Exothermic and endothermic reactions and calorimetryIB MYP Chemistry: Revision notes
Section 1
Exothermic and endothermic reactions
Chemical reactions transfer energy between the reacting substances and the surroundings. We see this as a temperature change.
- Exothermic: energy is released to the surroundings, so the temperature rises. Examples: combustion, neutralisation, rusting of iron.
- Endothermic: energy is taken in from the surroundings, so the temperature falls. Examples: thermal decomposition (such as heating calcium carbonate) and photosynthesis, which takes in energy from sunlight.
Everyday uses:
- Hand warmers use an exothermic reaction (iron reacting with oxygen) to release heat.
- Instant cold packs use an endothermic change that takes in heat from the injury.
Do not say a cold pack 'gives out cold'. It feels cold because it takes energy in from your hand.
Section 2
Calorimetry: measuring energy changes
Calorimetry measures the energy transferred in a reaction by the temperature change of water. The energy is:
Q = m × c × ΔT
- Q is the energy transferred in joules (J)
- m is the mass of water in grams
- c is the specific heat capacity of water, 4.18 J/g/°C
- ΔT is the temperature change in °C (final minus start)
Worked example: 200 g of water is heated from 20.0 °C to 45.0 °C.
Q = 200 × 4.18 × 25.0 = 20 900 J = 20.9 kJ (divide by 1000 to convert J to kJ).
Use the mass of the water, not the mass of the fuel, in Q = m × c × ΔT.
Section 3
Comparing fuels
To compare fuels, burn each one in a spirit burner under a can of water. Measure the water's mass, its start and highest temperatures, and the mass of the burner before and after. Keep the water mass, the can and the distance to the flame the same.
To compare fairly, work out the energy per gram of fuel:
energy per gram = Q ÷ mass of fuel burned
In the example above, if 0.90 g of ethanol was burned, the energy per gram is 20.9 ÷ 0.90 = 23.2 kJ/g.
To find the energy per mole, work out moles = mass ÷ Mr, then divide the energy by the moles. For ethanol (Mr = 46), 0.90 g is 0.0196 mol, so the energy per mole is 20.9 ÷ 0.0196 = about 1070 kJ/mol.
Moles = mass ÷ Mr. Keep your units clear: J, kJ, kJ/g or kJ/mol.
Section 4
Sources of error and improvements
The energy found in the lab is usually lower than the data book value. Reasons and improvements:
- Heat lost to the surroundings and absorbed by the can: use a draught shield, insulate the can, put a lid on it, or use a copper can that conducts heat well.
- Incomplete combustion (soot on the can): make sure there is plenty of air.
- Uneven temperature in the water: stir the water and record the highest temperature.
- Single result: repeat the experiment and calculate a mean.
These ideas help in criterion B (designing a fair, safe method) and criterion C (evaluating it).
'Human error' is not an acceptable source of error. Name a specific problem, such as heat loss to the surroundings.
Must know
- Exothermic: energy released, temperature rises. Endothermic: energy taken in, temperature falls
- Examples: combustion and neutralisation (exothermic); thermal decomposition and photosynthesis (endothermic)
- Q = m × c × ΔT, with c of water = 4.18 J/g/°C; divide by 1000 for kJ
- Energy per gram = Q ÷ mass of fuel; energy per mole = Q ÷ moles of fuel
- Main error: heat loss, so results are lower than data book values; use a draught shield or insulation
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Exothermic and endothermic reactions and calorimetry
- A hiker in Chile carries a disposable hand warmer. When she opens the sachet, the iron powder inside reacts with oxygen from the air and the sachet warms to about 55 °C for several hours. Her friend carries an instant cold pack for sports injuries. When the pack is squeezed, a solid dissolves in water inside it and the pack quickly becomes cold.Explain, in terms of energy transfer, why the hand warmer feels warm but the cold pack feels cold.2 marks
- A student in Kuala Lumpur burns ethanol in a spirit burner to heat 100 g of water in a metal can. The temperature of the water rises from 20.0 °C to 52.0 °C. The mass of the spirit burner falls from 85.40 g to 84.65 g. The specific heat capacity of water is 4.18 J/g/°C.Calculate the energy transferred to the water, in joules.2 marks
- Students at a school in Lagos compare two liquid fuels, ethanol and propan-1-ol, to find which releases more energy per gram. They burn each fuel in a spirit burner to heat 100 g of water in a metal can. Burning 0.80 g of ethanol raises the temperature of the water by 30.0 °C. Burning 0.70 g of propan-1-ol raises the temperature of the water by 33.0 °C. The specific heat capacity of water is 4.18 J/g/°C. The data book value for propan-1-ol is 33.6 kJ/g.State two variables that the students should control to make the comparison fair, and one safety precaution they should take.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).