Question

The equilibrium constant, Kp, for the following reaction is 10.5 at 350 K: 2CH2Cl2(g) CH4(g) CCl4(g) Calculate the equilibrium partial pressures of

Answer 1

The complete question is as follows: The equilibrium constant, Kp, for the following reaction is 10.5 at 350 K: 2CH2Cl2(g) CH4(g) + CCl4(g) Calculate the equilibrium partial pressures of all species when CH2Cl2(g) is introduced into an evacuated flask at a pressure of 0.968 atm at 350 K. PCH2Cl2 = atm PCH4 = atm PCCl4 = atm

Answer: The equilibrium partial pressures of all species, that is, $CH_{4}$, $CCl_{4}$ and $CH_{2}Cl_{2}$ is 0.420 atm, 0.420 atm and 0.128 atm.

Explanation:

For the given reaction equation, the initial and equilibrium concentration of involved species is as follows.

                       $2CH_{2}Cl_{2}(g) \rightarrow CH_{4}(g) + CCl_{4}(g)$

Initial:                0.968 atm         0               0

Equilibrium:  (0.968 - 2x)           x                x

Now, $K_{p}$ for this reaction is as follows.

$K_{p} = \frac{P_{CH_{4}}P_{CCl_{4}}}{P^{2}_{CH_{2}Cl_{2}}}\\10.5 = \frac{x \times x}{(0.968 - 2x)^{2}}\\x = 0.420$

$P_{CH_{4}} = x = 0.420 atm\\P_{CCl_{4}} = x = 0.420 atm\\P_{CH_{2}Cl_{2}} = (0.968 atm - 2x) = (0.968 atm - 2(0.420)) = 0.128 atm$

Thus, we can conclude that the equilibrium partial pressures of all species, that is, $CH_{4}$, $CCl_{4}$ and $CH_{2}Cl_{2}$ is 0.420 atm, 0.420 atm and 0.128 atm.