Most chemistry happens in open containers, at the constant pressure of the air, where some energy goes into pushing back the atmosphere. The quantity that tracks heat under those conditions is the enthalpy, and it is written right into a chemical equation, balanced with Mole Ratios.
Key idea
The enthalpy change $\Delta H$ is the heat absorbed by the system at constant pressure: $\Delta H = q_p$. In a thermochemical equation, $\Delta H$ belongs to the reaction as written, for those mole amounts: $$\mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \to 2\,\mathrm{NH_3(g)} \qquad \Delta H = -92.2\ \text{kJ}.$$ Reverse the equation and $\Delta H$ changes sign; multiply the coefficients by a number and multiply $\Delta H$ by the same number. States matter.
Why it works
$H = U + PV$, and at constant pressure $\Delta H = \Delta U + P\Delta V = q_p + w + P\Delta V = q_p$, since $w = -P\Delta V$: the PV work is exactly what $H$ accounts for. Enthalpy is a state function, so going forward and then back must return the same total: the reverse reaction has $-\Delta H$. It is also extensive: making 4 mol of ammonia releases twice the heat of making 2 mol. Water as liquid and as vapor have different enthalpies (vaporization absorbs 44.0 kJ/mol at 25 °C), so $\mathrm{H_2O(l)}$ and $\mathrm{H_2O(g)}$ give different $\Delta H$ values.
Worked example: Reverse and scale
Given $\mathrm{N_2} + 3\,\mathrm{H_2} \to 2\,\mathrm{NH_3}$, $\Delta H = -92.2\ \text{kJ}$, find $\Delta H$ for (a) $2\,\mathrm{NH_3} \to \mathrm{N_2} + 3\,\mathrm{H_2}$ and (b) $\tfrac{1}{2}\mathrm{N_2} + \tfrac{3}{2}\mathrm{H_2} \to \mathrm{NH_3}$.
- (a) The reverse reaction: change the sign: $+92.2\ \text{kJ}$.
- (b) Every coefficient is halved: $\tfrac{1}{2}(-92.2) = -46.1\ \text{kJ}$.
Answer: (a) $+92.2\ \text{kJ}$; (b) $-46.1\ \text{kJ}$.
Worked example: States matter
$\mathrm{CH_4(g)} + 2\,\mathrm{O_2(g)} \to \mathrm{CO_2(g)} + 2\,\mathrm{H_2O(l)}$ has $\Delta H = -890.3\ \text{kJ}$. Condensing water releases $44.0\ \text{kJ/mol}$. Find $\Delta H$ when the water is formed as vapor.
- Making water vapor instead of liquid skips the condensation of 2 mol of water, which would have released $2 \times 44.0 = 88.0\ \text{kJ}$.
- $\Delta H = -890.3 + 88.0 = -802.3\ \text{kJ}$.
Answer: $-802.3\ \text{kJ}$
Worked example: Halving and reversing
$2\,\mathrm{NO_2(g)} \to \mathrm{N_2O_4(g)}$ has $\Delta H = -57.2\ \text{kJ}$. Find $\Delta H$ for $\mathrm{N_2O_4(g)} \to 2\,\mathrm{NO_2(g)}$ and for $\mathrm{NO_2(g)} \to \tfrac{1}{2}\mathrm{N_2O_4(g)}$.
- Reversed: $+57.2\ \text{kJ}$; breaking the dimer apart absorbs energy.
- Halved (same direction as given): $\tfrac{1}{2}(-57.2) = -28.6\ \text{kJ}$.
Answer: $+57.2\ \text{kJ}$ and $-28.6\ \text{kJ}$.
Common mistake
Keeping the sign of $\Delta H$ when the reaction is reversed, or forgetting to double $\Delta H$ when the coefficients are doubled. Check: an exothermic reaction run backward must be endothermic, and twice the reaction must exchange twice the heat.