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Hess's law questions are puzzles: you are handed three or four equations and asked for the enthalpy of a fourth. Students who struggle usually know the law but have no method for building the cycle. This post gives that method, works through three examples, and lists the sign and state errors that cause most wrong answers.
What does Hess's law say?
Germain Hess published his thermochemical work in Annalen der Physik in 1840. In modern terms, as the NCERT Class 11 chapter on Thermodynamics puts it: if a reaction takes place in several steps, its standard enthalpy change is the sum of the enthalpy changes of the steps. It works because enthalpy is a state function; only the start and end matter.
The JEE (Main) 2026 syllabus lists "Hess's law of constant heat summation" with enthalpies of bond dissociation, combustion, formation, atomisation, sublimation, phase transition, hydration, ionisation and solution. The NEET (UG)-2026 syllabus has the same. Check the syllabus for your exam year.
How do you build a cycle from given equations?
Three rules:
Reverse an equation and the sign of ΔH flips.
Multiply an equation by a number and ΔH is multiplied by the same number.
Add equations and add their ΔH values. Species on both sides cancel.
Method: look at the target equation. For each substance in it, find the given equation where it appears, then reverse and scale that equation so the substance is on the correct side with the correct coefficient.
Example 1 (NCERT's cycle). Burning carbon in limited oxygen always gives some CO₂ as well, so the enthalpy of formation of CO cannot be measured directly. It is found from two combustions instead:
C(graphite) + O₂(g) → CO₂(g), ΔH = −393.5 kJ mol⁻¹
CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ mol⁻¹
Target: C(graphite) + ½O₂(g) → CO(g). Keep the first equation, reverse the second (ΔH = +283.0), and add. CO₂ cancels, and half an O₂ is left on the left. ΔH = −393.5 + 283.0 = −110.5 kJ mol⁻¹.
How do you use enthalpies of formation?
When formation values are given, ΔrH = Σ ΔfH(products) − Σ ΔfH(reactants), each multiplied by its coefficient. Elements in their standard states have ΔfH = 0.
NCERT's example: for CaCO₃(s) → CaO(s) + CO₂(g), ΔrH = (−635.1) + (−393.5) − (−1206.9) = +178.3 kJ mol⁻¹. Positive, so you must heat limestone to decompose it.
Example 2. NCERT gives the enthalpy of combustion of liquid ethanol as −1367 kJ mol⁻¹, with ΔfH of −393.5 kJ mol⁻¹ for CO₂(g) and −285.83 kJ mol⁻¹ for H₂O(l). Find ΔfH of ethanol.
C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l).
−1367 = [2(−393.5) + 3(−285.83)] − [ΔfH(ethanol) + 0].
−1367 = −1644.49 − ΔfH(ethanol).
ΔfH(ethanol) = −1644.49 + 1367 ≈ −277.5 kJ mol⁻¹.
O₂ contributes zero because it is an element in its standard state.
How do you use bond enthalpies?
For gas-phase reactions, ΔrH ≈ Σ (bonds broken) − Σ (bonds formed). Note the order is the reverse of the formation formula, because breaking bonds always takes energy.
Example 3. Estimate ΔH for CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g), using NCERT's mean bond enthalpies: C–H 414, Cl–Cl 243, C–Cl 330 and H–Cl 431 kJ mol⁻¹.
Broken: one C–H and one Cl–Cl: 414 + 243 = 657 kJ mol⁻¹.
Formed: one C–Cl and one H–Cl: 330 + 431 = 761 kJ mol⁻¹.
ΔH ≈ 657 − 761 = −104 kJ mol⁻¹.
Only the bonds that change need counting; the three untouched C–H bonds cancel. NCERT points out that mean bond enthalpies are averages, so the result is an estimate, valid when all species are gases.
Why do energy questions trip students up?
Boo (1998) interviewed 48 Grade 12 students about bonds and energy in five familiar reactions after instruction, and found a range of ideas at odds with the chemist's view, some not reported before. One idea worth checking in yourself: breaking a bond never releases energy. Energy is released when new bonds form, and ΔH is the balance of the two.
Which mistakes should you avoid?
Not flipping the sign when reversing an equation.
Not scaling ΔH when multiplying an equation.
Ignoring states. H₂O(l) and H₂O(g) have different enthalpies of formation.
Giving elements a non-zero ΔfH, or using diamond instead of graphite for carbon.
Bond enthalpy formula reversed, using products minus reactants.
A practice routine
Day 1: four Hess's cycles from given equations, writing the target first each time.
Day 2: four formation-enthalpy problems, including one where you find an unknown ΔfH.
Day 3: four bond-enthalpy estimates, listing only the bonds that change.
Day 4: mixed set of eight. Mark every error as sign, scaling, state or method.
Day 5: redo the errors. Our post on why JEE scores plateau shows how to keep an error log.
The physics side of heat, mixing and phase changes is in our guide to calorimetry and heat transfer. When a cycle will not close and you cannot see why, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, use Example 1's method to find ΔH for CO(g) + ½O₂(g) → CO₂(g) from the formation values of CO and CO₂, and check you get −283.0 kJ mol⁻¹.
