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Hess's Law and Enthalpies of Formation

How do you use Hess's law and enthalpies of formation to calculate ΔH?

IntermediateThermochemistry & ThermodynamicsLast reviewed 4 October 2026
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What is it?

Hess’s law: the enthalpy change of a reaction is the same whatever route is taken from the reactants to the products. Enthalpy is a state function: it depends only on the start and end, not on the path.

This means you can find ΔH\Delta H for a reaction that is hard to measure directly by adding up the ΔH\Delta H values of other reactions that combine to give it.

Key idea

If equations add up to the target equation, their ΔH\Delta H values add up to the target ΔH\Delta H. Reverse an equation: change the sign of ΔH\Delta H. Multiply an equation: multiply ΔH\Delta H by the same number.

Why does it matter?

  • Impossible measurements. The formation of CO from carbon can’t be measured cleanly, because some COX2\ce{CO2} always forms. Hess’s law gets it from two easy combustions.
  • Tables instead of experiments. With a table of enthalpies of formation, you can calculate ΔH\Delta H for thousands of reactions without doing any of them.
  • Energy planning. Fuel values, metabolic energy and industrial heat balances all rely on these calculations.

How does it work?

1. Combining equations

To reach a target equation:

  1. Write the given equations so that each target reactant is on the left and each target product is on the right; reverse an equation if needed (and change the sign of its ΔH\Delta H).
  2. Multiply equations so the amounts match the target (and multiply their ΔH\Delta H).
  3. Add the equations; substances that appear on both sides cancel. Add the ΔH\Delta H values.

2. Standard enthalpy of formation

The standard enthalpy of formation, ΔHf°\Delta H_\text{f}°, is the enthalpy change when 1 mol of a compound forms from its elements in their standard states (the most stable form at 1 bar and 25 °C):

C(s,graphite)+OX2(g)→COX2(g)ΔHf°=−393.5 kJ/mol\begin{gathered} \ce{C(s, graphite) + O2(g) -> CO2(g)} \\[4pt] \Delta H_\text{f}° = -393.5\ \text{kJ/mol} \end{gathered}

For an element in its standard state, such as OX2(g)\ce{O2(g)}, HX2(g)\ce{H2(g)} or C(graphite), ΔHf°=0\Delta H_\text{f}° = 0.

SubstanceΔHf°\Delta H_\text{f}° (kJ/mol)
COX2(g)\ce{CO2(g)}−393.5
CO(g)\ce{CO(g)}−110.5
HX2O(l)\ce{H2O(l)}−285.8
HX2O(g)\ce{H2O(g)}−241.8
CHX4(g)\ce{CH4(g)}−74.6
CX2HX5OH(l)\ce{C2H5OH(l)}−277.6
OX2(g)\ce{O2(g)}, HX2(g)\ce{H2(g)}, C(graphite)0

3. ΔH from enthalpies of formation

ΔH°rxn=∑n ΔHf°(products)−∑n ΔHf°(reactants)\begin{aligned} &\Delta H°_\text{rxn} = \sum n\,\Delta H_\text{f}°(\text{products}) \\[4pt] &\qquad\quad - \sum n\,\Delta H_\text{f}°(\text{reactants}) \end{aligned}

where nn is each coefficient (in mol) in the balanced equation. “Products minus reactants.”

Think of it like this

Climbing a mountain: your change in height between the car park and the summit is the same whether you take the steep direct path or the long zig-zag trail. Enthalpy works the same way: only the start and the end matter.

More precisely

Hess’s law follows from the first law of thermodynamics (energy is conserved). The formation-enthalpy formula is Hess’s law applied to a two-step route: break the reactants down into their elements (the reverse of formation), then build the products from those elements. Average bond enthalpies give another estimate of ΔH\Delta H, but because they are averages, the result is only approximate.

Visualise it

Enthalpy level diagram drawn to scale. C(s) + O2(g) is at 0 kJ, CO(g) + ½O2(g) at −110.5 kJ, and CO2(g) at −393.5 kJ. The direct route from C + O2 to CO2 releases 393.5 kJ. The indirect route goes first to CO (−110.5 kJ) and then to CO2 (−283.0 kJ). The two steps add up to the same −393.5 kJ.
Both routes from C + O₂ to CO₂ give the same total enthalpy change.

Worked example

Worked example: Combining equations

Question: Find ΔH\Delta H for C(s)+12 OX2(g)→CO(g)\ce{C(s) + 1/2 O2(g) -> CO(g)} from:

(1) C(s)+OX2(g)→COX2(g)\ce{C(s) + O2(g) -> CO2(g)}, ΔH1=−393.5\Delta H_1 = -393.5 kJ

(2) CO(g)+12 OX2(g)→COX2(g)\ce{CO(g) + 1/2 O2(g) -> CO2(g)}, ΔH2=−283.0\Delta H_2 = -283.0 kJ

  1. Keep (1): it has C on the left, as in the target.

  2. Reverse (2), so that CO is on the right: COX2(g)→CO(g)+12 OX2(g)\ce{CO2(g) -> CO(g) + 1/2 O2(g)}, ΔH=+283.0\Delta H = +283.0 kJ

  3. Add: COX2\ce{CO2} cancels, and OX2\ce{O2} on the left becomes 1 − ½ = ½ mol:

    ΔH=−393.5 kJ+283.0 kJ=−110.5 kJ\begin{aligned} &\Delta H \\[4pt] &= -393.5\ \text{kJ} + 283.0\ \text{kJ} \\[4pt] &= -110.5\ \text{kJ} \end{aligned}

This matches ΔHf°\Delta H_\text{f}° of CO in the table.

Worked example: Using enthalpies of formation

Question: Calculate ΔH°\Delta H° for the combustion of methane: CHX4(g)+2 OX2(g)→COX2(g)+2 HX2O(l)\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}

  1. Products: (1 mol)(−393.5 kJ/mol)+(2 mol)(−285.8 kJ/mol)=−965.1 kJ(1\ \text{mol})(-393.5\ \text{kJ/mol}) + (2\ \text{mol})(-285.8\ \text{kJ/mol}) = -965.1\ \text{kJ}

  2. Reactants: (1 mol)(−74.6 kJ/mol)+(2 mol)(0 kJ/mol)=−74.6 kJ(1\ \text{mol})(-74.6\ \text{kJ/mol}) + (2\ \text{mol})(0\ \text{kJ/mol}) = -74.6\ \text{kJ}

  3. Subtract:

    ΔH°=−965.1 kJ−(−74.6 kJ)=−890.5 kJ\begin{aligned} &\Delta H° \\[4pt] &= -965.1\ \text{kJ} \\[4pt] &\quad - (-74.6\ \text{kJ}) \\[4pt] &= -890.5\ \text{kJ} \end{aligned}

  4. The mol units cancel in each product, leaving kJ for the reaction as written (−890.5 kJ per mole of CHX4\ce{CH4}).

Common mistake

Common mistake: Forgetting to change the sign

When you reverse an equation, ΔH\Delta H changes sign: if A → B has ΔH=−100\Delta H = -100 kJ, then B → A has ΔH=+100\Delta H = +100 kJ.

Common mistake: Forgetting the coefficients

Multiply each ΔHf°\Delta H_\text{f}° by its coefficient. In methane combustion there are 2 mol of water: 2 × (−285.8 kJ/mol) = −571.6 kJ.

Common mistake: Reactants minus products

The formula is products minus reactants. Reversing the order gives the right size with the wrong sign.

Common mistake: Mixing up the states of water

ΔHf°\Delta H_\text{f}° of HX2O(l)\ce{H2O(l)} (−285.8 kJ/mol) and of HX2O(g)\ce{H2O(g)} (−241.8 kJ/mol) differ by 44.0 kJ/mol, the enthalpy of vaporization. Use the state in the equation.

Notation note

  • ΔHf°\Delta H_\text{f}° is read “standard enthalpy of formation”; the ° means standard conditions (1 bar, usually 25 °C).
  • Σ (“sigma”) means “the sum of”.
  • Equations may contain fractional coefficients (such as ½ OX2\ce{O2}) when the target is “per mole” of one substance.

Remember this

Remember this

  • Hess’s law: ΔH is the same by any route (enthalpy is a state function).
  • Reverse an equation: change the sign of ΔH. Multiply it: multiply ΔH.
  • ΔHf°\Delta H_\text{f}° = 1 mol of compound from elements in standard states; elements have ΔHf°=0\Delta H_\text{f}° = 0.
  • ΔH°=∑nΔHf°(products)−∑nΔHf°(reactants)\Delta H° = \sum n\Delta H_\text{f}°(\text{products}) - \sum n\Delta H_\text{f}°(\text{reactants}).

Test yourself

Check your understanding before moving on.

Flashcards

10 cards

  1. Question
    State Hess's law.
    Answer

    The enthalpy change of a reaction is the same whatever route is taken from reactants to products.

  2. Question
    Why does Hess's law work?
    Answer

    Enthalpy is a state function: it depends only on the start and end, not the path (energy is conserved).

  3. Question
    An equation is reversed. What happens to ΔH?
    Answer

    It changes sign.

  4. Question
    An equation is multiplied by 3. What happens to ΔH?
    Answer

    ΔH is multiplied by 3.

  5. Question
    Define the standard enthalpy of formation, ΔHf°.
    Answer

    The enthalpy change when 1 mol of a compound forms from its elements in their standard states (1 bar, usually 25 °C).

  6. Question
    What is ΔHf° of O₂(g)?
    Answer

    0 kJ/mol: it is an element in its standard state.

  7. Question
    Formula for ΔH° from enthalpies of formation?
    Answer

    ΔH°=∑nΔHf°(products)−∑nΔHf°(reactants)\Delta H° = \sum n\Delta H_\text{f}°(\text{products}) - \sum n\Delta H_\text{f}°(\text{reactants})

  8. Question
    ΔH° for CH₄ + 2O₂ → CO₂ + 2H₂O(l)? (ΔHf°: CH₄ −74.6, CO₂ −393.5, H₂O(l) −285.8 kJ/mol)
    Answer

    [−393.5 kJ + 2(−285.8 kJ)] − (−74.6 kJ) = −890.5 kJ

  9. Question
    Why do ΔHf° of H₂O(l) and H₂O(g) differ?
    Answer

    By the enthalpy of vaporization: −241.8 kJ/mol − (−285.8 kJ/mol) = +44.0 kJ/mol.

  10. Question
    Use C + O₂ → CO₂ (−393.5 kJ) and CO + ½O₂ → CO₂ (−283.0 kJ) to find ΔH for C + ½O₂ → CO.
    Answer

    −393.5 kJ + 283.0 kJ = −110.5 kJ (the second equation is reversed)

Quiz

7 questions

  1. Question 1EasyWhich substance has a standard enthalpy of formation of 0 kJ/mol?
    Show answer

    Answer: N₂(g)

    N₂(g) is an element in its standard state. O₃ is not the standard state of oxygen (O₂ is), so its ΔHf° is not zero.

  2. Question 2MediumA → B has ΔH = −50 kJ. What is ΔH for 2B → 2A?
    Show answer

    Answer: +100 kJ

    Reversing changes the sign (+50 kJ), and doubling multiplies by 2: +100 kJ.

  3. Question 3MediumN₂ + O₂ → 2NO, ΔH = +182.6 kJ; 2NO + O₂ → 2NO₂, ΔH = −116.2 kJ. What is ΔH for N₂ + 2O₂ → 2NO₂?
    Show answer

    Answer: +66.4 kJ

    The two equations add directly (2NO cancels): +182.6 kJ + (−116.2 kJ) = +66.4 kJ.

  4. Question 4EasyWhat is the formula for ΔH° from enthalpies of formation?
    Show answer

    Answer: Σ nΔHf°(products) − Σ nΔHf°(reactants)

    Products minus reactants, each multiplied by its coefficient n.

  5. Question 5MediumCalculate ΔH° for C₂H₄(g) + H₂(g) → C₂H₆(g). (ΔHf°: C₂H₄ +52.4 kJ/mol, C₂H₆ −84.0 kJ/mol)
    Show answer

    Answer: −136.4 kJ

    ΔH° = (−84.0 kJ) − (+52.4 kJ + 0 kJ) = −136.4 kJ. H₂ is an element, so its ΔHf° is 0.

  6. Question 6MediumWhy is Hess's law useful for C(s) + ½O₂(g) → CO(g)?
    Show answer

    Answer: Some CO₂ always forms, so the heat cannot be measured cleanly

    Hess's law gets ΔH from two reactions that can be measured: the combustions of C and of CO.

  7. Question 7HardΔH° for CH₄ + 2O₂ → CO₂ + 2H₂O(l) is −890.5 kJ. Which ΔHf° values does the calculation need?
    Show answer

    Answer: CH₄, CO₂ and H₂O (O₂ is zero)

    [−393.5 kJ + 2(−285.8 kJ)] − [−74.6 kJ + 2(0 kJ)] = −890.5 kJ. O₂ is an element, so it contributes 0.

Notes and downloads

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

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