A student dissolves 4.00 g of anhydrous copper(II) sulfate in 50.0 cm³ of water and measures a temperature rise of 14.2 °C. Explain how the student would determine the enthalpy change of dissolution per mole of copper(II) sulfate from this experiment, and identify one significant source of error that would make the calculated value less negative than the true value.

IB DP Chemistry Higher Level (2023 syllabus) — R1.1 Measuring enthalpy changes · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Specific heat capacity of the solution = 4.18 J g⁻¹ K⁻¹. Molar mass of CuSO₄ = 159.6 g mol⁻¹. Assume the density of the solution is 1.00 g cm⁻³.

Model answer (4 marks)

1. Calculate the heat released by the dissolution using q = mcΔT, where m ≈ 50.0 g (density 1.00 g cm⁻³ × 50.0 cm³), c = 4.18 J g⁻¹ K⁻¹ and ΔT = 14.2 K.
2. Find the number of moles of CuSO₄: n = 4.00 g ÷ 159.6 g mol⁻¹ ≈ 0.02506 mol.
3. The molar enthalpy change is ΔH = –q/n (negative because the temperature rises, indicating an exothermic process). Convert the result to kJ mol⁻¹.
4. One significant source of error that would make the calculated ΔH less negative is heat loss from the solution to the surroundings (e.g. to the air or the calorimeter), which reduces the observed temperature rise.

Examiner tips

  • Use the correct formula q = mcΔT and include units; remember to convert mass to grams. Show the calculation of moles and the division to obtain ΔH, then change J to kJ. State the sign convention clearly: exothermic → negative ΔH. Mention heat loss as the error that makes ΔH less negative.

Common mistakes

  • Treating the solution mass as 50.0 g without justification; students often forget to add the mass of the solute. Using ΔT = 14.2 °C but not converting to Kelvin. Failing to include the negative sign for an exothermic process. Ignoring the heat loss error or citing an unrelated error such as incorrect molar mass.

Mark scheme (4 marks)

  1. The heat released by the reaction (q) is calculated using q = mcΔT, where m is the mass of the solution (approximately 50.0 g), c is the specific heat capacity, and ΔT is the temperature change.
  2. The enthalpy change per mole is obtained by dividing q by the number of moles of CuSO₄ dissolved (n = 4.00/159.6 ≈ 0.02506 mol), giving ΔH in J mol⁻¹, then converting to kJ mol⁻¹.
  3. Since the temperature rises, the dissolution is exothermic, so ΔH is assigned a negative sign (ΔH = −q/n).
  4. A significant source of error is heat loss to the surroundings (e.g. to the air or the calorimeter/container), which means less heat is recorded by the solution, so the measured temperature rise is smaller than it should be, making the calculated ΔH less negative than the true value.

Key terms in this question

enthalpy change of dissolution

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