Question

Nitric oxide reacts with chlorine gas according to the reaction: 2 NO( g) + Cl2( g) ∆ 2 NOCl( 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 is 115 torr. What were the initial partial pressures of NO and Cl2 ?

Answer 1

Answer:  The initial partial pressures of $NO$ and $Cl_2$ is 139.4 torr.

Explanation:

The equilibrium reaction is

                           $2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

initially conc.         p            p                  0

At eqm.                p-2x        p-x               2x

Given: $p_{NOCl}= 2x= 115torr$

$x=57.5 torr$

The expression for equilibrium constant is :

$K_p=\frac{p_{NOCl}^2}{p_{NO}^2\times p_{Cl_2}}$

$p_{NOCl}$ = 2x =115 torr

$p_{NO}$ = (p-2x) = p-115 torr

$p_{Cl_2}$ = (p-x)= p-57.5 torr

Now put all the given values in this expression, we get

$0.27=\frac{(115)^2}{(p-115)^2\times (p-57.5)}$

$p=139.4torr$

Therefore, the initial partial pressures of $NO$ and $Cl_2$ is 139.4 torr.