What does it mean when a reaction is spontaneous?

Jul 11, 2018

It just means the reaction is energetically downhill, i.e. energy is released and the system overall becomes more thermodynamically stable as a result.

Spontaneity is a measure of how energetically feasible the reaction is, assuming that it would occur quickly if it can occur. That is, if the kinetics are fast enough for us to observe in our lifetime, AND the reaction is spontaneous, then we can see it happen.

Spontaneity is measured by the change in the Gibbs' free energy, $\Delta G$, the maximum amount of non-pressure/volume work that can be done by the system.

A common equation that is true at constant temperature is:

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

where $\Delta H$ is the change in enthalpy (heat flow at constant pressure) and $\Delta S$ is the change in entropy (tendency of energy dispersal) at a certain temperature $T$ in $\text{K}$.

$\Delta H$ and $\Delta S$ are commonly tabulated at $T = \text{298.15 K}$ and $\text{1 atm}$ as standard enthalpies of formation $\Delta {H}_{f}^{\circ}$ and standard molar entropies ${S}^{\circ}$, and thus, the spontaneity of known reactions at $\text{298.15 K}$ and $\text{1 atm}$ can be (re-)predicted.

• If $\Delta {G}^{\circ} < 0$, the reaction is spontaneous at $\text{298.15 K}$ and $\text{1 atm}$.
• If $\Delta {G}^{\circ} > 0$, the reaction is nonspontaneous at $\text{298.15 K}$ and $\text{1 atm}$.
• If $\Delta {G}^{\circ} = 0$, the reaction is neither, and is at dynamic equilibrium instead (at $\text{298.15 K}$ and $\text{1 atm}$).

If the reaction is at equilibrium but is NOT at $\text{298.15 K}$ and $\text{1 atm}$, then in general, $\Delta G \ne \Delta {G}^{\circ}$, and

${\cancel{\Delta G}}^{0} = \Delta {G}^{\circ} + R T \ln {\cancel{Q}}^{K}$

$\Delta {G}^{\circ} = - R T \ln K$

And thus, at that temperature that is not $\text{298.15 K}$, the equilibrium constant can be estimated.