Reaction enthalpy, enthalpy of formation and Hess’s law

What is reaction enthalpy?

In plain words: every substance carries a certain enthalpy, like money in an account. When reactants become products, at constant pressure with only pressure–volume work, the difference between the products’ total and the reactants’ total is paid to the surroundings as heat (exothermic, negative) or drawn from them (endothermic, positive). Absolute balances cannot be read, so the standard formation enthalpy of an element in its reference state is zero by convention, and other formation enthalpies are listed relative to those states. This does not mean the actual internal energy or enthalpy of an element is zero. Products minus reactants gives the reaction enthalpy, and the route taken does not matter.

Reaction enthalpy

For a reaction written as reactants → products (NCERT pages 146–147):

Here each H denotes a molar enthalpy (or the appropriate partial molar enthalpy in a mixture), and a and b are the coefficients in the balanced equation. For CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), Δ_rH = [H_m(CO₂, g) + 2H_m(H₂O, l)] − [H_m(CH₄, g) + 2H_m(O₂, g)], with H_m the molar enthalpy. Engineers need reaction enthalpies to plan the heating or cooling that keeps an industrial reaction at a steady temperature; they are also needed to find how equilibrium constants change with temperature (page 147).

Standard enthalpy of reaction

A reaction enthalpy depends on the conditions, so standard conditions are agreed (NCERT Section 5.4(a), page 147).

The standard enthalpy of reaction is the enthalpy change for a reaction when all the participating substances are in their standard states.

NCERT Chemistry Part I (Class 11), Section 5.4, page 147

The standard state of a substance at a specified temperature is its pure form at 1 bar.

NCERT Chemistry Part I (Class 11), Section 5.4, page 147

For gases the standard state is the hypothetical ideal gas at 1 bar; for solutes it is a specified concentration or molality with ideal dilute behaviour (IUPAC standard-state convention). The NCERT sentence above is the introductory pure-substance shorthand.

The standard state of liquid ethanol at 298 K is pure liquid ethanol at 1 bar; that of solid iron at 500 K is pure iron at 1 bar. Data are usually given at 298 K, and the superscript ⊖ (often printed °) marks standard conditions: ΔH^⊖.

Enthalpy changes in phase transformations

Melting, boiling and sublimation also involve enthalpy changes, and an equilibrium transition of a pure substance can occur at constant temperature at its transition pressure (NCERT Section 5.4(b), pages 147–148):

• Fusion: H₂O(s) → H₂O(l); Δ_fusH^⊖ = 6.00 kJ mol⁻¹ at 273 K (NCERT’s text value; its Table 5.1 gives 6.01 and Exercise 5.10 uses 6.03 kJ mol⁻¹: use the value a question gives). The standard enthalpy of fusion is the enthalpy change on melting one mole of a solid in its standard state. Melting is endothermic, so Δ_fusH^⊖ is positive for the ordinary melting transitions considered here; freezing gives the same amount of heat back.
• Vaporisation: H₂O(l) → H₂O(g); Δ_vapH^⊖ = +40.79 kJ mol⁻¹ at the boiling point. The standard enthalpy of vaporisation is the heat needed to vaporise one mole of a liquid at constant temperature and 1 bar.
• Sublimation, solid straight to vapour: dry ice (solid CO₂) sublimes at 195 K with Δ_subH^⊖ = 25.2 kJ mol⁻¹; naphthalene sublimes slowly, Δ_subH^⊖ = 73.0 kJ mol⁻¹. The standard enthalpy of sublimation is the enthalpy change when one mole of a solid sublimes at constant temperature and 1 bar. At one temperature, Δ_subH = Δ_fusH + Δ_vapH (Hess’s law; figure below).

The size of these enthalpies reflects the strength of the forces between the molecules. Strong hydrogen bonds hold water molecules together in the liquid, while the dipole–dipole forces in acetone are much weaker, so it takes less heat to vaporise a mole of acetone than a mole of water (page 148). In physics the same quantities appear, per kilogram, as latent heats.

Some values from NCERT Table 5.1 (page 147), in kJ mol⁻¹, each at the substance’s own melting point (fusion) or boiling point (vaporisation), e.g. 273 K and 373 K for water. At 298 K water’s Δ_vapH^⊖ is larger, 44.01 kJ mol⁻¹ (NCERT Problem 5.7).

Standard enthalpy of formation

NCERT defines the standard molar enthalpy of formation, Δ_fH^⊖, as the standard enthalpy change for forming one mole of a substance from its elements in their most stable states of aggregation, also called their reference states (NCERT Section 5.4(c), page 149). The reference states used in these examples at 25 °C and 1 bar are H₂(g), O₂(g), C(graphite) and S(rhombic) (page 149). By convention, Δ_fH^⊖ of an element in its reference state is zero (page 150). Examples:

• H₂(g) + ½O₂(g) → H₂O(l); Δ_fH^⊖ = −285.8 kJ mol⁻¹
• C(graphite, s) + 2H₂(g) → CH₄(g); Δ_fH^⊖ = −74.81 kJ mol⁻¹
• 2C(graphite, s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l); Δ_fH^⊖ = −277.7 kJ mol⁻¹

Δ_fH^⊖ is a special case of Δ_rH^⊖. These are not enthalpies of formation (page 150):

• CaO(s) + CO₂(g) → CaCO₃(s); Δ_rH^⊖ = −178.3 kJ mol⁻¹, because calcium carbonate is made from compounds, not from its elements;
• H₂(g) + Br₂(l) → 2HBr(g), because two moles of product form. Halving the equation gives ½H₂(g) + ½Br₂(l) → HBr(g); Δ_fH^⊖ = −36.4 kJ mol⁻¹, so the reaction as written has Δ_rH^⊖ = 2 × (−36.4) = −72.8 kJ mol⁻¹. NCERT prints −178.3 kJ mol⁻¹ for this equation, the value of the CaCO₃ line above it; that is a slip.

Some values from NCERT Table 5.2 (page 149), in kJ mol⁻¹ at 298 K:

Reaction enthalpies from enthalpies of formation. For any reaction (NCERT Eq. 5.15, page 150):

For CaCO₃(s) → CaO(s) + CO₂(g): Δ_rH^⊖ = (−635.1) + (−393.5) − (−1206.9) = +178.3 kJ mol⁻¹. The decomposition is endothermic, so limestone must be heated to make lime (page 150).

Thermochemical equations

A balanced equation together with its Δ_rH is a thermochemical equation; the physical states (and allotropes) must be given (NCERT Section 5.4(d), page 150). For example, C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l); Δ_rH^⊖ = −1367 kJ mol⁻¹, the combustion of liquid ethanol. The conventions (pages 150–151):

1. For the quoted molar reaction enthalpy, interpret coefficients as molar amounts per mole of reaction. The equation also expresses particle-number ratios, but a kJ mol^⁻¹ value is not the heat for a single molecular event.
2. Δ_rH^⊖ refers to the amounts in the equation as written, “per mole of reaction”. For Fe₂O₃(s) + 3H₂(g) → 2Fe(s) + 3H₂O(l), Δ_rH^⊖ = 3(−285.83) − (−824.2) = −33.3 kJ mol⁻¹; written with half these coefficients, it is −16.6 kJ mol⁻¹. Enthalpy is extensive. (NCERT labels these two results Δ_fH^⊖; they are reaction enthalpies, Δ_rH^⊖.)
3. Reversing an equation reverses the sign: N₂(g) + 3H₂(g) → 2NH₃(g) has Δ_rH^⊖ = −91.8 kJ mol⁻¹, so 2NH₃(g) → N₂(g) + 3H₂(g) has +91.8 kJ mol⁻¹.

Hess’s law of constant heat summation

Enthalpy is a state function, so the enthalpy change between reactants and products does not depend on the path (NCERT Section 5.4(e), page 151).

If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.

NCERT Chemistry Part I (Class 11), Section 5.4, page 151

Example: the enthalpy of formation of CO. Partial combustion of carbon can give a mixture of CO and CO₂, making a clean direct measurement for C(graphite) + ½O₂(g) → CO(g) difficult. Use two reactions that can be (pages 151–152):

• (i) C(graphite) + O₂(g) → CO₂(g); Δ_rH^⊖ = −393.5 kJ mol⁻¹

• (ii) CO(g) + ½O₂(g) → CO₂(g); Δ_rH^⊖ = −283.0 kJ mol⁻¹

• (iii), which is (ii) reversed: CO₂(g) → CO(g) + ½O₂(g); Δ_rH^⊖ = +283.0 kJ mol⁻¹

Add (i) and (iii): CO₂ cancels and ½O₂ is left on the left: C(graphite) + ½O₂(g) → CO(g); Δ_rH^⊖ = −393.5 + 283.0 = −110.5 kJ mol⁻¹. In general, if A → B in one step has Δ_rH, and by another route in steps with Δ_rH₁, Δ_rH₂, …, then Δ_rH = Δ_rH₁ + Δ_rH₂ + … (NCERT Eq. 5.16).

Solved examples

Common mistakes to avoid

Quick revision summary

• Δ_rH = Σ(coefficient × H)_products − Σ(coefficient × H)_reactants.
• Standard state: a defined reference at specified temperature and usually 1 bar; pure condensed phases, ideal-gas reference, or specified ideal-dilute solute reference; written ⊖.
• Fusion, vaporisation and sublimation enthalpies are positive; Δ_subH = Δ_fusH + Δ_vapH at the same temperature.
• Δ_fH^⊖: one mole of substance from elements in reference states; zero for elements in their reference states.
• Δ_rH^⊖ = ΣaΔ_fH^⊖(products) − ΣbΔ_fH^⊖(reactants).
• Thermochemical equations: states given; ΔH scales with the coefficients; reversing changes the sign.
• Hess’s law: Δ_rH is the same by any route.

Questions students ask

What is the difference between enthalpy of formation and enthalpy of reaction?

The enthalpy of formation is the reaction enthalpy of one particular reaction, forming one mole of a substance from its elements in their reference states, while an enthalpy of reaction, Δ_rH, belongs to any balanced equation, for the amounts as written. So every Δ_fH^⊖ is a Δ_rH^⊖, but not the reverse: NH₃(g) + HCl(g) → NH₄Cl(s) starts from compounds, so its Δ_rH^⊖ is not an enthalpy of formation. It follows from enthalpies of formation (standard data), −314.4 − (−46.1) − (−92.3) = −176.0 kJ mol^⁻¹, which is why tables list Δ_fH^⊖ values rather than every reaction.

Is the standard enthalpy of formation always negative?

No, the standard enthalpy of formation is negative for most compounds but positive for some substances: ethyne, C₂H₂(g), has Δ_fH^⊖ = +226.7 kJ mol^⁻¹ and ozone, O₃(g), +142.7 kJ mol^⁻¹ (standard data), and NO(g) +90.25 kJ mol^⁻¹ (NCERT Table 5.2). A positive value means the substance lies above its elements in enthalpy, so it forms endothermically and is energy-rich: ethyne gives out a great deal of heat when it burns, which is why it is used in welding torches. Such substances need not decompose in practice, since the change can be very slow, and thermodynamic stability at constant temperature and pressure requires the Gibbs energy of the relevant reaction at the actual composition. Standard formation Gibbs energy tests the specified formation reaction under standard conditions, not every possible decomposition.

What are the applications of Hess’s law?

Hess’s law lets you find an enthalpy change that cannot easily be measured by combining reactions whose ΔH values are known, since enthalpy is a state function. Its main uses are enthalpies of formation from enthalpies of combustion; lattice enthalpies from a Born–Haber cycle; mean bond enthalpies from enthalpies of atomisation; enthalpies of phase and allotropic changes; and the enthalpy of forming a hydrate, as for copper sulphate. Each time, known equations are added, reversed or multiplied until they give the target equation.

Does the enthalpy of a reaction depend on temperature or on a catalyst?

The enthalpy of a reaction changes with temperature but not with a catalyst: a catalyst only provides a faster route, lowering the activation energy both ways, and since H is a state function the route does not change Δ_rH. Temperature acts through the heat capacities: Δ_rH(T₂) ≈ Δ_rH(T₁) + Δ_rC_p(T₂ − T₁), Kirchhoff’s relation with Δ_rC_p taken constant and no intervening phase changes (beyond NCERT). For H₂(g) + ½O₂(g) → H₂O(l), Δ_rC_p = 75.3 − 28.8 − ½ × 29.4 = +31.8 J K^⁻¹ mol^⁻¹, so Δ_rH goes from −285.8 kJ mol^⁻¹ at 298 K to about −284.2 kJ mol^⁻¹ at 348 K.

Is Hess’s law in the JEE Main and NEET syllabus?

Yes, Hess’s law of constant heat summation is named in Chemistry Unit 4, Chemical Thermodynamics, of both the JEE Main 2027 syllabus (NTA) and the NEET (UG) 2026 syllabus (NMC), next to the enthalpies of bond dissociation, combustion, formation, atomisation, sublimation, phase transition, hydration, ionisation and solution. With Hess’s law, thermochemical equations, or enthalpies of formation or combustion, are combined to find a reaction enthalpy, as in the solved examples on this page.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. The enthalpies of combustion of methane, graphite and dihydrogen at 298 K are −890.3, −393.5 and −285.8 kJ mol⁻¹. Is Δ_fH of CH₄(g) (i) −74.8; (ii) −52.27; (iii) +74.8; (iv) +52.26 kJ mol⁻¹ (NCERT Exercise 5.5, page 165)? Decide by calculation from the combustion data.

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   C + 2H₂ → CH₄: Δ_fH = −393.5 + 2(−285.8) − (−890.3) = −74.8 kJ mol⁻¹. Answer (i).

2. Q2. Given N₂(g) + 3H₂(g) → 2NH₃(g); Δ_rH^⊖ = −92.4 kJ mol⁻¹, what is the standard enthalpy of formation of NH₃ gas (NCERT Exercise 5.13, page 166)?

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   The equation forms 2 mol of NH₃, so Δ_fH^⊖ = −92.4/2 = −46.2 kJ mol⁻¹. (NCERT Table 5.2 lists −46.11 kJ mol⁻¹, and page 151 uses −91.8 kJ mol⁻¹ for this equation; NCERT’s data differ slightly, so use the value given.)

3. Q3. The enthalpy of combustion of carbon to CO₂ is −393.5 kJ mol⁻¹. Calculate the heat released on forming 35.2 g of CO₂ from carbon and dioxygen (NCERT Exercise 5.11, page 166).

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   n(CO₂) = 35.2/44.0 = 0.800 mol. Heat released = 0.800 × 393.5, about 315 kJ.