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)=([NO]^2[br_2])/([NOBr]^2)\\\\K_(eq)=(0.18^2* 0.9)/(0.82^2)\\\\K_(eq)=0.043\ atm`

Hence , this is the required solution .