Hess’s Law

Learning Objectives (OCR Spec 3.2.1)

By the end of this lesson you should be able to:

• State Hess’s Law and explain it as a consequence of the conservation of energy
• Construct enthalpy cycles linking reactants, products and elements (or combustion products)
• Use standard enthalpy changes of formation to calculate ΔrH°
• Apply Hess’s law to reactions that cannot be measured directly
• Apply the algebraic shortcut ΔrH° = ΣΔfH°(products) − ΣΔfH°(reactants)

Statement of Hess’s Law

The total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions (and therefore the initial and final states) are the same.

This is a direct consequence of the first law of thermodynamics: energy cannot be created or destroyed. Because enthalpy is a state function, the difference in enthalpy between two states depends only on those states and not on the path between them.

Enthalpy Cycles

A Hess cycle is a diagram with two routes from reactants to products. One route is the direct reaction; the other goes via a common intermediate set of substances (usually the constituent elements or the combustion products).

Route 1 - Direct

Reactants → Products (ΔrH°, what we want)

Route 2 - Indirect

Reactants → Intermediates → Products

By Hess’s law, ΔH(direct) = ΔH(indirect), so:

ΔrH° = ΣΔH(indirect route)

Using Enthalpies of Formation

When the intermediate substances are the elements in their standard states, the connecting arrows are enthalpies of formation. The cycle looks like this:

graph TD
  R[Reactants] -->|ΔrH° = ?| P[Products]
  E[Elements in standard states] -->|ΔfH° reactants| R
  E -->|ΔfH° products| P

The arrows leaving the elements point towards reactants and products - both are formation arrows.

Deriving the Formula

To go from reactants to products via elements, we must go backwards from reactants to elements (i.e. reverse the formation of reactants, changing the sign) and then forwards from elements to products:

ΔrH° = −ΣΔfH°(reactants) + ΣΔfH°(products)

Rearranged:

ΔrH° = ΣΔfH°(products) − ΣΔfH°(reactants)

This is the most important calculation in OCR thermochemistry. Memorise it.

Remember: ΔfH° of elements in their standard states is zero, so elements contribute nothing to the sum.

Worked Example 1 – Forming Ammonia

Q: Calculate ΔrH° for N2(g) + 3H2(g) → 2NH3(g), given ΔfH°(NH3(g)) = −46.1 kJ mol⁻¹.

A:

ΔrH° = ΣΔfH°(products) − ΣΔfH°(reactants) = [2 × ΔfH°(NH3)] − [ΔfH°(N2) + 3 × ΔfH°(H2)] = [2 × (−46.1)] − [0 + 3 × 0] = −92.2 kJ mol⁻¹

The reaction is exothermic (Haber process). Notice how the stoichiometric coefficients multiply the formation enthalpies.

Worked Example 2 – Combustion of Methane via ΔfH°

Q: Given ΔfH°(CH4(g)) = −74.8, ΔfH°(CO2(g)) = −393.5 and ΔfH°(H2O(l)) = −285.8 kJ mol⁻¹, calculate ΔcH°(CH4).

A:

Equation: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

ΔrH° = [(−393.5) + 2 × (−285.8)] − [(−74.8) + 2 × 0] = [−393.5 + (−571.6)] − [−74.8] = −965.1 − (−74.8) = −965.1 + 74.8 = −890.3 kJ mol⁻¹

Textbook value −5 kJ. Notice: oxygen contributes 0 because it is an element in its standard state.

Worked Example 3 – Thermal Decomposition of Calcium Carbonate

Q: Calculate ΔrH° for the decomposition CaCO3(s) → CaO(s) + CO2(g), given:

Substance	ΔfH° / kJ mol⁻¹
CaCO3(s)	−1207
CaO(s)	−635
CO2(g)	−394

A:

ΔrH° = [ΔfH°(CaO) + ΔfH°(CO2)] − [ΔfH°(CaCO3)] = [−635 + (−394)] − [−1207] = −1029 + 1207 = +178 kJ mol⁻¹

Endothermic, as expected - this is why we need to heat limestone strongly to manufacture quicklime (CaO) for the cement industry.