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JEE Mains Chemistry · Chemical Thermodynamics

Hess's Law: Formation and Combustion Enthalpies

A reaction's enthalpy from formation enthalpies (products minus reactants), from combustion enthalpies (reactants minus products), or by adding, reversing and scaling given equations.

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Why this matters

Seventeen PYQs, nine of them numerical, and two from 2026. Four sum formation enthalpies, six work from combustion enthalpies, and seven add, reverse and scale given equations, several with the heat written on the product side.

Concept 1 of 3: Reaction enthalpy from formation enthalpies

Enthalpy is a state function, so any reaction can be taken apart into its elements and rebuilt. Each compound's formation enthalpy is the cost of making it from elements; products minus reactants gives the reaction. Elements in their reference state cost nothing.

Definition

  • ΔrH∘=∑ν ΔfH∘(products)−∑ν ΔfH∘(reactants)\Delta_r H^\circ = \sum\nu\,\Delta_f H^\circ(\text{products}) - \sum\nu\,\Delta_f H^\circ(\text{reactants}), with the coefficients ν\nu of the balanced equation.
  • ΔfH∘=0\Delta_f H^\circ = 0 for an element in its reference state at ANY temperature: O2(g)\mathrm{O_2(g)}, H2(g)\mathrm{H_2(g)}, C(graphite). It is not zero for O(g), C(diamond) or O3(g)\mathrm{O_3(g)}.
  • Standard state: the pure substance at 1 bar, at whatever temperature is stated. No temperature is built in.
  • ΔfH∘\Delta_f H^\circ is per mole of compound; scale it to match the equation.
  • Heat per gram of a mixture: divide the equation's heat by the total mass of the reactants in it.

Reaction enthalpy from formation enthalpies

ΔrH∘=∑ν ΔfHproducts∘−∑ν ΔfHreactants∘\Delta_r H^\circ = \sum \nu\,\Delta_f H^\circ_{\mathrm{products}} - \sum \nu\,\Delta_f H^\circ_{\mathrm{reactants}}

Worked example

Find ΔrH∘\Delta_r H^\circ for 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)\mathrm{4NH_3(g) + 5O_2(g) \to 4NO(g) + 6H_2O(g)}. ΔfH∘\Delta_f H^\circ (kJ mol⁻¹): NH3\mathrm{NH_3} −46, NO +90, H2O(g)\mathrm{H_2O(g)} −242.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q49Moderate

Example 1 · Chemical Thermodynamics · Hess's Law: Formation and Combustion Enthalpies

Consider the reaction :
2H2 S( g)+3O2( g)→2H2O(l)+2SO2( g)2H_{2}\text{ }S(\text{ }g) + 3O_{2}(\text{ }g) \rightarrow 2H_{2}O(l) + 2{SO}_{2}(\text{ }g)
The magnitude of enthalpy change for the reaction in kJmol−1kJ{mol}^{- 1} is ____\_\_\_\_ . (Nearest integer) Given : ΔfH⊖(H2 S)=−20.1 kJ mol−1\Delta_{f}H^{\ominus}\left( H_{2}\text{ }S \right) = - 20.1\text{ }kJ{\text{ }mol}^{- 1}
ΔfHΘ(H2O)=−286.0 kJ mol−1\Delta_{f}H^{\Theta}\left( H_{2}O \right) = - 286.0\text{ }kJ{\text{ }mol}^{- 1}
ΔfHΘ(SO2)=−297.0 kJ mol−1\Delta_{f}H^{\Theta}\left( {SO}_{2} \right) = - 297.0\text{ }kJ{\text{ }mol}^{- 1}

The standard state is not 0 °C

'Standard' fixes the pressure at 1 bar and says nothing about temperature. So ΔfH500∘\Delta_f H^\circ_{500} of O2(g)\mathrm{O_2(g)} is zero too, and options that say 273 K are wrong.

One mole of everything

Multiply each ΔfH∘\Delta_f H^\circ by its coefficient in the balanced equation before adding. Using one mole of each species is the most common slip on this page.
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Concept 2 of 3: Formation enthalpy from combustion enthalpies

Most compounds cannot be made directly from their elements, but almost all of them burn. Burning the elements and burning the compound both end at the same carbon dioxide and water. So the difference between the two heats of combustion is the heat of formation.

Definition

  • ΔfH(compound)=∑ν ΔcH(elements)−ΔcH(compound)\Delta_f H(\text{compound}) = \sum\nu\,\Delta_c H(\text{elements}) - \Delta_c H(\text{compound}).
  • ΔcH\Delta_c H(C, graphite) =ΔfH(CO2)= \Delta_f H(\mathrm{CO_2}); ΔcH(H2)=ΔfH(H2O)\Delta_c H(\mathrm{H_2}) = \Delta_f H(\mathrm{H_2O}).
  • Any reaction: ΔrH=∑ΔcH(reactants)−∑ΔcH(products)\Delta_r H = \sum\Delta_c H(\text{reactants}) - \sum\Delta_c H(\text{products}). Reactants minus products: the reverse of the formation rule.
  • The same answer comes from writing ΔcH=∑ΔfH(products)−∑ΔfH(reactants)\Delta_c H = \sum\Delta_f H(\text{products}) - \sum\Delta_f H(\text{reactants}) and solving for the unknown ΔfH\Delta_f H.

Formation enthalpy from combustion enthalpies

ΔfH=∑ν ΔcHelements−ΔcHcompound\Delta_f H = \sum \nu\,\Delta_c H_{\mathrm{elements}} - \Delta_c H_{\mathrm{compound}}

Worked example

ΔcH\Delta_c H (kJ mol⁻¹): CH4\mathrm{CH_4} −890, C(graphite) −393, H2\mathrm{H_2} −286. Find ΔfH\Delta_f H of methane.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 25 July 2022 · Q52Moderate

Example 2 · Chemical Thermodynamics · Hess's Law: Formation and Combustion Enthalpies

The enthalpy of combustion of propane, graphite and dihydrogen at 298 K298\text{ }K are: −2220.0 kJ mol−1-2220.0\text{ }kJ{\text{ }mol}^{-1}, −393.5 kJ mol−1-393.5\text{ }kJ{\text{ }mol}^{-1} and −285.8 kJ mol−1-285.8\text{ }kJ{\text{ }mol}^{-1} respectively. The magnitude of enthalpy of formation of propane (C3H8)\left( C_{3}H_{8} \right) is ____ kJ mol−1kJ\ mol^{-1}. (Nearest integer)

Flipping the combustion rule

With combustion enthalpies it is reactants minus products. Using products minus reactants gives the right size with the wrong sign, and that sign-flipped value is always an option.

Which water?

ΔcH(H2)\Delta_c H(\mathrm{H_2}) is about −286 kJ mol⁻¹ when the water is liquid and about −242 kJ mol⁻¹ when it is vapour. Use the value the question gives and match the water's state in the target equation.
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Concept 3 of 3: Hess's law: adding, reversing and scaling equations

Treat the given equations like algebra. Reverse one and its ΔH changes sign; multiply it and ΔH is multiplied; add equations and their ΔH values add. Arrange them so that everything except the target cancels.

Definition

  • Reverse an equation: ΔH→−ΔH\Delta H \to -\Delta H. Multiply it by kk: ΔH→k ΔH\Delta H \to k\,\Delta H. Add equations: add their ΔH\Delta H.
  • Heat written on the product side ("… + 400 kJ") is heat released: ΔH=−400\Delta H = -400 kJ. On the reactant side it means ΔH>0\Delta H > 0.
  • Keep physical states apart: a solid, its aqueous solution and a gas are different species.
  • Mixed fuels: find the moles of each fuel, then add moles × heat for each.
  • Gas volumes at 25 °C and 1 atm: 24.47 L mol⁻¹. The value 22.4 L mol⁻¹ is for 0 °C.

Hess's law

ΔHtarget=∑iki ΔHi\Delta H_{\mathrm{target}} = \sum_i k_i\,\Delta H_i

Worked example

Given C(graphite)+O2(g)→CO2(g)\mathrm{C(graphite) + O_2(g) \to CO_2(g)}, ΔH=−393.5\Delta H = -393.5 kJ, and CO(g)+12O2(g)→CO2(g)\mathrm{CO(g) + \tfrac{1}{2}O_2(g) \to CO_2(g)}, ΔH=−283.0\Delta H = -283.0 kJ, find ΔH\Delta H for C(graphite)+12O2(g)→CO(g)\mathrm{C(graphite) + \tfrac{1}{2}O_2(g) \to CO(g)}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q29Moderate

Example 3 · Chemical Thermodynamics · Hess's Law: Formation and Combustion Enthalpies

Consider the following data (i) 2Al(s)+6HCl(aq)→Al2Cl6(aq)+3H2( g)+1200 kJ/mol2Al(s) + 6HCl(aq) \rightarrow {Al}_{2}{Cl}_{6}(aq) + 3H_{2}(\text{ }g) + 1200\text{ }kJ/mol. (ii) H2( g)+Cl2( g)→2HCl(g)+164 kJ/molH_{2}(\text{ }g) + {Cl}_{2}(\text{ }g) \rightarrow 2HCl(g) + 164\text{ }kJ/mol. (iii) HCl(g)+aq→HCl(aq)+83 kJ/molHCl(g) + aq \rightarrow HCl(aq) + 83\text{ }kJ/mol. (iv) Al2Cl6( s)+aq→Al2Cl6(aq)+663 kJ/mol{Al}_{2}{Cl}_{6}(\text{ }s) + aq \rightarrow {Al}_{2}{Cl}_{6}(aq) + 663\text{ }kJ/mol. The enthalpy of formation of anhydrous solid Al2Cl6Al_{2}Cl_{6} is :

Heat on the product side

'C(s)+O2(g)→CO2(g)+400\mathrm{C(s) + O_2(g) \to CO_2(g)} + 400 kJ' means ΔH=−400\Delta H = -400 kJ. Taking it as +400 flips every sign that follows.

22.4 L at room temperature

At 25 °C and 1 atm one mole of gas fills 24.47 L, not 22.4 L. With 22.4 the moles, and so the heat, come out about 9% too large.
Drill 6 more on hess's law: adding, reversing and scaling equations

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Reaction enthalpy from formation enthalpies

    Reaction enthalpy from formation enthalpies

    ΔrH∘=∑ν ΔfHproducts∘−∑ν ΔfHreactants∘\Delta_r H^\circ = \sum \nu\,\Delta_f H^\circ_{\mathrm{products}} - \sum \nu\,\Delta_f H^\circ_{\mathrm{reactants}}
  • Formation enthalpy from combustion enthalpies

    Formation enthalpy from combustion enthalpies

    ΔfH=∑ν ΔcHelements−ΔcHcompound\Delta_f H = \sum \nu\,\Delta_c H_{\mathrm{elements}} - \Delta_c H_{\mathrm{compound}}
  • Hess's law: adding, reversing and scaling equations

    Hess's law

    ΔHtarget=∑iki ΔHi\Delta H_{\mathrm{target}} = \sum_i k_i\,\Delta H_i

Watch out for (6)

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