MODULE A12

Energetics

Exothermic, endothermic, bond energies, Hess's Law, calorimetry — the energy story behind every chemical reaction! 🔥❄️

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1. Exothermic & Endothermic Reactions

TypeEnergy TransferΔHTemperature changeExamples
ExothermicHeat released to surroundingsNegative (−)Surroundings get HOTTERCombustion, neutralisation, respiration, rusting
EndothermicHeat absorbed from surroundingsPositive (+)Surroundings get COLDERPhotosynthesis, dissolving NH₄NO₃, thermal decomposition, electrolysis
🔥 ΔH (enthalpy change) = heat energy change at constant pressure.
ΔH negative → exothermic → energy released → products are at LOWER energy than reactants.
ΔH positive → endothermic → energy absorbed → products are at HIGHER energy than reactants.

2. Bond Breaking and Bond Making

All chemical reactions involve breaking bonds in reactants and forming new bonds in products.

ProcessEnergyExplanation
Bond breakingENDOTHERMIC — requires energy input (+)You must put energy in to pull bonded atoms apart
Bond makingEXOTHERMIC — releases energy (−)Energy is released when new bonds form
🔑 Overall ΔH from bond energies:
ΔH = Σ(bonds broken) − Σ(bonds formed)
= (energy to break ALL bonds in reactants) − (energy released making ALL bonds in products)
If ΔH is negative → more energy released than absorbed → exothermic
If ΔH is positive → more energy absorbed than released → endothermic
📝 Key Bond Energy Values (kJ/mol) — memorise these:
H−H: 436  |  C−H: 412  |  O=O: 496  |  C=O (CO₂): 743  |  O−H: 463  |  C−C: 347  |  C=C: 612  |  N≡N: 944  |  N−H: 391

3. Calculating ΔH from Bond Energies — Step by Step

Example: CH₄ + 2O₂ → CO₂ + 2H₂O

  1. Draw out structural formulas showing all bonds
  2. List all bonds BROKEN in reactants and their energies
  3. List all bonds FORMED in products and their energies
  4. Apply ΔH = Σ(broken) − Σ(formed)
CH₄ + 2O₂ → CO₂ + 2H₂O worked example:
Bonds broken: 4(C−H) + 2(O=O) = 4(412) + 2(496) = 1648 + 992 = 2640 kJ
Bonds formed: 2(C=O in CO₂) + 4(O−H in 2H₂O) = 2(743) + 4(463) = 1486 + 1852 = 3338 kJ
ΔH = 2640 − 3338 = −698 kJ/mol (exothermic — combustion releases energy ✓)

4. Hess's Law

Hess's Law: The total enthalpy change for a reaction is independent of the route taken. It depends only on the initial and final states.

This means you can calculate ΔH for a reaction that is difficult to measure directly by using an energy cycle with known values.

Energy cycle rule: ΔH₁ = ΔH₂ + ΔH₃ (going via an alternative route must give the same total ΔH)
Enthalpy TermSymbolDefinition
Enthalpy of combustionΔHᶜHeat released when 1 mole of substance burns completely in excess oxygen under standard conditions
Enthalpy of neutralisationΔHₙHeat released when 1 mole of water is formed in a neutralisation reaction under standard conditions
Enthalpy of formationΔHᶠHeat change when 1 mole of compound is formed from its elements under standard conditions

5. Calorimetry — Q = mcΔT

Calorimetry measures the heat energy change of a reaction by monitoring the temperature change of a known mass of water (or solution).

Q = m × c × ΔT
Q = heat energy (J or kJ)  |  m = mass of solution (g)  |  c = specific heat capacity of water = 4.18 J g⁻¹ °C⁻¹  |  ΔT = temperature change (°C)

To find ΔH per mole: divide Q by the moles of reactant used.
Exothermic: temperature rises → Q is positive → ΔH is negative (−).
📝 Calorimetry Worked Example:
50 cm³ of 1 mol/dm³ HCl is neutralised by NaOH. Temperature rises from 22°C to 29°C.
m = 50 g (assume density = 1 g/cm³) | ΔT = 7°C | c = 4.18 J/g°C
Q = 50 × 4.18 × 7 = 1463 J = 1.463 kJ
n(HCl) = 0.05 mol | ΔH = −1.463 / 0.05 = −29.26 kJ/mol (negative = exothermic)

6. Energy Profile Diagrams

An energy profile diagram (reaction profile) shows how energy changes as reactants convert to products. The y-axis is potential energy (enthalpy); the x-axis is reaction progress.

⚡ Interactive Energy Profile Diagram

Toggle between exothermic and endothermic, show/hide the catalysed pathway, and watch the animated curve draw itself!

📈 Calorimetry Graph & Fuel ΔH Comparison

🃏 Flashcards — Energetics

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Answer
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❓ Quiz — Energetics

🔢 Worked Example — ΔH from Bond Energies

Problem: Calculate ΔH for the combustion of hydrogen: H₂ + ½O₂ → H₂O
Bond energies: H−H = 436 kJ/mol, O=O = 496 kJ/mol, O−H = 463 kJ/mol

Step 1: Energy to break bonds in reactants

Reactants: 1 mol H₂ (one H−H bond) + ½ mol O₂ (half an O=O bond).
Energy to break H−H = 436 kJ. Energy to break ½ O=O = ½ × 496 = 248 kJ.
Total energy to break bonds in reactants = 436 + 248 = ?

Step 2: Energy released forming bonds in products

Products: 1 mol H₂O with 2 O−H bonds formed.
Energy released making 2 O−H bonds = 2 × 463 = ?

Step 3: Apply ΔH = bonds broken − bonds formed

ΔH = 684 − 926 = ?

Step 4: Interpret — exothermic or endothermic?

ΔH = −242 kJ/mol. Is this exothermic or endothermic? Type one word.

🔗 Matching — Energetics Concepts

Term / Formula

Definition / Description

📝 CSEC-Style Questions

Q1. Distinguish between exothermic and endothermic reactions. Give ONE example of each. State the sign of ΔH in each case. [5 marks]+
Mark Scheme

1 Exothermic: heat energy is RELEASED to the surroundings ✓; temperature of surroundings increases ✓; ΔH is negative (−) ✓. Example: combustion, neutralisation, respiration ✓.

2 Endothermic: heat energy is ABSORBED from the surroundings ✓; temperature of surroundings decreases ✓; ΔH is positive (+) ✓. Example: photosynthesis, dissolving ammonium nitrate, thermal decomposition ✓.

Award max 5 marks total: 2 for each type (definition + example) + 1 for both ΔH signs.

Q2. Use the following bond energies to calculate ΔH for the reaction: N₂ + 3H₂ → 2NH₃. Bond energies: N≡N = 944 kJ/mol, H−H = 436 kJ/mol, N−H = 391 kJ/mol. [5 marks]+
Mark Scheme

1 Bonds broken: 1(N≡N) + 3(H−H) = 944 + 3(436) = 944 + 1308 = 2252 kJ ✓✓

2 Bonds formed: 2 × NH₃ has 2 × 3 N−H bonds = 6(N−H) = 6(391) = 2346 kJ ✓✓

3 ΔH = 2252 − 2346 = −94 kJ/mol ✓ (exothermic — this is the Haber Process!)

Q3. 100 cm³ of 0.5 mol/dm³ H₂SO₄ is neutralised by NaOH solution. The temperature rises from 21.0°C to 27.4°C. Calculate the enthalpy of neutralisation (specific heat capacity of solution = 4.18 J g⁻¹ °C⁻¹; density = 1 g/cm³). [5 marks]+
Mark Scheme

1 m = 100 g (density = 1 g/cm³) ✓ | ΔT = 27.4 − 21.0 = 6.4°C ✓

2 Q = m × c × ΔT = 100 × 4.18 × 6.4 = 2675.2 J = 2.675 kJ ✓

3 n(H₂SO₄) = c × V = 0.5 × 0.1 = 0.05 mol → produces 0.1 mol H₂O (H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O) ✓

4 ΔH = −2.675 / 0.1 = −26.75 kJ/mol ✓ (negative sign = exothermic)

Q4. Sketch an energy profile diagram for an exothermic reaction. Label: reactants, products, transition state, activation energy (Eₐ), ΔH, and the catalysed pathway. Explain how a catalyst affects Eₐ and ΔH. [6 marks]+
Mark Scheme

1 Correct shape: reactants at higher energy than products (exothermic) ✓; hump above reactants ✓

2 Reactants and products labelled at correct positions ✓

3 Eₐ = vertical arrow from reactants to peak ✓; ΔH = vertical arrow from reactants to products (downward, negative) ✓

4 Catalysed pathway: lower hump/peak; same reactant and product energy levels; labelled ✓

5 Catalyst LOWERS Eₐ ✓; ΔH is UNCHANGED (same energy difference between reactants and products) ✓

⭐ Key Formulas & Concepts

Exo vs Endo

Exothermic: ΔH < 0 (negative) Endothermic: ΔH > 0 (positive) Bond breaking = endothermic Bond making = exothermic

Bond Energy Formula

ΔH = Σ(bonds broken) − Σ(bonds formed) ΔH < 0 → exothermic ΔH > 0 → endothermic

Calorimetry

Q = m × c × ΔT c(water) = 4.18 J g⁻¹ °C⁻¹ ΔH = −Q / n (kJ/mol)

Hess's Law

ΔH independent of route ΔH₁ = ΔH₂ + ΔH₃ (energy cycle)

Energy Profile

Eₐ = reactants → peak ΔH = reactants → products Catalyst: lowers Eₐ only

Bond Energies

H−H: 436 | O=O: 496 C−H: 412 | O−H: 463 C=O: 743 | N≡N: 944

📚 Resources

📖 ChemGuide: Bond Energies 🔬 PhET: Reaction Energy ▶ YouTube: Bond Energies ▶ YouTube: Calorimetry ▶ YouTube: Energy Profiles
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