Question

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

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

Answer 1

Answer: (C)

The enthalpy change for a given reaction at $298\, K$ is $- x J / mol$.
For the reaction to be spontaneous at $298 \, K$, 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 $\frac{ x }{298}$.