1.2.1 · Bond enthalpies SL core
Breaking bonds absorbs energy; forming them releases it. Everything in this section follows from that one sentence and a table in the data booklet.
If the bonds formed are stronger than the bonds broken, more comes out than went in and ΔH is negative. Average bond enthalpy values are given in the data booklet, and they are always positive — the table lists the energy needed to break each bond, so the direction is yours to supply.
The short route is faster and safer, because every bond you do not write down is a bond you cannot look up wrongly. But you must be certain a bond really is unchanged. If you are not, count them all: the answer is identical, and the arithmetic is only longer.
Three separate published schemes, on three different bond-enthalpy items, cap a reversed sign at part marks:
| Award [1 max] for +65 «kJ mol-1» | out of 2 |
| Award [2 max] for +663 «kJ mol-1» | out of 3 |
| Award [1 max] for +45 «kJ» | out of 2, on formation data |
The arithmetic scores; the sign is separate. Getting it backwards never loses everything — and never scores everything either.
Carbon monoxide reacts with chlorine to form phosgene. Determine ΔH from bond enthalpies.
| Bonds broken | C≡O 1077 + Cl–Cl 242 = 1319 kJ mol–1 |
| Bonds formed | C=O 804 + 2 × C–Cl (2 × 324) = 1452 kJ mol–1 |
| ΔH | 1319 − 1452 = −133 kJ mol–1 |
Carbon monoxide has a triple bond. A report on this very item records the (misconception CO having a C=O bond enthalpy). Using 804 instead of 1077 shifts the answer by 273 kJ mol–1 — and the mistake is invisible in the working.
Examiner feedback on one calculation is unusually reassuring and unusually specific at the same time: All candidates approached the question in the correct way, adding up the bond enthalpies of the bonds — and then The most common mistake was using the incorrect bond enthalpy for one or more bonds.
Other sessions name the same family of faults: candidates lost a mark for a missing bond broken in ethanol (often C-C or C-O) and using the single bond enthalpy where a double bond was present, and elsewhere There were all types of mistakes. From students using the wrong table to wrong bond enthalpies. Draw the structures before you look anything up. Every one of these is a counting error, not a chemistry error.
Feedback on one item names incorrect cross-referencing, an unbalanced equation, and the misuse of values together. An unbalanced equation makes every bond count wrong before a single value is read.
The syllabus asks for this explicitly: Include explanation of why bond enthalpy data are average values and may differ from those measured experimentally. A published scheme lists four accepted reasons, and asks for any two:
| 1 | bond enthalpies are average values |
| 2 | enthalpies of formation are specific to the compounds |
| 3 | bond enthalpies apply to gases AND this reaction involves a liquid |
| 4 | bond enthalpies do not account for the change of state |
Two of the four are about state, not about averaging — and that is the half students miss. Feedback records: Most students stated that bond enthalpies are average values gaining the first mark. Only a small number gained the second.
One examination question puts the whole idea in a single sentence. A student using bond enthalpy data correctly calculated an enthalpy change of 0 kJ mol–1. Another student, using enthalpy of formation data, correctly calculated −4 kJ mol–1 for the same reaction. The question then asks how the two students can carry out a calculation for the same reaction and obtain different results when both calculations are correct.
The published scheme accepts any one of: bond enthalpy values are averages / not specific to this reaction, or bond enthalpies apply to gases but this reaction involves liquids, or the second set of data was specific to the reaction.
Notice what the question concedes. It does not say one student was wrong. Both methods are correct; they answer slightly different questions, because one uses averages taken across many compounds and the other uses values measured for these compounds. The 4 kJ mol–1 gap is smaller than a 1 % error on a single average bond enthalpy.
1.2.2 · Hess's law SL core
The enthalpy change for a reaction is independent of the pathway between the initial and final states. Two routes between the same start and finish must give the same answer.
You almost never measure the reaction you want. Hess's law lets you build it from reactions you can measure — or from tabulated data — by going the long way round. Every method in the rest of this document is an application of it: the two routes in Figure 1, the summation equations of 1.2.4, and the Born–Haber cycle of 1.2.5 are all the same statement wearing different clothes.
Hess's law is a named syllabus outcome, and you should be able to state it. But it is worth knowing how it actually appears on papers.
Across every question paper and mark scheme in the evidence base behind this document, the word “Hess” appears exactly once — and that once is a note to examiners about error carried forward, not a question. The familiar textbook task, here are three equations, combine them, does not appear at all. What appears instead is 1.2.4 and 1.2.5, which are applications of Hess's law under other names. Learn the law as the reason the other methods work, and practise it through them, rather than hunting for standalone Hess's law questions.
A scheme on a cycle question notes: Award ECF for M3 for correct application of Hess's law based on the candidate's values/cycle. Error carried forward. A value misread early does not destroy the marks for the reasoning that follows — provided the reasoning is visible on the page. Show the cycle.
1.2.3 · Enthalpies of combustion and formation HL only
Two tabulated quantities, two precise definitions. Questions test the definitions directly, as equations.
Standard enthalpy of formation, ΔHf — the enthalpy change when one mole of a compound is formed from its elements in their standard states.
Standard enthalpy of combustion, ΔHc — the enthalpy change when one mole of a substance burns completely in oxygen.
An element in its standard state has ΔHf = 0 by definition. It takes no energy to make something out of itself.
A multiple-choice question asks which equation correctly shows both enthalpy of formation, ΔHf, of a compound and enthalpy of combustion, ΔHc, of its elements. The options were:
| A | CO(g) + ½O2(g) → CO2(g) | combustion of CO, but CO is a compound, not an element |
| B | ½N2(g) + 1½H2(g) → NH3(g) | formation of ammonia — but nothing is burned |
| C | 2S(s) + 3O2(g) → 2SO3(g) | combustion of sulfur — but two moles of product, so not a formation equation |
| D | H2(g) + ½O2(g) → H2O(l) | one mole of water from its elements and one mole of hydrogen burned |
The published key is D. Examiner feedback records that candidates chose C as well. Only D is both enthalpy of formation and enthalpy of combustion. C fails on a coefficient — a formation equation makes exactly one mole.
On another item, candidates failed to identify the formation equation even though it was the only equation that started with elements as their standard states. The test is mechanical and takes two seconds: elements on the left, one mole of one compound on the right. Nothing else qualifies.
1.2.4 · Calculating from formation or combustion data HL only
Two equations, given in the data booklet, that run in opposite directions. This is the single most confusable pair in the topic.
Ask where the shared reference state sits. For formation data it is the elements, underneath everything — so you go up to the products and up to the reactants, and take products minus reactants. For combustion data it is the combustion products, above everything — so the order flips to reactants minus products. If you can reconstruct that, you never have to remember which is which.
Given ΔHc values of −1561 for C2H6, −1411 for C2H4 and −286 for H2 (kJ mol–1), which expression gives ΔH for C2H4(g) + H2(g) → C2H6(g)?
which is printed on the paper as −1411 − 286 + 1561, and equals −136 kJ mol–1. That is the published key. Applying the formation rule instead — products minus reactants — gives +136: the same magnitude, the wrong sign, and one of the printed options.
A published scheme sets out a formation-data calculation as
A harder version turns it around. Given ΔHr = −4 kJ mol–1 for ethanol + ethanoic acid → ethyl ethanoate + water, and formation enthalpies of −286 (water), −278 (ethanol) and −484 (ethanoic acid), find ΔHf of the ester. The scheme writes it out:
The first mark is for the expression alone — the scheme says M1 for the correct expression or rearrangement. Write the equation out with the unknown in it before you start rearranging, and the first mark is banked whatever happens next. The sign trap is capped again: Award [1] for +480 «kJ mol−1».
Examiner feedback names using the enthalpy of formation of H2O(l) instead of H2O(g), errors in transferring the values and reversing the subtraction. Liquid and gaseous water have different enthalpies of formation, and the equation tells you which one you need. Copy the state symbols out of the equation before you open the data booklet.
1.2.5 · Born–Haber cycles HL only
Lattice enthalpy cannot be measured directly. A Born–Haber cycle reaches it the long way round — which is Hess's law again, applied to an ionic solid.
The guide is explicit: The construction of a complete Born–Haber cycle will not be assessed. What is assessed is interpreting and determining values from a cycle you are given, for compounds of univalent and divalent ions. Both examples in the current-syllabus papers print the cycle and ask you to fill it in or read from it.
| Step | Direction | Why |
|---|---|---|
| Atomization / sublimation | up | separating a solid into gaseous atoms costs energy |
| Bond dissociation | up | breaking a bond always costs energy |
| Ionization energy | up | removing an electron from a positive nucleus costs energy |
| Electron affinity | down | the one step that usually releases energy |
| Lattice enthalpy (IB definition) | up | pulling the ions apart from the solid costs energy, so the value is positive |
Examiner feedback records that 56% of the students knew which stages in a Born-Haber cycle had positive enthalpy changes — and that most students realise that atomisation is endothermic. Knowing the directions is a question in its own right, before any arithmetic.
The question supplies four values and expects three more from the data booklet:
| From the question | From the data booklet |
|---|---|
| ΔHf(CaBr2) = −648 2nd ionization energy of Ca = +1145 atomization of Ca = +178 vaporization of Br2(l) = +31 |
Br–Br bond enthalpy = 193 1st ionization energy of Ca = 590 electron affinity of Br = −325 |
Go up from the solid to the gaseous ions. The formation enthalpy is reversed, because the cycle runs up from the compound:
Then the two bromine atoms each gain an electron, releasing energy twice over:
Three things earn the three marks: finding the right three values in the booklet, doubling the electron affinity, and closing the cycle. The scheme adds M1 and M2 can be scored on the diagram — annotate the printed cycle and you are already scoring. And once more, Award [2 max] for -2100 «kJ mol-1»: the sign.
Examiner feedback names the commonest Born–Haber fault precisely: candidates failed at using the correct atomisation energy or bond enthalpy, or just omitted terms. An omitted step, not a wrong one. Count the arrows in the printed cycle and check you have used every one.
On an annotation question, the scheme instructs Do not accept ionization energy for Processes while accepting ionization energies. Calcium loses two electrons, so the plural is the answer. On the same question, explaining why calcium chloride has the lower lattice enthalpy needs Ca2+ has larger ionic radius «than Mg2+» — with the note Must be clear reference to ionic for M1. “Bigger atom” does not score.
Energy cycles — the Born–Haber cycle
Five of the six steps are measured; the sixth is what you want. Hide any one of them and the cycle will give it back, because energy round a closed loop must sum to zero.
Sign discipline is where the marks go. Ionisation energies and atomisation are endothermic and positive; electron affinity for a halogen and lattice formation are exothermic and negative. The unknown step is computed by closing the cycle, never looked up.
Each rung is drawn to scale in energy. The unknown step is the dashed one — its length is whatever it takes to close the loop.
Putting it together
Every method in this topic reaches the same quantity by a different road, and Hess's law is the reason they arrive at the same place.
| Outcome | Level | The method | The mark that is separately lost |
|---|---|---|---|
| 1.2.1 | SL core | Bonds broken minus bonds formed | The sign; and reading the wrong bond order out of the table |
| 1.2.2 | SL core | Hess's law — any path, same answer | Not shown as a task of its own; it earns the method mark inside the others |
| 1.2.3 | HL only | Recognising the two definitions as equations | The coefficient — a formation equation makes exactly one mole |
| 1.2.4 | HL only | Products − reactants, or reactants − products | The direction, and with it the sign; and the state symbol |
| 1.2.5 | HL only | Reading values off a printed cycle | An omitted step; the undoubled electron affinity; the sign |
Write the expression before you write any number. One scheme gives a mark for the expression alone, another awards marks scored on the diagram, a third carries the error forward through a correct method. In every case the reasoning is what is being paid for, and it only earns anything if it is visible.
- Bonds broken total 1450 kJ mol–1 and bonds formed total 1600. What is ΔH, and is the reaction endothermic or exothermic?
- Why can you count only the bonds that change, and still get the right answer?
- Give two reasons a bond-enthalpy result differs from one calculated with formation data.
- Write the equation whose enthalpy change is ΔHf of CH3OH(l).
- Using ΔHf data, ΔH = −110 for the product and −30 and −20 for the two reactants. What is ΔH for the reaction?
- The same three numbers were combustion data instead. Now what is ΔH?
- In a Born–Haber cycle for MgCl2, how many times does the electron affinity of chlorine appear?
- Which single step in a Born–Haber cycle is normally exothermic?
1. 1450 − 1600 = −150 kJ mol–1; exothermic. 2. Unchanged bonds appear on both sides in equal number, so they cancel exactly — which is Hess's law. 3. Any two of: bond enthalpies are averages; formation data are specific to the compounds; bond enthalpies apply to gases while the reaction involves a liquid; bond enthalpies do not account for a change of state. 4. C(s) + 2H2(g) + ½O2(g) → CH3OH(l) — elements in their standard states, one mole of product. 5. Products − reactants = −110 − (−30 − 20) = −60 kJ mol–1. 6. Reactants − products = (−30 − 20) − (−110) = +60 kJ mol–1 — same size, opposite sign. 7. Twice — there are two chloride ions. 8. The electron affinity.