Topic 6.8 Notes – Enthalpy of Formation

1. What Standard Enthalpy of Formation Is

The standard enthalpy of formation, written ΔHf°, is the enthalpy change when 1 mole of a compound forms from its elements in their most stable forms at 25°C and 1 atm.

Units are kJ/mol.

Break that definition apart:

• “1 mole” → The reaction must produce exactly one mole of the compound.
• “From elements only” → You cannot start from compounds.
• “Most stable (standard) state” at 25°C and 1 atm
  • $\ce{O2(g)}$, $\ce{N2(g)}$, $\ce{H2(g)}$
  • $\ce{C(s, graphite)}$ (not diamond)
  • $\ce{Br2(l)}$, $\ce{Hg(l)}$

Here’s what a formation reaction looks like:

$$\ce{Na(s) + 1/2 Cl2(g) -> NaCl(s)}$$

That reaction forms 1 mole of NaCl(s) from elements in their standard states.

Elements in Their Standard State

For any element in its standard state,
$$\Delta H_f^\circ = 0$$

Examples:

• $\ce{O2(g)}$
• $\ce{H2(g)}$
• $\ce{C(s, graphite)}$

Students lose easy points by looking these up in tables. They are always zero.

Sign Meaning

• ΔHf° < 0 → Exothermic formation (compound is lower energy than elements)
• ΔHf° > 0 → Endothermic formation

If it’s strongly negative, that compound is very stable relative to its elements.

2. How to Calculate ΔH°rxn from ΔHf° Values

This is the formula you need instantly in your head:

$$\Delta H^\circ_{\text{rxn}} = \sum n\Delta H_f^\circ(\text{products}) - \sum m\Delta H_f^\circ(\text{reactants})$$

It’s always:

Products − Reactants

And you must:

• Multiply each ΔHf° by its stoichiometric coefficient
• Use the correct physical state

Think of it like comparing stored energy:

• Add up energy stored in products
• Subtract energy stored in reactants

The difference is the reaction enthalpy.

3. How to Set Up the Calculation Correctly

Step-by-Step

1. Balance the equation.
2. Write the formula symbolically.
3. Substitute values with coefficients.
4. Do products minus reactants.
5. Include units (kJ for the reaction as written).

Let’s do a quick example:

$$\ce{2SO2(g) + O2(g) -> 2SO3(g)}$$

Using sample values:

• ΔHf°($\ce{SO2(g)}$) = −297 kJ/mol
• ΔHf°($\ce{SO3(g)}$) = −396 kJ/mol
• ΔHf°($\ce{O2(g)}$) = 0

$$\Delta H^\circ_{\text{rxn}} = [2(-396)] - [2(-297) + 1(0)]$$

$$= (-792) - (-594)$$

$$= -198 \text{ kJ}$$

Negative → exothermic.

Details the AP Loves Testing

• Physical states matter.
  $\ce{H2O(l)}$ and $\ce{H2O(g)}$ have different values.
• Do not forget coefficients.
• Keep negatives inside parentheses.
• Elements are zero.

One common trap: forgetting that multiplying by 2 doubles the ΔHf° contribution.

Formation vs Bond Energies

Students mix these up constantly.

Different formulas. Opposite subtraction order.

4. Why This Formula Works

This method comes from Hess’s Law.

Enthalpy is a state function, meaning it depends only on:

• Initial state
• Final state

Not on the pathway.

Standard enthalpies of formation act like building blocks from elements. When you:

• Add up products’ formation values
• Subtract reactants’ formation values

You’re mathematically constructing the net enthalpy change between initial and final states.

That’s why the formula works every time.

5. Interpreting the Final Answer

After calculating:

• Negative ΔH°rxn
  • Exothermic
  • Heat released
  • Products lower in enthalpy
• Positive ΔH°rxn
  • Endothermic
  • Heat absorbed

If comparing reactions, the more negative value releases more heat.

For example:

• −1200 kJ vs −850 kJ
  → −1200 kJ releases more energy.

On free response, you must show:

• The formula
• Substitution with coefficients
• The final comparison statement

They award points for setup, not just the number.