Question

At 500 K the reaction 2 NO(g) + Cl2(g) ⇌ 2 NOCl(g) has Kp = 51 In an equilibrium mixture at 500 K, the partial pressure of NO is 0.125 atm and Cl2 is 0.165 atm. What is the partial pressure of NOCl in the equilibrium mixture?

Answer 1

Answer:

p3=0.36atm (partial pressure of NOCl)

Explanation:

2 NO(g) + Cl2(g) ⇌ 2 NOCl(g)  Kp = 51

lets assume the partial pressure of NO,Cl2 , and NOCl at eequilibrium are P1 , P2,and P3 respectively

$Kp=\frac{[NOCl]^{2} }{[NO]^{2} [Cl_2] }$

$Kp=\frac{[p3]^{2} }{[p1]^{2} [p2] }$

p1=0.125atm;

p2=0.165atm;

p3=?

Kp=51;

On solving;

p3=0.36atm (partial pressure of NOCl)