Question

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

CH4(g) + CCl4(g) 2CH2Cl2(g)

Calculate the equilibrium partial pressures of all species when CH4 and CCl4, each at an intitial partial pressure of 1.12 atm, are introduced into an evacuated vessel at 350 K.

PCH4 = ________ atm

PCCl4 = ________atm

PCH2Cl2 = ________ atm

Answer 1

Let's prepare the ICE table

p(CH4) p(CCl4) p(CH2Cl2)

initial 1.12 1.12 0

change -1x -1x +2x

equilibrium 1.12-1x 1.12-1x +2x

Equilibrium constant expression is

Kp = p(CH2Cl2)^2/p(CH4)*p(CCl4)

0.0952 = (2*x)^2/(1.12-1*x)^2

sqrt(0.0952) = (2*x)/(1.12-1*x)

0.3085 = (2*x)/(1.12-1*x)

0.34557-0.30854*x = 2*x

0.34557-2.30854*x = 0

x = 0.150

At equilibrium:

p(CH4) = 1.12-1x = 1.12-1* 0.150 = 0.97 atm

p(CCl4) = 1.12-1x = 1.12-1* 0.150 = 0.97 atm

p(CH2Cl2) = +2x = +2* 0.150 = 0.300 atm

p(CH4) = 0.97 atm

p(CCl4) = 0.97 atm

p(CH2Cl2) = 0.300 atm