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

Jan 2, 2017

Yes, it is certainly nonspontaneous, at all temperatures, if there is more order in the products, AND the reaction absorbs energy from the surroundings.

If there is less order in the products, AND the reaction absorbs energy from the surroundings, it is not obvious without knowing the temperature.

Recall the relationship between Gibbs' free energy, enthalpy, and entropy:

$\Delta G = \Delta H - T \Delta S$

We know that:

• when $\Delta G < 0$, the reaction is spontaneous.
• when $\Delta G = 0$, the reaction is at equilibrium.
• when $\Delta G > 0$, the reaction is nonspontaneous.

When a reaction is endothermic, energy is absorbed, and thus, $\Delta H > 0$. Furthermore, when a reaction becomes more ordered (less "positionally random"), the entropy has decreased (entropy increases with more motion in the system and vice versa), meaning that $\Delta S < 0$.

Thus, $\Delta G$ is necessarily positive since $T \ge \text{0 K}$, always. Let us assume that $T > 0$:

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

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

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

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

Therefore, the reaction is nonspontaneous at all temperatures if $\Delta H > 0$ AND $\Delta S < 0$.

However, if $\Delta S > 0$ (i.e. the particles become more "positionally random"), then it matters what the temperature is...

$\textcolor{red}{\Delta G} = \left(+\right) - \left(+\right) \left(+\right)$

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

= color(red)(???)

• $\Delta G < 0$ if $| T \Delta S | > | \Delta H |$
• $\Delta G > 0$ if $| T \Delta S | < | \Delta H |$

Thus, this kind of reaction is nonspontaneous at low temperatures, and spontaneous at high temperatures.