3.1.8.2 - Entropy and Gibbs Free Energy

Why enthalpy is not enough

If $\Delta H$ were the whole story, every exothermic change would be feasible and every endothermic change would not be. Chemistry is more subtle than that. A system also tends to move toward a more dispersed, less ordered arrangement of particles and energy.

The bigger the spread of particles and energy, the higher the entropy. That is why gases usually have much larger entropy values than liquids, and liquids have larger entropy values than solids. Heating a substance also increases entropy because particles have more energy and can occupy a wider range of positions and motions.

The diagram below shows the particle arrangements in a solid, liquid, and gas, with the largest increase in entropy happening on boiling.

[DIAGRAM: asset_name: 1.8.2 - Gibbs Free-Energy Change, Delta G, and Entropy Change, Delta S - Diagram 1; asset_slug: 1.8.2 - Gibbs Free-Energy Change, Delta G, and Entropy Change, Delta S - Diagram 1; recommended_method: matplotlib_exact_plot; description: Twelve non-overlapping particles in each solid, liquid and gas panel; ordered versus irregular arrangements and explicit not-to-scale note.]

Physical changes show this clearly. When ice melts, the particles gain freedom of movement, so entropy increases. When liquid water becomes steam, the increase is even larger because gas particles are much more widely separated and randomly arranged. Chemical changes can also increase or decrease entropy. Reactions that produce more moles of gas often have a positive $\Delta S$, while reactions that remove gas molecules from the system often have a negative $\Delta S$.

One important idea is that a feasible reaction does not have to have a positive reaction entropy change. A reaction can still be feasible with a negative $\Delta S$ if the enthalpy term is favorable enough. That is exactly why we need Gibbs free energy.

Calculating entropy changes from absolute entropy values

Chemists can calculate the entropy change for a reaction from tabulated absolute entropy values. These are usually given in $\mathrm{J}\, \mathrm{K}^{-1}\, \mathrm{mol}^{-1}$ under standard conditions. Unlike enthalpies of formation, the entropy values of elements in their standard states are not zero, so you must use the data values exactly as given.

For a reaction:

The method is always the same:

1. Add the entropy values for all products, taking account of stoichiometric coefficients.
2. Add the entropy values for all reactants.
3. Subtract reactants from products.
4. Give the answer with a sign and units.

Worked example:

Using $S^\circ$ values of $93$, $40$, and $214\,\mathrm{J}\,\mathrm{K}^{-1}\, \mathrm{mol}^{-1}$ respectively:

• sum of product entropies $= 40 + 214 = 254\,\mathrm{J}\,\mathrm{K}^{-1}\, \mathrm{mol}^{-1}$
• sum of reactant entropies $= 93\,\mathrm{J}\,\mathrm{K}^{-1}\, \mathrm{mol}^{-1}$
• $\Delta S = 254 - 93 = +161\,\mathrm{J}\,\mathrm{K}^{-1}\, \mathrm{mol}^{-1}$

The answer is positive because the reaction forms a gas from a solid, so the particles become much more dispersed.

When you predict the sign before calculating, you are less likely to lose a sign error. More gas particles usually means $\Delta S$ is positive. Fewer gas particles usually means $\Delta S$ is negative.

Using Gibbs free energy to judge feasibility

Entropy helps, but enthalpy still matters. Gibbs free energy combines both effects:

Here, $T$ must be in kelvin. $\Delta H$ and $\Delta G$ are usually in $\mathrm{kJ}\, \mathrm{mol}^{-1}$, while $\Delta S$ is often given in $\mathrm{J}\, \mathrm{K}^{-1}\, \mathrm{mol}^{-1}$. That means you usually need to convert $\Delta S$ into $\mathrm{kJ}\, \mathrm{K}^{-1}\, \mathrm{mol}^{-1}$ by dividing by $1000$ before substituting into the equation.