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s2008m [1.1K]
3 years ago
7

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 o

f 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 =
Chemistry
1 answer:
statuscvo [17]3 years ago
8 0

<u>Answer:</u> The value of K_{eq} is 0.044

<u>Explanation:</u>

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)

<u>Initial:</u>          0.0410    0.0651

<u>At eqllm:</u>     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

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