For a reaction at $25^{\circ} C$ enthalpy change ( $△ H$ ) and entropy change ( $△ S$ ) are -11.7 K J $m o l^{- 1}$ and -105 J $m o l^{- 1}$ $K^{- 1}$ , respectively. It is a spontaneous reaction

True

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False

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Solution

The correct option is B
False

Given, $△ H$ = $11.7 \times 103 J m o l^{- 1}$
$△ S$ = $- 105 J m o l^{- 1} K^{- 1}$ , T = 298 K
$△ G$ = $△ H$ - T $△ S$ = $- 11700 - 298 \times (- 105)$
= 19590 J = + 19.59 kJ
Since $△ G$ = +ve
Therefore, reaction is not spontaneous.

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For a reaction at $25^{\circ} C$ enthalpy change ( $△ H$ ) and entropy change ( $△ S$ ) are -11.7 K J $m o l^{- 1}$ and -105 J $m o l^{- 1}$ $K^{- 1}$ , respectively. It is a spontaneous reaction

For a reaction at $25^{\circ} C$ enthalpy change $(△ H)$ and entropy change $(△ S)$ are $- 11.7 \times 103 J m o l^{- 1}$ and $- 105 J m o l^{- 1} K^{- 1}$ respectively. Find out whether this reaction is spontaneous or not?