DeltaG^(@)for a reactionis 46.06kcal//mole , K_(p) for the reaction at

1.	DeltaG^(@)for a reactionis 46.06kcal//mole , K_(p) for the reaction at 300K is
Answer» <P>`10^(-8)` ` 10^(22.22)` `10^(-33.33)` NONE of these Solution :`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)`

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