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Equilibrium Constant and Standard Free Energy Change

The standard ...

Question

The standard free energy change of a reaction is $△ G^{o} = - 115 k J a t 298 K .$ Calculate the equilibrium constant $K_{P}$ in $l o g K_{P} . (R = 8.314 J K^{- 1} m o l^{- 1})$

A

20.16

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B

2.303

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C

2.016

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D

13.83

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Solution

The correct option is A 20.16
$l o g K_{p} = \frac{- △ G^{o}}{2.303 R T}$
$= \frac{- (- 115 \times 1000)}{2.303 \times 8.314 \times 298}$
$= 20.16$

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Standard XIII Chemistry

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