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Standard Enthalpy Changes

UNIT 2 CHEMISTRY · PEARSON AAQ · D1.3.3

Standard Enthalpy Changes

FREE CHEMISTRY REVISION · TOPIC 40

Standard Enthalpy Changes

Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · D1.3.3

What you will learn

Define standard enthalpy changes of reaction, formation and combustion, recognise their equations, and explain the roles of standard states and stated conditions. Distinguish the one mole formed from the one mole burned.

1. Standard conditions and standard states

The standard-state symbol θ (often printed ⦵) indicates a change defined using standard states. Standard pressure is 100 kPa (1 bar). A temperature must also be specified; school tables commonly quote 298 K (approximately 25 °C). The symbol alone does not mean every datum must be at 298 K: standard-state quantities can be quoted at other stated temperatures.

For this course, use pure substances in their standard states at the specified temperature and 100 kPa; standard solution data use the specified standard concentration, conventionally 1 mol dm⁻³ at this level. The rigorous solution standard state is an idealised reference. State any concentration relevant to an aqueous reaction.

An element’s reference standard state is its most stable form under the stated conditions: for example C(s, graphite), O₂(g), H₂(g), Br₂(l) and I₂(s) at 298 K. Diamond and gaseous carbon atoms are not carbon’s reference form at 298 K. Standard states are not the same as a substance merely being solid, and standard enthalpy conditions are not the 273 K often used in definitions of STP.

Physical-state symbols matter: H₂O(l) and H₂O(g) have different enthalpies. Quote temperature, pressure and the chemical and physical states consistently.

2. Symbols, units and definitions

QuantitySymbolDefinition and amount
Standard reaction enthalpyΔᵣHθEnthalpy change when reactants in the amounts specified by the balanced equation react to form products, all in standard states under stated standard conditions.
Standard formation enthalpyΔfHθEnthalpy change when exactly one mole of a substance is formed from its constituent elements in their reference standard states under standard conditions.
Standard combustion enthalpyΔcHθEnthalpy change when exactly one mole of a substance is completely burned in oxygen under standard conditions, with reactants and products in their specified standard states.

Pearson uses the general symbol ΔHθ, with f or c identifying formation or combustion. The subscript r specifies reaction when needed. Molar values are usually in kJ mol⁻¹; the mole refers to the defined process and its equation. A sample’s total enthalpy change is measured in kJ, so do not attach mol⁻¹ automatically.

One mole: identify what is countedFormation counts one mole of product; combustion counts one mole of the substance burned.FORMATIONCOMBUSTIONElements in standard statesExactly 1 mol of productExactly 1 mol of substance + O₂Complete oxidation products
The one-mole requirement applies to different sides of the equations. A formation equation may use fractional elemental coefficients. A combustion equation may produce more than one mole of CO₂ or H₂O.

3. Formation: one mole of the named product

A formation equation starts with elements, each in its reference standard state, and produces exactly one mole of the named substance. Balance atoms, retaining coefficient 1 for that product. Fractional elemental coefficients are allowed because coefficients express molar ratios.

Named substanceFormation equation at 298 KReason
Water(l)H₂(g) + ½O₂(g) → H₂O(l)One mole of liquid water from reference elements.
Carbon dioxideC(s, graphite) + O₂(g) → CO₂(g)One mole of CO₂.
Ammonia½N₂(g) + ³⁄₂H₂(g) → NH₃(g)Fractions preserve one mole of NH₃.
Magnesium oxideMg(s) + ½O₂(g) → MgO(s)One mole of the oxide.
Ethanol(l)2C(s, graphite) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)One mole of ethanol; all reactants are reference elements.

CO(g) + ½O₂(g) → CO₂(g) is not a formation equation for CO₂ because CO is already a compound. 2H₂(g) + O₂(g) → 2H₂O(l) is balanced but forms two moles, so its enthalpy change is twice ΔfHθ of water. Divide every coefficient and the enthalpy change by two. Formation can be exothermic or endothermic; its definition does not force a negative sign.

4. Why reference elements have zero formation enthalpy

ΔfHθ = 0 kJ mol⁻¹ for an element in its reference standard state, by the chosen enthalpy reference convention. Forming the reference element from itself produces no change of state or composition: O₂(g) → O₂(g), for example, has zero enthalpy change.

This is not a claim that oxygen or graphite has no energy. It assigns a reference for formation enthalpies. O(g) is not O₂(g), and diamond is not graphite. Their formation enthalpies relative to the reference elements need not be zero. Nor does zero formation enthalpy mean zero combustion enthalpy: graphite burns exothermically.

5. Combustion: one mole of the substance burned

Write one mole of the fuel or named substance on the reactant side, add sufficient oxygen, then balance the complete oxidation products. For the C/H/O substances below at 298 K, the usual products are CO₂(g) and H₂O(l). Other elements can give different oxidation products, so state them rather than assuming every combustion equation produces CO₂.

SubstanceComplete combustion equationWhat is counted
MethaneCH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)One mole CH₄ burned; two moles water formed.
EthanolC₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)One mole ethanol burned; not one mole product.
HydrogenH₂(g) + ½O₂(g) → H₂O(l)One mole H₂ burned.
Carbon monoxideCO(g) + ½O₂(g) → CO₂(g)One mole CO completely oxidised.

Incomplete combustion of a carbon fuel to CO or soot is not its standard enthalpy change of complete combustion. For these fuels, combustion is exothermic and ΔcHθ is negative. Some equations fit two definitions: H₂(g) + ½O₂(g) → H₂O(l) describes both formation of water and combustion of hydrogen under the stated conditions. Always name the substance associated with the symbol.

6. Worked recognition and interpretation

Example 1 — correct the amount formed

An equation shows N₂(g) + 3H₂(g) → 2NH₃(g), with a supplied change of −92 kJ for the equation. Divide everything by 2: ½N₂(g) + ³⁄₂H₂(g) → NH₃(g). The supplied formation enthalpy is therefore −46 kJ mol⁻¹, if the standard states and conditions apply. This value is rounded teaching data, not a precision tabulation.

Example 2 — classify a combustion equation

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l) consumes one mole of ethanol completely. It defines ΔcHθ of ethanol under the stated conditions. Producing two moles CO₂ does not disqualify it: the one-mole rule counts ethanol consumed.

Example 3 — reject a non-element reactant

CO(g) + ½O₂(g) → CO₂(g) forms one mole CO₂, but is not ΔfHθ of CO₂. CO is a compound. It is the combustion of CO and a standard reaction enthalpy if the required conditions apply.

Example 4 — distinguish water states

A methane equation producing H₂O(g) has a different enthalpy change from one producing H₂O(l). Condensing the product water releases extra heat. Therefore the liquid-water combustion value is more negative under comparable conditions; do not silently mix the two values.

Example 5 — reference-state reasoning

ΔfHθ of C(s, graphite) is zero by convention at 298 K, while diamond need not be zero. “Both are elements” is insufficient: the reference allotrope and conditions must match.

Example 6 — scale a standard reaction

For 2H₂(g) + O₂(g) → 2H₂O(l), suppose the supplied reaction change is −572 kJ per mole of reaction as written. The equation H₂(g) + ½O₂(g) → H₂O(l) has half the value: −286 kJ mol⁻¹. This is both formation of water(l) and combustion of hydrogen; the original doubled equation is not the one-mole definition.

7. Why conditions must be stated

Enthalpy depends on temperature, composition and physical state, and can depend on pressure. Dissolution and aqueous reactions can depend on concentration. Stated standard conditions make values comparable and allow consistent use of tabulated data. “Standard” does not mean “whatever the laboratory happens to be like”. A reaction need not actually be carried out in one direct experimental step for its standard enthalpy change to be defined.

When recognising an equation, check four things: the named substance, what one mole refers to, whether the reactants meet the definition, and whether states and conditions match. Keep detailed calculations using tables and energy cycles for D1.3.5.

Common misconceptions

  • Formation means any reaction producing a compound: it must start from reference elements and form one mole.
  • Combustion must make one mole of product: it consumes one mole of the named substance.
  • All elemental forms have zero formation enthalpy: only the reference form at the stated conditions does.
  • Standard automatically means STP: use 100 kPa and the stated temperature, commonly 298 K.
  • State symbols are optional: liquid and gaseous products differ in enthalpy.

Exam tips

  • Write the full definition, including the one-mole requirement and standard states.
  • Use coefficient 1 for the formation product or combustion substance.
  • Allow fractional O₂, H₂ or N₂ coefficients; keep atoms balanced.
  • Name graphite when specifying carbon’s reference form at 298 K.
  • Include θ, the correct process subscript and kJ mol⁻¹.

Check your understanding

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Question 1 of 5 · Unit 2 D1.3.3

Which equation defines formation of one mole of liquid water at 298 K under standard conditions?

Question 2 of 5 · Unit 2 D1.3.3

In ΔcHθ of ethanol, exactly one mole refers to what?

Question 3 of 5 · Unit 2 D1.3.3

Why is ΔfHθ of O₂(g) zero at 298 K?

Question 4 of 5 · Unit 2 D1.3.3

Which standard pressure is used with these definitions?

Question 5 of 5 · Unit 2 D1.3.3

CO(g) + ½O₂(g) → CO₂(g) under standard conditions represents what?

Scope checked against Pearson BTEC Level 3 AAQ Applied Science, Issue 7, May 2026, Unit 2 D1.3.3, printed page 40. Official Pearson specification (PDF).

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