Enthalpy & Energy Profile Diagrams
In January 1986, NASA engineers reviewing pre-launch calorimetry data for the Space Shuttle Challenger measured the enthalpy released by the solid rocket booster's ammonium perchlorate propellant at −3·5 MJ kg⁻¹ under standard conditions — but at the launch-morning temperature of −8°C, the O-ring elastomer's stored elastic energy was outside its designed range. Understanding enthalpy and energy profile diagrams is understanding why the numbers on paper must match the conditions on the day.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
You've probably held a hand warmer on a cold day — it heats up without any battery or flame. And you've seen an instant cold pack snap cold the moment you crack it. Both are just chemicals reacting. Why does one reaction heat up and the other cool down?
Before we name anything — write down what you think the reaction must be doing with energy in each case. What is the system doing with energy when it gets hot? When it gets cold?
Key Facts
- The definition of enthalpy (H) and enthalpy change (ΔH)
- The signs of ΔH for exothermic and endothermic reactions
- The five labelled features of an energy profile diagram
Concepts
- Why exothermic reactions have ΔH < 0
- How activation energy (Ea) relates to the transition state
- Why reversing an equation flips the sign of ΔH
Skills
- Classify reactions as exo/endothermic from ΔH or a diagram
- Sketch and label a correct energy profile diagram
- Write and manipulate thermochemical equations
Enthalpy (H) is a measure of the total energy stored in the bonds of a chemical system at constant pressure — and ΔH tells us how much of that energy is transferred when a reaction occurs.
We can't measure the absolute enthalpy of any substance — but we can measure the change in enthalpy (ΔH) when a reaction occurs. ΔH is defined as the enthalpy of products minus the enthalpy of reactants:
ΔH = H(products) − H(reactants)
If products have less stored energy than reactants, the difference is released to the surroundings as heat — the reaction feels warm. If products have more stored energy, the reaction must absorb heat from the surroundings — it feels cold. Enthalpy is a state function, meaning ΔH depends only on the initial and final states, not the pathway taken. Standard enthalpy changes are measured at 25°C and 100 kPa (standard conditions), denoted ΔH°.
Enthalpy change (ΔH = Hproducts − Hreactants) is a state function measuring heat energy transferred at constant pressure; negative ΔH = exothermic (heat released to surroundings), positive ΔH = endothermic (heat absorbed). Always write the sign explicitly.
Pause — copy the highlighted definition into your book before moving on.
Mini-task: A reaction has ΔH = −286 kJ mol⁻¹. (a) Is this exothermic or endothermic? (b) What does the negative sign tell you about the enthalpy of the products compared to the reactants? Write 1–2 sentences.
We just saw that ΔH = H(products) − H(reactants) defines the direction of energy transfer. That raises a question: how do we classify reactions as releasing or absorbing heat? This card answers it → by the sign of ΔH and the relative enthalpy of products versus reactants.
Whether a reaction releases or absorbs heat depends entirely on whether the products sit at lower or higher enthalpy than the reactants.
In an exothermic reaction, energy is released to the surroundings — ΔH is negative because H(products) < H(reactants). The surroundings heat up; a thermometer in the solution rises. Examples include combustion of fuels, neutralisation of strong acids and bases, and cellular respiration.
In an endothermic reaction, energy is absorbed from the surroundings — ΔH is positive because H(products) > H(reactants). The surroundings cool down; a thermometer drops. Examples include dissolving ammonium nitrate, photosynthesis, and thermal decomposition.
| Feature | Exothermic | Endothermic |
|---|---|---|
| ΔH sign | Negative (< 0) | Positive (> 0) |
| Energy flow | System → surroundings | Surroundings → system |
| Surroundings temperature | Increases (warms up) | Decreases (cools down) |
| Product energy vs reactants | Lower | Higher |
| Common examples | Combustion, neutralisation, respiration | Dissolving NH₄NO₃, photosynthesis, thermal decomposition |
Exothermic reactions (ΔH < 0) release heat to surroundings — products sit at lower enthalpy than reactants. Endothermic reactions (ΔH > 0) absorb heat — products sit higher. Examples: combustion (exothermic); dissolving NH₄NO₃ (endothermic).
Add the highlighted point to your notes before the check below.
Explain it: In your own words, explain why the surroundings of an exothermic reaction heat up. Use the terms "enthalpy", "products", "reactants", and "energy flow" in your answer.
We just saw that exo/endothermic reactions are classified by the sign of ΔH. That raises a question: can we visualise the energy pathway of a reaction, including the barrier that must be overcome? This card answers it → with the energy profile diagram, mapping enthalpy versus reaction progress.
An energy profile diagram is a graph of enthalpy versus reaction progress — it makes the invisible energy landscape of a reaction visible at a glance.
The x-axis is "reaction coordinate" (progress of reaction, not time). The y-axis is enthalpy (kJ mol⁻¹). Reactants start at one enthalpy level; products end at another. Between them is a peak — the transition state — representing the activation energy (Ea), the minimum energy required to break bonds and start the reaction.
Five features to label on every diagram:
- Reactants (left side, labelled with formula or "Reactants")
- Products (right side, labelled)
- Transition state (peak — highest point on the curve)
- Ea — arrow from reactant level to the peak
- ΔH — arrow from reactant level to product level
An energy profile diagram plots enthalpy vs reaction coordinate; the peak is the transition state, and activation energy (Ea) is the arrow from reactant level to the peak — never from zero. Five required labels: Reactants, Products, Transition state, Ea arrow, ΔH arrow.
Pause — write the highlighted definition into your book.
Match it:
We just saw that energy profile diagrams visualise Ea and ΔH. That raises a question: how do we communicate ΔH values precisely in written equations? This card answers it → with thermochemical equations that include state symbols and a ΔH value tied to the exact equation as written.
A thermochemical equation is a balanced chemical equation that also states the enthalpy change for the reaction as written — change either the equation or the coefficients, and ΔH changes too.
Thermochemical equations must include:
- State symbols — (s), (l), (g), (aq) — for every species
- The ΔH value in kJ mol⁻¹ written after the equation
- An understanding that ΔH refers to the equation exactly as written
Example:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = −890 kJ mol⁻¹
This means 890 kJ is released per mole of methane burned. Two rules apply universally:
- Scale the equation → scale ΔH by the same factor (2 mol CH₄ burned: ΔH = −1780 kJ mol⁻¹)
- Reverse the equation → flip the sign of ΔH (endothermic decomposition: ΔH = +890 kJ mol⁻¹)
A thermochemical equation states ΔH for the reaction as written, with state symbols for every species. Scaling coefficients scales ΔH by the same factor; reversing the equation flips the sign of ΔH. ΔH belongs to the equation, not the substance.
Add the highlighted point to your notes before the check below.
Mini-task: The combustion of propane has ΔH = −2220 kJ mol⁻¹. Write the thermochemical equation for: (a) burning 0.5 mol of propane, and (b) the reverse reaction. Include state symbols in both.
Worked examples · reveal as you go
Thermochemical equations. The thermochemical equation for the combustion of propane is:
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l) ΔH = −2220 kJ mol⁻¹
(a) Is this reaction exothermic or endothermic? (b) How much energy is released when 2.00 mol of propane is burned? (c) Write the thermochemical equation for the reverse reaction.
ΔH = −2220 kJ mol⁻¹ → negative sign → exothermic. Energy is released to the surroundings.
Energy = 2.00 × 2220 = 4440 kJ released. ΔH is per mole of equation as written — scale proportionally.
3CO₂(g) + 4H₂O(l) → C₃H₈(g) + 5O₂(g) ΔH = +2220 kJ mol⁻¹
(a) Exothermic (ΔH < 0) | (b) 4440 kJ released | (c) ΔH = +2220 kJ mol⁻¹
Energy profile diagrams. A reaction has ΔH = +85 kJ mol⁻¹ and a forward activation energy of 120 kJ mol⁻¹.
(a) Is the reaction exo- or endothermic? (b) What is the activation energy for the reverse reaction? (c) Describe the key features of the energy profile diagram for this reaction.
ΔH = +85 kJ mol⁻¹ → positive → endothermic. Products are at higher enthalpy than reactants.
Ea(reverse) = Ea(forward) − ΔH = 120 − 85 = 35 kJ mol⁻¹
Reactants at baseline. Peak is 120 kJ mol⁻¹ above reactants (Ea = 120 kJ mol⁻¹, arrow up from reactant level to peak). Products are 85 kJ mol⁻¹ above reactants (ΔH = +85 kJ mol⁻¹, arrow pointing upward). Transition state at the peak. x-axis: "Reaction coordinate". y-axis: "Enthalpy (kJ mol⁻¹)".
(a) Endothermic | (b) Ea(rev) = 35 kJ mol⁻¹ | (c) Products sit 85 kJ mol⁻¹ above reactants; peak is 120 kJ mol⁻¹ above reactants
A 1 g sample of glucose is burned completely in excess oxygen. Before you calculate anything — predict whether the ΔH value will be positive or negative, and explain your reasoning.
How close was your prediction?
Good instinct — combustion is the classic exothermic reaction.
The key is the sign convention: heat released to surroundings = negative ΔH. Lock this in.
Formula Reference — This Lesson
Common errors · the 3 traps that cost marks
"The reaction gets hot/cold"
Students write "the exothermic reaction gets hot" or "the endothermic reaction gets cold."
Fix: The surroundings change temperature. The system (the chemicals reacting) loses or gains enthalpy. Precision in this language is an HSC marker requirement. Write: "the surroundings increase in temperature" not "the reaction gets hot."
"Activation energy is the height of the peak from the x-axis"
Students draw the Ea arrow starting from the x-axis (zero) rather than from the reactant enthalpy level.
Fix: Ea is the energy difference from the reactant level to the transition state only. The position of the reactant level on the y-axis is arbitrary — only the differences between levels matter.
"Reversing the equation just removes the negative sign"
Students think reversing an equation turns −890 kJ mol⁻¹ into 890 kJ mol⁻¹ (unsigned) rather than +890 kJ mol⁻¹.
Fix: Reversing flips the sign — so a negative becomes positive and vice versa. ΔH = −890 becomes ΔH = +890 kJ mol⁻¹ when the equation is reversed. You must write the + sign explicitly.
Quick-fire practice · 5 reps +2 XP per reveal
Classify each reaction as exothermic or endothermic and state the ΔH sign:
(a) Combustion of ethanol: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)
(b) Photosynthesis: 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)
(c) Dissolving NaOH(s) in water — the solution warms up
(d) Dissolving NH₄Cl(s) in water — the solution cools down
(b) Endothermic | ΔH > 0 | Surroundings (light) → system (energy stored in glucose)
(c) Exothermic | ΔH < 0 | System → surroundings (solution warms up)
(d) Endothermic | ΔH > 0 | Surroundings → system (solution cools down)
Student A says: "For a reaction with ΔH = −60 kJ mol⁻¹ and Ea = 100 kJ mol⁻¹, the activation energy for the reverse reaction is 100 − (−60) = 160 kJ mol⁻¹." Is this correct? What is the conceptual error if any?
Student C says: "I doubled the equation 2CH₄(g) + 4O₂(g) → 2CO₂(g) + 4H₂O(l) but kept ΔH = −890 kJ mol⁻¹ because the substances are the same." Identify the error and write the correct thermochemical equation.
Correction: 2CH₄(g) + 4O₂(g) → 2CO₂(g) + 4H₂O(l) ΔH = −1780 kJ mol⁻¹
Student B says: "I drew my energy profile diagram with the transition state peak touching the top of the y-axis to show it has maximum energy. The Ea arrow goes from the x-axis to the top of the peak." Identify the two errors.
Error 2: The Ea arrow must start from the reactant level, not the x-axis. Ea is the energy difference between reactants and the transition state.
The extension question: The combustion of glucose is C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l), ΔH = −2803 kJ mol⁻¹. What does this tell you about the enthalpy change for photosynthesis? Use Hess's Law reasoning.
Go back to your Think First response. Now that you've studied enthalpy and energy profile diagrams, return to the January 1986 NASA Challenger calorimetry scenario:
- The propellant's ΔH = −3·5 MJ kg⁻¹ is a state function — it doesn't change with temperature. What does change is the activation energy of the O-ring's elastic recovery. Sketch a labelled energy profile diagram showing how lowering temperature raises Ea for O-ring deformation.
- Is the propellant combustion exothermic or endothermic? Write the sign of ΔH and explain what the negative sign tells you about the relative energies of reactants and products.
- What is the difference between ΔH and Ea? Why can the same reaction have a fixed ΔH but a temperature-dependent rate?
Pick your answer, then rate your confidence — that tells the system what to drill next.
Wrong: An exothermic reaction has ΔH > 0 because it releases heat.
Right: Exothermic reactions have ΔH < 0 (negative) because the system loses energy to the surroundings. Endothermic reactions have ΔH > 0 (positive) because the system gains energy. The sign convention refers to the system, not the surroundings.
Wrong: "The reaction gets hot/cold."
Right: "The surroundings change temperature." The system (reaction mixture) loses or gains enthalpy — language precision is an HSC marker requirement.
Wrong: "Reversing the equation removes the negative sign."
Right: Reversing flips the sign — ΔH = −890 becomes ΔH = +890 kJ mol⁻¹. The + must be written explicitly.
Q1. Distinguish between exothermic and endothermic reactions in terms of (a) the sign of ΔH, (b) the direction of energy flow, and (c) how each would appear on an energy profile diagram. Use a specific example of each. 4 MARKS
Q2. The thermochemical equation for the combustion of ethane is:
2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l) ΔH = −3120 kJ mol⁻¹
(a) Calculate the enthalpy change when 0.500 mol of ethane is burned. (1 mark)
(b) Write the thermochemical equation for the reverse of this reaction. (1 mark)
(c) Explain why water being produced as liquid (l) rather than gas (g) affects the value of ΔH. (2 marks)
4 MARKS
Q3. Real-World Application: Instant cold packs used by sports trainers contain solid ammonium nitrate (NH₄NO₃) separated from water by a thin inner bag. When the bag is cracked, NH₄NO₃ dissolves in the water and the pack becomes very cold.
(a) Is the dissolution of ammonium nitrate exothermic or endothermic? Justify using the observation described. (2 marks)
(b) Draw a labelled energy profile diagram for the dissolution of NH₄NO₃. Include: reactants, products, transition state, Ea, and ΔH arrows. Show whether products sit above or below reactants. (3 marks)
5 MARKS
Show comprehensive answers ▼
Activity 1 — Classifying Reactions
Row 1 (ethanol combustion): Exothermic | ΔH < 0 (negative) | System → surroundings (heat released)
Row 2 (photosynthesis): Endothermic | ΔH > 0 (positive) | Surroundings (light) → system (energy stored in glucose)
Row 3 (NaOH dissolving): Exothermic | ΔH < 0 | System → surroundings (solution warms up)
Row 4 (NH₄Cl dissolving): Endothermic | ΔH > 0 | Surroundings → system (solution cools down)
Extension: Photosynthesis and respiration have equal and opposite ΔH values — they are the reverse of each other. Combustion of glucose (respiration) ΔH = −2803 kJ mol⁻¹; photosynthesis ΔH = +2803 kJ mol⁻¹. This is a direct consequence of Hess's Law (path independence of enthalpy), which you'll formalise in Lesson 8.
Activity 2 — Diagram Errors
1. Student A: The arithmetic is actually correct but the conceptual reasoning is confused. Ea(rev) = Ea(fwd) − ΔH = 95 − (−40) = 135 kJ mol⁻¹. For an exothermic reaction (ΔH = −40), products sit below reactants, so the reverse reaction must climb a larger hill: 95 + 40 = 135 kJ mol⁻¹.
2. Student B: (1) Peak height from x-axis is meaningless — only differences between levels matter. (2) Ea arrow must start from the reactant enthalpy level, not from zero or the x-axis.
3. Student C: Doubling the equation doubles ΔH. Correct thermochemical equation: 2CH₄(g) + 4O₂(g) → 2CO₂(g) + 4H₂O(l) ΔH = −1780 kJ mol⁻¹.
Multiple Choice
1. B — Endothermic: ΔH > 0 (positive), energy flows from surroundings into the system, surroundings cool down.
2. D — Ea(rev) = 95 + 40 = 135 kJ mol⁻¹. For the reverse reaction, you start from the products (40 kJ mol⁻¹ lower than reactants since ΔH = −40) and climb to the same peak — a larger gap.
3. C — Doubling the equation doubles ΔH: 2 × (−572) = −1144 kJ mol⁻¹.
4. A — Exothermic: products are below reactants. ΔH < 0 because H(products) < H(reactants).
5. B — The reverse reaction absorbs 890 kJ mol⁻¹. Reversing an equation flips the sign of ΔH: the decomposition of CO₂ and H₂O back to CH₄ and O₂ is endothermic (ΔH = +890 kJ mol⁻¹). Energy is not "released the same way" — the direction of flow reverses.
Short Answer Model Answers
Q1 (4 marks): Exothermic: ΔH < 0; energy flows from system to surroundings; on energy profile diagram, products sit below reactants, ΔH arrow points downward; example — combustion of methane, ΔH = −890 kJ mol⁻¹ [2 marks]. Endothermic: ΔH > 0; energy flows from surroundings to system; products sit above reactants, ΔH arrow points upward; example — dissolving ammonium nitrate, ΔH > 0 [2 marks].
Q2 (4 marks): (a) Equation shows 2 mol ethane releases 3120 kJ. For 0.500 mol: ΔH = (0.500/2) × 3120 = 780 kJ released [1]. (b) 4CO₂(g) + 6H₂O(l) → 2C₂H₆(g) + 7O₂(g), ΔH = +3120 kJ mol⁻¹ (sign flips) [1]. (c) Converting water from liquid to gas requires energy (latent heat of vaporisation = 44 kJ mol⁻¹ per mol H₂O) [1]. If water were produced as gas, less energy would be released to the surroundings — ΔH would be less negative. The liquid (l) state releases the full amount including the condensation energy [1].
Q3 (5 marks): (a) Endothermic [1]. Justification: the pack becomes cold, meaning heat is being transferred from the surroundings (including the injured area) into the system — the reaction absorbs heat rather than releasing it. This means ΔH > 0 [1]. (b) Diagram requirements: reactants (solid NH₄NO₃ + liquid H₂O) at a lower enthalpy level [1]; products (NH₄⁺(aq) + NO₃⁻(aq)) at a higher enthalpy level — ΔH arrow points upward [1]; transition state at the peak above products; Ea arrow from reactant level to peak [1]. Accept: x-axis labelled "Reaction coordinate", y-axis "Enthalpy (kJ mol⁻¹)".
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