Question

Consider the reaction: 2 NO(g) + Br2(g) ∆ 2 NOBr(g) Kp = In a reaction mixture at equilibrium, the partial pressure of NO is 108 torr and that of Br2 is 126 torr. What is the partial pressure of NOBr in this mixture?

Answer 1

The question is incomplete, here is the complete question:

Consider the following reaction:

2NO(g) +Br₂(g) ⇔ 2NOBr(g)   Kp= 28.4 at 298 K

In a reaction mixture at equilibrium, the partial pressure of NO is 107 torr and that of Br₂ is 166 torr. What is the partial pressure of NOBr in this mixture?

Answer: The equilibrium partial pressure of NOBr in the mixture is 6460.53 torr

Explanation:

We are given:

Equilibrium partial pressure of NO = 108 torr

Equilibrium partial pressure of bromine gas = 126 torr

The given chemical equation follows:

$2NO(g)+Br_2(g)\rightleftharpoons 2NOBr(g)$

The expression of $K_p$ for above equation follows:

$K_p=\frac{(p_{NOBr})^2}{(p_{NO})^2\times p_{Br_2}}$

We are given:

$K_p=28.4$

Putting values in above expression, we get:

$28.4=\frac{(p_{NOBr})^2}{(108)^2\times 126}\\\\p_{NOBr}=6460.53,-6460.53$

Neglecting the negative value because partial pressure cannot be negative.

Hence, the equilibrium partial pressure of NOBr in the mixture is 6460.53 torr