For the reaction: 2 A (g) + B (s) ⇌ 2 C (s) + D (g) At 298 K in a 10.0 L vessel, the equilibrium values are as follows: 0.721 atm of A, 4.18 mol of B, 6.25 mol of C, and 8.75 atm of D.

Question

What is the value of the equilibrium constant?

Answer 1

2A(g) + B(s) <==> 2C(s) + D(g)

In the equilibrium expression, we ignore solids as they are considered to have a value of unity. Thus, we can write the equilibrium expression for the above reaction as:

K = (P_D) / (P_A)² = 8.75 / (0.721)²

Kp = 16.8

Kc = Kp / (RT)^∆n where ∆n = 3 - 3 = 0

Kc = 16.8 / (0.0821 * 298)^0 = 16.8 / 24.5

Kc = 0.686