Question

The enthalpy change for a given reaction at 298 K is – x J mol–1 (x being positive). If the reaction occurs spontaneously at 298 K, the entropy change at that temperature

• A. Can be negative but numerically larger than x/298
• B. Can be negative but numerically smaller than x/298
• C. Cannot be negative
• D. Cannot be positive

Answer 1

Answer: (B)

Can be negative but numerically smaller than x/298

Answer 2

It is because of the fact that for spontaneity, the value of $\Delta$G = ($\Delta$H – T$\Delta$S) should be < 0. If $\Delta$S is – ve, the

value of T$\Delta$S shall have to be less than $\Delta$H or the value of $\Delta$S has to be less than $\frac{\Delta H}{T}$ i.e., $\frac{x}{298}$.