Grade 11 · Chemistry · Lesson 5

Energy & Chemical Change

Understand exothermic and endothermic reactions, interpret energy diagrams, define enthalpy change, apply Hess’s Law, and relate bond energy to heat of reaction.

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Exothermic Reactions

In an exothermic reaction, energy is released to the surroundings. The products have less stored chemical energy (lower enthalpy) than the reactants. The surroundings therefore gain energy and become warmer.

Hproducts < Hreactants   →   ΔH < 0 (exothermic)

Endothermic Reactions

In an endothermic reaction, energy is absorbed from the surroundings. The products have more stored chemical energy than the reactants. The surroundings lose energy and become cooler.

Hproducts > Hreactants   →   ΔH > 0 (endothermic)

Enthalpy and ΔH

Enthalpy (H) is the total heat energy stored in a substance at constant pressure. We cannot measure absolute enthalpy, but we can measure the change in enthalpy:

ΔH = Hproducts − Hreactants

Enthalpy is measured in kJ·mol−1. The sign of ΔH tells us the direction of energy flow relative to the system.

Potential Energy Diagrams

A potential energy (PE) diagram (also called an energy profile) shows how the potential energy of the reacting system changes as the reaction proceeds. Key features:

FeatureExothermicEndothermic
Reactant vs product energyReactants higherProducts higher
ΔH valueNegative (< 0)Positive (> 0)
Energy gapReactants − productsProducts − reactants
Activation energy (Ea)Energy from reactants to peakEnergy from reactants to peak

The activated complex (transition state) sits at the peak of the energy profile. It is an unstable, high-energy arrangement of atoms that exists momentarily between reactants and products.

Activation energy (Ea): The minimum energy that colliding particles must have for a reaction to occur. It represents the energy needed to break existing bonds and initiate the formation of new ones. Ea is always positive.

Effect of a Catalyst on the PE Diagram

A catalyst provides an alternative reaction pathway with a lower activation energy. This means more particles have sufficient energy to react, so the reaction is faster.

On a PE diagram: catalyst lowers the peak (activated complex) but reactant and product baselines stay fixed. The ΔH arrow is unchanged.

Standard Enthalpy of Formation (ΔH°f)

The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states at 298 K and 101.3 kPa.

Hess’s Law

Hess’s Law states that the total enthalpy change for a reaction is independent of the pathway — it depends only on the initial reactants and final products. This allows us to calculate ΔH for reactions that cannot be measured directly.

ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants)

Rules for manipulating thermochemical equations:

Bond Energy Approach

Energy is required to break bonds (endothermic process) and energy is released when bonds form (exothermic process). The heat of reaction can be estimated using average bond energies:

ΔH ≈ Σ(bond energies broken) − Σ(bond energies formed)

If more energy is released forming bonds than is needed to break bonds, ΔH < 0 (exothermic). Bond energy calculations give approximate answers because average values are used.

Calorimetry: Measuring Heat of Reaction

A calorimeter measures the heat released or absorbed by a chemical reaction by monitoring the temperature change of a known mass of water (or solution). The formula used is:

Q = mcΔT
SymbolQuantityUnits
QHeat energy transferredJ or kJ
mMass of solution (usually taken as mass of water)g
cSpecific heat capacity of water4.18 J·g−1·°C−1
ΔTChange in temperature (Tfinal − Tinitial)°C or K

To find ΔH per mole: calculate moles of limiting reactant, then ΔH = −Q / n (the negative sign because if the solution heats up, the reaction was exothermic).

⭐ IEB Extension — Born-Haber Cycles

The Born-Haber cycle is a Hess’s Law cycle used to calculate the lattice energy of ionic compounds. It connects several enthalpy changes:

  • Enthalpy of atomisation: forming gaseous atoms from the element
  • Ionisation energy: removing electrons from a gaseous atom to form cations
  • Electron affinity: energy change when an electron is added to a gaseous atom to form anions
  • Lattice energy: energy released when gaseous ions combine to form the ionic lattice
  • Enthalpy of formation: the overall enthalpy change forming the ionic compound from elements

By Hess’s Law: ΔH°f = sum of all steps in the cycle. Lattice energy is large and negative for ionic compounds, indicating strong electrostatic attraction between ions.

⭐ IEB Extension — Gibbs Free Energy

Entropy (S) is a measure of disorder or randomness in a system. Reactions tend toward higher entropy (ΔS > 0 means entropy increases).

Gibbs Free Energy (ΔG) combines enthalpy and entropy to predict spontaneity:

ΔG = ΔH − TΔS
  • If ΔG < 0: reaction is spontaneous (thermodynamically favourable)
  • If ΔG > 0: reaction is non-spontaneous
  • If ΔG = 0: system is at equilibrium

T is temperature in Kelvin. A reaction can be spontaneous at high temperatures if TΔS is large enough to overcome a positive ΔH.

Energy Diagram Builder

Diagram Controls
100 kJ
60 kJ
Calculated Values
ΔH
--kJ/mol
Type
--
Ea (fwd)
--kJ
Ea (cat)
--kJ
0/8
Quiz complete! Review the explanations to strengthen your understanding.
IEB Extension Questions
Answer all questions in your notebook. Show all working including substitution into formulae and units. Draw diagrams neatly and label fully.
Question 1 — Label the PE Diagram
Draw a potential energy diagram for an exothermic reaction with activation energy Ea = 80 kJ and ΔH = −50 kJ. Label the following on your diagram: reactants, products, activated complex (transition state), Ea (forward), Ea (reverse), ΔH, and the effect of adding a catalyst (show the new activated complex). Explain in one sentence why the catalyst does not change ΔH. (6 marks)
Question 2 — Hess’s Law Calculation
Given the following thermochemical equations:

(1) C(s) + O₂(g) → CO₂(g)    ΔH = −394 kJ·mol−1
(2) H₂(g) + ½O₂(g) → H₂O(l)    ΔH = −286 kJ·mol−1
(3) C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)    ΔH = −1367 kJ·mol−1

Calculate ΔH for the formation of ethanol: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l).
Show which equations you reverse or multiply and why. (5 marks)
Question 3 — Bond Energy Calculation
Use the average bond energies below to calculate ΔH for the reaction: H₂(g) + Cl₂(g) → 2HCl(g)

Bond energies: H–H = 436 kJ·mol−1; Cl–Cl = 242 kJ·mol−1; H–Cl = 432 kJ·mol−1

(a) Identify which bonds are broken and which are formed.
(b) Calculate the total energy required to break bonds.
(c) Calculate the total energy released when bonds form.
(d) Calculate ΔH. Is the reaction exothermic or endothermic? (5 marks)
Question 4 — Calorimetry Calculation
A student dissolves 4.00 g of NaOH(s) in 100 g of water in a polystyrene cup calorimeter. The temperature rises from 22.0°C to 36.4°C. (cwater = 4.18 J·g−1·°C−1; M(NaOH) = 40 g·mol−1)

(a) Calculate Q for the water (in kJ).
(b) Calculate the number of moles of NaOH dissolved.
(c) Calculate ΔH in kJ·mol−1 for dissolving NaOH. State whether the process is exothermic or endothermic.
(d) Identify one source of error in this experiment and explain how it would affect your result. (6 marks)
Question 5 — Catalyst Effect
The catalytic decomposition of hydrogen peroxide (H₂O₂) is described by: 2H₂O₂(l) → 2H₂O(l) + O₂(g)    ΔH = −196 kJ·mol−1

(a) Is this reaction exothermic or endothermic? Justify your answer.
(b) Draw and label two potential energy diagrams on the same axes: one without a catalyst and one with MnO₂ as catalyst. Indicate Ea and ΔH on both diagrams.
(c) Explain, using the concept of activation energy, why adding MnO₂ speeds up the reaction.
(d) After the reaction is complete, would you expect to find MnO₂ in the products? Explain. (6 marks)
Question 6 — Reading a Calorimetry Graph (Data Table)
A learner investigates the endothermic dissolving of ammonium nitrate (NH₄NO₃) by adding 5.60 g of the solid to 50 g of water at 22.0°C in a polystyrene cup. Because stirring and dissolving take time, the first reliable thermometer reading is only taken 1 minute after mixing. Temperature is then recorded every minute as the mixture slowly warms back toward room temperature:

Time after mixing (min)Temperature (°C)
114.0
214.6
315.2
415.8
516.4

(a) Plot temperature (y-axis) against time (x-axis) for t = 1 to 5 minutes, and draw the best-fit straight line through the points.
(b) Calculate the gradient of this line (in °C·min−1). What does this gradient represent physically?
(c) The mixture actually reached its lowest temperature at t = 0 (the instant of mixing), before any readings could be taken. Use the gradient from (b) to extrapolate the line back to t = 0 and determine this lowest (corrected) temperature.
(d) Hence calculate the corrected temperature change, ΔT, for the experiment (using the initial temperature of 22.0°C).
(e) Calculate Q, the heat absorbed by the water (M(NH₄NO₃) = 80 g·mol−1; cwater = 4.18 J·g−1·°C−1; assume the mass of solution ≈ mass of water = 50 g).
(f) Calculate the number of moles of NH₄NO₃ used, and hence the experimental ΔH (in kJ·mol−1) for the dissolving process. Explain why extrapolating back to t = 0 gives a more accurate ΔH than simply using the lowest measured temperature (14.0°C) directly. (8 marks)