Question

The equilibrium constant, K, for the following reaction is 10.5 at 350 K.

2CH2Cl2(g) CH4(g) + CCl4(g)

An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.09E-2 M CH2Cl2, 0.165 M CH4 and 0.165 M CCl4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.82E-2 mol of CH2Cl2(g) is added to the flask?

[CH2Cl2] = M
[CH4] = M
[CCl4] = M

Answer 1

The reaction,

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

Intial equilibrium             5.09 x 10^-2                               0.165                     0.165

added                            +3.82 x 10^-2                            +0.0191                  +0.0191

let x be the amount reacted then,

K = x^2/3.82 x 10^-2-2x = 10.5

x = 0.0191 M

the concentrations would be thus,

[CH2Cl2] = 0.0891 - 2 x 0.0191 = 0.051 M

[CH4] = 0.1841 M

[CCl4] = 0.1841 M