Hess's Law and Enthalpy Cycles

This lesson covers Hess's law and its application to calculating enthalpy changes using formation enthalpies, combustion enthalpies and mean bond enthalpies. Constructing enthalpy cycle diagrams is a key skill for AQA specification 3.1.4.

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Hess's Law

Key Definition: Hess's law states that the total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.

This is a consequence of the law of conservation of energy. Whether a reaction happens in one step or multiple steps, the overall ΔH is the same.

Route 1:  A ———————————→ D      ΔH₁

Route 2:  A → B → C → D         ΔH₂ + ΔH₃ + ΔH₄

Hess's law:  ΔH₁ = ΔH₂ + ΔH₃ + ΔH₄

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Method 1: Using Standard Enthalpies of Formation (ΔfH°)

The general formula:

ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants)

Enthalpy Cycle Diagram

    Reactants  ————ΔrH°————→  Products
         \                      /
    Σ ΔfH°(reactants)    Σ ΔfH°(products)
           \                  /
            ↘              ↙
             Elements in
            standard states

Following the cycle: ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants)

Worked Example 1

Question: Calculate the standard enthalpy change for the following reaction:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Given: ΔfH°(CH₄) = −74.8 kJ mol⁻¹, ΔfH°(CO₂) = −393.5 kJ mol⁻¹, ΔfH°(H₂O(l)) = −285.8 kJ mol⁻¹

Answer:

ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants)

Products: ΔfH°(CO₂) + 2 × ΔfH°(H₂O) = (−393.5) + 2(−285.8) = −965.1 kJ mol⁻¹

Reactants: ΔfH°(CH₄) + 2 × ΔfH°(O₂) = (−74.8) + 2(0) = −74.8 kJ mol⁻¹

Note: ΔfH°(O₂) = 0 because O₂ is an element in its standard state.

ΔrH° = (−965.1) − (−74.8) = −965.1 + 74.8 = −890.3 kJ mol⁻¹

Worked Example 2

Question: Calculate ΔfH° for ethanol, given:

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l) ΔcH° = −1367 kJ mol⁻¹

ΔfH°(CO₂) = −393.5 kJ mol⁻¹, ΔfH°(H₂O(l)) = −285.8 kJ mol⁻¹

Answer:

ΔcH° = Σ ΔfH°(products) − Σ ΔfH°(reactants)

−1367 = [2(−393.5) + 3(−285.8)] − [ΔfH°(C₂H₅OH) + 3(0)]

−1367 = [−787.0 + (−857.4)] − ΔfH°(C₂H₅OH)

−1367 = −1644.4 − ΔfH°(C₂H₅OH)

ΔfH°(C₂H₅OH) = −1644.4 + 1367 = −277.4 kJ mol⁻¹

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Method 2: Using Standard Enthalpies of Combustion (ΔcH°)

The general formula:

ΔrH° = Σ ΔcH°(reactants) − Σ ΔcH°(products)

Note: The formula is reversed compared with formation enthalpies — reactants minus products.

Enthalpy Cycle Diagram

    Reactants  ————ΔrH°————→  Products
         \                      /
    Σ ΔcH°(reactants)    Σ ΔcH°(products)
           \                  /
            ↘              ↙
            Combustion
             products
            (CO₂ + H₂O)

Worked Example 3

Question: Calculate the enthalpy change for the hydrogenation of ethene:

C₂H₄(g) + H₂(g) → C₂H₆(g)

Given: ΔcH°(C₂H₄) = −1411 kJ mol⁻¹, ΔcH°(H₂) = −286 kJ mol⁻¹, ΔcH°(C₂H₆) = −1560 kJ mol⁻¹

Answer:

ΔrH° = Σ ΔcH°(reactants) − Σ ΔcH°(products)

Reactants: ΔcH°(C₂H₄) + ΔcH°(H₂) = (−1411) + (−286) = −1697 kJ mol⁻¹

Products: ΔcH°(C₂H₆) = −1560 kJ mol⁻¹

ΔrH° = (−1697) − (−1560) = −1697 + 1560 = −137 kJ mol⁻¹

The reaction is exothermic, as expected for the conversion of a weaker π bond to stronger σ bonds (C—H and C—C).

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Method 3: Using Mean Bond Enthalpies

Key Definition: The mean bond enthalpy is the average energy required to break one mole of a particular type of bond in the gaseous state, averaged over a range of different compounds.

The general formula:

ΔrH° ≈ Σ (bond enthalpies of bonds broken) − Σ (bond enthalpies of bonds formed)

"Energy in minus energy out" — breaking bonds requires energy (endothermic); forming bonds releases energy (exothermic).

Key Mean Bond Enthalpies