Question

The enthalpy change for a given reaction at 298 K is $-xJmo{l}^{-1}$ (x being positive). If the reaction occurs spontaneously at 298 K, the entropy change at that temperature

• A. can be negative but numerically larger than $\frac{x}{298}$
• B. can be negative but numerically smaller than $\frac{x}{298}$
• C. cannot be negative
• D. cannot be positive

Answer 1

The correct option is B can be negative but numerically smaller than $\frac{x}{298}$

It is because of the fact that for spontanity of raction, the value of $△G<0$
Also, $△G=(△H-T△S)$ .
If value of $△S$ is -ve, then the value of $-T△S$ will be positive. To get over all value less than zero, the value of $T△S$ should be less than $△H$ as $△H$ has a negative value or the numerical value of $△S$ has to be less than $\frac{△H}{T},i.e.,\frac{x}{298}.$