Thermochemistry: enthalpy and Hess's law with exercises

End of September 2026, Year 4 of the Italian liceo scientifico (science-focused high school; it applies identically to the applied-sciences track and to the chemistry technical institute): after the chapter on solutions and concentration, the chemistry programme moves on to thermochemistry, usually between the start and the middle of October, before reaching kinetics and equilibrium. The first test on enthalpy typically falls between early November and early December, five to nine weeks from now.

If you haven’t yet revised concentrations, which come in useful when an exercise needs you to calculate moles from a solution, you will find the method in Solution concentration: molarity and dilutions. Here I start from enthalpy instead: what it is, how it is calculated with Hess’s law, and where the mistake that loses the most marks in tests hides.

System, surroundings and the meaning of enthalpy

In thermochemistry you always distinguish the system (the chemical reaction being studied) from the surroundings (everything else: the container, the air, the laboratory). Enthalpy $H$ is a quantity that measures the energy content of a system at constant pressure, the most common condition in the laboratory, where reactions take place in contact with the atmosphere. What is actually measured is never $H$ itself, but its change during a reaction:

The sign of $\Delta H$ tells you whether the reaction releases or absorbs energy, and from this the two fundamental categories arise.

Exothermic and endothermic reactions

An exothermic reaction releases heat to the surroundings: the energy of the products is lower than that of the reactants, so $\Delta H<0$. An endothermic reaction absorbs heat from the surroundings: the products have higher energy than the reactants, so $\Delta H>0$.

On the energy diagram (energy on the vertical axis, progress of the reaction on the horizontal axis):

• in the exothermic one, the line of the products lies below that of the reactants;
• in the endothermic one, the line of the products lies above that of the reactants.

An example of an exothermic reaction you already know from everyday life is any combustion (burning methane, petrol, wood): the excess energy is released as heat, and that is why you get warm near a fire. An example of an endothermic reaction is the dissolving of ammonium nitrate in water, which cools the solution: the same principle as “instant ice” packs.

The thermochemical equation

A thermochemical equation is a balanced chemical equation with the value of $\Delta H$ alongside it, together with the physical state of each species (because going from liquid to gas costs energy, and changes the balance):

Two conventions to always respect:

• $\Delta H$ refers to the reaction as written, with those coefficients: if you double all the coefficients, $\Delta H$ doubles.
• The unit is kJ (or kJ/mol, referring to one mole of the reaction as written); do not confuse it with the bond energy of a single bond, which is expressed per mole of bonds, not per mole of reaction.

Standard enthalpy of formation

The standard enthalpy of formation $\Delta H_f^{\circ }$ of a substance is the enthalpy change of the reaction that forms one mole of that substance from the elements in their standard state (the most stable form at 25 °C and 1 atm: ${\mathrm{O}}_2(\mathrm{g})$, ${\mathrm{H}}_2(\mathrm{g})$, $\mathrm{C}(\mathrm{graphite})$, and so on).

A convention that almost every exercise uses without saying so explicitly: the enthalpy of formation of an element in its standard state is, by definition, zero. $\Delta H_f^{\circ }({\mathrm{O}}_2(\mathrm{g}))=0$, $\Delta H_f^{\circ }({\mathrm{H}}_2(\mathrm{g}))=0$: they are already “in their starting form”, there is no formation reaction to carry out.

The formula for calculating $\Delta H^{\circ }$ of a reaction

From tabulated enthalpies of formation you calculate the standard enthalpy of any reaction with this formula:

where $n_p$ and $n_r$ are the stoichiometric coefficients of the balanced equation. The point almost the whole class trips on: the coefficients multiply each value of $\Delta H_f^{\circ }$ before adding or subtracting. If you have $2{\mathrm{H}}_2\mathrm{O}$ in the reaction, the formula contains $2\times \Delta H_f^{\circ }({\mathrm{H}}_2\mathrm{O})$, not $\Delta H_f^{\circ }({\mathrm{H}}_2\mathrm{O})$ just once.

Hess’s law

Hess’s law says that $\Delta H$ of a reaction depends only on the initial state (reactants) and the final state (products), not on the path followed to get there: enthalpy is a state function. Practical consequence: if a reaction can be obtained by adding two or more intermediate reactions, its $\Delta H$ is the sum of the $\Delta H$ of those intermediate reactions.

To use Hess’s law you need two rules for manipulating equations:

1. If you reverse a reaction (swap reactants and products), the sign of $\Delta H$ changes: what used to release energy now absorbs it in the same amount.
2. If you multiply all the coefficients of a reaction by a factor $k$, $\Delta H$ is multiplied by the same factor $k$.

The formula with enthalpies of formation seen above is, in fact, a special case of Hess’s law: the reaction of interest is broken down into the path “reactants → elements → products”, going through the formation of each species from its elements.

A note on bond energies

An alternative method, less precise but useful when you don’t have enthalpies of formation, estimates $\Delta H$ from bond energies (the energy needed to break one type of bond in the gas phase):

Here the sign is reversed compared with the formula for enthalpies of formation, because breaking bonds costs energy (it is always positive) while forming them releases energy. At this level it is enough to recognise the method, without going into it further.

Heat from mass or moles

Once $\Delta H$ is known (in kJ/mol, referring to the balanced reaction), the heat exchanged by any quantity of reactant is found with a direct proportion:

where $n$ is the moles of the reactant to which the reaction refers (watch the coefficient: if $\Delta H$ is given for the reaction with 2 mol of a reactant, the value per mole of that reactant is $\Delta H/2$). If you have a mass instead of moles, convert first with $n=m/M$, the same step used in stoichiometry (I talk about it in Mole and stoichiometry: the method in Year 3).

Typical mistakes

Three worked exercises

1. Calculating $\Delta H_r^{\circ }$ from enthalpies of formation. Calculate the standard enthalpy of combustion of methane:

knowing that $\Delta H_f^{\circ }(\mathrm{C}{\mathrm{H}}_4)=-74.8$ kJ/mol, $\Delta H_f^{\circ }(\mathrm{C}{\mathrm{O}}_2)=-393.5$ kJ/mol, $\Delta H_f^{\circ }({\mathrm{H}}_2\mathrm{O}(\mathrm{l}))=-285.8$ kJ/mol and $\Delta H_f^{\circ }({\mathrm{O}}_2)=0$ (element in its standard state).

I apply the formula, multiplying each value by its own coefficient:

The negative sign confirms that the combustion of methane is exothermic, as you would expect from any combustion.

There is no dedicated practice test on this topic on the site yet: you will find all the ones available on the other subjects at /simulazioni/liceo/, with the correct answer free and the guided solution unlocking with the first lesson.

2. Hess’s law with intermediate reactions. Knowing the two reactions:

find $\Delta H_3$ for the overall reaction $\mathrm{C}(\mathrm{graphite})+{\mathrm{O}}_2(\mathrm{g})\to \mathrm{C}{\mathrm{O}}_2(\mathrm{g})$.

Adding the two intermediate equations line by line, $\mathrm{CO}(\mathrm{g})$ appears as a product in the first and as a reactant in the second: it cancels out, exactly as in a system of equations. Adding the two $\frac{1}{2}{\mathrm{O}}_2$ as well gives a whole ${\mathrm{O}}_2$, and the net result is precisely the required reaction. By Hess’s law, $\Delta H_3$ is the sum of the two values:

The value found coincides, not by chance, with $\Delta H_f^{\circ }(\mathrm{C}{\mathrm{O}}_2)$ used in exercise 1: the reaction “carbon plus oxygen gives carbon dioxide” is exactly the formation reaction of $\mathrm{C}{\mathrm{O}}_2$ from its elements, and Hess’s law guarantees that the result is identical regardless of the path (direct, or going via carbon monoxide).

3. Heat released by a given mass. Using the $\Delta H_r^{\circ }=-890.3$ kJ/mol found in exercise 1, how much heat is released by completely burning $8.0$ g of methane (molar mass $\mathrm{C}{\mathrm{H}}_4=16$ g/mol)?

I convert the mass into moles:

The $\Delta H_r^{\circ }$ given refers to 1 mol of $\mathrm{C}{\mathrm{H}}_4$ (the coefficient in the balanced reaction), so the heat released by $0.5$ mol is:

Burning $8.0$ g of methane releases about $445$ kJ of heat to the surroundings (the sign is omitted because the question asks “how much heat is released”, already implicitly positive for an exothermic reaction).

Before the test

For this topic there is no dedicated chemistry practice test on the site yet: you will find timed tests on other subjects at /simulazioni/liceo/. For the rest of the science programme, and for the chapters that follow thermochemistry (kinetics and equilibrium), you will find the full picture at /materie/scienze/; I talk about them specifically in Chemical kinetics: reaction rate and exercises and in Chemical equilibrium and Le Chatelier in Year 4. Heat exchange in physics, with specific heat and thermal equilibrium, is in Specific heat and thermal equilibrium: worked exercises.

If enthalpy and Hess’s law are creating more doubts than certainties for you, the first introductory call is free and lasts half an hour.