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Enthalpy Changes and Energy Profiles

Learn and apply Enthalpy Changes and Energy Profiles in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Enthalpy Changes and Energy Profiles: Orientation

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.

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 H⦵

A standard enthalpy change, Δ H⦵, 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).

Detailed Explanations

A. Reading an energy profile (workflow)

  1. Identify the reactants energy level, products energy level, and the peak.
  2. Work out Δ H using products − reactants (sign matters).
  3. Work out Eₐ(forward) using peak − reactants.
  4. 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 = H_products - H_reactants.)

Worked Examples

Modelled example 1

Compare exothermic and endothermic profiles

Core

Problem

State two differences between an exothermic and an endothermic energy profile.
Study the worked solution

  1. Compare endpoints

    Method

    Compare product and reactant enthalpy levels.

    Reason

    Δ H = H_products-H_reactants.

    Working

    Exothermic products lie lower; endothermic products lie higher.

  2. Compare signs

    Method

    Assign the sign from the endpoint difference.

    Reason

    Lower products give a negative change; higher products give a positive change.

    Working

    Exothermic: Δ H < 0; endothermic: Δ H > 0.

Guided practice 2

Read three quantities from an energy profile

About 7 min

Problem

Reactants are at 25, products at -60, and the peak at 140 kJ mol⁻¹. Calculate Δ H, Eₐ(forward), and Eₐ(reverse).

Try this before viewing the solution

Hints

Hint 1: enthalpy change
Use products minus reactants.
Hint 2: activation energies
Subtract the relevant starting level from the same peak for each direction.
View solution step by step

  1. Calculate enthalpy change

    Method

    Subtract reactant energy from product energy.

    Reason

    The sign records the endpoint direction.

    Working

    Δ H = -60-25 = -85 kJ mol⁻¹.

  2. Calculate forward barrier

    Method

    Subtract reactants from the peak.

    Reason

    The forward path begins at the reactant level.

    Working

    E_(a,f) = 140-25 = 115 kJ mol⁻¹.

  3. Calculate reverse barrier

    Method

    Subtract products from the peak.

    Reason

    The reverse path begins at the product level.

    Working

    E_(a,r) = 140-(-60) = 200 kJ mol⁻¹.

Common misconception 3

Correct a catalyst profile

Find and correct the mistake

Learner diagram

A learner draws a catalysed profile with a lower peak and a lower product endpoint, claiming that both Eₐ and Δ H decrease. Diagnose the diagram.

Choose what changes

A catalyst changes

View solution step by step

  1. Lower the pathway peak

    Method

    Keep the alternative catalysed route below the uncatalysed peak.

    Reason

    A catalyst provides a pathway with lower activation energy.

    Working

    Eₐ decreases.

  2. Preserve endpoints

    Method

    Place reactants and products at their original levels.

    Reason

    Δ H is a state-function difference independent of pathway.

    Working

    Δ H remains unchanged.

Examiner practice 4

Define a standard enthalpy change

4 marks

Problem

Explain what is meant by a standard enthalpy change and state what a negative value shows. [4 marks]

Try this before viewing the solution

View solution step by step

  1. State standard conditions

    1 mark

    Method

    Give the specified temperature and pressure convention.

    Reason

    Standard enthalpy values require comparable reference conditions.

    Working

    Typically 298 K and 100 kPa.

  2. State standard states

    1 mark

    Method

    Require each substance in its standard state.

    Reason

    Physical form affects enthalpy.

    Working

    Standard states under the stated conditions.

  3. Treat solutions

    1 mark

    Method

    State 1.00 mol dm⁻³ where solution concentration is relevant.

    Reason

    This is the reviewed solution convention.

    Working

    c = 1.00 mol dm⁻³.

  4. Interpret the sign

    1 mark

    Method

    State that negative Δ H⦵ means heat is released at constant pressure.

    Reason

    Products have lower enthalpy than reactants.

    Working

    Negative value = exothermic change.

Challenge 5

Reverse an energy profile

Minimal support

Problem

A forward reaction has Δ H = -48 kJ mol⁻¹ and E_(a,f) = 72 kJ mol⁻¹. Determine Δ H and Eₐ for the reverse reaction, and explain the profile geometry.

Try this before viewing the solution

Hints

Hint 1: reverse the endpoint change
Reversing swaps reactant and product levels, so the enthalpy-change sign reverses.
Hint 2: measure from lower products
For the forward exothermic profile, the reverse route starts 48 kJ mol⁻¹ below the original reactants.
View solution step by step

  1. Reverse ΔH

    Method

    Change the sign when swapping reaction direction.

    Reason

    Products-minus-reactants becomes the negative of the original difference.

    Working

    Δ Hᵣₑᵥₑᵣₛₑ = +48 kJ mol⁻¹.

  2. Find reverse activation energy

    Method

    Add the 48 kJ mol⁻¹ endpoint gap to the forward barrier.

    Reason

    The reverse path begins at the lower original-product level but reaches the same peak.

    Working

    E_(a,r) = 72-(-48) = 120 kJ mol⁻¹.

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

E_(a,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.