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 5.47×10-2 M CH2Cl2, 0.177 M CH4 and 0.177 M CCl4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.44×10-2 mol of CH2Cl2(g) is added to the flask?

[CH2Cl2] =_____ M

[CH4] = ________M

[CCl4] =________ M

Answer 1

Ans. Balanced Reaction:       2CH₂Cl₂(g) ----------> CH₄(g) + CCl₄(g)

Before addition of additional CH₂Cl₂, the equilibrium concentrations would act as initial concertation for the change.

Given, Volume of reaction vessel = 1.0 L.

Initial moles of CH₂Cl₂ = 5.47 x 10^-2 M = 0.0547 M

[CH₂Cl₂] added to reaction vessel = Moles / Volume of reaction vessel

                        = 3.44 x 10^-2 mol / 1.0 L

                        = 0.0344 M

Total initial [CH₂Cl₂] after initiation = 0.0547 M + 0.0344 M = 0.0891 M

Ans,

            Initial [CH₄] = 0.177 M

            Initial [CCl₄] = 0.177 M

# Create an ICE table with total initial [CH₂Cl₂] and initial concentrations of CH4 an CCl4 as shown in picture-

Now,

            Equilibrium constant, Kc = [CH­₄] [CCl₄] / [CH₂Cl₂]²

            Or, 10.5 = (0.177 + 0.5X) (0.177 + 0.5X) / (0.0891 - X)²

            Or, 10.5 (0.00793881 + X² - 0.1782X) = 0.031329 + 0.25X² + 0.177X

            Or, 10.5X² + 0.083357505 - 1.8711X - 0.031329 - 0.25X² - 0.177X = 0

            Or, 10.25X² -2.0481X + 0.052028505 = 0

Solving the quadratic equation, we get following two roots-

            X1 = 0.1699 ; X2 = 0.0299

Since X can’t be greater than 0.0891, reject X1.

Hence, X =0.0299

# New Equilibrium [CH₂Cl₂] = 0.0891 – X = 0.0891 – 0.0299 = 0.0592 M

            [CH₄] = 0.177 + 0.5 (0.0299) = 0.19195 M

            [CCl₄] = 0.177 + 0.5 (0.0299) = 0.19195 M