ChemistryISC Class 11 · Chemical Thermodynamics

Entropy, Gibbs Energy and the Criterion of Spontaneity

Q: Can an endothermic process be spontaneous?

Yes, if the entropy increase is large enough that TΔS exceeds ΔH, so that ΔG is negative.

Q: Why is ΔG used instead of ΔS_total?

ΔG depends only on the properties of the system (ΔH and ΔS at constant T and p), so it is easier to apply.

Q: Does a negative ΔG mean the reaction will happen at once?

No. It means the reaction is thermodynamically allowed; the rate depends on the activation energy.

Updated 17 October 2026 · Fact-checked

Entropy S measures disorder; the second law says ΔS(universe) > 0 for a spontaneous process. At constant T and p the criterion is ΔG = ΔH − TΔS < 0. The standard Gibbs energy is related to the equilibrium constant by ΔG° = −2.303 RT log K.

Find the signs of ΔH and ΔS in ΔG = ΔH − TΔS; when they match, temperature decides, switching at T = ΔH/ΔS.

Illustration drawn for QuestionBuddy.

Understand Entropy, Gibbs Energy and Spontaneity

Spontaneity. A spontaneous process proceeds on its own once started, without continuous external help. Energy lowering alone (ΔH < 0) is not the whole story: some endothermic processes (dissolving NH₄Cl, melting ice at room temperature) are spontaneous because of increased disorder.

Entropy. S is a state function that measures randomness; for a reversible process ΔS = q_rev/T (J K⁻¹ mol⁻¹). It increases when a solid melts or a liquid vaporises (ΔS = ΔH_fus/T_m, ΔH_vap/T_b), when a gas expands or when the number of moles of gas increases in a reaction. The second law states that the total entropy of the system and surroundings increases in a spontaneous process: ΔS_total = ΔS_sys + ΔS_surr > 0. The third law states that the entropy of a perfect crystal is zero at absolute zero.

Gibbs energy. G = H − TS. At constant T and p, ΔG = ΔH − TΔS, and ΔG = −TΔS_total. So ΔG < 0: spontaneous; ΔG = 0: equilibrium; ΔG > 0: non-spontaneous (the reverse is spontaneous).

Effect of temperature. (ΔH −, ΔS +): spontaneous at all T. (ΔH +, ΔS −): never spontaneous. (ΔH −, ΔS −): spontaneous at low T (T < ΔH/ΔS). (ΔH +, ΔS +): spontaneous at high T (T > ΔH/ΔS). The changeover temperature is T = ΔH/ΔS.

Link with equilibrium. ΔG° = −RT ln K = −2.303 RT log K. A negative ΔG° means K > 1 (products favoured); a positive ΔG° means K < 1. The actual ΔG = ΔG° + RT ln Q.

Remember. ΔG predicts whether a reaction can occur, not how fast: a spontaneous reaction may be very slow without a catalyst.

Entropy, Gibbs Energy and Spontaneity: formulas and key facts

Entropy: $Δ S = q_{r} e v / T ∣Δ S_{f} u s = Δ H_{f} u s / T_{m} ∣Δ S_{v} a p = Δ H_{v} a p / T_{b}$; Units J K⁻¹ mol⁻¹.
Gibbs energy: $Δ G = Δ H - T Δ S$; Constant T and p.
Changeover temperature: $T = Δ H /Δ S$; When ΔG = 0.
Link with K: $Δ G ° = - 2.303 R T l o g K$; R = 8.314 J mol⁻¹ K⁻¹.

Entropy, Gibbs Energy and Spontaneity in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Does limestone decompose at a kiln temperature?

Suppose a cement-plant chemist uses CaCO₃(s) → CaO(s) + CO₂(g) with ΔH = +178 kJ mol⁻¹ and ΔS = +160 J K⁻¹ mol⁻¹.

Find the minimum temperature at which the decomposition is spontaneous.

Try it first, then show the solutionSolution

ΔG = ΔH − TΔS = 0 at the changeover temperature.
T = ΔH/ΔS = 178 000/160 = 1112 K.
Above about 1112 K (about 840 °C) ΔG is negative, so a kiln must run hotter than this.

Answer: About 1112 K.

Memory hook: Endothermic with a gas formed: entropy drives it at high temperature.

Tricks to remember Entropy, Gibbs Energy and Spontaneity

Short memory aids. Say each one aloud once, then use it on a question the same day.

Four sign cases
(−, +) always; (+, −) never; (−, −) low T; (+, +) high T.
Use it when: Spontaneity versus temperature.
Gases up, entropy up
More moles of gas in the products means ΔS positive.
Use it when: Predicting the sign of ΔS.

Entropy, Gibbs Energy and Spontaneity: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Spontaneity criterion · 1 mark · MCQBoard level

A process is spontaneous at constant temperature and pressure if:
(a) ΔG > 0
(b) ΔG < 0
(c) ΔG = 0
(d) ΔH > 0 only

Try it first, then show the solutionSolution

ΔG < 0 means the total entropy increases.

Answer: (b) ΔG < 0

Example 2: Entropy of fusion · 3 marks · Short answerBoard level

The enthalpy of fusion of ice is 6.0 kJ mol⁻¹ at 273 K. Find the entropy change on melting 1 mol of ice, and say why it is positive.

Try it first, then show the solutionSolution

At the melting point, ΔS = ΔH_fus/T = 6000/273. (1 mark)
= 22.0 J K⁻¹ mol⁻¹. (1 mark)
Positive because the ordered solid becomes a more disordered liquid. (1 mark)

Answer: 22.0 J K⁻¹ mol⁻¹

Example 3: Ammonia synthesis · 4 marks · Short answerHarder board level

For N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92.2 kJ and ΔS = −198.7 J K⁻¹ at 298 K. Find ΔG at 298 K and the temperature above which the reaction is no longer spontaneous.

Try it first, then show the solutionSolution

ΔG = ΔH − TΔS = −92 200 − 298 × (−198.7). (1 mark)
= −92 200 + 59 213 = −32 987 J ≈ −33.0 kJ. (1 mark)
Changeover: T = ΔH/ΔS = 92 200/198.7. (1 mark)
= 464 K; above this ΔG > 0 (entropy decrease dominates). (1 mark)

Answer: ΔG = −33.0 kJ; non-spontaneous above 464 K.

Example 4: ΔG° and K · 3 marks · Short answerHarder board level

The standard Gibbs energy change of a reaction is −30 kJ mol⁻¹ at 298 K. Find the equilibrium constant. (R = 8.314 J mol⁻¹ K⁻¹)

Try it first, then show the solutionSolution

ΔG° = −2.303 RT log K, so log K = −ΔG°/(2.303 RT). (1 mark)
log K = 30 000/(2.303 × 8.314 × 298) = 30 000/5706 = 5.26. (1 mark)
K = 10^5.26 = 1.8 × 10⁵. (1 mark)

Answer: K ≈ 1.8 × 10⁵

Example 5: Decomposition of limestone · 5 marks · Long answerHOTS / competency

For CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ and ΔS = +160 J K⁻¹. Find ΔG at 298 K and at 1500 K, the minimum temperature for spontaneity, and K at 1500 K.

Try it first, then show the solutionSolution

At 298 K: ΔG = 178 000 − 298 × 160 = 178 000 − 47 680 = +130.3 kJ, non-spontaneous. (1 mark)
At 1500 K: ΔG = 178 000 − 1500 × 160 = −62 000 J = −62.0 kJ, spontaneous. (1 mark)
T_min = ΔH/ΔS = 178 000/160 = 1112 K. (1 mark)
log K = 62 000/(2.303 × 8.314 × 1500) = 62 000/28 720 = 2.16. (1 mark)
K = 10^2.16 = 1.4 × 10². (1 mark)

Answer: ΔG(298 K) = +130.3 kJ; ΔG(1500 K) = −62.0 kJ; T_min = 1112 K; K ≈ 1.4 × 10².

Where students lose marks in Entropy, Gibbs Energy and Spontaneity

The slips examiners see most often, and what to write instead.

Mistake: Mixing J and kJ
Subtracting TΔS in J from ΔH in kJ.
Do this: Convert ΔS to kJ K⁻¹ or ΔH to J before combining.
Usually costs: 2 marks.
Mistake: Using °C for T
Substituting 25 instead of 298.
Do this: Use kelvin in ΔG = ΔH − TΔS.
Usually costs: 2 marks.
Mistake: Equating spontaneity with speed
Claiming a reaction with negative ΔG is fast.
Do this: ΔG tells direction, not rate.
Usually costs: 1 mark.

Exam tips

Write ΔG = ΔH − TΔS with units, converting J to kJ.
State the four sign cases when asked about temperature dependence.
Quote ΔG = 0 at equilibrium.
Use log K = −ΔG°/(2.303 RT).

Practice questions on Entropy, Gibbs Energy and Spontaneity

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The entropy of a system increases when:
(a) ice melts
(b) water vapour condenses
(c) a gas is compressed
(d) a crystal forms from solution
Hint
More disorder means more entropy.
Answer
(a) ice melts
Q2 · 2 marks · Short answerBoard level
Predict the sign of ΔS for 2H₂(g) + O₂(g) → 2H₂O(l).
Hint
Count the gas moles.
Answer
Negative: three moles of gas become a liquid, so the disorder falls.
Q3 · 3 marks · Short answerHarder board level
For a reaction ΔH = +30 kJ and ΔS = +100 J K⁻¹. Find ΔG at 400 K and the temperature above which it is spontaneous.
Hint
ΔG = ΔH − TΔS.
Answer
ΔG = 30 000 − 400 × 100 = −10 000 J = −10 kJ (spontaneous). T = ΔH/ΔS = 300 K.
Q4 · 3 marks · Short answerHarder board level
The enthalpy of vaporisation of a liquid is 40.7 kJ mol⁻¹ and its boiling point is 373 K. Find ΔS_vap.
Hint
ΔS = ΔH/T at the boiling point.
Answer
ΔS = 40 700/373 = 109 J K⁻¹ mol⁻¹.
Q5 · 3 marks · Short answerHarder board level
A reaction has ΔG° = +20 kJ mol⁻¹ at 298 K. Find K and comment on the extent of reaction. (R = 8.314 J mol⁻¹ K⁻¹)
Hint
log K = −ΔG°/(2.303 RT).
Answer
log K = −20 000/5706 = −3.505, so K = 3.1 × 10⁻⁴. K < 1: reactants are favoured; very little product forms.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose four reactions A, B, C and D have (ΔH in kJ, ΔS in J K⁻¹): A (−50, +50), B (+50, −50), C (−50, −100) and D (+50, +100).
(i) Which is spontaneous at all temperatures? (1 mark)
(ii) Which is never spontaneous? (1 mark)
(iii) Find the changeover temperature of C and D. (1 mark)
(iv) Find ΔG of C and D at 298 K. (1 mark)
Hint
Use ΔG = ΔH − TΔS and T = ΔH/ΔS.
Answer
(i) A. (ii) B. (iii) T = 50 000/100 = 500 K for both C (spontaneous below 500 K) and D (spontaneous above 500 K). (iv) C: −50 000 + 29 800 = −20.2 kJ; D: 50 000 − 29 800 = +20.2 kJ.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Gibbs–Helmholtz form of the spontaneity criterion?Show answer
ΔG = ΔH − TΔS < 0.
2.Third law of thermodynamics?Show answer
The entropy of a perfect crystal is zero at 0 K.
3.ΔG at equilibrium?Show answer
Zero.
4.Sign of ΔS when a gas is produced?Show answer
Positive.

Entropy, Gibbs Energy and Spontaneity: frequently asked questions

Can an endothermic process be spontaneous?

Yes, if the entropy increase is large enough that TΔS exceeds ΔH, so that ΔG is negative.

Why is ΔG used instead of ΔS_total?

ΔG depends only on the properties of the system (ΔH and ΔS at constant T and p), so it is easier to apply.

Does a negative ΔG mean the reaction will happen at once?

No. It means the reaction is thermodynamically allowed; the rate depends on the activation energy.