AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/6/6-8)
Unit 6 · Topic 6.8
6.8 Enthalpy of Formation
The standard enthalpy of formation, ΔH°f, is the enthalpy change for making one mole of a compound from its elements in their standard states. With a table of ΔH°f values you can calculate ΔH° for almost any reaction: products minus reactants, each multiplied by its coefficient.
Key terms
- standard enthalpy of formation (ΔH°f)
- standard state
- ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants)
What a formation reaction is
A formation reaction makes exactly one mole of a compound from its elements, each in its standard state. Standard state means the most stable form of the substance at 1 atm (or 1 bar) and the stated temperature, usually 25 °C. Solutions in standard state are 1 M.
Examples: C(s, graphite) + O₂(g) → CO₂(g), ΔH°f = −393.5 kJ/mol. H₂(g) + ½O₂(g) → H₂O(l), ΔH°f = −285.8 kJ/mol. Fractional coefficients are fine here, because the product must be exactly one mole.
A negative ΔH°f means the compound is lower in energy than the elements it's made from.
Elements in their standard states are zero
Forming an element from itself in the same state is no change at all, so ΔH°f = 0 for any element in its standard state. Examples: O₂(g), N₂(g), H₂(g), Cl₂(g), Br₂(l), I₂(s), Hg(l), Fe(s), Na(s) and C(s, graphite).
Other forms of an element are not zero. O₃(g), Br₂(g), I₂(g) and C(s, diamond) all have nonzero ΔH°f values because they are not the most stable form.
Writing formation equations correctly
A true formation equation has exactly one mole of the compound as its only product, and only elements in their standard states as reactants. For ammonia: ½N₂(g) + 3/2 H₂(g) → NH₃(g), ΔH°f = −46.1 kJ/mol. For ethanol: 2C(s, graphite) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l).
Watch for equations that look similar but aren't formation reactions. 2H₂(g) + O₂(g) → 2H₂O(l) makes two moles of water, so its ΔH° is 2 × (−285.8) = −571.6 kJ, not ΔH°f. H(g) + Cl(g) → HCl(g) starts from single atoms, which aren't the standard states of hydrogen and chlorine, so it isn't a formation reaction either.
The products-minus-reactants formula
ΔH°rxn = ΣnΔH°f(products) − ΣnΔH°f(reactants), where each n is the coefficient from the balanced equation.
Why it works: imagine breaking every reactant back down into its elements (the reverse of their formation reactions, so the sign flips), then building the products from those elements. By Hess's law (topic 6.9), adding those steps gives the overall ΔH°.
Phases matter. H₂O(l) is −285.8 kJ/mol but H₂O(g) is −241.8 kJ/mol. Use the value that matches the phase in the equation.
| Substance | ΔH°f (kJ/mol) |
|---|---|
| CO₂(g) | −393.5 |
| H₂O(l) | −285.8 |
| H₂O(g) | −241.8 |
| C₃H₈(g) | −103.8 |
| C₂H₅OH(l) | −277.6 |
| Fe₂O₃(s) | −824.2 |
| Al₂O₃(s) | −1675.7 |
| O₂(g), Al(s), Fe(s) | 0 |
Using the formula in reverse
If you know ΔH° for a reaction and every ΔH°f but one, you can solve for the missing one. This is how many ΔH°f values are actually found: by burning a compound in a calorimeter and working backward.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Combustion of propane
Use the table to calculate ΔH° for C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l).
Show the solutionHide the solution
- Step 1: Products: 3(−393.5) + 4(−285.8) = −1180.5 + (−1143.2) = −2323.7 kJ.
- Step 2: Reactants: 1(−103.8) + 5(0) = −103.8 kJ. O₂ is an element in its standard state.
- Step 3: ΔH° = products − reactants = −2323.7 − (−103.8) = −2219.9 kJ/mol.
- Step 4: Watch the double negative in the last step; it's where sign errors happen.
Answer: ΔH° ≈ −2220 kJ/mol
- Example 2Calculator allowed
Finding an unknown ΔH°f
For C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH° = −1366.8 kJ/mol. Using ΔH°f values for CO₂(g) and H₂O(l) from the table, find ΔH°f of ethanol.
Show the solutionHide the solution
- Step 1: Write the formula with x for ethanol's ΔH°f: −1366.8 = [2(−393.5) + 3(−285.8)] − [x + 3(0)].
- Step 2: Products: −787.0 + (−857.4) = −1644.4 kJ.
- Step 3: −1366.8 = −1644.4 − x, so x = −1644.4 + 1366.8 = −277.6 kJ/mol.
Answer: ΔH°f(C₂H₅OH, l) = −277.6 kJ/mol
- Example 3Calculator allowed
Trap: elements count as zero
The thermite reaction is Fe₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Fe(s). Calculate ΔH° using the table.
Show the solutionHide the solution
- Step 1: Al(s) and Fe(s) are elements in their standard states, so their ΔH°f values are 0. Don't hunt for nonzero values for them, and don't leave them out of the setup without knowing why.
- Step 2: Products: −1675.7 + 2(0) = −1675.7 kJ. Reactants: −824.2 + 2(0) = −824.2 kJ.
- Step 3: ΔH° = −1675.7 − (−824.2) = −851.5 kJ/mol. Strongly exothermic, which is why thermite produces molten iron.
Answer: ΔH° = −851.5 kJ/mol
Common mistakes
- Subtracting products from reactants. For ΔH°f values the order is products minus reactants.
- Forgetting to multiply each ΔH°f by its coefficient.
- Giving an element in its standard state a nonzero ΔH°f, or giving O₃ or diamond a value of zero.
- Using ΔH°f for H₂O(g) when the equation has H₂O(l).
On the exam
- The exam supplies a table of ΔH°f values. Show the setup with coefficients before plugging in; partial credit often depends on it.
- Be ready to compare an answer from ΔH°f values with one from bond enthalpies and explain why they differ.
Connected topics
Videos
Check yourself
4 questions on 6.8 Enthalpy of Formation. Pick an answer to see if you got it, and why.
| Substance | ΔH°f (kJ/mol) |
|---|---|
| CH₄(g) | −74.8 |
| CO₂(g) | −393.5 |
| H₂O(l) | −285.8 |
Standard enthalpies of formation at 25 °C, for the reaction CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
What is ΔH° for the combustion of methane?
O₂(g) does not appear in the table. Why is it not needed for the calculation?
Which of the following equations has a ΔH° equal to the standard enthalpy of formation of H₂O(l)?
| Substance | ΔH°f (kJ/mol) |
|---|---|
| NH₃(g) | −46.1 |
| NO(g) | +90.3 |
| H₂O(g) | −241.8 |
Standard enthalpies of formation at 25 °C
Use the data to calculate ΔH° for 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g).
0 of 4 answered