Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Chemistry guide (first assessment 2025).
Learn
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
R1.2.1 Bond enthalpy
Bond enthalpy
The enthalpy change when 1 mol of a particular covalent bond is broken, with the substance and the fragments formed all in the gaseous state, e.g. H₂(g) → 2H(g), +436 kJ mol⁻¹. Units kJ mol⁻¹. Bond enthalpies are always positive: energy must be absorbed to overcome the attraction between the bonded nuclei and the shared pair of electrons.
Bond breaking and bond forming
Bond breaking is endothermic: energy is absorbed to separate bonded atoms. Bond forming is exothermic: the same amount of energy is released when that bond forms. A reaction is exothermic overall when the energy released in forming the bonds of the products is greater than the energy absorbed in breaking the bonds of the reactants, and endothermic when it is less.
Average bond enthalpy
The bond enthalpy of a type of bond (e.g. C–H) averaged over a range of compounds. The energy needed to break a given bond depends on the other atoms and bonds in the molecule, so its value in any one compound differs slightly from the average. Average bond enthalpies refer to substances in the gaseous state.
Enthalpy change from average bond enthalpies
ΔH = Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds formed). Only bonds that actually change need be counted. The result is an estimate that may differ from an experimental value, because (1) the bond enthalpies are averages rather than the values in the particular compounds, and (2) they apply to gaseous species, so no enthalpy change of state is included when a reactant or product is a liquid or solid under the conditions of the experiment.
Students often think Chemical bonds store energy, and the energy is released when the bonds break, as when a fuel burns. In fact Absorbed. Breaking a bond requires energy to overcome the attraction between the bonded nuclei and the shared electrons, so bond breaking is always endothermic. Energy is released only when bonds form.
Students often think An exothermic reaction is one in which more bonds are formed than broken, so counting bonds shows whether energy is released. In fact No. What matters is the total bond enthalpy of the bonds formed compared with that of the bonds broken, not the number of bonds. When methane burns, six bonds are broken and six are formed, yet the reaction is strongly exothermic because the C=O and O–H bonds formed are stronger in total.
R1.2.2 Hess’s law
Hess’s law
The enthalpy change for a reaction is independent of the pathway between the initial and final states. Whether a reaction takes place in one step or several, and whatever the route, ΔH is the same provided the reactants and products, their amounts and physical states, and the conditions are the same. It is a consequence of the conservation of energy, and it allows enthalpy changes that cannot be measured directly to be calculated.
Combining equations in a Hess’s law calculation
Step equations are combined to give the target equation. Reversing an equation changes the sign of its ΔH; multiplying an equation by a factor multiplies its ΔH by the same factor; adding equations adds their ΔH values. Species that appear on both sides of the combined equation cancel, and the result must be exactly the target equation.
Students often think Each step of a route adds its own enthalpy change, so a route with more steps has a larger total enthalpy change. In fact No. By Hess’s law the enthalpy change depends only on the initial and final states. The individual steps have their own ΔH values, but they sum to the same total as the direct route.
Students often think A reaction that is faster or more vigorous releases more energy per mole of reaction. In fact No. How fast a reaction proceeds (kinetics) and how much energy it transfers per mole (energetics) are independent. A faster reaction releases the same energy in a shorter time.
R1.2.3 Standard conditions and standard states HL
Standard conditions and standard states
Standard conditions are a pressure of 100 kPa and a stated temperature, usually 298 K (and 1 mol dm⁻³ for solutions). The standard state of a substance is its pure, most stable form under these conditions, e.g. H₂O(l), Br₂(l), O₂(g), C(s, graphite). The symbol ⦵ shows that every substance is in its standard state.
Standard enthalpy change of formation, ΔH⦵f
The enthalpy change when 1 mol of a substance is formed from its elements in their standard states, under standard conditions. Units kJ mol⁻¹. Fractional coefficients are used where needed so that exactly 1 mol of the substance forms, e.g. ½N₂(g) + 1½H₂(g) → NH₃(g). For an element in its standard state, ΔH⦵f = 0 by definition.
Standard enthalpy change of combustion, ΔH⦵c
The enthalpy change when 1 mol of a substance is burned completely in oxygen, under standard conditions with all substances in their standard states. Units kJ mol⁻¹; always negative. Complete combustion of a compound of carbon, hydrogen and oxygen gives CO₂(g) and H₂O(l), e.g. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH⦵c = −890 kJ mol⁻¹.
Students often think Formation of a compound means forming it from its separate gaseous atoms, so the equation starts from C(g), H(g) and O(g). In fact No. The reactants are the elements in their standard states, e.g. C(s, graphite), H₂(g), O₂(g), not gaseous atoms.
Students often think A correct chemical equation must have whole-number coefficients, so the definitions refer to whatever amount the smallest whole-number equation gives, e.g. 2 mol of butane. In fact No. ΔH⦵f refers to 1 mol of the substance formed and ΔH⦵c to 1 mol of the substance burned, so fractional coefficients such as ½O₂(g) or 6½O₂(g) are used where needed.
R1.2.4 Enthalpy change of reaction from ΔH⦵f data HL
Enthalpy change of reaction from ΔH⦵f data
ΔH⦵ = ΣΔH⦵f(products) − ΣΔH⦵f(reactants), with each ΔH⦵f multiplied by the coefficient of that substance in the equation; elements in their standard states contribute zero. It is Hess’s law applied to a cycle in which the reactants and the products are both formed from the same elements.
Enthalpy change of reaction from ΔH⦵c data
ΔH⦵ = ΣΔH⦵c(reactants) − ΣΔH⦵c(products), with each ΔH⦵c multiplied by its coefficient. The order is the reverse of the ΔH⦵f equation because in this cycle the reactants and the products both burn to the same combustion products. Elements that burn, such as C(s) and H₂(g), have non-zero ΔH⦵c values; O₂(g) and fully oxidized substances such as H₂O(l) and CO₂(g) contribute zero.
Students often think Both kinds of data are used the same way round, so one equation (products minus reactants, or reactants minus products) is used for both. In fact No. With ΔH⦵f data, ΔH⦵ = ΣΔH⦵f(products) − ΣΔH⦵f(reactants). With ΔH⦵c data, ΔH⦵ = ΣΔH⦵c(reactants) − ΣΔH⦵c(products).
Students often think Elements always have enthalpy values of zero, so ΔH⦵c of an element such as carbon is zero in a calculation. In fact No. Zero applies to ΔH⦵f of an element in its standard state. Elements that burn have large, negative ΔH⦵c values, e.g. C(s, graphite) −393.5 and H₂(g) −285.8 kJ mol⁻¹.
R1.2.5 Born–Haber cycle HL
Born–Haber cycle
An energy cycle, an application of Hess’s law, that links the enthalpy of formation of an ionic compound to the steps that form it through gaseous ions: atomization of the metal, ionization of the metal atoms, atomization of the non-metal, electron gain by the non-metal atoms, and the lattice enthalpy. It is used to determine one value, often the lattice enthalpy, from all the others.
Enthalpy of atomization
The enthalpy change when 1 mol of gaseous atoms is formed from the element in its standard state, e.g. Na(s) → Na(g) (sublimation) or ½Cl₂(g) → Cl(g). Units kJ mol⁻¹; always positive. For a diatomic gaseous element it is half the bond enthalpy, because one bond gives two atoms.
First and second ionization energies
The first ionization energy is the energy required to remove 1 mol of electrons from 1 mol of gaseous atoms, M(g) → M⁺(g) + e⁻. The second ionization energy removes 1 mol of electrons from 1 mol of gaseous 1+ ions, M⁺(g) → M²⁺(g) + e⁻. Both are endothermic, so forming M²⁺(g) from M(g) requires IE1 + IE2. Units kJ mol⁻¹.
Electron affinity
The enthalpy change when 1 mol of electrons is added to 1 mol of gaseous atoms (first electron affinity, X(g) + e⁻ → X⁻(g)) or to 1 mol of gaseous 1− ions (second electron affinity, X⁻(g) + e⁻ → X²⁻(g)). First electron affinities of non-metals such as chlorine and oxygen are exothermic; second electron affinities are endothermic, because the incoming electron is repelled by the negative ion. Units kJ mol⁻¹.
Lattice enthalpy
In the IB convention, the enthalpy change when 1 mol of an ionic solid is separated into its gaseous ions, e.g. NaCl(s) → Na⁺(g) + Cl⁻(g). It is endothermic (positive) and measures the strength of the electrostatic attractions between all the ions in the lattice. Units kJ mol⁻¹. The reverse process, forming the solid from its gaseous ions, releases the same amount of energy.
Students often think Lattice enthalpy is the energy released when gaseous ions come together to form the lattice, so it is always negative. In fact No. In the IB convention lattice enthalpy is the enthalpy change for separating 1 mol of an ionic solid into its gaseous ions, e.g. NaCl(s) → Na⁺(g) + Cl⁻(g), which is endothermic, so its value is positive.
Students often think The atomization of chlorine is the bond enthalpy of Cl–Cl, so the full bond enthalpy is used for each mole of Cl atoms. In fact No. NaCl needs 1 mol of Cl(g) atoms, which is ½Cl₂(g) → Cl(g), so the enthalpy of atomization is half the Cl–Cl bond enthalpy: ½ × 242 = +121 kJ mol⁻¹.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Methane burns in oxygen: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). The reaction is exothermic. Which statement explains why?
Answer and reasoning
Forming the bonds in CO₂ and H₂O releases more energy than breaking the bonds in CH₄ and O₂ absorbs. — Breaking the C–H and O=O bonds absorbs energy; forming the C=O and O–H bonds releases energy. Because the bonds formed are stronger in total, more energy is released than absorbed, and the difference is transferred to the surroundings.
Breaking the bonds in CH₄ and O₂ releases the stored energy, which passes to the surroundings. — A student who thinks bonds store energy that is released when they break chooses this. Breaking any bond, in CH₄ or in O₂, absorbs energy; the energy released comes from forming the bonds in CO₂ and H₂O.
More bonds are formed in CO₂ and H₂O than are broken in CH₄ and O₂, so energy is released. — A student who judges by counting bonds chooses this. In fact 6 bonds are broken (4 C–H, 2 O=O) and 6 are formed (2 C=O, 4 O–H); the reaction is exothermic because the bonds formed are stronger, not more numerous.
The energy is stored in the CH₄; the O₂ only lets it be released and has no part in the change. — A student who sees the fuel as the energy store chooses this. The O=O bonds must be broken too, absorbing energy, and the energy released comes from forming bonds in CO₂ and H₂O, which contain the oxygen atoms.
2 An exothermic reaction P → Q takes place in a single step. With a catalyst, the same reaction takes place by a different route of three steps with a lower activation energy, and is much faster. Both routes start from the same reactants and give the same products under the same conditions. How does ΔH of the catalysed reaction compare with ΔH of the uncatalysed reaction?
Answer and reasoning
It is the same, because ΔH depends only on the initial and final states, not the route. — By Hess’s law the enthalpy change is independent of the pathway. The catalysed route has three steps and a lower activation energy, but it starts and ends at the same states, so the ΔH values of its steps sum to the same total.
It is smaller in size, because the catalyst lowers the activation energy of the reaction. — A student who treats the activation energy as part of ΔH chooses this. A catalyst lowers the energy barrier but leaves the energies of the reactants and products, and so ΔH, unchanged.
It is more negative, because each of the three steps adds its own enthalpy change to the total. — A student who thinks more steps means a larger energy change chooses this. The steps’ ΔH values (some may be positive) sum to the same total as the one-step route.
It is more negative, because a faster reaction transfers more energy to the surroundings per mole. — A student who links speed with the amount of energy chooses this. The catalysed reaction releases the same energy per mole, only in a shorter time.
3 Which equation represents the standard enthalpy change of formation, ΔH⦵f, of methanol, CH₃OH(l)? HL
Answer and reasoning
C(g) + 4H(g) + O(g) → CH₃OH(l) — A student who thinks formation starts from gaseous atoms chooses this. The reactants must be the elements in their standard states, C(s), H₂(g) and O₂(g); forming from atoms would include atomization energies.
2C(s) + 4H₂(g) + O₂(g) → 2CH₃OH(l) — A student who insists on whole-number coefficients chooses this. It forms 2 mol of methanol, so its enthalpy change is 2ΔH⦵f; ΔH⦵f refers to exactly 1 mol of product.
C(s) + 2H₂(g) + ½O₂(g) → CH₃OH(l) — ΔH⦵f is for 1 mol of the compound formed from its elements in their standard states: carbon as C(s) (graphite), hydrogen as H₂(g), oxygen as O₂(g). One mole of CH₃OH needs ½O₂(g), so a fractional coefficient is correct.
CO₂(g) + 2H₂O(l) → CH₃OH(l) + 1½O₂(g) — A student who takes 'formation' to mean any reaction that makes the compound chooses this. It is the overall equation for converting carbon dioxide and water into 1 mol of methanol, but its reactants are compounds and O₂ is a second product; ΔH⦵f starts from the elements C(s), H₂(g) and O₂(g) only, with the compound as the only product.
4 Use the data to calculate the standard enthalpy change of combustion of ethane, C₂H₆(g). ΔH⦵f / kJ mol⁻¹: C₂H₆(g) −84.0; CO₂(g) −393.5; H₂O(l) −285.8; H₂O(g) −241.8. HL
Answer and reasoning
+1560.4 kJ mol⁻¹ — A student who uses reactants − products, the order for ΔH⦵c data, with ΔH⦵f data gets +1560.4 kJ mol⁻¹. With ΔH⦵f data, ΔH⦵ = ΣΔH⦵f(products) − ΣΔH⦵f(reactants); a combustion cannot be endothermic.
−1428.4 kJ mol⁻¹ — A student who takes the water from combustion as steam uses H₂O(g): [2(−393.5) + 3(−241.8)] + 84.0 = −1428.4 kJ mol⁻¹. In the standard state water is H₂O(l).
−3120.8 kJ mol⁻¹ — A student who writes the equation with whole numbers, 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O, gets −3120.8 kJ for 2 mol of ethane. ΔH⦵c refers to 1 mol of ethane burned: −1560.4 kJ mol⁻¹.
−1560.4 kJ mol⁻¹ — C₂H₆(g) + 3½O₂(g) → 2CO₂(g) + 3H₂O(l). ΔH⦵c = ΣΔH⦵f(products) − ΣΔH⦵f(reactants) = [2(−393.5) + 3(−285.8)] − [−84.0 + 0] = −1644.4 + 84.0 = −1560.4 kJ mol⁻¹. O₂ is an element in its standard state, so ΔH⦵f(O₂) = 0.
5 Lattice enthalpy is defined here as the enthalpy change when 1 mol of an ionic solid is separated into its gaseous ions. The Born–Haber cycle for potassium chloride forms KCl(s) from K(s) and Cl₂(g) via K(g), Cl(g), K⁺(g) and Cl⁻(g). Data / kJ mol⁻¹: ΔH⦵f(KCl, s) −437; enthalpy of atomization of K(s) +89; first ionization energy of K +419; Cl–Cl bond enthalpy +242; first electron affinity of Cl −349. What is the lattice enthalpy of KCl? HL
Answer and reasoning
−717 kJ mol⁻¹ — A student who takes lattice enthalpy as the energy released when the lattice forms gives −717 kJ mol⁻¹. As defined here, lattice enthalpy is for separating the solid into gaseous ions, which absorbs energy, so its value is positive.
−157 kJ mol⁻¹ — A student who goes from KCl(s) back to K(s) and ½Cl₂(g) against the ΔH⦵f arrow but keeps its sign adds −437: 89 + 121 + 419 − 349 − 437 = −157 kJ mol⁻¹. Going against an arrow reverses the sign, so this step is +437, giving +717 kJ mol⁻¹.
+628 kJ mol⁻¹ — A student who applies the ionization energy directly to K(s) omits atomization: 121 + 419 − 349 + 437 = +628 kJ mol⁻¹. Ionization energy is for gaseous atoms, so K(s) → K(g) (+89) must be included.
6 Ethene reacts with hydrogen: C₂H₄(g) + H₂(g) → C₂H₆(g). Average bond enthalpies / kJ mol⁻¹: C=C 614; C–C 346; C–H 414; H–H 436. What is the enthalpy change of this reaction calculated from these data?
Answer and reasoning
+124 kJ mol⁻¹ — A student who calculates bonds formed − bonds broken gets 1174 − 1050 = +124 kJ mol⁻¹. Bond breaking absorbs energy, so ΔH = Σ(broken) − Σ(formed) = −124 kJ mol⁻¹.
+290 kJ mol⁻¹ — A student who counts only one C–H bond formed gets 1050 − (346 + 414) = +290 kJ mol⁻¹. Each H atom from H₂ forms its own C–H bond, so two C–H bonds (2 × 414) are formed.
−124 kJ mol⁻¹ — Bonds broken: C=C + H–H = 614 + 436 = 1050 kJ mol⁻¹. Bonds formed: C–C + 2 C–H = 346 + 2(414) = 1174 kJ mol⁻¹. ΔH = 1050 − 1174 = −124 kJ mol⁻¹. The four C–H bonds already in ethene are unchanged and can be left out.
−560 kJ mol⁻¹ — A student who treats H₂ as adding on intact leaves out the H–H bond broken: 614 − 1174 = −560 kJ mol⁻¹. The H–H bond (436 kJ mol⁻¹) must be broken for the H atoms to bond to carbon.
7 Use the enthalpy changes of reactions 1 and 2 to calculate ΔH for the decomposition of hydrogen peroxide: 2H₂O₂(l) → 2H₂O(l) + O₂(g). Reaction 1: H₂(g) + O₂(g) → H₂O₂(l), ΔH₁ = −187.8 kJ mol⁻¹. Reaction 2: 2H₂(g) + O₂(g) → 2H₂O(l), ΔH₂ = −571.6 kJ mol⁻¹.
Answer and reasoning
−947.2 kJ mol⁻¹ — A student who reverses and doubles reaction 1 but keeps its negative sign gets −571.6 + 2(−187.8) = −947.2 kJ mol⁻¹. Reversing an equation changes the sign of ΔH: −2ΔH₁ = +375.6 kJ mol⁻¹.
+196.0 kJ mol⁻¹ — A student who reverses reaction 2 instead of reaction 1 gets 2(−187.8) − (−571.6) = +196.0 kJ mol⁻¹, which is ΔH for the reverse reaction, forming H₂O₂ from H₂O and O₂. In the target, H₂O₂ is a reactant, so reaction 1 is the one to reverse.
−383.8 kJ mol⁻¹ — A student who reverses reaction 1 but does not double it gets −571.6 + 187.8 = −383.8 kJ mol⁻¹. The target has 2H₂O₂, so reversed reaction 1 must be multiplied by 2.
8 Which statement about the standard enthalpy change of combustion of butane, ΔH⦵c(C₄H₁₀), is correct? HL
Answer and reasoning
It includes the energy supplied to ignite the butane, as well as that released. — A student who treats the activation energy as part of the enthalpy change chooses this. The energy supplied to ignite the fuel overcomes the activation energy and is returned as the products form; ΔH⦵c depends only on the reactant and product states.
It refers to 1 mol of butane burned, so the equation includes 6½O₂(g). — ΔH⦵c is for 1 mol of the substance burned completely with all substances in their standard states: C₄H₁₀(g) + 6½O₂(g) → 4CO₂(g) + 5H₂O(l). For exactly 1 mol of butane, O₂ needs the fractional coefficient 6½.
It refers to the water formed as H₂O(g), since the flame is far hotter than 100 °C. — A student who thinks of the conditions in the flame chooses this. A standard enthalpy change compares substances in their standard states; water is H₂O(l) under standard conditions.
It includes any reaction of butane with oxygen, whether CO or CO₂ forms. — A student who counts incomplete burning as combustion chooses this. ΔH⦵c refers to complete combustion, giving CO₂(g); forming CO releases less energy and would give a different value.
9 Carbon cannot be burned to carbon monoxide without some carbon dioxide also forming, so ΔH⦵f(CO) is found indirectly. ΔH⦵c / kJ mol⁻¹: C(s, graphite) −393.5; CO(g) −283.0. What is ΔH⦵f of CO(g)? HL
Answer and reasoning
−110.5 kJ mol⁻¹ — ΔH⦵f(CO) is ΔH⦵ for C(s) + ½O₂(g) → CO(g). With ΔH⦵c data, ΔH⦵ = ΣΔH⦵c(reactants) − ΣΔH⦵c(products) = −393.5 − (−283.0) = −110.5 kJ mol⁻¹. O₂ does not burn, so it contributes zero.
+110.5 kJ mol⁻¹ — A student who uses products − reactants, the order for ΔH⦵f data, gets −283.0 − (−393.5) = +110.5 kJ mol⁻¹. With ΔH⦵c data the order is reactants − products.
+283.0 kJ mol⁻¹ — A student who sets ΔH⦵c of the element carbon to zero gets 0 − (−283.0) = +283.0 kJ mol⁻¹. 'Elements are zero' applies to ΔH⦵f only; carbon burns, and ΔH⦵c(C) = −393.5 kJ mol⁻¹.
−676.5 kJ mol⁻¹ — A student who adds both combustion values, following the CO combustion arrow backwards without changing its sign, gets −393.5 + (−283.0) = −676.5 kJ mol⁻¹. Going from CO₂ back to CO reverses the combustion of CO, contributing +283.0 kJ mol⁻¹.
Working Target: C(s) + ½O₂(g) → CO(g). ΔH⦵ = ΣΔH⦵c(reactants) − ΣΔH⦵c(products) = [ΔH⦵c(C) + ½ × 0] − ΔH⦵c(CO) = −393.5 − (−283.0) = −110.5 kJ mol⁻¹. Cycle: C + O₂ → CO₂ (−393.5) equals C + ½O₂ → CO (ΔH⦵f) followed by CO + ½O₂ → CO₂ (−283.0), so ΔH⦵f = −393.5 + 283.0.
10 Lattice enthalpy is defined here as the enthalpy change when 1 mol of an ionic solid is separated into its gaseous ions. Data / kJ mol⁻¹ for the Born–Haber cycle of magnesium chloride: ΔH⦵f(MgCl₂, s) −641; enthalpy of atomization of Mg(s) +147; first ionization energy of Mg +738; second ionization energy of Mg +1451; enthalpy of atomization of chlorine, ½Cl₂(g) → Cl(g), +121; first electron affinity of Cl −349. What is the lattice enthalpy of MgCl₂? HL
Answer and reasoning
+2749 kJ mol⁻¹ — A student who counts the chlorine steps once, as for NaCl, gets 147 + 738 + 1451 + 121 − 349 + 641 = +2749 kJ mol⁻¹. MgCl₂ contains two Cl⁻ ions, so the atomization and electron affinity of chlorine are each needed twice.
+1783 kJ mol⁻¹ — A student who takes the second ionization energy as the whole energy for Mg(g) → Mg²⁺(g) omits IE1: 147 + 1451 + 242 − 698 + 641 = +1783 kJ mol⁻¹. IE2 is Mg⁺(g) → Mg²⁺(g) + e⁻; both IE1 and IE2 are needed.
+3917 kJ mol⁻¹ — A student who treats electron gain as endothermic enters +349 for each Cl: 147 + 738 + 1451 + 242 + 698 + 641 = +3917 kJ mol⁻¹. The first electron affinity of chlorine is exothermic, −349 kJ mol⁻¹.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
3 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 Using average bond enthalpies (C–H 414, O=O 498, C=O 804, O–H 463 kJ mol⁻¹), a student calculates ΔH = −808 kJ mol⁻¹ for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). The experimental enthalpy change for burning methane under standard conditions (100 kPa, 298 K) is −890 kJ mol⁻¹. Which statement explains why the experimental value is more exothermic than the calculated one?
Answer and reasoning
Heat is lost to the surroundings in the experiment, which makes the measured value more exothermic. — A student who explains every discrepancy by heat loss chooses this. Heat loss lowers the temperature rise, so it makes a measured exothermic value less negative, not more.
The measured value also includes the activation energy supplied to ignite the gas, which is then released. — A student who treats the activation energy as part of ΔH chooses this. Energy supplied to overcome the activation energy is returned as the products form; ΔH depends only on the reactant and product states.
Breaking the C–H and O=O bonds releases energy, which the calculation wrongly treats as absorbed. — A student who thinks bond breaking releases energy chooses this. Breaking bonds always absorbs energy, so the calculation is right to enter them as positive terms; treating them as releases would make the calculated value far more negative than −890 kJ mol⁻¹.
At 298 K the water forms as H₂O(l); its condensation releases energy that gas-phase bond enthalpies omit. — Bond enthalpies apply only to gases, so the calculation gives ΔH for forming H₂O(g). Under standard conditions water is a liquid, and condensing 2 mol of water vapour releases extra energy, making the experimental value more negative. (The bond enthalpies being averages also contributes a small difference.)
2 Why is the C–H bond enthalpy in a table of bond enthalpies described as an average value?
Answer and reasoning
It is the mean of repeated measurements made on a single compound, taken to reduce the random error in the data. — A student who reads 'average' as in a practical write-up chooses this. The averaging is over the same bond in different compounds, not over repeat trials on one compound.
The energy to break a C–H bond varies from one compound to another, so a mean over many compounds is given. — A C–H bond in methane, ethane or ethanol has a slightly different bond enthalpy because the rest of the molecule affects it. The tabulated value is the mean over many compounds, which is why calculations with it give estimates.
Values measured in the solid, liquid and gas are pooled into one mean, as the same C–H bonds are present in all three. — A student who thinks bond enthalpies apply in any state, because the same bonds are present in solid, liquid and gas, chooses this. Bond enthalpies are defined for the gaseous state only; changes of state involve intermolecular forces, which are not included, and the averaging is over different compounds.
Every C–H bond has the same strength; 'average' only means the single measured value has been rounded. — A student who thinks each bond type has one fixed strength chooses this. The strength of a C–H bond depends on its molecular environment, so values differ between compounds.
3 Magnesium oxide, MgO(s), has ΔH⦵f = −602 kJ mol⁻¹. Which statement explains why the formation of MgO(s) from its elements is exothermic? HL
Answer and reasoning
The lattice enthalpy of MgO is so large that more energy is released as the gaseous ions form the solid than is absorbed in forming them. — Forming Mg²⁺(g) and O²⁻(g) from the elements is strongly endothermic overall. The small, doubly charged ions attract all their neighbours strongly, so the lattice enthalpy is very large; the energy released when the gaseous ions form the solid exceeds the energy absorbed, making ΔH⦵f negative.
Mg²⁺ and O²⁻ each gain a stable full outer shell of electrons, and reaching this arrangement is what releases the energy. — A student who credits the octet with releasing energy chooses this. Forming these ions absorbs energy: Mg(g) → Mg²⁺(g) needs IE1 + IE2 = +2189 kJ mol⁻¹, and O(g) → O²⁻(g) is endothermic overall because the second electron affinity is endothermic. The energy comes from the gaseous ions forming the lattice.
Each Mg²⁺ ion bonds only to the O²⁻ ion that took its electrons, and forming that one ionic bond releases the energy. — A student who pictures an ionic bond between electron-transfer partners chooses this. In the lattice each ion attracts several oppositely charged neighbours, and it is these attractions summed over the whole lattice that release the energy.
Breaking the O=O bond when oxygen is atomized releases energy, which outweighs the energy absorbed in ionizing Mg. — A student who thinks bond breaking releases energy chooses this. Atomizing oxygen absorbs energy (half the O=O bond enthalpy), adding to the endothermic steps rather than offsetting them.
That was your twenty minutes. Real practice on R1.2 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Chemistry guide (first assessment 2025) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
DP Science Cafe