Reactivity 1.4 Entropy and Spontaneity

IB DP CHEMISTRY Reactivity 1: What Drives Chemical Reactions? Reactivity 1.4 Entropy and Spontaneity

Revision Notes · Higher Level Fahad H. Ahmad

+92 323 509 4443 | Megalecture.com

Original notes prepared for the IB Diploma Programme Chemistry course (first assessment 2025)

What the syllabus requires

Reactivity 1.4 asks a single, deep question: why does one reaction happen on its own while another, though it releases energy, refuses to go? The answer combines the enthalpy change you met in R1.2 with a new quantity, entropy, through the Gibbs energy equation. Almost all of this material is examined at Higher Level.

Understanding You should be able to...

Entropy, S Describe entropy as a measure of the dispersal of energy and matter, and predict the sign of ΔS for a physical or chemical change.

Absolute entropies (HL) Use standard entropy values to calculate ΔS° for a reaction as Σ S°(products) − Σ S°(reactants).

Gibbs energy (HL) Apply ΔG = ΔH − TΔS to decide whether a reaction is spontaneous.

Temperature dependence (HL) Determine how spontaneity depends on temperature for each combination of signs of ΔH and ΔS, and find the crossover temperature.

Gibbs energy and K (HL) Relate ΔG° to the equilibrium constant through ΔG° = −RT ln K.

Exam note: The term 'spontaneous' and the whole Gibbs-energy treatment (sections 3 onward) are HL only. SL students still need the idea of entropy and predicting the sign of ΔS. Every HL section below is flagged.

1. Entropy: the dispersal of energy and matter

Entropy, symbol S, measures how spread out the energy and matter of a system are among the ways they can be arranged. A system with many accessible arrangements (microstates) has high entropy; one with few has low entropy. Nature tends towards states of higher total entropy simply because they are overwhelmingly more probable. The units of entropy are J K−1 mol−1.

The physical picture

Think of energy quanta being shared among particles. The more particles there are, the more freely they move, and the more volume they occupy, the more ways there are to distribute that energy — so the higher the entropy. This is why the state of matter matters so much:

• Solid: particles locked in a lattice, few arrangements → lowest entropy.
• Liquid: particles mobile but still close together → intermediate entropy.
• Gas: particles far apart and fast-moving → much the highest entropy.

So for a given substance S(gas) > S(liquid) > S(solid), and the jump in entropy on boiling is far larger than the jump on melting, because the gaseous state opens up an enormous number of new arrangements.

Predicting the sign of ΔS

You are not expected to guess a numerical value, but you must judge whether entropy rises (ΔS > 0) or falls (ΔS < 0). Work through this checklist, in order of importance — the number of moles of gas almost always decides the answer:

Look for... Effect on ΔS Why

An increase in the moles of gas ΔS positive Gas dominates entropy; more gas particles means far more ways to disperse energy.

A decrease in the moles of gas ΔS negative Gas is consumed, removing disorder.

Change of state s → l → g ΔS positive Particles gain freedom of movement.

A solid dissolving in water usually ΔS positive Ordered lattice breaks up and ions spread through the solution.

More moles of product than reactant (same states)

ΔS positive More particles over which to share the energy.

Worked example 1 − predicting the sign of ΔS

Predict, with a reason, the sign of ΔS for each change.

• 2H2(g) + O2(g) → 2H2O(l)

3 mol of gas become 2 mol of liquid: gas moles fall sharply, so ΔS is negative.

• CaCO3(s) → CaO(s) + CO2(g)

A gas is produced from a solid: gas moles rise from 0 to 1, so ΔS is positive.

• N2(g) + 3H2(g) → 2NH3(g)

4 mol of gas become 2 mol of gas: gas moles fall, so ΔS is negative.

• NH4Cl(s) → NH3(g) + HCl(g)

1 mol solid becomes 2 mol gas: a large increase in disorder, so ΔS is positive.

2. Absolute entropies and ΔS° (HL)

Unlike enthalpy, entropy has a natural zero: the third law of thermodynamics states that a perfect crystal at absolute zero (0 K) has S = 0. Because of this we can assign every substance an absolute standard entropy, S°, measured at 298 K and 100 kPa. These are always positive (even for elements) and are tabulated in the data booklet in J K−1 mol−1.

Some representative values (J K−1 mol−1):

Substance S° Substance S°

H2(g) 130.6 H2O(l) 69.9

O2(g) 205.0 H2O(g) 188.7

N2(g) 191.6 CO2(g) 213.7

NH3(g) 192.5 CaCO3(s) 92.9

SO2(g) 248.2 CaO(s) 39.7

Substance S° Substance S°

SO3(g) 256.8 C(graphite) 5.7

Notice the trend: gases (large values) far exceed liquids and solids; the hard, highly ordered graphite lattice sits right at the bottom at only 5.7 J K−1 mol−1.

Calculating the standard entropy change

For any reaction the standard entropy change is the total entropy of the products minus that of the reactants, each multiplied by its stoichiometric coefficient:

ΔS° = Σ S°(products) − Σ S°(reactants) where Σ means 'the sum over each species, weighted by its coefficient'. The answer comes out in J K−1 mol−1 — keep this unit firmly in mind, because it will clash with the kJ of enthalpy in the next section.

Worked example 2 − calculating ΔS°

Calculate ΔS° for the thermal decomposition of calcium carbonate:

CaCO3(s) → CaO(s) + CO2(g)

ΔS° = [S°(CaO) + S°(CO2)] − S°(CaCO3)

= (39.7 + 213.7) − 92.9 = +160.5 J K−1 mol−1

The positive sign is exactly what we predicted in section 1: a gas is generated from a solid, so the system becomes more disordered.

Worked example 3 − a reaction that loses entropy

Calculate ΔS° for the Haber process, N2(g) + 3H2(g) → 2NH3(g).

ΔS° = 2 × S°(NH3) − [S°(N2) + 3 × S°(H2)]

= (2 × 192.5) − (191.6 + 3 × 130.6)

= 385.0 − 583.4 = −198.4 J K−1 mol−1

Negative, as expected: 4 mol of gas collapse into 2 mol of gas.

Worked example 4 − entropy of vaporisation

Calculate ΔS° when one mole of water evaporates: H2O(l) → H2O(g).

ΔS° = S°(H2O, g) − S°(H2O, l) = 188.7 − 69.9 = +118.8 J K−1 mol−1

A large positive value: going from the liquid to the gas hugely increases the freedom of the molecules. Every simple s→l→g change follows the same pattern.

Figure. Standard molar entropies for CaCO3(s) → CaO(s) + CO2(g). A gas is produced, so ΔS° = Σ S°(products) − Σ

S°(reactants) = +160.5 J K−1 mol−1 (positive).

3. Spontaneity and Gibbs energy (HL)

A spontaneous reaction is one that, once started, proceeds without any continuous outside help — it is thermodynamically 'downhill'. Whether a reaction is spontaneous depends on both the enthalpy change and the entropy change, combined into a single quantity, the Gibbs energy change:

ΔG = ΔH − TΔS where T is the absolute temperature in kelvin. Under standard conditions this is written ΔG° = ΔH° − TΔS°. The sign of ΔG is the single test for spontaneity:

Sign of ΔG Meaning Reaction...

ΔG < 0 (negative) products lower in Gibbs energy is spontaneous (feasible) in the forward direction

ΔG = 0 no net driving force is at equilibrium

ΔG > 0 (positive) products higher in Gibbs energy is non-spontaneous forward; the reverse is spontaneous

Critical unit warning: ΔH is quoted in kJ mol−1 but ΔS in J K−1 mol−1. Before subtracting, convert ΔS to kJ by dividing by 1000 (or convert ΔH to J). Forgetting this is the single most common mistake in this whole topic.

Two opposing tendencies

Read the equation as a competition. The ΔH term rewards reactions that release energy (exothermic, ΔH < 0). The −TΔS term rewards reactions that increase disorder (ΔS > 0), and its influence grows with temperature. Sometimes they pull the same way; sometimes they conflict, and then temperature decides the winner — the subject of the next section.

Worked example 5 − deciding spontaneity

The dissolving of ammonium nitrate, NH4NO3(s) → NH4

+(aq) + NO3

−(aq), has ΔH° = +25.7 kJ mol−1 and ΔS° = +108.7 J K−1 mol−1. Is it spontaneous at 298 K?

ΔG° = ΔH° − TΔS° = 25.7 − 298(0.1087)

= 25.7 − 32.4 = −6.7 kJ mol−1

ΔG° is negative, so dissolving is spontaneous even though it is endothermic — the cold pack still works. The favourable entropy of mixing outweighs the unfavourable enthalpy.

4. The four cases and the crossover temperature (HL)

Because ΔG = ΔH − TΔS, the way spontaneity depends on temperature is fixed entirely by the signs of ΔH and ΔS. There are four possible combinations.

ΔH ΔS −TΔS Spontaneous when? Example

− (exo) + always negative at all temperatures (ΔG always < 0) C(s) + O2(g) → CO2(g)

+ (endo) − always positive at no temperature (ΔG always > 0) 3O2(g) → 2O3(g)

− (exo) − positive only at low temperature N2 + 3H2 → 2NH3

+ (endo) + negative only at high temperature CaCO3(s) → CaO(s) + CO2(g)

Figure. The four sign combinations of ΔH and ΔS. Where enthalpy and entropy agree the outcome is fixed (always / never);

where they conflict, temperature decides via ΔG = ΔH − TΔS.

The first two cases are decided outright, with enthalpy and entropy agreeing. The last two are the interesting ones: enthalpy and entropy disagree, so there is a particular temperature at which the reaction switches between spontaneous and non-spontaneous.

Figure. ΔG = ΔH − TΔS for CaCO3 decomposition: a straight line of slope −ΔS and intercept ΔH. ΔG = 0 at the crossover T

= ΔH/ΔS ≈ 1109 K; above it decomposition is spontaneous.

Finding the crossover temperature

At the switchover point ΔG = 0, so setting ΔH − TΔS = 0 gives the crossover temperature:

T = ΔH / ΔS

Use consistent units (both in J, or both in kJ). For an endothermic reaction with ΔS > 0 (like limestone decomposing), the reaction becomes spontaneous above this temperature. For an exothermic reaction with ΔS < 0 (like the Haber process), it is spontaneous only below it.

Worked example 6 − crossover temperature for limestone

For CaCO3(s) → CaO(s) + CO2(g), ΔH° = +178 kJ mol−1 and ΔS° = +160.5 J K−1 mol−1. Above what temperature does decomposition become spontaneous?

Set ΔG° = 0: T = ΔH° / ΔS°

Convert ΔS° to kJ: 160.5 / 1000 = 0.1605 kJ mol−1 K−1.

T = 178 / 0.1605 = 1109 K (about 836 °C).

Below 1109 K, ΔG° > 0 and limestone is stable; above it, ΔG° < 0 and it decomposes. This is why a lime kiln must be run red-hot.

Worked example 7 − crossover temperature for the Haber process

For N2(g) + 3H2(g) → 2NH3(g), ΔH° = −92.2 kJ mol−1 and ΔS° = −198.4 J K−1 mol−1. Below what temperature is the forward reaction spontaneous?

T = ΔH° / ΔS° = (−92.2) / (−0.1984) = 465 K (about 192 °C).

Both terms are negative, so this is a 'low-temperature' case: ΔG° < 0 (spontaneous) below 465 K, and ΔG° > 0 (non-spontaneous) above it. Industry still runs the reaction near 450 °C for an acceptable rate, accepting a lower equilibrium yield.

5. Calculating ΔG° (HL)

There are two standard routes to ΔG°. Choose whichever matches the data you are given.

Route A: from ΔH° and ΔS°

Substitute directly into ΔG° = ΔH° − TΔS°, remembering to convert ΔS° into kJ and to use T in kelvin. Unless told otherwise, take T = 298 K.

Worked example 8 − ΔG° from ΔH° and ΔS°

For the Haber process ΔH° = −92.2 kJ mol−1 and ΔS° = −198.4 J K−1 mol−1. Find ΔG° at 298 K and comment.

ΔG° = ΔH° − TΔS°

= −92.2 − (298 × (−198.4/1000))

= −92.2 − (−59.1) = −92.2 + 59.1 = −33.1 kJ mol−1

ΔG° is negative, so ammonia synthesis is spontaneous at 298 K. (It is nonetheless run hot to gain an acceptable rate — see section 6.)

Route B: from standard Gibbs energies of formation

Just as with enthalpy, we can tabulate the standard Gibbs energy of formation ΔGf

° of each compound (zero for elements in their standard states). Then, by Hess's law,

ΔG° = Σ ΔGf

°(products) − Σ ΔGf

°(reactants)

Worked example 9 − ΔG° from ΔGf

° values

Find ΔG° for the complete combustion of methane at 298 K:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

Data (kJ mol−1): ΔGf

°[CH4] = −50.8; [CO2] = −394.4; [H2O(l)] = −237.1; O2 = 0.

ΔG° = [(−394.4) + 2(−237.1)] − [(−50.8) + 0]

= −868.6 − (−50.8) = −817.8 kJ mol−1

Strongly negative, confirming that methane combustion is highly spontaneous.

6. Gibbs energy, equilibrium and rate (HL)

Spontaneity is about the position a reaction settles into, not how fast it gets there. Two further ideas complete the picture.

Linking ΔG° to the equilibrium constant

The standard Gibbs energy change is quantitatively tied to the equilibrium constant K by

ΔG° = −RT ln K where R = 8.31 J K−1 mol−1. The relationship tells us at a glance which side of an equilibrium is favoured:

ΔG° ln K K Position of equilibrium negative positive K > 1 products favoured zero K = 1 neither side favoured positive negative K < 1 reactants favoured

Worked example 10 − from ΔG° to K

For the Haber process ΔG° = −33.1 kJ mol−1 at 298 K. Estimate K.

Rearrange: ln K = −ΔG° / (RT). Put ΔG° in joules: −33100 J mol−1.

ln K = −(−33100) / (8.31 × 298) = 33100 / 2476 = 13.4

K = e13.4 ≈ 6 × 105.

A large K: at equilibrium the mixture is rich in ammonia, consistent with the negative ΔG°.

Spontaneous does not mean fast

This is the distinction examiners test most often. Gibbs energy is a thermodynamic quantity: a negative ΔG° tells you a reaction can happen and how far it will go, but says nothing about the rate, which is a kinetic question governed by activation energy (R2.2). A reaction can be strongly spontaneous yet immeasurably slow.

Classic illustration: The conversion of diamond to graphite has ΔG° ≈ −2.9 kJ mol−1 at room temperature, so it is spontaneous — yet diamonds do not visibly change, because the activation energy is enormous and the rate is effectively zero. 'Spontaneous' means feasible, not quick.

7. Bringing it all together

The Contact process step 2SO2(g) + O2(g) → 2SO3(g) lets us use every tool in this topic on a single reaction. Given ΔH° = −197.8 kJ mol−1.

Worked example 11 − a full thermodynamic analysis

• Predict the sign of ΔS°: 3 mol of gas become 2 mol of gas, so gas moles fall and ΔS° should be negative.
• Calculate ΔS°: = 2(256.8) − [2(248.2) + 205.0] = 513.6 − 701.4 = −187.8 J K−1 mol−1. The sign confirms the prediction.

(iii) Calculate ΔG° at 298 K: = −197.8 − 298(−0.1878) = −197.8 + 56.0 = −141.8 kJ mol−1 → spontaneous.

• Crossover temperature: T = ΔH°/ΔS° = (−197800)/(−187.8) = 1053 K. Being an exo/− case, the reaction is spontaneous below 1053 K.
• Estimate K at 298 K: ln K = 141800/(8.31 × 298) = 57.3, so K = e57.3 ≈ 8 × 1024 — essentially complete. In practice the plant runs near 450 °C (723 K), still well below the crossover, sacrificing a little yield to gain a workable rate.

8. Common pitfalls

• Unit mismatch: subtracting ΔS in J directly from ΔH in kJ. Always convert ΔS to kJ (divide by 1000) first.
• Temperature in °C: the T in ΔG = ΔH − TΔS must be in kelvin. Add 273 to any Celsius value.
• Confusing feasibility with speed: ΔG < 0 means the reaction is possible, not that it is observably occurring.
• Sign slips: a double negative in the − TΔS term (when ΔS is negative) becomes a plus − track it carefully.
• Forgetting states: the sign of ΔS hinges on gas moles, so you must include state symbols before judging it.
• Treating ΔH° and ΔS° as temperature-independent is an approximation the IB accepts, but ΔG° itself varies strongly with T.

9. Quick reference

Idea Statement

Entropy S = dispersal of energy and matter; units J K−1 mol−1; S(g) > S(l) > S(s)

Sign of ΔS positive if gas moles increase, a solid dissolves, or the particle count rises

Entropy change ΔS° = Σ S°(products) − Σ S°(reactants)

Gibbs energy ΔG = ΔH − TΔS (ΔH in kJ, ΔS in kJ, T in K)

Spontaneity test ΔG < 0 spontaneous; = 0 equilibrium; > 0 non-spontaneous

Crossover T T = ΔH / ΔS (the temperature where ΔG = 0)

Gibbs from formation ΔG° = Σ ΔGf

°(products) − Σ ΔGf

°(reactants)

Gibbs and K ΔG° = −RT ln K; ΔG° < 0 → K > 1

10. Test yourself

Attempt these before checking the fully worked answers below. Assume T = 298 K and use the entropy data of section 2 where needed.

• Predict the sign of ΔS for: (a) 2SO2(g) + O2(g) → 2SO3(g); (b) NaCl(s) → Na+(aq) + Cl−(aq); (c) H2O(l) → H2O(s).
• Calculate ΔS° for 2H2(g) + O2(g) → 2H2O(l) and comment on its sign.
• A reaction has ΔH° = +178 kJ mol−1 and ΔS° = +160.5 J K−1 mol−1. Find ΔG° at (a) 298 K and (b) 1500 K, and state whether each is spontaneous.
• For an exothermic reaction with ΔH° = −92.2 kJ mol−1 and ΔS° = −198.4 J K−1 mol−1, find the temperature above which it becomes non-spontaneous.
• Using ΔGf °[N2O4(g)] = +97.9 and [NO2(g)] = +51.3 kJ mol−1, find ΔG° for N2O4(g) → 2NO2(g). Is it spontaneous at 298 K?
• For the reaction in Q5, calculate K at 298 K using ΔG° = −RT ln K.
• Explain why the decomposition of limestone is not carried out at room temperature even though it produces useful quicklime.
• A student claims that because diamond → graphite has a negative ΔG°, a diamond ring will turn grey within a year. Comment on the chemistry of this claim.

Worked answers

• (a) Negative — gas moles fall from 3 to 2. (b) Positive — an ordered ionic lattice disperses into mobile hydrated ions. (c) Negative — a liquid becomes a more ordered solid.
• ΔS° = 2(69.9) − [2(130.6) + 205.0] = 139.8 − 466.2 = −326.4 J K−1 mol−1. Strongly negative, because 3 mol of gas condense into 2 mol of liquid.
• Convert ΔS° = 0.1605 kJ mol−1 K−1. (a) ΔG° = 178 − 298(0.1605) = 178 − 47.8 = +130.2 kJ mol−1, non-spontaneous. (b) ΔG° = 178 − 1500(0.1605) = 178 − 240.8 = −62.8 kJ mol−1, spontaneous. The reaction turns spontaneous only at high temperature.
• At the crossover ΔG° = 0, so T = ΔH° / ΔS° = (−92.2) / (−0.1984) = 465 K (192 °C). Since both terms are negative the reaction is spontaneous below 465 K and non-spontaneous above it.
• ΔG° = 2(+51.3) − (+97.9) = 102.6 − 97.9 = +4.7 kJ mol−1. Positive, so the forward reaction is not spontaneous under standard conditions at 298 K (though it lies close to equilibrium).
• ln K = −ΔG°/(RT) = −4700 / (8.31 × 298) = −4700 / 2476 = −1.90. So K = e−1.90 = 0.15. As ΔG° > 0, K < 1 and reactants (N2O4) are favoured, consistent with Q5.
• Decomposition is endothermic with ΔS° > 0, so ΔG° is positive (non-spontaneous) at room temperature and only becomes negative above about 1100 K (worked example 6). The kiln must be heated strongly to drive the reaction.
• The claim confuses thermodynamics with kinetics. A negative ΔG° means the change is feasible, not that it is fast. The activation energy for rearranging the diamond lattice is enormous, so the rate at room temperature is effectively zero and the ring is safe. 'Spontaneous' does not mean observable within any human timescale.