Types of enthalpy

Enthalpy $H$ is the state function $H = U + PV$, where $U$ is internal energy, $P$ is pressure and $V$ is volume. At constant pressure — the usual laboratory condition, open to the atmosphere — the enthalpy change of a process equals the heat exchanged with the surroundings:

where: $q_p$ is the heat absorbed (positive) or released (negative) at constant pressure. This single line is why $\Delta H$, not $\Delta U$, is the number quoted for almost every everyday reaction. For reactions involving gases, $\Delta H$ and $\Delta U$ differ by a small correction:

where: $\Delta n_g$ is (moles of gaseous products) minus (moles of gaseous reactants), $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ is the gas constant, and $T$ is the absolute temperature. Standard values are quoted at 1 bar and usually 298 K, written $\Delta H^\circ$.

The named types, each tied to a specific defining process:

• Enthalpy of combustion ($\Delta_c H$): 1 mole of a substance burnt completely in excess oxygen. Always exothermic.
• Enthalpy of formation ($\Delta_f H$): 1 mole of a compound formed from its elements in their standard states. $\Delta_f H^\circ$ of an element in its most stable form is defined as zero.
• Enthalpy of atomisation ($\Delta_a H$): 1 mole of a substance broken completely into gaseous atoms. Always endothermic.
• Bond dissociation enthalpy: 1 mole of a specific covalent bond broken in the gas phase. Always endothermic.
• Enthalpy of sublimation / fusion / vaporisation: solid to vapour directly, solid to liquid at the melting point, or liquid to gas at the boiling point, all for 1 mole. All endothermic.
• Enthalpy of ionisation: removing an electron from 1 mole of gaseous atoms or ions. Endothermic.
• Enthalpy of solution ($\Delta_{sol}H$): 1 mole of solute dissolving in a specified or excess solvent; sign depends on the balance between lattice enthalpy (energy to separate the solid) and hydration/solvation enthalpy (energy released as ions or molecules are surrounded by solvent).
• Enthalpy of dilution: the further heat change when an already-formed solution is diluted with more solvent.

Calorimetry provides the experimental link: a bomb calorimeter runs at constant volume, so it measures $q_v = \Delta U$ directly (typical for combustion), while a constant-pressure calorimeter measures $q_p = \Delta H$. Only this qualitative distinction is needed, not instrument design.

For $\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)$, $\Delta U^\circ = -2202\ \text{kJ mol}^{-1}$ at 298 K. Find $\Delta H^\circ$ for the combustion of propane, then use it to classify the reaction and check the size of the correction against the total.

Step 1. Identify gaseous species only — water here is liquid, so it is excluded from the gas count. Gaseous products: $3\ \text{CO}_2$. Gaseous reactants: $1\ \text{C}_3\text{H}_8 + 5\ \text{O}_2 = 6$ moles of gas.

Step 2. Compute $\Delta n_g$ = (gaseous products) − (gaseous reactants) $= 3 - 6 = -3$.

Step 3. Apply $\Delta H = \Delta U + \Delta n_g RT$ with $R = 8.314\times10^{-3}\ \text{kJ K}^{-1}\text{mol}^{-1}$ and $T = 298\ \text{K}$: $\Delta n_g RT = (-3)(8.314\times10^{-3})(298) = -7.432\ \text{kJ mol}^{-1}$.

Step 4. Add the correction: $\Delta H^\circ = -2202 + (-7.432) = -2209.4\ \text{kJ mol}^{-1}$. This is the standard enthalpy of combustion of propane, and it is exothermic, as combustion always is. The correction is only about $0.3\%$ of the total, because it moves relatively few gas moles compared with the huge bond-energy change of complete combustion. $\Delta H^\circ_c(\text{C}_3\text{H}_8) = -2209.4\ \text{kJ mol}^{-1}$.

Competitive stretch

Given the enthalpy of atomisation of $\text{CH}_4(g)$ is $1665\ \text{kJ mol}^{-1}$, find the average C–H bond dissociation enthalpy, and separately classify $\text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(g)$ at 373 K.

Step 1. Atomisation of methane, $\text{CH}_4(g) \rightarrow \text{C}(g) + 4\text{H}(g)$, breaks all 4 equivalent C–H bonds, so the average bond dissociation enthalpy per bond is $\dfrac{1665}{4} = 416.25\ \text{kJ mol}^{-1}$. This is an average — the four real stepwise bond enthalpies in $\text{CH}_4 \rightarrow \text{CH}_3 \rightarrow \text{CH}_2 \rightarrow \text{CH} \rightarrow \text{C}$ are not identical to each other.

Step 2. For $\text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(g)$, heat must be supplied to overcome hydrogen bonding and convert liquid to vapour, so $\Delta H > 0$: the process is endothermic, and since it occurs at the boiling point it is the enthalpy of vaporisation, $\Delta_{vap}H$.

• Forgetting the sign convention in $\Delta n_g$ — count only gaseous species, and take products minus reactants, not the other way round.
• Assuming every enthalpy of formation is exothermic — many are not; formation of $\text{NO}(g)$ from $\text{N}_2$ and $\text{O}_2$ has a positive $\Delta_f H^\circ$.
• Confusing enthalpy of solution with enthalpy of dilution — solution is dissolving in fresh or excess solvent, dilution is the extra heat change on adding more solvent to a solution that already exists.
• Treating tabulated bond dissociation enthalpies for molecules with several identical bonds (like the four C–H bonds in methane) as exact stepwise values rather than averages.

Ask the tutor: If you are unsure which named enthalpy applies to a given process, or how to find $\Delta n_g$ from a balanced equation, ask the AI tutor for a quick drill matching definitions to examples.