Hess's law, standard enthalpies of formation, and Born-Haber cycles

Intuition Beginner

Enthalpy is a state function -- it depends only on the current state of the system, not on how the system reached that state. This means the total enthalpy change for a chemical reaction is the same no matter what route you take from reactants to products. That single fact is Hess's law.

Why does this matter? Many reactions are hard or impossible to measure directly. You cannot easily measure the enthalpy change for converting graphite to diamond, because the reaction is vanishingly slow. But you can measure the enthalpy of combustion for both graphite and diamond. Hess's law lets you subtract the two measured values to obtain the unmeasurable one. The same trick works for any reaction where you can find an alternative path through measurable steps.

The practical tool is the standard enthalpy of formation $\Delta H_f^{\circ }$, defined as the enthalpy change when one mole of a compound forms from its constituent elements in their standard states at 1 bar and a specified temperature (usually 298 K). By convention, $\Delta H_f^{\circ }=0$ for any element in its reference form -- ${\text{O}}_2(g)$, ${\text{N}}_2(g)$, C(graphite), $\text{Na}(s)$, and so on.

Once you have a table of formation enthalpies, the enthalpy change for any reaction is a simple subtraction:

$\Delta H_{\text{rxn}}^{\circ }$ = (total formation enthalpy of all products) $-$ (total formation enthalpy of all reactants)

This is Hess's law in its most useful form. You add up the formation enthalpies of the products, subtract the formation enthalpies of the reactants, and you are done.

Visual Beginner

Picture an energy-level diagram. Reactants sit at one level, products at another. The vertical distance between them is $\Delta H_{\text{rxn}}^{\circ }$. You can connect the two levels with a straight arrow (the direct reaction) or with a zigzag path through intermediate compounds. Both paths give the same vertical distance because enthalpy is a state function.

The same diagram works for formation enthalpies. Every compound has an energy level relative to its elements (which sit at zero by definition). A reaction enthalpy is just the difference between product and reactant levels.

Worked example Beginner

Calculate $\Delta H^{\circ }$ for the combustion of methane: ${\text{CH}}_4(g)+2{\text{O}}_2(g)\to {\text{CO}}_2(g)+2{\text{H}}_2\text{O}(l)$.

Step 1. Look up standard enthalpies of formation at 298 K (all in kJ/mol):

Species	$\Delta H_f^{\circ }$
${\text{CH}}_4(g)$	$-74.8$
${\text{O}}_2(g)$	$0$ (element)
${\text{CO}}_2(g)$	$-393.5$
${\text{H}}_2\text{O}(l)$	$-285.8$

Step 2. Apply Hess's law:

$$\Delta H^{\circ }=[\Delta H_f^{\circ }({\text{CO}}_2)+2\Delta H_f^{\circ }({\text{H}}_2\text{O})]-[\Delta H_f^{\circ }({\text{CH}}_4)+2\Delta H_f^{\circ }({\text{O}}_2)]$$ $$\Delta H^{\circ }=[(-393.5)+2(-285.8)]-[(-74.8)+2(0)]$$ $$\Delta H^{\circ }=-965.1-(-74.8)=-890.3\text{kJ/mol}$$

The combustion of one mole of methane releases 890.3 kJ -- strongly exothermic, which is why methane (natural gas) is a useful fuel.

Check your understanding Beginner

Formal definition Intermediate+

Hess's law

Hess's law states that the total enthalpy change for a chemical reaction is independent of the route by which the reaction is carried out, provided the initial and final states are the same. Formally, if a reaction can be written as the sum of two or more steps, the overall enthalpy change is the sum of the enthalpy changes of the individual steps:

$$\Delta H_{\text{overall}}=\Delta H_1+\Delta H_2+\cdots +\Delta H_n.$$

This is a direct consequence of enthalpy being a state function. For a state function $H$, the integral over any closed path vanishes:

$$\oint dH=0.$$

Thus any cycle of reactions that returns to the starting composition must satisfy ${\sum }_i\Delta H_i=0$.

Standard enthalpies of formation

The standard enthalpy of formation ${\Delta }_f{H}^{\circ }$ of a compound is the enthalpy change when one mole of that compound is formed from its constituent elements in their standard states at a specified temperature (conventionally 298.15 K) and a pressure of 1 bar. The standard state of an element is its most stable physical form under these conditions: C(graphite), ${\text{O}}_2(g)$, ${\text{H}}_2(g)$, $\text{Na}(s)$, ${\text{Br}}_2(l)$, $\text{S}$(rhombic), and so on.

The reaction enthalpy at standard conditions follows from

$$\boxed{{\Delta }_r{H}^{\circ }=\sum _i{\nu }_i{\Delta }_f{H}_i^{\circ },}$$

where ${\nu }_i$ are the stoichiometric coefficients (positive for products, negative for reactants). This is the working form of Hess's law: the tabulated formation enthalpies encode all the thermochemical information needed for any reaction.

Bond dissociation enthalpies

The bond dissociation enthalpy (BDE) is the enthalpy required to break a particular bond in a gaseous molecule, producing two fragments in the gas phase. For a diatomic $A\text{-}B$:

$$A\text{-}B(g)\to A(g)+B(g),\Delta H=D(A\text{-}B).$$

The mean bond enthalpy is an average over all bonds of a given type across many molecules. For C-H bonds, the mean bond enthalpy is about 413 kJ/mol, but individual C-H bond dissociation enthalpies range from about 360 kJ/mol (in ${\text{CH}}_3\text{-H}$) to about 440 kJ/mol (in H-CF${}_3$). Mean bond enthalpies provide quick estimates of reaction enthalpies when formation data are unavailable:

$${\Delta }_r{H}^{\circ }\approx \sum D(\text{bonds broken})-\sum D(\text{bonds formed}).$$

This approximation works well for gas-phase reactions but fails when significant non-bonded interactions (hydrogen bonding, ion-dipole, lattice energy) differ between reactants and products.

Calorimetry

The enthalpy data in thermochemical tables come from calorimetry -- the measurement of heat flow during a chemical or physical process.

Constant-pressure calorimetry (coffee-cup calorimeter) measures $q_p=\Delta H$ directly, because at constant pressure the heat exchanged equals the enthalpy change. The measurement uses $q=m{c}_s\Delta T$, where $m$ is the mass of the solution, $c_s$ is its specific heat capacity, and $\Delta T$ is the observed temperature change.

Constant-volume calorimetry (bomb calorimeter) measures $q_V=\Delta U$, the internal energy change. The relationship to enthalpy is $\Delta H=\Delta U+\Delta (PV)$. For reactions involving only solids and liquids, $\Delta (PV)\approx 0$ and $\Delta H\approx \Delta U$. For reactions producing or consuming gases, $\Delta (PV)=\Delta n_{\text{gas}}RT$, where $\Delta n_{\text{gas}}$ is the change in moles of gas. A bomb calorimeter gives $\Delta U_c^{\circ }$ for combustion; the conversion to $\Delta H_c^{\circ }$ requires the $\Delta n_{\text{gas}}RT$ correction.

Key theorem with proof Intermediate+

Theorem (Hess's law from the state-function property of enthalpy). Consider a reaction that can proceed from reactants to products via two different paths consisting of thermochemical steps. Then the enthalpy changes along the two paths are equal.

Proof. Let Path 1 consist of steps $A\to B\to C$ and Path 2 consist of steps $A\to D\to C$. The enthalpy change along Path 1 is $\Delta H_1+\Delta H_2$ where $\Delta H_1=H(B)-H(A)$ and $\Delta H_2=H(C)-H(B)$. Their sum is $H(C)-H(A)$. The enthalpy change along Path 2 is $\Delta H_3+\Delta H_4$ where $\Delta H_3=H(D)-H(A)$ and $\Delta H_4=H(C)-H(D)$. Their sum is also $H(C)-H(A)$. Since $H$ is a state function, $H(C)-H(A)$ is the same regardless of path, and the two sums are equal. $\square$

Corollary (Formation-enthalpy formula). For any reaction ${\sum }_i{\nu }_i{A}_i=0$ at standard conditions, the reaction enthalpy is ${\Delta }_r{H}^{\circ }={\sum }_i{\nu }_i{\Delta }_f{H}_i^{\circ }$.

Proof. Each compound $A_i$ can be formed from its elements: ${\text{elements}}_i\to A_i$ with $\Delta H={\Delta }_f{H}_i^{\circ }$. The reaction ${\sum }_i{\nu }_i{A}_i=0$ is equivalent to: (dissociate all reactants to elements, then reassemble elements into products). The dissociation of reactant $j$ (with ${\nu }_j<0$) contributes $|{\nu }_j|\cdot (-{\Delta }_f{H}_j^{\circ })$ to the enthalpy, since the reverse of formation absorbs the negative of the formation enthalpy. The formation of product $k$ (with ${\nu }_k>0$) contributes ${\nu }_k\cdot {\Delta }_f{H}_k^{\circ }$. Summing: ${\Delta }_r{H}^{\circ }={\sum }_i{\nu }_i{\Delta }_f{H}_i^{\circ }$. The elements cancel because they appear on both sides of the two-step path. $\square$

Worked example: indirect determination via Hess's law

The enthalpy of formation of carbon monoxide cannot be measured directly because burning carbon in limited oxygen produces a mixture of CO and ${\text{CO}}_2$. Use two measurable reactions:

Reaction 1 (combustion of C to ${\text{CO}}_2$): $\text{C}(s)+{\text{O}}_2(g)\to {\text{CO}}_2(g)$, $\Delta H_1=-393.5$ kJ/mol.

Reaction 2 (combustion of CO to ${\text{CO}}_2$): $\text{CO}(g)+\frac{1}{2}{\text{O}}_2(g)\to {\text{CO}}_2(g)$, $\Delta H_2=-283.0$ kJ/mol.

Target: $\text{C}(s)+\frac{1}{2}{\text{O}}_2(g)\to \text{CO}(g)$, $\Delta H_f^{\circ }(\text{CO})=?$

Reverse Reaction 2: ${\text{CO}}_2(g)\to \text{CO}(g)+\frac{1}{2}{\text{O}}_2(g)$, $\Delta H=+283.0$ kJ/mol.

Add to Reaction 1: $\text{C}(s)+{\text{O}}_2(g)+{\text{CO}}_2(g)\to {\text{CO}}_2(g)+\text{CO}(g)+\frac{1}{2}{\text{O}}_2(g)$.

Cancel ${\text{CO}}_2$ and $\frac{1}{2}{\text{O}}_2$: $\text{C}(s)+\frac{1}{2}{\text{O}}_2(g)\to \text{CO}(g)$.

$$\Delta H_f^{\circ }(\text{CO})=-393.5+283.0=-110.5\text{kJ/mol}.$$

Exercises Intermediate+

The Born-Haber cycle for ionic compounds Master

Hess's law finds one of its most powerful applications in determining lattice energies of ionic solids through the Born-Haber cycle. The lattice energy $U_{\text{latt}}$ is the enthalpy change when one mole of an ionic crystal is formed from its constituent gaseous ions:

$${\text{M}}^{+}(g)+{\text{X}}^{-}(g)\to \text{MX}(s),\Delta H=-U_{\text{latt}}.$$

(The lattice energy is defined as a positive quantity; the formation of the crystal from ions is exothermic, $\Delta H<0$.)

The lattice energy cannot be measured directly because gaseous ions cannot be produced and combined in a calorimeter in a single controlled step. The Born-Haber cycle circumvents this by constructing a closed thermodynamic loop from measurable quantities.

The NaCl Born-Haber cycle

The formation of sodium chloride from its elements is:

$$\text{Na}(s)+\frac{1}{2}{\text{Cl}}_2(g)\to \text{NaCl}(s),\Delta H_f^{\circ }=-411.2\text{kJ/mol}.$$

The Born-Haber cycle decomposes this into five measurable steps:

1. Sublimation of sodium: $\text{Na}(s)\to \text{Na}(g)$, $\Delta H_{\text{sub}}=+108.4$ kJ/mol.

2. Ionization of sodium: $\text{Na}(g)\to {\text{Na}}^{+}(g)+e^{-}$, $I{E}_1=+495.8$ kJ/mol.

3. Dissociation of chlorine: $\frac{1}{2}{\text{Cl}}_2(g)\to \text{Cl}(g)$, $\frac{1}{2}D(\text{Cl-Cl})=+122.6$ kJ/mol.

4. Electron affinity of chlorine: $\text{Cl}(g)+e^{-}\to {\text{Cl}}^{-}(g)$, $EA=-349.0$ kJ/mol.

5. Lattice formation: ${\text{Na}}^{+}(g)+{\text{Cl}}^{-}(g)\to \text{NaCl}(s)$, $\Delta H=-U_{\text{latt}}$.

By Hess's law, the sum of all five steps equals the formation enthalpy:

$$\Delta H_f^{\circ }=\Delta H_{\text{sub}}+I{E}_1+\frac{1}{2}D+EA+(-U_{\text{latt}}).$$

Solving for the lattice energy:

$$U_{\text{latt}}=\Delta H_{\text{sub}}+I{E}_1+\frac{1}{2}D+EA-\Delta H_f^{\circ }$$ $$U_{\text{latt}}=108.4+495.8+122.6+(-349.0)-(-411.2)$$ $$U_{\text{latt}}=789.0\text{kJ/mol}.$$

The lattice energy of NaCl is 789 kJ/mol -- a large value reflecting the strong electrostatic attraction between the sodium and chloride ions in the crystal. This value can be compared with the theoretical prediction from the Born-Lande equation or the Born-Mayer equation, which compute $U_{\text{latt}}$ from the crystal geometry (Madelung constant), ionic charges, and interionic distance.

The Born-Lande equation

The Born-Lande equation provides a theoretical estimate of the lattice energy from first principles:

$$U_{\text{latt}}=\frac{N_{A}M{z}^{+}{z}^{-}{e}^2}{4\pi {\epsilon }_0{r}_0}\left(1-\frac{1}{n}\right),$$

where $N_A$ is Avogadro's number, $M$ is the Madelung constant (1.7476 for the NaCl structure), $z^{+}$ and $z^{-}$ are the ionic charges, $e$ is the elementary charge, ${\epsilon }_0$ is the permittivity of free space, $r_0$ is the distance between ion centres, and $n$ is the Born exponent (typically 8-10 for NaCl-type crystals, related to the repulsive part of the interionic potential).

For NaCl: $r_0=282$ pm, $M=1.7476$, $n=8$. Substituting gives $U_{\text{latt}}\approx 766$ kJ/mol. The agreement with the Born-Haber value of 789 kJ/mol is within 3%, with the discrepancy arising from the simplified repulsive potential and the neglect of van der Waals contributions.

Born-Haber cycles beyond NaCl

The Born-Haber approach generalises to any ionic compound. For an ${\text{M}}^{2+}{\text{X}}^{2-}$ salt such as MgO, the cycle requires both the first and second ionization energies of magnesium ($I{E}_1=738$, $I{E}_2=1451$ kJ/mol) and both electron affinities of oxygen ($E{A}_1=-141$, $E{A}_2=+798$ kJ/mol). The second electron affinity of oxygen is positive because adding a second electron to ${\text{O}}^{-}$ requires energy to overcome the electrostatic repulsion. The resulting lattice energy for MgO is about 3795 kJ/mol -- nearly five times that of NaCl -- reflecting the doubled charges and shorter interionic distance.

The Born-Haber cycle also serves as a diagnostic tool. If the experimental lattice energy (from the cycle) agrees with the theoretical Born-Lande value, the ionic model is appropriate. Large discrepancies indicate significant covalent character. For AgCl, the Born-Haber lattice energy exceeds the Born-Lande prediction by about 80 kJ/mol, signalling partial covalent bonding consistent with silver's polarizable $d^{10}$ electron configuration. This was one of the earliest quantitative measures of covalent character in "ionic" compounds.

Calorimetry in depth Master

Constant-pressure calorimetry

A coffee-cup calorimeter operates at constant (atmospheric) pressure and measures $\Delta H$ directly. The reaction occurs in a solution of known mass and specific heat, insulated from the surroundings. The temperature change $\Delta T$ of the solution gives the heat:

$$q_{\text{soln}}=m_{\text{soln}}{c}_{s,\text{soln}}\Delta T.$$

For an exothermic reaction, the solution warms ($\Delta T>0$, $q_{\text{soln}}>0$). Since the calorimeter is insulated, $q_{\text{rxn}}+q_{\text{soln}}=0$, giving $q_{\text{rxn}}=-q_{\text{soln}}$. At constant pressure, $q_{\text{rxn}}=\Delta H$.

The accuracy of this method depends on the insulation quality (minimising heat loss), the precision of the temperature measurement, and the assumption that the solution's specific heat equals that of pure water (4.184 J/(g K)). For dilute aqueous solutions this is a good approximation.

Constant-volume calorimetry

A bomb calorimeter operates at constant volume. The sample is placed in a heavy-walled steel container (the "bomb"), pressurised with oxygen, and ignited electrically. The bomb sits in a water bath of known heat capacity $C_{\text{cal}}$ (determined by calibration with a standard such as benzoic acid, $\Delta U_c^{\circ }=-3227$ kJ/mol). The temperature rise of the water bath gives the heat released:

$$q_V=-C_{\text{cal}}\Delta T.$$

Because volume is constant, $q_V=\Delta U$, not $\Delta H$. The conversion requires:

$$\Delta H=\Delta U+\Delta n_{\text{gas}}RT.$$

For the combustion of benzoic acid, ${\text{C}}_6{\text{H}}_5\text{COOH}(s)+\frac{15}{2}{\text{O}}_2(g)\to 7{\text{CO}}_2(g)+3{\text{H}}_2\text{O}(l)$: $\Delta n_{\text{gas}}=7-7.5=-0.5$ mol. The correction is $\Delta n_{\text{gas}}RT=(-0.5)(8.314)(298.15)=-1.24$ kJ/mol, small relative to the $-3227$ kJ/mol total but necessary for high-precision work.

Differential scanning calorimetry (DSC)

Modern thermochemistry uses differential scanning calorimetry, which measures the heat flow required to keep a sample and a reference at the same temperature as both are heated or cooled at a controlled rate. DSC detects phase transitions (melting, crystallisation, glass transitions), measures heat capacities, and quantifies reaction enthalpies on milligram samples. It is the workhorse technique for polymer science, pharmaceutical development, and materials characterization.

Temperature dependence of reaction enthalpy Master

Kirchhoff's law gives the temperature dependence of the reaction enthalpy. Differentiating ${\Delta }_r{H}^{\circ }(T)={\sum }_i{\nu }_i{\Delta }_f{H}_i^{\circ }(T)$ with respect to $T$ at constant pressure:

$$\frac{d{\Delta }_r{H}^{\circ }}{dT}=\sum _i{\nu }_i{\left(\frac{\partial H_i}{\partial T}\right)}_P=\sum _i{\nu }_i{C}_{P,i}={\Delta }_r{C}_P,$$

where $C_{P,i}$ are the molar heat capacities at constant pressure. This is Kirchhoff's law:

$$\boxed{{\Delta }_r{H}^{\circ }(T_2)={\Delta }_r{H}^{\circ }(T_1)+{\int }_{T_1}^{T_2}{\Delta }_r{C}_{P}dT.}$$

If ${\Delta }_r{C}_P$ is approximately constant over the temperature range (valid for small $\Delta T$ or when heat capacities change slowly), the integral simplifies to ${\Delta }_r{C}_P\cdot (T_2-T_1)$.

For the water-gas shift reaction at 800 K, using ${\Delta }_r{H}^{\circ }(298)=-41.2$ kJ/mol and ${\Delta }_r{C}_P\approx -2.3$ J/(mol K):

$${\Delta }_r{H}^{\circ }(800)\approx -41200+(-2.3)(800-298)=-41200-1155=-42.4\text{kJ/mol}.$$

The reaction becomes slightly more exothermic at higher temperature because the product heat capacity is lower than the reactant heat capacity.

Connections Master

• Chemical thermodynamics and equilibrium 14.06.01 extends the enthalpy picture to Gibbs free energy $G=H-TS$, connecting reaction enthalpy to equilibrium constants through $\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }=-RT\ln K$. Hess's law applies equally to $G$ as to $H$.

• Lewis structures and VSEPR 14.02.01 determines the bonding topology needed to identify which bonds break and form when estimating reaction enthalpies from bond dissociation data.

• Periodic trends 14.01.02 provides the ionization energies, electron affinities, and atomic radii that are inputs to the Born-Haber cycle and the Born-Lande equation.

• Solid-state chemistry [inorgchem.solid-state-ionic-covalent-metallic] extends lattice energy calculations to complex crystal structures, Madelung constants for non-NaCl lattices, and the Kapustinskii equation for estimating lattice energies without detailed structural data.

Historical notes Master

Germain Hess formulated his law of constant heat summation in 1840, based on calorimetric measurements of various reactions. Hess worked before the first law of thermodynamics was formally stated (Joule's equivalence of heat and mechanical work, 1843; Helmholtz's conservation of energy, 1847). Hess's experimental observation that reaction enthalpies are additive was in fact one of the key empirical inputs that guided the development of the first law.

The Born-Haber cycle was developed independently by Max Born and Fritz Haber around 1919. Born approached the problem from the physics of ionic crystals, deriving the lattice energy from electrostatic theory (the Born-Lande equation). Haber approached it from thermochemistry, recognizing that Hess's law could close the gap between measurable thermochemical quantities and the lattice energy. The synthesis of both approaches -- using the cycle to extract the lattice energy experimentally and comparing it to the theoretical value -- provided the first quantitative test of the ionic model of solids.

The Born-Haber cycle revealed that many nominally "ionic" compounds have lattice energies significantly larger than the purely electrostatic prediction, indicating covalent contributions. This insight was central to the development of the modern understanding that the ionic/covalent distinction is a continuum, not a dichotomy.

Haber's 1905 book Thermodynamik technischer Gasreaktionen was a landmark in applied thermochemistry, developing the thermodynamic analysis of gas-phase reactions relevant to the synthesis of ammonia and other industrial processes. His work laid the groundwork for the Haber-Bosch process, which combines nitrogen and hydrogen to produce ammonia -- a reaction whose thermodynamic feasibility and optimal conditions were analysed using the same Hess's-law tools developed in this unit.

Bibliography Master

• Hess, G. H., "Thermochemische Untersuchungen", Ann. Phys. Chem. 50 (1840), 385--404.
• Born, M., "Verhandl. Deut. Physik. Ges." 21 (1919), 13.
• Haber, F., Thermodynamik technischer Gasreaktionen (Oldenbourg, Munich, 1905).
• Born, M. & Lande, A., "Uber die Berechnung der Kompressibilitat regulärer Kristalle aus der Gittertheorie", Sitzungsber. Preuss. Akad. Wiss. (1918), 1045--1052.
• Kapustinskii, A. F., "Lattice energy of ionic crystals", Quart. Rev. Chem. Soc. 10 (1956), 283--294.
• Zumdahl, S. S. & DeCoste, D. J., Chemical Principles, 8e (Cengage, 2017), Ch. 6.
• Atkins, P. & de Paula, J., Physical Chemistry, 12e (Oxford, 2023), Ch. 2.
• Engel, T. & Reid, P., Thermodynamics, Statistical Thermodynamics, and Kinetics, 4e (Pearson, 2019), Ch. 4.