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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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instantaneous

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none of these

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Solution

The correct option is B non-spontaneous
Given, $T = 25^{\circ} C$ $Δ H = - 11.7 \times 10^{3} J m o l^{- 1}$ $Δ S = - 105 J m o l^{- 1} K^{- 1}$
We know for a reaction,
$Δ G = Δ H - T Δ S$
where, $Δ H$ is enthalpy change and $Δ S$ is entropy change
putting the values in formula, we get;
$∴ Δ G = - 11700 - 298 \times (- 105) = + 19590 J = 19.59 k J$
As $Δ G$ is positive,
Thus, reaction is non - spontaneous.

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Similar questions

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?

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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

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Gibbs Free Energy & Spontaneity

Standard XII Chemistry

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For a reaction at 25∘C enthalpy change (ΔH) and entropy change (ΔS) are −11.7×103Jmol−1 and −105 Jmol−1K−1 respectively. The reaction is:

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: