Q.A mixture of 1.57 mol of N 2, 1.92 mol of H 2 and 8.13 mol of NH 3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K c for the reaction N 2

The key idea is to compare the reaction quotient $Q_c$ with $K_c$ to determine the direction of net reaction.

Step 1: Calculate the initial concentrations.

Volume = 20 L.

$[N_2] = \frac{1.57}{20} = 0.0785$ M

$[H_2] = \frac{1.92}{20} = 0.096$ M

$[NH_3] = \frac{8.13}{20} = 0.4065$ M

Step 2: Write the expression for $Q_c$ and substitute.

For $N_2 + 3H_2 \rightleftharpoons 2NH_3$,

Step 3: Compute $Q_c$.

$(0.096)^3 = 8.847 \times 10^{-4}$

Denominator = $0.0785 \times 8.847 \times 10^{-4} = 6.945 \times 10^{-5}$

Numerator = $0.1652$

$Q_c = \frac{0.1652}{6.945 \times 10^{-5}} \approx 2.38 \times 10^3$ …

Approach

For $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$, evaluate $Q_c$ from the current concentrations. If $Q_c < K_c$ the reaction goes forward; if $Q_c > K_c$ it goes reverse; if equal, it is at equilibrium.

Step-by-step solution

1. Concentrations (volume $= 20\ \text{L}$)

2. Reaction quotient

$Q_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(0.4065)^2}{(0.0785)(0.096)^3} = \frac{0.1652}{6.95 \times 10^{-5}} = 2.38 \times 10^3$ …