The enthalpy change for a given reaction at 298 K is x J mol. If the reaction occurs spontaneously at 298 K, the entropy change at that temperature

Question

The enthalpy change for a given reaction at 298 K is x J mol. 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

Tardigrade Admin · Accepted Answer

It is because of the fact that for spontaneity the value of

Δ G = (Δ H - T Δ S)

should be < 0.
. If

Δ

S is -ve the value of T

Δ

S has to be numerically less than

Δ

H.
or Value of

Δ

S has to be numerically less than

Δ

H/T i.e., x/ 298

Q. The enthalpy change for a given reaction at 298 K is x J mol. If the reaction occurs spontaneously at 298 K, the entropy change at that temperature

can be negative but numerically larger than x/298

can be negative but numerically smaller than x/298

cannot be negative

cannot be positive

It 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 has to be numerically less than ΔH. or Value of ΔS has to be numerically less than ΔH/T i.e., x/ 298

It is because of the fact that for spontaneity the value of $Δ G = (Δ H - T Δ S)$ should be < 0.
. If $Δ$ S is -ve the value of T $Δ$ S has to be numerically less than $Δ$ H.
or Value of $Δ$ S has to be numerically less than $Δ$ H/T i.e., x/ 298