Question

A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction $N_2 (g) + 3H_2 (g) ⇋ 2NH_3$ (g) is 1.7 × 102. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Answer 1

The given reaction is:
$N_2(g) + 3H_2(g) ↔ 2NH_3(g)$
The given concentration of various species is:
$[N_2] = \frac {1.57}{20}\ mol\ L^{-1}$

$[H_2] = \frac {1.92}{20}\ mol\ L^{-1}$

$[NH_3] = \frac {8.13}{20}\ mol\ L^{-1}$
Now, reaction quotient Qc is:
$Q_c = \frac {[NH_3]^2}{[N_2][H_2]^3}$

$Q_c= \frac {(\frac {8.13}{20})^2}{(\frac {1.57}{20})(\frac {1.92}{20})^3 }$

$Q_c= 2.4×10^3$

Since $Q_c ≠ K_c$ , the reaction mixture is not at equilibrium.
Again, $Q_c > K_c$. Hence, the reaction will proceed in the reverse direction.