Question

Calculate the equilibrium constant (K_c) for the formation of NH₃ in the following reaction: 

\[\ce{N2_{(g)} + 3H2_{(g)} <=> 2NH3_{(g)}}\]

At equilibrium, the concentration of NH₃, H₂ and N₂ are 1.2 × 10⁻², 3.0 × 10⁻² and 1.5 × 10⁻² M respectively.

Answer 1

\[\ce{N2 + 3H2 <=> 2NH3}\]

K_c = `(["NH"_3]^2)/(["N"_2]["H"_2]^3)`

= `((1.2 xx 10^-2)^2)/((1.5 xx 10^-2)(3 xx 10^-2)^3)`

= `(1.2 xx 1.2 xx 10^-4)/(1.5 xx 10^-2 xx 3 xx 3 xx 3 xx 10^-6)`

= `(1.2 xx 1.2 xx 10^4)/(1.5 xx 9 xx 3)`

= `(1.44 xx 10^4)/(1.5 xx 9 xx 3)`

= 400