What entropy is really measuring

Entropy, $S$, is a state property. It measures how widely matter and energy are dispersed in a system, and its SI unit is joule per kelvin, $\mathrm{J\,K^{-1}}$. In chemistry, the value used most often is standard molar entropy, $S^\circ$: the entropy per mole of a substance in its standard state, with unit $\mathrm{J\,K^{-1}\,mol^{-1}}$.

“More disordered means higher entropy” is a handy classroom shortcut, but don’t treat it as the definition. A better way to think about entropy is the number of possible arrangements. When particles and energy can be spread out in more ways, entropy is higher. So, under the same conditions, gases have higher entropy than liquids, and liquids have higher entropy than solids. Gas particles can occupy many more positions and move more freely.

A system is the chemical or physical process being studied. The surroundings are everything outside that system that can exchange energy or matter with it. The total entropy change is written as

A spontaneous change increases total entropy; at equilibrium, the total entropy is no longer increasing.

Predicting the sign of $\Delta S$

For a physical change, look at freedom of movement. Melting, vaporization and sublimation usually increase entropy. Freezing, condensation and deposition usually decrease it. Dissolving a solid often increases entropy because the particles spread through the solvent, although situations with highly ordered solvent shells can be less straightforward.

For a chemical change, begin with the states of matter, then check the balanced equation. Gases tend to control entropy predictions because gaseous particles have far more possible positions than particles in solids or liquids. If the number of moles of gas increases, $\Delta S$ is usually positive. If the number of moles of gas decreases, $\Delta S$ is usually negative. When no gas is involved, compare how dispersed the particles become, for example through dissolving, dissociation or formation of more separate particles.

The balanced coefficients matter. One mole of solid forming two moles of gas gives a large entropy increase. Several moles of gas forming fewer moles of gas is usually an entropy decrease, even if liquid products are formed as well.

Calculating standard entropy changes

Standard conditions are agreed reference conditions used for tabulated thermodynamic data, with substances in their standard states. Entropy is a state function, so the route taken does not matter. Only the initial and final states do.

For a reaction,

Standard entropy values are provided in the data booklet.

When calculating, keep the physical state attached to the formula. $S^\circ$ for $H_2O(g)$ is not the same as $S^\circ$ for $H_2O(l)$, because gas and liquid water have different energy dispersal and particle freedom. Estimating the sign before calculating is a good way to catch arithmetic slips.

The perfect crystal at 0 K

The Structure 1.1 link is the third law idea. A perfect crystal is a solid in which every particle occupies a regular, repeating lattice position with no defects. At 0 K, a perfect crystal is predicted to have zero entropy because there is only one possible arrangement of the particles and their energy in the lowest-energy state. In the language of this topic, there is no additional dispersal to count.

Why Gibbs energy is needed

Enthalpy by itself doesn't decide which way a chemical change will go. Plenty of exothermic reactions are spontaneous, but not every one is; some endothermic processes happen spontaneously because entropy increases a lot. So we use a single quantity that brings together the enthalpy change, entropy change and temperature.

Gibbs energy, $G$, is a thermodynamic state function that shows the energy available to do useful work at constant temperature and pressure; its unit is joule, J. The change in Gibbs energy, $\Delta G$, is the Gibbs energy change for a process; in this topic, it is usually reported per mole of reaction in $\mathrm{kJ\,mol^{-1}}$.

That last unit point is the one I underline on the board. The data booklet gives $\Delta S^\circ$ values in $\mathrm{J\,K^{-1}\,mol^{-1}}$, while $\Delta H^\circ$ and $\Delta G^\circ$ are normally in $\mathrm{kJ\,mol^{-1}}$. Divide entropy changes by 1000 before multiplying by temperature if the other quantities are in $\mathrm{kJ\,mol^{-1}}$.

Rearranging the equation

You can use the same equation to find any unknown term. For example, $\Delta H^\circ = \Delta G^\circ + T\Delta S^\circ$, and $\Delta S^\circ = (\Delta H^\circ - \Delta G^\circ) / T$. The algebra is straightforward; the chemistry is in keeping the units consistent and making sense of the sign.

A graph of $\Delta G^\circ$ against $T$ is useful because the equation has the form of a straight line. The vertical intercept is $\Delta H^\circ$, and the gradient is $-\Delta S^\circ$. If the line crosses $\Delta G^\circ = 0$, that temperature marks the change between spontaneous and non-spontaneous behaviour under standard conditions.

What temperature does in the equation

Temperature changes the size of the entropy term, $T\Delta S^\circ$. At low temperature, $\Delta H^\circ$ may dominate. At high temperature, the entropy term can dominate instead. That is why some processes are only spontaneous above a certain temperature, while others are only spontaneous below a certain temperature.

Interpreting $\Delta G$

A spontaneous change is a process that goes in the stated direction under the given conditions without continual external driving. It may go to completion, or it may only go as far as equilibrium. Spontaneous does not mean fast. A reaction can be thermodynamically favourable but kinetically slow because it has a large activation energy.

At constant pressure, the sign of $\Delta G$ shows the thermodynamic direction:

• $\Delta G < 0$: the forward process is spontaneous under those conditions.
• $\Delta G = 0$: the system is at equilibrium; there is no net driving force.
• $\Delta G > 0$: the forward process is non-spontaneous; the reverse process is spontaneous.

$\Delta G$ combines two entropy effects. One is direct: the entropy change of the chemicals themselves as bonds, states and particle numbers change. The other comes from the surroundings, because heat transfer changes their entropy. In an exothermic reaction, heat released to the surroundings tends to increase the surroundings’ entropy. In an endothermic reaction, heat absorbed from the surroundings tends to decrease it.

That is why an endothermic reaction can still be spontaneous. If the system entropy increases enough, the direct entropy gain can outweigh the entropy decrease of the surroundings. A reaction between two solids that produces gas and aqueous ions is a good example of the kind of change where matter becomes much more dispersed, even though heat is absorbed.

Combining signs of $\Delta H$ and $\Delta S$

Learn the four sign combinations, since they let you predict temperature dependence before doing a calculation.

Four ΔH° and ΔS° sign combinations and their effect on ΔG° and spontaneity.

If $\Delta H^\circ$ is negative and $\Delta S^\circ$ is positive, $\Delta G^\circ$ is always negative: exothermic and more dispersed is the most favourable combination. If $\Delta H^\circ$ is positive and $\Delta S^\circ$ is negative, $\Delta G^\circ$ is always positive: endothermic and less dispersed is unfavourable at all temperatures.

The other two cases change with temperature. If both $\Delta H^\circ$ and $\Delta S^\circ$ are positive, the reaction becomes spontaneous at high temperature because $T\Delta S^\circ$ eventually becomes larger than $\Delta H^\circ$. If both $\Delta H^\circ$ and $\Delta S^\circ$ are negative, the reaction is spontaneous at low temperature but becomes non-spontaneous at high temperature because subtracting a negative entropy term makes $\Delta G^\circ$ larger.

Finding the temperature where spontaneity changes

At the boundary between spontaneous and non-spontaneous behaviour, $\Delta G^\circ = 0$. So $0 = \Delta H^\circ - T\Delta S^\circ$, giving

This threshold only has physical meaning when the signs allow a crossing. For $\Delta H^\circ > 0$ and $\Delta S^\circ > 0$, the reaction is spontaneous above the calculated temperature. For $\Delta H^\circ < 0$ and $\Delta S^\circ < 0$, it is spontaneous below the calculated temperature.

Link to electrochemical data

Electrochemical data can predict spontaneity too. For a cell reaction under standard conditions,

A positive $E^\circ_{\text{cell}}$ gives a negative $\Delta G^\circ$, so the cell reaction is spontaneous as written. It’s the same thermodynamic test, but using electrical data rather than enthalpy and entropy data.

$\Delta G$ changes as the mixture changes

A reaction’s $\Delta G$ changes as the reaction runs. Early on, the mixture may be mostly reactants, so the forward reaction can have a large negative $\Delta G$. As products build up and reactants are consumed, that driving force drops. In a reversible reaction, $\Delta G$ becomes less negative, then reaches zero at equilibrium.

A reaction quotient, $Q$, is a dimensionless ratio with the same form as the equilibrium expression, except it uses the mixture’s current composition rather than its equilibrium composition. An equilibrium constant, $K$, is a dimensionless ratio describing the composition of a reversible system at equilibrium at a specified temperature.

For the general reaction $aA + bB \rightleftharpoons cC + dD$,

Pure solids and pure liquids are left out, just as they are in equilibrium expressions.

Compare $Q$ with $K$ to decide which way the system must shift to reach equilibrium:

• $Q < K$: too many reactants compared with equilibrium, so the forward reaction is favoured.
• $Q = K$: the system is at equilibrium.
• $Q > K$: too many products compared with equilibrium, so the reverse reaction is favoured.

The $\Delta G$, $Q$ and $K$ equations

Under non-standard conditions,

Since $R$ is in joules, put $\Delta G$ and $\Delta G^\circ$ in $\mathrm{J\ mol^{-1}}$ in this equation, unless you have deliberately converted $R$ to $\mathrm{kJ\ K^{-1}\ mol^{-1}}$.

At equilibrium, $\Delta G = 0$ and $Q = K$. Substituting these gives

Rearranged, $K = e^{-\Delta G^\circ / RT}$. These equations are in the data booklet.

What $\Delta G^\circ$ says about the equilibrium mixture

The sign of $\Delta G^\circ$ tells you where the equilibrium lies under standard-state comparison:

• $\Delta G^\circ < 0$ means $K > 1$, so products are favoured at equilibrium.
• $\Delta G^\circ = 0$ means $K = 1$, so neither side is strongly favoured.
• $\Delta G^\circ > 0$ means $K < 1$, so reactants are favoured at equilibrium.

So the Reactivity 2.3 linking answer is: when $\Delta G^\circ$ is positive, the equilibrium mixture is likely to contain more reactants than products, because the forward reaction is not favourable under standard-state comparison.

So the equilibrium position lies to the reactant side.

Calculation habits for this section

Use kelvin for temperature, not degrees Celsius. With $R = 8.31\ \mathrm{J\ K^{-1}\ mol^{-1}}$, use $\mathrm{J\ mol^{-1}}$ unless every energy term has been consistently converted to $\mathrm{kJ\ mol^{-1}}$. And watch the logarithm: ln means natural logarithm, not log base 10. These are unit consistency points, not exam tricks.