Question

The decomposition of NOCl to form NO and Cl₂, 2 NOCl(g) D 2 NO(g) + Cl₂(g), has a K_p value of          

1.6 ´ 10^-5 at some temperature. If 0.500 atm of NOCl is placed in a closed vessel and allowed to come to equilibrium, what is its approximate equilibrium partial pressure?

Answer 1

Ans :- P_{NOCl (g)} = 0.48 atm

Explanation :-

ICE table is :

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

Initial (I)...................0.500 atm..............................0.0 atm...............................................0.0 atm

Change (C)..............- 2y..........................................+2y.......................................................+y

Equilibrium (E)........(0.500 - 2y) atm.........................2y atm................................................y atm

Expression of K_p is :

K_p = P_NO².P_Cl2 / P_NOCl²

1.6 x 10^-5 = 4y³ / (0.500 - 2y)²

On solving for y, we have

y = 0.0097385

So,

Equilibrium partial pressure of NOCl (g) = P_{NOCl (g)} = 0.500 - 2x0.0097385 = 0.48 atm