Enthalpy of Formation: Study Guide (Chemistry 6.8)

Why we need a table of formation enthalpies

In Topic 6.7 you estimated ΔH from average bond enthalpies. The estimates were close, but not exact, and the method works only for gases. Chemists want accurate values for thousands of reactions, including reactions of solids and liquids. Measuring every one would take for ever.

The solution is to measure one number per substance and store it in a table: its standard enthalpy of formation. From that table you can calculate ΔH for any reaction whose substances are listed, without doing the experiment.

What a standard enthalpy of formation is

The standard enthalpy of formation, ΔH°f, of a substance is the enthalpy change when one mole of the substance forms from its elements in their standard states.

• The symbol ° means standard conditions: a pressure of 1 bar (almost the same as 1 atm; the difference is too small to matter here) and, in data tables, usually 25 °C (298 K).
• The standard state of an element is its most stable form under those conditions: O₂(g), H₂(g), N₂(g), Cl₂(g), Br₂(l), Hg(l), C(graphite), Na(s), Fe(s), and so on.

Each ΔH°f refers to a specific equation with exactly 1 mol of product, so some equations need fractions:

• ΔH°f of CO₂(g): C(graphite) + O₂(g) → CO₂(g), ΔH° = −393.5 kJ mol⁻¹
• ΔH°f of NH₃(g): ½N₂(g) + 3/2 H₂(g) → NH₃(g), ΔH° = −46.1 kJ mol⁻¹
• ΔH°f of NO(g): ½N₂(g) + ½O₂(g) → NO(g), ΔH° = +90.3 kJ mol⁻¹

Most compounds have negative values: energy is released when they form from their elements. A few, such as NO(g), have positive values.

Elements in their standard state are zero

Forming O₂(g) from O₂(g) is no change at all, so ΔH°f[O₂(g)] = 0. The same is true of every element in its standard state. This is the reference point of the whole table: every other value tells you how far a substance lies above or below its elements.

Be careful: other forms of an element are not zero. Ozone, O₃(g), has ΔH°f = +143 kJ mol⁻¹, and diamond has ΔH°f = +1.9 kJ mol⁻¹, because graphite, not diamond, is the standard state of carbon.

The values used on this page

Standard enthalpies of formation at 298 K, in kJ mol⁻¹, from a standard data table:

Substance	ΔH°f	Substance	ΔH°f
CO₂(g)	−393.5	CaCO₃(s)	−1206.9
H₂O(l)	−285.8	CaO(s)	−635.1
H₂O(g)	−241.8	NH₃(g)	−46.1
C₂H₄(g)	+52.5	NO(g)	+90.3
O₂(g), N₂(g), H₂(g)	0	C(graphite)	0

The formula and where it comes from

To find the standard enthalpy change of a reaction:

ΔH°reaction = ΣnΔH°f(products) − ΣnΔH°f(reactants)

Here Σ means “add up” and n is the coefficient of each substance in the balanced equation. Because each ΔH°f is per mole, you multiply it by the number of moles in the equation.

Why it works. Imagine taking the reaction by a detour through the elements. First, turn the reactants back into their elements: that is the reverse of forming them, so its enthalpy change is −ΣnΔH°f(reactants). Then build the products from the same elements: +ΣnΔH°f(products). Enthalpy depends only on the start and the end, so the detour gives the same ΔH as the direct reaction. Topic 6.9 (Hess’s law) develops this idea fully; for now, Figure 1 shows it as an energy diagram.

A method that avoids sign slips

1. Write the balanced equation with state symbols.
2. Make a table: substance, coefficient n, ΔH°f, and n × ΔH°f. Put 0 for elements in their standard states.
3. Add the products column and add the reactants column separately.
4. Subtract: products total − reactants total. Put brackets around a negative reactants total so the double negative is clear.
5. Check the sign makes sense and the unit is kJ per mole of the equation as written (written kJ mol⁻¹ or kJ mol_rxn⁻¹).

Worked example 1: heating limestone

Question. Calculate ΔH° for CaCO₃(s) → CaO(s) + CO₂(g). Is the reaction exothermic or endothermic?

Substance	n	ΔH°f (kJ mol⁻¹)	n × ΔH°f (kJ)
CaO(s)	1	−635.1	−635.1
CO₂(g)	1	−393.5	−393.5
Products total			−1028.6
CaCO₃(s)	1	−1206.9	−1206.9
Reactants total			−1206.9

ΔH° = (−1028.6) − (−1206.9) = −1028.6 + 1206.9 = +178.3 kJ mol⁻¹.

Interpretation. ΔH° is positive, so the reaction is endothermic. That matches the fact that limestone must be heated strongly to decompose. Figure 1 shows the same calculation as levels.

Check. A common slip is to calculate reactants − products, which gives −178.3 kJ mol⁻¹. Ask yourself: do the products lie above or below the reactant? Here CaO + CO₂ (−1028.6) is above CaCO₃ (−1206.9), so ΔH must be positive.

Worked example 2: coefficients and a zero

Question. The first step in making nitric acid is 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g). Calculate ΔH°, and the enthalpy change per mole of NH₃.

Substance	n	ΔH°f (kJ mol⁻¹)	n × ΔH°f (kJ)
NO(g)	4	+90.3	+361.2
H₂O(g)	6	−241.8	−1450.8
Products total			−1089.6
NH₃(g)	4	−46.1	−184.4
O₂(g)	5	0	0
Reactants total			−184.4

ΔH° = (−1089.6) − (−184.4) = −905.2 kJ mol⁻¹ for the equation as written.

Per mole of NH₃: −905.2 ÷ 4 = −226.3 kJ per mol NH₃.

Points to notice.

• The state matters. Water is formed as a gas here, so you use −241.8, not −285.8.
• O₂ contributes 0, but you still list it so you do not forget anything.
• Ignoring the coefficients gives 90.3 + (−241.8) − (−46.1) = −105.4 kJ, which is far from the right answer.

Worked example 3: working backwards to a missing value

Question. The standard enthalpy of combustion of ethene is −1411.1 kJ mol⁻¹: C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l). Use the values for CO₂(g) and H₂O(l) to find ΔH°f of C₂H₄(g).

1. Products total: 2(−393.5) + 2(−285.8) = −787.0 + (−571.6) = −1358.6 kJ.
2. Reactants total: ΔH°f(C₂H₄) + 3(0) = x.
3. Set up the equation: −1411.1 = −1358.6 − x.
4. Solve: x = −1358.6 + 1411.1 = +52.5 kJ mol⁻¹.

Interpretation. Ethene has a positive enthalpy of formation: it lies above its elements on an energy diagram. That is why burning it releases even more energy than burning the same carbon and hydrogen as the elements would. This “working backwards” method is how tables are built for substances that cannot be made directly from their elements.

Physical processes too

The formula works for any process whose substances are in the table, not only reactions. For vaporising water:

H₂O(l) → H₂O(g): ΔH° = (−241.8) − (−285.8) = +44.0 kJ mol⁻¹

The value is positive because separating water molecules needs energy to overcome the hydrogen bonds between them, as you saw for phase changes in Topic 6.5. This is also why the “state” column matters: using the wrong state for water in a combustion reaction changes the answer by 44.0 kJ for every mole of water.

Formation enthalpies and bond enthalpies compared

In Topic 6.7 you estimated ΔH for H₂(g) + Cl₂(g) → 2HCl(g) as −183 kJ mol⁻¹ from bond enthalpies. With formation data, ΔH°f of HCl(g) is −92.3 kJ mol⁻¹, and both reactants are elements in their standard states:

ΔH° = 2(−92.3) − (0 + 0) = −184.6 kJ mol⁻¹

The two methods agree well here. They will not always agree so closely, and when they differ, trust the formation value. The table below sums up the differences.

	Bond enthalpies (6.7)	Formation enthalpies (6.8)
Formula	Σ(broken) − Σ(formed)	ΣnΔH°f(products) − ΣnΔH°f(reactants)
Which side first	reactants	products
Accuracy	estimate (average values)	accurate (measured for each substance)
States allowed	gases only	any state listed in the table
Best use	explaining why a reaction releases or absorbs energy	calculating how much

Both methods give a ΔH per mole of the equation as written, and both describe the same start and end points. They differ only in the reference they use: separate gaseous atoms for bond enthalpies, elements in their standard states for formation enthalpies.

Common misconceptions

• Reactants minus products. That reverses the sign. It is always products − reactants for formation enthalpies.
• Forgetting coefficients. Each ΔH°f is per mole; multiply by n from the balanced equation.
• Giving elements a non-zero value, or giving every form of an element zero. O₂(g) is 0; O₃(g) is not. C(graphite) is 0; C(diamond) is not.
• Using the wrong state. H₂O(l) and H₂O(g) differ by 44.0 kJ mol⁻¹.
• Mixing up the two methods. Bond enthalpies: broken − formed (reactants first). Formation enthalpies: products − reactants.
• Thinking ΔH°f equations can make more than 1 mol. The definition is per mole of the compound formed, so fractions such as ½N₂ are normal.
• Dropping a minus sign when subtracting a negative. Use brackets: −1028.6 − (−1206.9) = +178.3.

Where this leads

Next, Topic 6.9, Hess’s law, explains why the detour through the elements in Figure 1 must give the same ΔH as the direct reaction, and shows how to combine known reactions to find unknown ones. Try the practice questions now, then use the revision notes and the checklist to consolidate.