The equilibrium constant, K, for the following reaction is 10.5 at 350 K. 2CH2Cl

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

The equilibrium constant, K, for the following reaction is 1.29E-2 at 600 K.

COCl2(g) CO(g) + Cl2(g)

An equilibrium mixture of the three gases in a 1.00 L flask at 600 K contains 0.192 M COCl2, 4.98E-2 M CO and 4.98E-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 2.93E-2 mol of Cl2(g) is added to the flask?

[COCl2]

=

M

[CO]

=

M

[Cl2]

=

M

Explanation / Answer

Given that;

Kc = [CH4] [CCl4] / [CH2Cl2]^2 = 10.5

If we add 3.82E-2 mol of CH2Cl2(g) is added to the flask then new contraction of CH2Cl2 =

3.82E-2 mol of CH2Cl2(g)

Now make ICE table:

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

I                       3.82*10^-2 M 0 M                  0 M

C                     -x                     +x                    +x

E          3.82*10^-2 M - x                  x M                    x M

Then;

Kc = [CH4] [CCl4] / [CH2Cl2]^2 = 10.5

10.5 = x*x / (3.82*10^-2 - x )^2

If x is very less then 3.82*10^-2 – x = 3.82*10^-2 M

Then;

10.5 = x*x / (3.82*10^-2 M)^2

10.5*(3.82*10^-2)^2 = x^2

x^2 = 0.015

x= 0.12

Then new concentration at equilibrium :

[CH2Cl2] =3.82*10^-2 + 5.09*10^-2 M = 8.91*10^-2M

[CH4] =0.165 M + 0.12 M = 0.285 M

[CCl4]= 0.165 M + 0.12 M = 0.285 M

If cross check then;

Kc = 0.285 M *0.285 M / 8.91*10^-2M = 10.5

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