3 higher-level sections hidden.
Educerie · IB Diploma · Chemistry
Reactivity 1 What drives chemical reactions? · R1.2 Energy cycles in reactions
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Explain that bond breaking absorbs energy and bond forming releases it | SL, HL | "Explain, in terms of bonds, why the reaction is exothermic" (2 marks) |
| Calculate ΔH of a reaction from average bond enthalpies | SL, HL | Paper 2: "Determine the enthalpy change using section … of the data booklet" (2–3 marks) |
| Explain why bond enthalpies are averages and why the calculated value differs from the experimental one | SL, HL | "Suggest why the value you calculated differs from the value in the data booklet" (1–2 marks) |
| State Hess's law and use it to find ΔH for a multistep reaction | SL, HL | Paper 1A cycle questions; Paper 2 "Determine ΔH for … using the equations below" (2–3 marks) |
| Combine two measured enthalpy changes into one you cannot measure directly | SL, HL | Paper 1B experiment on dissolving two salts (4–6 marks across parts) |
| Define standard enthalpies of formation and combustion and write the equations they refer to | HL only | "Write the equation for the standard enthalpy change of formation of ethanol" (1 mark) |
| Calculate ΔH from ΔHf⦵ or ΔHc⦵ data | HL only | Paper 2: 2–3 marks, often with a ΔHf⦵ table given |
| Interpret a Born–Haber cycle and determine a missing value for univalent and divalent ions | HL only | Paper 2: "Determine the lattice enthalpy of calcium chloride" (2–3 marks) |
Before you start
You need R1.1: what ΔH means, that exothermic reactions have a negative ΔH and endothermic ones a positive ΔH, and what an energy profile looks like. You need covalent bonds, single, double and triple, from Structure 2.2, and you must be able to draw a molecule's structural formula so you can count its bonds. HL students also need ionization energies from Structure 1.3 and the ionic lattice from Structure 2.1.
1The idea in one paragraph
Energy is never created or destroyed, so the energy change of a reaction depends only on where it starts and where it ends. That one fact gives you two ways to work out ΔH without a thermometer. You can take the reaction apart bond by bond: pay the energy to break every bond in the reactants, collect the energy released when every bond in the products forms, and the difference is ΔH. Or you can find another route from the same start to the same finish, made of steps whose enthalpy changes you already know, and add them up. That second idea is Hess's law, and at HL it becomes a machine for turning tables of data into the enthalpy change of any reaction, including the energy that holds an ionic crystal together.
2Breaking bonds costs energy; making bonds pays it back
A covalent bond is two nuclei held by a shared pair of electrons. Pulling the atoms apart means working against that attraction, so bond breaking is always endothermic: it absorbs energy. The reverse is always true as well. When two atoms come together to form a bond they fall into a lower-energy arrangement and the difference is released, so bond forming is always exothermic.
The size of that energy is the bond enthalpy: the energy needed to break one mole of a particular bond, with the molecule and the fragments all in the gaseous state. The value is always positive, because it is a cost. The same number, with a negative sign, is what you get back when the bond forms.
Every reaction breaks some bonds and makes others. Picture it happening in two stages, even though real molecules do not react this way: first every reactant bond breaks, leaving a cloud of separate gaseous atoms; then those atoms join into the products. Figure 1 draws methane burning in exactly that way.
The climb up is the energy absorbed in breaking bonds. The fall down is the energy released in making them. If the fall is bigger than the climb, the products end up lower than the reactants, the difference leaves as heat, and the reaction is exothermic. If the climb is bigger, the reaction is endothermic. That is the whole reason some reactions give out heat and others take it in: it depends on whether the bonds made are, in total, stronger than the bonds broken.
ΔH = Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds formed)
Read the order carefully: broken minus formed. It looks backwards compared with "products minus reactants", and it is, because bond enthalpies are costs of breaking. The energy of the products' bonds is released, so it is subtracted.
3Calculating ΔH from bond enthalpies
The method never changes, and the marks sit on the steps, so do all of them every time.
- Write the balanced equation with every species a gas. Bond enthalpies are defined for gases, and this method only works for them.
- Draw the structural formula of every molecule, so no bond is hidden.
- Count each type of bond broken and each type formed, multiplied by the coefficients in the equation.
- Look up each bond enthalpy in the data booklet and add up both columns.
- ΔH = broken − formed, with a sign and the unit kJ mol⁻¹.
Worked example 1: methane burning. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). Figure 2 draws the molecules so that every bond can be counted.
The bond enthalpies used on this page, in kJ mol⁻¹, are C–H 414, O=O 498, C=O 804, O–H 463, C–C 346, C=C 614, H–H 436, N≡N 945, N–H 391, Cl–Cl 242 and C–Cl 324. Always use the values in the question or your data booklet if they are different.
The two C=O bonds are counted as two, and the "2" in front of H₂O doubles its two O–H bonds to four. Missing a coefficient is the most common way to lose the second mark.
Worked example 2: adding hydrogen to ethene. CH₂=CH₂(g) + H₂(g) → CH₃–CH₃(g). You could break all six bonds in ethene and form all seven in ethane, and it would work. There is a faster way: bonds that appear unchanged on both sides cancel out. Ethene has four C–H bonds and ethane has six, so only two C–H bonds are really new.
The full count gives 2706 − 2830, which is the same −124 kJ mol⁻¹. Use whichever you trust; if you use the shortcut, make sure the cancelled bonds really are identical on both sides.
Worked example 3: the Haber process. N₂(g) + 3H₂(g) → 2NH₃(g).
Breaking the N≡N triple bond, at 945 kJ mol⁻¹, uses up most of the energy released. It is also why nitrogen "burning" in oxygen is endothermic: the N≡N and O=O bonds broken are, in total, stronger than the bonds in the nitrogen oxide formed.
4Why bond enthalpies are averages, and why your answer is a little wrong
The C–H bond in methane is not quite the same as the C–H bond in ethanol: each sits in a slightly different electronic neighbourhood, so each takes a slightly different energy to break. Even in one CH₄ molecule, the first C–H bond costs a different amount from the second, because the fragment left behind is different. So the data booklet gives an average bond enthalpy, a mean over a range of compounds.
That is why an answer from bond enthalpies is an estimate. When you compare it with an experimental value, there are two reasons to give, and a good answer gives both.
- Bond enthalpies are averages. The actual bonds in these particular molecules are not exactly average strength.
- Bond enthalpies apply to gases only. If any reactant or product is really a liquid or a solid, the energy of changing state is left out of the bond-enthalpy sum.
The second one explains most of the gap for combustion. The data booklet gives the enthalpy of combustion of methane as about −890 kJ mol⁻¹, and that value forms liquid water. Section 3 made gaseous water and got −808 kJ mol⁻¹. Condensing two moles of steam to liquid releases roughly 88 kJ more, which brings the estimate close to −890. What remains is the averaging.
This is a nature of science point: a bond enthalpy is a model, one number standing for a family of slightly different bonds, useful precisely because you know where it breaks down.
Bond enthalpy also connects back to structure. Shorter bonds are stronger, because the shared electrons sit closer to both nuclei and are held more tightly. A double bond is shorter and stronger than a single bond between the same atoms; a triple bond is shorter and stronger again. Down group 17, the halogen atom gets bigger, so a carbon–halogen bond gets longer and weaker. Figure 3 plots it.
Two things follow that the course asks about elsewhere. A C–I bond breaks far more easily than a C–F bond, so iodoalkanes react fastest in the substitution reactions of Reactivity 3.4. And ultraviolet light high in the atmosphere has enough energy to break the C–Cl bond in a CFC but not the much stronger C–F bond, which is why CFCs release chlorine atoms and not fluorine atoms. Bond polarity matters too: a polar bond often has extra strength from the attraction between its partial charges, so a very polar bond such as C–F or O–H tends to be stronger than its length alone suggests.
5Hess's law
Hess's law: the enthalpy change for a reaction is independent of the pathway between the initial and final states.
The reason is conservation of energy. If one route from A to B released 100 kJ and another route back cost only 90 kJ, going round the loop would create 10 kJ from nothing. So every route from A to B has the same total ΔH.
That lets you reach an enthalpy change you cannot measure. You cannot burn carbon so that it makes only CO, but you can measure carbon burning to CO₂ and CO burning to CO₂. Figure 4 puts them in a triangle.
Both routes start with C(s) + O₂(g) and end with CO₂(g), so ΔH₃ = ΔH₁ + ΔH₂.
The rule for walking round a cycle. Go from start to finish along the arrows you know. If you walk along an arrow in its own direction, add its ΔH. If you walk against it, reverse its sign. Nothing else is needed.
Doing it with equations instead. When a question gives you a list of equations rather than a diagram, rearrange them until they add up to the target equation. Three moves are allowed.
| What you do to the equation | What you do to its ΔH |
|---|---|
| Reverse it | Change the sign |
| Multiply every coefficient by a number | Multiply ΔH by the same number |
| Add two equations together | Add their ΔH values |
Worked example 4: a multistep reaction. Find ΔH for 2C(s) + H₂(g) → C₂H₂(g), the formation of ethyne. Nobody can make ethyne this way in a calorimeter, but three combustions can be measured:
Work from the target. It needs 2C on the left, so take equation (1) twice. It needs H₂ on the left, so take (2) once. It needs C₂H₂ on the right, but (3) has it on the left, so reverse (3).
Always check that everything you did not want cancels. If a CO₂ or an O₂ is left over, one of your multipliers is wrong.
Hess's law in the laboratory. Turning anhydrous sodium carbonate into the hydrated crystal is too slow and uncontrolled to measure directly. But both solids dissolve quickly to give the same solution, and each dissolving can be measured by calorimetry with Q = mcΔT from R1.1. Figure 5 shows the cycle.
The direct route is the hydration. The indirect route is to dissolve the anhydrous salt (ΔH₁) and then go backwards along the hydrated salt's dissolving arrow (−ΔH₂). So ΔH(hydration) = ΔH₁ − ΔH₂. Question 3 in Try it gives you the data.
6HLStandard enthalpies of formation and combustion
SL students can skip to section 9.
Tables of data can only work if everyone measures under the same conditions. Standard conditions are a pressure of 100 kPa and, unless stated, a temperature of 298 K, with every substance in its standard state: its normal physical state under those conditions. The symbol ⦵ means "standard".
Standard enthalpy change of formation, ΔHf⦵: the enthalpy change when one mole of a compound is formed from its elements in their standard states, under standard conditions.
Standard enthalpy change of combustion, ΔHc⦵: the enthalpy change when one mole of a substance burns completely in oxygen, under standard conditions.
The exam asks you to write the equations these refer to, and the definitions dictate every detail.
| Change | Equation | What the definition forces |
|---|---|---|
| ΔHf⦵ of ethanol | 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l) | exactly 1 mol of product, so ½O₂ is correct; elements as they are at 298 K: C(s), H₂(g), O₂(g) |
| ΔHf⦵ of sodium chloride | Na(s) + ½Cl₂(g) → NaCl(s) | Cl₂ is diatomic, so ½Cl₂, never Cl(g) |
| ΔHc⦵ of ethanol | C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l) | 1 mol of fuel; complete combustion; water liquid at 298 K |
| ΔHc⦵ of hydrogen | H₂(g) + ½O₂(g) → H₂O(l) | this is also ΔHf⦵ of water |
Three consequences are worth knowing by heart.
ΔHf⦵ of an element in its standard state is zero. Forming O₂(g) from O₂(g) is no change at all. That is why oxygen never appears in a ΔHf⦵ sum.
Allotropes do not both get zero. Only the standard state of the element does. For carbon that is graphite, so ΔHf⦵ of graphite is 0 and ΔHf⦵ of diamond is about +1.9 kJ mol⁻¹. The two have different bonding, so they have different energies, and converting one mole of graphite to diamond takes a little energy in.
Some reactions are two definitions at once. The combustion of hydrogen and the formation of water are one equation. The same goes for carbon: ΔHc⦵ of graphite equals ΔHf⦵ of CO₂.
7HLCalculating ΔH from formation and combustion data
Both methods are Hess's law with a fixed choice of the middle step. Figure 6 draws the two cycles side by side.
In panel (a), the middle step is the elements. You can make both the reactants and the products from them, so the arrows point up from the elements. Walking from reactants to products means going down the reactants' arrow backwards and up the products' arrow forwards:
ΔH⦵ = ΣΔHf⦵(products) − ΣΔHf⦵(reactants)
In panel (b), the middle step is CO₂ and water. Both the reactants and the products burn to them, so the arrows point down. Walking from reactants to products means going down the reactants' arrow forwards and up the products' arrow backwards:
ΔH⦵ = ΣΔHc⦵(reactants) − ΣΔHc⦵(products)
The subtraction flips because the arrows flip. Both equations are in the data booklet; sketch Figure 6 in ten seconds and you will never mix them up.
Worked example 5: formation data. Iron is extracted in a blast furnace by Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g). Values of ΔHf⦵ in kJ mol⁻¹: Fe₂O₃(s) −824.2, CO(g) −110.5, CO₂(g) −393.5, Fe(s) 0.
Multiply each value by its coefficient, write zero for the element so the marker sees you meant it, and bracket every negative number before you subtract it.
Worked example 6: combustion data. Find ΔHf⦵ of ethanol, which cannot be measured directly because carbon, hydrogen and oxygen will not simply combine to give ethanol. The target is 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l). Values of ΔHc⦵ in kJ mol⁻¹: C(s) −393.5, H₂(g) −285.8, C₂H₅OH(l) −1367.3, and O₂ has none because oxygen does not burn in oxygen.
Combustion is easy to measure, formation often is not, and Hess's law turns one into the other.
8HLBorn–Haber cycles
The ΔHf⦵ of an ionic compound can be measured; the energy holding its ions together cannot. A Born–Haber cycle is a Hess's law cycle that splits the formation into measurable steps, leaving the lattice enthalpy as the only unknown.
The steps, each one defined per mole:
| Step | What it is | Sign |
|---|---|---|
| Enthalpy of atomization of the metal | solid metal to gaseous atoms: Na(s) → Na(g). For a metal this is the sublimation enthalpy | positive |
| Enthalpy of atomization of the non-metal | half a mole of X₂ to one mole of X(g): ½Cl₂(g) → Cl(g). Equal to half the bond enthalpy | positive |
| Ionization energy | removing electrons from gaseous atoms: Na(g) → Na⁺(g) + e⁻. For Mg²⁺ you need the first and the second | positive |
| Electron affinity | adding an electron to a gaseous atom: Cl(g) + e⁻ → Cl⁻(g) | first one negative |
| Lattice enthalpy | separating one mole of the solid into its gaseous ions: NaCl(s) → Na⁺(g) + Cl⁻(g) | positive |
| Enthalpy of formation | Na(s) + ½Cl₂(g) → NaCl(s) | usually negative |
Lattice enthalpy is defined here, as in the IB data booklet, as the energy to break the lattice apart, so it is always positive. Some textbooks use the reverse direction, with a negative sign; stay with the booklet.
Figure 7 is the cycle for sodium chloride, drawn as energy levels. Arrows up absorb energy; arrows down release it.
Two routes lead from the elements at the bottom left to the gaseous ions. One goes up the staircase of atomizations, ionization and electron affinity. The other goes down to NaCl(s) by ΔHf⦵ and then up by the lattice enthalpy. Both reach the same level, so by Hess's law they must total the same:
The guide says you will not have to build a whole cycle from nothing, but you must read one and find any missing value, for univalent ions like Na⁺ and Cl⁻ and for divalent ions like Mg²⁺ and O²⁻. Figure 8 is the divalent case.
Three things are new with 2+ and 2− ions.
- Two ionization energies. Mg(g) → Mg⁺(g) + e⁻ is IE₁, then Mg⁺(g) → Mg²⁺(g) + e⁻ is IE₂. Both go in.
- Two electron affinities. The first is exothermic, because a neutral oxygen atom attracts an electron. The second is endothermic (+753 kJ mol⁻¹ here), because the incoming electron is being pushed onto an ion that is already negative, and the repulsion has to be overcome. Students very often give the second electron affinity a negative sign.
- Atomization of oxygen is half the O=O bond enthalpy, 498 ÷ 2 = 249 kJ mol⁻¹.
If you forget the second of anything, the answer is wrong by more than a thousand kilojoules. If the compound has two anions per metal, as in MgCl₂ or CaCl₂, the chlorine steps are doubled: two atomizations (the whole Cl–Cl bond, not half of it) and two electron affinities.
The lattice enthalpy of MgO is nearly five times that of NaCl, as Structure 2.1 predicts: the ions carry double the charge and are smaller, so they attract far more strongly.
9Where marks are lost
Subtracting the wrong way round with bond enthalpies. For bond enthalpies it is broken minus formed. For ΔHf⦵ it is products minus reactants. For ΔHc⦵ it is reactants minus products. Picture the cycle, never guess the order.
Missing a coefficient. Two molecules of water have four O–H bonds, not two. Multiply every bond count and every ΔHf⦵ or ΔHc⦵ value by the coefficient in the equation.
Using bond enthalpies for liquids and solids without comment. Bond enthalpies are for gases. If a species is liquid, the calculated answer leaves out the energy of changing state, and "explain the difference" wants you to say so.
Giving "heat loss" as the reason a bond-enthalpy value differs from the data booklet. No experiment was done, so nothing was lost. The reasons are averages and states.
Forgetting to reverse the sign when an equation is reversed. In a Hess's law question, write the rearranged equation with its new ΔH next to it before you add anything.
Giving an element a ΔHf⦵ value. O₂(g), Fe(s) and C(graphite) are all zero. Diamond is not, because it is not the standard state of carbon.
Getting the second electron affinity wrong. EA₂ of oxygen is positive. And for MgCl₂, both the atomization of chlorine and its electron affinity are needed twice.
Mixing lattice enthalpy conventions. In IB data it is the endothermic separation of the solid into gaseous ions, so the value is positive. An answer of −786 kJ mol⁻¹ for NaCl will be marked wrong unless the question uses the other direction.
10Draw it right
This subtopic draws two kinds of diagram: Hess's law cycles and energy-level diagrams, including Born–Haber cycles.
- Every species written with a state symbol. In a Hess cycle, the same species must appear at the same place on both routes.
- Every arrow points the way the change goes, and its ΔH is written beside it with a sign.
- In a Hess cycle, the target reaction is one arrow and the known route is two or more arrows sharing its start and finish. If the start and finish do not match, it is not a cycle.
- In an energy-level diagram, up means endothermic and down means exothermic. The vertical axis is enthalpy, even if it is not to scale.
- In a Born–Haber cycle, electrons are shown on the levels where they have been removed but not yet added (Na⁺(g) + e⁻ + Cl(g)).
- Every level is balanced: the same atoms and the same total charge on each one.
- For a divalent ion, two ionization steps or two electron affinity steps, each labelled, and the second electron affinity pointing up.
- When a calculation follows, write the Hess's law equation you read from the diagram before you substitute numbers. That line is the method mark.
11Try it
Marks in brackets. Answers and marker's notes are at the end. Use the bond enthalpies from section 3 unless the question gives others.
Q1. Ethene reacts with chlorine: CH₂=CH₂(g) + Cl₂(g) → CH₂ClCH₂Cl(g).
(a) Determine the enthalpy change for this reaction using average bond enthalpies. 3 marks
(b) The experimental value is different. Suggest one reason why. 1 mark
Q2. Determine the enthalpy change for S(s) + H₂(g) + 2O₂(g) → H₂SO₄(l) from these equations. 3 marks
S(s) + O₂(g) → SO₂(g), ΔH = −297 kJ mol⁻¹
2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = −198 kJ mol⁻¹
H₂(g) + ½O₂(g) → H₂O(l), ΔH = −286 kJ mol⁻¹
SO₃(g) + H₂O(l) → H₂SO₄(l), ΔH = −133 kJ mol⁻¹
Q3. (Paper 1B style.) A student wants the enthalpy change for Na₂CO₃(s) + 10H₂O(l) → Na₂CO₃·10H₂O(s). She dissolves 5.30 g of anhydrous Na₂CO₃ (M = 105.99 g mol⁻¹) in 50.0 g of water in a polystyrene cup: the temperature rises from 21.2 °C to 27.5 °C. She then dissolves 14.31 g of Na₂CO₃·10H₂O (M = 286.19 g mol⁻¹) in a fresh 50.0 g of water: the temperature falls from 21.4 °C to 5.2 °C. Assume the solution has the mass and specific heat capacity of the water, 4.18 J g⁻¹ K⁻¹.
(a) Calculate the enthalpy change of solution of anhydrous Na₂CO₃, in kJ mol⁻¹. 2 marks
(b) Calculate the enthalpy change of solution of Na₂CO₃·10H₂O, in kJ mol⁻¹. 1 mark
(c) Determine the enthalpy change of hydration of Na₂CO₃. 1 mark
(d) Heat is exchanged with the surroundings in both experiments. Explain its effect on each measured value. 2 marks
Q4 (HL). Ammonia is oxidized in the first step of making nitric acid: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g).
(a) Write the equation for the standard enthalpy change of formation of NO(g). 1 mark
(b) Calculate ΔH⦵ for the oxidation of ammonia. ΔHf⦵ in kJ mol⁻¹: NH₃(g) −46.1, NO(g) +90.3, H₂O(g) −241.8. 2 marks
Q5 (HL). Determine the standard enthalpy change of formation of propane, C₃H₈(g), from these ΔHc⦵ values in kJ mol⁻¹: C(s) −393.5, H₂(g) −285.8, C₃H₈(g) −2219. 3 marks
Q6 (HL). For calcium chloride, CaCl₂, the following values are given in kJ mol⁻¹: atomization of Ca +178; first ionization energy of Ca +590; second ionization energy of Ca +1145; Cl–Cl bond enthalpy +242; electron affinity of Cl −349; ΔHf⦵ of CaCl₂(s) −796.
(a) Determine the lattice enthalpy of CaCl₂. 3 marks
(b) Explain why the full Cl–Cl bond enthalpy is used here, and not half of it as for NaCl. 1 mark
12In one breath
Breaking a bond absorbs energy and making one releases it, so ΔH = bonds broken − bonds formed, using average bond enthalpies for gases; the answer is an estimate because the values are averages over many compounds and because liquids and solids are treated as gases. Shorter bonds are stronger, so C–F is much stronger than C–I. Hess's law says ΔH depends only on the start and the finish, because energy is conserved, so you can reach any reaction by another route: reverse an equation and flip the sign, multiply it and multiply ΔH, then add. That is how two easy calorimetry experiments give an enthalpy change you cannot measure directly. HL: ΔHf⦵ is one mole of compound from its elements in their standard states (elements themselves score zero), ΔHc⦵ is one mole burning completely; ΔH = ΣΔHf⦵(products) − ΣΔHf⦵(reactants), and ΔH = ΣΔHc⦵(reactants) − ΣΔHc⦵(products). A Born–Haber cycle is Hess's law for an ionic solid: atomize, ionize, add electrons, and the lattice enthalpy (positive, solid to gaseous ions) is whatever closes the loop. Divalent ions need two ionization energies or two electron affinities, and the second electron affinity is endothermic.
Answers
Q1. (a) Bonds broken: C=C + Cl–Cl = 614 + 242 = 856 kJ. Bonds formed: C–C + 2 × C–Cl = 346 + 2(324) = 994 kJ. ΔH = 856 − 994 = −138 kJ mol⁻¹. The four C–H bonds are unchanged and cancel; counting them on both sides gives the same answer. (b) Bond enthalpies are average values taken over many compounds, so the bonds in these molecules are not exactly the tabulated strength; or, if the experiment is done at room temperature, 1,2-dichloroethane is a liquid, and the bond-enthalpy method treats it as a gas. M1 for the correct bonds broken and their total, M1 for the correct bonds formed and their total, A1 for −138 kJ mol⁻¹ with sign and unit; 1 for either reason in (b). "Heat was lost to the surroundings" scores 0 in (b): no experiment was done.
Q2. Keep the first equation as it is: −297. Halve the second: SO₂ + ½O₂ → SO₃, −99. Keep the third: −286. Keep the fourth: −133. Adding them, SO₂, SO₃ and H₂O cancel and the oxygen totals 1 + ½ + ½ = 2O₂, leaving the target. ΔH = −297 − 99 − 286 − 133 = −815 kJ mol⁻¹. M1 for halving the second equation and its ΔH, M1 for combining all four correctly, A1 for −815 kJ mol⁻¹. Using −198 unhalved gives −914 and loses the final mark but keeps the combination mark.
Q3. (a) Q = mcΔT = 50.0 × 4.18 × 6.3 = 1317 J = 1.317 kJ. Amount = 5.30 ÷ 105.99 = 0.0500 mol. The temperature rose, so the process is exothermic: ΔH₁ = −1.317 ÷ 0.0500 = −26.3 kJ mol⁻¹. (b) Q = 50.0 × 4.18 × 16.2 = 3386 J; amount = 14.31 ÷ 286.19 = 0.0500 mol; the temperature fell, so ΔH₂ = +67.7 kJ mol⁻¹. (c) ΔH(hydration) = ΔH₁ − ΔH₂ = −26.3 − 67.7 = −94.0 kJ mol⁻¹. (d) In the first experiment heat escapes to the surroundings, so the measured temperature rise is too small and ΔH₁ is less negative than the true value. In the second, heat flows in from the warmer surroundings, so the measured temperature fall is too small and ΔH₂ is less positive than the true value. M1 for Q in (a), A1 for −26.3 kJ mol⁻¹ with the negative sign; A1 for +67.7 kJ mol⁻¹ in (b); A1 for −94.0 kJ mol⁻¹ in (c), and ECF from your (a) and (b); 1 for each experiment's effect correctly explained in (d). An answer with the signs of ΔH₁ and ΔH₂ reversed scores 0 for those values.
Q4 (HL). (a) ½N₂(g) + ½O₂(g) → NO(g). (b) ΣΔHf⦵(products) = 4(+90.3) + 6(−241.8) = 361.2 − 1450.8 = −1089.6. ΣΔHf⦵(reactants) = 4(−46.1) + 5(0) = −184.4. ΔH⦵ = −1089.6 − (−184.4) = −905.2 kJ mol⁻¹. 1 for (a) with one mole of NO and halves of both elements with state symbols; M1 for the two sums with coefficients, A1 for −905.2 kJ mol⁻¹. N(g) or O(g) in (a) scores 0, because they are not the standard states.
Q5 (HL). Target: 3C(s) + 4H₂(g) → C₃H₈(g). ΣΔHc⦵(reactants) = 3(−393.5) + 4(−285.8) = −1180.5 − 1143.2 = −2323.7. ΣΔHc⦵(products) = −2219. ΔHf⦵ = −2323.7 − (−2219) = −104.7 kJ mol⁻¹. M1 for the correct target equation or cycle, M1 for reactants − products with coefficients, A1 for −104.7 kJ mol⁻¹ (accept −105). Products − reactants gives +104.7 and scores M1 only.
Q6 (HL). (a) Route through the gaseous ions: +178 + 590 + 1145 + 242 + 2(−349) = +1457. The lattice enthalpy takes the solid to the ions, so ΔHf⦵ + lattice enthalpy = 1457, and lattice enthalpy = 1457 − (−796) = +2253 kJ mol⁻¹. (b) CaCl₂ contains two chloride ions, so two moles of Cl atoms are needed, and forming them breaks one whole mole of Cl–Cl bonds. M1 for including both ionization energies, M1 for twice the electron affinity with the full bond enthalpy, A1 for +2253 kJ mol⁻¹; 1 for (b). A negative lattice enthalpy scores the method marks only.
Educerie · written from the published IB Diploma Programme Chemistry guide, first assessment 2025, section Reactivity 1.2 Energy cycles in reactions. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.