ChemistryCBSE Class 11 · Thermodynamics

Entropy, Gibbs Energy and Spontaneity Class 11

Q: Can an endothermic process be spontaneous?

Yes, if the entropy increase is large enough at that temperature, such as the melting of ice above 0 °C.

Q: What does a negative ΔG mean?

The reaction can proceed spontaneously at that temperature and pressure; it says nothing about its speed.

Q: Why is ΔS_surr = −ΔH/T?

The heat released by the system flows into the surroundings at temperature T, raising their entropy.

Q: How are ΔG° and K related?

K = e^(−ΔG°/RT): a negative ΔG° gives K > 1.

Updated 17 October 2026 · Fact-checked

A process is spontaneous when the total entropy of system plus surroundings increases. At constant temperature and pressure this is expressed by ΔG = ΔH − TΔS < 0. ΔG° is linked to the equilibrium constant by ΔG° = −RT ln K = −2.303RT log K, and a reaction with ΔH and ΔS of the same sign changes spontaneity at T = ΔH/ΔS.

Understand Entropy, Gibbs Energy and Spontaneity

Entropy. $S$ measures the disorder (the number of microscopic arrangements) of a system and is a state function. $Δ S_{sy s} = \frac{q _{r e v}}{T}$ . It increases when a solid melts or vaporises, when a solid dissolves, when the number of gas moles increases and with temperature. Examples: $Δ_{f u s} S = Δ_{f u s} H / T_{f}$ , $Δ_{v a p} S = Δ_{v a p} H / T_{b}$ (about 88 J K⁻¹ mol⁻¹ for many liquids, Trouton's rule), and for an ideal gas expanding isothermally $Δ S = n R ln \frac{V _{2}}{V _{1}}$ .

Second and third laws. In a spontaneous process, $Δ S_{t o t a l} = Δ S_{sy s} + Δ S_{s u r r} > 0$ , with $Δ S_{s u r r} = - \frac{Δ H _{sy s}}{T}$ at constant pressure. The entropy of a perfect crystal is zero at 0 K (third law).

Gibbs energy. $G = H - T S$ , so at constant $T$ and $P$ :
$ $Δ G = Δ H - T Δ S = - T Δ S_{t o t a l} .$ $
$Δ G < 0$ spontaneous; $Δ G = 0$ equilibrium; $Δ G > 0$ non-spontaneous (the reverse is spontaneous). Effect of signs:

$Δ H < 0, Δ S > 0$ : spontaneous at all temperatures;
$Δ H > 0, Δ S < 0$ : never spontaneous;
$Δ H < 0, Δ S < 0$ : spontaneous at low T (below $T = Δ H /Δ S$ );
$Δ H > 0, Δ S > 0$ : spontaneous at high T (above $T = Δ H /Δ S$ ).

Equilibrium. $Δ G^{⊖} = - R T ln K = - 2.303 R T lo g K$ ; $Δ G^{⊖} < 0$ means $K > 1$ and products are favoured. At a phase-change temperature $Δ G = 0$ , so $T_{b} = Δ_{v a p} H / Δ_{v a p} S$ .

Note. $Δ G^{⊖}$ tells whether a reaction can go; it does not tell how fast.

Entropy, Gibbs Energy and Spontaneity: formulas and key facts

Gibbs energy change: $Δ G = Δ H - T Δ S$; Constant T and P.
Surroundings entropy: $Δ S_{s u r r} = - \frac{Δ H}{T}$; Constant pressure.
Equilibrium: $Δ G^{⊖} = - R T ln K = - 2.303 R T lo g K$; R = 8.314 J mol⁻¹ K⁻¹.
Crossover temperature: $T = \frac{Δ H}{Δ S}$; When $Δ G = 0$ .
Isothermal expansion: $Δ S = n R ln \frac{V _{2}}{V _{1}}$; Ideal gas.
Phase change: $Δ S = \frac{Δ H _{t r an s}}{T _{t r an s}}$; At the transition temperature.

How to solve Entropy, Gibbs Energy and Spontaneity questions

Get ΔH and ΔS in the same units (kJ and J do not mix), find the sign of ΔG at the required temperature, and then relate it to K if needed.

1List ΔH and ΔS with signs and convert to the same energy unit.
2Compute ΔG = ΔH − TΔS at the given T (in kelvin).
3Decide spontaneity from the sign of ΔG.
4For the crossover temperature set ΔG = 0: T = ΔH/ΔS.
5For K, use ΔG° = −2.303RT log K with R in J and ΔG° in J.
6Interpret the result: K above 1 means products are favoured.

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.

Ice cube in a drink

Suppose an ice cube melts in a warm drink: ΔfusH = +6.01 kJ/mol at 273 K.

Find ΔS for melting one mole of ice at 273 K.

Try it first, then show the solutionSolution

ΔS = ΔH/T = 6010/273 = 22.0 J K⁻¹ mol⁻¹.

Answer: +22.0 J K⁻¹ mol⁻¹.

Memory hook: More disorder in the liquid.

Rusting at room temperature

Suppose a reaction has ΔH = −400 kJ and ΔS = −100 J/K at 298 K.

Is it spontaneous?

Try it first, then show the solutionSolution

ΔG = −400 − 298 × (−0.100) = −400 + 29.8 = −370.2 kJ.

Answer: Yes, ΔG = −370 kJ.

Memory hook: The enthalpy term dominates.

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: Predicting spontaneity.
J versus kJ
ΔS is in J, ΔH in kJ: convert before T ΔS.
Use it when: Numericals.
ΔG = 0 at equilibrium
Phase changes and equilibria sit at ΔG = 0.
Use it when: Transition temperatures.
log K
ΔG° = −2.303RT log K.
Use it when: K from ΔG°.

Entropy, Gibbs Energy and Spontaneity: solved examples

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

Example 1: Entropy change of the surroundings · 2 marks · Very short answerBoard level

The reaction N₂(g) + 3H₂(g) → 2NH₃(g) has ΔH = −92.4 kJ at 298 K. Find the entropy change of the surroundings.

Try it first, then show the solutionSolution

ΔS_surr = −ΔH/T = 92400/298. (1)
= +310 J K⁻¹. (1)

Answer: +310 J K⁻¹.

Example 2: Spontaneity and the limiting temperature · 3 marks · Short answerHarder board level

A reaction has ΔH = −46 kJ and ΔS = −100 J K⁻¹ at 298 K. (a) Find ΔG and say whether it is spontaneous. (b) Find the temperature above which it ceases to be spontaneous.

Try it first, then show the solutionSolution

(a) ΔG = −46 − 298 × (−0.100) = −46 + 29.8 = −16.2 kJ: spontaneous. (1)
(b) ΔG = 0 when T = ΔH/ΔS = −46000/−100 = 460 K. (1)
Because both ΔH and ΔS are negative, the reaction is spontaneous only below 460 K. (1)

Answer: ΔG = −16.2 kJ; spontaneous below 460 K.

Example 3: Entropy of expansion · 3 marks · Short answerHarder board level

Two moles of an ideal gas expand isothermally and reversibly from 10 L to 40 L at 300 K. Find ΔS of the gas, and say what ΔS is for the surroundings and in a free expansion between the same states (R = 8.314 J/mol K).

Try it first, then show the solutionSolution

ΔS_sys = nR ln(V₂/V₁) = 2 × 8.314 × ln 4 = 2 × 8.314 × 1.386 = 23.1 J/K. (1)
For the reversible process ΔS_surr = −23.1 J/K, so ΔS_total = 0. (1)
S is a state function: the free expansion between the same states also has ΔS_sys = 23.1 J/K, but ΔS_total > 0 (irreversible). (1)

Answer: 23.1 J/K; reversible total 0; free expansion total > 0.

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

At 298 K (a) find K for a reaction with ΔG° = −28.5 kJ/mol, (b) find ΔG° for a reaction with K = 0.010, and (c) find ΔG° and K for a reaction with ΔH° = −60 kJ and ΔS° = −120 J K⁻¹ (R = 8.314 J/mol K).

Try it first, then show the solutionSolution

(a) log K = 28500/(2.303 × 8.314 × 298) = 5.0, so K = 1 × 10⁵. (1)
(b) ΔG° = −2.303RT log K = −5706 × (−2) = +11.4 kJ/mol. (1)
(c) ΔG° = −60 − 298 × (−0.120) = −24.2 kJ/mol. (1)
log K = 24240/5706 = 4.25, K ≈ 1.8 × 10⁴. (1)

Answer: (a) 1 × 10⁵; (b) +11.4 kJ/mol; (c) −24.2 kJ/mol and K ≈ 1.8 × 10⁴.

Example 5: The Haber process · 4 marks · Long answerHOTS / competency

For N₂(g) + 3H₂(g) → 2NH₃(g), ΔH° = −92 kJ and ΔS° = −199 J K⁻¹. (a) Find ΔG° at 298 K and at 700 K. (b) Find the temperature at which ΔG° = 0. (c) Find K at 298 K. (d) Comment on why industry uses a high temperature despite the result.

Try it first, then show the solutionSolution

(a) 298 K: ΔG° = −92 + 298 × 0.199 = −32.7 kJ; 700 K: −92 + 700 × 0.199 = +47.3 kJ. (1)
(b) T = ΔH/ΔS = 92000/199 = 462 K. (1)
(c) log K = 32700/5706 = 5.73, K ≈ 5 × 10⁵ at 298 K. (1)
(d) The reaction is favourable at low temperature but too slow there; a higher temperature and a catalyst give a workable rate at the cost of a lower equilibrium yield. (1)

Answer: (a) −32.7 kJ and +47.3 kJ; (b) 462 K; (c) 5 × 10⁵; (d) rate versus yield.

Where students lose marks in Entropy, Gibbs Energy and Spontaneity

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

Mistake: Mixing kJ and J in ΔH − TΔS.
Different units.
Do this: Convert ΔS to kJ K⁻¹ or ΔH to J.
Usually costs: The numerical.
Mistake: Using °C for T.
Not converting.
Do this: T must be in kelvin.
Usually costs: The numerical.
Mistake: Saying ΔS_sys > 0 means the process is spontaneous.
Ignoring the surroundings.
Do this: Spontaneity needs ΔS_total > 0 or ΔG < 0.
Usually costs: Concept mark.
Mistake: Using ln instead of log (or the reverse) in ΔG° = −RT ln K.
Mixing the two forms.
Do this: 2.303RT log K = RT ln K.
Usually costs: The K value.

Exam tips

Write ΔG = ΔH − TΔS and the sign table for the four cases.
State the unit conversion of ΔS in a separate line.
For the crossover temperature, show ΔG = 0.
Quote K with units where appropriate (dimensionless).
Interpret the result in words: spontaneous, at equilibrium or not spontaneous.

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
A reaction is spontaneous at all temperatures when
(a) ΔH > 0 and ΔS < 0 (b) ΔH < 0 and ΔS > 0 (c) ΔH > 0 and ΔS > 0 (d) ΔH < 0 and ΔS < 0
Hint
Both terms make ΔG negative.
Answer
(b).
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The entropy of a perfect crystal at 0 K is zero.
Reason (R): At 0 K a perfect crystal has only one possible arrangement of its particles.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Third law.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
Which has the higher entropy: 1 mol of ice at 0 °C, 1 mol of liquid water at 0 °C or 1 mol of steam at 100 °C? Give a reason.
Hint
Gases have the most freedom.
Answer
Steam: its molecules are far apart and move randomly, giving the most disorder.
Q4 · 3 marks · Short answerHarder board level
Benzene boils at 353 K with ΔvapH = 30.8 kJ/mol; water boils at 373 K with ΔvapH = 40.7 kJ/mol. Find ΔvapS for each and comment (Trouton's rule value is about 88 J K⁻¹ mol⁻¹).
Hint
ΔS = ΔH/T_b.
Answer
Benzene: 30800/353 = 87.3 J K⁻¹ mol⁻¹ (close to 88). Water: 40700/373 = 109 J K⁻¹ mol⁻¹, higher because hydrogen bonding makes liquid water more ordered.
Q5 · 4 marks · Long answerHOTS / competency
For a liquid, ΔvapH = 29.4 kJ/mol and ΔvapS = 85 J K⁻¹ mol⁻¹. (a) Find its normal boiling point. (b) Find ΔG at 298 K and at 400 K for vaporisation at 1 bar and say what happens. (c) What does ΔG = 0 represent?
Hint
At the boiling point ΔG = 0.
Answer
(a) T_b = 29400/85 = 346 K. (b) 298 K: ΔG = 29.4 − 298 × 0.085 = +4.1 kJ (not spontaneous, the liquid persists); 400 K: ΔG = 29.4 − 34 = −4.6 kJ (spontaneous, it vaporises). (c) Liquid and vapour in equilibrium at the boiling point.

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.ΔG formula?Show answer
ΔH − TΔS.
2.ΔG° and K?Show answer
ΔG° = −2.303RT log K.
3.ΔG at equilibrium?Show answer
Zero.
4.Entropy of a perfect crystal at 0 K?Show answer
Zero.
5.ΔS of surroundings?Show answer
−ΔH/T.
6.Trouton's constant?Show answer
About 88 J K⁻¹ mol⁻¹.

Entropy, Gibbs Energy and Spontaneity: frequently asked questions

Can an endothermic process be spontaneous?

Yes, if the entropy increase is large enough at that temperature, such as the melting of ice above 0 °C.

What does a negative ΔG mean?

The reaction can proceed spontaneously at that temperature and pressure; it says nothing about its speed.

Why is ΔS_surr = −ΔH/T?

The heat released by the system flows into the surroundings at temperature T, raising their entropy.

How are ΔG° and K related?

K = e^(−ΔG°/RT): a negative ΔG° gives K > 1.

What is a state function?

A property depending only on the state of the system, such as H, S and G.