At 400 K. an equilibrium mixture of H_2, I_2 and HI consists of 0.068 mol H_2, 0

Question

At 400 K. an equilibrium mixture of H_2, I_2 and HI consists of 0.068 mol H_2, 0.075 mol I_2, and 0.13 mol HI in a 1.00-L flask. What is the value of K_P for the following equilibrium? (R = 0.0821 L middot atm (K middot mol)) 2HI(g) H_2(g) + I_2(g) A) 0.039 B) 3.4 C) 26 D) 0.29 E) 8.2 If K_c = 0.l50 for A_2 + 2B 2AB, what is the value of K_c for the reaction 4AB 2A_2 + 4B? A) 3.33 B) 0.300 C) 44.4 D) -0.150 E) 0.150 For which of the following equilibria does K_e = K_p? A) CaO(s) + CO_2(g) CaCO_3(s) B) N_2(g) + 3H_2(g) 2NH_3(g) C) CO(g) + 3H_2(g) CH_4(g) + H_2O(g) D) CO(g) + H_2O(g) CO_2(g) + H_2(g) E) HBr(g) 1/2H_2(g) + 1/2Br2(l)

Explanation / Answer

1) Kc = [H2] [I2] / [HI]^2

Kc = 0.068 X 0.075 / (0.13)^2 = 0.301

n = 0

Kp = Kc(RT)^n

Kp = 0.301) Or (0.29)

2) Kc for

A2 +2B --> 2AB   is 0.150

Kc = [AB]^2 / [A2]B]^2

For

4AB --> 2A2 + 4B

It is reverse of above reaction multiplied by two

So new Kc will be = Kc' = [B]^4[A2]^2 / [AB]^4 = 1 / Kc^2

Kc' = 1 / 0.15 X 0.15 = 44.44

3) the relation between Kp and Kc is

Kp = Kc(RT)^n

n = number of moles of products - number of moles of reactants

A) n = -1

B) n = -2

C) n = -2

D) n = 2-2 = 0 Therefore Kp =Kc

E) n = 1/2 - 1 = -0.5

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