Hess's law and enthalpy cyclesAQA A-Level Chemistry: Revision notes

Hess's law states that the enthalpy change for a reaction is the same whichever route is taken, provided the initial and final conditions (and states) are the same. It follows from the conservation of energy: if two routes gave different enthalpy changes, energy could be created or destroyed.

Hess's law lets us find enthalpy changes that cannot be measured directly, for example because the reaction is too slow, does not go to completion or gives a mixture of products.

It is used with an enthalpy cycle: the direct route from reactants to products is equal to the sum of the enthalpy changes along an alternative route. If a reaction is reversed, the sign of ΔH is reversed. If an equation is multiplied, ΔH is multiplied by the same factor.

Both are defined under standard conditions: a pressure of 100 kPa, a stated temperature (usually 298 K) and every substance in its standard state.

The standard enthalpy of formation, ΔfH°, is the enthalpy change when one mole of a compound is formed from its elements in their standard states. The ΔfH of an element in its standard state is zero (for example O₂(g), C(graphite), H₂(g)). Diamond is not the standard state of carbon, so its ΔfH is not zero.

The standard enthalpy of combustion, ΔcH°, is the enthalpy change when one mole of a substance burns completely in oxygen. It is always negative.

The ΔcH of graphite equals the ΔfH of carbon dioxide, because both describe C(graphite) + O₂(g) → CO₂(g).

Imagine the reactants breaking down into their elements and the elements building up the products. Then:

ΔrH = ΣΔfH(products) − ΣΔfH(reactants)

Multiply each ΔfH by the number of moles in the balanced equation, and use zero for elements.

Worked example: CaCO₃(s) → CaO(s) + CO₂(g), with ΔfH values of −1207.6, −634.9 and −393.5 kJ mol⁻¹.

ΔrH = [(−634.9) + (−393.5)] − (−1207.6) = +179.2 kJ mol⁻¹, so the reaction is endothermic.

To find the energy change for a given amount, multiply ΔH by the number of moles.

Here the reactants and products are both burned to give the same combustion products, so the cycle runs the other way:

ΔrH = ΣΔcH(reactants) − ΣΔcH(products)

Worked example: the enthalpy of formation of methane, C(s) + 2H₂(g) → CH₄(g), using ΔcH values of C −393.5, H₂ −285.8 and CH₄ −890.3 kJ mol⁻¹.

ΔfH = ΔcH(C) + 2ΔcH(H₂) − ΔcH(CH₄) = (−393.5) + 2(−285.8) − (−890.3) = −74.8 kJ mol⁻¹.

This cannot be measured directly because carbon and hydrogen do not react together to give only methane.

Draw the enthalpy cycle first: put the reaction across the top, the common substances (elements or combustion products) at the bottom, and label each arrow with its data. Follow the arrows: with the arrow, add the value; against the arrow, subtract it.

Checks that save marks:

• balance the equation first, and use the correct number of moles of each species
• state symbols matter, for example H₂O(l) and H₂O(g) have different enthalpies of formation
• give the sign and the units (kJ mol⁻¹)
• for a reaction that is the reverse of one with known ΔH, just change the sign