1.4.1 · Entropy HL only
Entropy, S, measures how the matter and energy of a system are spread out — the more ways the energy can be distributed, the higher the entropy. Two things follow, and the examination tests both.
Absolute entropies are never zero and never negative, unlike enthalpies of formation. That is why the entropy change of a reaction is always a difference and never one of the values you looked up.
For predicting a sign, one question settles almost every examination item: does the number of moles of gas go up or down? Gases dominate the entropy of any reaction that has one. Four published multiple-choice items in the evidence behind these notes are answered by that count alone.
Its limit, stated honestly. The count cannot decide a reaction with no gas on either side. Dissolving a solid — KBr(s) → KBr(aq) — raises the entropy, because an ordered lattice becomes dispersed ions, and no gas is involved at all. None of the published keys turns on such a case, but know why the rule stops there.
| The question asked | The published key | Gas moles |
|---|---|---|
| Which change results in a decrease in entropy? | 2H2(g) + O2(g) → 2H2O(g) | 3 → 2 |
| In which reaction does entropy decrease? | NH3(g) + HCl(g) → NH4Cl(s) | 2 → 0 |
| Which reaction has the largest decrease in entropy? | 6CO2(g) + 6H2O(g) → C6H12O6(s) + 6O2(g) | 12 → 6 |
In the first item every other option increases the gas count or dissolves a solid. In the third, all three plausible options decrease it and the key is simply the one that falls furthest — by six moles. Count both sides before reading any option twice.
A structured question gave the standard entropy of chlorine as 223 and of phosgene as 284 J mol–1 K–1, with carbon monoxide (198) to be found in the data booklet, for CO(g) + Cl2(g) → COCl2(g).
The sign is a check on your arithmetic. Two moles of gas become one, so it must come out negative. If it does not, you have combined the terms the wrong way round.
Published examiner feedback names both. First, the order: About 80% of the candidates could correctly calculate the entropy of reaction from absolute entropies and, again, many others lost a mark because they combined the terms in the wrong order.
Second, and worse: A significant proportion of students used the entropy of the product as the value for ΔS. An absolute entropy is not a change. In the worked example above that mistake would report +284 where the answer is −137 — wrong in size and in sign, and it reports an increase where there is a decrease.
Feedback on one item: Most students related the calculated decrease in entropy to the decrease in the moles of gas. The moles-of-gas argument is the answer, not merely the route to it. When a question asks you to outline why an entropy change is negative, count the gas moles out loud on the page.
1.4.2 · The Gibbs equation, and its units HL only
One equation, three quantities, and two of them in kilojoules while the third is in joules. That single line is where this topic is won and lost.
On one structured item the published scheme splits two marks like this:
| M1 | conversion to common units |
| M2 | the correct value — the scheme notes M1 is for conversion to common units M2 is for correct value |
Half the marks, before any arithmetic. Write the conversion down as a separate line — ΔS = −220 J K–1 mol–1 = −0.220 kJ K–1 mol–1 — and it is banked whatever happens next.
A published scheme, working in kilojoules throughout: ΔH = −92.0 kJ mol–1, ΔS = −198.1 J K–1 mol–1, T = 773 K.
Another published scheme, on a different item, working in joules throughout:
Notice what changes and what does not. The first divides ΔS by 1000; the second multiplies ΔH by 1000. Both reach the same physics. Leaving ΔS in joules while ΔH is in kilojoules makes the temperature term one thousand times too large — in the first calculation that would turn +61.1 into a five-figure number, which is the size of error that should stop you.
Among the advice to future candidates in one published report:
Check units: this is the most common mistake in calculations involving entropy, enthalpy and ΔG.
Another lists Using correct units: enthalpy and entropy, conversion between joules and kilojoules. And a third, reporting on a two-mark item: most students attempted to use the correct equation for determining ΔG, but many of them did not convert the units. The average mark on this question was 0.7 out of 2 marks.
0.7 out of 2, with the right equation. That is what the units cost.
Feedback records candidates who tried to add in an entropy term, apparently not realising Tx0 = 0! — the exclamation mark is the examiners'. If the entropy change is zero, ΔG = ΔH at every temperature, and there is nothing to calculate.
1.4.3 · Spontaneity, and the temperature that changes it HL only
At constant pressure a change is spontaneous when ΔG is negative. Whether temperature can change that answer depends on one thing only.
If ΔH and ΔS have the same sign, there is a crossover temperature. The two terms pull against each other, and at T = ΔH / ΔS they cancel exactly. If the signs disagree, one answer holds at every temperature and there is nothing to calculate. Figure 1 sets out all four cases; this is the question that sorts them.
Hydrogenating ethyne, C2H2(g) + 2H2(g) → C2H6(g), with ΔH = −312 kJ mol–1 from the previous part and ΔS supplied as −233 J K–1 mol–1. Calculate the temperature at which this reaction is no longer spontaneous.
| Both signs negative | so a crossover exists, and the reaction is spontaneous when cold |
| Set ΔG = 0 | 0 = ΔH − TΔS, so T = ΔH / ΔS |
| Convert | −233 J K–1 mol–1 = −0.233 kJ K–1 mol–1 |
The scheme awards M1 for the conversion alone, accepting either ΔS = -233 x 10-3 «kJ K-1 mol-1» or ΔH = -312000 «J mol-1» — both routes again. And it adds a refusal worth memorising: Do not award ECF for M2 if the answer is a negative Kelvin temperature. A negative absolute temperature is not a wrong answer; it is an impossible one, and it forfeits even the carry-forward mark.
Published examiner feedback on a recent item: A minority of candidates recognised that spontaneity was a continuum, whereas the majority incorrectly stated that temperature would have no effect, treating it as a binary property. One report lists the point among its advice: Recognition that spontaneity is not a digital effect; some reactions are more spontaneous than others.
The same commentary records candidates who chose 273 K as standard temperature instead of 298 K. Standard temperature for thermodynamic data is 298 K, not the 273 K of gas calculations.
Asked to explain the effect of raising the temperature, one scheme requires:
| M1 | the −TΔS expression becomes more negative / so Gibbs energy/ΔG becomes more negative |
| M2 | increase in spontaneity «as temperature is increased» |
With the note: Do not accept “spontaneous” without reference to increases. The answer is a change, not a state. Say more spontaneous, and say which term moved.
Entropy and spontaneity — the crossover temperature
ΔG = ΔH − TΔS. Two of the four sign combinations depend on temperature and two do not, and this is the fastest way to see which is which.
ΔG against temperature is a straight line: intercept ΔH, gradient −ΔS. It crosses zero only if ΔH and ΔS share a sign.
The units are the question. ΔS is in joules and ΔH in kilojoules, so one of them must be converted before they are subtracted. Both routes are shown in the readout.
1.4.4 · ΔG, Q and K HL only
As a reaction approaches equilibrium, ΔG becomes less negative and finally reaches zero. Two equations connect the Gibbs energy to the equilibrium constant, and both are in the data booklet.
The second follows from the first: set ΔG to zero and Q to K, and rearrange. R is 8.31 J K–1 mol–1 — joules again, so the answer comes out in joules and a question asking for kJ mol–1 needs one more step.
A multiple-choice question: H2(g) + I2(g) ⇌ 2HI(g) with K = 40.2 at 30 °C. The question supplies ΔG = −RT ln K and R = 8.31 J K–1 mol–1, and asks for the answer in kJ mol–1. The printed options were −9300, −921, −9.30 and −0.921.
| Option | What produces it |
|---|---|
| −9.30 | the answer — 30 °C converted to 303 K, joules converted to kJ |
| −9300 | correct arithmetic, left in joules |
| −921 | 30 used as the temperature, and left in joules |
| −0.921 | 30 used as the temperature, converted to kJ — both slips |
The chemistry is one line and the question is entirely about units. Feedback on this item records that Over 50% of students correctly calculated the Gibbs energy change from the equation, formula and data supplied — which means nearly half did not, with every step given to them.
A published report on a ΔG = −RT ln Kc calculation lists what went wrong: using an incorrect value for T (500°C instead of 773K), not carry out the ln of Kc or not dividing by 1000 to convert J (in R value).
A temperature left in Celsius, a logarithm not taken, a factor of a thousand. Not one of the three is about equilibrium or energy. Another report notes candidates who did not change ΔG from kJ to J in the equation for Kc — and prefaces it with Predictably.
On one item, 10% of students opted for ΔH = 0, evidencing a misunderstanding of Gibbs free energy and equilibrium. At equilibrium ΔG is zero. ΔH is not — it is whatever the chemistry makes it, and an exothermic and an endothermic reaction can both sit at equilibrium, with entropy changes of opposite sign making up the difference.
Putting it together
Four outcomes, one equation, and a single recurring cause of lost marks.
| Outcome | What is assessed | The mark that is separately lost |
|---|---|---|
| 1.4.1 | Predicting the sign of ΔS; calculating it from absolute entropies | Combining the terms the wrong way round; using one absolute entropy as the change |
| 1.4.2 | ΔG = ΔH − TΔS | The units — a published scheme gives half its marks for the conversion alone |
| 1.4.3 | The sign of ΔG; the crossover temperature | Treating spontaneity as a switch; using 273 K; a negative Kelvin answer |
| 1.4.4 | ΔG = −RT ln K and its rearrangements | Celsius not converted; the logarithm not taken; joules not converted |
(1) Write every quantity with its unit. ΔH in kJ, ΔS in J, ΔG in kJ, R in J, T in K. (2) Choose one system and convert into it — all kilojoules or all joules, and say which on the page. (3) Check the temperature is in kelvin and that you added 273, not used 273. Every trap in this topic is caught by those three lines, and one published scheme pays a mark for the second of them.
- Predict the sign of ΔS for 2NO2(g) → N2O4(g), and say why.
- Standard entropies: A = 210, B = 205, AB = 240 J K–1 mol–1. What is ΔS for A + B → AB?
- ΔH = −40.0 kJ mol–1 and ΔS = −120 J K–1 mol–1. Calculate ΔG at 298 K.
- For the same reaction, at what temperature does it stop being spontaneous?
- A reaction has ΔH positive and ΔS negative. At what temperature is it spontaneous?
- K = 0.50 at 298 K. Is ΔG positive or negative, and why?
- What is zero at equilibrium: ΔG, ΔG, or ΔH?
- Why does an answer of −150 K to a crossover calculation score nothing at all?
1. Negative. Two moles of gas become one, so the matter and energy are less spread out. 2. 240 − (210 + 205) = −175 J K–1 mol–1. 3. −40.0 − 298 × (−0.120) = −40.0 + 35.8 = −4.2 kJ mol–1 — spontaneous, but only just. 4. Both signs negative, so a crossover exists: T = −40.0 / −0.120 = 333 K. Above that it is no longer spontaneous. 5. At no temperature. The signs disagree, so both terms push ΔG positive at every T. 6. K < 1, so ln K is negative, so −RT ln K is positive: reactants are favoured. 7. ΔG. ΔG is a fixed property of the reaction and is only zero when K = 1; ΔH is whatever the chemistry makes it. 8. A negative absolute temperature is impossible, so the working must contain an error. A published scheme refuses even the carry-forward mark for it.