Question

For the reaction

2NH

3

(g)↽−

−⇀

 3H

2

(g)+N

2

(g)

2⁢NH3⁢(g)⁢↽−−⇀ 3⁢H2⁡(g)+N2⁡(g)

the equilibrium concentrations were found to be [NH

3

]=0.250 M

[NH3]=0.250 M, [H

2

]=0.450 M

[H2]=0.450 M, and [N

2

]=0.800 M

[N2]=0.800 M. What is the equilibrium constant for this reaction?

Answer 1

Ans:-

Reaction is

$2⁢NH_3⁢(g)⁢\iff 3⁢H_2⁡(g)+N_2⁡(g)$

We have $[NH_3]=0.250 M$ $[H_2]=0.450 M$ $[N_2]=0.800 M$

$\Rightarrow$ Equilibrium constant ($K$)$=\dfrac{[H_2]^3\times[N_2]}{[NH_3]^2}$

$\Rightarrow$ Equilibrium constant ($K$)=$\dfrac{(0.450)^3\times(0.800)}{(0.250)^2}=1.1664$

$\therefore$ Equilibrium constant ($K$)=$1.1664$