Question

The equilibrium constant for the reaction: 

`N_(2(g)) + 3H_(2(g)) ⇌ 2NH_(3(g)) at 715 K, is 6.0 xx 10^(-2)`

If, in a particular reaction, there ate 0·25 mol L^-1 of H₂. and 0·06 mol L^-1 of NH₃ present, calculate the concentration of N₂ at equilibrium

Answer 1

`N_(2(g)) + 3H_(2(g)) ⇌ 2NH_(3(g))`

K (equilibrium constant) = `6.0 xx 10^(-2) mol L^(-1)`

Temperature (T) = 715 K

`[H_2] = 0.25 mol L^(-1)`

`[NH_3] = 0.06 mol L^(-1)`

`[N_2] = ?`

`K= [NH_3]^2/([N_2]xx[H_2]^3)`

`[N_2] = [NH_3]^2/[Kxx[H_2]^3]`

`= (0.06)^2/(6.0 xx 10^(-2) xx (0.25)^3)`

`= 0.0036/(6.0 xx 10^(-2)xx 0.015625)`

`[N_2] = 3.84  mol L^(-1)`