Standard Enthalpy Changes of Combustion (ΔH_c^⦵) and Formation (ΔH_f^⦵) HL Only

Specification Reference R1.2.3

Quick Notes

ΔH_c^⦵ = enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions.
ΔH_f^⦵ = enthalpy change when 1 mole of a compound is formed from its elements in their standard states.
Both measured under standard conditions (298 K, 100 kPa, 1 mol dm⁻³).
Use Hess’s Law with combustion or formation data to calculate enthalpy changes.
Values are in the data booklet, units are kJ mol⁻¹.

Full Notes

Hess’s Law for Enthalpy of Combustion

The enthalpy of combustion (ΔH_c) is the enthalpy change when one mole of a substance is completely burned in oxygen.

Hess’s law can be used to construct cycles that calculate enthalpy of formation from the enthalpies of combustion of the reactants and products.

Worked Example

WorkedExample: Find the enthalpy of formation (ΔHf) of propane (C₃H₈) using the following data.

Reaction:

ΔH_꜀ for C(s) = −394 kJ mol⁻¹
ΔH_꜀ for H₂(g) = −286 kJ mol⁻¹
ΔH_꜀ for C₃H₈(g) = −2220 kJ mol⁻¹
Draw a Hess Cycle:
Showing two possible routes - we can combust the 3C(s) and 4H₂(g) directly to form 3CO₂(g) and 4H₂O(g) or we can can combust the C₃H₈(g) to also form 3CO₂(g) and 4H₂O(g).
Apply Hess’s Law:
Using Hess’s Law we know route 1 = route 2
Meaning
ΔH₁ = ΔH? + ΔH₂
ΔH? = ΔH₁ − ΔH₂
ΔH? = (−2326) − (−2220) = −106 kJ mol⁻¹

Matt’s exam tip

When drawing or using enthalpy cycles, you must ensure equations are balanced and include molar ratios when calculating enthalpy changes for each route.

Hess’s Law for Enthalpy of Formation

The enthalpy of formation (ΔH_f) is the enthalpy change when one mole of a compound is formed from its elements in their standard states.

Elements in their standard states (for example, O₂(g)) have an enthalpy of formation, ΔH_f, of zero – this is really important for calculations involving standard enthalpies of formation.

Hess’s law can be used to determine an enthalpy change of a reaction using standard enthalpies of formation of reactants and products.

Formula for Hess’s Cycle using enthalpy of formation:
ΔHᵣ = ΣΔH_f(products) − ΣΔH_f(reactants)

Worked Example

Calculation for the formation of CO₂
Calculate the enthalpy change for the combustion of carbon, based on the following.

Reaction
C (s) + O₂ (g) → CO₂ (g)
Given Data
ΔH_f^° (CO₂) = −393 kJ mol⁻¹
ΔH_f^° (O₂) = 0 (since elements in their standard states have ΔH_f^° = 0)
Using Hess’s Law
ΔH_r = ΣΔH_f^° (products) − ΣΔH_f^° (reactants)
ΔH_r = ΔH_f^° (CO₂) − (ΔH_f^° (C) + ΔH_f^° (O₂))
ΔH_r = −393 − (0 + 0) = −393 kJ mol⁻¹

Summary

ΔH_c^⦵ is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions.
ΔH_f^⦵ is the enthalpy change when one mole of a compound is formed from its elements in their standard states.
Both are measured under standard conditions (298 K, 100 kPa, 1 mol dm⁻³).
Hess’s Law allows enthalpy changes to be calculated from combustion or formation data.

Structure 2.2 – Linked Course Question

Would you expect allotropes of an element, such as diamond and graphite, to have different ΔH_f^⦵ values?

Yes, allotropes like diamond and graphite have different ΔH_f^⦵ (standard enthalpy of formation) values because they have different structures and bond energies. Even though they are made of the same element (carbon), the energy required to form each allotrope from its standard state is different. In the IB data booklet, graphite is the standard state of carbon, so ΔH_f^⦵ = 0 for graphite, while diamond has a small positive ΔH_f^⦵.