Question

The equilibrium constant, Kc, for the following reaction is 9.52×10-2 at 350 K.

CH4 (g) + CCl4 (g) <-> 2 CH2Cl2 (g)

Calculate the equilibrium concentrations of reactants and product when 0.281 moles of CH4 and 0.281 moles of CCl4 are introduced into a 1.00 L vessel at 350 K.

[CH4] =

[CCl4] =

[CH2Cl2] =

Answer 1

since volume is 1 L, the number of mol and concentration will be same

CH4 (g) + CCl4 (g) <-> 2 CH2Cl2 (g)
0.281        0.281                     0              (initial)
0.281-x     0.281-x               2x             (at equilibrium)

Kc = [CH2Cl2]^2 / {[CH4][CCl4]}
9.52*10^-2 = (2x)^2 / (0.281-x)^2
sqrt(9.52*10^-2) = 2x/(0.281-x)
0.3085 = 2x/(0.281-x)
0.0867 - 0.3085*x = 2x
x = 0.0376 M

[CH4] = 0.281-x = 0.281-0.0376 = 0.243 M
[CCl4] = 0.281-x = 0.281-0.0376 = 0.243 M
[CH2Cl2] = 2x = 2*0.0376 M = 0.0752 M