(I once asked a class, "Why does a fire feel hot?" Most students said, "Because it is burning." But that is not an explanation — it is a restatement. The real answer is that the chemical reaction of combustion releases energy. That energy is what we feel as heat. Understanding why reactions release or absorb energy is what this chapter is about. It is the difference between knowing that something happens and knowing why it happens. And that difference is the whole point of chemistry.) Every chemical reaction involves energy. Some reactions release energy — they feel hot. Think of a burning match, a car engine, or the metabolism of food in your body. These are called exothermic reactions. Other reactions absorb energy — they feel cold. Think of an instant ice pack, photosynthesis, or the decomposition of calcium carbonate. These are called endothermic reactions. Understanding energy changes is essential to chemistry, because it tells us which reactions will happen, how fast, and how much energy they will produce. It is also crucial for industry — chemical engineers need to know how much energy a reaction will release or absorb so they can design safe and efficient processes. In this chapter, we are going to look at energy changes in detail. We will learn about enthalpy, exothermic and endothermic reactions, energy level diagrams, bond energies, and Hess's law. By the end, you will understand the energy that drives every chemical reaction. Let us begin.
Enthalpy (H) is the total heat content of a system at constant pressure. We cannot measure the absolute enthalpy of a substance — only the change in enthalpy during a process. This change is called the enthalpy change (ΔH).
Sign convention:
The unit of enthalpy change is kilojoules per mole (kJ/mol).
(The sign convention is one of those things that trips students up. Here is a way to remember it: from the point of view of the chemicals, if they lose energy, they have less energy left, so ΔH is negative. If they gain energy, they have more, so ΔH is positive. The chemicals are the system; we are the surroundings.)
An exothermic reaction is one that releases energy to the surroundings, usually as heat. The temperature of the surroundings increases.
Characteristics:
Examples:
| Reaction | Explanation |
|---|---|
| Combustion of fuels | Burning methane, petrol, or wood releases a large amount of heat. |
| Respiration | The breakdown of glucose in cells releases energy for the body. |
| Neutralisation | Acid + base → salt + water releases heat. |
| Displacement reactions | Zinc + copper sulphate releases heat. |
| Corrosion of metals | Rusting of iron releases heat slowly. |
| Addition of water to concentrated acid | Dissolving concentrated sulphuric acid in water releases a lot of heat. |
An endothermic reaction is one that absorbs energy from the surroundings, usually as heat. The temperature of the surroundings decreases.
Characteristics:
Examples:
| Reaction | Explanation |
|---|---|
| Photosynthesis | Plants absorb light energy to convert CO₂ and water into glucose. |
| Thermal decomposition | Heating calcium carbonate to form calcium oxide and CO₂ absorbs heat. |
| Melting and boiling | Changes of state absorb energy. |
| Dissolving ammonium nitrate in water | The temperature drops — this is the principle of instant cold packs. |
| Electrolysis | Electrical energy is absorbed to break down compounds. |
An energy level diagram shows the enthalpy of the reactants and products, and the enthalpy change of the reaction.
The reactants are at a higher energy level than the products. The difference in energy is released as heat.
| Feature | Description |
|---|---|
| Reactants | Higher energy level |
| Products | Lower energy level |
| ΔH | Negative (energy is released) |
| Arrow | Points downwards from reactants to products |
The reactants are at a lower energy level than the products. The difference in energy is absorbed from the surroundings.
| Feature | Description |
|---|---|
| Reactants | Lower energy level |
| Products | Higher energy level |
| ΔH | Positive (energy is absorbed) |
| Arrow | Points upwards from reactants to products |
(If you remember only one thing about energy level diagrams, remember this: exothermic reactions go downhill (energy is released), and endothermic reactions go uphill (energy is absorbed). Once you see the diagram, the sign of ΔH is obvious.)
Every chemical reaction involves breaking bonds in the reactants and forming new bonds in the products.
The overall enthalpy change of a reaction depends on the balance between these two processes:
ΔH = Energy absorbed to break bonds − Energy released when new bonds form
Example: Calculate the enthalpy change for the reaction:
H₂ + Cl₂ → 2HCl
Bond energies (kJ/mol): H–H = 436, Cl–Cl = 243, H–Cl = 431.
Step 1 — Energy absorbed to break bonds:
H–H + Cl–Cl = 436 + 243 = 679 kJ/mol
Step 2 — Energy released when new bonds form:
2 × H–Cl = 2 × 431 = 862 kJ/mol
Step 3 — Calculate ΔH:
ΔH = 679 − 862 = −183 kJ/mol
Since ΔH is negative, the reaction is exothermic.
To compare enthalpy changes fairly, chemists define standard conditions: 298 K (25°C) and 1 atmosphere pressure. Under these conditions, we can define several standard enthalpy changes.
| Type | Symbol | Definition |
|---|---|---|
| Standard enthalpy of reaction | ΔH°r | The enthalpy change when molar quantities of reactants react completely under standard conditions. |
| Standard enthalpy of formation | ΔH°f | The enthalpy change when one mole of a compound is formed from its elements in their standard states. |
| Standard enthalpy of combustion | ΔH°c | The enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions. |
| Standard enthalpy of neutralisation | ΔH°n | The enthalpy change when one mole of water is formed from the reaction of an acid and a base under standard conditions. |
| Standard enthalpy of solution | ΔH°sol | The enthalpy change when one mole of a substance is dissolved in a large excess of water. |
Hess's law states that the total enthalpy change for a reaction is independent of the route taken, as long as the initial and final conditions are the same. In other words, the enthalpy change is the same whether the reaction happens in one step or several steps.
Hess's law is useful because it allows us to calculate enthalpy changes that are difficult to measure directly.
Example: Calculate the enthalpy of formation of carbon monoxide (CO), given:
Solution: We want the enthalpy change for:
C(s) + ½O₂(g) → CO(g)
Using Hess's law, we can reverse the second reaction and add it to the first:
C(s) + O₂(g) → CO₂(g) ΔH = −393 kJ/mol
CO₂(g) → CO(g) + ½O₂(g) ΔH = +283 kJ/mol
Adding these gives:
C(s) + ½O₂(g) → CO(g) ΔH = −393 + 283 = −110 kJ/mol
So the enthalpy of formation of CO is −110 kJ/mol.
Energy changes are not limited to chemical reactions. Physical processes also involve energy changes:
| Process | Energy Change | Explanation |
|---|---|---|
| Melting | Endothermic | Energy is needed to break the forces holding particles in the solid lattice. |
| Boiling | Endothermic | Energy is needed to overcome the forces between particles in the liquid. |
| Freezing | Exothermic | Energy is released as particles settle into a more ordered arrangement. |
| Condensation | Exothermic | Energy is released as gas particles come together to form a liquid. |
| Dissolving | Varies | Can be exothermic (e.g., NaOH) or endothermic (e.g., NH₄NO₃) depending on the substance. |
Understanding energy changes is important in many areas of everyday life and industry:
| Application | How It Uses Energy Changes |
|---|---|
| Fuels | The energy released by combustion is used for heating, cooking, and power generation. |
| Hand warmers | Use exothermic reactions (e.g., rusting of iron) to produce heat. |
| Instant cold packs | Use endothermic reactions (e.g., dissolving ammonium nitrate) to absorb heat. |
| Food and nutrition | The energy content of food is measured in calories or joules, based on the energy released during respiration. |
| Industrial processes | Chemical engineers must account for energy changes to design safe and efficient reactors. |
| Batteries | Store chemical energy and release it as electrical energy through exothermic reactions. |
(Energy changes are the hidden engine of the world. Every reaction, every process, every change in your body and in industry involves energy. Once you understand enthalpy, you understand the energy that drives the universe.)
Enthalpy changes can be measured experimentally using a calorimeter. The principle is simple: the heat released or absorbed by a reaction is transferred to a known mass of water, and the temperature change of the water is measured.
Formula: Q = m × c × ΔT
Where:
The enthalpy change per mole is then calculated: ΔH = −Q / n (where n is the number of moles).
Question: When ammonium nitrate is dissolved in water, the temperature of the solution decreases. Is the process exothermic or endothermic? Explain.
Solution: The process is endothermic. The temperature of the surroundings decreases, meaning heat is absorbed from the surroundings. The dissolution of ammonium nitrate absorbs energy, so ΔH is positive. This is why ammonium nitrate is used in instant cold packs.
Question: Calculate the enthalpy change for the reaction: CH₄ + 2O₂ → CO₂ + 2H₂O. Bond energies (kJ/mol): C–H = 413, O=O = 498, C=O = 745, O–H = 464.
Solution:
Bonds broken: 4 × C–H + 2 × O=O = 4(413) + 2(498) = 1652 + 996 = 2648 kJ/mol.
Bonds formed: 2 × C=O + 4 × O–H = 2(745) + 4(464) = 1490 + 1856 = 3346 kJ/mol.
ΔH = 2648 − 3346 = −698 kJ/mol. The reaction is exothermic.
Question: Given:
1. C(s) + O₂(g) → CO₂(g) ΔH = −393 kJ/mol
2. H₂(g) + ½O₂(g) → H₂O(l) ΔH = −286 kJ/mol
3. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = −890 kJ/mol
Calculate the enthalpy of formation of methane (CH₄).
Solution: We want: C(s) + 2H₂(g) → CH₄(g).
Reverse reaction 3: CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ΔH = +890 kJ/mol.
Add reaction 1 and 2 × reaction 2 to this:
C(s) + O₂(g) → CO₂(g) ΔH = −393
2H₂(g) + O₂(g) → 2H₂O(l) ΔH = 2(−286) = −572
CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ΔH = +890
Sum: C(s) + 2H₂(g) → CH₄(g) ΔH = −393 − 572 + 890 = −75 kJ/mol.
Question: When 0.05 moles of a fuel is burned, the heat released raises the temperature of 200 g of water by 30°C. Calculate the enthalpy of combustion of the fuel. (c = 4.2 J/g°C)
Solution:
Q = m × c × ΔT = 200 × 4.2 × 30 = 25,200 J = 25.2 kJ.
ΔH = −Q / n = −25.2 / 0.05 = −504 kJ/mol.
1. Define enthalpy change.
Enthalpy change (ΔH) is the heat energy transferred during a chemical reaction at constant pressure. It is negative for exothermic reactions and positive for endothermic reactions.
2. Distinguish between exothermic and endothermic reactions.
Exothermic reactions release energy to the surroundings (ΔH negative, temperature rises). Endothermic reactions absorb energy from the surroundings (ΔH positive, temperature falls).
3. Draw an energy level diagram for an exothermic reaction.
Reactants are at a higher energy level; products are at a lower energy level. An arrow points downwards from reactants to products, labelled ΔH negative.
4. Is bond breaking endothermic or exothermic?
Bond breaking is endothermic — energy is needed to break bonds.
5. Calculate the enthalpy change for a reaction where the energy absorbed to break bonds is 800 kJ/mol and the energy released when new bonds form is 600 kJ/mol.
ΔH = 800 − 600 = +200 kJ/mol. The reaction is endothermic.
6. State Hess's law.
Hess's law states that the total enthalpy change for a reaction is independent of the route taken, as long as the initial and final conditions are the same.
7. Why does dissolving ammonium nitrate in water produce a cold pack?
Dissolving ammonium nitrate is endothermic — it absorbs heat from the surroundings. This lowers the temperature, producing a cold pack.
Question 1 [10 marks]
(a) Distinguish between exothermic and endothermic reactions, giving one example of each. [4]
(b) Draw energy level diagrams for both types of reaction, showing the sign of ΔH. [4]
(c) Explain why bond breaking is endothermic and bond making is exothermic. [2]
(a) Exothermic: releases energy to the surroundings, ΔH negative, temperature rises. Example: combustion of methane. Endothermic: absorbs energy from the surroundings, ΔH positive, temperature falls. Example: photosynthesis. (4 marks: 2 for each type with example.)
(b) Exothermic: reactants at higher energy, products at lower energy, downward arrow, ΔH negative. Endothermic: reactants at lower energy, products at higher energy, upward arrow, ΔH positive. (4 marks: 2 for each diagram.)
(c) Bond breaking is endothermic because energy is needed to overcome the forces holding the atoms together. Bond making is exothermic because energy is released as new bonds form. (2 marks: 1 for each.)
Question 2 [10 marks]
(a) Define the standard enthalpy of combustion. [2]
(b) The combustion of 0.1 moles of ethanol releases enough heat to raise the temperature of 500 g of water by 20°C. Calculate the enthalpy of combustion of ethanol. (c = 4.2 J/g°C) [4]
(c) Explain why the experimental value for the enthalpy of combustion is often less than the standard value. [4]
(a) The standard enthalpy of combustion is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions (298 K, 1 atm). (2 marks: 1 for definition, 1 for standard conditions.)
(b) Q = m × c × ΔT = 500 × 4.2 × 20 = 42,000 J = 42 kJ. ΔH = −Q / n = −42 / 0.1 = −420 kJ/mol. (4 marks: 2 for Q calculation, 2 for ΔH.)
(c) The experimental value is often less than the standard value because some heat is lost to the surroundings, some heat is absorbed by the calorimeter, and the combustion may be incomplete. Standard values assume perfect insulation and complete combustion. (4 marks: 1 for heat loss, 1 for calorimeter absorption, 1 for incomplete combustion, 1 for "standard assumes ideal conditions".)
Question 3 [10 marks]
(a) State Hess's law. [2]
(b) Calculate the enthalpy of formation of carbon monoxide (CO) given:
C(s) + O₂(g) → CO₂(g) ΔH = −393 kJ/mol
CO(g) + ½O₂(g) → CO₂(g) ΔH = −283 kJ/mol [4]
(c) Explain why Hess's law is useful in chemistry. [4]
(a) Hess's law states that the total enthalpy change for a reaction is independent of the route taken, as long as the initial and final conditions are the same. (2 marks: 1 for "independent of route", 1 for "same initial and final conditions".)
(b) We want: C(s) + ½O₂(g) → CO(g). Reverse the second reaction: CO₂(g) → CO(g) + ½O₂(g) ΔH = +283 kJ/mol. Add to the first reaction: C(s) + O₂(g) → CO₂(g) ΔH = −393. Sum: C(s) + ½O₂(g) → CO(g) ΔH = −393 + 283 = −110 kJ/mol. (4 marks: 2 for reversal, 2 for calculation.)
(c) Hess's law is useful because it allows us to calculate enthalpy changes that are difficult or impossible to measure directly, such as the enthalpy of formation of compounds that cannot be synthesised directly from their elements, or reactions that are too slow or dangerous to carry out. It also helps us understand the energy relationships between different reactions. (4 marks: 2 for "difficult to measure directly", 2 for "understanding relationships".)
Challenge 1: Explain why a reaction can be spontaneous even if it is endothermic. Use an example in your answer.
A reaction can be spontaneous even if it is endothermic because the spontaneity of a reaction depends on both enthalpy and entropy (disorder). The second law of thermodynamics states that a reaction is spontaneous if the total entropy of the universe increases. Some endothermic reactions are spontaneous because they produce a large increase in entropy — for example, the dissolution of ammonium nitrate in water is endothermic but spontaneous because the ions become more disordered in solution. Another example is the evaporation of water, which is endothermic but spontaneous at room temperature because water vapour has much higher entropy than liquid water. So enthalpy is not the only factor — entropy matters too.
Challenge 2: Explain why the enthalpy of neutralisation for a strong acid and a strong base is always approximately −57 kJ/mol, regardless of which acid and base are used.
The enthalpy of neutralisation for a strong acid and a strong base is always approximately −57 kJ/mol because the reaction is essentially the same in all cases: H⁺(aq) + OH⁻(aq) → H₂O(l). Strong acids and strong bases completely dissociate in water, so the only reaction that occurs is the combination of hydrogen ions and hydroxide ions to form water. Since this is the same reaction every time, the enthalpy change is the same. The identity of the spectator ions (e.g., Na⁺, Cl⁻, K⁺, NO₃⁻) does not affect the enthalpy change. This is why the value is constant. For weak acids or bases, the value is slightly different because some energy is used to ionise the weak acid or base.
Challenge 3: Explain the difference between the standard enthalpy of formation and the standard enthalpy of combustion, using an example of each.
The standard enthalpy of formation (ΔH°f) is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions. For example, the enthalpy of formation of water is the enthalpy change for H₂(g) + ½O₂(g) → H₂O(l) = −286 kJ/mol. The standard enthalpy of combustion (ΔH°c) is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions. For example, the enthalpy of combustion of methane is the enthalpy change for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) = −890 kJ/mol. The key difference is that formation starts from elements and forms a compound, while combustion starts with a compound and burns it in oxygen. Both are exothermic for most compounds, but they describe different processes.
"Energy is the currency of the universe. Every reaction either spends it or earns it. Now you know how to read the balance sheet."
