Applications of Hess’s law — AQA A-Level Chemistry
Test yourself on Applications of Hess’s law with AQA A-Level practice questions.
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Applications of Hess’s law explained
Hess's law states the enthalpy change of a reaction is independent of the route taken, as enthalpy is a state function.
Read the full explanation
This allows calculation of values that cannot be measured directly. Draw the target reaction across the top, then build a triangle down to a common set of substances. With enthalpies of formation, arrows point up from elements, so ∆H = Σ∆fH(products) − Σ∆fH(reactants). With enthalpies of combustion, arrows point down to combustion products, so ∆H = Σ∆cH(reactants) − Σ∆cH(products). For example, calculating the enthalpy of formation of ethanol using its combustion data. Multiply each value by the number of moles in the balanced equation, and remember that going against an arrow reverses the sign.
Your focus
- Construct a labelled Hess cycle linking a target reaction to a set of given enthalpies of formation or combustion.
- Calculate an enthalpy change from formation data using Σ∆fH(products) − Σ∆fH(reactants), including the balancing numbers.
- Calculate an enthalpy change from combustion data and explain why the subtraction is the other way round.
Applications of Hess’s law exam tips
Marking Points
- One mark for the correct cycle or expression, products minus reactants for formation data or reactants minus products for combustion data.
- One mark for multiplying each enthalpy value by its balancing number from the equation.
- One mark for reversing the sign when travelling against the direction of an arrow in the cycle.
- One mark for the final answer with correct sign and units kJ mol⁻¹.
Examiner Tips
- 💡Draw the cycle as your working; a correct Hess cycle can often earn the method mark on its own even without a written expression.
- 💡Write the balanced equation first and put the number of moles above each substance before you substitute.
Common Mistakes
- Using the formation rule when the data given are enthalpies of combustion, which reverses the whole answer.
- Forgetting that ∆fH of an element in its standard state is zero and leaving it out of the count of substances.
- Ignoring the balancing numbers, so 2 CO₂ is counted once rather than twice.