Equilibrium constant, $K_{C}$ for the reaction
$N_{2} (g) + 3 H_{2} (g) ⇋ 2 N H_{3} (g)$ at 500 K is 0.061 $L^{2} m o l^{- 2}$ .
At a particular time, the analysis shows that composition of the reaction mixure is $3.0 m o l L^{- 1}$ $N_{2}, 2.0$ $m o l L$ $^{- 1} H_{2}$ and $0.5 m o l L^{- 1}$ $N H_{3}$ . What will be the direction of equilibrium?

at equilibrium

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not at equilibrium, backward shift

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not at equilibrium, forward shift

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can not be predicted

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Solution

The correct option is C not at equilibrium, forward shift
$N_{2} (g) + 3 H_{2} (g) ⇋ 2 N H_{3} (g) t = t_{e q} 3 M 2 M 0.5 M Q = \frac{0.5 \times 0.5}{3 \times 2^{3}} = \frac{1}{3 \times 8 \times 4} = \frac{1}{96}$
$∵$ Q < K, system is not at equilibrium and equilibrium will be reached by shifting in forward direction

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Similar questions

Q. Equilibrium constant,

K_{C}

for the reaction

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

500 K

0.061

At a particular time, the analysis shows that composition of the reaction mixture is

3.0 m o l L^{- 1} N_{2}, 2.0 m o l L^{- 1} H_{2}

and

0.5 m o l L^{- 1} N H_{3} .

Is the reaction at equilibirium and if not in which direction does the reaction tend to proceed to reach equilibrium?

Q. Equilibrium constant,

K_{C}

for the reaction at

500 K

0.061

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

. At a particular time, the analysis shows that composition of the reaction mixture is

3.0 m o l L^{- 1}

N_{2},

2.0 m o l L^{- 1} H_{2}

and

0.5 m o l L^{- 1} N H_{3}

. Is the reaction at equilibrium and if not in which direction does the reaction tend to proceed to reach equilibrium?

Equilibrium constant, K_cfor the reaction

at 500 K is 0.061.

At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L^–1 N₂, 2.0 mol L^–1 H₂ and 0.5 mol L^–1 NH₃. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

Equilibrium constant, K_c for the reaction

at 500 K is 0.061.