Question

A student ran the following reaction in the laboratory at 311 K:CH4(g) + CCl4(g) 2CH2Cl2(g)When she introduced 4.10×10-2 moles of CH4(g) and 6.51×10-2 moles of CCl4(g) into a 1.00 liter container, she found the equilibrium concentration of CCl4(g) to be 6.03×10-2 M.Calculate the equilibrium constant, Kc, she obtained for this reaction.Kc =

Answer 1

Answer: The value of $K_{eq}$ is 0.044

Explanation:

We are given:

Initial moles of methane = $4.10\times 10^{-2}mol=0.0410moles$

Initial moles of carbon tetrachloride = $6.51\times 10^{-2}mol=0.0651moles$

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of methane = $\frac{0.0410}{1.00}=0.0410M$

Concentration of carbon tetrachloride = $\frac{0.0651}{1.00}=0.0651M$

The given chemical equation follows:

                    $CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)$

Initial:          0.0410    0.0651

At eqllm:     0.0410-x   0.0651-x       2x

We are given:

Equilibrium concentration of carbon tetrachloride = $6.02\times 10^{-2}M=0.0602M$

Evaluating the value of 'x', we get:

$\Rightarrow (0.0651-x)=0.0602\\\\\Rightarrow x=0.0049M$

Now, equilibrium concentration of methane = $0.0410-x=[0.0410-0.0049]=0.0361M$

Equilibrium concentration of $CH_2Cl_2=2x=[2\times 0.0049]=0.0098M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[CH_2Cl_2]^2}{[CH_4]\times [CCl_4]}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.0098)^2}{0.0361\times 0.0603}\\\\K_{eq}=0.044$

Hence, the value of $K_{eq}$ is 0.044