Question

Consider the following reaction:

I2(g)+Cl2(g)⇌2ICl(g), Kp=81.9 (at 25∘C)

A reaction mixture at 25∘C initially contains PI2 = 0.110 atm , PCl2 = 0.110 atm , and PICl = 0.00 atm.

Find the equilibrium partial pressure of I2 at this temperature.

Find the equilibrium partial pressure of Cl2 at this temperature.

Find the equilibrium partial pressure of ICl at this temperature.

Answer 1

ICE Table:

Equilibrium constant expression is

Kp = p(ICl)^2/p(I2)*p(Cl2)

81.9 = (2*x)^2/(0.11-1*x)^2

sqrt(81.9) = (2*x)/(0.11-1*x)

9.0499 = (2*x)/(0.11-1*x)

0.9955-9.05*x = 2*x

0.9955-11.05*x = 0

x = 0.0901

At equilibrium:

p(I2) = 0.11-1x = 0.11-1* 0.0901 = 0.0199 atm

p(Cl2) = 0.11-1x = 0.11-1* 0.0901 = 0.0199 atm

p(ICl) = +2x = +2* 0.0901 = 0.180 atm

Answer:

p(I2) = 0.0199 atm

p(Cl2) = 0.0199 atm

p(ICl) = 0.180 atm