Hess's Law

Some enthalpy changes simply cannot be measured on a bench. You cannot burn graphite and hydrogen gas together in a clean, controlled reaction to watch methane form — side reactions and incomplete combustion get in the way. Yet chemists confidently quote a number for that very enthalpy of formation. How? Enthalpy HH is a state function — it depends only on where you start and where you end, never on the road taken. Hess's law turns that single fact into a calculating tool: add, reverse and scale thermochemical equations like ordinary algebra, and their enthalpies add, reverse and scale right along with them, landing on the enthalpy of a reaction you could never run directly. This article rebuilds the rule from the state-function idea, applies it to the classic methane and carbon-monoxide problems, and flags the sign and scaling slips that cost the most marks.

The core insight

Because enthalpy is a state function, the total ΔH\Delta H for a reaction is the same by any route, so thermochemical equations can be reversed, scaled and added like algebra to build an unmeasurable ΔH\Delta H from measurable ones.

The core idea

Hess's law of constant heat summation states that the enthalpy change for a reaction is the same whether it happens in one step or through several intermediate steps, as long as the initial and final states are identical. This is not a separate law of nature — it is a direct consequence of HH being a state function. If enthalpy depended on the path, two different routes between the same reactants and products could give two different ΔH\Delta H values, and energy could be created or destroyed by simply choosing a longer route. Since that violates conservation of energy, ΔH\Delta H for the direct route must equal the sum of ΔH\Delta H for any indirect route between the same states. Three algebraic rules follow, and they are all Hess's law is in practice:

  • Reverse a reaction and the sign of ΔH\Delta H flips.
  • Scale a reaction by a factor nn and ΔH\Delta H scales by nn too.
  • Add reactions together, cancelling any species common to both sides, and the corresponding ΔH\Delta H values add.

For a target reaction A→B\text{A} \rightarrow \text{B} reached either directly (path 1, ΔH1\Delta H_1) or via an intermediate C (path 2, steps ΔH2\Delta H_2 and ΔH3\Delta H_3), Hess's law reads:

ΔH1=ΔH2+ΔH3\Delta H_1 = \Delta H_2 + \Delta H_3

where: ΔH1\Delta H_1 is the enthalpy change of the single-step (direct) path, and ΔH2\Delta H_2, ΔH3\Delta H_3 are the enthalpy changes of the two steps of the alternative path through C — the equality holds because both paths start at A and end at B. This is exactly why enthalpies of formation that cannot be measured directly are still tabulated: they are built from combustion or other measurable enthalpies by algebraically combining equations until the unmeasurable target falls out.

Figure
An enthalpy cycle showing two routes from reactants A to products B one direct step and one through an intermediate C with the two step enthalpies summing to the direct enthalpy
Hess cycle — direct path equals sum of indirect steps A B C ΔH1 (direct) ΔH2 ΔH3 ΔH1 = ΔH2 + ΔH3, because enthalpy depends only on start and end states
Derivation ABuilding an unmeasurable enthalpy from measurable ones

The goal is to obtain the standard enthalpy of formation of methane, ΔfH∘[CH4]\Delta_f H^\circ[\text{CH}_4], purely by combining three separately measurable combustion enthalpies through the reverse-scale-add rules of Hess's law, without ever running the formation reaction itself in the laboratory. Given: C(graphite)+O2(g)→CO2(g)\text{C(graphite)} + \text{O}_2(g) \rightarrow \text{CO}_2(g), ΔH1=−393.5 kJ mol−1\Delta H_1 = -393.5\ \text{kJ mol}^{-1}; H2(g)+12O2(g)→H2O(l)\text{H}_2(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l), ΔH2=−285.8 kJ mol−1\Delta H_2 = -285.8\ \text{kJ mol}^{-1}; and CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l), ΔH3=−890.3 kJ mol−1\Delta H_3 = -890.3\ \text{kJ mol}^{-1}. Target: C(graphite)+2H2(g)→CH4(g)\text{C(graphite)} + 2\text{H}_2(g) \rightarrow \text{CH}_4(g).

Step 1. Look at what the target needs: one C on the left and two H₂ on the left, with CH₄ alone on the right. Equation 1 already supplies the one C on the left, unchanged, contributing ΔH1=−393.5 kJ mol−1\Delta H_1 = -393.5\ \text{kJ mol}^{-1}.

Step 2. Scale equation 2 by 2 so it supplies two H₂ on the left: 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}, with 2ΔH2=−571.6 kJ mol−12\Delta H_2 = -571.6\ \text{kJ mol}^{-1}, since scaling a reaction scales its ΔH\Delta H by the same factor.

Step 3. Equation 3 has CH₄ on the wrong side, so reverse it to place CH₄ among the products: CO2+2H2O→CH4+2O2\text{CO}_2 + 2\text{H}_2\text{O} \rightarrow \text{CH}_4 + 2\text{O}_2, with −ΔH3=+890.3 kJ mol−1-\Delta H_3 = +890.3\ \text{kJ mol}^{-1}, since reversing a reaction flips the sign of ΔH\Delta H.

Step 4. Add all three modified equations. On the left: C+O2+2H2+O2+CO2+2H2O\text{C} + \text{O}_2 + 2\text{H}_2 + \text{O}_2 + \text{CO}_2 + 2\text{H}_2\text{O}; on the right: CO2+2H2O+CH4+2O2\text{CO}_2 + 2\text{H}_2\text{O} + \text{CH}_4 + 2\text{O}_2. The species CO2\text{CO}_2, 2H2O2\text{H}_2\text{O} and 2O22\text{O}_2 appear on both sides and cancel, leaving exactly the target: C+2H2→CH4\text{C} + 2\text{H}_2 \rightarrow \text{CH}_4.

Step 5. Sum the corresponding enthalpies, since adding equations adds their ΔH\Delta H values: ΔfH∘[CH4]=ΔH1+2ΔH2+(−ΔH3)=−393.5−571.6+890.3=−74.8 kJ mol−1\Delta_f H^\circ[\text{CH}_4] = \Delta H_1 + 2\Delta H_2 + (-\Delta H_3) = -393.5 - 571.6 + 890.3 = -74.8\ \text{kJ mol}^{-1}, which matches the experimentally accepted value.

Worked example

Problem. Given ΔfH∘[CO2(g)]=−393.5 kJ mol−1\Delta_f H^\circ[\text{CO}_2(g)] = -393.5\ \text{kJ mol}^{-1} and ΔfH∘[CO(g)]=−110.5 kJ mol−1\Delta_f H^\circ[\text{CO}(g)] = -110.5\ \text{kJ mol}^{-1}, find ΔH\Delta H for CO(g)+12O2(g)→CO2(g)\text{CO}(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g).

Step 1. Write the two given formation equations: C+O2→CO2\text{C} + \text{O}_2 \rightarrow \text{CO}_2 with ΔHa=−393.5 kJ mol−1\Delta H_a = -393.5\ \text{kJ mol}^{-1}, and C+12O2→CO\text{C} + \tfrac{1}{2}\text{O}_2 \rightarrow \text{CO} with ΔHb=−110.5 kJ mol−1\Delta H_b = -110.5\ \text{kJ mol}^{-1}.

Step 2. The target has CO as a reactant, but equation b has CO as a product, so reverse it: CO→C+12O2\text{CO} \rightarrow \text{C} + \tfrac{1}{2}\text{O}_2, with −ΔHb=+110.5 kJ mol−1-\Delta H_b = +110.5\ \text{kJ mol}^{-1}.

Step 3. Add this reversed equation to equation a. Left side gains CO+12O2+C+O2\text{CO} + \tfrac{1}{2}\text{O}_2 + \text{C} + \text{O}_2; right side gains C+12O2+CO2\text{C} + \tfrac{1}{2}\text{O}_2 + \text{CO}_2. Cancelling the C and 12O2\tfrac{1}{2}\text{O}_2 common to both sides leaves exactly CO+12O2→CO2\text{CO} + \tfrac{1}{2}\text{O}_2 \rightarrow \text{CO}_2, the target.

Step 4. Sum the enthalpies: ΔH=ΔHa+(−ΔHb)=−393.5+110.5=−283.0 kJ mol−1\Delta H = \Delta H_a + (-\Delta H_b) = -393.5 + 110.5 = -283.0\ \text{kJ mol}^{-1}. The enthalpy of combustion of carbon monoxide is −283.0 kJ mol−1-283.0\ \text{kJ mol}^{-1}.

Competitive stretch

The same cycle shows why bond-enthalpy and lattice-enthalpy problems (Born–Haber cycles) are also "Hess's law" underneath: any set of equations that algebraically add up to a target reaction gives a valid route, whether the intermediate is a compound like CO₂ or a hypothetical state like gaseous ions. Practice spotting which given equation must be reversed and which must be scaled before adding — that is the entire skill.

Common traps

  • Forgetting to reverse the sign of ΔH\Delta H when an equation is flipped to match the target reaction.
  • Forgetting to multiply ΔH\Delta H by the same factor used to scale an equation's coefficients.
  • Cancelling species incorrectly across several equations — write out everything on each side first, then cancel systematically rather than by eye.
  • Thinking Hess's law only applies to combustion-to-formation problems; it applies to any set of reactions that can be algebraically combined into the target, including bond and lattice enthalpy cycles.

Ask the tutor: Stuck on which equations to reverse or scale for a particular Hess's law problem — ask the AI tutor to walk through the cycle with you step by step.

Why can't we just measure the enthalpy of formation of methane directly?

Because graphite and hydrogen gas do not react cleanly to give pure methane under controlled laboratory conditions — side reactions, incomplete conversion and other products get in the way, so no calorimeter experiment can isolate exactly this ΔH\Delta H. Combustion reactions, by contrast, go essentially to completion and are easy to measure calorimetrically. Hess's law lets us start from those clean, measurable combustion numbers and algebraically reconstruct the formation enthalpy we cannot measure directly, using only the fact that enthalpy does not care about the path taken.

How do I know which given equation to reverse or scale?

Look only at where each species needs to end up in the target equation. If a species sits on the wrong side of a given equation compared to the target, reverse that equation (and flip its ΔH\Delta H sign). If a species needs more or fewer moles than the given equation supplies, scale the whole equation by the required factor (and scale ΔH\Delta H the same way). Do this species by species, then add all the modified equations and check that everything except the target's reactants and products cancels out.

Is Hess's law only useful for combustion and formation reactions?

No — it is a completely general tool for any reaction whose ΔH\Delta H is hard to measure directly, as long as you can find other reactions with known ΔH\Delta H that combine algebraically to give the target. Bond-enthalpy calculations, ionisation and lattice enthalpy cycles (Born–Haber cycles), and enthalpies of neutralisation or solution built from other steps all rely on exactly the same reverse-scale-add logic introduced here.

Before this

Where this leads