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VCE Chemistry (Year 12) · Unit 3 — Supplying Energy: Fuels, Cells and Electrolysis · 14m read

Enthalpy and Calorimetry

Ever wondered why some reactions release heat (like burning fuel) while others absorb it (like instant ice packs)? Enthalpy is simply the total heat energy a substance holds, and calorimetry lets you measure exactly how much heat flows in or out during a chemical reaction using the formula q = mcΔT (where m is mass, c is heat capacity, and ΔT is temperature change). You'll use these ideas constantly in VCE Chemistry to predict reaction behavior and analyze lab data.

Learning objective: Use ΔH values in thermochemical equations and apply the calorimetry formula q = mcΔT to experimental data.

By the end of this lesson you can:

  • I distinguish exothermic (ΔH < 0) from endothermic (ΔH > 0) reactions.
  • I apply q = mcΔT to calculate heat released or absorbed.
  • I scale ΔH per mole when given mass or volume data.

Calorimetry is weighing heat. Measure the temperature change of water to calculate how much energy a reaction released or absorbed.

Enthalpy change and calorimetry

Enthalpy change (ΔH) is the heat absorbed or released at constant pressure. ΔH < 0 means the reaction releases heat to the surroundings — exothermic (combustion, neutralisation, most oxidations). ΔH > 0 means the reaction absorbs heat — endothermic (photosynthesis, dissolving ammonium nitrate).

A thermochemical equation pairs a balanced equation with its ΔH, always for the moles shown. Doubling the equation doubles ΔH; reversing the direction flips its sign.

In a calorimetry experiment, heat transferred to a known mass of water is calculated by q = mcΔT, where m is mass of water (in grams), c is the specific heat capacity of water (4.18 J g⁻¹ °C⁻¹), and ΔT is the temperature change in °C. To convert to molar enthalpy: divide by moles of the limiting reactant — and flip the sign if the reaction released heat (q was gained by water, but lost by the system).

Thermochemistry rules

  • ΔH sign: Negative = exothermic; positive = endothermic.
  • q = mcΔT: c of water = 4.18 J g⁻¹ °C⁻¹.
  • Molar ΔH: ΔH = ± q / n (kJ/mol); flip sign to match system.
  • Scaling: Double the equation → double ΔH.
  • Reversing: Reversed reaction has opposite sign ΔH.

Quick reference

| Formula | Meaning |
|---|---|
| q = mcΔT | Heat (J) |
| c (water) | 4.18 J g⁻¹ °C⁻¹ |
| ΔH | Enthalpy change per mol |
| Exothermic | ΔH negative |
| Endothermic | ΔH positive |
| Calibration factor | CF = E ÷ ΔT |

Worked example (Medium)

Burning 0.50 g of methane raised the temperature of 200 g of water by 30.0 °C. Calculate the enthalpy of combustion of methane in kJ/mol. (M(CH₄) = 16.0 g/mol)

  1. Calculate heat absorbed. q = mcΔT. q = 200 × 4.18 × 30.0 = 25 080 J = 25.08 kJ
  2. Moles of CH₄ burned. n = m/M. n = 0.50 / 16.0 = 0.03125 mol
  3. Molar enthalpy. ΔH per mole, sign negative (exothermic). ΔH = −25.08 / 0.03125 = −802.6 kJ/mol

Answer: ΔH(combustion of CH₄) ≈ −803 kJ/mol (exothermic).

Worked example (Hard)

A student burns 0.80 g of ethanol (M = 46.07 g/mol, literature ΔH = −1367 kJ/mol) under 200 g of water. Temperature rises 18.5°C. Calculate experimental ΔH and the percentage accuracy.

  1. Heat absorbed. q = mcΔT = 200 × 4.18 × 18.5 = 15466 J = 15.47 kJ.
  2. Moles. n = 0.80/46.07 = 0.01736 mol.
  3. Experimental ΔH. ΔH = −15.47/0.01736 = −891 kJ/mol.
  4. Accuracy. |−891|/|−1367| × 100 = 65.2%. Heat loss accounts for the 34.8% discrepancy.

Answer: Experimental ΔH = −891 kJ/mol (65.2% of literature value). The discrepancy is due to heat lost to surroundings — a systematic error in simple calorimetry.

Common mistakes

  • Forgetting to flip the sign of ΔH (water gains heat but the reaction releases it) This is a commonly penalised error in VCAA exams. Fix: Review the underlying concept and practise similar problems.
  • Not converting joules to kilojoules before reporting ΔH in kJ/mol Incorrect application leads to wrong answers. Fix: Always return to first principles and verify your working.
  • Failing to explain WHY the experimental value is lower than the literature value A frequent source of confusion in VCE Chemistry. Fix: Distinguish between related concepts through targeted practice.

Hess's Law and enthalpy cycles

Hess’s Law states that enthalpy change depends only on initial and final states, not the pathway. This allows calculation of ΔH for reactions that cannot be directly measured. Three methods: (1) Using formation enthalpies: ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants). (2) Using combustion enthalpies: ΔH°rxn = ΣΔH°c(reactants) − ΣΔH°c(products). (3) Algebraic manipulation of known equations. Bond energy calculations give approximate values: ΔH ≈ Σ(bonds broken) − Σ(bonds formed). VCAA frequently tests Hess’s Law calculations.

Practice questions

  1. An exothermic reaction has a ΔH that is:

    1. Negative
    2. Positive
    3. Zero
    4. Always in J

    (2 marks)

  2. If 50.0 g of water warms by 10.0 °C, how much heat (in kJ) did it absorb?

    1. 2.09 kJ
    2. 0.21 kJ
    3. 50.0 kJ
    4. 20.9 kJ

    (3 marks)

  3. The thermochemical equation for combustion of ethane is 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O, ΔH = −3120 kJ. What is ΔH per mole of C₂H₆?

    (3 marks)

Summary

  • Distinguish exothermic (ΔH < 0) from endothermic (ΔH > 0) reactions.
  • Apply q = mcΔT to calculate heat released or absorbed.
  • Scale ΔH per mole when given mass or volume data.

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