Answer: The relation between the forward and reverse reaction is [tex]K_f=\frac{1}{K_r}[/tex]
Explanation:
For the given chemical equation:
[tex]2NH_3(g)\rightleftharpoons N_2 (g)+3H_2(g)[/tex]
The expression of equilibrium constant for above equation follows:
[tex]K_f=\frac{[N_2][H_2]^3}{[NH_3]^2}[/tex] .......(1)
The reverse equation follows:
[tex]N_2 (g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex]
The expression of equilibrium constant for above equation follows:
[tex]K_r=\frac{[NH_3]^2}{[N_2][H_2]^3}[/tex] .......(2)
Relation expression 1 and expression 2, we get:
[tex]K_f=\frac{1}{K_r}[/tex]
Hence, the relation between the forward and reverse reaction is [tex]K_f=\frac{1}{K_r}[/tex]