Chemistry, asked by rukshar91, 1 year ago

the equilibrium constant for the reaction n2(g)+3h2(g)⇌2nh3(g) is K. what will be the equilibrium constant for the reaction involving decomposition of one mole of ammonia?

Answers

Answered by IlaMends
25

Answer: The equilibrium constant for the reaction involving decomposition of one mole of ammonia is:

K'=\frac{1}{K}

Explanation:

N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

The equilibrium constant will be given as :

K=\frac{[NH_3]^2}{[N_2][H_2]^3}

Now the equilibrium constant for the reaction of decomposition of 1 mole of ammonia:

2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

K'=\frac{[N_2][H_2]^3}{[NH_3]^2}=\frac{1}{K}

The equilibrium constant for the reaction involving decomposition of one mole of ammonia is:

K'=\frac{1}{K}

Answered by pkdhas73
0

Explanation:

A+3B-----2X,K=x what will be the equilibrium constant for the decomposition of one mol of X

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