Enthalpy, calorimetry and Hess’s law

An enthalpy change, ΔH, is the heat energy transferred in a reaction at constant pressure. It is negative for exothermic reactions and positive for endothermic ones. It can be measured in a calorimeter from a temperature change, or worked out from other enthalpy changes using Hess’s law.

Key facts

  • Heat taken in by water or a solution: q = mcΔT, with m in g, c = 4.18 J g⁻¹ K⁻¹ for water and ΔT in K. Then ΔH = −q ÷ (moles that reacted), in kJ/mol.
  • Standard conditions: 298 K, 100 kPa, solutions 1 mol dm⁻³ and every substance in its standard state. ΔH° means an enthalpy change under standard conditions.
  • Standard enthalpy of formation, ΔHf°: the enthalpy change when one mole of a compound forms from its elements in their standard states. For an element in its standard state ΔHf° = 0.
  • Standard enthalpy of combustion, ΔHc°: the enthalpy change when one mole of a substance burns completely in oxygen.
  • Hess’s law: the enthalpy change of a reaction is the same whatever route is taken, because energy is conserved.
  • From formation data: ΔH = ΣΔHf(products) − ΣΔHf(reactants). From combustion data: ΔH = ΣΔHc(reactants) − ΣΔHc(products).
  • Values from simple calorimetry are usually less exothermic than data-book values: heat is lost to the surroundings and some fuel burns incompletely.

Worked examples

Example 1

Burning 0.46 g of ethanol (Mr = 46) heats 100 g of water from 20.0 °C to 44.0 °C. Calculate the enthalpy change of combustion of ethanol. (c = 4.18 J g⁻¹ K⁻¹)

  1. Temperature rise. ΔT = 44.0 − 20.0 = 24.0 K
  2. Heat taken in by the water. q = 100 × 4.18 × 24.0 = 10 032 J = 10.0 kJ
  3. Moles of ethanol burned. n = 0.46 ÷ 46 = 0.0100 mol
  4. Divide and make it negative, since combustion gives out heat. ΔH = −10.0 ÷ 0.0100 = −1.00 × 10³ kJ/mol
  5. Heat lost to the air and incomplete combustion make this much less exothermic than the data-book value.

Answer: −1.00 × 10³ kJ/mol (the data-book value is −1367 kJ/mol)

Example 2

Calculate ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). ΔHf° in kJ/mol: CH₄ −75, CO₂ −394, H₂O(l) −286.

  1. O₂ is an element in its standard state: ΔHf° = 0.
  2. Products. −394 + 2(−286) = −966
  3. Reactants. −75 + 0 = −75
  4. Subtract. ΔH = −966 − (−75) = −891 kJ/mol

Answer: −891 kJ/mol

Example 3

Find the enthalpy of formation of propane, 3C(s) + 4H₂(g) → C₃H₈(g). ΔHc° in kJ/mol: C −394, H₂ −286, C₃H₈ −2220.

  1. Both sides burn to the same carbon dioxide and water, so use ΔH = ΣΔHc(reactants) − ΣΔHc(products).
  2. Reactants burning. 3(−394) + 4(−286) = −2326 kJ
  3. Subtract the combustion of propane. ΔHf = −2326 − (−2220) = −106 kJ/mol

Answer: −106 kJ/mol