Entropy and free energy

An instant cold pack gets colder while it dissolves, which means something other than energy is deciding what happens, and that something can be measured and tabulated as precisely as an enthalpy.

The previous lesson made reaction enthalpies computable and then produced three reactions that absorb heat and happen anyway. What they had in common was that the products are more spread out than the reactants. This lesson turns that observation into a number.

Counting the arrangements

Take a crystal of a salt and a beaker of water. Before dissolving, every ion sits at a fixed lattice site; afterwards each ion can be anywhere in the liquid, and the water molecules around it have been reorganised too. The number of microscopic arrangements consistent with the second situation is unimaginably larger than the number consistent with the first.

Ludwig Boltzmann's proposal, engraved on his tombstone, is that this count is a thermodynamic quantity:

S = k_{B} \ln Ω

where $Ω$ is the number of microscopic states compatible with the macroscopic condition and $k_{B} = 1.381 \times 10^{- 23}$ J K⁻¹. The logarithm is not decoration. Put two independent systems side by side and their state counts multiply, while any sensible extensive property must add, and the logarithm is what turns one into the other.

Entropy also has a purely thermodynamic definition that never mentions atoms: for a reversible transfer of heat, $d S = δ Q_{rev} / T$ . The two definitions agree, which is one of the deeper results in physics, and for chemistry the second is what makes entropy measurable. Melting ice at $273.15$ K absorbs $6.01$ kJ mol⁻¹ reversibly, so the entropy of fusion is $6010 / 273.15 = 22.0$ J mol⁻¹ K⁻¹, a number obtained with a thermometer and a heater.

The Second Law is then the statement that the entropy of an isolated system never falls. Applied to the universe as a whole, that is the criterion we were missing: a change happens if the total entropy of system plus surroundings increases.

The Third Law makes entropy absolute

Enthalpy has no zero, but entropy does. The Third Law says that the entropy of a perfect crystalline substance approaches zero as the temperature approaches absolute zero: at $T = 0$ a perfect crystal has one arrangement, $Ω = 1$ , so $S = 0$ . That fixes the origin, and it means entropies can be tabulated as absolute values rather than as changes.

Measuring one means integrating the heat capacity from as near zero kelvin as an experiment can reach, adding the entropy of each phase change on the way up. The result is the standard molar entropy $S^{ominus}$ , in J mol⁻¹ K⁻¹, and unlike a formation enthalpy it is never zero for an element: graphite is $5.74$ , hydrogen gas $130.68$ , oxygen $205.14$ .

The values behave as the counting argument predicts. Gases greatly exceed liquids, which exceed solids: water is $69.91$ as a liquid and $188.83$ as a vapour, and that difference of $119$ J mol⁻¹ K⁻¹ is the biggest single lever in most reactions involving water. Bigger and floppier molecules have more entropy, since there are more ways to distribute energy among their vibrations: methane is $186.26$ and octane vapour $467$ . Harder substances have less, because stiff bonds mean widely spaced vibrational levels and fewer accessible states, which is why diamond sits at $2.38$ .

The Third Law also has known exceptions, and they are informative rather than embarrassing. Carbon monoxide freezes into a lattice where each molecule can point either way, and the mismatch is too small for the crystal to sort itself out before it stops moving. That leaves $2^{N}$ arrangements at absolute zero, predicting a residual entropy of $R \ln 2 = 5.76$ J mol⁻¹ K⁻¹, against a measured $4.6$ . Ice has a similar disorder in its hydrogen bonds, predicted at about $3.37$ and measured at $3.4$ . The law is a statement about perfect crystals, and a frozen-in disorder is a failure to be perfect.

The entropy of a reaction

Because $S^{ominus}$ values are absolute, a reaction entropy is a stoichiometric sum in exactly the way a reaction enthalpy is, with no reference state to cancel:

Δ_{r} S^{ominus} = \sum ν_{i} S^{ominus} (i)

The sign is usually predictable before any arithmetic, from the change in the number of moles of gas. Gases dominate the sum, so a reaction that consumes gas has a negative $Δ_{r} S$ and one that produces gas a positive one.

Example. Find $Δ_{r} S^{ominus}$ for the combustion of methane, $C(g) + 2 (g) \to C(g) + 2 O(l)$ , using $S^{ominus} = 186.26$ , $205.14$ , $213.74$ and $69.91$ J mol⁻¹ K⁻¹.

Products: $213.74 + 2 (69.91) = 353.56$ . Reactants: $186.26 + 2 (205.14) = 596.54$ . So $Δ_{r} S^{ominus} = 353.56 - 596.54 = - 242.98$ J mol⁻¹ K⁻¹. Three moles of gas become one, and two of the products are liquid, so the mixture ends far more ordered than it began. A reaction everyone agrees is spontaneous has a strongly negative entropy change, which already shows that system entropy alone is not the criterion either.

Now you. Find $Δ_{r} S^{ominus}$ for $2 (g) + (g) \to 2 O(l)$ , using $S^{ominus} = 130.68$ for hydrogen, $205.14$ for oxygen and $69.91$ for liquid water.

Answer

Products: $2 (69.91) = 139.82$ . Reactants: $2 (130.68) + 205.14 = 466.50$ . So $Δ_{r} S^{ominus} = - 326.68$ J mol⁻¹ K⁻¹. Three moles of gas become two moles of liquid, which is about as large a decrease as ordinary chemistry offers.

The surroundings are part of the account

Methane burns, and its entropy change is $- 243$ J K⁻¹ per mole. The Second Law is not violated, because the system is not isolated: the reaction dumps $890$ kJ into the room.

Heat arriving in the surroundings raises their entropy. If the surroundings are large enough to stay at constant temperature and pressure, the heat they receive is $- Δ H$ and they receive it reversibly, so

Δ S_{surr} = - \frac{Δ H}{T}

The division by $T$ carries the physical content: the same joule of heat buys more entropy in a cold place than a hot one, because it makes a larger relative difference to a system with little thermal energy already. For burning methane at $298$ K, $Δ S_{surr} = 890500 / 298.15 = 2987$ J K⁻¹, which dwarfs the system's $- 243$ , and the total is strongly positive.

The same accounting settles the melting of ice, which the previous lesson left as a puzzle. At $283$ K the ice gains $6010 / 273.15 = 22.0$ J K⁻¹ per mole as it melts while the room loses $6010 / 283.15 = 21.2$ , so the total is $+ 0.78$ J mol⁻¹ K⁻¹ and the ice melts. At $263$ K the surroundings would lose $22.9$ , the total would be negative, and it does not. The melting point is where the two terms cancel exactly.

Free energy: the criterion in system quantities alone

Requiring $Δ S_{total} > 0$ is correct but inconvenient, because it asks about the universe when the experimenter only has a flask. Substituting the expression for the surroundings fixes that:

Δ S_{total} = Δ S - \frac{Δ H}{T}

Multiply by $- T$ , which is negative and so reverses the inequality, and define the result as the Gibbs free energy change:

Δ G = Δ H - T Δ S = - T Δ S_{total}

A process at constant temperature and pressure happens spontaneously when $Δ G < 0$ . Every quantity on the right belongs to the system, which is why this combination and not the entropy is the working criterion of chemistry. Josiah Willard Gibbs published it in 1876 in a journal so obscure that Maxwell had to draw European attention to it.

$Δ G$ has a second reading worth carrying: it is the maximum non-expansion work a process can deliver. A reaction with $Δ G = - 474$ kJ mol⁻¹ can in principle supply that much electrical work in a fuel cell, and this is exactly the quantity a battery converts, as the last lesson of this course will use.

Example. For $2 (g) + (g) \to 2 O(l)$ , with $Δ_{r} H^{ominus} = - 571.6$ kJ mol⁻¹ and $Δ_{r} S^{ominus} = - 326.68$ J mol⁻¹ K⁻¹, find $Δ_{r} G^{ominus}$ at $298.15$ K.

Convert the entropy to kJ before combining, a step worth checking every time: $T Δ S = 298.15 \times (- 0.32668) = - 97.4$ kJ mol⁻¹. So $Δ_{r} G^{ominus} = - 571.6 - (- 97.4) = - 474.2$ kJ mol⁻¹, which agrees with twice the tabulated formation free energy of liquid water, $- 237.1$ . The reaction is strongly favoured despite its large entropy penalty, because the enthalpy term is larger still.

Now you. For $(g) + 3 (g) \to 2 N(g)$ , $Δ_{r} H^{ominus} = - 91.8$ kJ mol⁻¹ and $Δ_{r} S^{ominus} = - 198.75$ J mol⁻¹ K⁻¹. Find $Δ_{r} G^{ominus}$ at $298.15$ K.

Answer

$T Δ S = 298.15 \times (- 0.19875) = - 59.3$ kJ mol⁻¹, so $Δ_{r} G^{ominus} = - 91.8 + 59.3 = - 32.5$ kJ mol⁻¹, matching twice the tabulated $- 16.4$ for ammonia. Negative, so at room temperature and standard conditions the synthesis is favoured.

Temperature picks the winner

The two terms of $Δ G = Δ H - T Δ S$ can agree or fight, which gives four cases. If $Δ H < 0$ and $Δ S > 0$ the reaction goes at every temperature; if the signs are reversed it goes at none. The interesting cases are the two where they conflict, because there the temperature decides, and the crossover is where $Δ G = 0$ :

T_{cross} = \frac{Δ H}{Δ S}

An endothermic reaction with a positive entropy change is impossible when cold and spontaneous when hot. That is every decomposition that releases a gas, every evaporation, and much of extractive metallurgy.

Example. Limestone decomposes as $CaC(s) \to CaO(s) + C(g)$ . Using $Δ_{f} H^{ominus} = - 1206.9$ , $- 635.1$ and $- 393.5$ kJ mol⁻¹, and $S^{ominus} = 92.9$ , $39.75$ and $213.74$ J mol⁻¹ K⁻¹, find $Δ_{r} G^{ominus}$ at $298$ K and the temperature above which the reaction becomes favourable.

$Δ_{r} H^{ominus} = - 635.1 - 393.5 + 1206.9 = + 178.3$ kJ mol⁻¹ and $Δ_{r} S^{ominus} = 39.75 + 213.74 - 92.9 = + 160.59$ J mol⁻¹ K⁻¹. At $298.15$ K, $Δ_{r} G^{ominus} = 178.3 - 298.15 (0.16059) = + 130.4$ kJ mol⁻¹, hopelessly unfavourable, which is why buildings made of limestone stay up. The crossover is $T = 178300 / 160.59 = 1110$ K, or $837$ °C. Real lime kilns run at $900$ to $1000$ °C, and the measured temperature at which carbon dioxide reaches one bar over the solid is about $1170$ K. The estimate is out by five per cent because it holds $Δ H$ and $Δ S$ fixed over eight hundred kelvin, which Kirchhoff's law says they are not.

Now you. Magnesium carbonate decomposes as $MgC(s) \to MgO(s) + C(g)$ , with $Δ_{r} H^{ominus} = + 100.7$ kJ mol⁻¹ and $Δ_{r} S^{ominus} = + 174.98$ J mol⁻¹ K⁻¹. Estimate the crossover temperature, and say whether magnesium carbonate or calcium carbonate is easier to decompose.

Answer

$T = 100700 / 174.98 = 575$ K, or $302$ °C. The entropy changes are similar, since both release one mole of gas, so the crossover is set by the enthalpy, and magnesium carbonate needs far less: its smaller cation binds the carbonate ion more strongly and polarises it, weakening it. Magnesium carbonate is much the easier to decompose.

What the criterion still does not deliver

Two things are missing, and both are the business of later lessons.

The first is time. $Δ_{r} G^{ominus}$ for the combustion of methane is $- 818$ kJ mol⁻¹, about as favourable as chemistry gets, and a mixture of methane and air in a sealed bottle will sit unchanged for a century. Free energy says which way the mixture would go if it went. It says nothing whatever about how long that takes, and nothing in this lesson can be repaired to make it say so. That is the sixth lesson.

The second is subtler and immediate. Take the criterion literally and a reaction with $Δ G < 0$ should run until a reactant is exhausted, since the sign does not change on the way. Reactions do not do this. Heat calcium carbonate at $1100$ K in a closed vessel and it stops with solid and gas both present; dissolve acetic acid in water and most of it stays intact. Something makes $Δ G$ climb to zero at a composition short of completion, and stop there.

The missing piece is that $Δ_{r} G^{ominus}$ is a standard value, referring to a specific composition, and the actual $Δ_{r} G$ depends on how much of each substance is present. Working out that dependence gives the equilibrium constant, and it is the next lesson.