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

Step-by-step 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=(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}=(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=([CH_2Cl_2]^2)/([CH_4][CCl_4])`

We are given:

Putting values in above equation, we get:

`9.52* 10^(-2)=([CH_2Cl_2]^2)/((0.377)* (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