The enthalpy change for a given reaction at `298 K` is `-x cal mol^(-1)`. If the reaction occurs spontaneously at `298 K`, the entropy change at that temperature

Options

A. Can be `-ve` but numerically latger than `-x//298 cal K^(-1)`
B. Can be `-ve`, but numerically smaller than `x//298 cal K^(-1)`
C. Cannot be negative
D. Cannot be positive

Explanation

To solve the problem, we need to analyze the relationship between enthalpy change (ΔH), entropy change (ΔS), and Gibbs free energy change (ΔG) for the reaction occurring at 298 K. The key equation we will use is:

\[ \Delta G = \Delta H - T \Delta S \]

Given:
- ΔH = -x cal mol\(^{-1}\) (where x is a positive value)
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