1.6.6Physical Chemistry
Applications of Hess’s Law — AQA AS Chemistry Revision Notes & Topical Questions
5 min read 5 parts Lesson 38 of 102
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Part 1 of 5Using Combustion Data
Formula
- Formula ΔrH = ΣΔcH(reactants) − ΣΔcH(products), which is the reverse of the pattern for formation data.
Cycle
- Cycle Reactants and products both burn to the same combustion products, so the arrows in the cycle point downwards to CO₂ and H₂O.
Why useful
- Why useful Many organic compounds cannot be made from their elements directly, but their combustion enthalpies are easily measured.
Coefficients
- Coefficients Every combustion enthalpy must be multiplied by the number of moles of that substance in the balanced equation.
Oxygen and products
- Oxygen and products Oxygen, carbon dioxide and water as products are not included in the sum because they are combustion products.
Part 2 of 5Where Else It Is Used
Where Else It Is Used
Formation of organics
- Formation of organics The enthalpy of formation of propane or ethanol is found from combustion data for carbon, hydrogen and the compound itself.
Hydration and solution
- Hydration and solution Hess’ law connects lattice enthalpies, hydration enthalpies and enthalpies of solution in an enthalpy cycle.
Bond enthalpies
- Bond enthalpies Cycles with atoms in the gas phase allow mean bond enthalpies to be calculated from enthalpies of formation.
Indirect reactions
- Indirect reactions Reactions that are too slow or give mixtures can be studied through a related reaction that is easy to measure.
| Substance | ΔcH° / kJ mol⁻¹ |
|---|---|
| C(s) | −394 |
| H₂(g) | −286 |
| C₃H₈(g) | −2219 |
Part 3 of 5Method for Combustion Cycles
1
Write the equation to be found, such as 3C(s) + 4H₂(g) → C₃H₈(g).
2
Draw both reactants and products down to the same combustion products, CO₂ and H₂O.
3
Equate the direct route with the indirect route to form an equation linking the unknown to the known values.
4
Rearrange to ΔrH = Σ ΔcH(reactants) − Σ ΔcH(products), using moles from the equation.
5
Substitute carefully with signs, then evaluate and include units of kJ mol⁻¹.
Part 4 of 5Worked Example
Question Calculate the standard enthalpy of formation of propane from 3C(s) + 4H₂(g) → C₃H₈(g), given ΔcH°: C = −394, H₂ = −286, C₃H₈ = −2219 kJ mol⁻¹.
- Use ΔfH = Σ ΔcH(reactants) − Σ ΔcH(products).
- Reactants3 × (−394) + 4 × (−286) = −1182 − 1144 = −2326 kJ.
- Products1 × (−2219) = −2219 kJ.
- ΔfH = −2326 − (−2219) = −107 kJ mol⁻¹.
Answer ΔfH°(C₃H₈) = −107 kJ mol⁻¹.
Part 5 of 5Exam Checklist & Tips
1.6.6 – Key Points for the Exam
- Enthalpy of combustion The enthalpy change when one mole of a substance burns completely in oxygen.
- Enthalpy cycle A diagram showing two routes between the same reactants and products.
- Indirect route A route using other reactions whose enthalpy changes are known.
- Mean bond enthalpy The average enthalpy needed to break one mole of a bond across a range of compounds in the gas phase.
Common Mistakes
- Using the formation formula pattern (products minus reactants) with combustion data, which gives the wrong sign.
- Leaving out the factor of 3 or 4 for carbon and hydrogen in the equation.
- Including the combustion enthalpy of CO₂ or H₂O, which are zero as they cannot burn further.
- Mixing standard enthalpies of formation and combustion within the same cycle.
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