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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^(-22.22)`

B

`10^(-8)`

C

`10^(-44.55)`

D

`10^(-35.54)`

Text Solution

Verified by Experts

The correct Answer is:
D
According to thermodynamics,
`Delta G^(@) = - 2.303 RT log k_(P)`
We have `Delta G^(@) = 46.06 kcal mol^(-1)`
`= (46.06) (1000) (4.184) J mol^(-1)`
`R = 8.314 J K^(-1) mol^(-1)`
`T = 300 K`
`:. Log K_(p) = (Delta G^(@))/(-2.303 RT)`
`= ((46.06)(1000)(4.184))/(-(2.303)(8.314)(300))`
`= (192715.04)/(5744.14)`
`= 33.54`
or `K_(P) = 10^(-33.54)`
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