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

Correct option (B) can be negative but numerically smaller than x/298

Explanation:

It I is because of the fact that for spontaneity, the value of ΔG = (ΔH – TΔS) should be < 0. If ΔS is – ve, the value of TΔS shall have to be less than ΔH or the value of ΔS has to be less than that of x