4.2.3Additional lessons
Hess’s Law Calculations — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 75 of 172
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 5Strategy
Target equation
- Target equation Write the equation whose enthalpy change is required, noting which species are reactants and which are products.
Matching
- Matching Find in the given equations the ones that contain each target species and check which side it is on.
Adjust
- Adjust Reverse an equation if a species is on the wrong side and multiply by a factor if the amount differs.
Combine
- Combine Add the adjusted equations so that unwanted species cancel and the target remains.
Add values
- Add values Add the adjusted ΔH values, remembering the sign changes and multipliers.
Part 2 of 5Step by Step
1
Label each given equation (1), (2), (3) with its ΔH value.
2
Decide which equations need to be reversed so that reactants and products match the target equation.
3
Multiply equations by integers where the number of moles differs from the target equation.
4
Add the equations and cross out species that appear on both sides.
5
Compare the sum with the target equation, then add the modified enthalpy values to find the answer.
| Operation on equation | Effect on ΔH |
|---|---|
| Reverse | Change sign |
| Multiply by n | Multiply by n |
| Add equations | Add the values |
Part 3 of 5Checking the Answer
Checking the Answer
Sign
- Sign An exothermic process gives a negative answer, so consider whether the sign is chemically sensible.
Size
- Size Compare with related reactions, such as combustion or formation, to check the magnitude.
Units
- Units Quote kJ mol⁻¹ for the reaction as written in the target equation.
Significant figures
- Significant figures Give the answer to the same number of decimal places as the least precise data.
Part 4 of 5Worked Example
Question Given (1) C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ mol⁻¹ and (2) CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ mol⁻¹, find ΔH for C(s) + ½O₂(g) → CO(g).
- Keep equation (1) as it is, because C(s) is on the left in the target equation.
- Reverse equation (2) so that CO is a product: CO₂(g) → CO(g) + ½O₂(g), ΔH = +283.0 kJ mol⁻¹.
- AddC + O₂ + CO₂ → CO₂ + CO + ½O₂, which simplifies to C + ½O₂ → CO.
- ΔH = −393.5 + 283.0 = −110.5 kJ mol⁻¹.
Answer ΔH = −110.5 kJ mol⁻¹
Part 5 of 5Exam Checklist & Tips
4.2.3 – Key Points for the Exam
- Target equation The reaction whose enthalpy change is to be found.
- Thermochemical equation A balanced equation that includes states and the enthalpy change.
- Cancel Remove a species that appears equally on both sides when equations are added.
- Multiplier A factor used to scale an equation so that mole amounts match.
Common Mistakes
- Reversing an equation but not changing the sign of ΔH.
- Cancelling species that are in different states.
- Forgetting to scale ΔH when an equation is multiplied.
- Adding the data without checking that the target equation is obtained.
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