# Question #d46e8

Nov 18, 2015

See explanation.

#### Explanation:

For spontaneous processes, the change on the free energy $\textcolor{g r e e n}{\Delta G}$ has to be negative.

The relationship between free energy $\textcolor{g r e e n}{\Delta G}$ and the enthalpy $\textcolor{b l u e}{\Delta H}$ is the following:

$\textcolor{g r e e n}{\Delta G} = \textcolor{b l u e}{\Delta H} - T \textcolor{red}{\Delta S}$

For endothermic processes, $\textcolor{b l u e}{\Delta H}$ is positive. Therefore, for the process to be spontaneous:
$\textcolor{g r e e n}{\Delta G} = \textcolor{b l u e}{\Delta H} - T \textcolor{red}{\Delta S} < 0$

$\implies \textcolor{b l u e}{\Delta H} < T \textcolor{red}{\Delta S}$

Since $\textcolor{b l u e}{\Delta H} > 0$, for the term $T \textcolor{red}{\Delta S}$ to be greater than $\textcolor{b l u e}{\Delta H}$, the change on the entropy has to be positive as well ($\textcolor{red}{\Delta S} > 0$).

$\implies T > \frac{\textcolor{b l u e}{\Delta H}}{\textcolor{red}{\Delta S}}$

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.

$\text{If " color(blue)(DeltaH)>0 and color(red)(DeltaS)<0 =>color(green)(DeltaG)>0 " Always}$

You can follow this figure to predict the spontaneity of the reaction based on the signs of $\textcolor{b l u e}{\Delta H} \mathmr{and} \textcolor{red}{\Delta S}$ Here is a video that further explains this topic:
Thermodynamics | Spontaneous Process & Entropy.