Revise the theory
This practical links to these notes pages:
- Energetics, Calorimetry & Hess Law Cycles (Unit 1 · Spec 3.1.4)
1. Practical Aim & Reaction Context
The aim of Required Practical 2 is to determine the molar enthalpy change (ΔH) for a chemical reaction in solution. Typical reactions tested in OxfordAQA examinations include:
- Metal Displacement Reaction: e.g.
Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)(strongly exothermic displacement). - Acid-Base Neutralisation: e.g.
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)(ΔHneut ~ -57 kJ mol⁻¹). - Dissolution of an Ionic Solid: e.g.
NH₄NO₃(s) → NH₄⁺(aq) + NO₃⁻(aq)(endothermic enthalpy of solution).
2. Polystyrene Cup Calorimeter Construction
To minimize heat exchange with the surrounding laboratory atmosphere, a simple polystyrene cup calorimeter is constructed:
Apparatus Components & Functions
- Expanded Polystyrene Cup: Polystyrene has very low thermal conductivity (excellent insulator) and a negligible heat capacity, ensuring heat is retained in the water.
- Outer Glass Beaker: Provides mechanical stability and creates an insulating dead air jacket around the cup.
- Plastic / Cardboard Lid: Prevents heat loss via evaporation and convection. Features two small holes: one for the thermometer and one for a stirrer.
- Stirrer: Passes through the second hole in the lid and mixes the solution so its temperature is the same throughout.
- Thermometer: A school thermometer with 1 °C divisions reads to ±0.5 °C. A thermometer graduated in 0.1 °C or 0.2 °C divisions, or a digital probe (±0.1 °C), gives a smaller uncertainty. A temperature change uses two readings, so its uncertainty is double the reading uncertainty.
Sources of Heat Loss & Prevention
- Convective Loss: Hot air escaping from surface → prevented by placing lid securely on cup.
- Conductive Loss: Heat transferring through walls → minimized by nested polystyrene cups and air gap.
- Cooling During Slow Reaction: Even with insulation, heat escapes while the reaction progresses → compensated by temperature-time extrapolation.
3. Sequential Temperature Logging Protocol
Simply recording the initial temperature and the maximum observed temperature introduces significant systematic error because heat begins dissipating to the surroundings before the maximum temperature is reached. To compensate for this, a strict temperature-time logging procedure is used:
- Using a volumetric pipette, measure 50.0 cm³ of copper(II) sulfate solution into the polystyrene cup.
- Start the stopwatch. Record the temperature every minute for 3 minutes (minutes 1, 2, and 3) with gentle, continuous stirring to establish a stable pre-reaction baseline. Stir with a stirrer, not the thermometer.
- At minute 4, add the second reactant (e.g. an excess of powdered zinc metal, approximately 3.0 g). Do NOT record the temperature at minute 4. Immediately place the lid on the cup and stir vigorously.
- Resume temperature logging at minute 5. Record the temperature every minute from minute 5 to minute 15.
- Plot a graph of temperature (vertical axis) against time (horizontal axis).
Safety
- Wear eye protection.
- Zinc powder is flammable, so keep it away from flames.
4. Temperature-Time Cooling Curve Extrapolation
Plotting the recorded data yields two distinct regions: an initial flat baseline (minutes 1 to 3), and, after mixing, a rise to the highest temperature followed by a cooling curve as heat is lost to the surroundings (up to minute 15). The minute 5 reading is often still rising, so it is not part of the cooling line.
2. Draw a line of best fit through the cooling points, from the highest temperature onwards, and extrapolate it back to minute 4.
3. The vertical distance between these two extrapolated lines at exactly minute 4 represents the theoretical maximum temperature change (ΔTtheoretical).
5. Quantitative Enthalpy Calculations (q = mcΔT)
To convert the measured temperature change into the molar enthalpy change of reaction (ΔH), follow this two-stage quantitative sequence:
where:
- q = heat energy transferred (in Joules, J)
- m = mass of solution (in g). Assume density = 1.00 g cm⁻³ (so 50.0 cm³ = 50.0 g)
- c = specific heat capacity of solution = 4.18 J g⁻¹ K⁻¹
- ΔT = theoretical temperature change from extrapolation (in K or °C)
where:
- nlimiting = moles of the limiting reactant that fully reacted (mol)
- Division by 1000 converts Joules (J) to kilojoules (kJ)
- The Negative Sign: Must be inserted manually for exothermic reactions (where temperature rises). Omitting the sign forfeits the final mark.
6. Key Calorimetric Assumptions & Examiner Pitfalls
Unit 5 questions frequently challenge candidates to state the approximations made in calorimetry calculations:
| Calorimetric Assumption | Physical Reality | Impact on Final ΔH Value |
|---|---|---|
| No heat is lost to the surroundings | Some heat escapes through polystyrene walls and air gap during reaction. | Measured ΔT is slightly too small; calculated ΔH is less negative (underestimated). |
| Specific heat capacity c = 4.18 J g⁻¹ K⁻¹ | The solution contains dissolved ions (e.g. Cu²⁺, SO₄²⁻) whose heat capacity is slightly lower than pure water. | Assumes 4.18; slight deviation in calculated q. |
| Density of solution = 1.00 g cm⁻³ | The 0.500 mol dm⁻³ solution is slightly denser than water (about 1.08 g cm⁻³). | So assuming 1.00 g cm⁻³ slightly underestimates m. |
| Heat capacity of polystyrene cup is zero | The cup and thermometer bulb absorb a tiny quantity of heat. | Calculated q is slightly lower than true energy released. |
7. Worked Calorimetry Problem: Zinc Displacement of Copper Sulfate
Problem: In an RP2 experiment, 50.0 cm³ of 0.500 mol dm⁻³ CuSO₄ solution is placed in a polystyrene cup. Powdered zinc (3.00 g, an excess) is added at minute 4. The temperature readings are:
| Time / min | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Temperature / °C | 20.2 | 20.2 | 20.2 | zinc added | 38.2 | 40.8 | 40.4 | 40.0 | 39.7 | 39.2 | 38.8 | 38.3 | 38.0 | 37.6 | 37.3 |
The baseline temperature is 20.2 °C. The highest temperature (40.8 °C) is at minute 6, so the cooling line is drawn through the points from minute 6 onwards. Extrapolated back to minute 4, it gives a theoretical maximum temperature of 41.6 °C. (Including the minute 5 reading, which was still rising, would pull this down to about 40.5 °C.)
Assuming the solution has a density of 1.00 g cm⁻³ and a specific heat capacity of 4.18 J g⁻¹ K⁻¹, calculate the molar enthalpy change for this displacement reaction in kJ mol⁻¹ to 3 significant figures.
Step 1: Calculate theoretical temperature rise (ΔT)
ΔT = 41.6 - 20.2 = 21.4 °C (or 21.4 K)
Step 2: Calculate heat energy transferred (q)
m = 50.0 cm³ × 1.00 g cm⁻³ = 50.0 g
q = m × c × ΔT = 50.0 × 4.18 × 21.4 = 4472.6 J = 4.4726 kJ
Step 3: Calculate moles of limiting reactant (CuSO₄)
Zinc is in excess.
Moles of CuSO₄ = c × V = 0.500 × (50.0 / 1000) = 0.0250 mol
Step 4: Calculate molar enthalpy change (ΔH)
ΔH = - (q / n) = - (4.4726 kJ / 0.0250 mol) = -178.9 kJ mol⁻¹
Final Answer: ΔH = -179 kJ mol⁻¹ (3 significant figures)
8. Practice Exam Questions
Question 1: Why is the temperature of the reaction mixture NOT recorded at exactly the fourth minute during RP2?
Show Answer & Explanation
Correct Answer: B
At minute 4, the solid reactant is tipped in, the lid replaced, and stirring begun. The temperature is changing rapidly and is not homogeneous throughout the liquid. Recording is resumed at minute 5 once heat has distributed evenly.
Question 2: In calculating the heat change q for the reaction of 25.0 cm³ of 1.0 mol dm⁻³ HCl with 25.0 cm³ of 1.0 mol dm⁻³ NaOH, what value of mass m should be substituted into q = mcΔT?
Show Answer & Explanation
Correct Answer: B
The total volume of liquid heated by the reaction is 25.0 + 25.0 = 50.0 cm³. Assuming a solution density of 1.00 g cm⁻³, the total mass m = 50.0 g.
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Exam-style question
2 marksA student dissolves anhydrous copper(II) sulfate in water in a polystyrene cup. State two assumptions made when calculating the enthalpy change.
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Any two from:
- The density of the solution is 1.00 g cm⁻³, so the mass of solution equals the volume of water [1]
- The specific heat capacity of the solution is the same as water, 4.18 J g⁻¹ K⁻¹ [1]
- All the heat released is transferred to the solution, with none lost to the surroundings or the cup [1]