Ch 1 · Lattice energy and Born–Haber cycles · AS & A Level Chemistry 9701
Topic 1 of 4 in Chemical energetics — AS & A Level Chemistry 9701 — 58 lessons.
Introduction
The lattice energy of an ionic solid cannot be measured directly, since it is impossible to form a crystal lattice from separate gaseous ions in a calorimeter. A Born–Haber cycle is a Hess's Law energy cycle that links this inaccessible quantity to enthalpy changes that can be measured, allowing lattice energy to be calculated indirectly and used to explain why some ionic solids are more stable than others.
Key ideas — Enthalpy of atomisation and lattice energy
The standard enthalpy change of atomisation, Δ H_at^ominus, is the enthalpy change when one mole of gaseous atoms is formed from an element in its standard state under standard conditions, for example Na(s) arrow Na(g). It is always endothermic, since bonds or metallic/lattice forces must be broken. The lattice energy, Δ H_latt^ominus, is the enthalpy change when one mole of solid ionic compound is formed from its constituent gaseous ions under standard conditions, for example Na^+(g) + Cl^-(g) arrow NaCl(s). Lattice energy is always exothermic (a negative value), since bringing oppositely charged gaseous ions together into a lattice releases electrostatic energy.
🧑🏫 Teacher's explanation
Imagine you are building a Lego castle. To get the individual bricks ready, you first have to take them out of their packaging and snap them apart; this requires energy (atomisation). Once you have the individual bricks, you snap them together to form the sturdy castle structure. The energy released when those bricks lock together is like lattice energy.
📖 Definitions
🔍 Worked example — Easy
Calculate the enthalpy change of atomisation for Magnesium, given that the enthalpy change of formation of MgO(s) is -602 kJ mol^-1, the lattice energy of MgO(s) is -3791 kJ mol^-1, the enthalpy of sublimation of Mg(s) is +148 kJ mol^-1, and the enthalpy of ionisation of Mg(g) is +739 kJ mol^-1, and the enthalpy of ionisation of O(g) is +1314 kJ mol^-1 and the first electron affinity of O(g) is -141 kJ mol^-1 and the second electron affinity of O(g) is +773 kJ mol^-1. (Note: Use the cycle to find Δ H_at for Mg)
🔍 Worked example — Standard
A hypothetical ionic compound AB has a lattice energy of -750 kJ mol^-1. The enthalpy of formation of AB(s) is -800 kJ mol^-1. The enthalpy of atomisation of A(s) is +150 kJ mol^-1 and the enthalpy of atomisation of B(s) is +100 kJ mol^-1. Calculate the sum of the first electron affinity of B and the first ionisation energy of A.
🔍 Worked example — Exam-style
Explain, in terms of electrostatic forces, why the lattice energy of MgO is significantly more exothermic than the lattice energy of NaCl.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — First electron affinity and its trends
The first electron affinity is the enthalpy change when one mole of gaseous atoms each gains one electron to form one mole of gaseous 1− ions, e.g. Cl(g) + e^- arrow Cl^-(g). It is exothermic for Group 16 and 17 atoms, since the incoming electron is attracted to the nucleus. Down Group 17 (F → Cl → Br → I) and down Group 16 (O → S → Se), first electron affinity generally becomes less exothermic, because atomic radius increases and shielding by inner shells increases, weakening the nuclear attraction for the incoming electron despite the rising nuclear charge. Fluorine is anomalous: its first electron affinity is less exothermic than chlorine's, because the incoming electron enters a very small 2p orbital already crowded with electrons, and inter-electron repulsion partly offsets the strong nuclear attraction.
🧑🏫 Teacher's explanation
Think of electron affinity like a person trying to sit down in a crowded bus seat. If the seat is empty and the person is close to the center (the nucleus), they settle in easily and release energy (exothermic). If the seat is already occupied by many people (small atoms like Fluorine), they have to push and shove to fit, which creates repulsion and makes the process less 'comfortable' or energetic.
📖 Definitions
🔍 Worked example — Easy
Calculate the enthalpy change for the reaction: Cl(g) + e^- arrow Cl^-(g) if the first electron affinity of Chlorine is -364 kJ mol^-1.
🔍 Worked example — Standard
Given the following data for Sodium Chloride: Δ H_atom Na = +104 kJ mol^-1, Δ H_atom Cl = +122 kJ mol^-1, IE_Na = +496 kJ mol^-1, EA_Cl = -364 kJ mol^-1, and Δ H_lattice = -787 kJ mol^-1. Calculate the enthalpy change of formation (Δ H_f) for NaCl(s).
🔍 Worked example — Exam-style
Explain why the magnitude of the lattice energy for MgO(s) is significantly greater than that of NaCl(s).
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
\ extLi(s) arrow Li(g) Δ H = +161 kJ mol^-1 F(s) arrow F(g) Δ H = +129 kJ mol^-1 Li(g) arrow Li^+(g) Δ H = +520 kJ mol^-1 F(g) + e^- arrow F^-(g) Δ H = -328 kJ mol^-1
Key ideas — Constructing a Born–Haber cycle
A Born–Haber cycle for an ionic solid MX starts from the elements in their standard states and ends at the solid lattice, with the standard enthalpy of formation, Δ H_f^ominus, as the direct route. The indirect route passes through gaseous atoms and then gaseous ions: atomisation of the metal, the appropriate ionisation energy or energies of the metal (a second ionisation energy is needed for a 2+ cation), atomisation of the non-metal, the appropriate electron affinity or affinities of the non-metal (a second, endothermic electron affinity is needed for a 2− anion), and finally the lattice energy.
🧑🏫 Teacher's explanation
Think of a Born-Haber cycle like a financial accounting ledger for a chemical reaction. If you want to know the total cost of building a complex Lego castle (the ionic lattice), you can either look at the receipt for the finished castle (the enthalpy of formation) or you can add up the individual costs of every single brick, the glue used to stick them together, and the energy spent assembling them (atomisation, ionisation, and electron affinity).
📖 Definitions
🔍 Worked example — Easy
Calculate the lattice energy (Δ H_latt^ominus) for NaCl(s) given the following data: Δ H_f^ominus [NaCl(s)] = -414 kJ mol^-1, Δ H_at [Na(s) arrow Na(g)] = +102 kJ mol^-1, Δ H_at [Cl_2(g) arrow Cl(g)] = +121 kJ mol^-1, IE_1 [Na(g)] = +496 kJ mol^-1, and EA_1 [Cl(g)] = -339 kJ mol^-1.
🔍 Worked example — Standard
A compound MgCl_2 has a lattice energy of -2540 kJ mol^-1. Calculate the enthalpy of formation Δ H_f^ominus given: Δ H_at(Mg) = +148 kJ mol^-1, IE_1(Mg) = +739 kJ mol^-1, IE_2(Mg) = +1450 kJ mol^-1, Δ H_at(Cl_2) = +122 kJ mol^-1, EA_1(Cl) = -349 kJ mol^-1, and EA_2(Cl) = +352 kJ mol^-1.
🔍 Worked example — Exam-style
Explain qualitatively why the lattice energy of MgO is significantly more exothermic than that of NaCl.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Calculating lattice energy from a Born–Haber cycle
By Hess's Law, the sum of enthalpy changes around the indirect route equals Δ H_f^ominus:
🧑🏫 Teacher's explanation
Think of a Born-Haber cycle as a financial accounting system for chemical energy. Imagine you are building a Lego castle. You can either buy the finished castle (the enthalpy of formation) or you can buy the individual bricks, glue them together, and then assemble the castle piece by piece (the sum of atomisation, ionisation, and electron affinity energies). Hess's Law tells us that the total cost is the same regardless of whether you buy the pre-built set or the individual bricks.
📖 Definitions
🔍 Worked example — Easy
Calculate the lattice energy (Δ H_latt^ominus) for NaCl(s) given: Δ H_f^ominus = -414 kJ mol^-1, Δ H_at^ominus(Na) = +102 kJ mol^-1, IE_1(Na) = +496 kJ mol^-1, Δ H_at^ominus(Cl) = +122 kJ mol^-1, and EA_1(Cl) = -339 kJ mol^-1.
🔍 Worked example — Standard
Calculate the lattice energy of MgCl_2(s) given: Δ H_f^ominus = -648 kJ mol^-1, Δ H_at^ominus(Mg) = +148 kJ mol^-1, IE_1(Mg) = +739 kJ mol^-1, IE_2(Mg) = +1450 kJ mol^-1, Δ H_at^ominus(Cl) = +122 kJ mol^-1, and EA_1(Cl) = -339 kJ mol^-1.
🔍 Worked example — Exam-style
A hypothetical ionic compound AX_2 has Δ H_f^ominus = -800 kJ mol^-1. The values for A are Δ H_at^ominus = +200 kJ mol^-1 and IE_1 = +500 kJ mol^-1, IE_2 = +1000 kJ mol^-1. The values for X are Δ H_at^ominus = +150 kJ mol^-1 and EA_1 = -300 kJ mol^-1. Calculate Δ H_latt^ominus.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Effect of ionic charge and radius on lattice energy
Lattice energy arises from the electrostatic attraction between oppositely charged ions, which by Coulomb's law is stronger when the ionic charges are larger and when the ions are closer together. A higher ionic charge increases the magnitude (more exothermic value) of the lattice energy, because the electrostatic force between ions of charge +2 and -2 is far greater than between ions of charge +1 and -1. A smaller ionic radius also increases the magnitude of lattice energy, because it allows the ions to pack closer together, reducing the inter-ionic distance and strengthening the attraction.
🧑🏫 Teacher's explanation
Imagine two magnets. If you have two small, powerful magnets, they snap together with a very strong force. If you replace them with two much larger, weaker magnets, they won't pull together nearly as strongly. Lattice energy works exactly like this, but with electric charges instead of magnetism.
📖 Definitions
🔍 Worked example — Easy
Predict whether the lattice energy of MgO is expected to be more or less exothermic than NaCl based on ionic charge and radius.
🔍 Worked example — Standard
Calculate the lattice energy (Δ H_latt) for NaCl using the following data: Δ H_atom, Na = +102 kJ mol^-1, Δ H_atom, Cl = +336 kJ mol^-1, IE_1(Na) = +496 kJ mol^-1, EA_1(Cl) = -339 kJ mol^-1, Δ H_formation(NaCl) = -413.5 kJ mol^-1.
🔍 Worked example — Exam-style
Explain why the lattice energy of CaCl_2 is significantly higher than that of NaCl using the concept of ionic charge.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked example
Question. Use the data below to calculate the lattice energy of magnesium chloride, MgCl_2. [5]
🔍 Worked example — Easy
Calculate the lattice energy of NaCl using the following data: Δ H_f^ominus[NaCl] = -413 kJ mol^-1; Δ H_at^ominus[Na] = +107 kJ mol^-1; IE_1[Na] = +496 kJ mol^-1; Δ H_at^ominus[Cl] = +122 kJ mol^-1; EA_1[Cl] = -349 kJ mol^-1.
🔍 Worked example — Standard
Calculate the lattice energy of MgO given: Δ H_f^ominus[MgO] = -602 kJ mol^-1; Δ H_at^ominus[Mg] = +148 kJ mol^-1; IE_1[Mg] = +738 kJ mol^-1; IE_2[Mg] = +1451 kJ mol^-1; Δ H_at^ominus[O] = +249 kJ mol^-1; EA_1[O] = -141 kJ mol^-1; EA_2[O] = -798 kJ mol^-1.
🔍 Worked example — Exam-style
Using the data from the card: Δ H_at^ominus[Mg] = +148 kJ mol^-1; IE_1[Mg] = +738 kJ mol^-1; IE_2[Mg] = +1451 kJ mol^-1; Δ H_at^ominus[Cl] = +121 kJ mol^-1; EA_1[Cl] = -349 kJ mol^-1; Δ H_f^ominus[MgCl_2] = -641 kJ mol^-1. Calculate the lattice energy of MgCl_2.
⚠️ Common errors
Common mistakes
Practice
Answers