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4.1.4Reactivity 3 — What are the mechanisms of chemical change?

Standard Enthalpy Change — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 12 of 13 Unit quiz

📚 IB Chemistry HL Hub / Topical Revision
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Part 1 of 5Standard Conditions

4.1.4 Standard Enthalpy Change
Pressure
  • Standard pressure is 100 kPa for reactions involving gases.
Temperature
  • Enthalpy values are usually quoted at 298 K (25 °C), although temperature is not part of the definition of standard state.
Concentration
  • Solutions are taken at 1 mol dm⁻³, and each substance is in its most stable physical state under these conditions.
Symbol
  • The superscript ° in ΔH° indicates standard conditions, and the unit is kJ mol⁻¹.
Why it matters
  • Fixed conditions allow enthalpy values from different experiments to be compared and combined.

Part 2 of 5Key Definitions

1
Standard enthalpy of combustion, ΔHc°, is the enthalpy change when one mole of a substance burns completely in oxygen under standard conditions.
2
Standard enthalpy of formation, ΔHf°, is the enthalpy change when one mole of a compound forms from its elements in their standard states.
3
The enthalpy of formation of an element in its standard state is zero by definition.
4
Combustion is always exothermic, while formation can be exothermic or endothermic.
5
For a reaction, ΔH° = ΣΔHf°(products) − ΣΔHf°(reactants), with each term multiplied by its coefficient.
SubstanceΔHf° / kJ mol⁻¹
CO₂(g)−393.5
H₂O(l)−285.8
O₂(g)0

Part 3 of 5Using the Definitions

Using the Definitions
Equation writing
  • The equation for formation has one mole of the compound as the only product, with fractions allowed for the elements.
State symbols
  • State symbols are essential, as ΔHf° differs between H₂O(l) and H₂O(g).
Combustion products
  • Complete combustion of an organic compound gives CO₂(g) and H₂O(l) under standard conditions.
Units
  • Values are per mole of the substance named in the definition, not per mole of reaction in general.

Part 4 of 5Worked Example

Question Calculate ΔH° for C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l), given ΔHf°(C₃H₈) = −104.7, ΔHf°(CO₂) = −393.5, ΔHf°(H₂O(l)) = −285.8 kJ mol⁻¹.

  1. ΣΔHf°(products) = 3(−393.5) + 4(−285.8) = −1180.5 − 1143.2 = −2323.7.
  2. ΣΔHf°(reactants) = −104.7 + 5(0) = −104.7.
  3. ΔH° = −2323.7 − (−104.7).
  4. ΔH° = −2219.0 kJ mol⁻¹.

Answer ΔH° = −2219 kJ mol⁻¹

Part 5 of 5Exam Checklist & Tips

4.1.4 – Key Points for the Exam

  • Standard conditions Pressure of 100 kPa, solutions at 1 mol dm⁻³, usually 298 K.
  • Standard enthalpy of formation Enthalpy change when one mole of a compound forms from its elements in their standard states.
  • Standard enthalpy of combustion Enthalpy change when one mole of a substance burns completely in oxygen under standard conditions.
  • Standard state The most stable form of a substance at 100 kPa and the stated temperature.

Common Mistakes

  • Writing the formation equation with more than one mole of the compound.
  • Omitting state symbols or using the wrong state for water.
  • Giving elements such as O₂(g) a non-zero formation enthalpy.
  • Forgetting that formation of carbon uses graphite as the standard state.

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