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Delta G^(@) for a reaction is 46.06 kcal...

`Delta G^(@)` for a reaction is `46.06 kcal mol^(-)`. `K_(P)` for the reaction at `300 K` is

A

`10^(-8)`

B

`10^(-22.22)`

C

`10^(-33.55)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
`DeltaG^(@)=46-.06 kcal mol^(-1)`
`=46.06xx1000xx4.814 J mol^(-1)`
`DeltaG^(@)=-RT ln K_(P)=-2.303 RT log K_(P)`
`46.04xx1000xx4.184`
`=-2.303xx8.31xx300 log K_(P)`
or `log K_(P)=-33.55` or `K_(P)=10^(-33.55)`
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