For a reversible reaction at $298$ K, the equilibrium constant K is $200$ . What is the value of $△ G^{0}$ at $298$ K?

- 13.13

kcal

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0.13

kcal

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3.158

kcal

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0.413

kcal

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Solution

The correct option is A - $3.158$ kcal
Applying $△ G^{0} = 2.303 R T \times l o g K$
$= - 2.303 \times 2 \times 298 \times l o g 200$
= $- 3158.4 c a l = - 3.158 k c a l$

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