Question

For the reaction:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + Heat; the equilibrium constant K = 0.50 at 673K

(a) Write the expression for K_c and K_p.

(b) What will be the equilibrium constant value for the reverse reaction at 673K?

(c) What is the effect if increasing temperature on the yield of NH₃?

(d) What is the effect adding N₂(g) and H₂(g) on the yield of NH₃?

(e) What is the effect of increasing pressure on the yield of NH₃?

Answer 1

(a) K_c = \(\frac{[NH_3]^2}{[N_2][H_2]^3}\)

And K_p = \(\frac{[P_{NH_3}]^2}{P_{N_2}\times(P_{H_2})^3}\)

(b) K_eq. (For reverse reaction) = \(\frac{1}{K}\) = \(\frac{1}{0.50}\) = 2

(c) As this is an exothermic reaction, on increasing the temperature, the reaction will proceed in reverse direction. This results in decrease in concentration of NH₃.

(d) On increasing the concentration of N₂(g) and H₂(g), the yield of NH₃ will increase.

(e) On increasing the pressure, the reaction will proceed in forward reaction in which number of moles of gas or pressure decreases.