Question #d46e8

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Question #d46e8

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Answer 1 · 2015-11-18T15:18:55

See explanation.

Explanation:

For spontaneous processes, the change on the free energy $Δ G$ $color(green)(DeltaG)$ has to be negative.

The relationship between free energy $Δ G$ $color(green)(DeltaG)$ and the enthalpy $Δ H$ $color(blue)(DeltaH)$ is the following:

$Δ G = Δ H - T Δ S$ $color(green)(DeltaG)=color(blue)(DeltaH)-Tcolor(red)(DeltaS)$

For endothermic processes, $Δ H$ $color(blue)(DeltaH)$ is positive. Therefore, for the process to be spontaneous:
$Δ G = Δ H - T Δ S < 0$ $color(green)(DeltaG)=color(blue)(DeltaH)-Tcolor(red)(DeltaS)<0$

$\Rightarrow Δ H < T Δ S$ $=>color(blue)(DeltaH)< Tcolor(red)(DeltaS)$

Since $Δ H > 0$ $color(blue)(DeltaH)>0$ , for the term $T Δ S$ $Tcolor(red)(DeltaS)$ to be greater than $Δ H$ $color(blue)(DeltaH)$ , the change on the entropy has to be positive as well ( $Δ S > 0$ $color(red)(DeltaS)>0$ ).

$\Rightarrow T > \frac{Δ H}{Δ S}$ $=>T > (color(blue)(DeltaH))/(color(red)(DeltaS))$

Therefore, we can say that endothermic processes can be spontaneous at high temperature.

Note that, endothermic reactions are never spontaneous if the change on entropy is negative.

$If Δ H > 0 and Δ S < 0 \Rightarrow Δ G > 0 Always$ $"If " color(blue)(DeltaH)>0 and color(red)(DeltaS)<0 =>color(green)(DeltaG)>0 " Always"$

You can follow this figure to predict the spontaneity of the reaction based on the signs of $Δ H and Δ S$ $color(blue)(DeltaH) and color(red)(DeltaS)$
enter image source here

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Question #d46e8

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