Hess's law

You want the enthalpy of forming carbon monoxide, C(graphite)+12O2(g)→CO(g)\text{C(graphite)} + \tfrac12\text{O}_2(g) \rightarrow \text{CO}(g). Try to burn carbon carefully enough to stop exactly at CO and you fail — the reaction refuses to stay put, sliding on to CO2\text{CO}_2 no matter how little oxygen you allow. The direct experiment is simply impossible to control. Yet chemists quote a precise number for this enthalpy change, −110.5 kJ mol−1-110.5\ \text{kJ mol}^{-1}, obtained without ever performing that reaction. The trick is Hess's law of constant heat summation: since enthalpy is a state function, the total ΔH\Delta H for going from reactants to products is the same however many steps you take to get there. So instead of measuring the awkward direct path, you measure two easy ones — burning carbon fully to CO2\text{CO}_2, and burning CO to CO2\text{CO}_2 — and combine them algebraically. This article rebuilds that combination rule from the state-function idea, generalises it into the formation-enthalpy and bond-enthalpy formulas used throughout thermodynamics, and reworks the CO and ethanol-combustion calculations end to end so the arithmetic pattern becomes automatic.

The core insight

Because enthalpy is a state function, the ΔH\Delta H of any reaction equals the sum of the ΔH\Delta H values of any set of steps — direct or indirect — connecting the same reactants to the same products, so unknown enthalpies can be built algebraically from known ones by reversing, scaling and adding thermochemical equations.

The core idea

1. Hess's law, stated precisely. The total enthalpy change of a reaction is the same whether it occurs in one step or via any number of intermediate steps, provided the initial and final states (temperature, pressure, physical states) are identical. It is named for Germain Hess, who established it experimentally in 1840, decades before the state-function idea itself was formalised.

2. Why it must be true. HH is a state function: ΔH\Delta H depends only on the initial and final states of a system, never on the path taken between them. A reaction can be pictured as one edge of a path diagram connecting reactants to products; Hess's law says every alternative path between the same two endpoints — however many arrows and intermediates it passes through — must sum to the identical net ΔH\Delta H, because HH itself has no memory of which route was taken.

3. The algebra of combining equations. If a target equation can be built by adding, subtracting, reversing, or scaling other equations whose ΔH\Delta H values are known, the same operations performed on the numbers give the target ΔH\Delta H:

  • Reversing a reaction flips the sign of ΔH\Delta H.
  • Scaling a reaction by a factor nn multiplies ΔH\Delta H by nn.
  • Adding reactions, with any species common to both sides cancelling, adds their ΔH\Delta H values.

4. Two standard, ready-made applications of this same idea, both examinable in their own right:

ΔHrxn∘=∑np ΔfH∘(products)−∑nr ΔfH∘(reactants)(1)\Delta H^\circ_{rxn} = \sum n_p\,\Delta_f H^\circ(\text{products}) - \sum n_r\,\Delta_f H^\circ(\text{reactants}) \qquad (1)

where: ΔfH∘\Delta_f H^\circ is the standard enthalpy of formation of a species from its elements in their standard states, and npn_p, nrn_r are the stoichiometric coefficients of products and reactants respectively. By definition ΔfH∘=0\Delta_f H^\circ = 0 for any element in its standard state.

ΔHrxn∘=∑(bond enthalpies broken in reactants)−∑(bond enthalpies formed in products)(2)\Delta H^\circ_{rxn} = \sum(\text{bond enthalpies broken in reactants}) - \sum(\text{bond enthalpies formed in products}) \qquad (2)

where: bond enthalpies are always positive (bond breaking is endothermic, bond formation is exothermic), so breaking is added and forming is subtracted; Equation (2) applies to gas-phase reactions where all bonds can be tabulated. Both formulas are Hess's law in disguise: routing every species through its elements (the common reference point) or through separated atoms is just another indirect path with the same endpoints as the direct reaction.

Figure
A path diagram showing carbon plus oxygen going directly to carbon dioxide with enthalpy H1, and separately to carbon monoxide then to carbon dioxide with enthalpies H and H2, illustrating that H1 equals H plus H2
Two paths, same endpoints, same total enthalpy change C + O2 (g) CO2 (g) direct: delta H1 = -393.5 kJ/mol CO (g) + half O2 delta H = -110.5 delta H2 = -283.0 delta H1 = delta H + delta H2, since both routes start and end the same
Derivation ACombining thermochemical equations by reversal and addition

Goal: starting from two measured enthalpy changes, ΔH1\Delta H_1 for C(graphite)+O2(g)→CO2(g)\text{C(graphite)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) and ΔH2\Delta H_2 for CO(g)+12O2(g)→CO2(g)\text{CO}(g) + \tfrac12\text{O}_2(g) \rightarrow \text{CO}_2(g), use Hess's law and the algebra of reversing, scaling and adding equations to derive ΔH\Delta H for the unmeasurable target reaction C(graphite)+12O2(g)→CO(g)\text{C(graphite)} + \tfrac12\text{O}_2(g) \rightarrow \text{CO}(g), showing every cancellation explicitly.

Step 1. Identify how the target relates to the two known equations. The target has C and 12O2\tfrac12\text{O}_2 as reactants and CO as product. Reaction 1 already has C and O2\text{O}_2 as reactants; reaction 2 has CO as a reactant, but the target needs CO as a product, so reaction 2 must be reversed before use.

Step 2. Reverse reaction 2. The reverse of CO(g)+12O2(g)→CO2(g)\text{CO}(g) + \tfrac12\text{O}_2(g) \rightarrow \text{CO}_2(g), ΔH2=−283.0 kJ mol−1\Delta H_2 = -283.0\ \text{kJ mol}^{-1}, is CO2(g)→CO(g)+12O2(g)\text{CO}_2(g) \rightarrow \text{CO}(g) + \tfrac12\text{O}_2(g) with the sign flipped: −ΔH2=+283.0 kJ mol−1-\Delta H_2 = +283.0\ \text{kJ mol}^{-1}.

Step 3. Add reaction 1 to the reversed reaction 2. Writing both equations with reactants on the left and products on the right and adding them side by side: C+O2+CO2→CO2+CO+12O2\text{C} + \text{O}_2 + \text{CO}_2 \rightarrow \text{CO}_2 + \text{CO} + \tfrac12\text{O}_2.

Step 4. Cancel species common to both sides. CO2\text{CO}_2 appears on both sides and cancels completely. O2\text{O}_2 appears as 1 O21\,\text{O}_2 on the left and 12O2\tfrac12\text{O}_2 on the right; subtracting leaves 12O2\tfrac12\text{O}_2 on the left. The net equation is C+12O2→CO\text{C} + \tfrac12\text{O}_2 \rightarrow \text{CO} — exactly the target.

Step 5. Because the combined equation matches the target exactly, Hess's law guarantees its ΔH\Delta H is the simple sum of the two enthalpies used to build it: ΔHtarget=ΔH1+(−ΔH2)\Delta H_{target} = \Delta H_1 + (-\Delta H_2). This is the general pattern: whatever arithmetic combination of equations reproduces the target equation, the same arithmetic combination of their ΔH\Delta H values reproduces the target's ΔH\Delta H.

Worked example

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}; and CO(g)+12O2(g)→CO2(g)\text{CO}(g) + \tfrac12\text{O}_2(g) \rightarrow \text{CO}_2(g), ΔH2=−283.0 kJ mol−1\Delta H_2 = -283.0\ \text{kJ mol}^{-1}. Find ΔH\Delta H for C(graphite)+12O2(g)→CO(g)\text{C(graphite)} + \tfrac12\text{O}_2(g) \rightarrow \text{CO}(g).

Step 1. From the derivation above, the target equation equals reaction 1 plus the reversed reaction 2, so ΔH=ΔH1+(−ΔH2)\Delta H = \Delta H_1 + (-\Delta H_2).

Step 2. Substitute the given values: ΔH=−393.5+(−(−283.0))=−393.5+283.0\Delta H = -393.5 + (-(-283.0)) = -393.5 + 283.0.

Step 3. Compute: ΔH=−110.5 kJ mol−1\Delta H = -110.5\ \text{kJ mol}^{-1}.

Step 4. Interpret: this is the standard enthalpy of formation of CO(g)\text{CO}(g), ΔH=−110.5 kJ mol−1\Delta H = -110.5\ \text{kJ mol}^{-1}, obtained entirely from two reactions that could be measured cleanly, with no need to control the tricky direct carbon-to-CO combustion.

Competitive stretch

Use the formation-enthalpy formula, Equation (1), on a full molecule. Given ΔfH∘(CO2,g)=−393.5\Delta_f H^\circ(\text{CO}_2, g) = -393.5, ΔfH∘(H2O,l)=−285.8\Delta_f H^\circ(\text{H}_2\text{O}, l) = -285.8, ΔfH∘(C2H5OH,l)=−277.6\Delta_f H^\circ(\text{C}_2\text{H}_5\text{OH}, l) = -277.6 (all kJ mol−1\text{kJ mol}^{-1}), find ΔH∘\Delta H^\circ for C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)\text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l). Products: 2(−393.5)+3(−285.8)=−1644.4 kJ mol−12(-393.5) + 3(-285.8) = -1644.4\ \text{kJ mol}^{-1}. Reactants: 1(−277.6)+3(0)=−277.6 kJ mol−11(-277.6) + 3(0) = -277.6\ \text{kJ mol}^{-1} (recall ΔfH∘(O2,g)=0\Delta_f H^\circ(\text{O}_2, g) = 0, since O2\text{O}_2 is an element in its standard state). So ΔHrxn∘=−1644.4−(−277.6)=−1366.8 kJ mol−1\Delta H^\circ_{rxn} = -1644.4 - (-277.6) = -1366.8\ \text{kJ mol}^{-1} — a large negative value consistent with ethanol's use as a fuel.

Common traps

  • Forgetting to flip the sign of ΔH\Delta H whenever a given equation is reversed to fit the target, as in Step 2 of the derivation above.
  • Forgetting to multiply ΔH\Delta H by the same factor used to scale a reaction's coefficients when the target needs more than one unit of a given equation.
  • Swapping products and reactants in Equation (1), a common slip that flips the final sign of ΔHrxn∘\Delta H^\circ_{rxn}.
  • Assigning a nonzero ΔfH∘\Delta_f H^\circ to an element in its standard state (e.g. O2(g)\text{O}_2(g), N2(g)\text{N}_2(g), graphite) — by definition it is zero.

Ask the tutor: Want to practise picking which given equations to reverse and scale for a trickier three-equation Hess's law combination?

Why can't we just measure the enthalpy of forming CO directly?

Burning carbon in a limited supply of oxygen does not stop cleanly at CO — some carbon inevitably goes all the way to CO2\text{CO}_2, and the exact mixture obtained is hard to control and reproduce. Hess's law sidesteps the problem: burning carbon fully to CO2\text{CO}_2, and separately burning CO to CO2\text{CO}_2, are both easy, reproducible measurements, and combining them algebraically gives the CO formation enthalpy without ever performing the awkward reaction.

Is the formation-enthalpy formula a different rule from Hess's law, or the same idea?

The same idea, applied through a fixed reference point. Imagine reactants and products both being broken down into their constituent elements and rebuilt: that detour is just another indirect path between the same initial and final states, so by Hess's law it must give the same total delta H as the direct reaction. Subtracting the reactants' formation enthalpies from the products' is exactly this detour written as one formula.

How do I know which equation to reverse or scale when combining several given reactions?

Work backward from the target equation. Check each given reaction's reactants and products against the target: if a species you need as a product appears as a reactant in a given equation, that equation must be reversed; if the target needs more (or fewer) moles of a species than a given equation supplies, that equation must be scaled by the matching factor. Do this species by species until every intermediate cancels and only the target's own reactants and products remain.

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