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.

Step-by-step 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 = (x * 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.