The standard Gibbs energy change at 300 K for the reaction 2A $\underset{}{⇌}$ B + C is 2494.2J. At a given time, the composition of the reaction mixture is [A] = $\frac{1}{2}$ , [B] = 2 and [C] = $\frac{1}{2}$ . The reaction proceeds in the : [R = 8.3 14 J/K/mol, e = 2.718]

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forward direction because Q < K_c

reverse direction because Q < K_c

forward direction because Q > K_c

reverse direction because Q > K_c

answer is D.

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Detailed Solution

$\begin{array}{l} {ΔG}^{\circ} = 2494 .2 J \\ 2 A ⇌ B + C . \\ [A] = \frac{1}{2}, [B] = 2, [C] = \frac{1}{2} \\ Q = \frac{[B] [C]}{[A]^{2}} = \frac{2 \times 1 / 2}{{(\frac{1}{2})}^{2}} = 4 \\ {ΔG}^{\circ} = - 2 .303 RTlog K_{C} \\ 2494 .2 J = - 2 .303 \times (8 .314 J / K / mol) \times (300 K) logK \\ \Rightarrow \log K_{C} = - \frac{2494 .2 J}{2 .303 \times 8 .314 J / K / mol \times 300 K} \\ \Rightarrow \log K_{C} = - 0 .4341; K_{C} = 0 .37; Q > K_{C} \end{array}$