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 ?

NO and Cl₂ initially at 139 torr react to form NOCl at 115 torr at equilibrium.

Chemical equilibrium is the state in which both the reactants and products are present in concentrations which have no further tendency to change with time.

2 NO(g) + Cl₂(g) ⇄ 2 NOCl(g)        Kp = 0.27 at 700 K

We will use letters for unknown pressures.

        2 NO(g) + Cl₂(g) ⇄ 2 NOCl(g)

I             x             x               0

C         -2y           -y             +2y

E         x-2y         x-y             2y

Since the pressure of NOCl at equilibrium is 115 torr, we can conclude that:

2y = 115 torr

y = 57.5 torr

The pressures at equilibrium are:

pNO = x - 115

pCl₂ = x - 57.5

pNOCl = 115

We will put these values in the equilbrium constant Kp.

Kp = 0.27 = 115² / (x - 115)² × (x - 57.5)

x = 139 torr

NO and Cl₂ initially at 139 torr react to form NOCl at 115 torr at equilibrium.

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 The initial partial pressures of and is 139.4 torr.

Explanation:

The equilibrium reaction is

initially conc.         p            p                  0

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

Given:

The expression for equilibrium constant is :

= 2x =115 torr

= (p-2x) = p-115 torr

= (p-x)= p-57.5 torr

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

Therefore, the initial partial pressures of and is 139.4 torr.

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