Question

Nitrosyl bromide decomposes according to the chemical equation below. 2NOBr(g) 2NO(g) + Br2(g) 1.00 atm of NOBr is sealed in a flask. At equilibrium, the partial pressure of NOBr is 0.82 atm. What is the equilibrium constant for the reaction?

Answer 1

Given :

2NOBr(g) - -> 2NO(g) + Br2(g)

Initial pressure of NOBr , 1 atm .

At equilibrium, the partial pressure of NOBr is 0.82 atm.

To Find :

The equilibrium constant for the reaction .

Solution :

             2NOBr(g) - -> 2NO(g) + Br2(g)

t=0 s           1 atm                 0             0

$t=t_{eqb}$       1( 1-2x)               2x           x

So ,

$1-2x=0.82\\\\x=0.09$

At equilibrium :

$K_{eq}=\dfrac{[NO]^2[br_2]}{[NOBr]^2}\\\\K_{eq}=\dfrac{0.18^2\times 0.9}{0.82^2}\\\\K_{eq}=0.043\ atm$

Hence , this is the required solution .