The enthalpy of combustion of propane, graphite and dihydrogen at 298 K are -2220.0 kJ mol⁻¹,-393.5 kJ mol⁻¹ and -285.8 kJ mol⁻¹ respectively. The magnitude of enthalpy of formation of propane (C₃ H₈) is _______________ kJ mol⁻¹. (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Burn the elements, then un-burn propane: reactant combustion enthalpies minus product combustion enthalpies.

1. Target reaction. Formation of propane from its elements in their standard states is 3C(graphite) + 4H₂(g) → C₃H₈(g). Propane has 8 H atoms, so you need 4 moles of H₂, not 8.
2. Hess's law route. Burn 3 mol C and 4 mol H₂ to CO₂ and H₂O. Then reverse the combustion of 1 mol propane to get it back from those same products. The CO₂, H₂O and O₂ all cancel.

   Δ_f H(C₃H₈) = 3 Δ_c H(C) + 4 Δ_c H(H₂) - Δ_c H(C₃H₈)

3. Put in numbers. 3(-393.5) + 4(-285.8) - (-2220.0) = -1180.5 - 1143.2 + 2220.0 = -103.7 kJ mol⁻¹.
4. So. The enthalpy of formation is -103.7 kJ mol⁻¹, so its magnitude to the nearest integer is 104.

The common slip: 1247. Multiplying the H₂ value by 8 (counting H atoms instead of H₂ molecules) gives -1246.9, a magnitude of 1247, not 104.

The idea to keep. When only combustion data are given: Δ H = Σ Δ_c H(reactants) - Σ Δ_c H(products). Write the balanced target equation first so the coefficients are right.

Answer from the NTA final answer key. Solution by PrepWiser.