Energy Change of Reactions (Higher Tier)

This lesson extends your understanding of energy changes by exploring how we quantify energy changes using $\Delta H$ΔH notation, how energy changes relate to bond breaking and bond forming at a deeper level, and how to interpret energy data from experiments. This content is targeted at Higher Tier students on the AQA GCSE Combined Science Trilogy specification (8464).

---

Enthalpy Change ($\Delta H$ΔH)

The enthalpy change ($\Delta H$ΔH) of a reaction is the overall energy change measured under standard conditions. It tells you how much energy is transferred to or from the surroundings per mole of reaction.

$\Delta H = H_{\text{products}} - H_{\text{reactants}}$ΔH=Hproducts​−Hreactants​

Value of $\Delta H$ΔH	Meaning
Negative ($\Delta H < 0$ΔH<0)	Exothermic — energy is released to the surroundings
Positive ($\Delta H > 0$ΔH>0)	Endothermic — energy is taken in from the surroundings

Units

$\Delta H$ΔH is measured in kilojoules per mole (kJ/mol).

For example:

• Combustion of methane: $\Delta H = -890 \text{ kJ/mol}$ΔH=−890 kJ/mol (exothermic)
• Thermal decomposition of calcium carbonate: $\Delta H = +178 \text{ kJ/mol}$ΔH=+178 kJ/mol (endothermic)

---

Energy Changes in Terms of Bond Breaking and Forming

At the molecular level, every reaction involves two energy processes:

1. Breaking bonds — requires energy input (endothermic step)
2. Forming bonds — releases energy (exothermic step)

The overall enthalpy change depends on which process transfers more energy:

graph TD
    A["REACTANTS"] -->|"Step 1: Break all bonds \n(energy IN)"| B["Individual Atoms"]
    B -->|"Step 2: Form new bonds \n(energy OUT)"| C["PRODUCTS"]
    D{"Compare energies"} --> E["If energy OUT > energy IN → Exothermic"]
    D --> F["If energy IN > energy OUT → Endothermic"]

$\Delta H = \sum \text{(bonds broken)} - \sum \text{(bonds formed)}$ΔH=∑(bonds broken)−∑(bonds formed)

Exam Tip: The sigma ($\sum$∑) notation simply means "the sum of." You add up all the energy values for bonds broken and subtract the sum of all energy values for bonds formed.

---

Worked Example: Interpreting $\Delta H$ΔH

Question: The enthalpy change for the reaction $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$N2​+3H2​→2NH3​ is $\Delta H = -92 \text{ kJ/mol}$ΔH=−92 kJ/mol. Explain what this tells you.

Answer:

Step 1: The negative sign means the reaction is exothermic.

Step 2: 92 kJ of energy is released to the surroundings for every mole of nitrogen that reacts.

Step 3: The energy released when N–H bonds form in ammonia is greater than the energy needed to break the N≡N and H–H bonds in the reactants.

Step 4: The surroundings would get hotter (temperature increase).

---

Energy Changes from Experimental Data

In the required practical, you measure the temperature change ($\Delta T$ΔT) of a solution. This can be used to estimate the energy change using:

$q = mc\Delta T$q=mcΔT

Where:

Symbol	Meaning	Units
$q$q	Energy transferred	Joules (J)
$m$m	Mass of solution	Grams (g)
$c$c	Specific heat capacity of water	4.18 J/g°C
$\Delta T$ΔT	Temperature change	°C

Worked Example: Calculating Energy Transfer

A student adds 50 cm³ of 1.0 mol/dm³ NaOH to 50 cm³ of 1.0 mol/dm³ HCl. The temperature rises by 6.8°C.

Step 1: Total volume of solution = $50 + 50 = 100 \text{ cm}^3$50+50=100 cm3

Assume density = 1 g/cm³, so mass = 100 g

Step 2: Calculate energy transferred:

$q = mc\Delta T = 100 \times 4.18 \times 6.8 = 2842 \text{ J} = 2.84 \text{ kJ}$q=mcΔT=100×4.18×6.8=2842 J=2.84 kJ

Step 3: The temperature rose, so this is an exothermic reaction.

Exam Tip (Higher): You may need to calculate the energy per mole. To do this, find the number of moles of the limiting reagent, then divide $q$q by the number of moles. $\Delta H = -q / n$ΔH=−q/n (negative for exothermic).

---

Assumptions and Limitations

When using $q = mc\Delta T$q=mcΔT:

Assumption	Reality
All heat goes into the solution	Some heat is lost to the surroundings and the cup
Specific heat capacity = 4.18 J/g°C	The solution is not pure water — it contains dissolved substances
Density of solution = 1 g/cm³	Concentrated solutions may have a higher density
No heat lost to evaporation	Some heat is always lost through evaporation

These assumptions mean the calculated $\Delta H$ΔH is only an estimate, not an exact value.

---

Comparing Theoretical and Experimental $\Delta H$ΔH Values

Source	Value for Neutralisation of HCl + NaOH
Theoretical (from bond energies / data book)	$\Delta H = -57.1 \text{ kJ/mol}$ΔH=−57.1 kJ/mol
Experimental (from calorimetry)	Typically $\Delta H \approx -50 \text{ to } -55 \text{ kJ/mol}$ΔH≈−50 to −55 kJ/mol

The experimental value is usually less negative (smaller magnitude) than the theoretical value because of heat loss to the surroundings.

---

Energy Level Diagrams with $\Delta H$ΔH

You can represent $\Delta H$ΔH on an energy level diagram:

graph TD
    subgraph "Exothermic Energy Level Diagram"
        A["Reactants: HCl + NaOH"] -->|"ΔH = -57.1 kJ/mol"| B["Products: NaCl + H₂O"]
    end
    subgraph "Endothermic Energy Level Diagram"
        C["Reactants: CaCO₃"] -->|"ΔH = +178 kJ/mol"| D["Products: CaO + CO₂"]
    end

---

Common Mistakes (Higher Tier)

Mistake	How to Avoid It
Forgetting the negative sign for exothermic $\Delta H$ΔH	Always check: temperature rise = exothermic = negative $\Delta H$ΔH
Using the wrong units for $q$q	$mc\Delta T$mcΔT gives Joules — divide by 1000 to convert to kJ
Not accounting for total mass of solution	Use the total mass of all solutions mixed, not just one
Confusing $q$q with $\Delta H$ΔH	$q$q is the actual energy transferred; $\Delta H$ΔH is per mole
Not explaining why experimental $\Delta H$ΔH differs from theoretical	Heat loss to surroundings is the main reason

---

Summary

• $\Delta H$ΔH is the enthalpy change — the overall energy change of a reaction, measured in kJ/mol.
• Negative $\Delta H$ΔH = exothermic; positive $\Delta H$ΔH = endothermic.
• $\Delta H = \sum \text{bonds broken} - \sum \text{bonds formed}$ΔH=∑bonds broken−∑bonds formed.
• Experimental energy changes can be calculated using $q = mc\Delta T$q=mcΔT.
• Experimental values are usually less than theoretical values due to heat loss.
• Always show working and include the correct sign for $\Delta H$ΔH in exam answers.