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 determine the entropy change (ΔS) for a reaction that has a given enthalpy change (ΔH) at a specific temperature (T). The reaction is spontaneous at 298 K, and the enthalpy change is given as -x cal mol^(-1).

### Step-by-Step Solution:

1. **Understand the relationship between ΔH, ΔS, and T**:
The Gibbs free energy change (ΔG) for a reaction is given by the equation:
\[
\Delta G = \Delta H - T \Delta S
...