Question

The equilibrium constant, Kc, for the following reaction is 9.52×10-2 at 350 K. CH4 (g) + CCl4 (g) 2 CH2Cl2 (g) Calculate the equilibrium concentrations of reactants and product when 0.377 moles of CH4 and 0.377 moles of CCl4 are introduced into a 1.00 L vessel at 350 K.

Answer 1

Answer: The equilibrium concentration of $CH_4\text{ and }CCl_4$ is 0.377 M and equilibrium concentration of $CH_2Cl_2$ is 0.116 M

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

• For Methane:

Volume of solution = 1 L

Moles of methane = 0.377 moles

Putting values in above equation, we get:

$\text{Molarity of }CH_4=\frac{0.377mol}{1L}\\\\\text{Molarity of }CH_4=0.377M$

• For carbon tetrachloride:

Volume of solution = 1 L

Moles of carbon tetrachloride = 0.377 moles

Putting values in above equation, we get:

$\text{Molarity of carbon tetrachloride}=\frac{0.377mol}{1L}\\\\\text{Molarity of carbon tetrachloride}=0.377M$

For the given chemical equation:

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

The expression for $K_c$ for the given equation follows:

$K_c=\frac{[CH_2Cl_2]^2}{[CH_4][CCl_4]}$

We are given:

$K_c=9.52\times 10^{-2}\\[CH_4]=0.377M\\[CCl_4]=0.377M$

Putting values in above equation, we get:

$9.52\times 10^{-2}=\frac{[CH_2Cl_2]^2}{(0.377)\times (0.377)}$

$[CH_2Cl_2]=0.116M$

Hence, the equilibrium concentration of $CH_4\text{ and }CCl_4$ is 0.377 M and equilibrium concentration of $CH_2Cl_2$ is 0.116 M