For a reaction at 25 -C enthalpy change -H and entropy change -S are -11-7 - 103 Jmol-1 and -105 J mol-1 K-1 respectively- The reaction is-

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Gibbs Free Energy & Spontaneity

For a reactio...

Question

For a reaction at 25 $^{\circ}$ C enthalpy change ( $Δ$ H) and entropy change ( $Δ$ S) are $- 11.7 \times 10^{3} J m o l^{- 1}$ and $- 105 J m o l^{- 1} K^{- 1}$ respectively. The reaction is:

spontaneous.

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non spontaneous.

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at equilibrium.

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can't say anything.

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Solution

The correct option is B non spontaneous.
Spontaneous reactions have negative free energy change and non spontaneous reactions have positive free energy change.
$Δ G = Δ H - T Δ S$ .
Substitute values in the above expression.

$Δ G = - 11.7 \times 10^{3} J m o l^{- 1} - 298 K \times - 105 J m o l^{- 1} K^{- 1} = 19590 J m o l^{- 1}$
Since, the gibbs free energy change is positive, the reaction is non spontaneous.

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