4.1.5Additional lessons
Calorimetry Experiments — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 72 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.
Section 1 of 5Solution Calorimetry
Method
- Method Reagents are mixed in an insulated container, such as a polystyrene cup with a lid, and the temperature is recorded over time.
Heat gained
- Heat gained The heat absorbed by the solution is q = mcΔT, taking the mass of solution as equal to the mass of water.
Enthalpy change
- Enthalpy change Divide q by the amount in moles of the limiting reactant and reverse the sign, so ΔH = −q ÷ n for a temperature rise.
Assumptions
- Assumptions The solution has density 1.00 g cm⁻³ and specific heat capacity of 4.18 J g⁻¹ K⁻¹, with no heat absorbed by the cup.
Graph
- Graph A temperature-time graph extrapolated back to the time of mixing gives a better ΔT than a single reading.
Section 2 of 5Combustion Calorimetry
1
A known mass of fuel is burned in a spirit burner beneath a metal can containing a measured mass of water.
2
The temperature rise of the water is recorded and q = mcΔT is calculated for the water.
3
The mass of fuel burned is found by weighing the burner before and after, and converted to moles.
4
Dividing the energy gained by the moles burned gives the enthalpy of combustion, with a negative sign.
5
The experimental value is usually smaller in magnitude than the data booklet value because of heat losses.
| Error | Effect on result | Improvement |
|---|---|---|
| Heat loss to surroundings | Value too small | Use insulation and a lid |
| Incomplete combustion | Value too small | Ensure good air supply |
| Evaporation of fuel | Mass error | Cap the burner after use |
Section 3 of 5Evaluating the Experiment
Evaluating the Experiment
Systematic errors
- Systematic errors Heat loss and calorimeter heat capacity cause results to be consistently low in magnitude.
Random errors
- Random errors Reading the thermometer and balance leads to scatter that can be reduced by repetition.
Uncertainty
- Uncertainty The uncertainty in ΔT is the biggest source, so a larger temperature change reduces the percentage uncertainty.
Digital probes
- Digital probes Temperature probes with data loggers give continuous readings and allow accurate extrapolation.
Section 4 of 5Worked Example
Question
- 50.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH and the temperature rises by 6.8 °C.
- Calculate ΔH for the neutralisation per mole of water.
- Mass of solution = 100.0 g, so q = 100.0 × 4.18 × 6.8 = 2842 J = 2.84 kJ.
- n(HCl) = n(NaOH) = 0.0500 dm³ × 1.00 = 0.0500 mol, so n(H₂O) = 0.0500 mol.
- ΔH = −q ÷ n = −2.842 ÷ 0.0500.What the symbols mean: ΔH = enthalpy change (kJ mol−1) · q = heat energy measured (J, divide by 1000 for kJ) · n = amount of the limiting reactant (mol); negative sign = exothermic
- ΔH = −56.8 kJ mol⁻¹.
Answer ΔH ≈ −56.8 kJ mol⁻¹
Section 5 of 5Exam Checklist & Tips
4.1.5 – Key Points for the Exam
- Calorimeter An insulated vessel in which temperature changes are measured to calculate energy transfer.
- Specific heat capacity The energy needed to raise the temperature of 1 g of a substance by 1 K.
- Limiting reactant The reactant that is completely used up first and determines the amount of product.
- Extrapolation Extending a graph line beyond measured points to find a value at a specific time.
Common Mistakes
- Using the mass of the reactants instead of the mass of the solution for q.
- Using the total moles of both reactants rather than the limiting reactant.
- Forgetting the negative sign in ΔH for a temperature rise.
- Using kJ for q with c in J g⁻¹ K⁻¹ without converting.
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