Question

22. The equilibrium constant Kc for the decomposition of phosgene, COCl2, is 4.63 × 10−3 at 527°C: COCl2(g) ⇆ CO(g) + Cl2(g) Calculate the equilibrium partial pressure of all the components, starting with pure phosgene at 0.360 atm.

PCOCl2 = atm

PCO = atm

PCl2 = atm

Answer 1

T = 527.0 oC =(527.0 + 273) K= 800.0 K

delta n = number of gaseous molecule in product - number of gaseous molecule in reactant
delta n = 1
Kp= Kc (RT)^delta n
Kp = 0.00463*(0.0821*800.0)^(1)
Kp = 0.3041

COCl2(g) <----> CO(g) + Cl2(g)
0.360           0        0    (initial)
0.360-x         x        x     (at equilibrium)

Kp = p(CO)*p(Cl2) / p(COCl2)
0.3041 = x*x/(0.360-x)
0.1095 - 0.3041*x = x^2
x^2 + 0.3041*x - 0.1095 = 0

This is quadratic equation (ax^2+bx+c=0)
a = 1.0
b = 0.3041
c = -0.1095

Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 0.530

roots are :
x = 0.212 and x = -0.516

x can't be negative

so, x = 0.212 atm

p(COCL2) = 0.360-x = 0.360-0.212 = 0.148 atm
p(CO) = x = 0.212 atm
p(Cl2) = x = 0.212 atm