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?

Spontaneous

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Non-spontaneous

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Solution

The correct option is B
Non-spontaneous

For checking spontaneity, we need to check the value of $△ G$ .
If $△ G$ = -ve $\to$ spontaneous
$△ G$ = +ve $\to$ Non-spontaneous
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
$∴$ reaction is not spontaneous.

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For a reaction at 25∘C enthalpy change (△H) and entropy change (△S) are −11.7×103Jmol−1 and −105Jmol−1K−1 respectively. Find out whether this reaction is spontaneous or not?

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?