Question

Consider the following reaction: Consider the 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 119 torr and that of Br2 is 151 torr . What is the partial pressure of NOBr in this mixture?

Answer 1

Answer:  partial pressure of NOBr is 7792 atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

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

Equilibrium constant is given as:

$K_{p}=\frac{[p_{NOBr}]^2}{[p_{NO}]^2\times [p_{Br_2}]^1}$

$28.4=\frac{[p_{NOBr}]^2}{[(119)^2\times (151)^1}$

$[p_{NOBr}]=7792$ atm

Partial pressure of NOBr is 7792 atm