Hess's Law works with any set of reactions. One set is chosen as the universal reference: the reactions that form each compound from its elements. With one table of those, you can find $\Delta H$ for thousands of reactions.
Key idea
The standard enthalpy of formation, $\Delta H^\circ_\text{f}$, is $\Delta H$ for making 1 mol of a substance from its elements in their standard states (1 bar, usually 298 K). For an element in its standard state, such as $\mathrm{O_2(g)}$, $\mathrm{C}$(graphite) or $\mathrm{Fe(s)}$, $\Delta H^\circ_\text{f} = 0$. Then $$\Delta H^\circ_\text{rxn} = \sum n\,\Delta H^\circ_\text{f}(\text{products}) - \sum n\,\Delta H^\circ_\text{f}(\text{reactants}).$$
Why it works
Imagine taking the reaction in two Hess's-law steps through the elements. First break every reactant down into its elements: that is the reverse of forming them, $-\sum n\Delta H^\circ_\text{f}(\text{reactants})$. Then build the products from those same elements: $+\sum n\Delta H^\circ_\text{f}(\text{products})$. The elements are the common starting line, set at zero. That is why "products minus reactants" works, and why an element in its standard state adds nothing.
| Substance | $\Delta H^\circ_\text{f}$ (kJ/mol) |
|---|---|
| $\mathrm{CO_2(g)}$ | −393.5 |
| $\mathrm{H_2O(l)}$ | −285.8 |
| $\mathrm{H_2O(g)}$ | −241.8 |
| $\mathrm{CH_4(g)}$ | −74.6 |
| $\mathrm{CaCO_3(s)}$ | −1207.6 |
| $\mathrm{CaO(s)}$ | −634.9 |
| $\mathrm{O_2(g)}$, $\mathrm{C}$(graphite) | 0 |
Watch the state symbols. $\Delta H^\circ_\text{f}$ is $-285.8\ \text{kJ/mol}$ for $\mathrm{H_2O(l)}$ but $-241.8\ \text{kJ/mol}$ for $\mathrm{H_2O(g)}$. The 44.0 kJ difference is the enthalpy of vaporizing a mole of water at 25 °C, so always use the value that matches the state written in the equation.
Worked example: Burning methane
Find $\Delta H^\circ$ for $\mathrm{CH_4(g)} + 2\,\mathrm{O_2(g)} \to \mathrm{CO_2(g)} + 2\,\mathrm{H_2O(l)}$.
- Products: $(-393.5) + 2(-285.8) = -965.1\ \text{kJ}$.
- Reactants: $(-74.6) + 2(0) = -74.6\ \text{kJ}$.
- $\Delta H^\circ = -965.1 - (-74.6) = -890.5\ \text{kJ}$.
Answer: $-890.5\ \text{kJ}$
Worked example: Decomposing limestone
Find $\Delta H^\circ$ for $\mathrm{CaCO_3(s)} \to \mathrm{CaO(s)} + \mathrm{CO_2(g)}$.
- Products: $-634.9 + (-393.5) = -1028.4\ \text{kJ}$. Reactant: $-1207.6\ \text{kJ}$.
- $\Delta H^\circ = -1028.4 - (-1207.6) = +179.2\ \text{kJ}$: endothermic, which is why lime kilns must be heated strongly.
Answer: $+179.2\ \text{kJ}$
Worked example: Working backward to a formation enthalpy
Ethanol burns by $\mathrm{C_2H_5OH(l)} + 3\,\mathrm{O_2} \to 2\,\mathrm{CO_2} + 3\,\mathrm{H_2O(l)}$, $\Delta H^\circ = -1366.8\ \text{kJ}$. Find $\Delta H^\circ_\text{f}$ of ethanol.
- Products: $2(-393.5) + 3(-285.8) = -1644.4\ \text{kJ}$.
- $-1366.8 = -1644.4 - \Delta H^\circ_\text{f}(\text{ethanol})$, since $\mathrm{O_2}$ contributes 0.
- $\Delta H^\circ_\text{f} = -1644.4 + 1366.8 = -277.6\ \text{kJ/mol}$.
Answer: $-277.6\ \text{kJ/mol}$
Common mistake
Subtracting in the wrong order (reactants minus products, which flips the sign), or giving $\mathrm{O_2(g)}$ a nonzero value. Check: combustion reactions should come out strongly negative; and any element in its standard state contributes exactly zero, however many moles of it there are.