This whole subtopic is higher level. Nothing in it is on an SL paper.
Educerie · IB Diploma · Chemistry
Reactivity 1 What drives chemical reactions? · R1.4 Entropy and spontaneity
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Explain entropy as the dispersal of matter and energy, and rank solids, liquids and gases | HL only | "Explain why the entropy of a gas is greater than that of a liquid" (2 marks) |
| Predict whether a physical or chemical change increases or decreases the entropy of the system | HL only | "Predict, giving a reason, the sign of ΔS for …" (1–2 marks) |
| Calculate ΔS⦵ from standard entropy values | HL only | Paper 2: "Calculate ΔS⦵ using section … of the data booklet" (2 marks) |
| Use ΔG⦵ = ΔH⦵ − TΔS⦵ to find any unknown term, with the units right | HL only | "Calculate ΔG⦵ at 298 K" (2 marks) |
| Interpret the sign of ΔG and say whether a change is spontaneous | HL only | "Deduce whether the reaction is spontaneous at 298 K" (1 mark) |
| Determine the temperature at which a reaction becomes spontaneous | HL only | "Determine the minimum temperature at which … is spontaneous" (2 marks) |
| Explain why ΔG becomes less negative and reaches zero at equilibrium | HL only | "Outline what happens to ΔG as the reaction approaches equilibrium" (2 marks) |
| Calculate with ΔG = ΔG⦵ + RT lnQ and ΔG⦵ = −RT lnK | HL only | "Calculate K at 298 K" or "Determine the direction of change" (2–3 marks) |
Before you start
You need ΔH and its sign from R1.1, and ΔHf⦵ calculations from R1.2, because almost every ΔG⦵ question begins with one. You need the particle picture of solids, liquids and gases (Structure 1.1) and absolute temperature in kelvin. For the last section you need the reaction quotient Q and the equilibrium constant K from Reactivity 2.3; if you have not met them yet, read sections 1 to 7 now and come back to section 8 later.
1The idea in one paragraph
Some changes happen by themselves and some do not, and the enthalpy change alone cannot tell you which, because some endothermic changes happen by themselves too. The missing piece is entropy, a measure of how spread out matter and energy are. Left alone, the universe moves towards more spread-out arrangements. A reaction changes entropy in two ways: directly, by changing the chemicals themselves (a solid turning into a gas spreads out enormously), and indirectly, by giving heat to the surroundings or taking it from them. Gibbs energy puts both into one number, ΔG = ΔH − TΔS. If ΔG is negative, the change is spontaneous. Because temperature multiplies the entropy term, heating can turn a reaction on or off, and the temperature at which it switches is ΔH ÷ ΔS. Finally, ΔG is not fixed during a reaction: it gets less negative as products build up, and reaches zero at equilibrium, which ties Gibbs energy to K.
2What "spontaneous" means, and why ΔH is not enough
A spontaneous change is one that, once started, will carry on by itself without continuous outside help: a hot drink cooling, iron rusting, ice melting on a warm day, methane burning. The reverse of each is non-spontaneous: it can be forced, but only by continually supplying energy.
Two warnings before anything else.
Spontaneous does not mean fast. Diamond turning into graphite is spontaneous at room temperature. It is also so slow that nobody has ever seen it happen. Methane and oxygen react spontaneously, yet a gas tap can stay open all day without a fire until a spark supplies the activation energy. Thermodynamics tells you whether a change can happen; kinetics (Reactivity 2.2) tells you how fast. They are separate questions.
Exothermic is not the rule. Most spontaneous reactions are exothermic, which is why it is tempting to think that "downhill in enthalpy" decides direction. But water evaporates from a puddle at 20 °C, ice melts at 5 °C, and ammonium nitrate dissolves in water so endothermically that the beaker turns icy cold, and all three are spontaneous. Something else is pushing them.
3Entropy
Entropy, S, is a measure of the dispersal or distribution of matter and energy in a system. The more ways the particles can be arranged and the energy shared out among them, the higher the entropy.
Look at the three states in Figure 1.
In a solid, each particle sits in a fixed place and can only vibrate, so there are few ways to arrange the particles and few ways to share out their energy. In a liquid the particles move past each other, so there are many more arrangements. In a gas the particles move freely through the whole container, far apart, with a wide range of speeds, and the number of possible arrangements is vastly larger again. Under the same conditions:
S(gas) ≫ S(liquid) > S(solid)
The jump from liquid to gas is much bigger than the jump from solid to liquid, because a liquid is still compact and a gas is not. Water shows it: S⦵ of H₂O(l) is 70.0 J K⁻¹ mol⁻¹, while S⦵ of H₂O(g) is 188.8 J K⁻¹ mol⁻¹.
Entropy also rises with temperature, because hotter particles have more energy to share out in more ways. Figure 2 sketches the whole story for one substance from absolute zero upwards.
At 0 K a perfect crystal has every particle in its exact place and no thermal energy to distribute: there is only one possible arrangement, so its entropy is predicted to be zero. That gives entropy an absolute starting point, which enthalpy does not have. Two consequences follow:
- Standard entropy values, S⦵, are absolute and always positive for any substance above 0 K, including elements. O₂(g) has S⦵ = 205.2 J K⁻¹ mol⁻¹, not zero. Do not carry over the "elements are zero" rule from ΔHf⦵.
- The unit is J K⁻¹ mol⁻¹: joules, not kilojoules. This one detail causes more lost marks than anything else in the subtopic.
4Predicting the sign of ΔS
For a change, ΔS = S(final) − S(initial). A positive ΔS means the system has become more dispersed. You can predict the sign without any data by asking what happens to the gas, because gases dominate entropy.
| What happens | Sign of ΔS for the system | Why |
|---|---|---|
| Amount of gas increases | positive | gas particles have far more arrangements |
| Amount of gas decreases | negative | |
| Solid → liquid, or liquid → gas | positive | more freedom of movement |
| Gas → liquid, or liquid → solid | negative | |
| A solid or liquid dissolves | usually positive | ions or molecules spread through the solvent |
| More particles form (with no gas change) | usually positive | more particles, more ways to arrange them |
Worked example 1: predictions. Count moles of gas on each side first.
| Change | Moles of gas: before → after | ΔS |
|---|---|---|
| CaCO₃(s) → CaO(s) + CO₂(g) | 0 → 1 | positive |
| N₂(g) + 3H₂(g) → 2NH₃(g) | 4 → 2 | negative |
| H₂O(l) → H₂O(g) | 0 → 1 | positive |
| NH₄NO₃(s) → NH₄⁺(aq) + NO₃⁻(aq) | 0 → 0; solid dissolves | positive |
| 2Mg(s) + O₂(g) → 2MgO(s) | 1 → 0 | negative |
When the moles of gas are equal on both sides, as in H₂(g) + Cl₂(g) → 2HCl(g), the change in entropy is small and you cannot confidently predict its sign without data. Say so; that answer earns the mark.
Now go back to the three puzzles in section 2. Evaporation, melting and dissolving all have a positive ΔS. That is what drives them uphill in enthalpy.
5Calculating ΔS⦵ from standard entropies
Because S⦵ values are absolute, the calculation is a straightforward "products minus reactants", with every value multiplied by its coefficient:
ΔS⦵ = ΣS⦵(products) − ΣS⦵(reactants)
Values used on this page, in J K⁻¹ mol⁻¹: N₂(g) 191.6, H₂(g) 130.7, NH₃(g) 192.8, CaCO₃(s) 92.9, CaO(s) 39.7, CO₂(g) 213.8, CH₄(g) 186.3, O₂(g) 205.2, H₂O(l) 70.0. In an exam use the values in the data booklet, which may differ in the last figure.
Worked example 2: the Haber process. N₂(g) + 3H₂(g) → 2NH₃(g).
Negative, as predicted: four moles of gas become two.
Worked example 3: decomposing limestone. CaCO₃(s) → CaO(s) + CO₂(g).
Positive, as predicted: a gas appears from a solid. Almost all of the increase comes from the CO₂.
6Gibbs energy: both entropy changes in one number
The second law of thermodynamics says that a change is spontaneous if it increases the total entropy of the universe: the system and its surroundings. A chemist measures the system easily. The surroundings seem harder, until you notice that the only thing a reaction does to them is give them heat or take it away. Figure 3 shows the two parts.
An exothermic reaction releases heat (−ΔH) into the surroundings, which disperses that energy among their particles and raises their entropy. The same amount of heat matters more when the surroundings are cold than when they are already hot, so the entropy change of the surroundings is −ΔH divided by T. The total is ΔS − ΔH/T. Multiply through by −T and you have the quantity chemists actually use:
ΔG⦵ = ΔH⦵ − TΔS⦵
Gibbs energy change, ΔG, is −T times the total entropy change. A total entropy increase, the condition for a spontaneous change, therefore means a negative ΔG. This is exactly the guide's point: ΔG accounts for the direct entropy change of the chemicals (ΔS) and the indirect entropy change of the surroundings caused by the transfer of heat (through ΔH). You do not need to reproduce the derivation, but it is the reason the equation has the shape it does.
ΔG also has a practical meaning. It relates the energy that can be obtained from a reaction to ΔH and ΔS: a reaction with ΔG = −100 kJ mol⁻¹ can, at most, supply 100 kJ of useful work per mole, such as electrical work in a cell. Not all of ΔH is available, because some of it must go to the surroundings as heat to pay for any entropy the system loses. Reactivity 3.2 uses this directly: ΔG⦵ = −nFE⦵ links Gibbs energy to the voltage of a cell.
The units. ΔH is in kJ mol⁻¹, ΔS is in J K⁻¹ mol⁻¹, and ΔG is in kJ mol⁻¹. Convert ΔS to kJ by dividing by 1000 before you substitute. T is always in kelvin.
Worked example 4: is ammonia formation spontaneous at 298 K? ΔHf⦵ of NH₃(g) is −46.1 kJ mol⁻¹, so for N₂ + 3H₂ → 2NH₃, ΔH⦵ = 2(−46.1) = −92.2 kJ mol⁻¹. From worked example 2, ΔS⦵ = −198.1 J K⁻¹ mol⁻¹.
Read it term by term. The enthalpy term (−92.2) favours the reaction. The entropy term (+59.0) opposes it, because gas is being lost. At 298 K the enthalpy term wins.
7The sign of ΔG, and the temperature that flips it
Once you know the signs of ΔH and ΔS, you know how ΔG behaves as the temperature changes. Plot ΔG = ΔH − TΔS against T and it is a straight line: it starts at ΔH when T = 0 and has a slope of −ΔS. Figure 4 shows all four possibilities.
| ΔH | ΔS | ΔG = ΔH − TΔS | Spontaneous? |
|---|---|---|---|
| negative | positive | always negative | at every temperature |
| positive | negative | always positive | never |
| negative | negative | negative at low T, positive at high T | only below T = ΔH ÷ ΔS |
| positive | positive | positive at low T, negative at high T | only above T = ΔH ÷ ΔS |
In the last two rows, the change switches at the temperature where ΔG = 0:
ΔG = 0 when T = ΔH ÷ ΔS (with ΔS in kJ K⁻¹ mol⁻¹)
Worked example 5: when does limestone decompose? From ΔHf⦵ values (CaCO₃ −1206.9, CaO −635.1, CO₂ −393.5 kJ mol⁻¹), ΔH⦵ = (−635.1 − 393.5) − (−1206.9) = +178.3 kJ mol⁻¹. From worked example 3, ΔS⦵ = +160.6 J K⁻¹ mol⁻¹. Both positive, so the reaction is spontaneous only above some temperature.
Figure 5 draws the line. Below about 1110 K, ΔG⦵ is positive and limestone is stable, which is why cliffs do not fizz. Above it, ΔG⦵ is negative, which is why a lime kiln runs at around 900 °C or more.
A check that the method works. For ice melting, ΔH = +6.01 kJ mol⁻¹ and ΔS = +22.0 J K⁻¹ mol⁻¹. T = 6.01 ÷ 0.0220 = 273 K. That is the melting point of ice, exactly where solid and liquid are in balance and neither change is favoured. Below 273 K freezing is spontaneous; above it, melting is.
The same idea for ammonia. ΔH and ΔS are both negative, so the reaction is spontaneous only below T = −92.2 ÷ −0.1981 ≈ 465 K. An industrial plant runs hotter than this, around 700 K, because at room temperature the reaction is far too slow. Using the same values, ΔG⦵ at 700 K is about +46 kJ mol⁻¹, so the equilibrium lies on the reactants' side and only a small fraction of the gases is converted on each pass. The plant accepts that for the sake of speed, and uses high pressure to push the equilibrium back towards ammonia (Reactivity 2.3).
Two assumptions sit behind every one of these temperatures, and a Paper 2 "suggest" question may ask for them: that ΔH and ΔS do not change with temperature, and that all substances stay in the states written in the equation. Neither is exactly true, so a calculated turning temperature is an estimate.
8ΔG during a reaction, and at equilibrium
ΔG⦵ is the Gibbs energy change under standard conditions: every substance at standard concentration or pressure. A real reaction mixture is rarely standard, and its composition changes as it reacts. The Gibbs energy change for the mixture as it actually is at any moment is
ΔG = ΔG⦵ + RT lnQ
where Q is the reaction quotient: the same expression as the equilibrium constant, but filled with the concentrations or pressures present now. R is the gas constant, 8.31 J K⁻¹ mol⁻¹, so RT lnQ comes out in joules: convert it to kJ, or ΔG⦵ to J, before you add.
Follow a reaction that starts with pure reactants. Q is tiny, so lnQ is large and negative, and ΔG is strongly negative: the forward reaction runs. As products build up, Q rises, lnQ becomes less negative, and ΔG becomes less negative. Eventually Q reaches K, and ΔG reaches zero. Nothing drives the reaction either way any more: that is equilibrium. Figure 6 shows the same thing as the total Gibbs energy of the mixture, which slides downhill from either end to a minimum.
This also answers which way a mixture moves before equilibrium. If Q < K, ΔG is negative and the forward reaction is favoured. If Q > K, ΔG is positive for the forward reaction, so the backward reaction is favoured. The mixture always moves towards the bottom of the curve.
At equilibrium, ΔG = 0 and Q = K. Put both into the equation:
ΔG⦵ = −RT lnK
This is a direct bridge from thermodynamic data to the composition of an equilibrium mixture. Figure 7 shows what the sign of ΔG⦵ says about K.
- ΔG⦵ negative: lnK positive, K > 1, products favoured at equilibrium.
- ΔG⦵ zero: K = 1.
- ΔG⦵ positive: lnK negative, K < 1, the equilibrium mixture is mostly reactants. That answers the guide's linking question: a positive ΔG⦵ does not mean "no reaction", it means "an equilibrium that lies to the left".
Because K depends exponentially on ΔG⦵, small energy differences make large differences to K. At 298 K, every 5.7 kJ mol⁻¹ makes K ten times larger or smaller.
Worked example 6: K for the Haber process at 298 K. From worked example 4, ΔG⦵ = −33.2 kJ mol⁻¹ = −33 166 J mol⁻¹ (unrounded).
A huge K: at equilibrium at room temperature, the mixture would be almost all ammonia. The only thing stopping it is the rate.
Worked example 7: which way will it go? For N₂O₄(g) ⇌ 2NO₂(g) at 298 K, ΔG⦵ = +4.7 kJ mol⁻¹. A mixture is made in which Q = 0.010. RT = 8.31 × 298 = 2476 J mol⁻¹ = 2.476 kJ mol⁻¹.
ΔG⦵ is positive, yet this mixture reacts forwards, because it starts with far less NO₂ than equilibrium needs. The sign of ΔG⦵ tells you where equilibrium lies; the sign of ΔG tells you which way a particular mixture will move.
9Where marks are lost
Not converting ΔS to kJ. ΔS is in J K⁻¹ mol⁻¹ and ΔH in kJ mol⁻¹. Substituting −198.1 straight into ΔH − TΔS gives a number a thousand times too big. Divide by 1000 first, and write the line that shows it.
Using °C. Every T in this subtopic is in kelvin. 25 °C is 298 K.
Giving elements a standard entropy of zero. That is the ΔHf⦵ rule, not the S⦵ rule. O₂, H₂ and N₂ all have positive S⦵ values and all go into the sum.
Forgetting coefficients in ΔS⦵. Three moles of H₂ means 3 × 130.7.
Confusing spontaneous with fast. A negative ΔG says the reaction can happen; it says nothing about how quickly. If asked why a reaction with ΔG < 0 is not observed, the answer is a high activation energy.
Saying "exothermic, so spontaneous". An exothermic reaction with a negative ΔS becomes non-spontaneous above T = ΔH ÷ ΔS. Always consider both terms.
Confusing ΔG with ΔG⦵. ΔG⦵ is a fixed value for the reaction at a given temperature and tells you where equilibrium lies. ΔG changes as the reaction proceeds and is zero at equilibrium. "ΔG⦵ is zero at equilibrium" is wrong.
Mixing joules and kilojoules in RT lnK. R is in J K⁻¹ mol⁻¹, so ΔG⦵ must be in J mol⁻¹ in this equation.
10Draw it right
Two graphs come up: ΔG against T, and G against the extent of reaction.
- A ΔG against T graph has T in kelvin on the horizontal axis and ΔG in kJ mol⁻¹ on the vertical axis, with the ΔG = 0 line drawn clearly.
- The line is straight (assuming ΔH and ΔS are constant). Its intercept at T = 0 is ΔH and its gradient is −ΔS. So a line sloping downwards means ΔS is positive.
- Mark and label the temperature where the line crosses ΔG = 0, and state that it equals ΔH ÷ ΔS.
- Label which side of the crossing is spontaneous (ΔG < 0).
- A G against extent of reaction curve has pure reactants at one end, pure products at the other, and a single minimum in between, labelled equilibrium, where the gradient (ΔG) is zero.
- On that curve, the difference in G between pure products and pure reactants is ΔG⦵. If the products end lower, ΔG⦵ is negative and the minimum lies nearer the products.
- When reading ΔS from a gradient, convert: a gradient of −0.10 kJ K⁻¹ mol⁻¹ is ΔS = +100 J K⁻¹ mol⁻¹.
11Try it
Marks in brackets. Answers and marker's notes are at the end. Use R = 8.31 J K⁻¹ mol⁻¹ and the S⦵ values in section 5.
Q1. Predict the sign of the entropy change of the system for each change, giving a reason. 4 marks
(a) H₂O(g) → H₂O(l)
(b) 2H₂O₂(l) → 2H₂O(l) + O₂(g)
(c) KCl(s) → K⁺(aq) + Cl⁻(aq)
(d) 2SO₂(g) + O₂(g) → 2SO₃(g)
Q2. Methane burns: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH⦵ = −890.3 kJ mol⁻¹.
(a) Calculate ΔS⦵ for the reaction. 2 marks
(b) Calculate ΔG⦵ at 298 K. 2 marks
(c) A mixture of methane and air can be kept at 298 K without reacting. Explain why, given your answer to (b). 1 mark
Q3. (Data-based.) A student finds ΔG⦵ for a reaction at four temperatures.
| T / K | 300 | 500 | 700 | 900 |
|---|---|---|---|---|
| ΔG⦵ / kJ mol⁻¹ | +40 | +20 | 0 | −20 |
(a) Determine ΔH⦵ and ΔS⦵ for the reaction, with units. 3 marks
(b) State the range of temperatures over which the reaction is spontaneous. 1 mark
Q4. For a reaction at 298 K, ΔG⦵ = −20.0 kJ mol⁻¹.
(a) Calculate the equilibrium constant, K. 2 marks
(b) In a mixture at 298 K, Q = 5.0 × 10³. Calculate ΔG and deduce the direction in which the reaction will proceed. 2 marks
Q5. For MgCO₃(s) → MgO(s) + CO₂(g), ΔH⦵ = +100.7 kJ mol⁻¹ and ΔS⦵ = +175.0 J K⁻¹ mol⁻¹. Determine the lowest temperature at which the decomposition is spontaneous, and state one assumption you have made. 3 marks
12In one breath
A spontaneous change carries on by itself once started, but it need not be fast, and it need not be exothermic, because entropy also matters. Entropy is the dispersal of matter and energy: gas far above liquid above solid, rising with temperature, zero only for a perfect crystal at 0 K, so every S⦵ value, including an element's, is positive and in J K⁻¹ mol⁻¹. Predict ΔS from moles of gas first; calculate it as ΣS⦵(products) − ΣS⦵(reactants). ΔG⦵ = ΔH⦵ − TΔS⦵ combines the entropy change of the chemicals with the entropy change of the surroundings caused by heat; convert ΔS to kJ first and use kelvin. Negative ΔG means spontaneous. Negative ΔH with positive ΔS is always spontaneous, the reverse never; when the signs match, the switch is at T = ΔH ÷ ΔS, below it for two negatives and above it for two positives. As a reaction runs, ΔG = ΔG⦵ + RT lnQ gets less negative and reaches zero at equilibrium, where Q = K, so ΔG⦵ = −RT lnK: negative ΔG⦵ means K > 1, positive means K < 1, and every 5.7 kJ mol⁻¹ at 298 K is a factor of ten in K.
Answers
Q1. (a) Negative: a gas becomes a liquid, so the particles have far fewer possible arrangements. (b) Positive: a gas is produced from liquids, 0 → 1 mol of gas. (c) Positive: a solid lattice breaks up into ions dispersed through the water. (d) Negative: three moles of gas become two. 1 for each correct sign with a valid reason. A sign with no reason scores 0 for that part.
Q2. (a) ΣS⦵(products) = 213.8 + 2(70.0) = 353.8. ΣS⦵(reactants) = 186.3 + 2(205.2) = 596.7. ΔS⦵ = 353.8 − 596.7 = −242.9 J K⁻¹ mol⁻¹. (b) ΔG⦵ = −890.3 − 298(−0.2429) = −890.3 + 72.4 = −817.9 kJ mol⁻¹. (c) The reaction is spontaneous, but it has a high activation energy, so at 298 K the rate is negligible until a spark or flame supplies the energy to start it. M1 for both sums with coefficients, A1 for −242.9 J K⁻¹ mol⁻¹; M1 for converting ΔS to kJ and substituting T = 298, A1 for −818 kJ mol⁻¹; 1 for high activation energy in (c). Using −242.9 without conversion gives about +71 500 and scores M0 A0 in (b); ECF from a wrong (a).
Q3. (a) ΔG⦵ = ΔH⦵ − TΔS⦵ is a straight line against T. Its gradient is (−20 − 40) ÷ (900 − 300) = −0.10 kJ K⁻¹ mol⁻¹, and the gradient is −ΔS⦵, so ΔS⦵ = +100 J K⁻¹ mol⁻¹. Substituting a point: 40 = ΔH⦵ − 300(0.10), so ΔH⦵ = +70 kJ mol⁻¹. Check: 70 ÷ 0.10 = 700 K, the temperature where ΔG⦵ = 0. (b) Above 700 K. M1 for using the gradient as −ΔS⦵, A1 for +100 J K⁻¹ mol⁻¹ (accept +0.10 kJ K⁻¹ mol⁻¹), A1 for +70 kJ mol⁻¹; 1 for "above 700 K" in (b). A ΔS⦵ with the wrong sign loses its A1 but the ΔH⦵ mark can still be scored by ECF.
Q4. (a) lnK = −ΔG⦵ ÷ RT = 20 000 ÷ (8.31 × 298) = 8.076, so K = exp(8.076) = 3.2 × 10³. (b) ΔG = ΔG⦵ + RT lnQ = −20.0 + 2.476 × ln(5.0 × 10³) = −20.0 + 2.476 × 8.517 = −20.0 + 21.1 = +1.1 kJ mol⁻¹. ΔG is positive (equivalently, Q > K), so the forward reaction is not spontaneous for this mixture and the reaction proceeds in the reverse direction until Q falls to K. M1 for ΔG⦵ in J with the correct rearrangement, A1 for K = 3.2 × 10³; A1 for ΔG ≈ +1.1 kJ mol⁻¹ with consistent units, 1 for the reverse direction with a reason. Using −20.0 in kJ with R in J gives K ≈ 1.01 and scores M0.
Q5. ΔS⦵ = +175.0 J K⁻¹ mol⁻¹ = +0.1750 kJ K⁻¹ mol⁻¹. At the turning point ΔG⦵ = 0, so T = ΔH⦵ ÷ ΔS⦵ = 100.7 ÷ 0.1750 = 575 K. Both ΔH⦵ and ΔS⦵ are positive, so the reaction is spontaneous above this temperature. Assumption: ΔH⦵ and ΔS⦵ do not change with temperature (or: all substances stay in the states shown). M1 for setting ΔG⦵ = 0 and converting ΔS⦵ to kJ, A1 for 575 K (accept 302 °C), 1 for a valid assumption. 0.575 K, from failing to convert, scores M0 A0.
Educerie · written from the published IB Diploma Programme Chemistry guide, first assessment 2025, section Reactivity 1.4 Entropy and spontaneity. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.