Enthalpy of Formation/Combustion, Calorimetry q = mcΔT, Hess's Law & Born-Haber Cycles
Chemical energetics quantifies the heat energy exchanged at constant pressure (ΔH) when chemical or ionic bonds are broken and formed under standard conditions (100 kPa and 298 K). In the laboratory, enthalpy changes of combustion and neutralisation are measured via spirit-burner or polystyrene-cup calorimetry using q = mcΔT and ΔH = -q / (1000 × n), accounting for systematic heat loss and cooling curve extrapolation. When direct measurement is impossible, Hess's Law states that total enthalpy change is independent of route, enabling calculation from standard enthalpies of formation (ΔfH°), combustion (ΔcH°), or Born-Haber lattice enthalpy cycles. Examiners also test why differences between experimental Born-Haber lattice enthalpies and theoretical electrostatic (perfectly ionic) lattice enthalpies reveal covalent character caused by cation polarisation.
- Standard Conditions & Enthalpy Definitions (ΔfH°, ΔcH°, ΔatH°)
- Standard conditions: pressure = 100 kPa, temperature = 298 K (25°C), solutions at 1.0 mol dm⁻³, all substances in their standard states. Standard Enthalpy of Formation (ΔfH°): enthalpy change when 1 mole of a compound is formed from its elements in their standard states (ΔfH° of any element in its standard state = 0 kJ mol⁻¹). Standard Enthalpy of Combustion (ΔcH°): enthalpy change when 1 mole of a substance is completely burned in excess oxygen under standard conditions. Standard Enthalpy of Atomisation (ΔatH°): enthalpy change when 1 mole of gaseous atoms is formed from an element in its standard state (e.g. ½Cl₂(g) → Cl(g)).
- Calorimetry Calculations: q = mcΔT & ΔH = ±q / (1000n)
- 1. Calculate heat energy absorbed/released by the water or solution in Joules: q = m × c × ΔT, where m = mass of water or solution being heated in grams (never the mass of fuel or solid reactant), c = specific heat capacity of water (4.18 J g⁻¹ K⁻¹), and ΔT = temperature change in K or °C. 2. Convert q from Joules to kilojoules (÷ 1000) and divide by the moles n of the limiting reactant/fuel burned: ΔH = -q / (1000 × n) (include a minus sign if temperature rose, i.e. exothermic; plus sign if temperature fell).
- Hess's Law Shortcut Formulas (Formation vs Combustion vs Bond Enthalpy)
- Hess's Law states that the enthalpy change of a chemical reaction is independent of the route taken, provided initial and final conditions are the same. 1. Given Formation data (arrows point UP from elements): ΔrH° = ΣΔfH°(Products) - ΣΔfH°(Reactants). 2. Given Combustion data (arrows point down to combustion products): ΔrH° = ΣΔcH°(Reactants) - ΣΔcH°(Products). 3. Given Mean Bond Enthalpies (gaseous state only): ΔrH = Σ(Bonds Broken in Reactants) - Σ(Bonds Formed in Products).
Enthalpy of Formation/Combustion, Calorimetry q = mcΔT, Hess's Law & Born-Haber Cycles: Reaction Profile (ΔH & E_a) + Le Chatelier Yield Simulator
Toggle a catalyst on the enthalpy profile diagram or adjust temperature and pressure to see how dynamic equilibrium shifts.
Modelled on N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (ΔH = −92 kJ/mol) (4 moles gas → 2 moles gas):
Higher pressure shifts equilibrium to the side with fewer gas moles (right, 4 → 2). Lower temperature favours the exothermic forward reaction.
Key Concepts & Exam Definitions (6)
- Standard Conditions & Enthalpy Definitions (ΔfH°, ΔcH°, ΔatH°)
- Standard conditions: pressure = 100 kPa, temperature = 298 K (25°C), solutions at 1.0 mol dm⁻³, all substances in their standard states. Standard Enthalpy of Formation (ΔfH°): enthalpy change when 1 mole of a compound is formed from its elements in their standard states (ΔfH° of any element in its standard state = 0 kJ mol⁻¹). Standard Enthalpy of Combustion (ΔcH°): enthalpy change when 1 mole of a substance is completely burned in excess oxygen under standard conditions. Standard Enthalpy of Atomisation (ΔatH°): enthalpy change when 1 mole of gaseous atoms is formed from an element in its standard state (e.g. ½Cl₂(g) → Cl(g)).
- Calorimetry Calculations: q = mcΔT & ΔH = ±q / (1000n)
- 1. Calculate heat energy absorbed/released by the water or solution in Joules: q = m × c × ΔT, where m = mass of water or solution being heated in grams (never the mass of fuel or solid reactant), c = specific heat capacity of water (4.18 J g⁻¹ K⁻¹), and ΔT = temperature change in K or °C. 2. Convert q from Joules to kilojoules (÷ 1000) and divide by the moles n of the limiting reactant/fuel burned: ΔH = -q / (1000 × n) (include a minus sign if temperature rose, i.e. exothermic; plus sign if temperature fell).
- Hess's Law Shortcut Formulas (Formation vs Combustion vs Bond Enthalpy)
- Hess's Law states that the enthalpy change of a chemical reaction is independent of the route taken, provided initial and final conditions are the same. 1. Given Formation data (arrows point UP from elements): ΔrH° = ΣΔfH°(Products) - ΣΔfH°(Reactants). 2. Given Combustion data (arrows point down to combustion products): ΔrH° = ΣΔcH°(Reactants) - ΣΔcH°(Products). 3. Given Mean Bond Enthalpies (gaseous state only): ΔrH = Σ(Bonds Broken in Reactants) - Σ(Bonds Formed in Products).
- Born-Haber Cycles & Lattice Enthalpy (ΔLEH°)
- Lattice Enthalpy of Formation is the exothermic enthalpy change when 1 mole of a solid ionic compound is formed from its gaseous ions: M⁺(g) + X⁻(g) → MX(s) (Lattice Enthalpy of Dissociation is the exact reverse endothermic process: MX(s) → M⁺(g) + X⁻(g)). A Born-Haber cycle equates ΔfH°(MX(s)) to the sum of: ΔatH°(metal) + ΔatH°(non-metal) + IE₁(+ IE₂) of metal + EA₁(+ EA₂) of non-metal + ΔLEH°(formation). Note: First Electron Affinity (EA₁, X(g) + e⁻ → X⁻(g)) is exothermic, whereas Second Electron Affinity (EA₂, e.g. O⁻(g) + e⁻ → O²⁻(g)) is endothermic due to repulsion between the negative ion and incoming electron.
- Theoretical (Perfectly Ionic) vs Experimental (Born-Haber) Lattice Enthalpy
- Theoretical lattice enthalpy assumes the crystal is 100% purely ionic with spherical, non-polarisable point charges. If the experimental Born-Haber lattice enthalpy is significantly larger in magnitude (more exothermic/endothermic) than the theoretical value (e.g. in AgI or MgCl₂), the compound has significant covalent character: a small, highly charged cation (high charge density, e.g. Mg²⁺, Al³⁺, Ag⁺) distorts (polarises) the electron cloud of a large, easily polarisable anion (e.g. I⁻, Br⁻), adding covalent bonding strength.
- Enthalpy of Solution (ΔsolH°) & Enthalpy of Hydration (ΔhydH°)
- Enthalpy of Hydration (ΔhydH°): exothermic enthalpy change when 1 mole of gaseous ions dissolves in excess water to form 1 mole of aqueous ions at infinite dilution: M⁺(g) + aq → M⁺(aq) (more exothermic for smaller ions with higher charge because they attract δ⁻ O / δ⁺ H of dipolar water molecules more strongly). Enthalpy of Solution (ΔsolH°): enthalpy change when 1 mole of an ionic solid dissolves in excess water: ΔsolH° = ΔLEH°(dissociation, positive) + ΣΔhydH°(gaseous ions, negative).
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Write the state-symbol equations for the standard enthalpy of formation of liquid ethanol (C₂H₅OH) and the standard enthalpy of atomisation of bromine.
State the three main sources of error in a spirit-burner combustion calorimetry experiment (apart from 'human error').
Why does the enthalpy of combustion calculated using mean bond enthalpies differ from the experimental standard enthalpy of combustion in a data book?
Why is the first electron affinity of oxygen exothermic (-141 kJ mol⁻¹), whereas the second electron affinity of oxygen is strongly endothermic (+798 kJ mol⁻¹)?
Why is the lattice dissociation enthalpy of MgO (+3791 kJ mol⁻¹) much greater than that of NaCl (+787 kJ mol⁻¹)?
What does it mean if the experimental Born-Haber lattice enthalpy of silver iodide (AgI) is 889 kJ mol⁻¹ whereas the theoretical (perfect ionic model) value is only 777 kJ mol⁻¹?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
In an experiment, 0.920 g of ethanol (C₂H₅OH, Mr = 46.0) was burned in a spirit burner underneath a copper calorimeter containing 150 cm³ of water. The temperature of the water rose from 21.5°C to 48.5°C. Calculate the experimental enthalpy of combustion of ethanol in kJ mol⁻¹ (c = 4.18 J g⁻¹ K⁻¹) and explain how a cooling curve graph is used to improve accuracy in polystyrene cup calorimetry. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Temperature change ΔT = 48.5 - 21.5 = 27.0°C (27.0 K); mass of water m = 150 g [1 mark]. 2. Heat absorbed by water q = mcΔT = 150 × 4.18 × 27.0 = 16,929 J = 16.929 kJ [1 mark]. 3. Moles of ethanol burned n = mass ÷ Mr = 0.920 ÷ 46.0 = 0.0200 mol [1 mark]. 4. Experimental ΔcH = -q ÷ n = -16.929 ÷ 0.0200 = -846 kJ mol⁻¹ (must include negative sign for exothermic reaction) [1 mark]. 5. In cooling curve extrapolation, temperature is recorded every minute for 3 minutes before mixing reactants at t = 4 min, and for 6-10 minutes after; extrapolating the linear cooling line back to the vertical line at t = 4 min gives the theoretical maximum temperature rise (ΔT) corrected for heat loss during the reaction [1 mark].
- Q2
Given the following standard enthalpies of formation (in kJ mol⁻¹): ΔfH°[C₃H₈(g)] = -104, ΔfH°[CO₂(g)] = -394, and ΔfH°[H₂O(l)] = -286. Write the balanced equation for the complete combustion of propane and use Hess's Law to calculate the standard enthalpy of combustion of propane. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Balanced equation: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l) (note that ΔfH°[O₂(g)] = 0 kJ mol⁻¹) [1 mark]. 2. State Hess's Law formula for formation enthalpies: ΔcH° = ΣΔfH°(Products) - ΣΔfH°(Reactants) [1 mark]. 3. Calculate ΣΔfH°(Products) = (3 × -394) + (4 × -286) = -1182 + (-1144) = -2326 kJ mol⁻¹ [1 mark]. 4. Calculate ΔcH° = -2326 - (-104) = -2222 kJ mol⁻¹ [1 mark].
- Q3
Use the following data (in kJ mol⁻¹) to calculate the lattice enthalpy of formation of magnesium chloride, MgCl₂(s): ΔfH°[MgCl₂(s)] = -642; ΔatH°[Mg(s)] = +148; IE₁[Mg(g)] = +738; IE₂[Mg(g)] = +1451; ΔatH°[Cl(g)] = +122; EA₁[Cl(g)] = -349. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Write the Born-Haber equation remembering there are 2 moles of Cl atoms and 2 moles of Cl⁻ ions per mole of MgCl₂: ΔfH°[MgCl₂(s)] = ΔatH°[Mg] + IE₁[Mg] + IE₂[Mg] + (2 × ΔatH°[Cl]) + (2 × EA₁[Cl]) + ΔLEH°[MgCl₂] [1 mark]. 2. Substitute the gaseous ion formation steps: Sum of endothermic/exothermic steps to form Mg²⁺(g) + 2Cl⁻(g) = +148 + 738 + 1451 + (2 × 122) + (2 × -349) [1 mark]. 3. Simplify sum = +2337 + 244 - 698 = +1883 kJ mol⁻¹ [1 mark]. 4. Rearrange for ΔLEH°: -642 = +1883 + ΔLEH° → ΔLEH° = -642 - 1883 = -2525 kJ mol⁻¹ [1 mark].
- Q4
The lattice dissociation enthalpy of calcium chloride, CaCl₂(s), is +2258 kJ mol⁻¹. The enthalpy of hydration of Ca²⁺(g) is -1650 kJ mol⁻¹ and the enthalpy of hydration of Cl⁻(g) is -364 kJ mol⁻¹. Calculate the enthalpy of solution of CaCl₂(s) and explain why the enthalpy of hydration of Ca²⁺ is much more exothermic than that of K⁺ (-322 kJ mol⁻¹). (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Write the expression for enthalpy of solution: ΔsolH°[CaCl₂] = ΔLEH°(dissociation) + ΔhydH°[Ca²⁺(g)] + (2 × ΔhydH°[Cl⁻(g)]) [1 mark]. 2. Substitute values: ΔsolH° = +2258 + (-1650) + (2 × -364) = +2258 - 1650 - 728 = -120 kJ mol⁻¹ [1 mark]. 3. Ca²⁺ has a 2+ charge whereas K⁺ only has a 1+ charge, and Ca²⁺ has a smaller ionic radius than K⁺ (higher charge density) [1 mark]. 4. Therefore, Ca²⁺ forms much stronger ion-dipole attractions with the δ⁻ oxygen atoms of surrounding dipolar water molecules, releasing more energy on hydration [1 mark].
Common Exam Questions (4)
How do I remember whether to do Products minus Reactants or Reactants minus Products in Hess's Law?
Use the acronym 'F-P-R, C-R-P, B-B-F': Formation data = Products minus Reactants; Combustion data = Reactants minus Products; Bond enthalpies = Broken (reactants) minus Formed (products). Even better, always draw the 3-corner Hess's cycle and follow the arrows.
Why do I have to multiply ΔatH°[Cl] and EA₁[Cl] by 2 in the Born-Haber cycle for MgCl₂, when bond dissociation enthalpy of Cl₂ is not multiplied by 2?
Because standard enthalpy of atomisation (ΔatH°) is defined per 1 mole of gaseous atoms formed (½Cl₂(g) → Cl(g)), so forming 2 moles of Cl(g) requires 2 × ΔatH°. In contrast, bond dissociation enthalpy (E(Cl-Cl)) is defined per 1 mole of Cl₂ molecules broken (Cl₂(g) → 2Cl(g)), which already produces 2 moles of Cl(g) atoms. Always check whether the question gives ΔatH° or Bond Dissociation Enthalpy.
In a neutralisation calorimetry experiment (e.g. 25 cm³ of 1.0 M HCl + 25 cm³ of 1.0 M NaOH), what mass 'm' do I put into q = mcΔT?
You must use the total combined volume/mass of the aqueous solution after mixing: 25 cm³ + 25 cm³ = 50 cm³ = 50 g (assuming solution density = 1.00 g cm⁻³), because the heat released by the reaction heats up the entire 50 cm³ of solution.
Why is the standard enthalpy of neutralisation of any strong acid with any strong alkali always approximately -57 kJ mol⁻¹?
Strong acids (e.g. HCl, HNO₃) and strong alkalis (e.g. NaOH, KOH) are 100% dissociated into aqueous spectator ions in solution, so the only actual chemical reaction occurring in every case is the formation of water: H⁺(aq) + OH⁻(aq) → H₂O(l). Weak acids give a slightly less exothermic value because endothermic energy is required to ionise the un-dissociated weak acid molecules.
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