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

Options

A. `10^(-8)`
B. ` 10^(22.22)`
C. `10^(-33.33)`
D. none of these

Answer

C

Explanation

`DeltaG^(@) = 46.06 kcal mol^(-1)`
`=46.06 xx 1000xx4.184J mol^(-1)`
`DeltaG^(@) = - RT lnK_(p) =- 2.303 RT logK_(p)`
` 46.06 xx 1000 xx 4.184 = - 2.303 xx 8.314 xx 300 log K_(p)`
or `log K_(p) = - 33.55 `or`K_(p) =10^(-33.55)`