A mixture of 1.57 mol of N₂, 1.92 mol of H₂ and 8.13 mol of NH₃ is…

Question 12 of 69intermediate🔧 ApplyNumerical3 marks

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 $N_2 (g) + 3H_2 (g) \rightleftharpoons 2NH_3 (g)$ is $1.7 \times 10^2$ . Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Correct Answer

The reaction mixture is not at equilibrium. The direction of the net reaction is in the reverse direction.

Exercise: EXERCISES | Q: 6.12 | (Page 42)

For More Understanding

Explanation

To determine if the reaction is at equilibrium, we calculate the reaction quotient, $Q_c$ , and compare it with the equilibrium constant, $K_c$ . If $Q_c > K_c$ , the reaction will proceed in the reverse direction to reach equilibrium.

Solution Steps

Step 1: Calculate the initial concentrations of the reactants and products. Volume of the vessel, $V = 20$ L
$[N_2] = \frac{1.57 \text{ mol}}{20 \text{ L}} = 0.0785 \text{ M}$ $[H_2] = \frac{1.92 \text{ mol}}{20 \text{ L}} = 0.096 \text{ M}$ $[NH_3] = \frac{8.13 \text{ mol}}{20 \text{ L}} = 0.4065 \text{ M}$
Step 2: Write the expression for the reaction quotient, $Q_c$ . For the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ , the reaction quotient is:
$Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$
Step 3: Substitute the concentration values into the $Q_c$ expression.
$Q_c = \frac{(0.4065)^2}{(0.0785) \times (0.096)^3}$ $Q_c = \frac{0.1652}{0.0785 \times 0.0008847}$ $Q_c = \frac{0.1652}{0.00006945} = 2.38 \times 10^3$
Step 4: Compare $Q_c$ with $K_c$ to determine the direction of the net reaction. Given $K_c = 1.7 \times 10^2 = 170$ Since $Q_c$ ( $2.38 \times 10^3$ ) is greater than $K_c$ ( $1.7 \times 10^2$ ), the reaction mixture is not at equilibrium. To reach equilibrium, the value of $Q_c$ must decrease, which means the reaction will proceed in the reverse direction (decomposing $NH_3$ into $N_2$ and $H_2$ ).

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