Standard enthalpy of formation is a key idea in thermochemistry and a foundation for many calculations in IB Chemistry. It appears frequently in Topic 5, especially in Hess’s law cycles and enthalpy of reaction problems. Understanding the definition and how to apply it correctly will help you solve energetics questions with confidence.
What Is the Standard Enthalpy of Formation?
Standard enthalpy of formation (ΔHf°) is the enthalpy change when one mole of a compound is formed from its elements in their standard states, under standard conditions.
Standard conditions for IB Chemistry are:
- Temperature: 298 K
- Pressure: 100 kPa
- Solutions: 1.0 mol dm⁻³
The symbol ΔHf° emphasizes formation (f) and standard conditions (°).
Key Features of ΔHf°
1. One Mole of Product
Formation equations must always create exactly one mole of the substance, no matter how many reactants it takes.
2. Elements Must Be in Their Standard States
Examples:
- Hydrogen: H₂(g)
- Oxygen: O₂(g)
- Carbon: C(graphite)
- Bromine: Br₂(l)
Standard state means the physically most stable form at 298 K and 100 kPa.
3. Pure Elements Have ΔHf° = 0
Because forming an element from itself requires no enthalpy change.
Examples:
- ΔHf°(O₂) = 0
- ΔHf°(N₂) = 0
- ΔHf°(C, graphite) = 0
This is crucial for Hess’s law cycles.
Example of a Formation Equation
Formation of water:
H₂(g) + ½O₂(g) → H₂O(l)
Key points:
- One mole of water produced
- Elements in standard states
- Fractional coefficients allowed
IB students sometimes avoid fractions, but they are perfectly acceptable in formation equations.
Why Standard Enthalpy of Formation Is Important
You use ΔHf° values to calculate the enthalpy change of a reaction using:
ΔH°reaction = ΣΔHf°(products) – ΣΔHf°(reactants)
This method is essential when:
- Experimental measurement is difficult
- Bond enthalpy values are unavailable
- You are performing Hess’s law calculations
- You need accurate thermodynamic predictions
ΔHf° values are consistent and reliable, making them ideal for theoretical calculations.
Using ΔHf° in a Hess’s Law Cycle
Formation data often appear in:
- Combustion cycles
- Formation cycles
- Decomposition cycles
- Born–Haber lattice cycles
For example, to find the enthalpy of a reaction, you add the enthalpies of formation of products and subtract those of reactants, ensuring stoichiometric coefficients match the equation.
Worked Example (IB-Style)
Find ΔH° for:
2CO(g) + O₂(g) → 2CO₂(g)
Given:
ΔHf°(CO₂) = –393 kJ/mol
ΔHf°(CO) = –110 kJ/mol
ΔHf°(O₂) = 0 kJ/mol
Apply the formula:
Products:
2(–393) = –786
Reactants:
2(–110) + 0 = –220
ΔH°reaction = –786 – (–220) = –566 kJ
This method is standard on Paper 2.
Common Mistakes to Avoid
- Forgetting to ensure one mole of the compound is formed in the equation
- Using incorrect physical states (e.g., water vapor instead of liquid)
- Forgetting that elements have ΔHf° = 0 only in their standard states
- Applying the wrong stoichiometric coefficients
- Confusing ΔHf° with ΔHc° (standard enthalpy of combustion)
Precision matters greatly in energetics.
FAQs
Why do elements have zero enthalpy of formation?
Because formation refers to creating a compound from its elements. An element is already in its simplest form, so forming it requires no energy change.
Why must the product be exactly one mole?
Standard enthalpy values are defined per mole to ensure consistent comparison across substances.
Can ΔHf° be positive?
Yes. Endothermic formation reactions—such as forming nitrogen monoxide (NO)—have positive ΔHf° values.
Conclusion
Standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from elements in their standard states. This definition forms the foundation for many thermodynamic calculations in IB Chemistry, especially in Hess’s law and enthalpy of reaction problems. Once you master ΔHf°, energetics becomes far more logical and systematic.
