H2 Chemistry (9476): Enthalpy Changes and Energy Profiles (A Level)
Key idea: Use energy profile diagrams to interpret ΔH and activation energy, and write correct exothermic/endothermic explanations.
By the end, you can
- Enthalpy Changes and Energy Profiles
This lesson sets up the “energy language” used everywhere else: Δ H signs, activation energy, catalysts, and how to read energy profile diagrams. If you can label a profile correctly, you can usually pick up most of the explanation marks in energetics questions.
Keep Hess Law and Cycles and the Energetics and Thermodynamics hub in view, because most questions mix definitions with cycle reasoning.
1. Definitions (Must Know)
A. Enthalpy change, Δ H
The enthalpy change, Δ H, is the heat energy change of a reaction at constant pressure.
- exothermic: Δ H < 0 (heat released)
- endothermic: Δ H > 0 (heat absorbed)
B. Standard enthalpy change, Δ Hominus
A standard enthalpy change, Δ Hominus, is measured under standard conditions (typically 298 K and 100 kPa; solutions at 1.00 mol dm⁻³ where relevant), with substances in their standard states.
C. Activation energy, Eₐ
The activation energy, Eₐ, is the minimum energy required for a reaction to occur (reach the transition state).
D. Energy profile diagram
An energy profile diagram shows energy (y-axis) against reaction progress (x-axis).
2. Key Ideas (What Earns Marks)
- For an energy profile:
- Δ H = Hₚᵣₒdᵤcₜₛ - Hᵣₑₐcₜₐₙₜₛ
- Eₐ(forward) = Hₚₑₐₖ - Hᵣₑₐcₜₐₙₜₛ
- Eₐ(reverse) = Hₚₑₐₖ - Hₚᵣₒdᵤcₜₛ
- Exothermic profile: products lower than reactants, so Δ H < 0.
- Endothermic profile: products higher than reactants, so Δ H > 0.
- A catalyst provides an alternative pathway with a lower Eₐ but does not change Δ H.
- Always include units for enthalpy/activation energy: kJ mol⁻¹.
- Δ H is the vertical difference between products and reactants.
- Eₐ (forward) is the vertical difference between reactants and the peak.
- Catalyst: lower peak, same start/end.
Quick visuals (schematic profiles):
Data table
| Exothermic | |
|---|---|
| Reaction progress | Enthalpy (kJ mol^-1) |
| 0 | 0 |
| 0.2 | 40 |
| 0.5 | 120 |
| 0.7 | 30 |
| 1 | -60 |
Data table
| Uncatalysed | |
|---|---|
| Reaction progress | Enthalpy (kJ mol^-1) |
| 0 | 0 |
| 0.2 | 40 |
| 0.5 | 120 |
| 0.7 | 30 |
| 1 | -60 |
| Catalysed | |
|---|---|
| Reaction progress | Enthalpy (kJ mol^-1) |
| 0 | 0 |
| 0.2 | 20 |
| 0.5 | 70 |
| 0.7 | 10 |
| 1 | -60 |
3. Detailed Explanations
A. Reading an energy profile (workflow)
- Identify the reactants energy level, products energy level, and the peak.
- Work out Δ H using products − reactants (sign matters).
- Work out Eₐ(forward) using peak − reactants.
- Work out Eₐ(reverse) using peak − products.
Mini example:
- Reactants: 0 kJ mol⁻¹
- Products: -50 kJ mol⁻¹
- Peak: 100 kJ mol⁻¹
So:
- Δ H = -50 - 0 = -50 kJ mol⁻¹ (exothermic)
- Eₐ(forward) = 100 - 0 = 100 kJ mol⁻¹
- Eₐ(reverse) = 100 - (-50) = 150 kJ mol⁻¹
B. Why a catalyst does not change Δ H
Because Δ H depends only on the energies of reactants and products (a state-function difference), changing the pathway cannot change Δ H.
Therefore a catalyst can lower Eₐ (lower peak) but must start/end at the same energy levels, so Δ H is unchanged.
C. Reverse activation energy shortcut
From the definitions above: Eₐ(reverse) = Eₐ(forward) - Δ H
(This works because Δ H = Hₚᵣₒdᵤcₜₛ - Hᵣₑₐcₜₐₙₜₛ.)
4. Common Mistakes
- Saying catalysts change Δ H (they don’t).
- Confusing endothermic/exothermic signs.
- Missing units (kJ mol⁻¹) or mixing up “per mole of reaction as written”.
- Labelling the profile with Δ H as a “peak height” (it is products − reactants).
When you can explain this confidently, use the Energetics Thermodynamics quiz and the Exam Skills hub to pressure-test exam wording.
5. Exam Tips
- State sign convention: exothermic Δ H<0, endothermic Δ H>0.
- Energy profile labels: activation energy is from reactants to peak; Δ H is products − reactants.
- Always include units: kJ mol⁻¹.
6. Worked Examples
Example 1Core
State two differences between an exothermic and an endothermic energy profile.
Show Answer
Mark scheme:
- Exothermic: products lower than reactants; Δ H negative.
- Endothermic: products higher than reactants; Δ H positive.
Example 2Core
An energy profile shows reactants at 25 kJ mol⁻¹, products at -60 kJ mol⁻¹, and the peak at 140 kJ mol⁻¹. Calculate Δ H, Eₐ(forward), and Eₐ(reverse).
Show Answer
Mark scheme:
- Δ H = -60 - 25 = -85 kJ mol⁻¹.
- Eₐ(forward) = 140 - 25 = 115 kJ mol⁻¹.
- Eₐ(reverse) = 140 - (-60) = 200 kJ mol⁻¹.
Example 3Core
Explain (using an energy profile diagram idea) what happens to Eₐ and Δ H when a catalyst is added.
Show Answer
Mark scheme:
- A catalyst provides an alternative pathway with a lower peak, so Eₐ decreases.
- The energies of reactants and products do not change, so Δ H stays the same.
7. Mind Stretchers
Mind stretcher 1Extension
A reaction has Eₐ(forward) = 75.0 kJ mol⁻¹ and Δ H = -40.0 kJ mol⁻¹. Calculate Eₐ(reverse).
Show Hint
Read the vertical energy differences; a catalyst changes the route and activation energy, not the reactant or product levels.
Show Answer
Mark scheme:
- Eₐ(reverse) = Eₐ(forward) - Δ H
- Eₐ(reverse) = 75.0 - (-40.0) = 115 kJ mol⁻¹
Mind stretcher 2: Recovering the reverse activation energyExtension
Question. A forward reaction has Eₐ = 92 kJ mol⁻¹ and Δ H = -37 kJ mol⁻¹. Determine the reverse activation energy and explain why a catalyst leaves Δ H unchanged.
Show Hint
For an exothermic forward reaction, the products lie 37 kJ mol⁻¹ below the reactants.
Show Answer
Eₐ,ᵣₑᵥₑᵣₛₑ = 92 + 37 = 129 kJ mol⁻¹. A catalyst lowers the maximum along an alternative pathway in both directions but does not alter the initial or final energy levels, so Δ H is unchanged.
8. Practice, Quiz and Next Step
Fixed task
A forward reaction has $E_a=92\ \text{kJ mol}^{-1}$ and $\Delta H=-37\ \text{kJ mol}^{-1}$. Determine the reverse activation energy and explain why a catalyst leaves $\Delta H$ unchanged.
Expected answer and marking feedback
$E_{a,\mathrm{reverse}}=92+37=129\ \text{kJ mol}^{-1}$. A catalyst lowers the maximum along an alternative pathway in both directions but does not alter the initial or final energy levels, so $\Delta H$ is unchanged.
Repair before continuing
Saying catalysts change $\Delta H$ (they don’t).
Open course practice Next registered lesson: Calorimetry Mc Delta T A Level