Question

The equilibrium system 2A ⇌ B + 2C has an equilibrium constant K = 2.6 x 10–6. Initially 3.0 moles of A are placed in a 1.5-L flask. Determine the concentration of B at equilibrium.

Answer 1

Initial concentration of A, [A] = mol/L = 3.0 mol / 1.5 L = 2.0 M

Since initially there is not product( B and C), the equilibrium will shift towards the product side till equilibrium is achieved. Hence the concentration of A will decrease and that of B and C will increase.

---2A <-----> B + 2C ; Kc = 2.6*10^-6

I: 2.0 M ------ 0 M, 0 M

C:- 2X ------- +X, +2X

E: (2.0 - 2X), X M, 2X M

Kc = 2.6*10^-6 = [B] * [C]² / [A]²

=> 2.6*10^-6 = X * (2X)² / (2.0 - 2X)²

=> 2.6*10^-6 = X * (2X)² / [2*(1.0 - X)]²

=>  2.6*10^-6 = X³ / (1.0 - X)²

=> (1.0 - X)² *  2.6*10^-6 = X³

Solving the above polynomial equaiton gives:

X = 0.01363 M

Hence concentration of B at equilibrium = X =  0.01363 M (Answer)