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 5.12×10^-2 M CH₂Cl₂, 0.166 M CH₄ and 0.166 M CCl₄. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.121 mol of CH₄(g) is added to the flask?

[CH2Cl2]	=	_________M
[CH4]	=	_________ M
[CCl4]	=	_________ M

Answer 1

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

The equilibrium constant is given as

K = [CH₄][CCl₄]/[CH₂Cl₂]² = 10.5 at 350°C.

Treat the given equilibrium concentrations as the initial concentrations before excess CH₄ was added.

Moles CH₂Cl₂ present initially = (volume of container in L)*(concentration in mol/L) = (1.00 L)*(5.12*10^-2 mol/L) = 5.12*10^-2 mol.

Moles CH₄ present initially = (1.0 L)*(0.166 mol/L) = 0.166 mol.

Moles CCl₄ present initially = (1.0 L)*(0.166 mol/L) = 0.166 mol.

When extra CH₄ is added, we know, as per Le-Chatilier principle, the backward reaction will be favoured. Write down the reverse reaction and evaluate the equilibrium constant as

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

K’ = 1/K = [CH₂Cl₂]²/[CH₄][CCl₄] = 1/10.5 = 0.0952

Now, moles CH₄ present at equilibrium = (0.166 + 0.121) mol = 0.287 mol.

[CH₄]_e = (0.287/1.00) mol/L = 0.287 M.

Let x be the change in concentration of CCl₄; the change in concentration of CH₂Cl₂ will be 2x. Set up the ICE chart as

CH₄ (g) + CCl₄ (g) <=====> 2 CH₂Cl₂

initial(new)                      0.287       0.166                      0.0512

change                                 -x               -x                           +2x

equilibrium                  (0.287 – x)(0.166 – x)               (0.0512 + 2x)

Since the volume is 1.00 L, we must have,

K^’ = [CH₂Cl₂]²/[CH₄][CCl₄]

====> 0.0952 = (0.0512 + 2x)²/(0.287-x)(0.166 – x)

Solve the binomial equation to obtain x = 0.007

Therefore, [CH₄] = (0.287 – 0.007)/1.00 mol/L = 0.280 M

[CCl₄] = (0.166 – 0.007)/1.00 mol/L = 0.159 M

[CH₂Cl₂] = (0.0512 + 2*0.007)/1.00 mol/L = 0.0652 M (ans).