Question

The equilibrium constant, Kp, for the following reaction is 1.57 at 600 K:

CO(g) + Cl2(g) COCl2(g)

Calculate the equilibrium partial pressures of all species when CO and Cl2, each at an intitial partial pressure of 1.65 atm, are introduced into an evacuated vessel at 600 K.

PCO = ______ atm

PCl2= _______ atm

PCOCl2 = _______ atm

Answer 1

Let's prepare the ICE table

p(CO) p(Cl2) p(COCl2)

initial 1.65 1.65 0

change -1x -1x +1x

equilibrium 1.65-1x 1.65-1x +1x

Equilibrium constant expression is

Kp = p(COCl2)/p(CO)*p(Cl2)

1.57 = (1*x)/((1.65-1*x)(1.65-1*x))

1.57 = (1*x)/(2.7225-3.3*x + 1*x^2)

4.27432-5.181*x + 1.57*x^2 = 1*x

4.27432-6.181*x + 1.57*x^2 = 0

Let's solve this quadratic equation

Comparing it with general form: (ax^2+bx+c=0)

a = 1.57

b = -6.181

c = 4.274

solution of quadratic equation is found by below formula

x = {-b + √(b^2-4*a*c)}/2a

x = {-b - √(b^2-4*a*c)}/2a

b^2-4*a*c = 11.36

putting value of d, solution can be written as:

x = {6.181 + √(11.36)}/3.14

x = {6.181 - √(11.36)}/3.14

solutions are :

x = 3.042 and x = 0.895

x can't be 3.042 as this will make the concentration negative.so,

x = 0.895

At equilibrium:

p(CO) = 1.65-1x = 1.65-1*0.89498 = 0.7550 atm

p(Cl2) = 1.65-1x = 1.65-1*0.89498 = 0.7550 atm

p(COCl2) = +1x = +1*0.89498 = 0.895 atm

p(CO) = 0.755 atm

p(Cl2) = 0.755 atm

p(COCl2) = 0.895 atm