For $A (g) + B (g) C (g) + D (g)$ , $K_{c} = 6.0$ at a given temperature. 1.0 M of A and 1.0 M of B are allowed to reach equilibrium in a rigid vessel. Then 1.0 M more of A is added at constant temperature and a new equilibrium is reached. Calculate the new equilibrium $[C]$ in mol/L, to two decimal places.

Numerical answer type, answer in mol/L, tolerance 0.005.

Answer: 0.88 mol/L

First equilibrium:

x^{2} / (1 - x)^{2} = 6

gives

x = 0.7101

M of C and of D. After adding A:

[A] = 1.2899

[B] = 0.2899

. With

y

reacting further,

(0.7101 + y)^{2} = 6 (1.2899 - y) (0.2899 - y)

, so

y = 0.1734

. New

[C] = 0.883

mol/L.