The enthalpy changes for given reaction at 298 K is - x cal/mol. If the reaction occurs spontaneously at 298 K, the entropy change at that temperature

Question

The enthalpy changes for given reaction at 298 K is - x cal/mol. If the reaction occurs spontaneously at 298 K, the entropy change at that temperature, can be negative, but numerically larger than x/298 cal mol − 1 K − 1, can be negative, but numerically smaller than x/298 cal mol − 1 K − 1, cannot be negative, cannot be positive

MARKS App · Accepted Answer

Δ G = Δ H - T Δ S For a spontaneous process, Δ H - T Δ S 0 . So, ΔS > ΔH T = − x cal mol 298 K Hence, Δ S is numerically smaller than x 298 cal mol − 1 K − 1 .

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