IB ChemistryReactivity 1R1.1R1.1.5

Hess's Law

The enthalpy change for a reaction is independent of the pathway. Energy cycles and indirect calculations.

Reactivity 1.1 SL & HL ⏱️ ~5 min revision
Fundamental Law

Hess's Law of Constant Heat Summation

The total enthalpy change for a reaction is independent of the pathway taken between initial reactants and final products, provided the initial and final conditions are identical. This is a direct consequence of enthalpy being a state function (conserving energy via the First Law of Thermodynamics).

Using Formation Data (ΔHf⦵)

Formation Cycle

Hess's Law Using Enthalpies of Formation

\( \Delta H_r^\ominus = \sum \Delta H_f^\ominus (\text{products}) - \sum \Delta H_f^\ominus (\text{reactants}) \)

Cycle direction: Reactants ← Elements → Products (Elements at the bottom)

Using Combustion Data (ΔHc⦵)

Combustion Cycle

Hess's Law Using Enthalpies of Combustion

\( \Delta H_r^\ominus = \sum \Delta H_c^\ominus (\text{reactants}) - \sum \Delta H_c^\ominus (\text{products}) \)

Note: Reversed order compared to formation. Reactants minus products (Combustion products at bottom)

Examiner Trap

Four Critical Hess's Law Calculation Traps

  • Stoichiometry: Always multiply ΔHf⦵ / ΔHc⦵ by the molar coefficient in the balanced equation.
  • Elements: ΔHf⦵ of elements in their standard state = 0 (do not look for them in data tables).
  • Combustion products: CO₂ and H₂O are complete combustion products: their ΔHc⦵ = 0 in a combustion cycle.
  • Arrow direction: When traversing an energy cycle against the direction of an arrow, reverse the sign of that ΔH step.
Worked Example

Enthalpy of Combustion from Formation Data

Calculate ΔH for: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

Given:
ΔHf°(CH4) = −74.8 kJ mol-1
ΔHf°(CO2) = −393.5 kJ mol-1
ΔHf°(H2O) = −285.8 kJ mol-1
ΔHf°(O2) = 0 (element in standard state)

Solution:
ΔH = ΣΔHf°(products) − ΣΔHf°(reactants)
ΔH = [(−393.5) + 2(−285.8)] − [(−74.8) + 2(0)]
ΔH = [−965.1] − [−74.8] = −890.3 kJ mol-1

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