1) The enthalpy change for a given reaction at 298 K is -x Jmol-1, For the reaction to be spontaneous at 298 K , the entropy change at that temperature A) can be negative , but numerically greater than $\frac{x}{298 }Jk^{-1}$ B) can be negative , but numerically smaller than $\frac{x}{298 }Jk^{-1}$ C) can not be negative D) can not be positive Answer: Option BExplanation:$\triangle G=\triangle H -T\triangle S$ For spontaneous reaction $\triangle H=-ve$ $T\triangle S > \triangle H$ $\triangle G=\frac{-q}{T}=-\frac{-X}{298}=\frac{X}{298}$ So, $\triangle G $ can be negative
