A (g) + B (g) ⇌ 2 C (g) + D (g) initially contains 1.5 M A and 1.0 M B. equilibrium concentration of C is 0.55 M, equilibrium constant (Kc) for the reaction?

Question

Consider the following reaction:

A (g) + B (g) ⇌ 2 C (g) + D (g)

A reaction mixture initially contains 1.5 M A and 1.0 M B. If the equilibrium concentration of C is 0.55 M, what is the value of the equilibrium constant (Kc) for the reaction?

Answer 1

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

1.5.......1.0...............0..............0.........Initial

-x.........-x................+2x..........+x........Change

1.5-x....1.0-x.............2x............x.........Equilibrium

[C] = 0.55 M = 2x

x = 0.55/2 = 0.275

Equilibrium concentrations are as follows:

[A] = 1.5 - x = 1.5 - 0.275 = 1.225

[B] = 1.0 - 0.275 = 0.725

[C] = 0.55

[D] = x = 0.275

Kc = [C]²[D] / [A][B]

Plug in the equilibrium concentrations above and solve for Kc.