Question

Consider the following reaction:
A(g)?2B(g)
Find the equilibrium partial pressures of A and Bfor each of the following different values of Kp. Assume that the initial partial pressure of B in each case is 1.0 atm and that the initial partial pressure of A is 0.0 atm. Make any appropriate simplifying assumptions.

Kp= 1.8

Kp= 1.6×10^?4

Kp= 1.8×10⁵

Answer 1

                              A(g)    $\rightleftharpoons$        2B(g)

Initial                     0                1

Equilibrium         x/2                1-x

Kp = P_B²/P_A

Part 1) kp = 1.8

1.8 = (1-x)²/(x/2)

(1.8x) = 2 x (1-x)²

1.8x = 2 + 2x² -4x

2x²-5.8x +2 = 0

x = 0.399

Therefore, at equilibrium,

P_B = 1-x =1-0.399 = 0.601 atm

P_A = x/2 =0.399/2 = 0.199 atm

Part 2) kp = 1.6 x 10^-4

1.6 x 10^-4 = (1-x)²/(x/2)

(1.6 x 10^-4 x) = 2 x (1-x)²

1.6 x 10^-4 x = 2 + 2x² -4x

2x² -4x +2 = 0

x = 1

Therefore, at equilibrium,

P_B = 1-1 = 0 atm

P_A = 1/2 = 0.5 atm

Part 3) kp = 1.8 x 10⁵

1.8 x 10⁵ = (1-x)²/(x/2)

(1.8 x 10⁵ x) = 2 x (1-x)²

1.8 x 10⁵ x = 2 + 2x² -4x

2x² -1.8 x 10⁵ x +2 = 0

x = 0.000011

Therefore, at equilibrium,

P_B = 1-0.000011 $\approx$ 1 atm

P_A = 0.000011 /2 = 5.5 x 10^-6 atm