Question

given:
2NO(g)+Cl2(g)⇌2NOCl(g)
Kp=0.27 at 700 K
A reaction mixture initially contains equal partial pressures of NO and Cl2. At equilibrium, the partial pressure of NOCl was measured to be 115 torr.

What were the initial partial pressures of NO and Cl2?

Answer 1

Partial pressure of NOCl = 115 torr = 0.151 atm

2 NO (g) + Cl2 (g) <----------> 2 NOCl(g)

   x               x                          0

x-2y x-y 2y

2y = 0.151

y = 0.0766

Kp = P2NOCl / P2NO x PCl2

0.27 = (0.151)^2 / (x-0.151)^2 x (x - 0.0766)

(x-0.151)^2 x (x - 0.0766) = 0.0848

x^2 + 0.0228 - 0.302 x ( x - 0.0766) = 0.0848

x = 0.567

initial partial pressure of NO and Cl2 = 0.567 atm