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9701 · 5.1

Enthalpy change, ΔH — practice questions

Practice and worked examples for 9701 Enthalpy change, ΔH. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

In an experiment, 50.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid was added to 50.0 cm³ of 1.00 mol dm⁻³ sodium hydroxide solution in a polystyrene cup. The initial temperature of both solutions was 19.5 °C. The maximum temperature reached was 26.3 °C. Calculate the standard enthalpy change of neutralisation, in kJ mol⁻¹. (Assume the density of the solution is 1.00 g cm⁻³ and the specific heat capacity of the solution is 4.18 J g⁻¹ K⁻¹).

Show solution outline
  1. Calculate the heat energy evolved (q): Total volume of solution = 50.0 + 50.0 = 100.0 cm³. Mass of solution (m) = volume × density = 100.0 cm³ × 1.00 g cm⁻³ = 100.0 g. Temperature change (ΔT) = T_{final} - T_{initial} = 26.3 - 19.5 = 6.8 °C (or 6.8 K). q = mcΔT = 100.0 g × 4.18 J g⁻¹ K⁻¹ × 6.8 K = 2842.4 J.
  2. Calculate the moles of water formed (n): Moles of HCl = concentration × volume = 1.00 mol dm⁻³ × (50.0/1000) dm³ = 0.0500 mol. Moles of NaOH = 1.00 mol dm⁻³ × (50.0/1000) dm³ = 0.0500 mol. The reaction is HCl + NaOH → NaCl + H₂O. The stoichiometry is 1:1, so 0.0500 mol of H₂O is formed.
  3. Calculate the molar enthalpy change (ΔH): The temperature increased, so the reaction is exothermic and ΔH is negative. ΔH = -q / n = -2842.4 J / 0.0500 mol = -56848 J mol⁻¹.
  4. Convert to kJ mol⁻¹ and give to 3 significant figures: ΔH = -56848 / 1000 = -56.848 kJ mol⁻¹. ΔH = -56.8 kJ mol⁻¹ (to 3 s.f.)

Worked example 2

A spirit burner containing propan-1-ol (CH₃CH₂CH₂OH) was used to heat 200.0 g of water in a copper calorimeter. The initial mass of the burner was 85.42 g and the final mass was 84.78 g. The temperature of the water rose from 20.5 °C to 55.8 °C. Calculate the enthalpy of combustion of propan-1-ol.

Show solution outline
  1. Calculate the heat energy absorbed by the water (q): Mass of water (m) = 200.0 g. Specific heat capacity of water (c) = 4.18 J g⁻¹ K⁻¹. Temperature change (ΔT) = 55.8 - 20.5 = 35.3 °C (or 35.3 K). q = mcΔT = 200.0 g × 4.18 J g⁻¹ K⁻¹ × 35.3 K = 29510.8 J.
  2. Calculate the moles of propan-1-ol burned (n): Mass of propan-1-ol burned = 85.42 g - 84.78 g = 0.64 g. Molar mass (Mᵣ) of CH₃CH₂CH₂OH = (3 × 12.0) + (8 × 1.0) + 16.0 = 60.0 g mol⁻¹. Moles (n) = mass / Mᵣ = 0.64 g / 60.0 g mol⁻¹ = 0.01067 mol.
  3. Calculate the molar enthalpy change (ΔH): Combustion is always exothermic, so ΔH is negative. We assume all heat from combustion is absorbed by the water, so heat released by reaction = -q. ΔH = -q / n = -29510.8 J / 0.01067 mol = -2765773 J mol⁻¹.
  4. Convert to kJ mol⁻¹ and give to appropriate significant figures: ΔH = -2765773 / 1000 = -2765.8 kJ mol⁻¹. The mass burned (0.64 g) is given to 2 s.f., so the answer should be too. ΔH = -2800 kJ mol⁻¹ (to 2 s.f.)
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