A mixture of 1.57 mol of $N_{2}$ ,1.92 mol of $H_{2}$ and 8.13 mol of ${N H}_{3}$ is introduced in 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 $1.7 \times 10^{2}$ .

Is the reaction mixture at equilibrium? If not what is the direction of net reaction?

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Solution

$N_{2} (g) + 3 H_{2} (g) = 2 N H_{3} (g)$
so, reaction quotient $(Q_{c}) = \frac{[N H_{3}]^{2}}{[H_{2}]^{3} [N_{2}]}$

Given ,
$[N H 3] = 8.13 m o l / 20 L$ { concentration = mole/volume }
= $0.4065 M$

$[N_{2}] = 1.57 m o l / 20 L = 0.0785 M$

$[H$ 2] = $1.92 m o l / 20 L = 0.096 M$

$Q_{c} = \frac{(0.4065 M)^{2}}{(0.096 M)^{3} (0.0785 M)}$
$= 2.379 \times 10^{3} / M^{2}$

Here, you can see that , $K_{c}$ is not equal to $Q_{c}$ , so the reaction is not in equilibrium.

because $Q_{c} > K_{c}$ it means that the reaction will proceed in backward direction or in the direction of reactants .

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Similar questions

Q. 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)

1.7 \times 10^{2}

. Is the reaction mixture at equilibrium? If not what is the direction of the net reaction?

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, K_c for the reaction

Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

A mixture of 1.57 mol of $N_{2}$ , 1.92 mol of $H_{2}$ and 8.13 mol of $N H_{3}$ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, $K_{c}$ for the reaction
$N_{2} (g) + 3 H_{2} (g) \leftrightarrow 2 N H_{3} (g) i s 1.7 \times 10^{2}$
Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?