IMAT Chemistry · Equilibrium, Kinetics and Energetics

Exothermic and Endothermic Reactions, Bond Enthalpies and Hess's Law

The enthalpy change ΔH is the heat a reaction gives out or takes in at constant pressure: negative when products are lower in energy than reactants, positive when they are higher.

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Why this matters

The 2026 paper asked what is true of the energy of the products in an exothermic reaction, and the 2025 paper asked what the activation energy represents on an energy profile. Bond enthalpy and Hess's law calculations are in the syllabus but have not been set yet.

Concept 1 of 4: Exothermic and endothermic reactions and the sign of ΔH

Chemicals store energy in their bonds. If the products hold less of it than the reactants, the difference escapes as heat and the surroundings warm up: the reaction is exothermic. If the products hold more, the reaction must pull heat in from the surroundings, which cool down: it is endothermic.

Definition

The system is the reacting chemicals; the surroundings are everything else (the water, the beaker, the air).
The enthalpy change $\Delta H$ is the heat exchanged at constant pressure, in kJ/mol: $\Delta H = H_\text{products} - H_\text{reactants}$ .
Exothermic: heat is released, $\Delta H < 0$ , the products have lower energy than the reactants.
Endothermic: heat is absorbed, $\Delta H > 0$ , the products have higher energy than the reactants.
Breaking bonds always takes in energy; making bonds always releases it. The sign of $\Delta H$ depends on which is larger.
$\Delta H$ says nothing about how fast a reaction goes.

Feature	Exothermic	Endothermic
Heat flow	Released to the surroundings	Absorbed from the surroundings
Sign of ΔH	Negative	Positive
Energy of products	Lower than the reactants	Higher than the reactants
Temperature of surroundings	Rises	Falls
Examples	Combustion, respiration, neutralisation, freezing, condensation	Photosynthesis, thermal decomposition of $\mathrm{CaCO_3}$ , melting, dissolving ammonium nitrate

Practice this conceptself-check · 4 quick reps

Exothermic does not mean fast

Rusting is exothermic but takes weeks, and wood does not catch fire at room temperature even though burning is exothermic. Whether a reaction gives out heat (ΔH) and how fast it goes (the activation energy and the conditions) are separate questions.

Breaking bonds never releases energy

Energy is released when bonds form, and must be supplied to break them. A reaction is exothermic when the bonds formed release more energy than the bonds broken took in.

Concept 2 of 4: Energy profiles and the activation energy

Even an exothermic reaction does not simply run downhill. Bonds must first start to break, which needs energy, so the reactants have to climb over an energy barrier before they can fall to the products. The height of that barrier, measured from the reactants, is the activation energy.

Definition

An energy profile plots the energy of the system against the progress of the reaction.

The peak is the transition state (activated complex).
The activation energy $E_a$ is the energy barrier the reactants must overcome to become products: from the reactants up to the peak.
$\Delta H$ is the vertical gap from the reactants to the products, not to the peak.
For the reverse reaction, the barrier is measured from the products up to the same peak.
A catalyst lowers the peak (so lowers $E_a$ both ways) but leaves the reactant and product levels, and so $\Delta H$ , unchanged.

Energy profile relations

\Delta H = H_{\text{products}} - H_{\text{reactants}} \qquad E_{a,\text{reverse}} = E_{a,\text{forward}} - \Delta H

$E_a$ activation energy, measured from the starting level up to the peak
$\Delta H$ enthalpy change of the forward reaction

Worked example

A reaction has a forward activation energy of 120 kJ/mol and

\Delta H = -80\ \text{kJ/mol}

. Describe its profile and find the activation energy of the reverse reaction.

Practice this conceptself-check · 3 quick reps

Activation energy is not the energy difference between products and reactants

The gap between reactants and products is

\Delta H

. The activation energy is the barrier from the reactants up to the transition state. Options describing

E_a

as energy released, or as the products minus the reactants, are wrong.

Concept 3 of 4: Estimating ΔH from bond enthalpies

Imagine the reaction in two steps: first break every bond in the reactants, then make every bond in the products. Breaking costs energy, making pays it back. The enthalpy change is the cost minus the pay-back, so a reaction that forms stronger bonds than it breaks is exothermic.

Definition

The bond enthalpy is the energy needed to break one mole of a given bond in gaseous molecules, in kJ/mol. It is always positive.
Tables give mean bond enthalpies (averaged over many compounds), so the answer is an estimate.
Count every bond: $\mathrm{CH_4}$ has four C-H bonds; two molecules of $\mathrm{H_2O}$ contain four O-H bonds.

ΔH from bond enthalpies

\Delta H \approx \sum E(\text{bonds broken}) - \sum E(\text{bonds formed})

$E$ mean bond enthalpy, in kJ/mol

Worked example

Estimate

\Delta H

for

\mathrm{H_2 + Cl_2 \rightarrow 2HCl}

. Bond enthalpies in kJ/mol: H-H 436, Cl-Cl 243, H-Cl 432.

Practice this conceptself-check · 3 quick reps

Bonds broken minus bonds formed, not the other way round

\Delta H \approx

(energy to break bonds) minus (energy released forming bonds). Subtracting in the other order gives the right size with the wrong sign, and that wrong sign is always one of the options.

Concept 4 of 4: Hess's law and enthalpies of formation

Energy cannot be created or destroyed, so the energy change between a starting point and an end point cannot depend on the route taken. If a reaction is hard to measure directly, you can add up the enthalpy changes of other reactions that together lead to the same place.

Definition

Hess's law: the enthalpy change of a reaction is the same whatever route is taken, as long as the start and end are the same.
Reversing an equation changes the sign of $\Delta H$ . Multiplying an equation by a number multiplies $\Delta H$ by it.
The standard enthalpy of formation $\Delta H_f^\circ$ is the enthalpy change when one mole of a compound forms from its elements in their standard states. It is zero for an element in its standard state ( $\mathrm{O_2(g)}$ , graphite).

ΔH from enthalpies of formation

\Delta H = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})

$\Delta H_f^\circ$ standard enthalpy of formation, in kJ/mol, multiplied by the coefficient in the equation

Worked example

Given

\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}

\Delta H = -394\ \text{kJ/mol}

, and

\mathrm{CO(g) + \tfrac{1}{2}O_2(g) \rightarrow CO_2(g)}

\Delta H = -283\ \text{kJ/mol}

, find

\Delta H

for

\mathrm{C(s) + \tfrac{1}{2}O_2(g) \rightarrow CO(g)}

Practice this conceptself-check · 3 quick reps

Formation enthalpy is zero only for the standard form of an element

\Delta H_f^\circ

is zero for

\mathrm{O_2(g)}

or graphite, but not for ozone or diamond, which are not the standard states. And it is products minus reactants, each multiplied by its coefficient.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Energy profiles and the activation energy
Energy profile relations
$\Delta H = H_{\text{products}} - H_{\text{reactants}} \qquad E_{a,\text{reverse}} = E_{a,\text{forward}} - \Delta H$
Estimating ΔH from bond enthalpies
ΔH from bond enthalpies
$\Delta H \approx \sum E(\text{bonds broken}) - \sum E(\text{bonds formed})$
Hess's law and enthalpies of formation
ΔH from enthalpies of formation
$\Delta H = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$

Reference tables (1)

Exothermic and endothermic reactions and the sign of ΔH5 rows

Feature	Exothermic	Endothermic
Heat flow	Released to the surroundings	Absorbed from the surroundings
Sign of ΔH	Negative	Positive
Energy of products	Lower than the reactants	Higher than the reactants
Temperature of surroundings	Rises	Falls
Examples	Combustion, respiration, neutralisation, freezing, condensation	Photosynthesis, thermal decomposition of $\mathrm{CaCO_3}$ , melting, dissolving ammonium nitrate

Watch out for (5)

Exothermic does not mean fast
→ Exothermic and endothermic reactions and the sign of ΔH
Breaking bonds never releases energy
→ Exothermic and endothermic reactions and the sign of ΔH
Activation energy is not the energy difference between products and reactants
→ Energy profiles and the activation energy
Bonds broken minus bonds formed, not the other way round
→ Estimating ΔH from bond enthalpies
Formation enthalpy is zero only for the standard form of an element
→ Hess's law and enthalpies of formation

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