R1.2 Energy cycles in reactions | IB DP Chemistry

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🎯What you need to be able to do

Calculate the enthalpy change of a reaction from average bond enthalpy data.
Explain why bond enthalpy data are average values and may differ from those measured experimentally.
State Hess’s law and apply it to calculate enthalpy changes in multistep reactions.
AHL Deduce equations and solve problems involving standard enthalpies of combustion and of formation.
AHL Calculate enthalpy changes using \( \Delta H^{\ominus} = \Sigma \Delta H_{\mathrm{f}}^{\ominus}(\text{products}) - \Sigma \Delta H_{\mathrm{f}}^{\ominus}(\text{reactants}) \) and the corresponding combustion expression.
AHL Interpret a Born–Haber cycle and determine values from it for compounds of univalent and divalent ions.

📚The chemistry

Bond breaking and bond forming

Two statements, and getting them the right way round is worth remembering for the rest of the course:

Bond breaking is endothermic
energy must be supplied to overcome the attraction holding the atoms together

Bond forming is exothermic
energy is released as the attraction is established

The overall enthalpy change is the balance:

\[ \Delta H = \Sigma(\text{bonds broken}) - \Sigma(\text{bonds formed}) \]

If the bonds formed are stronger than the bonds broken, more energy is released than absorbed and the reaction is exothermic. That is the whole of combustion in one sentence: the C=O and O–H bonds in the products are much stronger than the C–C, C–H and O=O bonds broken.

“Broken minus formed” is the direction that matters. Reverse it and every sign in your answer is wrong. Fix it by reasoning rather than memorising: breaking costs energy (positive), forming releases it (negative), so the sum is (energy in) − (energy out). And note that bond enthalpies are always quoted as positive values — the sign comes from which side of the subtraction they sit on, not from the data.

Why bond enthalpies are averages

The data booklet lists average bond enthalpies, and the syllabus wants you to be able to say why they are averages and what follows.

A C–H bond in methane is not identical to a C–H bond in ethanol or in benzene: the strength depends on what else is attached to the carbon and on the rest of the molecule. Even within methane, the four C–H bonds require different energies to break one after another. The tabulated value is therefore a mean taken over many different compounds.

Three consequences:

A value calculated from average bond enthalpies is approximate, and will differ somewhat from the experimental \( \Delta H \).
The method applies strictly to substances in the gaseous state, because it ignores intermolecular forces. If a reactant or product is a liquid, the calculation omits the enthalpy of vaporization and the answer is further out.
It cannot be used at all for species where the bonding is not simple covalent — you cannot compute the enthalpy of formation of sodium chloride from bond enthalpies. And it fails badly for delocalised systems: benzene is a standard example, because the delocalization energy is not captured by any table of localised C–C and C=C values.

Hess’s law

The enthalpy change for a reaction is independent of the pathway between the initial and final states. It follows directly from the conservation of energy: if two routes from the same reactants to the same products gave different enthalpy changes, you could run one forwards and the other backwards and create energy from nothing.

In practice Hess’s law is what lets you find enthalpy changes that cannot be measured directly — the enthalpy of formation of carbon monoxide, for instance, which cannot be made cleanly from its elements without also forming \( \mathrm{CO_2} \).

The mechanics: write the target equation, then manipulate the given equations until they add to it. When you reverse an equation, change the sign of its \( \Delta H \); when you multiply an equation by a factor, multiply its \( \Delta H \) by the same factor. Then add.

AHL Standard enthalpies of formation and combustion

\( \Delta H_{\mathrm{f}}^{\ominus} \)
the enthalpy change when one mole of a compound is formed from its elements in their standard states.
It is zero for any element in its standard state, by definition.

\( \Delta H_{\mathrm{c}}^{\ominus} \)
the enthalpy change when one mole of a substance undergoes complete combustion in excess oxygen.
It is always negative.

Both sets of data are in the booklet, and each gives a route to \( \Delta H \) for any reaction. The two expressions look confusingly similar and the order is reversed between them, so learn the reason rather than the letters:

\[ \Delta H^{\ominus} = \Sigma \Delta H_{\mathrm{f}}^{\ominus}(\text{products}) - \Sigma \Delta H_{\mathrm{f}}^{\ominus}(\text{reactants}) \]

\[ \Delta H^{\ominus} = \Sigma \Delta H_{\mathrm{c}}^{\ominus}(\text{reactants}) - \Sigma \Delta H_{\mathrm{c}}^{\ominus}(\text{products}) \]

Why they differ: formation arrows point from the elements up to both reactants and products, so the elements are the common floor and you go up to the products and back down from the reactants — products minus reactants. Combustion arrows point down from both reactants and products to the same combustion products, so the common floor is at the bottom and the subtraction reverses. Sketching the cycle takes ten seconds and removes the need to remember which is which.

Remember to multiply each value by the stoichiometric coefficient in the balanced equation before summing.

AHL Born–Haber cycles

A Born–Haber cycle is Hess’s law applied to the formation of an ionic compound. It exists because lattice enthalpy cannot be measured directly — you cannot take a mole of solid sodium chloride and pull it apart into gaseous ions in a calorimeter — so it is obtained indirectly from quantities that can be measured.

The steps of the cycle, all of which the syllabus names:

Enthalpy of atomization — converting the elements to gaseous atoms (sublimation for a metal, half a bond enthalpy for a diatomic gas). Endothermic.
Ionization energy — removing electrons from the gaseous metal atoms; successive values for a divalent ion. Endothermic.
Electron affinity — adding electrons to the gaseous non-metal atoms. First one exothermic; any second one endothermic, because you are forcing an electron onto a species that is already negative.
Lattice enthalpy — the gaseous ions coming together into the solid lattice (or the reverse, depending on the definition used; be consistent, and read the sign the question implies).
Enthalpy of formation — the direct route from elements to compound, which closes the cycle.

The guide states explicitly that the construction of a complete Born–Haber cycle will not be assessed. What is assessed is your ability to interpret one you are given and determine a missing value from it — usually the lattice enthalpy, by equating the direct and indirect routes.

✏️Worked example

(a) Use the average bond enthalpies below to calculate \( \Delta H \) for the complete combustion of methane: \( \mathrm{CH_4}(g) + 2\mathrm{O_2}(g) \rightarrow \mathrm{CO_2}(g) + 2\mathrm{H_2O}(g) \). C–H 414, O=O 498, C=O 804, O–H 463 kJ mol⁻¹.
(b) The accepted \( \Delta H_{\mathrm{c}}^{\ominus} \) of methane is −891 kJ mol⁻¹. Give two reasons for the difference.
(c) Use Hess’s law and the following data to find \( \Delta H \) for \( \mathrm{C}(s) + \tfrac{1}{2}\mathrm{O_2}(g) \rightarrow \mathrm{CO}(g) \):
\( \mathrm{C}(s) + \mathrm{O_2}(g) \rightarrow \mathrm{CO_2}(g) \), \( \Delta H = -394\ \mathrm{kJ\,mol^{-1}} \);
\( \mathrm{CO}(g) + \tfrac{1}{2}\mathrm{O_2}(g) \rightarrow \mathrm{CO_2}(g) \), \( \Delta H = -283\ \mathrm{kJ\,mol^{-1}} \).
(d) AHL Calculate \( \Delta H^{\ominus} \) for \( \mathrm{C_2H_5OH}(l) + 3\mathrm{O_2}(g) \rightarrow 2\mathrm{CO_2}(g) + 3\mathrm{H_2O}(l) \) using \( \Delta H_{\mathrm{f}}^{\ominus} \): ethanol −278, \( \mathrm{CO_2} \) −394, \( \mathrm{H_2O}(l) \) −286 kJ mol⁻¹.

(a) Count the bonds carefully on each side.

Broken: 4 × C–H in methane, plus 2 × O=O.

\[ \Sigma(\text{broken}) = 4(414) + 2(498) = 1656 + 996 = 2652\ \mathrm{kJ\,mol^{-1}} \]

Formed: 2 × C=O in \( \mathrm{CO_2} \), plus 4 × O–H (each water has two O–H bonds, and there are two waters).

\[ \Sigma(\text{formed}) = 2(804) + 4(463) = 1608 + 1852 = 3460\ \mathrm{kJ\,mol^{-1}} \]

\[ \Delta H = 2652 - 3460 = -808\ \mathrm{kJ\,mol^{-1}} \]

(b) Two reasons: the values used are average bond enthalpies taken over many compounds, so they do not describe the specific bonds in methane and carbon dioxide exactly; and the calculation assumes all species are gaseous, whereas the accepted \( \Delta H_{\mathrm{c}}^{\ominus} \) refers to liquid water. Condensing the water would release additional energy, making the true value more exothermic — which is the direction of the discrepancy here.

(c) The target has CO on the right, and the second given equation has it on the left, so reverse that equation and change the sign of its \( \Delta H \):

\[ \mathrm{CO_2}(g) \rightarrow \mathrm{CO}(g) + \tfrac{1}{2}\mathrm{O_2}(g), \qquad \Delta H = +283\ \mathrm{kJ\,mol^{-1}} \]

Add it to the first equation. The \( \mathrm{CO_2} \) cancels and one of the \( \tfrac{1}{2}\mathrm{O_2} \) cancels against the \( \mathrm{O_2} \):

\[ \Delta H = -394 + 283 = -111\ \mathrm{kJ\,mol^{-1}} \]

(d) Use the formation expression, remembering that \( \Delta H_{\mathrm{f}}^{\ominus}(\mathrm{O_2}) = 0 \) because oxygen is an element in its standard state, and multiplying by the coefficients:

\[ \Delta H^{\ominus} = \left[2(-394) + 3(-286)\right] - \left[(-278) + 3(0)\right] \]

\[ = (-788 - 858) - (-278) = -1646 + 278 = -1368\ \mathrm{kJ\,mol^{-1}} \]

Which matches the accepted enthalpy of combustion of ethanol, −1367 kJ mol⁻¹, to within rounding.

Check it. Every combustion of a hydrocarbon or an alcohol must come out negative and large — hundreds to thousands of kJ mol⁻¹. A small or positive answer means a sign error or a missed coefficient. In (a) the commonest slip is counting two O–H bonds instead of four; check by counting atoms: two water molecules contain four hydrogens, each bonded to oxygen, so four O–H bonds. In (d), check that oxygen contributed nothing — if a value for \( \mathrm{O_2} \) appears anywhere in your working, you have missed the definition of standard enthalpy of formation.

Not multiplying by the stoichiometric coefficients. In (d) it is \( 2 \times (-394) \) and \( 3 \times (-286) \), not \( -394 \) and \( -286 \). This single omission changes −1368 to −402, and because the answer is still negative and still looks like an enthalpy of combustion, nothing about it announces the error. The habit that prevents it: write the balanced equation first, then write the coefficient in front of every value before you do any arithmetic.

📝Practise

Work through these on paper, then reveal the answer.

1. Explain why bond breaking is endothermic and bond forming exothermic, and state the relationship used to calculate \( \Delta H \) from bond enthalpies.

A covalent bond exists because of an electrostatic attraction between the shared pair of electrons and the two positively charged nuclei. To separate the atoms this attraction must be overcome, which requires energy to be supplied to the system — so bond breaking is endothermic. Conversely, when atoms come together and a bond is established, the system moves to a lower energy state and the surplus energy is released to the surroundings — so bond forming is exothermic. The overall enthalpy change is the balance of the two: \( \Delta H = \Sigma(\text{bond enthalpies of bonds broken}) - \Sigma(\text{bond enthalpies of bonds formed}) \). If the bonds formed are stronger in total, more energy is released than absorbed and \( \Delta H \) is negative. Note that tabulated bond enthalpies are always positive numbers; the direction is supplied by which side of the subtraction they occupy.

2. Calculate \( \Delta H \) for \( \mathrm{N_2}(g) + 3\mathrm{H_2}(g) \rightarrow 2\mathrm{NH_3}(g) \) using bond enthalpies: N≡N 945, H–H 436, N–H 391 kJ mol⁻¹.

Bonds broken: one N≡N and three H–H. \( \Sigma = 945 + 3(436) = 945 + 1308 = 2253\ \mathrm{kJ\,mol^{-1}} \). Bonds formed: each \( \mathrm{NH_3} \) contains three N–H bonds and there are two molecules, so six N–H bonds. \( \Sigma = 6(391) = 2346\ \mathrm{kJ\,mol^{-1}} \). \( \Delta H = 2253 - 2346 = \mathbf{-93\ kJ\,mol^{-1}} \). The reaction is exothermic, consistent with the accepted value of about −92 kJ mol⁻¹, and this modest exothermicity is exactly why the Haber process is run at a compromise temperature — see R2.3. The error to avoid is writing \( 2 \times 391 \) instead of \( 6 \times 391 \).

3. State Hess's law and explain why it follows from the conservation of energy. Give one enthalpy change that can only be found using it.

Hess’s law: the enthalpy change for a reaction is independent of the pathway between the initial and final states, depending only on those states. It follows from the conservation of energy by a simple argument: suppose two different routes from the same reactants to the same products released different amounts of energy. You could then run the more exothermic route forwards and the less exothermic route backwards, returning the system to exactly where it started while leaving a net surplus of energy behind — energy created from nothing, which is impossible. Therefore the two routes must give identical enthalpy changes. An enthalpy change that cannot be measured directly: the enthalpy of formation of carbon monoxide, because burning carbon in a limited supply of oxygen always produces a mixture of CO and \( \mathrm{CO_2} \), so the heat measured does not correspond to a single reaction. Other good answers: the enthalpy of formation of an unstable compound, or the lattice enthalpy of an ionic solid, which is why Born–Haber cycles exist.

4. Use these data to calculate \( \Delta H \) for \( 2\mathrm{C}(s) + 3\mathrm{H_2}(g) \rightarrow \mathrm{C_2H_6}(g) \): \( \Delta H_{\mathrm{c}}^{\ominus} \) of C(s) = −394, of H₂(g) = −286, of C₂H₆(g) = −1560 kJ mol⁻¹.

Use the combustion expression, in which reactants come first: \( \Delta H^{\ominus} = \Sigma \Delta H_{\mathrm{c}}^{\ominus}(\text{reactants}) - \Sigma \Delta H_{\mathrm{c}}^{\ominus}(\text{products}) \). Reactants: \( 2 \times (-394) + 3 \times (-286) = -788 - 858 = -1646 \). Products: \( 1 \times (-1560) = -1560 \). \( \Delta H = -1646 - (-1560) = \mathbf{-86\ kJ\,mol^{-1}} \). Two checks. First, this is an enthalpy of formation — one mole of a compound from its elements in their standard states — and the accepted \( \Delta H_{\mathrm{f}}^{\ominus} \) of ethane is −84 kJ mol⁻¹, so the answer is right. Second, note the reversed order compared with the formation expression: here the common products (\( \mathrm{CO_2} \) and \( \mathrm{H_2O} \)) lie at the bottom of the cycle rather than the elements lying at the top, which is precisely why the subtraction runs the other way.

5. AHL Calculate \( \Delta H^{\ominus} \) for \( \mathrm{CaCO_3}(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO_2}(g) \) given \( \Delta H_{\mathrm{f}}^{\ominus} \): CaCO₃ −1207, CaO −635, CO₂ −394 kJ mol⁻¹. Comment on the sign.

\( \Delta H^{\ominus} = \Sigma \Delta H_{\mathrm{f}}^{\ominus}(\text{products}) - \Sigma \Delta H_{\mathrm{f}}^{\ominus}(\text{reactants}) = \left[(-635) + (-394)\right] - \left[-1207\right] = -1029 + 1207 = \mathbf{+178\ kJ\,mol^{-1}} \). The value is positive, so the thermal decomposition of calcium carbonate is endothermic — which is why limestone must be heated strongly and continuously in a lime kiln, and why the reaction does not occur spontaneously at room temperature. It is worth noticing that this reaction nevertheless does go at high temperature despite being endothermic, because the large positive entropy change from producing a gas eventually dominates: see R1.4.

6. AHL A Born–Haber cycle for NaCl gives: enthalpy of atomization of Na +107, first ionization energy of Na +496, atomization of ½Cl₂ +122, electron affinity of Cl −349, and \( \Delta H_{\mathrm{f}}^{\ominus}(\mathrm{NaCl}) \) −411 kJ mol⁻¹. Determine the lattice enthalpy, defined as the energy released when gaseous ions form the solid.

Hess’s law says the direct route (elements → solid NaCl, which is \( \Delta H_{\mathrm{f}}^{\ominus} \)) must equal the indirect route (elements → gaseous atoms → gaseous ions → solid lattice). So \( \Delta H_{\mathrm{f}}^{\ominus} = \text{atomization(Na)} + \text{IE(Na)} + \text{atomization(Cl)} + \text{EA(Cl)} + \Delta H_{\text{lattice}} \). Substituting: \( -411 = (+107) + (+496) + (+122) + (-349) + \Delta H_{\text{lattice}} \). The four known terms sum to \( +376 \), so \( \Delta H_{\text{lattice}} = -411 - 376 = \mathbf{-787\ kJ\,mol^{-1}} \). The sign is negative because, on this definition, energy is released as the gaseous ions come together into the lattice. Two checks: the magnitude should be large — several hundred kJ mol⁻¹ — because ionic bonding is strong, and it must be numerically larger than the sum of the endothermic steps, or the compound would never form at all. Be careful with the definition: some sources define lattice enthalpy as the energy required to separate the lattice into gaseous ions, in which case the same number is quoted as \( +787 \). Read the wording the question gives you.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

Your data booklet — the average bond enthalpy table, the enthalpies of formation and combustion, and both Hess’s law expressions. Everything on this page except the reasoning is given to you in the examination.
RSC Learn Chemistry — the classic Hess’s law practical determining the enthalpy change of the thermal decomposition of sodium hydrogencarbonate indirectly, via two reactions with acid.
Khan Academy — a slower treatment of Born–Haber cycles than the syllabus requires, useful if the arrangement of the steps is not yet obvious.