Lesson 4 · 25 min

Gibbs energy: temperature decides

Count the surroundings’ entropy too and out comes ΔG = ΔH − TΔS. Four sign cases, one ΔH–ΔS map and a temperature slider explain why water freezes below 0 °C and limestone needs a hot kiln.

In this lesson

Include the surroundings’ entropy change −ΔH/T in the total and derive ΔG = ΔH − TΔS
Use the four sign combinations of ΔH and ΔS to decide at which temperatures a reaction is spontaneous
Compute ΔG° and the crossover temperature T* = ΔH°/ΔS° from tabulated data
Tell “thermodynamically spontaneous” apart from “fast”

Water freezes by itself at −10 °C, melts by itself at +10 °C, and at 0 °C the two coexist. Yet on freezing, water molecules file into an orderly lattice and the system’s entropy falls — so how can freezing be spontaneous, after the last lesson? The answer: we forgot the surroundings. The heat released on freezing flows into the air around, raising the surroundings’ entropy, and the colder it is, the more entropy the same heat brings. Whether a change is spontaneous depends on the sum of both accounts.

Predict first

Ammonia synthesis $\ce{N2(g) + 3H2(g)} \longrightarrow \ce{2NH3(g)}$ : ΔH° = −91.9 kJ, ΔS° = −198 J/K. Raising the temperature from 300 K to 700 K, the reaction (under standard conditions) becomes…

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

More favourable: higher temperature helps Less and less favourable, with ΔG° > 0 above about 460 K Unaffected: ΔH and ΔS do not change with temperature

Counting the surroundings too

The second law demands that the entropy of the universe increase: $\Delta S_\text{univ} = \Delta S_\text{sys} + \Delta S_\text{surr} > 0$ . At constant temperature and pressure the heat released by the system, $-\Delta H$ , flows into the surroundings; they are large and stay at essentially constant temperature, so $\Delta S_\text{surr} = -\Delta H/T$ . Substitute and multiply by $-T$ :

-T\Delta S_\text{univ} = \Delta H - T\Delta S

The right-hand side contains only properties of the system. Gibbs defined it as the change in Gibbs energy, $G = H - TS$ . An increase in the entropy of the universe is the same as $\Delta G < 0$ : at constant temperature and pressure, a change with ΔG < 0 is spontaneous, one with ΔG > 0 runs spontaneously in reverse, and ΔG = 0 means equilibrium.

Take freezing: at −10 °C, $\Delta S_\text{sys} \approx -22.0$ J/K while $\Delta S_\text{surr} = +6010/263 = +22.9$ J/K, so the total is positive; at +10 °C the surroundings gain only $6010/283 = 21.2$ J/K, the total turns negative, and the ice melts instead.

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

ΔH is the energy account and TΔS the “number of ways” account; temperature sets the weight of the second.

$\Delta G$: Gibbs energy change; < 0 spontaneous, = 0 equilibrium
$\Delta H$: enthalpy change / kJ
$T$: thermodynamic temperature / K
$\Delta S$: entropy change of the system / (kJ/K) — convert J to kJ

Four sign combinations

The signs of ΔH and ΔS decide what temperature does (values in the examples computed from 298 K data)
ΔH	ΔS	Spontaneous when	Example
−	+	at every temperature	$\ce{2H2O2(l)} \longrightarrow \ce{2H2O(l) + O2(g)}$ : −196.1 kJ, +125.9 J/K
+	−	never	$\ce{3O2(g)} \mathrel{\ce{->}} \ce{2O3(g)}$ : +285.4 kJ, −137.7 J/K
−	−	below T* = ΔH/ΔS	ammonia synthesis: T* ≈ 464 K; water freezing: 273 K
+	+	above T* = ΔH/ΔS	limestone: T* ≈ 1119 K; water boiling: about 370 K

Source: ΔH° and ΔS° from the atlas table “Standard thermodynamic data” (298.15 K); water freezing from the enthalpy of fusion at 273.15 K (CRC Handbook).

ΔG = ΔH − TΔS · every point computed from data

The ΔH–ΔS map: temperature decides

Temperature T · 298 K (25 °C)

7 of 18 processes have ΔG° < 0 at 298 K

Choose a process

$\ce{CaCO3}\text{(s, calcite)} \longrightarrow \ce{CaO(s)} + \ce{CO2(g)}$

ΔH°: 179.2 kJ
ΔS°: 160.2 J/K
Crossover T* = ΔH°/ΔS°: 1119 K
At this temperature · ΔG°(298 K): 131.4 kJ

ΔG° > 0: not favoured under standard conditions

Each point is a process, plotted at (ΔS°, ΔH°). The green slanted line is ΔG = 0 (ΔH = TΔS), and points below it have ΔG° < 0. Drag the temperature from 100 K to 2000 K: the line swings anticlockwise, the upper-right points (endothermic, entropy up) turn green one by one, and the lower-left points (exothermic, entropy down) turn red.

The standard Gibbs energy change $\Delta_\mathrm{r}G^\circ$ is ΔG when every species is in its standard state (gases at 1 bar, solutes at 1 mol/kg). At 298 K it follows from standard Gibbs energies of formation $\Delta_\mathrm{f}G^\circ$ , just like enthalpy; at other temperatures one usually approximates $\Delta_\mathrm{r}G^\circ(T) \approx \Delta_\mathrm{r}H^\circ - T\Delta_\mathrm{r}S^\circ$ , treating the 298 K values of ΔH° and ΔS° as constants. When ΔH° and ΔS° have the same sign there is a crossover temperature $T^* = \Delta H^\circ/\Delta S^\circ$ at which the reaction changes direction.

Worked example

How hot must a lime kiln be?

$\ce{CaCO3(s)} \longrightarrow \ce{CaO(s) + CO2(g)}$ . From ΔfH° (kJ/mol): CaCO₃ −1207.6, CaO −634.92, CO₂ −393.51, and S° (J K⁻¹ mol⁻¹): CaCO₃ 91.7, CaO 38.1, CO₂ 213.79, find ΔG° at 298 K and the crossover temperature.

Concept check

At 25 °C ammonium nitrate dissolves spontaneously although the process is endothermic (ΔH° = +25.7 kJ/mol). Why?

ΔH is too small to matter TΔS° ≈ +32.4 kJ exceeds ΔH°, so ΔG° < 0 The solution cools, and cooling is always spontaneous Dissolving is fast

Try the calculation

Using ΔH° = −91.88 kJ and ΔS° = −198.1 J/K, find the crossover temperature T* (K) of ammonia synthesis, $\ce{N2(g) + 3H2(g)} \longrightarrow \ce{2NH3(g)}$ .

Key takeaways

ΔS_surr = −ΔH/T; adding it to the total gives ΔG = ΔH − TΔS.
At constant T and p: ΔG < 0 spontaneous, ΔG = 0 equilibrium, ΔG > 0 spontaneous in reverse.
When ΔH and ΔS share a sign there is a crossover T* = ΔH/ΔS; with opposite signs the direction does not depend on temperature.
Spontaneous means thermodynamically allowed; the rate is set by the activation energy.

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Data and references

P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed. (2018), Topics 3D–3E (Gibbs energy).
J. D. Cox, D. D. Wagman, V. A. Medvedev, CODATA Key Values for Thermodynamics (1989); W. M. Haynes (ed.), CRC Handbook of Chemistry and Physics, 97th ed. (2016).
R. A. Robie, B. S. Hemingway, Thermodynamic properties of minerals and related substances at 298.15 K and 1 bar and at higher temperatures, U.S. Geological Survey Bulletin 2131 (1995) — calcite/lime equilibrium.
J. W. Gibbs, “On the equilibrium of heterogeneous substances”, Trans. Connecticut Acad. 3, 108–248 (1876).