Question

The enthalpy change for a given reaction at 298 K is -x J mol⁻¹ (x being positive). If the reaction 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

Answer: (B)

Can be negative but numerically smaller than $\frac{x}{298}$

Answer 2

For a reaction to be spontaneous, the Gibbs Free Energy change ($\Delta G$) must be negative. The Gibbs Free Energy equation is $\Delta G = \Delta H - T\Delta S$. Given $\Delta H = -x$ (where $x > 0$) and $T = 298$ K. So, $\Delta G = -x - 298\Delta S$. For spontaneity, $\Delta G < 0$, which means $-x - 298\Delta S < 0$. Rearranging the inequality: $-298\Delta S < x$, which simplifies to $298\Delta S > -x$, or $\Delta S > -\frac{x}{298}$. This inequality implies that if $\Delta S$ is negative, its numerical value must be less than $\frac{x}{298}$.