Some reactions can't be run cleanly in a calorimeter: burning carbon gives a mix of CO and $\mathrm{CO_2}$, so the heat of making pure CO can't be measured directly. Because enthalpy is a state function, the rules from Enthalpy and Thermochemical Equations let us compute it from reactions we can measure.
Key idea
Hess's law: if a reaction is the sum of other reactions, its $\Delta H$ is the sum of their $\Delta H$ values. Arrange the given equations so they add up to the target: reverse an equation (and change the sign of its $\Delta H$) or multiply it (and its $\Delta H$) as needed. Species that appear on both sides cancel.
Why it works
Enthalpy depends only on the state, not on the route. Going from C and $\mathrm{O_2}$ to $\mathrm{CO_2}$ directly, or by way of CO, starts and ends at the same place, so the total enthalpy change is the same. That is just bookkeeping: adding equations adds the enthalpies of everything on each side, so the $\Delta H$ values add too. Reversing a step undoes it; doubling it does it twice.
Worked example: Making carbon monoxide
Find $\Delta H$ for $\mathrm{C(s)} + \tfrac{1}{2}\mathrm{O_2} \to \mathrm{CO}$ from (1) $\mathrm{C} + \mathrm{O_2} \to \mathrm{CO_2}$, $\Delta H = -393.5\ \text{kJ}$ and (2) $\mathrm{CO} + \tfrac{1}{2}\mathrm{O_2} \to \mathrm{CO_2}$, $\Delta H = -283.0\ \text{kJ}$.
- We need C on the left (equation 1 as written) and CO on the right (equation 2 reversed: $\mathrm{CO_2} \to \mathrm{CO} + \tfrac{1}{2}\mathrm{O_2}$, $\Delta H = +283.0$).
- Add: $\mathrm{C} + \mathrm{O_2} + \mathrm{CO_2} \to \mathrm{CO_2} + \mathrm{CO} + \tfrac{1}{2}\mathrm{O_2}$; cancel $\mathrm{CO_2}$ and $\tfrac{1}{2}\mathrm{O_2}$.
- $\Delta H = -393.5 + 283.0 = -110.5\ \text{kJ}$.
Answer: $-110.5\ \text{kJ}$
Worked example: Three steps
Find $\Delta H$ for $2\,\mathrm{C(s)} + \mathrm{H_2(g)} \to \mathrm{C_2H_2(g)}$ from: (1) $\mathrm{C} + \mathrm{O_2} \to \mathrm{CO_2}$, $-393.5\ \text{kJ}$; (2) $\mathrm{H_2} + \tfrac{1}{2}\mathrm{O_2} \to \mathrm{H_2O(l)}$, $-285.8\ \text{kJ}$; (3) $\mathrm{C_2H_2} + \tfrac{5}{2}\mathrm{O_2} \to 2\,\mathrm{CO_2} + \mathrm{H_2O(l)}$, $-1299.6\ \text{kJ}$.
- Need 2 C on the left: equation (1) × 2: $-787.0$.
- Need $\mathrm{H_2}$ on the left: equation (2) as is: $-285.8$.
- Need $\mathrm{C_2H_2}$ on the right: equation (3) reversed: $+1299.6$.
- Check that $\mathrm{CO_2}$, $\mathrm{H_2O}$ and $\mathrm{O_2}$ cancel ($\tfrac{5}{2}\,\mathrm{O_2}$ on each side). Sum: $-787.0 - 285.8 + 1299.6 = +226.8\ \text{kJ}$.
Answer: $+226.8\ \text{kJ}$
Common mistake
Reversing or scaling an equation but forgetting to do the same to its $\Delta H$. In the first example, adding $-393.5$ and $-283.0$ gives $-676.5$, a wrong answer that came from keeping equation (2)'s sign after reversing it. Check: write the adjusted $\Delta H$ beside each adjusted equation before adding anything.