A mixture of 1.57 mol of $N_{2}$ , 1.92 mol of H $_{2}$ and 8.13 mol of $N H_{3}$ is introduces into a 20L reaction vessel at 500K. At this temperature, the equilibrium constant, $K_{C}$ for the reaction $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ is $1.7 \times 1 0^{2}$ . Is the reaction mixture at equilibrium? If not what is the direction of the net reaction?

Question

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Answer 1

The reaction for the formation of ammonia is $N_{2} + 3 H_{2} ⇋ 2 N H_{3}$ . The expression for the reaction quotient is $Q = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}$ .

Substitute values in the above expression.

$Q = \frac{( \frac{8 . 1 3}{2 0} ) ^{2}}{( \frac{1 . 5 7}{2 0} ) ( \frac{1 . 9 2}{2 0} ) ^{3}} = 2.38 \times 1 0^{3}$

But the value of $K_{c}$ is $1.7 \times 1 0^{2}$ . Thus, the value of the reaction quotient is greater than the value of the equilibrium constant $(Q > K_{c})$ .

Thus, the reaction is not at equilibrium. To attain an equilibrium, the reaction will shift backward.

Answer 2

The reaction for the formation of ammonia is

N_{2} + 3 H_{2} ⇋ 2 N H_{3}

. The expression for the reaction quotient is

Q = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}

.

Substitute values in the above expression.

Q = \frac{( \frac{8 . 1 3}{2 0} ) ^{2}}{( \frac{1 . 5 7}{2 0} ) ( \frac{1 . 9 2}{2 0} ) ^{3}} = 2.38 \times 1 0^{3}

But the value of

K_{c}

is

1.7 \times 1 0^{2}

. Thus, the value of the reaction quotient is greater than the value of the equilibrium constant

(Q > K_{c})

.

Thus, the reaction is not at equilibrium. To attain an equilibrium, the reaction will shift backward.