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Where does the energy of a reaction come from and how is it measured?

Enthalpy changes, exothermic and endothermic reactions, calorimetry, standard enthalpies of formation and combustion, Hess's law and bond enthalpies.

A focused answer to WJEC A-Level Chemistry Unit 2, covering enthalpy changes, exothermic and endothermic reactions, calorimetry, standard enthalpies of formation and combustion, Hess's law cycles and bond-enthalpy calculations.

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What this dot point is asking

WJEC wants you to define enthalpy changes, classify reactions as exothermic or endothermic, measure enthalpy by calorimetry, and calculate enthalpy changes using Hess's law (from formation or combustion data) and from bond enthalpies.

The answer

Enthalpy and its sign

Calorimetry

Hess's law

Bond enthalpies

Standard enthalpy definitions

WJEC expects precise definitions, each "per mole" and under standard conditions. The standard enthalpy of formation is the enthalpy change when one mole of a compound forms from its elements in their standard states. The standard enthalpy of combustion is the enthalpy change when one mole of a substance burns completely in oxygen. The standard enthalpy of neutralisation is the enthalpy change when an acid and base react to form one mole of water. Quoting the exact wording, including "one mole" and "standard states", is often worth a mark in itself.

Choosing the right Hess cycle

Which Hess cycle you draw depends on the data given. With formation enthalpies, arrows point up from the elements to both reactants and products, so ΔH=ΣΔHf(products)ΣΔHf(reactants)\Delta H = \Sigma \Delta H_f(\text{products}) - \Sigma \Delta H_f(\text{reactants}). With combustion enthalpies, arrows point down to the combustion products, so the formula reverses: ΔH=ΣΔHc(reactants)ΣΔHc(products)\Delta H = \Sigma \Delta H_c(\text{reactants}) - \Sigma \Delta H_c(\text{products}). Matching the cycle to the type of data given is the single most important decision in these calculations.

Examples in context

Comparing fuels. Calorimetry of ethanol, petrol and hydrogen lets engineers compare energy released per gram and per mole, informing the choice of transport fuels in the wider-impact content. Designing cold packs. Endothermic dissolving of ammonium nitrate in water powers instant cold packs, a deliberate use of a positive ΔH\Delta H.

Try this

Q1. State whether a reaction with ΔH=286\Delta H = -286 kJ mol1^{-1} is exothermic or endothermic. [1 mark]

  • Cue. Exothermic, because ΔH\Delta H is negative.

Q2. Write the expression for the heat transferred in a calorimetry experiment. [1 mark]

  • Cue. q=mcΔTq = mc\Delta T.

Q3. State why bond-enthalpy calculations give only approximate values. [1 mark]

  • Cue. Bond enthalpies are mean (average) values across many compounds, not exact for a specific molecule.

Q4. Define the standard enthalpy of formation. [1 mark]

  • Cue. The enthalpy change when one mole of a compound forms from its elements in their standard states.

Q5. State the Hess's law expression for ΔH\Delta H using combustion data. [1 mark]

  • Cue. ΔH=ΣΔHc(reactants)ΣΔHc(products)\Delta H = \Sigma \Delta H_c(\text{reactants}) - \Sigma \Delta H_c(\text{products}).

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC 20194 marksUse the standard enthalpies of formation, ΔHf\Delta H_f for CH4\text{CH}_4 = -74.8, for CO2\text{CO}_2 = -393.5 and for H2O(l)\text{H}_2\text{O(l)} = -285.8 kJ mol1\text{kJ mol}^{-1}, to calculate the standard enthalpy of combustion of methane. CH4+2O2CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}.
Show worked answer →

Use ΔH=ΣΔHf(products)ΣΔHf(reactants)\Delta H = \Sigma \Delta H_f(\text{products}) - \Sigma \Delta H_f(\text{reactants}).

Products: ΔHf(CO2)+2×ΔHf(H2O)=393.5+2(285.8)=965.1\Delta H_f(\text{CO}_2) + 2 \times \Delta H_f(\text{H}_2\text{O}) = -393.5 + 2(-285.8) = -965.1 kJ mol-1.

Reactants: ΔHf(CH4)+2×ΔHf(O2)=74.8+0=74.8\Delta H_f(\text{CH}_4) + 2 \times \Delta H_f(\text{O}_2) = -74.8 + 0 = -74.8 kJ mol-1 (oxygen is an element, so zero).

ΔH=965.1(74.8)=890.3\Delta H = -965.1 - (-74.8) = -890.3 kJ mol-1.

Markers reward the formula, oxygen as zero, correct signs, and the answer with units.

WJEC 20214 marksIn a calorimetry experiment, burning 0.500 g of ethanol (molar mass 46.0 g mol-1) raised the temperature of 200 g of water by 25.0 degrees C. Calculate the enthalpy of combustion of ethanol. The specific heat capacity of water is 4.18 J g-1 K-1.
Show worked answer →

Heat absorbed by water: q=mcΔT=200×4.18×25.0=20900q = mc\Delta T = 200 \times 4.18 \times 25.0 = 20900 J =20.9= 20.9 kJ.

Moles of ethanol burned: n=0.500/46.0=0.01087n = 0.500 / 46.0 = 0.01087 mol.

Enthalpy of combustion: ΔH=q/n=20.9/0.01087=1923\Delta H = -q/n = -20.9 / 0.01087 = -1923 kJ mol-1 (negative, exothermic).

Markers reward q=mcΔTq = mc\Delta T, correct moles, dividing to get per mole, and the negative sign for an exothermic reaction.

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