For a reaction:
$2 A (g) + 4 B (g) ⟶ 5 C (g) + 2 D (l) Δ E^{\circ} = 40 kcal / mole$
and $Δ S^{\circ} = + 200 cal / Kmole$ . The value of $Δ G_{200}^{\circ}$ will be:

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+ 0.4 kcal/mole

- 0.4 kcal/mole

- 360 kcal/mole

answer is C.

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Detailed Solution

$\begin{array}{l} Δ H^{\circ} = 40 - \frac{1 \times 2 \times 200}{1000} = 39.6 \\ Δ G_{200}^{\circ} = 39.6 - \frac{200 \times 200}{1000} = - 0.4 kcal / mol \end{array}$

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