K 3.61 13.0 M 9. Given the following: 2 NO(g) U N 2 (g) + O 2 (g) K 1 = 2.4 x 10 30 NO(g) + 1 / 2 Br 2 (g) U NOBr(g) K 2 = 1.4 Determine K for 1 / 2 N 2 (g) + 1 / 2 O 2 (g) + 1 / 2 Br 2 (g) U NOBr(g) 1 / 2 N 2 (g) + 1 / 2 O 2 (g) U NO(g) K 1 -1/2 = (2.4 x 10 -30 ) -1/2 = 6.5 x 10 16 NO(g) + 1 / 2 Br 2 (g) U NOBr(g) K 2 = 1.4 NO(g) cancels, giving net equation. By rule 3, K = (K 1 -1/2 ) x K 2 = 9.0 x 10 -16 11. Lead fluoride dissolves in strong acid by the following reaction: PbF 2 (s) + 2H 3 O 1+ (aq) U Pb 2+ (aq) + 2HF(aq) +2H 2 O(l) a) What is the equilibrium constant expression for the reaction? + = 1+ 2 23 [Pb ][HF] K [PbF ][H O ]
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Chemical Equilibrium 5-2 b) Use the following to determine the value of the equilibrium constant of the above reaction: PbF 2 (s) U Pb 2+ (aq) + 2F 1- (aq) K 1 = 3.7 x 10 -8 HF(aq) + H 2 O(l) U H 3 O 1+ + F 1- (aq) K 2 = 7.2 x 10 -4 Reverse the second reaction and multiply it by two 2H 3 O 1+ (aq) + 2F 1- (aq) U 2HF(aq) + 2H 2 O(l) K = K 2 -2 Addition of the above equation and the first equation yields the reaction of interest. Therefore, the equilibrium constant is K = × = × -8 -2 1 2- 4 2 2 K3 . 7 1 0 = 7.1 10 K( 7 . 2 1 0 ) 13. Equal numbers of moles of Cl 2 and NO are placed in a vessel at some temperature where they reach the following equilibrium: 2 NO(g) + Cl 2 (g) U 2 ClNO(g). Indicate whether each of the following statements about the resulting equilibrium mixture is true, false, or depends upon the value of the equilibrium constant. 2 NO(g) + Cl 2 (g) U 2 ClNO(g) Initial a a 0 Δ - 2x - x + 2x Equilibrium a - 2x a - x 2x a) [NO] > [ClNO] (a - 2x) > (2x) is true only if x is small, so it depends upon K b) [Cl 2 ] < [NO] (a-x) < (a-2x) cannot be true. It is always false.
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SQSM-05 - Chapter 5 Chemical Equilibrium 1 What are the...

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