Question

in an evacuated 2.0 liter container, 1.8 moles of N2 and 2.2 moles of H2 are placed. at equilibrium 0.6 M of NH3 are present. solve for Kc

Answer 1

$\begin{aligned} N_2 + 3H_2 \to 2NH_3 \end{aligned}$

Volume of cylinder = 2.0litre = 2.0$dm^3$

Concentration of $N_2$ = 1.8moles/2 $dm^3$ = 0.9M

Concentration of $H_2$ = 2.2moles/2 $dm^3$ = 1.10M

Before reaction;

0.9(N) + 1.10(H) = 0(NH3)

During reaction

-x(N) -3x(H) = 2x(NH3)

After reaction;

0.9-x(N) + 1.10-3x(H) = 0.6(NH3)

$\therefore$ x = 0.3M

Therefore, final conc. of $N_2= 0.9-0.3= 0.6M$

Final conc. of $H_2= 1.10-3(0.3) = 0.2M$

$K_c= \dfrac{[NH_3]^2}{[N_2][H_2]^3}$

$K_c= \dfrac{[0.6]^2}{[0.6][0.2]^3}$

$K_c = 75$