# Is a reaction that is exothermic and becomes more positionally random spontaneous or non spontaneous? What about less positionally random? Is there enough information to tell?

Jan 2, 2017

See this answer for the endothermic case.

For the exothermic case, the change in enthalpy $\Delta H < 0$, and if the reaction becomes more "positionally random", then the change in entropy $\Delta S > 0$ since there is more motion.

Therefore, the Gibbs' free energy is:

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

$= \left(-\right) - \left(+\right) \left(+\right)$

$= \left(-\right) - \left(+\right)$

$= \textcolor{b l u e}{\left(-\right)}$

which if you recall, is negative for a spontaneous reaction.

So, an exothermic reaction with increased motion (more "positional randomness") is spontaneous at all temperatures.

Less "positionally random" implies less motion and thus $\Delta S < 0$. I think you are at the point where you can guess what the sign of $\Delta G$ would conditionally be (hint: it's analogous to having $\Delta H > 0$ and $\Delta S > 0$; how would the magnitude of $T$ affect the sign of $\Delta G$?).