IB DP Chemistry Reactivity 1.4: Entropy and spontaneity

Reactivity 1.4, “Entropy and spontaneity”, answers one question: which way will a chemical change go on its own? This page teaches it from scratch for IB DP Chemistry. Its content and numbering come from the IB’s 2023 guide to Diploma Programme Chemistry, whose first exams ran in 2025, and each of understandings Reactivity 1.4.1 to 1.4.4 gets its own worked examples. Every understanding in this sub-topic sits in the additional higher level (AHL) material, so the whole page is HL only. It is written for exam sessions from 2025 onwards.

Use it with the entropy and spontaneity revision notes for short-form recall and the entropy and spontaneity practice set for marked questions. Every DP Chemistry unit is listed on the DP Chemistry hub, the printable DP Chemistry checklist lets you tick off each understanding, and our free diagnostic quizzes suggest where to begin. For paper formats, see DP Chemistry exam preparation.

Scope of Reactivity 1.4

Statement	Skills expected	Level
Reactivity 1.4.1	Describe entropy as the spread of matter and energy; predict whether a change increases or decreases the entropy of a system; calculate ΔS⦵ from S⦵ values	HL only
Reactivity 1.4.2	Use ΔG⦵ = ΔH⦵ − TΔS⦵ to find any unknown term, with the right units	HL only
Reactivity 1.4.3	Interpret the sign of ΔG; find the threshold temperature for spontaneity; link ΔG to the entropy change of the system and the surroundings	HL only
Reactivity 1.4.4	Explain why ΔG becomes less negative and reaches zero at equilibrium; use ΔG = ΔG⦵ + RT lnQ and ΔG⦵ = −RT lnK	HL only

The guide gives this sub-topic about 5 hours of AHL teaching time. According to the guide, standard entropy values, the other thermodynamic data and the two ΔG equations in 1.4.4 appear in the data booklet. Here, each example supplies the values it uses.

Reactivity 1.4.1 Entropy

Entropy, S, tells you how widely the matter and energy of a system are dispersed. When there are more ways to share the available energy among the particles, entropy is higher. S is measured in J K⁻¹ mol⁻¹.

Compare one substance in its three states under the same conditions:

S(gas) > S(liquid) > S(solid)

Gas particles roam the whole container and have many ways to hold energy, so a gas has far more possible arrangements than a liquid. Particles in a solid only vibrate about fixed sites, which gives the fewest arrangements. A perfect crystal at 0 K has a single arrangement, so its entropy is predicted to be zero. It follows that any real substance above 0 K, elements included, has a positive S⦵. Do not carry over the enthalpy rule that elements score zero.

Predicting the sign of ΔS

Check the system for these changes:

• Moles of gas. This is usually the deciding factor. If the products contain more moles of gas than the reactants, ΔS is positive.
• Change of state. Melting, boiling and sublimation raise entropy. Freezing, condensing and deposition lower it.
• Dissolving and precipitation. A soluble solid breaking up into scattered aqueous ions usually gains entropy. Ions clicking together into a precipitate lose it.
• Temperature. Warming a substance raises its entropy, because the particles have more energy to spread around.

Worked example 1: predicting ΔS

Decide whether ΔS is positive or negative for each change.

1. Water vapour from a kettle condenses on a cold window: H₂O(g) → H₂O(l). Gas turns into liquid, so ΔS is negative.
2. Heated ammonium dichromate breaks down: (NH₄)₂Cr₂O₇(s) → Cr₂O₃(s) + N₂(g) + 4H₂O(g). One solid produces 5 mol of gas, so ΔS is positive.
3. Ag⁺(aq) + Br⁻(aq) → AgBr(s). Mobile ions become fixed in a lattice, so ΔS is negative.
4. Propanone burns: CH₃COCH₃(l) + 4O₂(g) → 3CO₂(g) + 3H₂O(l). Gas falls from 4 mol to 3 mol, so ΔS is negative.

Item 4 depends on state symbols. Written with H₂O(g), the products would hold 6 mol of gas against 4 mol, and ΔS would come out positive.

Calculating ΔS⦵ from S⦵ values

Multiply each S⦵ by its coefficient, add up each side, then take reactants from products:

ΔS⦵ = ΣS⦵(products) − ΣS⦵(reactants)

Worked example 2: ammonium hydrogencarbonate

When warmed, ammonium hydrogencarbonate decomposes completely into gases: NH₄HCO₃(s) → NH₃(g) + CO₂(g) + H₂O(g). Use these S⦵ values in J K⁻¹ mol⁻¹: NH₄HCO₃(s) 120.9, NH₃(g) 192.8, CO₂(g) 213.8, H₂O(g) 188.8.

ΣS⦵(products)  = 192.8 + 213.8 + 188.8 = 595.4 J K⁻¹ mol⁻¹
ΣS⦵(reactants) = 120.9 J K⁻¹ mol⁻¹
ΔS⦵ = 595.4 − 120.9 = +474.5 J K⁻¹ mol⁻¹

A quick prediction agrees: one solid becomes 3 mol of gas, so expect a large positive value. If the sign disagrees, look for a missed coefficient or state symbol.

Reactivity 1.4.2 Gibbs energy

The Gibbs energy change, ΔG, ties together the enthalpy change, the entropy change and the absolute temperature. It relates to the energy a reaction can make available to do useful work.

ΔG⦵ = ΔH⦵ − TΔS⦵

Watch the units. ΔH and ΔG are quoted in kJ mol⁻¹, ΔS in J K⁻¹ mol⁻¹, and T must be in kelvin. Divide ΔS by 1000 before you put it into the equation.

Worked example 3: ΔG⦵ for the decomposition at 298 K

Use ΔH⦵ = +168.2 kJ mol⁻¹ and ΔS⦵ = +474.5 J K⁻¹ mol⁻¹ (from worked example 2).

ΔS⦵ = +474.5 ÷ 1000 = +0.4745 kJ K⁻¹ mol⁻¹
TΔS⦵ = 298 × 0.4745 = 141.4 kJ mol⁻¹
ΔG⦵ = 168.2 − 141.4 = +26.8 kJ mol⁻¹

Worked example 4: finding an unknown term

For ethanol evaporating, C₂H₅OH(l) → C₂H₅OH(g), at 298 K: ΔH⦵ = +42.3 kJ mol⁻¹ and ΔG⦵ = +6.0 kJ mol⁻¹. Find ΔS⦵.

TΔS⦵ = ΔH⦵ − ΔG⦵ = 42.3 − 6.0 = 36.3 kJ mol⁻¹
ΔS⦵ = 36.3 ÷ 298 = 0.122 kJ K⁻¹ mol⁻¹ = +122 J K⁻¹ mol⁻¹

A positive answer makes sense for a liquid becoming a gas.

Reactivity 1.4.3 Spontaneity and the sign of ΔG

When pressure is constant, a negative ΔG means the change is spontaneous. A positive ΔG means the change as written is not spontaneous, though its reverse is. ΔG = 0 means the system is at equilibrium.

Spontaneity is not the same as speed. A change with a strongly negative ΔG may barely move at room temperature when its activation energy is large.

Why ΔG includes the surroundings

ΔG combines two entropy changes:

• the direct entropy change of the system, ΔS, from turning reactants into products;
• the indirect entropy change of the surroundings, from heat flowing in or out. When a reaction gives out heat, that energy spreads through the surroundings and their entropy rises. For heat passed at temperature T, ΔS(surroundings) = −ΔH/T.

A change is spontaneous when the combined entropy of system and surroundings goes up. Multiplying that total by −T produces ΔG:

ΔS(total) = ΔS − ΔH/T
−TΔS(total) = ΔH − TΔS = ΔG

So “ΔG is negative” and “total entropy increases” are the same condition.

The four sign combinations

ΔH	ΔS	ΔG = ΔH − TΔS	Spontaneous?
negative	positive	always negative	at all temperatures
positive	negative	always positive	at no temperature
negative	negative	negative only while TΔS is smaller in size than ΔH	at low temperature
positive	positive	negative only once TΔS outweighs ΔH	at high temperature

Finding the threshold temperature for spontaneity

For the last two rows, put ΔG = 0 and rearrange for T. This treats ΔH and ΔS as roughly constant over the temperature range.

T = ΔH ÷ ΔS     (ΔS in kJ K⁻¹ mol⁻¹)

Worked example 5: when does the decomposition become spontaneous?

T = 168.2 ÷ 0.4745 = 354 K

ΔH and ΔS are both positive, so the decomposition is spontaneous above 354 K. Test a point: at 400 K, ΔG = 168.2 − (400 × 0.4745) = −21.6 kJ mol⁻¹, which is negative as predicted. This is why the solid gives off gas when it is heated in an oven, which is how it acts as a raising agent in baking.

Worked example 6: forming nitrosyl chloride

For 2NO(g) + Cl₂(g) → 2NOCl(g), ΔH⦵ = −77.2 kJ mol⁻¹. Use S⦵ values NO(g) 210.8, Cl₂(g) 223.1 and NOCl(g) 261.7 J K⁻¹ mol⁻¹.

ΔS⦵ = (2 × 261.7) − ((2 × 210.8) + 223.1) = 523.4 − 644.7 = −121.3 J K⁻¹ mol⁻¹
ΔG⦵ at 298 K = −77.2 − (298 × −0.1213) = −77.2 + 36.15 = −41.1 kJ mol⁻¹
T = −77.2 ÷ −0.1213 = 636 K

Both ΔH and ΔS are negative, so forming NOCl is spontaneous below 636 K. At 700 K, ΔG = −77.2 + 84.91 = +7.71 kJ mol⁻¹, so above the threshold it is the decomposition of NOCl that is spontaneous.

Now view the same result through the surroundings at 298 K. ΔS(surroundings) = +77 200 ÷ 298 = +259.1 J K⁻¹ mol⁻¹. Add the system’s −121.3 to get a total of +137.8 J K⁻¹ mol⁻¹. Then −298 × 0.1378 = −41.1 kJ mol⁻¹, matching ΔG⦵.

Reactivity 1.4.4 Gibbs energy and equilibrium

As a reaction runs, reactants are consumed and products accumulate. ΔG, which drives the forward change, gets less negative, and at equilibrium it reaches zero. From then on there is no net change either way.

ΔG for any mixture depends on its composition through the reaction quotient, Q:

ΔG = ΔG⦵ + RT lnQ

With R = 8.31 J K⁻¹ mol⁻¹, this equation is in J mol⁻¹, so convert ΔG⦵ from kJ before substituting. At equilibrium, ΔG = 0 and Q = K, so:

ΔG⦵ = −RT lnK        so        K = e^(−ΔG⦵/RT)

ΔG⦵	K	Equilibrium mixture
negative	greater than 1	mainly products
zero	1	comparable amounts
positive	less than 1	mainly reactants

Worked example 7: K from ΔG⦵ at two temperatures

Use the NOCl results from worked example 6, keeping the unrounded ΔG⦵ = −41 053 J mol⁻¹ at 298 K.

lnK = −ΔG⦵ ÷ RT = 41 053 ÷ (8.31 × 298) = 16.58
K = e^16.58 = 1.6 × 10⁷

At 700 K, ΔG⦵ = +7710 J mol⁻¹:

lnK = −7710 ÷ (8.31 × 700) = −1.325
K = e^−1.325 = 0.266

At 298 K the equilibrium lies far towards NOCl. Above 636 K, where ΔG⦵ turns positive, K falls below 1 and reactants dominate.

Worked example 8: which way will a mixture shift?

For an isomerisation A(aq) ⇌ B(aq) at 298 K, ΔG⦵ = −5.00 kJ mol⁻¹. A mixture contains [A] = 0.0200 mol dm⁻³ and [B] = 0.300 mol dm⁻³.

Q = [B] ÷ [A] = 0.300 ÷ 0.0200 = 15.0
RT lnQ = 8.31 × 298 × ln 15.0 = +6706 J mol⁻¹
ΔG = −5000 + 6706 = +1706 J mol⁻¹ = +1.71 kJ mol⁻¹

Because ΔG is positive, the forward change is not spontaneous for this mixture, and the reverse reaction runs until ΔG = 0. Cross-check: K = e^(5000 ÷ (8.31 × 298)) = 7.53. Q (15.0) exceeds K, so the mixture must move back towards A, which agrees.

Common errors

• Putting ΔS in J K⁻¹ mol⁻¹ straight into ΔH − TΔS, which inflates TΔS by a factor of 1000.
• Substituting a Celsius temperature where kelvin is needed.
• Forgetting coefficients when adding S⦵ values, or giving an element an S⦵ of zero.
• Describing a reaction with negative ΔG as fast.
• Giving T = ΔH/ΔS without saying whether the reaction is spontaneous above or below it.
• Keeping ΔG⦵ in kJ inside lnK = −ΔG⦵/RT, which makes K look close to 1 every time.
• Treating ΔG and ΔG⦵ as the same: ΔG⦵ is fixed at a given temperature, while ΔG shifts as the mixture changes.

Next steps

Condense the unit with the revision notes, then try the practice questions. Plan the rest of your revision from the course hub.

Official syllabus

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