Energetics

Chemistry 1110 · CCEA A Level · 10 min read · Updated: 31 Aug 2026

Welcome to Energetics!

Have you ever wondered why lighting a match gives off a burst of heat, or why an instant cold pack gets icy cold when you snap it? The answer lies in Energetics (also known as Thermochemistry). In this chapter, we will explore how chemical reactions exchange energy with their surroundings, how we measure these changes in the lab, and how we can calculate energy values that are impossible to measure directly.

Don't worry if calculations in chemistry sometimes feel intimidating. We will break down every formula step-by-step with clear examples, memory aids, and real-world analogies to make sure you feel confident for your AS exams!

1. Fundamentals of Enthalpy

What is Enthalpy?

In chemistry, every substance contains stored chemical energy within its bonds and intermolecular forces. We call this heat content Enthalpy, represented by the symbol \(H\).
We cannot directly measure the absolute enthalpy \(H\) of a single substance, but we can measure the change in enthalpy (\(\Delta H\)) when a reaction takes place.

The enthalpy change is the heat energy transferred between the system (the chemical reaction) and the surroundings (everything else, including the thermometer, water, and beaker) at constant pressure:

\(\Delta H = H_{\text{products}} - H_{\text{reactants}}\)

The standard unit for enthalpy change is \(\text{kJ mol}^{-1}\) (kilojoules per mole).

Exothermic vs Endothermic Reactions

Chemical reactions fall into two main categories depending on whether they release or absorb energy:

1. Exothermic Reactions:

• What happens: Heat energy is released from the system to the surroundings.
• Temperature change: The temperature of the surroundings increases.
• Sign of \(\Delta H\): Negative (\(\Delta H < 0\)) because the products have less energy than the reactants.
• Everyday examples: Combustion of fuels, hand warmers, respiration, neutralisation of acids and alkalis.
• Memory Aid: EXothermic = Heat EXits the system!

2. Endothermic Reactions:

• What happens: Heat energy is absorbed by the system from the surroundings.
• Temperature change: The temperature of the surroundings decreases (feels cold).
• Sign of \(\Delta H\): Positive (\(\Delta H > 0\)) because the products have more energy than the reactants.
• Everyday examples: Thermal decomposition (e.g., heating limestone, \(\text{CaCO}_3\)), instant sports injury ice packs, photosynthesis.
• Memory Aid: ENdothermic = Heat goes ENters (IN)!

Enthalpy Profile Diagrams

An enthalpy profile diagram shows the relative enthalpy levels of reactants and products over the course of a reaction:

• Exothermic Profile: Reactants start higher up; products sit lower down. The arrow for \(\Delta H\) points downwards from reactants to products (indicating a negative value).
• Endothermic Profile: Reactants start lower down; products sit higher up. The arrow for \(\Delta H\) points upwards from reactants to products (indicating a positive value).
• Activation Energy (\(E_a\)): Both profiles feature an energy "hump." The height from the reactant level to the peak of the hump represents the Activation Energy—the minimum energy required to start the reaction.

Key Takeaway: If heat is given out, \(\Delta H\) is negative. If heat is taken in, \(\Delta H\) is positive.

2. Standard Enthalpy Definitions

To compare reactions fairly, chemists measure enthalpy changes under standard conditions, shown by the standard symbol \(^\ominus\) (e.g., \(\Delta H^\ominus\)).

Standard Conditions

• Pressure: \(100\text{ kPa}\) (approximately \(1\text{ atm}\))
• Temperature: \(298\text{ K}\) (\(25^\circ\text{C}\))
• Concentration: \(1.0\text{ mol dm}^{-3}\) (for aqueous solutions)
• Standard state: The physical state (solid, liquid, or gas) of a substance under these standard conditions (e.g., \(\text{H}_2\text{O}\) is liquid, \(\text{O}_2\) is gas, \(\text{C}\) is solid graphite).

Definitions You Must Memorise

1. Standard Enthalpy of Formation (\(\Delta_f H^\ominus\))
"The enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions."
• Example: \(2\text{C(s, graphite)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{C}_2\text{H}_5\text{OH(l)}\)
• Crucial Rule: By definition, the \(\Delta_f H^\ominus\) of any element in its standard state is always zero (e.g., \(\Delta_f H^\ominus[\text{O}_2\text{(g)}] = 0\text{ kJ mol}^{-1}\)).

2. Standard Enthalpy of Combustion (\(\Delta_c H^\ominus\))
"The enthalpy change when one mole of a substance is burned completely in excess oxygen under standard conditions, with all reactants and products in their standard states."
• Example: \(\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}\)
• Note: Combustion reactions are always exothermic, so \(\Delta_c H^\ominus\) values are always negative!

3. Standard Enthalpy of Neutralisation (\(\Delta_{\text{neut}} H^\ominus\))
"The enthalpy change when an acid and an alkali react to form one mole of water under standard conditions."
• Example: \(\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}\) or simply \(\text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{H}_2\text{O(l)}\)
• Did you know? For any strong acid reacting with any strong base, \(\Delta_{\text{neut}} H^\ominus \approx -57\text{ kJ mol}^{-1}\). This is because the same fundamental reaction is happening every time: \(\text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{H}_2\text{O(l)}\).

Common Mistake to Avoid: When writing thermochemical equations for definitions, ALWAYS ensure the balancing leaves one mole of the specified substance (e.g., 1 mole of product for \(\Delta_f H^\ominus\), or 1 mole of fuel for \(\Delta_c H^\ominus\)), even if it means using fractions like \(\frac{1}{2}\) or \(\frac{7}{2}\) for other reactants.

3. Measuring Enthalpy: Calorimetry

We measure energy changes experimentally using an apparatus called a calorimeter.

The Heat Energy Equation

The heat energy transferred (\(q\)) to or from a solution is calculated using:

\(q = mc\Delta T\)

• \(q\) = heat energy transferred (in Joules, \(\text{J}\))
• \(m\) = mass of the solution being heated or cooled (in grams, \(\text{g}\)). For aqueous solutions, we assume a density of \(1.0\text{ g cm}^{-3}\), so \(1\text{ cm}^3 \approx 1\text{ g}\).
• \(c\) = specific heat capacity of water (\(4.18\text{ J g}^{-1}\text{ K}^{-1}\) or \(\text{J g}^{-1}\text{ }^\circ\text{C}^{-1}\))
• \(\Delta T\) = temperature change (\(T_{\text{final}} - T_{\text{initial}}\) in \(\text{K}\) or \(^\circ\text{C}\))

Step-by-Step: Converting \(q\) to \(\Delta H\)

Follow these 3 simple steps for any calorimetry exam question:

Step 1: Calculate the heat transferred in Joules: \(q = mc\Delta T\).
Step 2: Convert Joules to Kilojoules: \(q\text{ (in kJ)} = \frac{q\text{ (in J)}}{1000}\).
Step 3: Find the moles (\(n\)) of the limiting reactant that reacted (\(n = \frac{m}{M_r}\) or \(n = c \times V\)).
Step 4: Calculate enthalpy change: \(\Delta H = -\frac{q\text{ (in kJ)}}{n}\).
Remember to add the correct sign! If temperature went UP, \(\Delta H\) is NEGATIVE. If temperature went DOWN, \(\Delta H\) is POSITIVE.

Worked Example: Spirit Burner Experiment

A student burns \(0.32\text{ g}\) of methanol (\(\text{CH}_3\text{OH}\), \(M_r = 32.0\)) in a spirit burner to heat \(100\text{ cm}^3\) of water. The temperature rises by \(15.0^\circ\text{C}\). Calculate the standard enthalpy of combustion of methanol.

1. Mass of water, \(m = 100\text{ g}\)
2. \(q = mc\Delta T = 100 \times 4.18 \times 15.0 = 6270\text{ J} = 6.27\text{ kJ}\)
3. Moles of methanol burned, \(n = \frac{0.32}{32.0} = 0.010\text{ mol}\)
4. \(\Delta_c H = -\frac{6.27\text{ kJ}}{0.010\text{ mol}} = -627\text{ kJ mol}^{-1}\)

Experimental Errors in Calorimetry

Experimental values for \(\Delta_c H^\ominus\) from simple calorimetry are almost always less negative (less exothermic) than data book values. Why?

• Heat loss to the surroundings and the calorimeter itself.
• Incomplete combustion of the fuel (seen as black soot on the copper can).
• Evaporation of fuel from the wick after weighing.
• Measurements taken under non-standard conditions.

How to improve accuracy:

• Use a polystyrene cup (an excellent insulator with low heat capacity).
• Put a lid on the cup to prevent heat loss through convection.
• Use cooling curves (graphical extrapolation): Record the temperature every minute for 3 minutes before adding the reactant, add it at minute 4 (do not measure), then record the temperature every minute from minute 5 to 10. Extrapolate the cooling curve back to minute 4 to find the true maximum temperature change that would have occurred without heat loss.

4. Hess's Law

What is Hess's Law?

"The total enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final conditions are the same."

Analogy: Imagine climbing from Base Camp to the peak of a mountain. Whether you climb straight up the cliff face (Route 1) or follow a winding scenic path (Route 2), your total change in elevation is identical!

Route 1: Using Enthalpies of Formation (\(\Delta_f H^\ominus\))

When you are given formation data, the elements in their standard states sit at the bottom of your cycle:

\(\Delta H_r^\ominus = \sum \Delta_f H^\ominus(\text{products}) - \sum \Delta_f H^\ominus(\text{reactants})\)

Quick Memory Aid: Formation = P - R (Products minus Reactants).

Route 2: Using Enthalpies of Combustion (\(\Delta_c H^\ominus\))

When you are given combustion data, the combustion products (\(\text{CO}_2\), \(\text{H}_2\text{O}\)) sit at the bottom of your cycle:

\(\Delta H_r^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products})\)

Quick Memory Aid: Combustion = R - P (Reactants minus Products).

Worked Example: Using Formation Data

Calculate the enthalpy change for the reaction:
\(\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}\)
Given:
\(\Delta_f H^\ominus[\text{CH}_4\text{(g)}] = -75\text{ kJ mol}^{-1}\)
\(\Delta_f H^\ominus[\text{CO}_2\text{(g)}] = -394\text{ kJ mol}^{-1}\)
\(\Delta_f H^\ominus[\text{H}_2\text{O(l)}] = -286\text{ kJ mol}^{-1}\)
\(\Delta_f H^\ominus[\text{O}_2\text{(g)}] = 0\text{ kJ mol}^{-1}\)

Calculation:
\(\sum \Delta_f H^\ominus(\text{products}) = [1 \times (-394)] + [2 \times (-286)] = -394 - 572 = -966\text{ kJ mol}^{-1}\)
\(\sum \Delta_f H^\ominus(\text{reactants}) = [1 \times (-75)] + [2 \times 0] = -75\text{ kJ mol}^{-1}\)
\(\Delta H_r^\ominus = \text{Products} - \text{Reactants} = (-966) - (-75) = -891\text{ kJ mol}^{-1}\)

5. Bond Enthalpies

What is Mean Bond Enthalpy?

Mean Bond Enthalpy is the enthalpy change required to break one mole of a specific covalent bond in a gaseous molecule, averaged over a wide range of different compounds.

• Breaking bonds: Requires energy \(\rightarrow\) Endothermic (\(\Delta H\) is positive).
• Making bonds: Releases energy \(\rightarrow\) Exothermic (\(\Delta H\) is negative).

Crucial Mnemonic: MEXO BENDO
• Making bonds = EXOthermic (\(-\))
• Breaking bonds = ENDOthermic (\(+\))

Calculating \(\Delta H\) from Bond Enthalpies

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

\(\Delta H = \text{Energy IN (breaking)} - \text{Energy OUT (making)}\)

Step-by-Step Worked Example

Calculate the enthalpy change for the combustion of methane:
\(\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)}\)
Given bond enthalpies:
\(\text{C}-\text{H} = 413\text{ kJ mol}^{-1}\)
\(\text{O}=\text{O} = 498\text{ kJ mol}^{-1}\)
\(\text{C}=\text{O} = 805\text{ kJ mol}^{-1}\)
\(\text{O}-\text{H} = 464\text{ kJ mol}^{-1}\)

Step 1: Draw out all the bonds so you don't miss any!
• Reactants: \(4 \times (\text{C}-\text{H})\) bonds and \(2 \times (\text{O}=\text{O})\) bonds.
• Products: \(2 \times (\text{C}=\text{O})\) bonds and \(4 \times (\text{O}-\text{H})\) bonds (since each \(\text{H}_2\text{O}\) has 2 \(\text{O}-\text{H}\) bonds, and there are 2 \(\text{H}_2\text{O}\) molecules).

Step 2: Calculate bonds broken (Reactants):
\(\text{Bonds broken} = (4 \times 413) + (2 \times 498) = 1652 + 996 = +2648\text{ kJ mol}^{-1}\)

Step 3: Calculate bonds formed (Products):
\(\text{Bonds formed} = (2 \times 805) + (4 \times 464) = 1610 + 1856 = 3466\text{ kJ mol}^{-1}\)

Step 4: Calculate \(\Delta H\):
\(\Delta H = 2648 - 3466 = -818\text{ kJ mol}^{-1}\)

Why do Bond Enthalpy Calculations Differ from Experimental Values?

Students are often asked why values calculated using bond enthalpies do not match exact experimental or Hess's law values. The two key reasons are:

1. Averaged values: Mean bond enthalpies are averaged across many different chemical environments (for example, the \(\text{C}-\text{H}\) bond in methane is slightly different from the \(\text{C}-\text{H}\) bond in ethanol).
2. Physical state: Bond enthalpies apply strictly to substances in the gaseous state. In real experiments, reactants or products may be liquids or solids, which involves additional enthalpy changes for changes of state (such as the enthalpy of vaporisation).

Summary & Revision Checklist

• Exothermic: Heat released, temperature rises, \(\Delta H < 0\).
• Endothermic: Heat absorbed, temperature falls, \(\Delta H > 0\).
• Calorimetry: \(q = mc\Delta T\); \(\Delta H = -\frac{q}{n}\) (watch units: J to kJ!).
• Hess's Law: Independent of route.
• Formation Data: \(\Delta H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants})\)
• Combustion Data: \(\Delta H = \sum \Delta_c H(\text{reactants}) - \sum \Delta_c H(\text{products})\)
• Bond Enthalpies: \(\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds formed})\) (BENDO - MEXO).

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is enthalpy change (\(\Delta H\)) and what is its standard unit?

\(\Delta H\) is the heat energy transferred between the system and surroundings at constant pressure (\(\Delta H = H_{\text{products}} - H_{\text{reactants}}\)), measured in \(\text{kJ mol}^{-1}\).

How do exothermic and endothermic reactions differ in terms of heat transfer and sign of \(\Delta H\)?

Exothermic reactions release heat to the surroundings and have a negative \(\Delta H\) (\(\Delta H < 0\)); endothermic reactions absorb heat and have a positive \(\Delta H\) (\(\Delta H > 0\)).

What are the key standard conditions used for measuring standard enthalpy changes (\(\Delta H^\ominus\))?

Standard conditions are a pressure of \(100\text{ kPa}\), a temperature of \(298\text{ K}\) (\(25^\circ\text{C}\)), and solutions at a concentration of \(1.0\text{ mol dm}^{-3}\).

What is the standard enthalpy of formation (\(\Delta_f H^\ominus\)) and what is its value for an element in its standard state?

It is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions; for any element in its standard state, \(\Delta_f H^\ominus = 0\text{ kJ mol}^{-1}\).

What is the standard enthalpy of combustion (\(\Delta_c H^\ominus\))?

It is the enthalpy change when one mole of a substance is burned completely in excess oxygen under standard conditions, with all reactants and products in standard states.

How is heat energy transferred (\(q\)) calculated in a calorimetry experiment?

Using \(q = mc\Delta T\), where \(m\) is mass in grams, \(c\) is specific heat capacity (\(4.18\text{ J g}^{-1}\text{ K}^{-1}\)), and \(\Delta T\) is temperature change.

What are the main experimental errors that cause simple calorimetry values to be less exothermic than theoretical values?

Heat loss to the surroundings, incomplete combustion of the fuel, fuel evaporation from the wick, and non-standard laboratory conditions.

What does Hess's Law state?

The total enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final conditions are the same.

How is \(\Delta H_r^\ominus\) calculated using standard enthalpies of formation versus standard enthalpies of combustion?

For formation data: \(\Delta H_r^\ominus = \sum \Delta_f H^\ominus(\text{products}) - \sum \Delta_f H^\ominus(\text{reactants})\); for combustion data: \(\Delta H_r^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products})\).

What is mean bond enthalpy, and is bond breaking exothermic or endothermic?

Mean bond enthalpy is the enthalpy required to break one mole of a covalent bond in a gaseous molecule averaged over various compounds; bond breaking is endothermic (\(\Delta H > 0\)).

Why do enthalpy changes calculated from mean bond enthalpies often differ from experimental values?

Mean bond enthalpies are averaged across many different compounds, and they apply strictly to gaseous states rather than liquids or solids.

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