Gibbs Free Energy and Thermodynamic Favorability: Study Guide (Chemistry 9.3)

Two drives, one decision

In Unit 6 you saw that many favoured reactions release energy (ΔH < 0). In Topic 9.2 you saw that many favoured changes spread matter and energy out (ΔS > 0). But neither rule works on its own:

• Ice melts in a warm room even though melting absorbs energy (ΔH > 0).
• Water freezes in a freezer even though freezing makes the particles more ordered (ΔS < 0).

So you need one quantity that weighs both effects at a given temperature. That quantity is the Gibbs free energy change, ΔG.

What ΔG° means

The standard Gibbs free energy change, ΔG°, is the free energy change when the reaction happens with every reactant and product in its standard state:

• pure solids and pure liquids;
• solutes at a concentration of 1 M;
• gases at a pressure of 1 bar (essentially the same as 1 atm).

The unit is kJ mol⁻¹, where “per mole” means per mole of reaction as written in the balanced equation. Double the equation and ΔG° doubles.

The sign of ΔG° gives the verdict:

Sign of ΔG°	Verdict
ΔG° < 0	thermodynamically favoured as written
ΔG° > 0	not favoured as written; the reverse process is favoured

A note on words. Older books call a process with ΔG° < 0 “spontaneous”. This course uses thermodynamically favoured (the exam spells it “favored”) because “spontaneous” sounds like “sudden” or “happens for no reason”. ΔG° tells you which direction is downhill in free energy. It tells you nothing about how fast the process goes. That idea is the whole of the next topic.

Method 1: from free energies of formation

The standard free energy of formation, ΔG°f, is ΔG° for making one mole of a substance from its elements in their standard states. As with ΔH°f in Topic 6.8, an element in its standard state, such as O₂(g) or C(graphite), has ΔG°f = 0.

The rule is the same “products minus reactants” rule you already know:

ΔG°reaction = ΣnΔG°f(products) − ΣnΔG°f(reactants)

Here n is each coefficient in the balanced equation.

Values used on this page (298 K, kJ mol⁻¹):

Substance	ΔG°f	Substance	ΔG°f
C₆H₁₂O₆(s), glucose	−910.56	NH₄Cl(s)	−202.97
C₂H₅OH(l), ethanol	−174.8	NH₃(g)	−16.4
CO₂(g)	−394.39	HCl(g)	−95.30

Worked example 1: is fermentation favoured?

Question. Yeast turns glucose into ethanol and carbon dioxide:

C₆H₁₂O₆(s) → 2C₂H₅OH(l) + 2CO₂(g)

Use the table to find ΔG° at 298 K. Is the process thermodynamically favoured?

1. Products: 2(−174.8) + 2(−394.39) = −349.6 + (−788.78) = −1138.38 kJ.
2. Reactants: 1(−910.56) = −910.56 kJ.
3. ΔG° = products − reactants = −1138.38 − (−910.56) = −227.82 kJ, which is −227.8 kJ mol⁻¹ of reaction (1 mol of glucose).

Answer. ΔG° = −227.8 kJ mol⁻¹. ΔG° < 0, so the fermentation of glucose is thermodynamically favoured at 298 K.

Check the traps. Reactants minus products gives +227.8 kJ, the wrong sign. Forgetting the coefficients gives +341.4 kJ, the wrong sign and the wrong size. Write the coefficient in front of each value before you add.

Method 2: from ΔH° and ΔS°

If you know ΔH° and ΔS° at a temperature T, you can calculate ΔG° directly:

ΔG° = ΔH° − TΔS°

Read the two terms as the two drives:

• ΔH° is the enthalpy term. A negative ΔH° (energy released to the surroundings) makes ΔG° more negative.
• −TΔS° is the entropy term. A positive ΔS° (matter or energy more dispersed) makes −TΔS° negative, which also pushes ΔG° down. Because ΔS° is multiplied by T, the entropy term grows as temperature rises.

Two rules prevent almost every lost mark:

1. Units. ΔH° is in kJ mol⁻¹, but tables give ΔS° in J K⁻¹ mol⁻¹. Divide ΔS° by 1000 before you multiply by T.
2. Kelvin. T is always in kelvin: T(K) = T(°C) + 273.15. Using °C can even give the wrong sign.

Worked example 2: heating ammonium chloride

Question. When solid ammonium chloride is heated strongly, it breaks down into two gases:

NH₄Cl(s) → NH₃(g) + HCl(g)

Data at 298 K:

Substance	ΔH°f (kJ mol⁻¹)	S° (J K⁻¹ mol⁻¹)
NH₄Cl(s)	−314.4	94.6
NH₃(g)	−45.9	192.8
HCl(g)	−92.3	186.9

(a) Calculate ΔH° and ΔS°. (b) Calculate ΔG° at 298 K and decide whether the decomposition is favoured. (c) Estimate the temperature above which it becomes favoured.

(a) ΔH° = (−45.9) + (−92.3) − (−314.4) = +176.2 kJ mol⁻¹ (endothermic). ΔS° = 192.8 + 186.9 − 94.6 = +285.1 J K⁻¹ mol⁻¹. This fits the particle picture: one mole of an ordered solid becomes two moles of freely moving gas, so matter is far more dispersed.

(b) Convert ΔS° first: +285.1 J K⁻¹ mol⁻¹ = +0.2851 kJ K⁻¹ mol⁻¹. TΔS° = 298 K × 0.2851 kJ K⁻¹ mol⁻¹ = 84.96 kJ mol⁻¹. ΔG° = 176.2 − 84.96 = +91.2 kJ mol⁻¹.

ΔG° > 0, so the decomposition is not favoured at 298 K. That matches experience: a bottle of ammonium chloride on the shelf does not turn into gas.

Cross-check with Method 1: ΔG° = (−16.4) + (−95.30) − (−202.97) = +91.3 kJ mol⁻¹. The two methods agree within rounding.

The unit trap: writing 176.2 − 298 × 285.1 gives −84 784 kJ mol⁻¹, which is far too large and has the wrong sign. If an answer is tens of thousands of kJ, check the units.

(c) ΔH° and ΔS° are both positive. The enthalpy term opposes the change; the entropy term favours it and grows with T. The changeover is where ΔG° = 0:

0 = ΔH° − TΔS°, so T = ΔH° ÷ ΔS° = 176 200 J mol⁻¹ ÷ 285.1 J K⁻¹ mol⁻¹ = 618 K (about 345 °C).

Above about 618 K, TΔS° is larger than ΔH° and ΔG° < 0. For example, at 700 K: ΔG° = 176.2 − 700(0.2851) = 176.2 − 199.6 = −23.4 kJ mol⁻¹, favoured.

Interpretation. This is an estimate: it treats ΔH° and ΔS° as constant and assumes the gases are at 1 bar. Even so, it lands close to the measured value: ammonium chloride is reported to decompose at about 338 °C at 1 atm.

The four sign combinations

You can often decide the verdict from the signs alone. ΔG° = ΔH° − TΔS° is a straight line when you plot ΔG° against T: it starts at ΔH° (when T = 0) and has a slope of −ΔS°.

ΔH°	ΔS°	Favoured (ΔG° < 0) when	Why
−	+	at all temperatures	both terms push ΔG° below zero; no calculation needed
+	−	at no temperature	both terms push ΔG° above zero; no calculation needed
+	+	at high temperature	the growing −TΔS° term eventually outweighs the positive ΔH°
−	−	at low temperature	ΔH° wins while T is small; at high T the positive −TΔS° term takes over

Only the last two cases need a number. For them, T = ΔH° ÷ ΔS° (in matching units) is the changeover temperature.

When enthalpy and entropy pull in opposite directions

Freezing water. For H₂O(l) → H₂O(s), ΔH° = −6.01 kJ mol⁻¹ (extra hydrogen bonds form between molecules as the solid builds, releasing energy) and ΔS° = −22.0 J K⁻¹ mol⁻¹ (molecules lose freedom to move). Both are negative, so freezing is favoured only at low temperature.

• At 263 K (−10 °C): ΔG° = −6.01 − 263(−0.0220) = −6.01 + 5.79 = −0.22 kJ mol⁻¹. Favoured: water freezes.
• At 283 K (+10 °C): ΔG° = −6.01 − 283(−0.0220) = −6.01 + 6.23 = +0.22 kJ mol⁻¹. Not favoured: ice melts instead.
• The changeover is 6.01 ÷ 0.0220 = 273 K, the normal melting point. At that temperature ice and water coexist.

Dissolving sodium nitrate. Solid NaNO₃ dissolves readily in water at room temperature, and the solution gets colder. The cooling shows that dissolving is endothermic (ΔH° > 0): more energy is needed to separate the ions from the lattice than is released when they are surrounded by water molecules. The process is still favoured because ΔS° > 0: ions held in fixed lattice positions spread out through the whole solution. At room temperature the −TΔS° term is large enough to outweigh ΔH°, so ΔG° < 0.

In both examples, looking at ΔH° alone or ΔS° alone gives the wrong prediction for one of the temperatures. You need both, combined through ΔG°.

Common misconceptions

• “Exothermic means favoured.” Only when ΔS° is not too negative. Freezing water is exothermic but is not favoured above 273 K.
• “Favoured means it happens quickly.” ΔG° says nothing about rate. A favoured process may take millions of years (see Topic 9.4).
• “ΔG° > 0 means nothing happens at all.” It means the process is not favoured as written under standard conditions; the reverse process is favoured.
• Mixing J and kJ. ΔS° in J K⁻¹ mol⁻¹ must be divided by 1000 before combining with ΔH° in kJ mol⁻¹.
• Using °C in ΔG° = ΔH° − TΔS°. At 25 °C, T = 298 K, not 25.
• Forgetting that ΔG°f of an element is zero, or giving a compound zero. Only elements in their standard states have ΔG°f = 0.
• Thinking ΔH° and ΔS° change sign with temperature. In this course they are treated as constant. It is the TΔS° term that changes with temperature.

Where this leads

Next, Topic 9.4 asks why some strongly favoured reactions do not happen at any measurable rate. Later in the unit you will link ΔG° to the equilibrium constant K and to cell potential. Try the practice questions now, then use the revision notes and the checklist to consolidate.