Enthalpy Changes of Reactions

Chemists rarely measure the enthalpy of a reaction by directly running it in a calorimeter — many reactions are too slow, too dangerous, or produce mixtures that are impossible to isolate. Instead they exploit a single deep fact: enthalpy is a state function, so the enthalpy change between reactants and products is fixed no matter which path connects them, direct or indirect. This is Hess's law, and it turns a table of standard enthalpies — of combustion, formation, atomisation, bond dissociation — into a toolkit for calculating enthalpy changes for reactions nobody has ever performed in a lab. This article names each type of standard enthalpy precisely, then shows how they combine through Hess's law to predict $\Delta_r H^\circ$ for any reaction built from tabulated pieces.

The core insight

Because enthalpy is a state function, $\Delta_r H^\circ$ for any reaction equals the sum of standard formation enthalpies of the products minus that of the reactants, regardless of the path taken to get there.

The core idea

Enthalpy is a state function. Its value depends only on the current state of a system (temperature, pressure, composition) — not on the route by which that state was reached. This single property is what makes Hess's law possible: if reactants can be converted to products by two different sequences of steps, both sequences must give the identical total enthalpy change, because both start and end at the same states.

Hess's law, stated formally. If a reaction can be written as the sum of two or more steps, the enthalpy change of the overall reaction equals the sum of the enthalpy changes of the individual steps:

\Delta_r H^\circ_{\text{overall}} = \Delta H^\circ_1 + \Delta H^\circ_2 + \cdots + \Delta H^\circ_n

where: each $\Delta H^\circ_i$ is the standard enthalpy change of one step in a sequence that starts at the same reactants and ends at the same products as the overall reaction, however that sequence is chosen.

The most useful consequence — the formation-enthalpy shortcut. Every reactant and product can, in principle, be built from its elements in their standard states (that is what $\Delta_f H^\circ$ measures). Treating "elements" as a common intermediate stage that every substance passes through gives a general formula for any reaction's enthalpy:

\Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})

where: each sum runs over every product or reactant weighted by its stoichiometric coefficient in the balanced equation, and $\Delta_f H^\circ$ of any element in its standard state is zero (as established in the Pass-1 definitions).

Figure

A Hess cycle showing reactants going to products by a direct arrow at the top and by an indirect route through elements at the bottom, with both paths having the same total enthalpy change

Derivation of the reaction from formation enthalpies

Derivation AReaction enthalpy from formation enthalpies

The goal is to show, using only the state-function property of enthalpy and simple algebra, that the enthalpy of any reaction can be computed purely from the tabulated standard formation enthalpies of its reactants and products, without ever running the reaction itself, and to arrive at the formula $\Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})$ .

Step 1. Set up the two competing routes between reactants and products. Route A is the direct reaction itself, with enthalpy change $\Delta_r H^\circ$ . Route B goes indirectly: first decompose every reactant back into its constituent elements in their standard states (this is the reverse of forming each reactant, so it costs $-\sum \Delta_f H^\circ(\text{reactants})$ ), then reassemble those same elements into the products (this is exactly the formation of each product, costing $+\sum \Delta_f H^\circ(\text{products})$ ).

Step 2. Invoke Hess's law. Since enthalpy is a state function, both routes start at the same reactants and end at the same products, so they must give the same total enthalpy change: $\Delta_r H^\circ = \Delta H^\circ_{\text{Route B}}$ .

Step 3. Write out Route B explicitly. $\Delta H^\circ_{\text{Route B}} = \big[-\sum \Delta_f H^\circ(\text{reactants})\big] + \big[\sum \Delta_f H^\circ(\text{products})\big]$ . Reordering the two terms gives $\Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})$ , the target formula.

Step 4. Note the stoichiometric weighting. Each formation enthalpy in the sums must be multiplied by the number of moles of that species appearing in the balanced equation, since $\Delta_f H^\circ$ is defined per mole of substance formed.

Worked example

Problem: Using $\Delta_f H^\circ[\text{CH}_4(g)] = -74.8\ \text{kJ mol}^{-1}$ , $\Delta_f H^\circ[\text{CO}_2(g)] = -393.5\ \text{kJ mol}^{-1}$ , and $\Delta_f H^\circ[\text{H}_2\text{O}(l)] = -285.8\ \text{kJ mol}^{-1}$ (the same value used for water's formation in Pass-1), calculate the standard enthalpy of combustion of methane, $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$ , and check it against the value $-890.3\ \text{kJ mol}^{-1}$ already stated for this reaction.

Step 1. List every species with its stoichiometric coefficient and its $\Delta_f H^\circ$ . Reactants: 1 mol $\text{CH}_4(g)$ at $-74.8$ , and 2 mol $\text{O}_2(g)$ at $0$ (element in its standard state). Products: 1 mol $\text{CO}_2(g)$ at $-393.5$ , and 2 mol $\text{H}_2\text{O}(l)$ at $-285.8$ .

Step 2. Sum the formation enthalpies of the products, weighted by coefficients. $\sum \Delta_f H^\circ(\text{products}) = (1)(-393.5) + (2)(-285.8) = -393.5 - 571.6 = -965.1\ \text{kJ mol}^{-1}$ .

Step 3. Sum the formation enthalpies of the reactants, weighted by coefficients. $\sum \Delta_f H^\circ(\text{reactants}) = (1)(-74.8) + (2)(0) = -74.8\ \text{kJ mol}^{-1}$ .

Step 4. Apply the formula and subtract. $\Delta_c H^\circ = \Delta_r H^\circ = -965.1 - (-74.8) = -965.1 + 74.8 = -890.3\ \text{kJ mol}^{-1}$ . This matches the combustion enthalpy of methane exactly, confirming that Hess's law and the formation-enthalpy shortcut reproduce the directly measured value. Answer: $\Delta_c H^\circ[\text{CH}_4] = -890.3\ \text{kJ mol}^{-1}$ , in agreement with the known value.

Competitive stretch

The same reaction can also be checked using bond enthalpies from Pass-1's second example: breaking 4 C–H bonds in $\text{CH}_4$ and 2 O=O bonds in $\text{O}_2$ , then forming 2 C=O bonds and 4 O–H bonds, must give the same $-890.3\ \text{kJ mol}^{-1}$ once tabulated bond enthalpies are supplied — this is the same Hess cycle drawn through bond-breaking and bond-forming steps instead of elements.

Common traps

Forgetting to multiply each $\Delta_f H^\circ$ by its stoichiometric coefficient before summing — a factor of 2 for water in the example above is easy to drop.
Reversing the subtraction order: it is always products minus reactants, never the other way round, or every sign in the answer flips.
Treating $\Delta_f H^\circ$ of $\text{O}_2(g)$ or any other standard-state element as a small nonzero number instead of exactly zero.
Assuming Hess's law only applies to reactions that can be physically decomposed into named steps — it applies to any two paths between the same states, including the purely formal formation-enthalpy route used here.

Ask the tutor: want to practice building a full Hess's law cycle for a reaction where one formation enthalpy is missing? Ask the AI tutor for a guided example.

Why does it matter that enthalpy is a state function for Hess's law to work?

If enthalpy depended on the path taken, two different routes between the same reactants and products could give different totals, and there would be no reason to expect a direct reaction and an indirect multi-step route to agree. Because enthalpy depends only on the initial and final states, any legitimate path — real or purely formal, like decomposing into elements and reassembling — must yield the same $\Delta H$ . That guarantee is exactly what lets chemists compute enthalpies for reactions they cannot run directly.

Can I use Hess's law with enthalpies of combustion instead of formation?

Yes. The same state-function argument works with any consistent set of tabulated enthalpies, including combustion data, as long as every reactant and product can be routed through a common intermediate (burnt to $\text{CO}_2$ and $\text{H}_2\text{O}$ , for combustion data). The general pattern is always $\Delta_r H^\circ = \sum(\text{reactant combustion enthalpies}) - \sum(\text{product combustion enthalpies})$ — note the subtraction order flips relative to the formation-enthalpy formula, because combustion moves substances toward the common endpoint rather than building them from it.

What if I do not have formation enthalpy data for one of the products?

Then the direct formula cannot be applied as written, and you need an alternative path built from whatever data you do have — perhaps combustion enthalpies, bond enthalpies, or a chain of other known reactions added and subtracted so that the missing species cancels out. This is the general skill Hess's law teaches: constructing any valid cycle of known steps that connects your actual reactants to your actual products, since the total is guaranteed to be the same however it is built.