Thermochemistry

🟢 Lite — Quick Review (1h–1d)

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Thermochemistry tracks the heat energy exchanged between a reaction system and its surroundings, expressed as the enthalpy change (ΔH) measured in kJ mol⁻¹. An exothermic reaction releases heat (ΔH is negative), while an endothermic reaction absorbs heat (ΔH is positive). The workhorse formula for calorimetry experiments is Q = mcΔT, where m is the mass of the solution (in g), c the specific heat capacity (≈ 4.2 J g⁻¹ K⁻¹ for water), and ΔT the temperature rise. To convert Q to molar enthalpy, use ΔH = mcΔT ÷ n, where n is the number of moles that reacted. Two exam essentials: Hess’s Law says ΔH is independent of the reaction pathway, and ΔH_f° of any element in its standard state is zero at 298 K.

🟡 Standard — Regular Study (2d–2mo)

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Definitions and Sign Convention

The enthalpy change (ΔH) is the heat exchanged at constant pressure. By IUPAC convention, energy released to the surroundings carries a negative sign (exothermic, e.g. combustion of methane: ΔH = −890 kJ mol⁻¹), while energy absorbed from the surroundings carries a positive sign (endothermic, e.g. thermal decomposition of calcium carbonate). Thermochemical equations must always show the physical state of every species (s, l, g, aq) because phase changes themselves involve heat.

Standard Enthalpies and Calorimetry

The standard enthalpy of formation (ΔH_f°) is the heat change when 1 mole of a compound is formed from its elements in their standard states (1 atm, 298 K). By definition, ΔH_f° of every element in its standard state = 0 — this is the reference point from which all other enthalpies are calculated. The standard enthalpy of combustion (ΔH_c°) is the heat released when 1 mole of a substance burns completely in excess O₂. In the laboratory, an insulated calorimeter (e.g. a copper calorimeter with a stirrer and thermometer) measures the temperature change; heat absorbed by the water or solution is Q = mcΔT, and the molar enthalpy is ΔH = −Q/n for exothermic reactions.

Hess’s Law and Cycle Calculations

Hess’s Law states that the total enthalpy change depends only on the initial and final states, not the route taken. This lets you calculate otherwise unmeasurable ΔH values. The three key working formulas are:

From formation enthalpies: ΔH_rxn = Σ ΔH_f°(products) − Σ ΔH_f°(reactants)
From combustion enthalpies: ΔH_rxn = Σ ΔH_c°(reactants) − Σ ΔH_c°(products)
From bond enthalpies: ΔH_rxn = Σ BE(bonds broken) − Σ BE(bonds formed)

Because energy is absorbed when bonds break and released when bonds form, reactions that form stronger bonds than they break are exothermic.

Neutralisation

For a strong acid + strong base in dilute solution, the enthalpy of neutralisation is approximately −57.1 kJ mol⁻¹, representing H⁺(aq) + OH⁻(aq) → H₂O(l). Values are smaller (less negative) for weak acids or weak bases because additional energy is needed to ionise them first.

Exam Pattern

NABTEB Chemistry typically tests thermochemistry through structured calculation questions (10–15 marks) on calorimetry, formation/combustion enthalpy tables, and Hess’s law cycle diagrams. Multiple-choice items commonly probe sign convention and the ΔH_f° = 0 rule for elements.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Lattice, Hydration, and Solution Enthalpies

For dissolution of an ionic solid, the enthalpy of solution (ΔH_sol) combines two opposing steps. A Born–Haber-style Hess cycle gives the relationship:

ΔH_solution = ΔH_lattice energy + ΔH_hydration

Lattice energy is endothermic when defined as the energy required to separate the solid into gaseous ions (or exothermic when defined as the formation of the solid from gaseous ions). Hydration is exothermic because ion–dipole interactions form with water molecules. When |ΔH_hydration| > |ΔH_lattice|, dissolution is exothermic and the solution warms; the reverse gives a cold-feeling salt like NH₄NO₃.

Bond Enthalpy vs ΔH_f° Calculations

Average bond dissociation enthalpies are tabulated as mean values over many compounds, so ΔH calculated from bond enthalpies is an estimate — typically within ±10 kJ mol⁻¹ of the value calculated from ΔH_f° (which is exact for the specific compound). Use bond enthalpies only when formation data is unavailable, and remember to multiply each bond energy by the number of such bonds in the balanced equation.

Calorimetry Pitfalls and Corrections

In a real calorimeter, heat loss to the vessel and surroundings means the measured ΔT is lower than the theoretical value. NABTEB questions may provide a calorimeter constant (C) so the corrected expression is Q = (mc + C)ΔT. Always quote ΔH in kJ mol⁻¹, converting from J by dividing by 1000, and remember to assign the negative sign explicitly for exothermic processes.

Common Mistakes

Forgetting that ΔH_f° of an element in its standard state is zero, not the element’s actual bonding energy.
Confusing system vs surroundings when assigning signs (heat leaving the system → ΔH negative).
Using ΔH_c° in the formation formula (should be combustion formula) or vice versa.
Forgetting to balance thermochemical equations — multiplying an equation by 2 also doubles its ΔH.

Worked Example

A 0.50 g sample of methanol (CH₃OH, M = 32 g mol⁻¹) is burned in a calorimeter containing 200 g of water. The temperature rises from 25.0 °C to 37.5 °C. With c = 4.2 J g⁻¹ K⁻¹:

Q = (200)(4.2)(12.5) = 10 500 J = 10.5 kJ n = 0.50 ÷ 32 = 0.0156 mol ΔH_c° = −10.5 ÷ 0.0156 = −673 kJ mol⁻¹

The accepted value is −726 kJ mol⁻¹; the discrepancy reflects heat loss to the calorimeter walls.

Practice Prompts

Calculate ΔH_rxn for the formation of ethanol C₂H₅OH(l) from C(s, graphite) and H₂(g), given ΔH_f°(ethanol) = −277 kJ mol⁻¹. (Answer: −277 kJ mol⁻¹)
Using bond enthalpies H–H = 436, O=O = 498, O–H = 463 kJ mol⁻¹, estimate ΔH for 2H₂(g) + O₂(g) → 2H₂O(g). (Answer: bonds broken = 2(436) + 498 = 1370; bonds formed = 4(463) = 1852; ΔH ≈ −482 kJ)

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Thermochemistry