Question

The enthalpy change for a given reaction at $298K$ is $-xJmo{l}^{-1}$. For the reaction to be spontaneous at $298K$, the entropy change at that temperature

• A. can be negative, but numerically greater than $\frac{x}{298}J{K}^{-1}$
• B. can be negative, but numerically smaller than $\frac{x}{298}J{K}^{-1}$
• C. can not be negative
• D. can not be positive

Answer 1

Answer: (C)

The enthalpy change for a given reaction at $298K$ is $-xJ/mol$.
For the reaction to be spontaneous at $298K$, the entropy change at that temperature cannot be negative.
For a spontaneous reaction
$\Delta G=\Delta H-T\Delta S<0$
We already know that $\Delta H$ is negative.
If $\Delta S$ is negative then $\Delta G$ as a whole might turn positive if it is numerically greater than $\frac{x}{298}$.