Respuesta :
Answer: The value of [tex]K_{eq}[/tex] is 0.044
Explanation:
We are given:
Initial moles of methane = [tex]4.10\times 10^{-2}mol=0.0410moles[/tex]
Initial moles of carbon tetrachloride = [tex]6.51\times 10^{-2}mol=0.0651moles[/tex]
Volume of the container = 1.00 L
Concentration of a substance is calculated by:
[tex]\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}[/tex]
So, concentration of methane = [tex]\frac{0.0410}{1.00}=0.0410M[/tex]
Concentration of carbon tetrachloride = [tex]\frac{0.0651}{1.00}=0.0651M[/tex]
The given chemical equation follows:
[tex]CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)[/tex]
Initial: 0.0410 0.0651
At eqllm: 0.0410-x 0.0651-x 2x
We are given:
Equilibrium concentration of carbon tetrachloride = [tex]6.02\times 10^{-2}M=0.0602M[/tex]
Evaluating the value of 'x', we get:
[tex]\Rightarrow (0.0651-x)=0.0602\\\\\Rightarrow x=0.0049M[/tex]
Now, equilibrium concentration of methane = [tex]0.0410-x=[0.0410-0.0049]=0.0361M[/tex]
Equilibrium concentration of [tex]CH_2Cl_2=2x=[2\times 0.0049]=0.0098M[/tex]
The expression of [tex]K_{eq}[/tex] for the above reaction follows:
[tex]K_{eq}=\frac{[CH_2Cl_2]^2}{[CH_4]\times [CCl_4]}[/tex]
Putting values in above expression, we get:
[tex]K_{eq}=\frac{(0.0098)^2}{0.0361\times 0.0603}\\\\K_{eq}=0.044[/tex]
Hence, the value of [tex]K_{eq}[/tex] is 0.044
