Worked examples: Thermochemistry

Work through each example by covering the answer, attempting the calculation yourself, then checking every line of your working. In this chapter the marks are in the steps, so we show them all: the heat change from Q = mcθ, then the division by the number of moles, then the unit and the sign. These are original questions in SPM style, not past-year papers. In every case the specific heat capacity of the solution is taken as 4.2 J g⁻¹ °C⁻¹ and the density of the solution as 1 g cm⁻³, the standard values a question provides.

Example 1, Heat of neutralisation

Question. 50 cm³ of 2.0 mol dm⁻³ hydrochloric acid is mixed with 50 cm³ of 2.0 mol dm⁻³ sodium hydroxide. The temperature rises from 29.0 °C to 42.6 °C. Calculate the heat of neutralisation.

Solution.

• Total volume of mixture = 50 + 50 = 100 cm³, so mass m = 100 g.
• Temperature change θ = 42.6 − 29.0 = 13.6 °C.
• Heat change Q = mcθ = 100 × 4.2 × 13.6 = 5712 J = 5.712 kJ.
• Moles of water formed = moles of HCl = (50 ÷ 1000) × 2.0 = 0.10 mol.
• Heat of neutralisation = 5.712 ÷ 0.10 = 57.1 kJ mol⁻¹. The temperature rose, so the reaction is exothermic: ΔH = −57.1 kJ mol⁻¹.

Common slip: using 50 g instead of 100 g for the mass, the heat warms the whole 100 cm³ of mixture.

Example 2, Heat of displacement

Question. Excess zinc powder is added to 100 cm³ of 0.50 mol dm⁻³ copper(II) sulfate solution. The temperature rises from 28.0 °C to 39.5 °C. Calculate the heat of displacement of copper.

Solution.

• Mass of solution m = 100 g.
• θ = 39.5 − 28.0 = 11.5 °C.
• Q = mcθ = 100 × 4.2 × 11.5 = 4830 J = 4.83 kJ.
• Moles of copper displaced = moles of CuSO₄ = (100 ÷ 1000) × 0.50 = 0.050 mol.
• Heat of displacement = 4.83 ÷ 0.050 = 96.6 kJ mol⁻¹, exothermic, so ΔH = −96.6 kJ mol⁻¹.

Common slip: using the mass of zinc in Q = mcθ. The mass in the formula is always the mass of the solution being warmed.

Example 3, Heat of precipitation

Question. 50 cm³ of 1.0 mol dm⁻³ silver nitrate solution is mixed with 50 cm³ of 1.0 mol dm⁻³ sodium chloride solution. A white precipitate forms and the temperature rises from 27.0 °C to 30.4 °C. Calculate the heat of precipitation of silver chloride.

Solution.

• Mass of mixture m = 100 g.
• θ = 30.4 − 27.0 = 3.4 °C.
• Q = mcθ = 100 × 4.2 × 3.4 = 1428 J = 1.428 kJ.
• Moles of AgCl formed = moles of AgNO₃ = (50 ÷ 1000) × 1.0 = 0.050 mol.
• Heat of precipitation = 1.428 ÷ 0.050 = 28.6 kJ mol⁻¹, exothermic, so ΔH = −28.6 kJ mol⁻¹.

Common slip: forgetting the 1 : 1 ratio and doubling the moles because two solutions were mixed.

Example 4, Heat of combustion

Question. When 1.15 g of ethanol (C₂H₅OH, molar mass 46 g mol⁻¹) is completely burnt, it raises the temperature of 200 cm³ of water from 28.0 °C to 60.0 °C. Calculate the heat of combustion of ethanol.

Solution.

• Mass of water m = 200 g (density 1 g cm⁻³).
• θ = 60.0 − 28.0 = 32.0 °C.
• Q = mcθ = 200 × 4.2 × 32.0 = 26 880 J = 26.88 kJ.
• Moles of ethanol burnt = 1.15 ÷ 46 = 0.025 mol.
• Heat of combustion = 26.88 ÷ 0.025 = 1075.2 kJ mol⁻¹, exothermic, so ΔH = −1075 kJ mol⁻¹ (3 s.f.).

Common slip: dividing the heat into the water’s mass instead of the fuel’s moles, or forgetting to work out moles of fuel from mass ÷ molar mass.

Example 5, Reading an energy level diagram

Question. In a reaction the temperature of the mixture falls from 30.0 °C to 24.5 °C. State whether the reaction is exothermic or endothermic, the sign of ΔH, and describe the energy level diagram.

Solution. The temperature falls, so heat is absorbed from the surroundings: the reaction is endothermic and ΔH is positive. On the energy level diagram, the products are drawn higher than the reactants, and the ΔH arrow points upward from the reactant level to the product level.

Common slip: linking a temperature fall to an exothermic reaction. A fall always means heat was taken in, so the reaction is endothermic.

Example 6, Comparing strong and weak acids

Question. The heat of neutralisation of hydrochloric acid with sodium hydroxide is higher than that of ethanoic acid with sodium hydroxide. Explain why.

Solution. Hydrochloric acid is a strong acid and is fully ionised in water, so all its hydrogen ions are already free to react. Ethanoic acid is a weak acid and is only partially ionised, so some heat energy is absorbed to complete the ionisation of the acid during the reaction. Less heat is therefore released overall, giving a lower heat of neutralisation.

Common slip: answering “because it is stronger” without explaining that energy is used to ionise the weak acid.

Example 7, Working backwards to a temperature rise

Question. The heat of neutralisation of a strong acid with a strong alkali is 57.0 kJ mol⁻¹. Predict the temperature rise when 100 cm³ of 1.0 mol dm⁻³ acid is mixed with 100 cm³ of 1.0 mol dm⁻³ alkali.

Solution.

• Moles of water formed = (100 ÷ 1000) × 1.0 = 0.10 mol.
• Heat released Q = 57.0 × 0.10 = 5.7 kJ = 5700 J.
• Total mass of mixture m = 200 g.
• Rearranging Q = mcθ: θ = Q ÷ (mc) = 5700 ÷ (200 × 4.2) = 6.8 °C (to 2 s.f.).

Common slip: forgetting that the mixture is 200 g, not 100 g, which halves the predicted temperature rise.

From examples to marks

Every example uses the same route: find Q with Q = mcθ, convert to kilojoules, work out the number of moles the definition names, divide, then attach the unit and the sign from the direction of the temperature change. Rehearse this until it is automatic, then attempt the practice questions for this chapter and mark your working line by line. A one-to-one teacher can check that each step of your calculation is shown and that no answer ever leaves out its unit or sign, the two habits that turn a right number into a full-mark SPM Chemistry answer.