Hess's Law: How to Calculate Enthalpy Change with Worked Examples
By ChemistryIQ Team · April 15, 2026
Direct Answer: Hess's Law in One Sentence
Hess's law states that the total enthalpy change of a reaction is independent of the pathway taken — meaning you can calculate ΔH for any reaction by combining a series of known reactions whose enthalpies are tabulated, as long as the steps add up to the target reaction.
The principle follows from enthalpy being a state function. A state function depends only on the initial and final states of the system, not on how you got there. Temperature is a state function. Altitude is a state function. Distance traveled is not a state function (it depends on the route). Hess's law is simply the statement that enthalpy behaves like altitude — the total change from start to finish is the same whether you drove up the mountain directly or took the scenic route.
Two practical tools come from Hess's law. First, you can add, subtract, reverse, or multiply known reactions to construct a target reaction and calculate its ΔH by applying the same operations to the enthalpy values. Second, you can use tabulated standard enthalpies of formation (ΔHf°) to calculate ΔH for any reaction using the formula: ΔH°reaction = Σ ΔHf°(products) − Σ ΔHf°(reactants), weighted by stoichiometric coefficients.
Hess's law is the workhorse of thermochemistry problems in general chemistry and shows up frequently on the ACS, MCAT, and similar standardized exams.
Enthalpy as a State Function — Why It Matters
A state function is any property whose value depends only on the current state of the system, not on the history of how the system arrived at that state. Enthalpy (H) is a state function. So are internal energy (U), entropy (S), Gibbs free energy (G), volume, pressure, and temperature.
For a reaction going from state A (reactants) to state B (products), the enthalpy change ΔH = H(B) − H(A). Because H depends only on the state, ΔH is the same whether the reaction happens in one step or fifty. This is the entire content of Hess's law.
Why this matters practically: many reactions of interest can't be carried out directly in a calorimeter. Combustion of graphite to form carbon monoxide (not CO₂) is hard to measure because some CO₂ always forms. But you can measure the combustion of graphite to CO₂ and the combustion of CO to CO₂ separately, then subtract — Hess's law gives you the ΔH for the unmeasurable step. This kind of indirect calculation is why the law matters.
The key insight for problems: treat chemical equations like algebraic equations. You can reverse them (changes the sign of ΔH), multiply them by a constant (multiplies ΔH by the same constant), and add them (adds the ΔH values). These three operations let you manipulate tabulated reactions into any target reaction.
Operations on Thermochemical Equations
Three operations are allowed when manipulating thermochemical equations:
1. **Reverse the reaction.** If you reverse a reaction, multiply ΔH by −1. An exothermic reaction (ΔH < 0) becomes endothermic (ΔH > 0) when reversed, and vice versa. This makes physical sense: breaking bonds requires energy equal to what was released when they formed.
Example: H₂(g) + ½O₂(g) → H₂O(l), ΔH = −285.8 kJ Reversed: H₂O(l) → H₂(g) + ½O₂(g), ΔH = +285.8 kJ
2. **Multiply the reaction by a constant.** Multiply ΔH by the same constant. Doubling the reaction doubles the enthalpy change; halving it halves the change. Enthalpy is an extensive property — it scales with the amount of material.
Example: H₂(g) + ½O₂(g) → H₂O(l), ΔH = −285.8 kJ Multiplied by 2: 2H₂(g) + O₂(g) → 2H₂O(l), ΔH = −571.6 kJ
3. **Add two or more reactions.** Add their ΔH values. When you add reactions, species that appear on both sides of the combined equation cancel out. Only net species remain, and those should match your target reaction.
Example: combine combustion of C(s) to CO(g) with combustion of CO(g) to CO₂(g) — intermediate CO cancels and you get the direct combustion of C(s) to CO₂(g).
These three operations together — reverse, multiply, add — let you construct virtually any target reaction from tabulated data.
Worked Example 1: Classic Two-Step Hess's Law Problem
Calculate ΔH for the reaction: C(s) + 2H₂(g) → CH₄(g)
Given: (1) C(s) + O₂(g) → CO₂(g), ΔH₁ = −393.5 kJ (2) H₂(g) + ½O₂(g) → H₂O(l), ΔH₂ = −285.8 kJ (3) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH₃ = −890.4 kJ
Strategy: figure out how to combine reactions (1), (2), and (3) to produce the target reaction. Look at where each species needs to be on the final equation.
Target left side: C(s) and 2H₂(g). Target right side: CH₄(g).
C(s) is on the left of (1) — keep (1) as is. H₂(g) is on the left of (2) — keep (2) but we need 2H₂(g), so multiply (2) by 2. CH₄(g) is on the LEFT of (3) but we need it on the RIGHT — reverse (3).
Apply the operations:
Keep (1): C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ Multiply (2) by 2: 2H₂(g) + O₂(g) → 2H₂O(l), ΔH = 2(−285.8) = −571.6 kJ Reverse (3): CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g), ΔH = +890.4 kJ
Add them all up and cancel species that appear on both sides: Left side: C(s) + O₂(g) + 2H₂(g) + O₂(g) + CO₂(g) + 2H₂O(l) Right side: CO₂(g) + 2H₂O(l) + CH₄(g) + 2O₂(g)
Cancel CO₂ (both sides), 2H₂O(l) (both sides), and 2O₂(g) from right side matches 2O₂(g) from the left side.
Net: C(s) + 2H₂(g) → CH₄(g) ✓
ΔH = −393.5 + (−571.6) + 890.4 = −74.7 kJ
The reaction is slightly exothermic. This matches the standard enthalpy of formation of methane, ΔHf°(CH₄) = −74.8 kJ/mol (the small difference is rounding).
Worked Example 2: Using Standard Enthalpies of Formation
A faster method when you have a table of standard enthalpies of formation (ΔHf°): use the formula directly:
ΔH°reaction = Σ n × ΔHf°(products) − Σ n × ΔHf°(reactants)
where n is the stoichiometric coefficient from the balanced equation.
Elements in their standard states (O₂ gas, H₂ gas, C graphite, etc.) have ΔHf° = 0 by definition. This simplifies many calculations.
Calculate ΔH for the combustion of propane: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)
From a thermochemistry table: ΔHf°(C₃H₈) = −103.8 kJ/mol ΔHf°(O₂) = 0 (element in standard state) ΔHf°(CO₂) = −393.5 kJ/mol ΔHf°(H₂O, liquid) = −285.8 kJ/mol
Apply the formula: ΔH = [3(−393.5) + 4(−285.8)] − [1(−103.8) + 5(0)] ΔH = [−1180.5 + (−1143.2)] − [−103.8] ΔH = −2323.7 − (−103.8) ΔH = −2323.7 + 103.8 ΔH = −2219.9 kJ
Propane combustion releases 2,219.9 kJ per mole — a very exothermic reaction, which is why propane is a good fuel.
This method is much faster than manipulating multiple reactions, as long as you have reliable ΔHf° data for every species in the equation. Always double-check signs and coefficients — a single sign error or missed coefficient is the most common source of mistakes.
Worked Example 3: Bond Enthalpy Estimation
When ΔHf° data isn't available, you can estimate ΔH using average bond enthalpies — a related but less accurate approach:
ΔH ≈ Σ bond enthalpies of bonds broken − Σ bond enthalpies of bonds formed
(Note: this is the reverse convention from the ΔHf° formula. Breaking bonds is endothermic; forming bonds is exothermic. The algebraic signs work out when you break bonds in reactants and form bonds in products.)
Calculate ΔH for the hydrogenation of ethene to ethane: C₂H₄(g) + H₂(g) → C₂H₆(g)
Draw out the structures: - C₂H₄: C=C double bond + 4 C−H bonds - H₂: 1 H−H bond - C₂H₆: C−C single bond + 6 C−H bonds
Bonds broken (reactants side): - 1 C=C double bond: +614 kJ - 4 C−H bonds: 4 × (+413) = +1,652 kJ - 1 H−H bond: +436 kJ - Total energy absorbed to break reactant bonds: +2,702 kJ
Bonds formed (products side): - 1 C−C single bond: −347 kJ (bond formation is exothermic) - 6 C−H bonds: 6 × (−413) = −2,478 kJ - Total energy released forming product bonds: −2,825 kJ
ΔH = +2,702 + (−2,825) = −123 kJ
The experimental value is about −137 kJ/mol. The bond enthalpy method gives a reasonable estimate but is less precise than ΔHf° calculations because bond enthalpies are averages across many compounds — the actual bond energy in a specific molecule varies slightly.
Use bond enthalpies when tabulated ΔHf° values aren't available. Otherwise, ΔHf° calculations are more accurate.
Common Mistakes and How to Avoid Them
Five mistakes dominate Hess's law problems:
1. **Forgetting to flip the sign when reversing a reaction.** If you reverse reaction (3) above, ΔH must change from −890.4 to +890.4. Missing this flip is the single most common error.
2. **Forgetting to multiply ΔH when scaling a reaction.** If you double the coefficients of reaction (2), you must double its ΔH too. Half-reactions halve the ΔH.
3. **Mixing up the formula signs.** The standard enthalpy formula is products minus reactants: ΔH = Σ ΔHf°(products) − Σ ΔHf°(reactants). Writing reactants minus products gives the wrong sign and a wrong answer.
4. **Including the ΔHf° of elements in standard states.** Elements in their standard states have ΔHf° = 0 by definition. O₂(g), N₂(g), H₂(g), Cl₂(g), C(graphite), and all other elements in their most stable form at 25°C contribute zero to the calculation. Forgetting this usually means adding a term you shouldn't have.
5. **Using the wrong physical state.** ΔHf° depends on the physical state. ΔHf°(H₂O, liquid) = −285.8 kJ/mol; ΔHf°(H₂O, gas) = −241.8 kJ/mol. The difference is the enthalpy of vaporization. If the problem specifies water vapor, use the gas-phase value. Check states carefully.
The defensive move: draw a horizontal line separating each reaction from the next, write the reaction clearly, and mark each manipulation (reverse, ×2, ×½) explicitly. Add the ΔH values column by column at the end, not in your head.
FAQs
Common questions about hess's law
Hess's law says the total enthalpy change for a reaction is the same regardless of the pathway. You can calculate the enthalpy change for any reaction by adding up the enthalpy changes of a series of known reactions that add up to the target reaction. This works because enthalpy is a state function — it only depends on the starting and ending states, not on the route taken to get there. Practically, Hess's law lets you calculate enthalpy changes for reactions that can't be measured directly.
Whenever you reverse a reaction, multiply its ΔH by −1. An exothermic reaction (ΔH < 0) becomes endothermic (ΔH > 0) when reversed, and vice versa. This reflects the fact that if forming water from hydrogen and oxygen releases 285.8 kJ, then splitting water back into hydrogen and oxygen must absorb 285.8 kJ. Forgetting to flip signs when reversing reactions is the single most common error in Hess's law problems.
ΔHf° (standard enthalpy of formation) is the enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states at 25°C and 1 atm. ΔHf° is a property of a compound, tabulated in reference books. ΔH°rxn (standard enthalpy of reaction) is the enthalpy change for a specific reaction under standard conditions — calculated from ΔHf° values using the formula ΔH°rxn = Σ ΔHf°(products) − Σ ΔHf°(reactants). Every ΔHf° is a specific ΔH°rxn (the formation reaction), but not every ΔH°rxn is a ΔHf°.
By convention. The enthalpy of formation measures the energy to form a compound from elements in their standard states. If the element is already in its standard state, no formation reaction is needed, so ΔHf° = 0 by definition. This choice of zero reference is arbitrary but universal — it lets us tabulate values for compounds without having to choose a different reference point for every element. Note that only the most stable form at 25°C and 1 atm gets ΔHf° = 0: O₂ gas is standard state (ΔHf° = 0), but ozone O₃ is not (ΔHf° = +142.7 kJ/mol).
The ΔHf° method is more accurate. Bond enthalpies are averages across many molecules containing the same bond type, so they don't capture subtle variations in bond strength based on the molecular environment. ΔHf° values are measured directly for specific compounds in specific states, so they reflect the actual thermochemistry of those compounds. Use bond enthalpies when ΔHf° data isn't available; otherwise, prefer ΔHf° calculations.
Yes. Snap a photo of any Hess's law problem and ChemistryIQ identifies whether to use reaction manipulation, standard enthalpies of formation, or bond enthalpies; shows each reversal and multiplication step with the sign change; cancels common species; and checks that the final equation matches the target. It handles combustion enthalpies, formation reactions, hydrogenation problems, and multi-step thermochemistry calculations at both the gen chem and ACS exam level.