Question

For the nitrogen fixation reaction 3H2(g) + N2(g) = 2NH3(g),

Kc = 6.0 x 10-2 at 500°C. If 0.253 M H2 and 0.044 M NH3 are
present at equilibrium, what is the equilibrium concentration
of N2?

Answer 1

The equilibrium concentration of N₂ : 1.992

Further explanation

Given

Kc = 6.0 x 10⁻² at 500°C

0.253 M H₂ and 0.044 M NH₃

Reaction

3H₂(g) + N₂(g) = 2NH₃(g)

Required

The equilibrium concentration of N₂

Solution

Kc for the reaction :

`\tt Kc=([NH_3]^2)/([H_2]^3[N_2])\\\\0.06=(0.044^2)/(0.253^3* [N_2])\\\\(N_2]=1.992`