Question

A student ran the following reaction in the laboratory at 316 K:CH4(g) + CCl4(g) 2CH2Cl2(g)When she introduced CH4(g) and CCl4(g) into a 1.00 L evacuated container, so that the initial partial pressure of CH4 was 0.596 atm and the initial partial pressure of CCl4 was 0.256 atm, she found that the equilibrium partial pressure of CCl4 was 0.218 atm.Calculate the equilibrium constant, Kp, she obtained for this reaction.

Answer 1

Answer : The equilibrium constant for this reaction is, 0.0475

Solution : Given,

Initial pressure of
`N_2` = 1.42 bar

Initial pressure of
`H_2` = 2.87 bar

`K_p` = 0.036

The given equilibrium reaction is,

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

Initially 0.596 0.256 0

At equilibrium (0.596-x) (0.256-x) 2x

The expression of
`K_p` will be,

`K_p=((p_(CH_2Cl_2))^2)/((p_(CH_4))(p_(CCl_4)))`

Thus, the partial pressure of
`CCl_4` at equilibrium = 0.218 = (0.256 - x)

That means,

(0.256 - x) = 0.218

x = 0.038 atm

The partial pressure of
`N_2` at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of
`CH_4` at equilibrium = (0.596-x) = (0.596-0.038) = 0.558 atm

The partial pressure of
`CH_2Cl_2` at equilibrium = 2x = 2 × 0.038 = 0.076 atm

Now put all the values of partial pressure in above expression, we get:

`K_p=((p_(CH_2Cl_2))^2)/((p_(CH_4))(p_(CCl_4)))`

`K_p=((0.076)^2)/((0.558)(0.218))`

`K_p=0.0475`

Therefore, the equilibrium constant for this reaction is, 0.0475