IB DP Chemistry Reactivity 1.4: Entropy and spontaneity – Revision Notes

Use these notes in the final weeks to recall Reactivity 1.4, “Entropy and spontaneity”, quickly. Statement numbers follow the DP Chemistry guide the IB issued in February 2023 (first assessment 2025): understandings Reactivity 1.4.1 to 1.4.4. All four understandings are additional higher level material, so this page is HL only, for sessions from 2025 onwards.

Anything here that feels too compressed is unpacked, with fuller examples, in the entropy and spontaneity study guide. When you are ready, work through the practice set. The DP Chemistry hub links every other unit, the printable checklist records what you have covered, and the free diagnostics flag topics that need more time.

Key definitions

• Entropy, S: how dispersed the matter and energy of a system are. The more ways the energy can be shared out, the higher S. Unit: J K⁻¹ mol⁻¹.
• Standard entropy, S⦵: the entropy value per mole for a substance under standard conditions. It is positive for every pure substance above 0 K, elements included.
• Perfect crystal at 0 K: only one possible arrangement, so its entropy is predicted to be zero.
• Gibbs energy change, ΔG: links ΔH, ΔS and the absolute temperature T; it relates to the energy a reaction can supply for work.
• Spontaneous change: at constant pressure, any change with a negative ΔG. It may still be slow.
• Reaction quotient, Q: the equilibrium-constant expression evaluated for any mixture, at equilibrium or not.

Formulas

Quantity	Equation	Units to use
Standard entropy change	ΔS⦵ = ΣS⦵(products) − ΣS⦵(reactants)	J K⁻¹ mol⁻¹
Gibbs energy	ΔG⦵ = ΔH⦵ − TΔS⦵	ΔH, ΔG in kJ mol⁻¹; ΔS in J K⁻¹ mol⁻¹ (divide by 1000); T in K
Threshold temperature	T = ΔH ÷ ΔS (set ΔG = 0)	ΔH and ΔS in matching units
Entropy change of surroundings	ΔS(surroundings) = −ΔH ÷ T	J K⁻¹ mol⁻¹ if ΔH in J mol⁻¹
ΔG for any mixture	ΔG = ΔG⦵ + RT lnQ	J mol⁻¹; R = 8.31 J K⁻¹ mol⁻¹
At equilibrium	ΔG⦵ = −RT lnK	J mol⁻¹

The guide says the data booklet gives standard entropy values, thermodynamic data and the two ΔG equations used in 1.4.4.

Reactivity 1.4.1 Predicting and calculating ΔS

Order under the same conditions: gas > liquid > solid.

ΔS of the system is positive when:

• the number of moles of gas increases (the strongest effect);
• a solid melts, a liquid boils or a solid sublimes;
• a solid dissolves to give aqueous ions, or two substances mix;
• a substance is heated.

ΔS is negative for the reverse of each.

Method in steps: ΔS⦵ from data

1. Write the balanced equation with state symbols.
2. Multiply each S⦵ by its coefficient.
3. Sum products; sum reactants; subtract (products − reactants).
4. Compare the sign with your gas-moles prediction.

Worked reminder. Titanium extraction, TiCl₄(l) + 2Mg(s) → Ti(s) + 2MgCl₂(s). S⦵ values: TiCl₄(l) 252.3, Mg(s) 32.7, Ti(s) 30.7, MgCl₂(s) 89.6 J K⁻¹ mol⁻¹. Products: 30.7 + (2 × 89.6) = 209.9. Reactants: 252.3 + (2 × 32.7) = 317.7. ΔS⦵ = 209.9 − 317.7 = −107.8 J K⁻¹ mol⁻¹. No gas is involved, but a liquid is turned into solids, so a negative value is expected.

Reactivity 1.4.2 Using ΔG⦵ = ΔH⦵ − TΔS⦵

Method in steps

1. Convert ΔS to kJ K⁻¹ mol⁻¹ (÷ 1000).
2. Put T in kelvin (add 273 to °C).
3. Substitute and keep the signs in brackets.
4. Rearrange for an unknown: ΔS = (ΔH − ΔG) ÷ T, or ΔH = ΔG + TΔS.

Worked reminder. A reaction has ΔH⦵ = −92.0 kJ mol⁻¹ and ΔG⦵ = −33.0 kJ mol⁻¹ at 298 K. ΔS⦵ = (−92.0 − (−33.0)) ÷ 298 = −0.198 kJ K⁻¹ mol⁻¹ = −198 J K⁻¹ mol⁻¹. Bracket the negative ΔG before subtracting it; that step is where sign slips creep in.

Reactivity 1.4.3 Spontaneity

ΔH	ΔS	Spontaneous
−	+	at all temperatures
+	−	never
−	−	below T = ΔH/ΔS
+	+	above T = ΔH/ΔS

Worked reminder. ΔH = +65.0 kJ mol⁻¹, ΔS = +160 J K⁻¹ mol⁻¹. T = 65.0 ÷ 0.160 = 406 K. With both signs positive, the change is spontaneous above 406 K.

Direct and indirect entropy. ΔG accounts for the system’s own entropy change (direct) and the surroundings’ change caused by heat transfer (indirect, −ΔH/T). Since ΔG = −TΔS(total), a negative ΔG means total entropy rises.

Reactivity 1.4.4 Gibbs energy and equilibrium

• As a reaction moves towards equilibrium, ΔG becomes less negative and reaches zero at equilibrium.
• If Q < K, ΔG is negative and the forward reaction proceeds. If Q > K, ΔG is positive and the reverse reaction proceeds.
• ΔG⦵ negative gives K > 1 (products favoured); ΔG⦵ positive gives K < 1 (reactants favoured).

Method in steps: ΔG for a given mixture

1. Write the expression for Q exactly as you would for K, using the mixture’s concentrations.
2. Convert ΔG⦵ to J mol⁻¹ and T to kelvin.
3. Calculate RT lnQ; it is negative when Q < 1 and positive when Q > 1.
4. Add it to ΔG⦵. The sign of ΔG gives the spontaneous direction for that mixture.

Worked reminder. At 310 K, ΔG⦵ = +3.00 kJ mol⁻¹ and Q = 0.0500. RT lnQ = 8.31 × 310 × ln 0.0500 = −7717 J mol⁻¹. ΔG = 3000 − 7717 = −4.72 kJ mol⁻¹. The forward reaction is spontaneous for this mixture, even though ΔG⦵ is positive.

Worked reminder. ΔG⦵ = +12.0 kJ mol⁻¹ at 298 K. lnK = −12 000 ÷ (8.31 × 298) = −4.85, so K = 7.86 × 10⁻³. At equilibrium, reactants make up most of the mixture.

Links to other parts of the course

• Structure 1.1: particle arrangements in solids, liquids and gases explain the order of S⦵ values and the zero-entropy perfect crystal.
• Reactivity 2.3: a positive ΔG⦵ means K < 1, so an equilibrium mixture contains mostly reactants; see the Reactivity 2.3 study guide.
• Reactivity 3.2: electrochemical data give a second route to deciding whether a redox reaction is spontaneous.

Must-know distinctions

• ΔG vs ΔG⦵: ΔG⦵ refers to standard conditions and is fixed at a given temperature; ΔG depends on the actual mixture through Q.
• Spontaneous vs fast: ΔG tells you whether a change is thermodynamically favoured; the activation energy controls its rate.
• System vs surroundings: ΔS⦵ from data is the system only. The surroundings’ change comes from −ΔH/T.
• Entropy vs enthalpy of formation for elements: an element in its standard state is given ΔfH⦵ = 0, but its S⦵ is always above zero.
• Threshold temperature: T = ΔH/ΔS is only meaningful when ΔH and ΔS have the same sign.

Quick self-test

1. State the sign of ΔS for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), with a reason.
2. Explain why a perfect crystal is expected to have zero entropy at absolute zero.
3. A change has ΔH = +25.0 kJ mol⁻¹ and ΔS = +95.0 J K⁻¹ mol⁻¹. Calculate ΔG at 298 K and say whether it is spontaneous.
4. ΔH = −120 kJ mol⁻¹ and ΔS = −150 J K⁻¹ mol⁻¹. Find the temperature range in which the reaction is spontaneous.
5. Which combination of signs of ΔH and ΔS gives a reaction that is never spontaneous?
6. ΔG⦵ = −10.0 kJ mol⁻¹ at 298 K. Calculate K.
7. What is ΔG⦵ for a reaction with K = 1?
8. ΔG⦵ = +15.0 kJ mol⁻¹ at 298 K, and a mixture has Q = 1.0 × 10⁻⁴. Calculate ΔG and state which direction is spontaneous.
9. An exothermic reaction releases 50.0 kJ mol⁻¹ at 250 K. Calculate ΔS(surroundings).
10. At 500 K, K = 2.0 × 10⁻³ for a reaction. Calculate ΔG⦵.
11. State what happens to the value of ΔG as a reaction mixture approaches equilibrium.

Answers

1. Negative: 3 mol of gas become 1 mol of gas (water is a liquid).
2. There is only one possible arrangement of the particles and their energy.
3. ΔG = 25.0 − (298 × 0.0950) = −3.31 kJ mol⁻¹; negative, so spontaneous.
4. T = −120 ÷ −0.150 = 800 K; spontaneous below 800 K.
5. ΔH positive and ΔS negative.
6. lnK = 10 000 ÷ (8.31 × 298) = 4.04; K = 56.7.
7. Zero, because ln 1 = 0.
8. ΔG = 15 000 + (8.31 × 298 × ln 1.0 × 10⁻⁴) = 15 000 − 22 808 = −7.81 kJ mol⁻¹; the forward reaction is spontaneous.
9. ΔS(surroundings) = +50 000 ÷ 250 = +200 J K⁻¹ mol⁻¹.
10. ΔG⦵ = −8.31 × 500 × ln(2.0 × 10⁻³) = +25.8 kJ mol⁻¹.
11. It becomes less negative and reaches zero at equilibrium.

Where marks are usually lost

• Mixing J and kJ in ΔH − TΔS: ΔS must be divided by 1000 first.
• Leaving T in °C.
• Dropping coefficients when summing S⦵, especially for O₂ and H₂.
• Missing state symbols, so liquid water is counted as a gas in a ΔS prediction.
• Giving T = ΔH/ΔS without “above” or “below”.
• Using kJ inside −RT lnK, so K comes out close to 1.
• Writing a sign-only answer for ΔS when the question asks for a reason; name the change in moles of gas.
• Treating ΔG = 0 as “no reaction” rather than as equilibrium.
• Stating that a spontaneous reaction must be fast.

Official syllabus

These notes are based on the Diploma Programme Chemistry guide (International Baccalaureate Organization), first assessment 2025, published February 2023, sub-topic Reactivity 1.4 (additional higher level).