The shape of the lesson
By the end of the lesson, learners can
- Heat the water: Run a neutralisation in the calorimetry simulator, then a combustion with a spirit burner.
- Combustion and displacement: For a fuel, the water in a copper can is heated by the flame; weigh the burner before and after to find the mass of fuel burned.
- Errors and improvements: The biggest error in simple calorimetry is heat exchanged with the surroundings.
- m is the mass of water or solution (1 cm3 = 1 g).
- c = 4.18 J g−1 K−1; q in J, divide by 1000.
- n is the limiting reagent (or water formed).
- Temperature rise means ΔH negative.
One picture every idea returns to
The heat transferred to or from the surroundings is q = mcΔT, where m is the mass of water or solution heated (not the mass of reactant), c = 4.18 J g−1 K−1 and ΔT is the temperature change. Divide by the amount in moles of the reactant that was used up (the limiting one) and convert J to kJ: ΔH = −q ÷ n. The minus sign means a temperature rise gives a negative (exothermic) ΔH. Heat loss to the surroundings makes experimental values less exothermic than data book values, most of all in combustion experiments.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Measure the temperature.. Nobody can measure an enthalpy change directly, but anyone can read a thermometer. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Heat the water | Run a neutralisation in the calorimetry simulator, then a combustion with a spirit burner. Project tab 2 of the lab (Chemical energetics lab, calorimetry) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Combustion and displacement | For a fuel, the water in a copper can is heated by the flame; weigh the burner before and after to find the mass of fuel burned. | Pairs |
| 21 to 30 min | Teach and check: Errors and improvements | The biggest error in simple calorimetry is heat exchanged with the surroundings. | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 16 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Errors and improvements, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| In q = mcΔT, m is the mass of the water or solution whose temperature changes, never the mass of fuel or solid added. | Remember too that q comes out in joules: divide by 1000 for kJ, and add the minus sign yourself when the temperature rose. |
Calorimetry: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
50.0 cm3 of 2.00 mol dm−3 HNO3 is added to 50.0 cm3 of 2.00 mol dm−3 KOH. The temperature rises from 21.0 °C to 34.6 °C. Calculate the enthalpy change of neutralisation.
0.320 g of methanol, CH3OH (Mr 32.0), is burned and heats 100 g of water from 20.0 °C to 32.5 °C. Calculate the experimental enthalpy change of combustion of methanol.
5.00 g of ammonium nitrate is dissolved in 50.0 g of water; the temperature falls by 7.2 K. Calculate the enthalpy change of solution of NH4NO3 (Mr 80.0).
Excess iron powder is added to 50.0 cm3 of 0.500 mol dm−3 CuSO4(aq); ΔT = +20.4 K. Calculate ΔH for Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s).
A student's value for ΔHc of propan-1-ol is −1250 kJ mol−1; the data book value is −2021 kJ mol−1. Give two reasons for the difference and one improvement.
Explain why a temperature against time graph, extrapolated back to the time of mixing, gives a more accurate ΔT than simply reading the highest temperature.
Answer key and marking notes
- q = 100 × 4.18 × 13.6 = 5685 J.
- n(H2O) = 2.00 × 0.0500 = 0.100 mol.
- ΔH = −5685 ÷ 0.100 = −56 850 J mol−1 = −56.8 kJ mol−1.
- q = 100 × 4.18 × 12.5 = 5225 J = 5.225 kJ.
- n = 0.320 ÷ 32.0 = 0.0100 mol.
- ΔHc = −5.225 ÷ 0.0100 = −523 kJ mol−1.
- q = 50.0 × 4.18 × 7.2 = 1505 J (use the mass of water).
- n = 5.00 ÷ 80.0 = 0.0625 mol.
- The temperature fell, so the process is endothermic: ΔH = +1505 ÷ 0.0625 = +24.1 kJ mol−1.
- q = 50.0 × 4.18 × 20.4 = 4264 J; n(CuSO4) = 0.0250 mol.
- ΔH = −4264 ÷ 0.0250 = −171 kJ mol−1.
- Heat lost to the surroundings (air, can, clamp) instead of heating the water.
- Incomplete combustion, shown by soot on the can; or evaporation of fuel from the wick.
- Improvement: a draught shield and lid, or a flame calorimeter with an oxygen supply.
- While the reaction mixture is heating up it is also losing heat, so the highest reading is lower than the true value.
- Extrapolating the cooling line back to the moment of mixing estimates the temperature reached if no heat were lost.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
150 g of water is heated from 21.0 °C to 29.0 °C. How much energy has the water gained? (c = 4.18 J g−1 K−1)
- A5.02 kJ
- B5.02 × 103 kJ
- C189 kJ
- D18.2 kJ
Using the apparatus shown, 50.0 cm3 of 1.00 mol dm−3 HCl is mixed with 50.0 cm3 of 1.00 mol dm−3 NaOH. The temperature rises by 6.8 K. What is the enthalpy change of neutralisation? (c = 4.18 J g−1 K−1; 1 cm3 of solution has a mass of 1 g)
- A−56.8 kJ mol−1
- B+56.8 kJ mol−1
- C−28.4 kJ mol−1
- D−5.68 × 104 kJ mol−1
Burning 0.460 g of ethanol (Mr = 46.0) in a spirit burner raises the temperature of 200 g of water in a copper can by 13.0 K. What is the experimental enthalpy change of combustion of ethanol? (c = 4.18 J g−1 K−1)
- A−1.09 × 103 kJ mol−1
- B+1.09 × 103 kJ mol−1
- C−2.50 kJ mol−1
- D−1.09 × 106 kJ mol−1
When solid ammonium nitrate dissolves in water in a polystyrene cup, the temperature of the solution falls from 20.5 °C to 14.0 °C. What can be concluded?
- Athe process is endothermic and ΔH is positive
- Bthe process is exothermic and ΔH is negative
- Cthe process is endothermic and ΔH is negative
- Dthe process is exothermic because energy is lost from the solution
The value of ΔHc of ethanol found with a spirit burner and a copper can is usually much less exothermic than the data book value of −1367 kJ mol−1. What is the main reason?
- Amuch of the energy released is lost to the surroundings instead of heating the water
- Bthe thermometer gives readings that are too high
- Cthe copper can absorbs no energy at all
- Dtoo much ethanol is burned, so the temperature rise is too large
Excess zinc powder (0.500 g, Mr = 65.4) is added to 50.0 cm3 of 0.200 mol dm−3 copper(II) sulfate solution. The temperature rises by 10.2 K. What is ΔH for Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)? (c = 4.18 J g−1 K−1)
- A−213 kJ mol−1
- B−279 kJ mol−1
- C+213 kJ mol−1
- D−2.13 × 105 kJ mol−1
The enthalpy change of neutralisation for HCl and NaOH is −57.1 kJ mol−1. Predict the temperature rise when 25.0 cm3 of 2.00 mol dm−3 HCl is mixed with 25.0 cm3 of 2.00 mol dm−3 NaOH in an insulated cup. (c = 4.18 J g−1 K−1)
- A13.7 K
- B27.3 K
- C6.83 K
- D0.0137 K
The enthalpy change of neutralisation of ethanoic acid with sodium hydroxide is −55.2 kJ mol−1, which is less exothermic than the value of −57.1 kJ mol−1 for hydrochloric acid. What is the best explanation?
- Aethanoic acid is a weak acid, so some energy is used to complete its ionisation as it reacts, and less energy is released overall
- Bethanoic acid forms less water per mole of acid than hydrochloric acid
- Cthe ethanoate ion forms stronger bonds with sodium ions than the chloride ion does
- Dethanoic acid is more concentrated, so it absorbs more heat from the solution
Quiz key
ΔT = 29.0 − 21.0 = 8.0 K (a change of 1 °C is the same as a change of 1 K). q = mcΔT = 150 × 4.18 × 8.0 = 5016 J = 5.02 kJ. 5.02 × 103 kJ is the value in joules with the unit changed to kJ without dividing by 1000. 189 kJ uses the final temperature in kelvin (302 K) instead of the temperature change. 18.2 kJ uses the final temperature, 29.0, as ΔT.
Total volume = 100 cm3, so m = 100 g. q = mcΔT = 100 × 4.18 × 6.8 = 2842 J = 2.842 kJ. n(H2O) = n(HCl) = 0.0500 × 1.00 = 0.0500 mol. ΔH = −q/n = −2.842 ÷ 0.0500 = −56.8 kJ mol−1, negative because the temperature rose (exothermic). +56.8 misses the sign. −28.4 uses only the 50 g of acid as the mass heated, but both solutions warm up. −5.68 × 104 leaves q in joules while quoting the answer in kJ.
q = mcΔT = 200 × 4.18 × 13.0 = 10 868 J = 10.87 kJ. n(ethanol) = 0.460 ÷ 46.0 = 0.0100 mol. ΔHc = −10.87 ÷ 0.0100 = −1087, so −1.09 × 103 kJ mol−1 to 3 s.f. The positive value has the wrong sign; combustion releases energy. −2.50 uses the mass of ethanol (0.460 g) in q = mcΔT instead of the mass of water. −1.09 × 106 uses q in joules without dividing by 1000.
Energy has been taken from the water (the surroundings) into the chemical system as the ionic lattice breaks up, so the process is endothermic and ΔH is positive. Using q = mcΔT with the temperature fall gives a positive ΔH = +q/n. Exothermic would make the solution warmer, not cooler. Endothermic with a negative ΔH mixes up the sign convention: a negative ΔH always means exothermic. The solution does lose energy, but it goes into the chemicals, which is the definition of an endothermic change, not an exothermic one.
Most of the energy from the flame heats the air, the can and the clamp instead of the water, so the measured temperature rise and q are too small and the calculated ΔHc is less negative than the true value. Incomplete combustion (soot on the can) and evaporation of ethanol from the wick add to the error. Thermometer readings that are too high would make the value more exothermic, not less. The copper can does absorb energy; ignoring it is another reason the result is low. Burning more ethanol raises both q and n together, so it does not explain a smaller value per mole.
q = 50.0 × 4.18 × 10.2 = 2132 J = 2.132 kJ. Copper(II) sulfate is the limiting reagent: n(Cu2+) = 0.0500 × 0.200 = 0.0100 mol, and the zinc that reacts is also 0.0100 mol. ΔH = −2.132 ÷ 0.0100 = −213 kJ mol−1. −279 divides by the total zinc added (0.500 ÷ 65.4 = 0.00765 mol), but some zinc is left over, so not all of it gave out energy. +213 has the wrong sign for a temperature rise. −2.13 × 105 leaves q in joules.
n(H2O) = 0.0250 × 2.00 = 0.0500 mol. q = 0.0500 × 57 100 = 2855 J. The mass heated is 25.0 + 25.0 = 50.0 g, so ΔT = 2855 ÷ (50.0 × 4.18) = 13.7 K. 27.3 K uses only 25.0 g of solution, but both solutions are warmed. 6.83 K uses n = 0.0250 mol, the volume of one solution multiplied by 1.00 instead of 2.00 mol dm−3. 0.0137 K uses 57.1 in kJ without converting to joules to match c in J g−1 K−1.
In a strong acid all the H+ ions are already free, so only H+ + OH− → H2O happens. Ethanoic acid is mostly un-ionised molecules; as H+ is removed the equilibrium shifts and more molecules ionise, and breaking the O–H bond needs energy, so the overall energy released per mole of water is smaller. Both acids are monoprotic and each forms 1 mol of water per mole of acid; the value is defined per mole of water anyway. Sodium ethanoate and sodium chloride are both fully dissociated in solution, so there is no difference in bonding with sodium ions. Concentration does not change the enthalpy change per mole.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.