contestada

For the reaction 2NH3(g)↽−−⇀ 3H2(g)+N2(g) the equilibrium concentrations were found to be [NH3]=0.250 M , [H2]=0.500 M , and [N2]=0.800 M . What is the equilibrium constant for this reaction?

Respuesta :

Answer: The equilibrium constant for this reaction is 1.6

Explanation:

The chemical equation for the decomposition of ammonia follows:

[tex]2NH_3(g)\rightleftharpoons 3H_2(g)+N_2(g)[/tex]

The expression for equilibrium constant is given as:

[tex]K_{eq}=\frac{[N_2][H_2]^3}{[NH_3]^2}[/tex]

We are given:

[tex][NH_3]=0.250M[/tex]

[tex][H_2]=0.500M[/tex]

[tex][N_2]=0.800M[/tex]

Putting values in above equation, we get:

[tex]K_{eq}=\frac{(0.800)\times (0.500)^3}{(0.250)^2}\\\\K_{eq}=1.6[/tex]

Hence, the equilibrium constant for this reaction is 1.6