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 x/298
• B. Can be negative but numerically smaller than x/298
• C. Cannot be negative
• D. Cannot be positive

Answer 1

The correct option is B

Can be negative but numerically smaller than x/298

It is because of the fact that for spontaneity the value of $\Delta G=(\Delta H-T\Delta S)$ should be $<0$. If $\Delta S$ is -ve, the value of $T\Delta S$ shall have to be less than $\Delta H$

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

at eq. $\Delta G=0$

$\Delta H=T\Delta S$

$\Delta S=\frac{\Delta H}{T}$

$\Delta S=\frac{-x}{298}$
$\Delta G=\Delta H(-ve)-T\Delta S(-ve)$
reaction to spontane.