Hess's Law

📘 Hess's Law

The total enthalpy change for a reaction is independent of the pathway between reactants and products, provided the initial and final conditions are the same. This is a consequence of enthalpy being a state function.

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

The cycle goes: Reactants → Elements → Products

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

Note: reversed compared to formation. Reactants minus products

⚠️ Examiner Traps

• Stoichiometry: Multiply ΔH_f⦵ by the coefficient in the balanced equation
• Elements: ΔH_f⦵ of elements in standard state = 0 (don't forget!)
• Combustion products: CO₂ and H₂O are combustion products. Their ΔH_c⦵ = 0 in the combustion cycle
• Arrow direction: In energy cycle diagrams, if you travel against an arrow, reverse the sign

📐 Worked Example: Applying Hess's Law

Using enthalpies of formation:

ΔH°_rxn = ΣΔH_f°(products) - ΣΔH_f°(reactants)

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

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

Solution:
ΔH = [(-393.5) + 2(-285.8)] - [(-74.8) + 2(0)]
ΔH = [-965.1] - [-74.8]
ΔH = -890.3 kJ mol^-1