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 $\frac{x}{298}$

$\Delta G=\Delta H-T\Delta S$

here $\Delta G$ has to be$-ve$ as for spontaneous reaction.

$\Delta H=-x$

Then

$\Delta S=-ve$ and the value will be $=(\Delta G-x)/298$