Question

He equilibrium constant, Kc, for the following reaction is 9.52×10-2 at 350 K:

CH4(g) + CCl4(g)→ 2CH2Cl2(g)

Calculate the equilibrium concentrations of reactants and product when 0.310 moles of CH4 and 0.310 moles of CCl4 are introduced into a 1.00 L vessel at 350 K.

Answer 1

Answer: The equilibrium concentration of methane, carbon tetrachloride and
`CH_2Cl_2` are 0.2686 M, 0.2686 M and 0.0828 M respectively.

Step-by-step explanation:

Molarity is calculated by using the equation:

`\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution}}`

• For methane:

Moles of methane = 0.310 moles

Volume of solution = 1.00 L

`\text{Molarity of methane}=(0.310)/(1)=0.310M`

• For carbon tetrachloride:

Moles of carbon tetrachloride = 0.310 moles

Volume of solution = 1.00 L

`\text{Molarity of }CCl_4=(0.310)/(1)=0.310M`

For the given chemical equation:

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

Initial: 0.310 0.310

At eqllm: 0.310-x 0.310-x 2x

The expression of
`K_c` for above equation follows:

`K_c=([CH_2Cl_2]^2)/([CH_4][CCl_4])`

We are given:

`K_c=9.52* 10^(-2)`

Putting values in above equation, we get:

`9.52* 10^(-2)=((2x)^2)/((0.310-x)* (0.310-x))\\\\x=-0.0565,0.0414`

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of methane = (0.310 - x) = [0.310 - 0.0414] = 0.2686 M

Equilibrium concentration of carbon tetrachloride = (0.310 - x) = [0.310 - 0.0414] = 0.2686 M

Equilibrium concentration of
`CH_2Cl_2` = 2x = (2 × 0.0414) = 0.0828 M

Hence, the equilibrium concentration of methane, carbon tetrachloride and
`CH_2Cl_2` are 0.2686 M, 0.2686 M and 0.0828 M respectively.