Consider the following reaction and its equilibrium constant:
$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g) K c = 5.85 \times 10^{- 3}$

A reaction mixture at a particular time, t contains: $[N O_{2}] = 0.025 M a n d [N_{2} O_{4}] = 0.033 M .$ Calculate $Q c$ , and determine the direction in which the reaction will proceed.

1.9 \times 10^{- 3}

, the reaction will proceed in forward direction

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1.9 \times 10^{- 2}

, the reaction will proceed in backward direction

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5.85 \times 10^{- 3}

, it will be equal to the equilibrium constant

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None of the above

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Solution

The correct option is B $1.9 \times 10^{- 2}$ , the reaction will proceed in backward direction
Given reaction is $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g) K c = 5.85 \times 10^{- 3}$

$[N O_{2}] = 0.025 M [N_{2} O_{4}] = 0.033 M$
Reaction Quotient $(Q_{c}) = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]}$
$Q_{c} = \frac{(0.025)^{2}}{(0.033)}$

$Q_{c} \approx 0.019 = 1.9 \times 10^{- 2}$

comparing $Q_{c}$ with $K_{c}$ we get $Q_{c} > K_{c}$
Hence, the reaction will proceed in backward direction.

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Similar questions

Q. Consider the following reaction and its equilibrium constant:

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g) K c = 5.85 \times 10^{- 3}

A reaction mixture at a particular time, t contains:

[N O_{2}] = 0.025 M a n d [N_{2} O_{4}] = 0.033 M .

Calculate

Q c

, and determine the direction in which the reaction will proceed.

The reaction N $_{2}$ O $_{4}$ (g) $⇌$ 2NO $_{2}$ (g) is
carried out at 298 K and 20 bar. Five moles of each of N $_{2}$ O $_{4}$ and
NO $_{2}$ , are taken initially. Given: $Δ$ $_{f}$ G $^{\circ}$
(N $_{2}$ O $_{4}$ ) = 100 kJ mol $^{- 1}$ and $Δ$ $_{f}$ G $^{\circ}$
(NO $_{2}$ ) = 50 kJ mol $^{- 1}$ . The reaction proceeds at an initial pressure
of 20 bar. Determine the direction in which the reaction proceeds to achieve
equilibrium. Also determine the amounts of N $_{2}$ O $_{4}$ and NO $_{2}$ when the reaction attains equilibrium.