1.2.4 (HL)—Hess's law applications

Syllabus
First assessment 2025
Objective
1.2.4
Level
HL

Hess-Law Applications

HL only

ΔH°=ΣΔHf°(products)ΣΔHf°(reactants)ΔH°=ΣΔHc°(reactants)ΣΔHc°(products)ΔH° = ΣΔHf°(products) − ΣΔHf°(reactants) ΔH° = ΣΔHc°(reactants) − ΣΔHc°(products)

Apply stoichiometric coefficients to every term, preserve signs, and use the formation or combustion formula that matches the supplied data.

For formation data, imagine every reactant and product connected to the same elements: product sums minus reactant sums gives the target. For combustion data the paths run toward common combustion products, reversing the subtraction. Write coefficients beside every tabulated value before calculating.

Worked example — formation data for pentane combustion: CX5HX12(l)X8+OX2(g)5COX2(g)X6+HX2O(l)\ce{C5H12(l)+8O2(g)->5CO2(g)+6H2O(l)}. Using ΔHf[CX5HX12(l)]=173\Delta H_f^\circ[\ce{C5H12(l)}]=-173, ΔHf[COX2(g)]=394\Delta H_f^\circ[\ce{CO2(g)}]=-394, ΔHf[HX2O(l)]=286\Delta H_f^\circ[\ce{H2O(l)}]=-286 and ΔHf[OX2(g)]=0kJmol1\Delta H_f^\circ[\ce{O2(g)}]=0\,\mathrm{kJ\,mol^{-1}}, ΔH=5(394)+6(286)[173+8(0)]=3513kJmol1\Delta H^\circ=5(-394)+6(-286)-[-173+8(0)]=-3513\,\mathrm{kJ\,mol^{-1}}. The negative sign identifies exothermic combustion.

Calculating with Standard Data

HL only

Assessment in practice

Representative question

Question 1

[Maximum number: 2]

Calculate the standard enthalpy change of formation, ΔHθf\Delta H^{\theta}{ }_{\mathrm{f}}, in kJmol1\mathrm{kJ} \mathrm{mol}^{-1}, of ethyl ethanoate, C4H8O2\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}. Use sections 1 and 13 of the data booklet and the value of 4 kJ mol1-4 \mathrm{~kJ} \mathrm{~mol}^{-1} for the standard enthalpy change of reaction, ΔHθr\Delta H^{\theta}{ }_{\mathrm{r}}.

Energy Cycles Summary

Retrieve the route: count bond breaking/forming, manipulate Hess equations, define standard formation/combustion values, apply product–reactant sums, and track every Born–Haber energy term.

Check equation direction, coefficients, signs, standard states, lattice enthalpy convention, and one-versus-two-electron steps.