Hess's Law
Section: 5 Chemical Energetics | Syllabus: Cambridge AS Level Physics 9702
Hess's Law Hess's Law The enthalpy change of a chemical reaction is independent of the route taken, provided the initial and final conditions are the same. Hess's law is a consequence of the law of conservation of energy.
It allows us to calculate enthalpy changes that cannot be measured directly by experiment - by finding an indirect route using known enthalpy values. Figure 5.3: Hess's Law - Two Routes, Same ΔH (Triangle energy cycle: top-left = Reactants (A), top-right = Products (B), bottom = Intermediate (C).
Route 1 (direct): A → B, labelled ΔH₁. Route 2 (indirect): A → C (ΔH₂) then C → B (ΔH₃). By Hess's Law: ΔH₁ = ΔH₂ + ΔH₃. Label both routes clearly.) Energy Cycles Using Enthalpies of Formation The most common type of Hess's law cycle links the enthalpy of reaction to enthalpies of formation of reactants and products: ΔH⦵ᵣ = Σ ΔH⦵f (products) − Σ ΔH⦵f (reactants) Figure 5.4: Hess's Law Cycle Using Enthalpies of Formation (Energy cycle: top row = Reactants → Products (ΔH⦵ᵣ, unknown, arrow going right).
Both reactants and products have downward arrows to the bottom row = Elements in standard states. Reactants arrow labelled −ΣΔH⦵f(reactants) (reversed formation); Products arrow labelled +ΣΔH⦵f(products).
Bottom of cycle completes the triangle. Label all three arrows clearly.) Calculating ΔH⦵ᵣ from ΔH⦵f values Find ΔH⦵ᵣ for: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH⦵f / kJ mol⁻¹: CH₄(g) = −74, O₂(g) = 0, CO₂(g) = −394, H₂O(l) = −286 ΔH⦵ᵣ = [−394 + 2(−286)] − [−74 + 2(0)] = [−394 − 572] − [−74] = −966 − (−74) = −890 kJ mol⁻¹ Energy Cycles Using Enthalpies of Combustion When enthalpies of combustion are given (rather than formation), the cycle is set up differently.
The combustion products (CO₂ and H₂O) are at the bottom: ΔH⦵ᵣ = Σ ΔH⦵c (reactants) − Σ ΔH⦵c (products) Figure 5.5: Hess's Law Cycle Using Enthalpies of Combustion (Energy cycle: top row = Reactants → Products (ΔH⦵ᵣ, unknown).
Both have downward arrows to the bottom row = Combustion products (CO₂ + H₂O). Reactants arrow labelled +ΣΔH⦵c(reactants); Products arrow labelled −ΣΔH⦵c(products) (reversed). Label all arrows. Note: the combustion arrows point downward - toward products of combustion.) Calculating ΔH⦵f of ethanol from combustion data Find ΔH⦵f for C₂H₅OH(l) using: ΔH⦵c(C) = −394, ΔH⦵c(H₂) = −286, ΔH⦵c(C₂H₅OH) = −1367 kJ mol⁻¹ Formation equation: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l) ΔH⦵f = Σ ΔH⦵c(reactants) − Σ ΔH⦵c(products) = [2(−394) + 3(−286) + 0] − [−1367] = [−788 − 858] − [−1367] = −1646 + 1367 = −279 kJ mol⁻¹ Enthalpy Changes That Cannot Be Found Directly Some reactions cannot be performed in a calorimeter - for example, forming carbon monoxide from its elements (some CO₂ always forms too).
Hess's law provides an indirect route: Finding ΔH⦵f of CO(g) - cannot be measured directly Given: (1) C(s) + O₂(g) → CO₂(g) ΔH = −394 kJ mol⁻¹ (2) CO(g) + ½O₂(g) → CO₂(g) ΔH = −283 kJ mol⁻¹ Target: C(s) + ½O₂(g) → CO(g) ΔH⦵f = ?
Route: (1) − (2): ΔH = −394 − (−283) = −111 kJ mol⁻¹ Exam Tip When manipulating equations in Hess's law: if you reverse an equation, change the sign of ΔH. If you multiply an equation by a factor, multiply ΔH by the same factor.
Always check that the final equation matches the target exactly.
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