The enthalpy change for a given reaction at `298 K` is `-x cal mol^(-1)`. If the reaction occurs spontaneously at `298 K`, the entropy change at that temperature

Options

A. can be negative but numerically larger than `x//298 cal K^(-1)`
B. can be negative but numerically smaller than `x//298 cal K^(-1)`
C. cannot be negative
D. cannot be positive

Answer

B

Explanation

`DeltaG=DeltaH-T DeltaS`
For spontaneity, `DeltaG lt 0`
As `DeltaH= -xkcal` is negative and `T=298 K`. For `DeltaG` to be `lt 0`, `DeltaS` can be negative but numerically less than `x//298`.