Standard Enthalpy of Reaction
The enthalpy of a reaction depends on the temperature and pressure at which it occurs, so to make values comparable we specify a standard state. The standard state of a substance at a given temperature is its pure form at 1 bar; for example, the standard state of liquid ethanol at 298 K is pure liquid ethanol at 1 bar, and that of solid iron at 500 K is pure iron at 1 bar. Data are usually tabulated at 298 K.
The standard enthalpy of reaction, written $\Delta_r H^\ominus$, is the enthalpy change for a reaction when every participating substance is in its standard state. The superscript $\ominus$ denotes standard conditions. The unit is $\text{kJ mol}^{-1}$, meaning "per mole of reaction" as written: once the equation is balanced in a particular way, that fixes one mole of reaction.
Because enthalpy is an extensive property, doubling the coefficients doubles $\Delta_r H^\ominus$, and reversing the equation reverses its sign. For ammonia synthesis $\ce{N2(g) + 3H2(g) -> 2NH3(g)}$ has $\Delta_r H^\ominus = -91.8~\text{kJ mol}^{-1}$, while the reverse decomposition $\ce{2NH3(g) -> N2(g) + 3H2(g)}$ has $\Delta_r H^\ominus = +91.8~\text{kJ mol}^{-1}$.
"Per mole" is per mole of reaction, not per mole of any one species
When NCERT balances $\ce{Fe2O3(s) + 3H2(g) -> 2Fe(s) + 3H2O(l)}$ one way it gets $\Delta_r H^\ominus = -33.3~\text{kJ mol}^{-1}$; halving every coefficient gives exactly half, $-16.6~\text{kJ mol}^{-1}$. The number is tied to the equation, so always read the coefficients before plugging in.
Change the stoichiometric coefficients → scale $\Delta_r H^\ominus$ by the same factor. Reverse the equation → flip the sign.
Exothermic vs Endothermic Profiles
A reaction that releases heat to the surroundings is exothermic and carries a negative $\Delta_r H$; one that absorbs heat is endothermic with a positive $\Delta_r H$ (NIOS §9.2). Burning methane, $\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$ with $\Delta H = -891~\text{kJ}$, is exothermic, whereas $\ce{H2(g) + I2(g) -> 2HI(g)}$ with $\Delta H = +52.2~\text{kJ}$ is endothermic. The sign of $\Delta_r H$ is the single fact a profile diagram has to communicate.
Reaction enthalpy profiles. In an exothermic reaction the products sit lower on the enthalpy axis than the reactants, so heat is released ($\Delta H < 0$). In an endothermic reaction the products sit higher, so heat is absorbed ($\Delta H > 0$). The vertical gap between the reactant and product levels is $\Delta_r H$.
Enthalpy of Formation
The standard molar enthalpy of formation, $\Delta_f H^\ominus$, is the standard enthalpy change when one mole of a compound is formed from its elements in their most stable reference states. The reference state of an element is its most stable state of aggregation at 25 °C and 1 bar — so the reference states of hydrogen, oxygen, carbon and sulphur are $\ce{H2(g)}$, $\ce{O2(g)}$, $\ce{C(graphite)}$ and $\ce{S(rhombic)}$ respectively. Three NCERT examples:
$\ce{H2(g) + 1/2 O2(g) -> H2O(l)}$; $\Delta_f H^\ominus = -285.8~\text{kJ mol}^{-1}$
$\ce{C(graphite) + 2H2(g) -> CH4(g)}$; $\Delta_f H^\ominus = -74.81~\text{kJ mol}^{-1}$
$\ce{2C(graphite) + 3H2(g) + 1/2 O2(g) -> C2H5OH(l)}$; $\Delta_f H^\ominus = -277.7~\text{kJ mol}^{-1}$
A formation enthalpy is just a special case of a reaction enthalpy in which exactly one mole of a single compound is formed from elements. By convention, the $\Delta_f H^\ominus$ of any element in its reference state is taken as zero. Two reactions illustrate what does not qualify: $\ce{CaO(s) + CO2(g) -> CaCO3(s)}$ is not a formation reaction because the carbonate is built from compounds, and $\ce{H2(g) + Br2(l) -> 2HBr(g)}$ produces two moles, so its enthalpy equals $2\,\Delta_f H^\ominus(\ce{HBr})$ — to get the formation value we divide through to write $\ce{1/2 H2(g) + 1/2 Br2(l) -> HBr(g)}$ with $\Delta_f H^\ominus = -36.4~\text{kJ mol}^{-1}$.
One mole of product, from elements, in reference states — all three must hold
Examiners disguise non-formation reactions by giving the right element but the wrong allotrope, or by forming two moles of product. If the equation makes more than one mole of the compound, or starts from a compound rather than an element, the listed $\Delta H$ is a reaction enthalpy, not $\Delta_f H^\ominus$.
Diamond is not the reference state of carbon — graphite is. So $\Delta_f H^\ominus(\ce{C, graphite}) = 0$ but $\Delta_f H^\ominus(\ce{C, diamond}) \neq 0$.
The ΔfH Summation Rule
Because enthalpy is a state function, the reaction enthalpy can be obtained from tabulated formation enthalpies through the central NEET formula:
$$\Delta_r H^\ominus = \sum_i a_i\,\Delta_f H^\ominus(\text{products}) - \sum_i b_i\,\Delta_f H^\ominus(\text{reactants})$$where $a_i$ and $b_i$ are the stoichiometric coefficients of products and reactants. Applying it to the decomposition of limestone, $\ce{CaCO3(s) -> CaO(s) + CO2(g)}$, with $\Delta_f H^\ominus$ values of $-635.1$, $-393.5$ and $-1206.9~\text{kJ mol}^{-1}$ for $\ce{CaO}$, $\ce{CO2}$ and $\ce{CaCO3}$:
$$\Delta_r H^\ominus = [(-635.1) + (-393.5)] - (-1206.9) = +178.3~\text{kJ mol}^{-1}$$The positive value confirms that decomposing calcium carbonate is endothermic — a chemical engineer must supply heat to drive it. This first worked example puts the rule to work on a multi-term reaction.
For $\ce{Fe2O3(s) + 3H2(g) -> 2Fe(s) + 3H2O(l)}$, calculate $\Delta_r H^\ominus$ given $\Delta_f H^\ominus(\ce{H2O, l}) = -285.83$ and $\Delta_f H^\ominus(\ce{Fe2O3, s}) = -824.2~\text{kJ mol}^{-1}$.
Step 1 — set element baselines. By convention $\Delta_f H^\ominus(\ce{Fe, s}) = 0$ and $\Delta_f H^\ominus(\ce{H2, g}) = 0$, since both are elements in their reference states.
Step 2 — apply the rule with coefficients. Products give $2(0) + 3(-285.83)$; reactants give $1(-824.2) + 3(0)$.
$$\Delta_r H^\ominus = 3(-285.83) - (-824.2) = -857.5 + 824.2 = -33.3~\text{kJ mol}^{-1}$$
Result. $\Delta_r H^\ominus = -33.3~\text{kJ mol}^{-1}$ — exothermic. Had the equation been written with half-coefficients, the answer would halve to $-16.6~\text{kJ mol}^{-1}$, demonstrating that enthalpy is extensive.
The summation rule is one face of a more general principle. See how indirect routes combine in Hess's Law of Constant Heat Summation.
Enthalpy of Combustion
The standard enthalpy of combustion, $\Delta_c H^\ominus$, is the enthalpy change per mole of a substance when it is completely burnt with all reactants and products in their standard states. Combustion reactions are always exothermic, which is why they dominate industry, rocketry and everyday fuels. The cooking gas in cylinders is mostly butane, and burning one mole releases 2658 kJ:
$\ce{C4H10(g) + 13/2 O2(g) -> 4CO2(g) + 5H2O(l)}$; $\Delta_c H^\ominus = -2658.0~\text{kJ mol}^{-1}$
Glucose combustion releases 2802 kJ per mole via $\ce{C6H12O6(g) + 6O2(g) -> 6CO2(g) + 6H2O(l)}$, $\Delta_c H^\ominus = -2802.0~\text{kJ mol}^{-1}$ — and the body extracts energy from food by the same overall oxidation, though through a long chain of enzyme-mediated steps. Combustion data are valuable precisely because they let us reach formation enthalpies that cannot be measured directly, as the benzene problem below shows.
Combustion of one mole of benzene at 298 K and 1 atm liberates 3267.0 kJ, forming $\ce{CO2(g)}$ and $\ce{H2O(l)}$. Find $\Delta_f H^\ominus$ of benzene. Take $\Delta_f H^\ominus(\ce{CO2, g}) = -393.5$ and $\Delta_f H^\ominus(\ce{H2O, l}) = -285.83~\text{kJ mol}^{-1}$. (NCERT Problem 5.9)
Target. $\ce{6C(graphite) + 3H2(g) -> C6H6(l)}$, $\Delta_f H^\ominus = ?$
Combustion (given). $\ce{C6H6(l) + 15/2 O2(g) -> 6CO2(g) + 3H2O(l)}$, $\Delta_c H^\ominus = -3267~\text{kJ mol}^{-1}$.
Apply the summation rule to the combustion. The combustion enthalpy equals products minus reactants, where benzene's formation enthalpy is the only unknown:
$$\Delta_c H^\ominus = \big[6\,\Delta_f H^\ominus(\ce{CO2}) + 3\,\Delta_f H^\ominus(\ce{H2O})\big] - \Delta_f H^\ominus(\ce{C6H6})$$
$$-3267 = \big[6(-393.5) + 3(-285.83)\big] - \Delta_f H^\ominus(\ce{C6H6})$$
$$-3267 = (-2361 - 857.49) - \Delta_f H^\ominus(\ce{C6H6}) = -3218.49 - \Delta_f H^\ominus(\ce{C6H6})$$
Solve. $\Delta_f H^\ominus(\ce{C6H6, l}) = -3218.49 - (-3267) = +48.5~\text{kJ mol}^{-1}$. NCERT obtains the value via Hess addition as $\Delta_f H^\ominus = +48.51~\text{kJ mol}^{-1}$ — endothermic, signalling that benzene is energetically less stable than its elements.
Phase-Transition Enthalpies
Phase changes occur at constant pressure and constant temperature, yet still involve heat. The standard enthalpy of fusion ($\Delta_{fus}H^\ominus$) accompanies melting one mole of solid, vaporisation ($\Delta_{vap}H^\ominus$) accompanies boiling one mole of liquid, and sublimation ($\Delta_{sub}H^\ominus$) accompanies a solid passing directly to vapour. All three are endothermic and positive, because each must overcome intermolecular attraction.
| Transition | Symbol | Example reaction | ΔH° (kJ mol⁻¹) |
|---|---|---|---|
| Fusion (melting) | $\Delta_{fus}H^\ominus$ | $\ce{H2O(s) -> H2O(l)}$ | +6.00 |
| Vaporisation (boiling) | $\Delta_{vap}H^\ominus$ | $\ce{H2O(l) -> H2O(g)}$ | +40.79 |
| Sublimation (dry ice) | $\Delta_{sub}H^\ominus$ | $\ce{CO2(s) -> CO2(g)}$ | +25.2 (at 195 K) |
| Sublimation (naphthalene) | $\Delta_{sub}H^\ominus$ | $\ce{C10H8(s) -> C10H8(g)}$ | +73.0 |
The magnitude reflects intermolecular force strength. Strong hydrogen bonds hold water molecules tightly, so its $\Delta_{vap}H^\ominus$ is large; acetone, held only by weaker dipole–dipole interactions, vaporises with much less heat. Sublimation is conceptually fusion plus vaporisation in one step, which is why $\Delta_{sub}H^\ominus$ values are comparatively large.
Atomisation & Bond Enthalpy
The enthalpy of atomisation, $\Delta_a H^\ominus$, is the enthalpy change on breaking one mole of bonds completely to give atoms in the gas phase. For dihydrogen $\ce{H2(g) -> 2H(g)}$, $\Delta_a H^\ominus = 435.0~\text{kJ mol}^{-1}$; for methane $\ce{CH4(g) -> C(g) + 4H(g)}$, $\Delta_a H^\ominus = 1665~\text{kJ mol}^{-1}$. For a metal it equals the sublimation enthalpy: $\ce{Na(s) -> Na(g)}$, $\Delta_a H^\ominus = 108.4~\text{kJ mol}^{-1}$.
For a diatomic molecule the bond dissociation enthalpy equals the atomisation enthalpy — for instance $\ce{Cl2(g) -> 2Cl(g)}$, $\Delta_{Cl-Cl}H^\ominus = 242~\text{kJ mol}^{-1}$ and $\ce{O2(g) -> 2O(g)}$, $\Delta_{O=O}H^\ominus = 428~\text{kJ mol}^{-1}$. In polyatomic molecules, the four successive C–H dissociations in methane cost different amounts (427, 439, 452 and 347 kJ mol⁻¹), so we use the mean bond enthalpy: $\Delta_{C-H}H^\ominus = \tfrac{1}{4}(1665) = 416~\text{kJ mol}^{-1}$.
The reaction enthalpy in the gas phase then follows from a simple balance of bonds broken against bonds formed:
$$\Delta_r H^\ominus = \sum (\text{bond enthalpies})_{\text{reactants}} - \sum (\text{bond enthalpies})_{\text{products}}$$In words, $\Delta_r H^\ominus$ equals the energy needed to break all reactant bonds minus the energy needed to break all product bonds. The relationship is approximate and holds only when every reactant and product is gaseous.
The bond dissociation energies of $\ce{X2}$, $\ce{Y2}$ and $\ce{XY}$ are in the ratio $1 : 0.5 : 1$. $\Delta H$ for the formation of $\ce{XY}$ is $-200~\text{kJ mol}^{-1}$. Find the bond dissociation energy of $\ce{X2}$.
Set up. Let the bond enthalpies be $\ce{X2} = a$, $\ce{Y2} = 0.5a$, $\ce{XY} = a$. The formation reaction is $\ce{1/2 X2 + 1/2 Y2 -> XY}$.
Apply the bond balance. Bonds broken in reactants $= \tfrac{1}{2}a + \tfrac{1}{2}(0.5a)$; bond formed in product $= a$.
$$\Delta_f H^\ominus = \left(\tfrac{a}{2} + \tfrac{0.5a}{2}\right) - a = \frac{1.5a}{2} - a = 0.75a - a = -0.25a$$
Solve. $-200 = -0.25a \;\Rightarrow\; a = \dfrac{-200}{-0.25} = 800~\text{kJ mol}^{-1}$.
Result. The bond dissociation energy of $\ce{X2}$ is $800~\text{kJ mol}^{-1}$ — matching the official key, option (3).
Lattice & Solution Enthalpy
The lattice enthalpy of an ionic compound is the enthalpy change when one mole of the solid dissociates into its gaseous ions: $\ce{NaCl(s) -> Na+(g) + Cl^-(g)}$, $\Delta_{lattice}H^\ominus = +788~\text{kJ mol}^{-1}$. It cannot be measured directly, so it is found indirectly through a Born–Haber cycle, which assembles sublimation, ionisation, dissociation and electron-gain enthalpies. Because enthalpy is a state function, the sum of enthalpy changes around the closed cycle is zero (Hess's law), and the lattice enthalpy emerges as $788$ kJ from $411.2 + 108.4 + 121 + 496 - 348.6$ for NaCl.
Born–Haber cycle for NaCl. The solid is built either directly (green) or via gaseous atoms and ions (purple, then lattice formation in red). Because enthalpy is a state function, the sum of changes around the closed cycle is zero — applying Hess's law gives $\Delta_{lattice}H^\ominus = 411.2 + 108.4 + 121 + 496 - 348.6 = +788~\text{kJ mol}^{-1}$ for the dissociation $\ce{NaCl(s) -> Na+(g) + Cl^-(g)}$.
When the solid dissolves, the lattice must break (endothermic) while the freed ions are hydrated (exothermic). The enthalpy of solution is the sum of these competing terms:
$$\Delta_{sol}H^\ominus = \Delta_{lattice}H^\ominus + \Delta_{hyd}H^\ominus$$For sodium chloride, $\Delta_{lattice}H^\ominus = +788$ and $\Delta_{hyd}H^\ominus = -784~\text{kJ mol}^{-1}$, so $\Delta_{sol}H^\ominus = +4~\text{kJ mol}^{-1}$ — barely any heat change, which is why NaCl dissolves so readily. For most ionic salts $\Delta_{sol}H^\ominus$ is positive and dissolution is endothermic, so solubility usually rises with temperature; if the lattice enthalpy is very large, the salt may scarcely dissolve at all.
Lattice enthalpy in NCERT is for dissociation, so it is positive
NCERT defines lattice enthalpy as the endothermic break-up $\ce{NaCl(s) -> Na+(g) + Cl^-(g)}$ with $\Delta_{lattice}H^\ominus = +788~\text{kJ mol}^{-1}$. Some other texts define it for the reverse (lattice formation), giving a negative number. Use the NCERT sign convention in $\Delta_{sol}H^\ominus = \Delta_{lattice}H^\ominus + \Delta_{hyd}H^\ominus$ or your solution enthalpy will come out wrong.
All Enthalpy Types at a Glance
Every named enthalpy is ultimately a labelled $\Delta_r H^\ominus$ for a specific kind of process. The table consolidates each type, its symbol, sign behaviour and a representative reaction drawn from NCERT Unit 5.
| Enthalpy type | Symbol | Defined for | Sign | Example reaction |
|---|---|---|---|---|
| Reaction | $\Delta_r H^\ominus$ | One mole of reaction as written | ± | $\ce{CaCO3(s) -> CaO(s) + CO2(g)}$ |
| Formation | $\Delta_f H^\ominus$ | 1 mol compound from elements | ± | $\ce{C(graphite) + 2H2(g) -> CH4(g)}$ |
| Combustion | $\Delta_c H^\ominus$ | Complete burning of 1 mol | − (always) | $\ce{C4H10(g) + 13/2 O2(g) -> 4CO2(g) + 5H2O(l)}$ |
| Fusion | $\Delta_{fus}H^\ominus$ | Melting 1 mol solid | + (always) | $\ce{H2O(s) -> H2O(l)}$ |
| Vaporisation | $\Delta_{vap}H^\ominus$ | Boiling 1 mol liquid | + (always) | $\ce{H2O(l) -> H2O(g)}$ |
| Sublimation | $\Delta_{sub}H^\ominus$ | Solid → vapour, 1 mol | + (always) | $\ce{CO2(s) -> CO2(g)}$ |
| Atomisation | $\Delta_a H^\ominus$ | 1 mol bonds → gaseous atoms | + (always) | $\ce{CH4(g) -> C(g) + 4H(g)}$ |
| Bond enthalpy | $\Delta_{bond}H^\ominus$ | Breaking 1 mol of a bond (gas) | + (always) | $\ce{H2(g) -> 2H(g)}$ |
| Lattice | $\Delta_{lattice}H^\ominus$ | Ionic solid → gaseous ions | + (NCERT) | $\ce{NaCl(s) -> Na+(g) + Cl^-(g)}$ |
| Solution | $\Delta_{sol}H^\ominus$ | 1 mol solute dissolved | ± (usually +) | $\ce{NaCl(s) ->[\text{aq}] Na+(aq) + Cl^-(aq)}$ |
Enthalpies of Different Reactions in one screen
- $\Delta_r H^\ominus$ is the standard reaction enthalpy; standard state = pure substance at 1 bar. Scale it with coefficients; reverse the equation → reverse the sign.
- $\Delta_f H^\ominus$ forms one mole of compound from elements in reference states; $\Delta_f H^\ominus$ of any element in its reference state is zero.
- Master formula: $\Delta_r H^\ominus = \sum \Delta_f H^\ominus(\text{products}) - \sum \Delta_f H^\ominus(\text{reactants})$, weighted by coefficients.
- $\Delta_c H^\ominus$ (combustion) is always negative; fusion, vaporisation, sublimation, atomisation and bond enthalpies are always positive.
- Bond-enthalpy route (gas phase only): $\Delta_r H^\ominus = \sum \text{B.E.(reactants)} - \sum \text{B.E.(products)}$.
- $\Delta_{sol}H^\ominus = \Delta_{lattice}H^\ominus + \Delta_{hyd}H^\ominus$; for NaCl, $+788 - 784 = +4~\text{kJ mol}^{-1}$.