Quick answer: Thermochemistry measures heat changes in reactions. This AP Chemistry lab on thermochemistry gives the aim, the equipment, the safety points, a step-by-step method, sample results and 7 analysis questions; the answers are on a separate page and the lab sheet is a free printable PDF.
Can you find the enthalpy change of a reaction you cannot measure directly? Here you use Hess’s law with two reactions that are easy to measure: sodium hydroxide dissolving in water and sodium hydroxide reacting with acid. The difference gives the enthalpy of neutralisation, which you compare with the accepted value.
NaOH pellets (2.00 g per run), 0.500 M hydrochloric acid, distilled water
Stopwatch and stirring rod
Safety
Safety
Sodium hydroxide is corrosive and the solution gets hot: wear eye protection and gloves, never touch pellets with bare hands, and wash spills at once with plenty of water. Hydrochloric acid at 0.5 M may irritate. Stir with the stirring rod, not the thermometer.
Method
Place 100.0 cm³ of water in the cup and measure the temperature for 2 minutes.
Add 2.00 g of NaOH, replace the lid and stir gently.
Record the highest temperature reached.
Repeat with 100.0 cm³ of 0.500 M HCl instead of water.
Use the same mass of NaOH and fresh solutions each time.
Set-up: the cup is insulated and covered so the temperature rise measures the heat released.
Results
Experiment
Initial temperature (°C)
Highest temperature (°C)
1: NaOH(s) in water
21.0
26.2
2: NaOH(s) in HCl(aq)
21.0
32.8
Take the solution mass as 102.0 g in both runs and c = 4.18 J/(g °C). M(NaOH) = 40.00 g/mol.
Analysis questions
Question 1 Fair test
Explain why both runs use the same mass of NaOH and the same volume of liquid.
Question 2 Dissolving
Calculate q for run 1 and ΔH₁ in kJ/mol of NaOH.
Question 3 Reaction with acid
Calculate q for run 2 and ΔH₂ in kJ/mol of NaOH. Show that the moles of acid equal the moles of NaOH.
Question 4 Hess’s law
Draw the energy cycle and calculate ΔH₃ for NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l).
Question 5 Comparison
The accepted value is −57.3 kJ/mol. Calculate the percentage difference.
Question 6 Direction of errors
Explain the effect of heat loss to the room on the magnitude of the result and name one other source of error.
Question 7 Improvement
Suggest two changes that would improve the accuracy.