Thermodynamics — Gibbs Energy & Spontaneity

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"Will this reaction happen on its own?" is the question this chapter answers. The answer is one sign: the sign of ΔG. Everything else (first law, enthalpy, entropy) is there to calculate it.

What you will be able to do

• Use the first law with the correct sign convention and calculate work for reversible and irreversible expansion
• Convert between ΔH and ΔU, and use Hess’s law and bond enthalpies to find ΔH
• Predict the sign of ΔS and calculate entropy changes for phase changes
• Decide spontaneity from ΔH, ΔS and T, and find the crossover temperature
• Link ΔG° to K and to E°, and explain the Joule–Thomson effect

The big picture

One sign answers the chapter’s main question.

In one line: A process at constant temperature and pressure is spontaneous when ΔG = ΔH − TΔS is negative; ΔG° also fixes the equilibrium constant through ΔG° = −RT ln K.

1. First law — energy is conserved: ΔU = q + w.
2. Enthalpy — heat at constant pressure; Hess’s law lets you add reactions like equations.
3. Entropy — the universe’s entropy rises in every spontaneous change.
4. Gibbs energy — combines both into one test that uses only the system.

Think of it like this: A ball rolls downhill on its own (energy falls, like negative ΔH). A drop of ink spreads through water on its own (disorder rises, like positive ΔS). Gibbs energy weighs both tendencies, with temperature deciding how much the spreading counts.

The language

Get these words right and the rest reads easily.

• System — the part you study. Surroundings — everything else.
• Open (exchanges matter and energy), closed (energy only), isolated (neither).
• State functions depend only on the initial and final states: U, H, S, G, p, V, T. Path functions depend on the route: q and w.
• Extensive properties depend on amount (U, H, S, G, V, heat capacity). Intensive do not (T, p, density, molar quantities).
• Isothermal (T constant), isobaric (p constant), isochoric (V constant), adiabatic (q = 0).

First law and work

Energy in, energy out — watch the signs.

ΔU = q + w

IUPAC/NCERT sign convention: q and w are positive when energy enters the system

Process	Work done	Notes
Expansion against constant pₑₓₜ (irreversible)	w = −pₑₓₜ ΔV	Gas does work → w negative
Free expansion (pₑₓₜ = 0)	w = 0	Ideal gas: q = 0, ΔU = 0 too
Reversible isothermal, ideal gas	w = −2.303 nRT log(V₂/V₁)	Maximum work for that expansion
Constant volume	w = 0	So ΔU = qᵥ
Adiabatic	q = 0	w = ΔU = nCᵥΔT

Key idea: For an ideal gas, U depends only on T. In any isothermal process of an ideal gas ΔU = 0, so q = −w: all the heat absorbed is turned into work.

Solved example 01 — Reversible vs irreversible work

2 mol of an ideal gas expand isothermally at 300 K from 10 L to 100 L. Find w, q and ΔU (R = 8.314 J K⁻¹ mol⁻¹).

1. w = −2.303 × 2 × 8.314 × 300 × log(100/10) = −2.303 × 4988.4 × 1.
2. Isothermal ideal gas → ΔU = 0, so q = −w.

Answer: w ≈ −11.5 kJ; q ≈ +11.5 kJ; ΔU = 0.

Solved example 02 — Work against constant pressure

1 mol of gas expands from 10 L to 20 L against a constant external pressure of 1 atm. Find w in joules (1 L atm = 101.3 J).

1. w = −pₑₓₜ ΔV = −1 atm × 10 L = −10 L atm.
2. Convert: −10 × 101.3 J.

Answer: w = −1013 J.

Enthalpy

Heat at constant pressure — the number in most thermochemical equations.

H = U + pV
ΔH = ΔU + Δn(g) RT

Δn(g) = moles of gaseous products − moles of gaseous reactants

• ΔH = qₚ (heat at constant pressure); ΔU = qᵥ (heat at constant volume, as in a bomb calorimeter).
• Exothermic: ΔH < 0. Endothermic: ΔH > 0.
• Cₚ − Cᵥ = R for one mole of ideal gas; heat q = nCΔT.
• The standard enthalpy of formation of an element in its most stable form is zero (graphite, O₂, H₂). Diamond and O₃ are not zero.
• Enthalpy of neutralisation of a strong acid by a strong base ≈ −57.1 kJ mol⁻¹ (it is just H⁺ + OH⁻ → H₂O). With a weak acid or base it is less negative — some energy goes into ionising it.

Solved example 03 — ΔH from ΔU

For N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92.4 kJ at 298 K. Find ΔU.

1. Δn(g) = 2 − 4 = −2.
2. ΔU = ΔH − Δn(g)RT = −92.4 − (−2 × 8.314 × 298 / 1000).
3. = −92.4 + 4.96.

Answer: ΔU ≈ −87.4 kJ.

Hess’s law and bond enthalpy

Add equations, add their ΔH values.

Hess’s law: because H is a state function, ΔH for a reaction is the same whatever route you take. Add or subtract equations, multiply them by numbers, and do exactly the same to their ΔH values. Reverse an equation → change the sign of ΔH.

ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants) ΔrH = Σ bond enthalpies (bonds broken) − Σ bond enthalpies (bonds formed)

Gas-phase reactions only; reactants minus products — the reverse of the formation-enthalpy formula

Solved example 04 — Hess’s law

Given C(graphite) + O₂ → CO₂, ΔH = −393.5 kJ and CO + ½O₂ → CO₂, ΔH = −283.0 kJ. Find ΔfH of CO.

1. Target: C(graphite) + ½O₂ → CO.
2. Take equation 1 and subtract equation 2 (reverse it, so its ΔH becomes +283.0).
3. ΔH = −393.5 + 283.0.

Answer: ΔfH(CO) = −110.5 kJ mol⁻¹.

Solved example 05 — Bond enthalpy

Find ΔH for H₂ + Cl₂ → 2HCl. Bond enthalpies: H–H 436, Cl–Cl 242, H–Cl 431 kJ mol⁻¹.

1. Bonds broken: 436 + 242 = 678 kJ.
2. Bonds formed: 2 × 431 = 862 kJ.
3. ΔH = 678 − 862.

Answer: ΔH = −184 kJ.

Check yourself

• C1. Which is larger in magnitude for the combustion of methane: ΔH or ΔU? (CH₄ + 2O₂ → CO₂ + 2H₂O(l))
• C2. Why is the enthalpy of neutralisation of acetic acid with NaOH less than 57.1 kJ mol⁻¹?

Answers at the end.

Entropy

The measure of how spread out energy and matter are.

ΔS = q(rev) / T
ΔS(total) = ΔS(system) + ΔS(surroundings) > 0

Second law: total entropy rises in every spontaneous process; = 0 at equilibrium

• ΔS increases when: a solid melts or a liquid boils; moles of gas increase; a solid dissolves; temperature rises; a molecule breaks into more pieces.
• Entropy of a phase change: ΔS = ΔH(transition) / T(transition).
• Trouton’s rule: for many liquids ΔS(vap) ≈ 88 J K⁻¹ mol⁻¹. Water is higher because of hydrogen bonding.
• Third law: the entropy of a perfectly crystalline substance at 0 K is zero, which allows absolute entropies.
• For the surroundings: ΔS(surr) = −ΔH(system) / T.

Solved example 06 — Entropy of vaporisation

ΔvapH of water at 373 K is 40.79 kJ mol⁻¹. Find ΔvapS.

1. ΔS = ΔH / T = 40 790 / 373.

Answer: ΔvapS ≈ 109.4 J K⁻¹ mol⁻¹ (above Trouton’s 88, because of hydrogen bonding).

Gibbs energy

The single test of spontaneity at constant T and p.

ΔG = ΔH − TΔS

ΔG < 0 spontaneous; ΔG = 0 equilibrium; ΔG > 0 non-spontaneous (the reverse is spontaneous)

ΔH	ΔS	ΔG	Spontaneous?	Example
−	+	Always −	At all temperatures	Combustion; 2O₃ → 3O₂
−	−	− at low T	Below T = ΔH/ΔS	Freezing of water; NH₃ formation
+	+	− at high T	Above T = ΔH/ΔS	Melting of ice; CaCO₃ decomposition
+	−	Always +	Never	3O₂ → 2O₃ on its own

Key idea: Crossover temperature T = ΔH / ΔS. Put ΔH and ΔS in the same energy unit (both J or both kJ) — the commonest numerical error in this chapter. −ΔG also equals the maximum useful (non-expansion) work the system can do.

Solved example 07 — Crossover temperature

For a reaction ΔH = +30 kJ mol⁻¹ and ΔS = +105 J K⁻¹ mol⁻¹. Above what temperature is it spontaneous?

1. Both signs positive → spontaneous at high T.
2. T = ΔH / ΔS = 30 000 J / 105 J K⁻¹.

Answer: Above about 286 K.

Check yourself

• C3. At 0 °C, ice and water are at equilibrium. What is ΔG for melting?
• C4. A reaction has ΔH < 0 and ΔS < 0. Is it spontaneous at high or low temperature?

Answers at the end.

ΔG°, K and E°

The bridge from thermodynamics to equilibrium and electrochemistry.

ΔG° = −2.303 RT log K
ΔG = ΔG° + 2.303 RT log Q

K > 1 ⇔ ΔG° < 0; K < 1 ⇔ ΔG° > 0

ΔG° = −nFE°cell

F = 96 500 C mol⁻¹; a positive E° means negative ΔG°

Solved example 08 — From K to ΔG°

K = 10 for a reaction at 300 K. Find ΔG°.

1. ΔG° = −2.303 × 8.314 × 300 × log 10.
2. = −5744 J mol⁻¹.

Answer: ΔG° ≈ −5.74 kJ mol⁻¹.

Solved example 09 — From E° to ΔG°

For the Daniell cell E° = 1.10 V and n = 2. Find ΔG°.

1. ΔG° = −nFE° = −2 × 96 500 × 1.10.

Answer: ΔG° = −212.3 kJ mol⁻¹.

Beyond the basics

Two ideas that appear in PGT and higher papers.

Joule–Thomson effect

• A real gas expanding adiabatically through a porous plug from high to low pressure: enthalpy stays constant (isenthalpic).
• Most gases cool, because work is done against intermolecular attractions. This is how gases are liquefied (Linde process).
• Every gas has an inversion temperature: below it the gas cools on expansion, above it the gas warms. H₂ and He have very low inversion temperatures, so they warm on expansion at room temperature and must be pre-cooled.
• An ideal gas shows no Joule–Thomson effect (no intermolecular forces).

Gibbs–Helmholtz equation

ΔG = ΔH + T (∂ΔG/∂T)ₚ

Shows how ΔG changes with temperature; since (∂ΔG/∂T)ₚ = −ΔS, it is ΔG = ΔH − TΔS in another form

Combined with ΔG = −nFE it gives the enthalpy of a cell reaction from its EMF and the temperature coefficient of EMF: ΔH = −nF [E − T(∂E/∂T)ₚ].

Exam trap gallery

Where marks are lost in thermodynamics.

#	What students write	What is actually correct	Why
1	Work done by the gas on expansion is positive.	In the IUPAC/NCERT convention it is negative: w = −pₑₓₜΔV.	Energy leaves the system.
2	ΔH = ΔU for every reaction.	ΔH = ΔU + Δn(g)RT; equal only when Δn(g) = 0.	Count only gaseous moles.
3	All exothermic reactions are spontaneous.	Spontaneity needs ΔG < 0; ΔH < 0 with large negative ΔS can be non-spontaneous at high T.	Entropy matters too.
4	ΔfH° of O₃ and diamond is zero.	Only for the most stable form: O₂, graphite.	O₃ and diamond are not reference states.
5	T = ΔH/ΔS with ΔH in kJ and ΔS in J K⁻¹.	Convert to the same unit first.	Otherwise the answer is off by 1000.
6	ΔG° < 0 means the reaction goes to completion.	It means K > 1 — products favoured at equilibrium.	ΔG° tells position of equilibrium, not completeness.
7	Entropy of the system always increases in a spontaneous change.	The total (system + surroundings) increases.	Freezing water below 0 °C: system ΔS < 0, still spontaneous.
8	Hydrogen cools on Joule–Thomson expansion at room temperature.	H₂ warms — room temperature is above its inversion temperature.	It must be pre-cooled before liquefaction.

Memory hook: “Energy in is plus” — q and w positive when energy enters. · “H minus TS” — ΔG = ΔH − TΔS. · “Broken minus formed” — bond enthalpy formula (opposite of products minus reactants). · “Big K, negative G” — ΔG° = −RT ln K.

Practice MCQs

Levels: Easy, Moderate, Tough.

Q01. Which is a state function? (Easy)

• (a) q
• (b) w
• (c) q + w
• (d) q × w

Q02. For an isothermal process of an ideal gas: (Easy)

• (a) q = 0
• (b) w = 0
• (c) ΔU = 0
• (d) ΔH ≠ 0

Q03. Which is intensive? (Easy)

• (a) enthalpy
• (b) volume
• (c) temperature
• (d) heat capacity

Q04. ΔfH° is zero for: (Easy)

• (a) C(diamond)
• (b) O₃(g)
• (c) Br₂(l)
• (d) H₂O(l)

Q05. A process is at equilibrium when: (Easy)

• (a) ΔG = 0
• (b) ΔH = 0
• (c) ΔS = 0
• (d) ΔU = 0

Q06. For 2SO₂(g) + O₂(g) → 2SO₃(g): (Moderate)

• (a) ΔH = ΔU
• (b) ΔH > ΔU
• (c) ΔH < ΔU
• (d) ΔH = 0

Q07. ΔS is positive for: (Moderate)

• (a) N₂ + 3H₂ → 2NH₃
• (b) H₂O(l) → H₂O(s)
• (c) CaCO₃(s) → CaO(s) + CO₂(g)
• (d) 2SO₂ + O₂ → 2SO₃

Q08. ΔH = +ve, ΔS = +ve. The reaction is spontaneous: (Moderate)

• (a) at all T
• (b) at low T
• (c) at high T
• (d) never

Q09. If K < 1, then ΔG° is: (Moderate)

• (a) negative
• (b) positive
• (c) zero
• (d) unpredictable

Q10. Enthalpy of neutralisation of HCl by NaOH is about: (Moderate)

• (a) −13.7 kJ mol⁻¹
• (b) −57.1 kJ mol⁻¹
• (c) −100 kJ mol⁻¹
• (d) +57.1 kJ mol⁻¹

Q11. ΔH = −10 kJ and ΔS = −40 J K⁻¹. The reaction is spontaneous below: (Moderate)

• (a) 250 K
• (b) 400 K
• (c) 25 K
• (d) 4 K

Q12. Free expansion of an ideal gas into vacuum under adiabatic conditions has: (Moderate)

• (a) q = 0, w = 0, ΔU = 0
• (b) q = 0, w < 0
• (c) ΔU < 0
• (d) ΔT > 0

Q13. For a cell with positive E°, which is true? (Tough)

• (a) ΔG° > 0, K < 1
• (b) ΔG° < 0, K > 1
• (c) ΔG° = 0, K = 1
• (d) ΔG° < 0, K < 1

Q14. Which gas warms on Joule–Thomson expansion at room temperature? (Tough)

• (a) N₂
• (b) CO₂
• (c) He
• (d) O₂

Q15. Molar entropy of vaporisation of a liquid boiling at 350 K with ΔvapH = 30.8 kJ mol⁻¹ is: (Tough)

• (a) 88 J K⁻¹ mol⁻¹
• (b) 11.4 J K⁻¹ mol⁻¹
• (c) 0.088 J K⁻¹ mol⁻¹
• (d) 880 J K⁻¹ mol⁻¹

Answer key with explanations

1. (c) q + w — q + w = ΔU.
2. (c) ΔU = 0 — U depends only on T.
3. (c) temperature — Independent of amount.
4. (c) Br₂(l) — Bromine is liquid in its standard state.
5. (a) ΔG = 0 — No net driving force.
6. (c) ΔH < ΔU — Δn(g) = −1, so ΔH = ΔU − RT.
7. (c) CaCO₃(s) → CaO(s) + CO₂(g) — A gas is produced.
8. (c) at high T — TΔS must exceed ΔH.
9. (b) positive — log K < 0.
10. (b) −57.1 kJ mol⁻¹ — H⁺ + OH⁻ → H₂O.
11. (a) 250 K — T = 10 000/40.
12. (a) q = 0, w = 0, ΔU = 0 — Nothing is pushed; no heat.
13. (b) ΔG° < 0, K > 1 — ΔG° = −nFE°.
14. (c) He — Very low inversion temperature.
15. (a) 88 J K⁻¹ mol⁻¹ — 30 800/350 = 88 (Trouton’s rule).

Check-yourself answers

• C1. ΔH. Δn(g) = 1 − 3 = −2, so ΔH = ΔU − 2RT: ΔH is more negative.
• C2. Part of the energy released is used to ionise the weak acid completely.
• C3. ΔG = 0 (equilibrium).
• C4. Low temperature (below T = ΔH/ΔS).

One-page revision

Read this page the night before the exam.

First law

• ΔU = q + w; energy in is +
• w = −pₑₓₜΔV (irreversible)
• w = −2.303nRT log(V₂/V₁) (reversible)
• Ideal gas isothermal: ΔU = 0, q = −w

Enthalpy

• ΔH = ΔU + Δn(g)RT
• Cₚ − Cᵥ = R
• ΔfH° = 0 for stable elements
• Neutralisation (strong) = −57.1 kJ

Hess and bonds

• Reverse equation → reverse sign
• ΔrH = ΣΔfH(prod) − ΣΔfH(react)
• Bond: broken − formed
• ΔfH(CO) = −110.5 kJ

Entropy

• ΔS = q(rev)/T
• ΔS(total) > 0 spontaneous
• ΔS = ΔH/T at phase change
• Trouton: ≈ 88 J K⁻¹ mol⁻¹

Gibbs

• ΔG = ΔH − TΔS
• T(crossover) = ΔH/ΔS (same units)
• −/+ always; +/− never
• −ΔG = max useful work

Links

• ΔG° = −2.303RT log K
• ΔG° = −nFE°
• J–T: isenthalpic; H₂, He warm
• ΔH = −nF[E − T(∂E/∂T)ₚ]