The standard Gibbs energy change at 300 K for the reaction $2 A ⇌ B + C$ is $2494.2 J {mol}^{- 1}$ At a given time, the composition of the reaction mixture is $[A] = (1 / 2) mol, [B] = 2 mol$ and $[C] = (1 / 2) mol$
The reaction proceeds in the ( Given : $R = 8.314 {JK}^{- 1} {mol}^{- 1}, e = 2.718$ )

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reverse direction because $Q > K_{c}$

forward direction because $Q > K_{c}$

reverse direction because $Q < K_{c}$

forward direction because $Q < K_{c}$

answer is B.

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Detailed Solution

From the expression $Δ_{r} G^{o} = - R T \ln K_{c}^{o}$ we get

$\ln K_{c}^{o} = - \frac{Δ_{r} G^{\circ}}{R T} = - \frac{(2492.2 J {mol}^{- 1})}{(8.314 {JK}^{- 1} {mol}^{- 1}) (300 K)} = - 1$ Thus $K_{c}^{o} = e^{- 1} = (2.718)^{- 1} = 0.3679$

Now $Q_{c} = \frac{[B] [C]}{[A]^{2}} = \frac{(2 mol) (1 mol / 2)}{(1 mol / 2)^{2}} = 4$

Since $Q_{c} > K_{c}^{\circ}$ , the reaction will proceed in the reverse direction.

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