Entropy and Gibbs energy

Entropy ($S$) measures how many different ways a system's energy and particles can be arranged - the number of accessible microstates. The more arrangements available, both for sharing the energy quanta among the particles and for positioning the particles themselves, the higher the entropy.

Predict the sign of $\Delta S$ by asking: does the number of accessible arrangements increase?

• State changes raise entropy: $S(\text{solid}) < S(\text{liquid}) < S(\text{gas})$. Gases have by far the highest entropy because their particles move freely.
• More gas molecules on the product side $\Rightarrow$ $\Delta S > 0$. Fewer $\Rightarrow$ $\Delta S < 0$. Counting moles of gas is the quickest check.
• Dissolving, mixing and increasing temperature generally increase entropy.

For a reaction, the standard entropy change is

$$\Delta S^{\ominus} = \Sigma S^{\ominus}(\text{products}) - \Sigma S^{\ominus}(\text{reactants})$$

remembering to multiply each absolute entropy by its stoichiometric coefficient. Absolute entropies $S^{\ominus}$ are tabulated in $\text{J K}^{-1}\text{mol}^{-1}$, and unlike $\Delta H_f^{\ominus}$ they are non-zero for elements (only a perfect crystal at 0 K has $S = 0$). For example, $\text{CaCO}_3(\text{s}) \rightarrow \text{CaO(s)} + \text{CO}_2(\text{g})$ makes a gas from a solid, so $\Delta S^{\ominus}$ is clearly positive.

Gibbs energy combines the enthalpy and entropy contributions into one quantity that decides spontaneity:

$$\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}$$

This equation is in the data booklet. $\Delta G^{\ominus}$ as a whole reflects the total entropy change: the $-T\Delta S^{\ominus}$ term accounts for the system and the $\Delta H^{\ominus}$ term for the surroundings: an exothermic reaction ($\Delta H < 0$) increases the entropy of the surroundings, while a positive $\Delta S$ increases the entropy of the system. Both push $\Delta G$ towards being negative.

Watch the units - the single most common error here:

• $\Delta H^{\ominus}$ is in kJ mol$^{-1}$
• $\Delta S^{\ominus}$ is in J K$^{-1}$ mol$^{-1}$

Divide $\Delta S^{\ominus}$ by 1000 (or multiply $\Delta H^{\ominus}$ by 1000) so both are in the same unit before subtracting. Always use $T$ in kelvin.

Gibbs energy is a state function, so you can also combine $\Delta G^{\ominus}$ values using Hess's law, or build $\Delta G^{\ominus}$ from standard Gibbs energies of formation in the same way as for enthalpies.

A reaction is spontaneous (the IB also says *feasible* or *thermodynamically favourable*) when

$$\Delta G < 0$$

The sign of $\Delta G$ in $\Delta G = \Delta H - T\Delta S$ can depend on temperature. There are four cases:

| $\Delta H$ | $\Delta S$ | Spontaneity |

|---|---|---|

| $-$ | $+$ | spontaneous at all $T$ |

| $+$ | $-$ | spontaneous at no $T$ |

| $-$ | $-$ | spontaneous at low $T$ |

| $+$ | $+$ | spontaneous at high $T$ |

In the two temperature-dependent cases the sign flips at the crossover temperature where $\Delta G^{\ominus} = 0$:

$$T = \dfrac{\Delta H^{\ominus}}{\Delta S^{\ominus}}$$

Two crucial cautions:

• Spontaneous says nothing about rate. A reaction can have $\Delta G < 0$ yet be extremely slow because of a high activation energy (e.g. diamond $\rightarrow$ graphite).
• $\Delta G^{\ominus}$ uses standard conditions; real conditions can make a reaction go even when $\Delta G^{\ominus} > 0$ (handled by $\Delta G = \Delta G^{\ominus} + RT\ln Q$).

The link between Gibbs energy and the equilibrium constant is

$$\Delta G^{\ominus} = -RT\ln K$$

which is in the data booklet. It tells you the *extent* of reaction:

• $\Delta G^{\ominus} < 0 \Rightarrow \ln K > 0 \Rightarrow K > 1$: products favoured.
• $\Delta G^{\ominus} > 0 \Rightarrow \ln K < 0 \Rightarrow K < 1$: reactants favoured.
• $\Delta G^{\ominus} = 0 \Rightarrow K = 1$.

The relationship is logarithmic, so even a modest $\Delta G^{\ominus}$ gives a very large or very small $K$.

Where does it come from? The general expression is the Gibbs energy isotherm:

$$\Delta G = \Delta G^{\ominus} + RT\ln Q$$

where $Q$ is the reaction quotient. As a reaction proceeds, $Q$ changes until $\Delta G = 0$ - that point is equilibrium, where $Q = K$. Substituting $\Delta G = 0$ and $Q = K$ gives $0 = \Delta G^{\ominus} + RT\ln K$, i.e. the booklet relationship $\Delta G^{\ominus} = -RT\ln K$. Remember to keep $\Delta G^{\ominus}$ in $\text{J mol}^{-1}$ with $R = 8.31$.