Module A12 · CSEC Chemistry
Energetics& Enthalpy

Exothermic and endothermic reactions, energy profiles, ΔH, calorimetry, heats of neutralisation and solution — the energy hidden inside every chemical reaction.

Section 1 — Exothermic & Endothermic Reactions

🔥 EXOTHERMIC
  • RELEASES energy to surroundings
  • Surroundings get HOTTER
  • ΔH is NEGATIVE (−ve)
  • Energy released forming bonds > absorbed breaking bonds
  • H(products) < H(reactants)
❄️ ENDOTHERMIC
  • ABSORBS energy from surroundings
  • Surroundings get COLDER
  • ΔH is POSITIVE (+ve)
  • Energy absorbed breaking bonds > released forming bonds
  • H(products) > H(reactants)
🧠 Memory Trick EXO = EXIT — energy EXITS to the surroundings → temperature rises → ΔH negative.
ENDO = ENTER — energy ENTERS the reaction from surroundings → temperature falls → ΔH positive.

Exothermic Examples

  • Burning fuels (combustion)
  • Neutralisation (acid + alkali)
  • Metals reacting with acids (Mg + HCl)
  • Cellular respiration (glucose + O₂)
  • Dissolving NaOH or H₂SO₄ in water

Endothermic Examples

  • Thermal decomposition (CaCO₃ → CaO + CO₂)
  • Photosynthesis (CO₂ + H₂O → glucose)
  • Dissolving NH₄Cl or KNO₃ in water
  • Evaporation of a liquid
🌍 Real WorldInstant cold packs use endothermic dissolution of NH₄NO₃ in water — the pack feels icy cold. Hand warmers use exothermic iron oxidation to release warmth. Refrigeration exploits endothermic evaporation to extract heat from your food!

Section 2 — Bond Breaking & Bond Forming

Every reaction involves breaking bonds in reactants (energy absorbed) and forming bonds in products (energy released).

Key Principle Breaking bonds: ALWAYS absorbs energy (endothermic step).
Forming bonds: ALWAYS releases energy (exothermic step).

Exothermic overall → energy released forming bonds > energy absorbed breaking bonds.
Endothermic overall → energy absorbed breaking bonds > energy released forming bonds.

Bond Energy Calculation

ΔH = Σ(bonds broken) − Σ(bonds formed)energies in kJ mol⁻¹ · positive result → endothermic · negative result → exothermic
Example: H₂(g) + Cl₂(g) → 2HCl(g) Bonds broken: H–H = 436 kJ/mol Cl–Cl = 242 kJ/mol Total absorbed = 678 kJ/mol Bonds formed: 2 × H–Cl = 2 × 431 = 862 kJ/mol ΔH = 678 − 862 = −184 kJ/mol NEGATIVE → EXOTHERMIC ✅

Section 3 — Enthalpy Change (ΔH)

Enthalpy (H) is the energy stored in a substance. We measure the enthalpy change (ΔH) — the difference in energy between products and reactants. Units: kJ mol⁻¹.

Formula: ΔH = H(products) − H(reactants) If ΔH is negative → exothermic (products are lower in energy).
If ΔH is positive → endothermic (products are higher in energy).

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

ΔH = −891 kJ mol⁻¹

891 kJ released when 1 mol CH₄ burns. Negative → exothermic.

H₂(g) + I₂(s) → 2HI(g)

ΔH = +26.5 kJ mol⁻¹

26.5 kJ absorbed per mole HI. For 2 mol HI (as equation): 53 kJ absorbed. Positive → endothermic.

💡 Reversible ReactionsIf the forward reaction is exothermic (ΔH −ve), the reverse is endothermic (+ve) with the same magnitude. e.g. N₂ + 3H₂ ⇌ 2NH₃   ΔH = −46.1 kJ mol⁻¹ forward.

Section 4 — Energy Profile Diagrams

An energy profile diagram shows the energy of reactants and products throughout the reaction, including the activation energy (Ea) — the energy barrier that must be overcome.

Exothermic Profile

Progress of reaction → Energy Reactants Products Ea ΔH (−ve)

Endothermic Profile

Progress of reaction → Energy Reactants Products Ea ΔH (+ve)

What MUST Appear on a Fully Labelled Energy Profile

  • Formulae of reactants and products labelled on the curve
  • Activation energy (Ea) arrow — from reactant level to the peak
  • ΔH arrow — from reactant level to product level, with numerical value and sign
  • Axis labels: "Energy content" (y-axis) and "Progress/Course of reaction" (x-axis)
💡 Effect of a Catalyst on the Energy Profile A catalyst lowers the PEAK (reduces activation energy). The reactant and product energy levels are unchanged. ΔH is unchanged. The reaction is faster because more particles now have sufficient energy to react.

Section 5 — Calorimetry: Measuring Energy Changes

In the lab, we measure energy changes by tracking temperature change in a solution using an insulated polystyrene cup (calorimeter).

q = m × c × ΔTq = heat (J) · m = mass (g) · c = 4.2 J g⁻¹ °C⁻¹ · ΔT = T₂ − T₁ (°C)
Three Assumptions 1. Density of dilute aqueous solution = 1 g cm⁻³ (mass = volume in cm³)
2. Specific heat capacity of dilute solution = 4.2 J g⁻¹ °C⁻¹ (same as water)
3. Negligible heat loss to surroundings during the reaction

Section 6 — Heat of Neutralisation

DefinitionThe heat of neutralisation is the heat change when 1 mole of water is produced in a neutralisation reaction between an acid and an alkali.

For any strong acid + strong alkali, ΔH ≈ −56.3 kJ mol⁻¹ — always the same because the ionic equation is always: H⁺(aq) + OH⁻(aq) → H₂O(l)

Worked Example — Heat of Neutralisation

Given: 50 cm³ NaOH (1.0 mol/dm³, T₁ = 26.0°C) + 50 cm³ HCl (1.0 mol/dm³, T₁ = 27.0°C) Maximum T₂ = 33.2°C Step 1: Average initial T = (26.0 + 27.0) / 2 = 26.5°C Step 2: ΔT = 33.2 − 26.5 = 6.7°C Step 3: Total mass = 50 + 50 = 100 g Step 4: q = 100 × 4.2 × 6.7 = 2814 J = 2.814 kJ Step 5: n(H₂O) = n(NaOH) = 1.0 × (50/1000) = 0.05 mol Step 6: ΔH = −2.814 / 0.05 = −56.28 kJ mol⁻¹ ✅ (Negative → exothermic — temperature INCREASED)

Section 7 — Heat of Solution

DefinitionThe heat of solution is the heat change when 1 mole of solute dissolves in sufficient solvent that further dilution produces no further heat change.
🔥 Exothermic DissolutionEnergy released during solvation > energy absorbed breaking bonds. Temperature rises. Examples: NaOH, H₂SO₄ dissolving in water — VERY exothermic! Always add acid to water, not water to acid.
❄️ Endothermic DissolutionEnergy absorbed breaking bonds > energy released during solvation. Temperature falls. Examples: NH₄Cl, KNO₃ dissolving in water — used in instant cold packs!

Worked Example — Heat of Solution (NH₄Cl)

Given: 5.35 g NH₄Cl dissolved in 100 cm³ water. T drops from 22.5°C to 16.4°C. Step 1: M(NH₄Cl) = 14+4+35.5 = 53.5 g/mol n = 5.35/53.5 = 0.10 mol Step 2: Mass of solution ≈ 100 g Step 3: ΔT = 22.5 − 16.4 = 6.1°C (temperature DECREASED) Step 4: q = 100 × 4.2 × 6.1 = 2562 J = 2.562 kJ Step 5: ΔH = +2.562/0.10 = +25.62 kJ mol⁻¹ ✅ (Positive → endothermic — temperature DECREASED)

Section 8 — Resources & Simulations

🔬 PhET — Reactions & Rates ▶ Energetics Videos ▶ Energy Profile Diagrams ▶ Calorimetry Experiments 📄 BBC Bitesize — Energy Changes

Section 9 — CSEC Practice Questions

Question 1 — Exo vs Endo
Classify each as exothermic or endothermic and explain your reasoning: (a) The combustion of methane (b) Dissolving ammonium nitrate in water (c) The reaction of magnesium with dilute HCl (d) Photosynthesis
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a Exothermic. Combustion always releases energy. The energy released forming bonds in CO₂ and H₂O is greater than the energy absorbed breaking bonds in CH₄ and O₂. Surroundings get hotter. ΔH is negative.

b Endothermic. The energy absorbed breaking the ionic bonds in NH₄NO₃ and overcoming water-water interactions is greater than the energy released when ions are solvated. The solution gets cold. ΔH is positive. (Used in instant cold packs!)

c Exothermic. The reaction produces heat — the test tube gets warm. Energy released forming MgCl₂ and H₂ bonds > energy absorbed breaking Mg and HCl bonds. ΔH is negative.

d Endothermic. Photosynthesis absorbs light energy from the sun to convert CO₂ and H₂O into glucose. Energy must be supplied — it doesn't happen in the dark. ΔH is positive.

✅ (a) Exothermic (b) Endothermic (c) Exothermic (d) Endothermic
Question 2 — Calorimetry Calculation
50.0 cm³ of 1.0 mol dm⁻³ NaOH is mixed with 50.0 cm³ of 1.0 mol dm⁻³ HNO₃ in a polystyrene cup. The temperature rises from 22.5°C to 29.1°C. Calculate the heat of neutralisation in kJ mol⁻¹.
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1 ΔT = 29.1 − 22.5 = 6.6°C

2 Total mass = 50.0 + 50.0 = 100.0 g (density = 1 g/cm³)

3 q = m × c × ΔT = 100.0 × 4.2 × 6.6 = 2772 J = 2.772 kJ

4 n(H₂O) = n(NaOH) = 1.0 × (50.0/1000) = 0.050 mol

5 ΔH = −2.772 ÷ 0.050 = −55.44 kJ mol⁻¹

Negative sign because temperature increased → reaction is exothermic → heat released to surroundings.

✅ ΔH = −55.4 kJ mol⁻¹ (approximately −56 kJ mol⁻¹ as expected for strong acid + strong base).
Question 3 — Energy Profile Diagrams
Sketch and label a fully correct energy profile diagram for an exothermic reaction. Then describe how adding a catalyst would change the diagram, listing every feature that changes and every feature that stays the same.
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1 Exothermic profile must include: Reactants labelled at a HIGH level. A peak (transition state) above the reactant level. Products labelled at a LOWER level than reactants. An Ea arrow from reactant level to peak. A ΔH arrow from reactant level DOWN to product level (−ve). Axis labels: Energy (y) and Progress of reaction (x).

2 After adding a catalyst:

Changes: The peak is LOWER (smaller activation energy). The curve reaches a lower peak before dropping to the products. The reaction rate is FASTER.

Stays the same: Reactant energy level (unchanged). Product energy level (unchanged). ΔH value (unchanged — same overall energy difference). The identities of reactants and products.

✅ Catalyst: only Ea (peak height) decreases. Reactant level, product level, and ΔH are ALL unchanged.
Question 4 — Bond Energy
Use bond energies to calculate ΔH for: N₂(g) + 3H₂(g) → 2NH₃(g). Bond energies: N≡N = 945 kJ/mol, H–H = 436 kJ/mol, N–H = 391 kJ/mol.
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1 Bonds broken (reactants):

1 × N≡N = 1 × 945 = 945 kJ/mol 3 × H–H = 3 × 436 = 1308 kJ/mol Total absorbed = 945 + 1308 = 2253 kJ/mol

2 Bonds formed (products): 2 NH₃ molecules, each with 3 N–H bonds = 6 N–H bonds total

6 × N–H = 6 × 391 = 2346 kJ/mol Total released = 2346 kJ/mol

3 ΔH = Σ(bonds broken) − Σ(bonds formed) = 2253 − 2346 = −93 kJ mol⁻¹

✅ ΔH = −93 kJ mol⁻¹ (negative → exothermic). Bonds formed release more energy than bonds broken absorb.
Question 5 — Heat of Solution
4.00 g of NaOH (M = 40 g/mol) is dissolved in 100 cm³ of water. The temperature rises from 20.0°C to 31.4°C. Calculate the heat of solution of NaOH in kJ mol⁻¹. Is the dissolution exothermic or endothermic?
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1 n(NaOH) = 4.00/40 = 0.100 mol

2 ΔT = 31.4 − 20.0 = 11.4°C (temperature INCREASED)

3 Mass ≈ 100 g (ignoring NaOH mass, or 104 g if including solute)

4 q = 100 × 4.2 × 11.4 = 4788 J = 4.788 kJ

5 ΔH = −4.788/0.100 = −47.88 kJ mol⁻¹

Negative (temperature INCREASED) → exothermic dissolution. Solvation energy released > energy absorbed breaking NaOH ionic bonds.

✅ ΔH ≈ −47.9 kJ mol⁻¹. Exothermic — temperature rose when NaOH dissolved.
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