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9476 / 2027
Chemical Energetics

Lesson 3 of 5

Hess cycles and bond energies

Choose a common endpoint and follow every sign and coefficient.

A-Level 9476 (2026-2027)

  • I can define the specified enthalpy terms and standard conditions.Syllabus

    Syllabus 9476, 7(c). Define the specified enthalpy terms and standard conditions.

  • I can explain lattice-energy magnitudes qualitatively.Syllabus

    Syllabus 9476, 7(e). Explain lattice-energy magnitudes qualitatively.

  • I can construct Hess and Born-Haber cycles and calculate energies.Syllabus

    Syllabus 9476, 7(f). Construct Hess and Born-Haber cycles and calculate energies.

Hess Law compares routes with identical endpoints

Enthalpy is a state function, so a direct route and an indirect route have the same total change.

Hess Law: the enthalpy change of a reaction depends on its initial and final states, not the route taken. Write the target equation first. Reverse any auxiliary equation that runs the wrong way and change its sign; multiply its enthalpy when multiplying its coefficients. Cancel identical species only when their states also match.

Worked exampleFormation enthalpy from combustion data

Find the formation enthalpy of C2H5OH(l). Use combustion enthalpies C(graphite) = -394, H2(g) = -286 and ethanol(l) = -1367 kJ mol-1, all giving CO2(g) and H2O(l).

Try it first. Then show the worked stepsHide the worked steps
  1. Target: 2C(graphite) + 3H2(g) + 1/2 O2(g) → C2H5OH(l).
  2. Combusting the elements to 2CO2(g) + 3H2O(l) releases 2(-394) + 3(-286) = -1646 kJ.
  3. The alternative route is formation of ethanol followed by its combustion: delta Hf - 1367 = -1646.
  4. Rearrange without changing the endpoints.
Answer

ΔHf = -279 kJ mol-1 for the supplied rounded data.

With formation data, ΔHreaction = ΣνΔHf(products) - ΣνΔHf(reactants), where ν is the stoichiometric coefficient. Elements in their standard states have standard formation enthalpy zero by definition. This does not mean they contain no energy or have zero atomisation enthalpy.

Worked exampleEstimate a gas-phase reaction using mean bond energies

Estimate ΔH for H2(g) + Cl2(g) → 2HCl(g), using H-H = 436, Cl-Cl = 243 and H-Cl = 431 kJ mol-1.

Try it first. Then show the worked stepsHide the worked steps
  1. Break one H-H and one Cl-Cl bond: energy input = 436 + 243 = 679 kJ.
  2. Form two H-Cl bonds: energy released = 2 × 431 = 862 kJ.
  3. Subtract energy released from energy absorbed.
Answer

ΔH ≈ 679 - 862 = -183 kJ mol-1 of reaction as written.

Try it step by step

Work it out in steps: Energy absorbed breaking bonds, then Energy released forming bonds, then Enthalpy change.

  1. Problem 1. Estimate delta H for H2(g) + Br2(g) -> 2HBr(g). Use H-H = 436, Br-Br = 193 and H-Br = 366 kJ mol-1. Give each answer as a whole number of kJ.
  2. Problem 2. Estimate delta H for CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g). Use C-H = 410, O=O = 496, C=O = 805 and O-H = 460 kJ mol-1. Give each answer as a whole number of kJ.
  3. Problem 3. Estimate delta H for N2(g) + 3H2(g) -> 2NH3(g). Use N-N triple bond = 944, H-H = 436 and N-H = 390 kJ mol-1. Give each answer as a whole number of kJ.
Show the answer

Problem 1.

Energy absorbed breaking bonds: Break one H-H and one Br-Br: 436 + 193 = 629 kJ absorbed.

Energy released forming bonds: Form two H-Br: 2 x 366 = 732 kJ released.

Enthalpy change: delta H = 629 - 732 = -103 kJ mol-1.

Problem 2.

Energy absorbed breaking bonds: Break four C-H and two O=O: 4(410) + 2(496) = 2632 kJ absorbed.

Energy released forming bonds: Form two C=O and four O-H: 2(805) + 4(460) = 3450 kJ released.

Enthalpy change: delta H = 2632 - 3450 = -818 kJ mol-1.

Problem 3.

Energy absorbed breaking bonds: Break one N-N triple bond and three H-H: 944 + 3(436) = 2252 kJ absorbed.

Energy released forming bonds: Form six N-H, three in each NH3: 6(390) = 2340 kJ released.

Enthalpy change: delta H = 2252 - 2340 = -88 kJ mol-1.

The bond-energy route passes through separated gaseous atoms. Include extra phase-change terms if the target uses liquids or solids. Mean bond energies give estimates because bond strength depends on molecular environment; formation enthalpies for the exact substances and states can give a more specific value.

Build an ionic solid through gaseous atoms and ions

Born-Haber cycles keep atomisation, ionisation, electron affinity and lattice formation distinct.

A first electron affinity is the enthalpy change when one mole of gaseous atoms each gains an electron to form gaseous singly negative ions: Cl(g) + e- → Cl-(g). It is often negative. Adding an electron to an already negative ion, as in O-(g) + e- → O2-(g), requires energy to overcome repulsion and is positive. Ionisation energies always refer to removing electrons from gaseous species and are positive.

The indirect route from Na(s) + 1/2 Cl2(g) to NaCl(s)
  1. Create gaseous atoms

    Na(s) → Na(g): atomisation +108. 1/2 Cl2(g) → Cl(g): atomisation +122 kJ per mole of NaCl formed.

  2. Create gaseous ions

    Na(g) → Na+(g) + e-: first ionisation +496. Cl(g) + e- → Cl-(g): electron affinity -349.

  3. Assemble the lattice

    Na+(g) + Cl-(g) → NaCl(s): the unknown lattice energy, L.

Construct the two routes of a Born-Haber cycle

The direct left-hand route forms solid sodium chloride from solid sodium and half a mole of chlorine molecules, with enthalpy minus 411 kilojoules per mole. The indirect route first makes gaseous atoms, then removes an electron from sodium, transfers it to chlorine, and forms the ionic lattice. Both routes have exactly the same initial and final states.

A reaction-route diagram, not a vertical energy scale. All arrow values are in kJ mol-1 of NaCl. To construct your own cycle, write the direct formation equation, insert the gaseous atom and ion states, label every arrow, then equate the two route totals.

Worked exampleCalculate the lattice energy with the syllabus convention

Using the cycle above and ΔHf(NaCl,s) = -411 kJ mol-1, find L.

Try it first. Then show the worked stepsHide the worked steps
  1. The two routes give -411 = 108 + 122 + 496 - 349 + L.
  2. The non-lattice steps total +377 kJ mol-1.
  3. L = -411 - 377 = -788 kJ mol-1.
Answer

L = -788 kJ mol-1. Reversing the final step to separate the lattice would require +788 kJ mol-1.

For MgO, include Mg atomisation, half an O=O bond dissociation, both the first and second ionisation energies of Mg, and both electron affinities of oxygen. Each coefficient follows the target formula. A highly exothermic lattice-formation step can compensate for costly ion formation; do not omit a positive second electron affinity to make a cycle look easier.

A simple electrostatic model predicts a larger lattice-energy magnitude for greater ionic charge product and smaller interionic separation. Thus MgO has a much more negative lattice energy than NaCl, while MgO is generally more negative than CaO because Mg2+ is smaller. Use these comparisons for similar structures; appreciable covalent character can cause departures from a purely ionic model.

Worked exampleDissolution separates the lattice, then hydrates the ions

Using L = -788 kJ mol-1 and total hydration enthalpy -784 kJ mol-1 for Na+ and Cl-, estimate the solution enthalpy of NaCl.

Try it first. Then show the worked stepsHide the worked steps
  1. Separating NaCl(s) into gaseous ions reverses lattice formation, so the change is +788.
  2. Hydrating both gaseous ions releases 784 kJ mol-1.
  3. Add the two steps: delta Hsol = -L + sum(delta Hhyd).
Answer

ΔHsol = +4 kJ mol-1. A slightly endothermic dissolution can still occur because enthalpy is only one contribution to free energy.

Exam-style question

Explain why the lattice energy of magnesium oxide is much more exothermic than that of sodium chloride. [2]

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