S4 Chemistry · Unit 18

Energy Changes &
Energy Profile Diagrams

Internal energy · First law of thermodynamics · Enthalpy changes · Exothermic & endothermic · Hess's law · Bond enthalpies · Energy profiles

18.1 Internal Energy 18.2 Enthalpy & ΔH 18.3 Exo & Endothermic 18.4 Standard Enthalpies 18.5 Hess's Law 18.6 Bond Enthalpies 18.7 Energy Profile Diagrams 18.8 Calorimetry Exercises Quiz Unit Test
18.1

Internal Energy and the First Law of Thermodynamics

Internal Energy (U)

The internal energy (U) of a system is the total energy stored within it — the sum of all kinetic energies (translational, rotational, vibrational) and potential energies (intermolecular, chemical bonds) of all particles in the system.

First Law of Thermodynamics Energy cannot be created or destroyed — it can only be transferred or converted from one form to another. The total energy of the universe is constant.

ΔU = q + w
where ΔU = change in internal energy, q = heat transferred to system, w = work done on system

In chemistry, most reactions occur at constant pressure (open to the atmosphere), so we use enthalpy (H) rather than internal energy.

18.2

Enthalpy and Enthalpy Change (ΔH)

Enthalpy (H)

Enthalpy is defined as: H = U + pV (internal energy + pressure × volume). For reactions at constant pressure, the heat exchanged equals the enthalpy change:

ΔH = H(products) − H(reactants) ΔH < 0 → exothermic (heat released to surroundings) ΔH > 0 → endothermic (heat absorbed from surroundings) Units: kJ mol⁻¹ (per mole of reaction as written)
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Sign convention — always from the system's perspective Exothermic: system loses energy → ΔH is NEGATIVE (−). The surroundings get warmer.
Endothermic: system gains energy → ΔH is POSITIVE (+). The surroundings get cooler.
18.3

Exothermic and Endothermic Reactions

Exothermic Reactions

Reactions that release heat to the surroundings. The temperature of the surroundings increases. Products have lower enthalpy than reactants.

Examples of exothermic reactions: • Combustion: CH₄ + 2O₂ → CO₂ + 2H₂O ΔH = −890 kJ/mol • Neutralisation: HCl + NaOH → NaCl + H₂O ΔH = −57.1 kJ/mol • Respiration: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O ΔH = −2803 kJ/mol • Formation MgO: 2Mg + O₂ → 2MgO ΔH = −1204 kJ/mol • Oxidation Fe: 4Fe + 3O₂ → 2Fe₂O₃ ΔH = −1648 kJ/mol

Endothermic Reactions

Reactions that absorb heat from the surroundings. The temperature of the surroundings decreases. Products have higher enthalpy than reactants.

Examples of endothermic reactions: • Thermal decomp: CaCO₃ → CaO + CO₂ ΔH = +178 kJ/mol • Photosynthesis: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ ΔH = +2803 kJ/mol • Dissolving NH₄NO₃: NH₄NO₃(s) → NH₄⁺(aq)+NO₃⁻(aq) ΔH = +25.7 kJ/mol • Evaporation: H₂O(l) → H₂O(g) ΔH = +44 kJ/mol
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Practical tests for exo/endothermic Exothermic: test tube feels HOT; thermometer rises. Endothermic: test tube feels COLD; thermometer drops. Dissolving ammonium nitrate in water is a classic endothermic demonstration — the beaker becomes ice-cold.
18.4

Standard Enthalpy Changes

Standard Conditions and Standard Enthalpy

Standard enthalpy changes (ΔH°) are measured under standard conditions: 298 K (25°C), 100 kPa pressure, all species in their standard states.

NameSymbolDefinitionExample
Standard Enthalpy of FormationΔH°fEnthalpy change when 1 mol of compound is formed from its elements in their standard statesC(s) + O₂(g) → CO₂(g) ΔH°f = −394 kJ/mol
Standard Enthalpy of CombustionΔH°cEnthalpy change when 1 mol of substance burns completely in excess oxygenCH₄ + 2O₂ → CO₂ + 2H₂O ΔH°c = −890 kJ/mol
Standard Enthalpy of NeutralisationΔH°nEnthalpy change when 1 mol of water is formed in an acid-base neutralisationH⁺ + OH⁻ → H₂O ΔH°n = −57.1 kJ/mol
Standard Enthalpy of AtomisationΔH°atEnthalpy change when 1 mol of gaseous atoms is formed from element in standard state½Cl₂(g) → Cl(g) ΔH°at = +121 kJ/mol
Standard Enthalpy of SolutionΔH°solEnthalpy change when 1 mol of substance dissolves in excess solventNaCl(s) → Na⁺(aq) + Cl⁻(aq) ΔH°sol = +3.9 kJ/mol
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Key rule: ΔH°f of any element in standard state = 0 The enthalpy of formation of H₂(g), O₂(g), C(graphite), Na(s), Fe(s) etc. is ZERO by definition — they are already in their standard states, so no formation is needed.
18.5

Hess's Law

Hess's Law (1840) The total enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final conditions are the same. ΔH is a state function — it depends only on start and end states, not on the pathway.

Using Hess's Law — Two Methods

Method 1: Enthalpy Cycle (Route A = Route B)

ΔH°reaction = ΣΔH°f(products) − ΣΔH°f(reactants) Example: CH₄ + 2O₂ → CO₂ + 2H₂O ΔH° = [ΔH°f(CO₂) + 2×ΔH°f(H₂O)] − [ΔH°f(CH₄) + 2×ΔH°f(O₂)] = [−394 + 2×(−286)] − [−74.8 + 2×(0)] = −966 − (−74.8) = −891.2 kJ/mol ✓ (matches direct measurement)

Method 2: Algebraic (adding equations)

Reactions can be added, reversed (flip sign of ΔH), or multiplied (scale ΔH) to give the target equation.

Worked Example 18.1 — Using Hess's Law

Find ΔH for: C(s) + 2H₂(g) → CH₄(g) — using combustion data:

Given: (1) C + O₂ → CO₂, ΔH₁ = −394 kJ/mol

Given: (2) H₂ + ½O₂ → H₂O, ΔH₂ = −286 kJ/mol

Given: (3) CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH₃ = −890 kJ/mol

1

Target: C + 2H₂ → CH₄. We need CH₄ as product, so reverse equation (3): CO₂ + 2H₂O → CH₄ + 2O₂, ΔH = +890

2

Add equation (1): C + O₂ → CO₂, ΔH = −394

3

Add equation (2) × 2: 2H₂ + O₂ → 2H₂O, ΔH = 2 × (−286) = −572

4

Sum: C + O₂ + 2H₂ + O₂ + CO₂ + 2H₂O → CO₂ + 2H₂O + CH₄ + 2O₂

5

Cancel species on both sides: C + 2H₂ → CH₄; ΔH = +890 − 394 − 572 = −76 kJ/mol ✓

18.6

Bond Enthalpies (Bond Energies)

What are Bond Enthalpies?

The bond enthalpy (bond dissociation enthalpy) is the energy required to break 1 mol of a specific bond in gaseous molecules — always endothermic (requires energy). Breaking bonds: endothermic (+). Making bonds: exothermic (−).

ΔH°reaction = Σ(bonds broken) − Σ(bonds formed) = Energy in (endothermic) − Energy out (exothermic) If bonds broken > bonds formed → endothermic (ΔH > 0) If bonds formed > bonds broken → exothermic (ΔH < 0)
BondBond Enthalpy (kJ/mol)BondBond Enthalpy (kJ/mol)
H–H+436C=C+614
C–H+413C≡C+839
C–C+347C=O+805
O–H+464O=O+498
C–O+358N≡N+945
N–H+391H–Cl+432
Cl–Cl+243H–F+568
C–Cl+339H–Br+366
Worked Example 18.2 — Bond Enthalpy Calculation

Calculate ΔH for: CH₄ + 2O₂ → CO₂ + 2H₂O using bond enthalpies.

1

Bonds broken (reactants): 4 × C–H = 4 × 413 = 1652; 2 × O=O = 2 × 498 = 996. Total broken = 2648 kJ

2

Bonds formed (products): 2 × C=O = 2 × 805 = 1610; 4 × O–H = 4 × 464 = 1856. Total formed = 3466 kJ

3

ΔH = 2648 − 3466 = −818 kJ/mol

Note: Bond enthalpies are average values → result (−818) differs slightly from experimental (−890 kJ/mol). This is normal — bond enthalpies are averages from many molecules.

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Bond enthalpy method only works for gaseous species Bond enthalpies assume all reactants and products are gases. If the question involves liquids or solids, you must account for the extra energy (enthalpy of vaporisation/sublimation) — or use Hess's law with formation enthalpies instead.
18.7

Energy Profile Diagrams

Energy Profile for Exothermic Reactions

In an exothermic reaction, the products have lower enthalpy than the reactants. The curve rises to a peak (the transition state / activated complex) then falls to a lower level.

Exothermic Reaction Energy Profile Enthalpy Reaction Progress → Reactants Products Transition state Ea(fwd) ΔH (−ve)
Exothermic: Products at lower enthalpy. ΔH is negative. Ea = activation energy (fwd).

Energy Profile for Endothermic Reactions

In an endothermic reaction, products have higher enthalpy than reactants. The curve rises to a peak then falls to a level still above the starting point.

Endothermic Reaction Energy Profile Enthalpy Reaction Progress → Reactants Products Transition state ΔH (+ve)
Endothermic: Products at higher enthalpy. ΔH is positive.

Effect of a Catalyst on the Energy Profile

A catalyst provides an alternative reaction pathway with a lower activation energy. It does NOT change ΔH (same reactants and products → same enthalpy difference). On the diagram, the peak is lower but the start and end levels stay the same.

Without catalyst: Ea = large → slow reaction With catalyst: Ea = smaller → faster reaction ΔH: UNCHANGED by catalyst Kc: UNCHANGED by catalyst (only affects rate, not equilibrium)
18.8

Calorimetry — Measuring Enthalpy Changes

The Calorimetry Equation

q = mcΔT where: q = heat energy transferred (J) m = mass of solution/water (g) — assume density = 1 g/cm³ for dilute solutions c = specific heat capacity of water = 4.18 J g⁻¹ °C⁻¹ ΔT = temperature change (°C) = T_final − T_initial Then: ΔH = −q / n (in kJ/mol) where n = moles of substance reacted Negative sign: exothermic reaction raises T (q positive) → ΔH negative
Worked Example 18.3 — Calorimetry Calculation

50 cm³ of 1.0 mol/dm³ HCl is mixed with 50 cm³ of 1.0 mol/dm³ NaOH. Temperature rises from 21.5°C to 28.2°C. Calculate ΔH of neutralisation.

1

Total volume = 100 cm³ → mass = 100 g (assuming density 1 g/cm³)

2

ΔT = 28.2 − 21.5 = 6.7°C

3

q = mcΔT = 100 × 4.18 × 6.7 = 2800.6 J = 2.80 kJ

4

Moles of water formed: n(HCl) = 0.050 × 1.0 = 0.050 mol (HCl + NaOH → NaCl + H₂O, 1:1)

5

ΔH = −q/n = −2.80 / 0.050 = −56.0 kJ/mol (theoretical = −57.1 kJ/mol)

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Sources of error in calorimetry (1) Heat lost to surroundings through the container → measured ΔT is less than true ΔT → ΔH underestimated. (2) Specific heat capacity of solution assumed = water — not exactly true for solutions. (3) Heat absorbed by calorimeter itself ignored. (4) Imperfect mixing. Use a polystyrene cup (good insulator) and lid to minimise heat loss.

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Exercises

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Multiple Choice Quiz — 25 Questions

Unit 18: Energy Changes

25 Questions
Q1

A reaction in which ΔH is negative is:

ΔH < 0 → exothermic. Products are at lower enthalpy than reactants. Energy is released to the surroundings → surroundings get warmer. Examples: combustion, neutralisation, respiration. Endothermic reactions have ΔH > 0.
Q2

The standard enthalpy of formation of any element in its standard state is:

ΔH°f of any element in its standard state = 0 by definition. Examples: H₂(g), O₂(g), N₂(g), C(graphite), Na(s), Fe(s) all have ΔH°f = 0. This is because no "formation" is needed — they are already in their standard state.
Q3

Hess's Law states that:

Hess's Law: ΔH depends only on start and end states, not the pathway. This makes enthalpy a STATE FUNCTION. It allows indirect calculation of enthalpy changes for reactions that cannot be measured directly (e.g. C + ½O₂ → CO, since CO₂ is always also formed).
Q4

In an energy profile diagram for an exothermic reaction, the products are:

Exothermic: energy released → products at LOWER enthalpy than reactants. ΔH = H(products) − H(reactants) < 0. The profile shows reactants at a higher plateau, rising to the transition state peak, then falling to products at a lower plateau.
Q5

Breaking a covalent bond is:

Breaking bonds ALWAYS requires energy (endothermic). You must put energy in to pull the bonded atoms apart. Forming bonds ALWAYS releases energy (exothermic). The overall ΔH = energy in (bonds broken) − energy out (bonds formed). Memory: "Break bonds — Break the bank (costs energy)".
Q6

The formula q = mcΔT is used in calorimetry. What does 'c' represent?

c = specific heat capacity = 4.18 J g⁻¹ °C⁻¹ for water/dilute aqueous solutions. It represents the energy needed to raise 1 g of substance by 1°C. q = heat transferred (J); m = mass (g); ΔT = temperature change (°C).
Q7

Given: ΔH°f(CO₂) = −394, ΔH°f(H₂O) = −286, ΔH°f(C₂H₆) = −85 kJ/mol. ΔH°c for ethane: C₂H₆ + 3½O₂ → 2CO₂ + 3H₂O is:

ΔH°c = ΣΔH°f(products) − ΣΔH°f(reactants). Products: 2(−394) + 3(−286) = −788 − 858 = −1646 kJ. Reactants: −85 + 3½(0) = −85 kJ. ΔH°c = −1646 − (−85) = −1646 + 85 = −1561 kJ/mol. Always negative for combustion.
Q8

When a catalyst is added to a reaction at equilibrium, which quantity changes?

A catalyst lowers the ACTIVATION ENERGY (Ea) by providing an alternative pathway. It does NOT change ΔH, Kc, the enthalpy of reactants or products, or the equilibrium position. Only temperature changes Kc. A catalyst makes the reaction reach equilibrium faster, but at the same position.
Q9

The enthalpy of neutralisation of a strong acid with a strong base (−57 kJ/mol) is always the same because:

Strong acid + strong base: both fully dissociated. The ONLY reaction is H⁺(aq) + OH⁻(aq) → H₂O(l), ΔH = −57.1 kJ/mol. The specific ions (Na⁺, Cl⁻, K⁺, NO₃⁻) are spectators and don't contribute to ΔH. For weak acids/bases, ΔH is less negative because energy is needed to ionise the weak electrolyte first.
Q10

Bond enthalpy calculations give slightly different values from Hess's law calculations because:

Bond enthalpies are AVERAGE values taken from many different molecules. The actual C–H bond energy in methane differs slightly from the C–H bond in ethane or propane. Hess's law uses actual enthalpies of formation (measured precisely), so gives more accurate results. Bond enthalpy method is useful for estimation when formation data is not available.
Q11

What is the activation energy (Ea) in an energy profile diagram?

Activation energy (Ea): the MINIMUM energy that colliding molecules must have for a reaction to occur. On the energy profile: Ea = energy of transition state − energy of reactants. A catalyst lowers Ea → more molecules exceed Ea → faster reaction. Ea is always positive (energy input needed to reach transition state).
Q12

100 cm³ of 0.5 mol/dm³ HNO₃ is mixed with 100 cm³ of 0.5 mol/dm³ NaOH. Temperature rises by 3.35°C. Heat released (q) is:

q = mcΔT = (100+100) × 4.18 × 3.35 = 200 × 4.18 × 3.35 = 2800.6 J ≈ 2802 J. n(H₂O) = 0.100 × 0.5 = 0.050 mol. ΔH = −q/n = −2802/0.050 = −56,040 J/mol ≈ −56.0 kJ/mol (close to theoretical −57.1 kJ/mol).
Q13

Which of these processes is endothermic?

CaCO₃ → CaO + CO₂: thermal decomposition is endothermic (ΔH = +178 kJ/mol). Energy must be supplied (high temperature in kilns). Combustion, neutralisation, and rusting are all exothermic (release energy). Test: endothermic reactions feel cold to touch or require continuous heating.
Q14

Use bond enthalpies: N₂ + 3H₂ → 2NH₃. (N≡N = 945; H–H = 436; N–H = 391 kJ/mol). ΔH is:

Broken: 1×N≡N = 945; 3×H–H = 3×436 = 1308. Total = 2253 kJ. Formed: 2×NH₃ = 2×3×N–H = 6×391 = 2346 kJ. ΔH = 2253 − 2346 = −93 kJ/mol ≈ −92 kJ/mol (experimental −92 kJ/mol). Exothermic — more energy released forming N–H bonds than used breaking N≡N and H–H.
Q15

The sign convention for ΔH is based on:

ΔH sign is from the SYSTEM's perspective. Exothermic: system loses heat → ΔH negative. Endothermic: system gains heat → ΔH positive. The surroundings do the opposite: if ΔH is negative, the surroundings gain heat (temperature rises). Note: ΔT positive ↑ means exothermic (surroundings gain heat), but ΔH is negative (system loses heat).
Q16

Photosynthesis is an example of which type of reaction?

Photosynthesis: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂, ΔH = +2803 kJ/mol. ENDOTHERMIC — absorbs light energy (from the sun) to convert low-energy CO₂ and H₂O into high-energy glucose. This is why plants need sunlight. Respiration (the reverse) is exothermic and releases the same energy.
Q17

In Hess's Law, if you reverse a reaction equation, you must:

Reversing a reaction equation changes the sign of ΔH (multiply by −1). If A → B has ΔH = −100 kJ/mol, then B → A has ΔH = +100 kJ/mol. Scaling: if you multiply the equation by n, multiply ΔH by n too. This is the foundation of Hess's law algebra.
Q18

Standard conditions for thermochemistry are:

Standard conditions (IUPAC): 298 K (25°C), 100 kPa pressure (1 bar), all substances in standard states (most stable physical form at 298 K and 100 kPa), solutions at 1 mol/dm³. Note: 100 kPa ≈ 1 bar (not 1 atm = 101.325 kPa). Standard conditions are marked with ° symbol: ΔH°.
Q19

Which of the following has the largest activation energy?

High Ea → slow reaction requiring much energy to get started. A reaction needing 800°C has a very high Ea — most molecules don't have enough energy at room temperature. Explosive reactions have low Ea (once started, chain reaction proceeds rapidly). Enzyme-catalysed reactions have low Ea (enzyme lowers it). Ionic reactions in solution (e.g. precipitation) have virtually zero Ea.
Q20

The enthalpy change calculated using bond enthalpies differs from Hess's law result because:

Two reasons: (1) Bond enthalpies are AVERAGE values from many different molecules — actual bond strength varies with molecular environment. (2) Bond enthalpy method assumes all species are in the GAS PHASE. If any reactants/products are liquid or solid, intermolecular forces must also be overcome/formed — but bond enthalpies ignore this. Hess's law uses precise experimental ΔHf° values.
Q21

Which statement about the enthalpy of atomisation is correct?

Enthalpy of atomisation is ALWAYS positive (endothermic). Energy must be supplied to convert a substance in its standard state to isolated gaseous atoms. For metals (Na, Fe, etc.): atomisation = sublimation → endothermic. For Cl₂: ½Cl₂(g) → Cl(g), ΔH°at = +121 kJ/mol (half the Cl–Cl bond enthalpy). Used in Born-Haber cycles.
Q22

A student measures ΔH for a reaction using a polystyrene cup. The value obtained is always less exothermic than the true value because:

Heat always leaks to the surroundings during a calorimetry experiment, even with a polystyrene cup (good insulator but not perfect). The measured ΔT is lower than the true ΔT → q calculated is smaller → ΔH is less negative than it should be. Ways to reduce this: use a lid, measure quickly, extrapolate the temperature-time graph to find the true maximum T.
Q23

The standard enthalpy of combustion is always:

Combustion is ALWAYS exothermic — burning releases energy. Therefore ΔH°c is always NEGATIVE. The products (CO₂, H₂O) are at lower enthalpy than the reactants (fuel + O₂). This is why fuels are useful — they release energy on burning. No fuel has a positive enthalpy of combustion.
Q24

What does the area under the Maxwell-Boltzmann distribution curve to the right of Ea represent?

The shaded area to the RIGHT of Ea on the Maxwell-Boltzmann distribution represents the fraction of molecules with energy ≥ Ea — these are the molecules that can successfully react when they collide. Higher temperature → the curve shifts right → larger area to the right of Ea → more successful collisions → faster rate. Catalyst lowers Ea → more of the curve is to the right of Ea → faster reaction at same temperature.
Q25

ΔH°f for water is −286 kJ/mol. This means:

ΔH°f(H₂O,l) = −286 kJ/mol means: H₂(g) + ½O₂(g) → H₂O(l), ΔH° = −286 kJ/mol. This refers to LIQUID water at standard conditions. 286 kJ of energy are RELEASED (exothermic, negative sign) per mole of liquid water formed. Note: ΔH°f(H₂O,g) = −242 kJ/mol (less negative because energy needed for vaporisation not released).
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Unit Test — 50 Marks

Section A — Short Answer

30 marks
Q1 [5 marks]

State and explain the First Law of Thermodynamics. Define internal energy and enthalpy. Explain why chemists use enthalpy rather than internal energy to describe most reactions. [5]

First Law: Energy cannot be created or destroyed — only transferred or converted. The total energy of the universe is constant. ΔU = q + w. [1] Internal energy (U): total of all kinetic and potential energies of all particles in a system (translational, rotational, vibrational KE + intermolecular PE + chemical bond PE). [1] Enthalpy (H = U + pV): at constant pressure, ΔH = heat exchanged. Most reactions occur at constant pressure (atmosphere), so heat measured = ΔH. [1] Why H not U: if a reaction is carried out in a sealed container (constant volume), ΔU = q. But most lab reactions are open to the atmosphere at constant pressure. Work done by gas expansion (pΔV) is automatically accounted for in ΔH, making it more practical. ΔH = ΔU + pΔV. [2]
Q2 [5 marks]

Using the following standard enthalpies of formation, calculate ΔH° for the complete combustion of propane: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. ΔH°f: C₃H₈(g) = −104, CO₂(g) = −394, H₂O(l) = −286 kJ/mol. [5]

ΔH°c = ΣΔH°f(products) − ΣΔH°f(reactants). [1] Products: 3 × ΔH°f(CO₂) + 4 × ΔH°f(H₂O) = 3(−394) + 4(−286) = −1182 + (−1144) = −2326 kJ. [1.5] Reactants: 1 × ΔH°f(C₃H₈) + 5 × ΔH°f(O₂) = −104 + 5(0) = −104 kJ. [1] ΔH°c = −2326 − (−104) = −2326 + 104 = −2222 kJ/mol. [1] Sign is negative (exothermic — combustion always releases energy). Propane is used in camping gas and BBQ fuel. [0.5]
Q3 [5 marks]

Use the following data and Hess's Law to find ΔH for the reaction: CO(g) + ½O₂(g) → CO₂(g). Given: C(s) + O₂(g) → CO₂(g), ΔH₁ = −394 kJ/mol; C(s) + ½O₂(g) → CO(g), ΔH₂ = −111 kJ/mol. [5]

Target: CO + ½O₂ → CO₂. Strategy: use ΔH₁ (C + O₂ → CO₂) and reverse ΔH₂. [1] Step 1: C + O₂ → CO₂, ΔH = −394 kJ (equation 1, forward). [1] Step 2: Reverse equation 2: CO(g) → C(s) + ½O₂(g), ΔH = +111 kJ. [1] Add the two equations: C + O₂ + CO → CO₂ + C + ½O₂. Cancel C and ½O₂ from both sides: CO + ½O₂ → CO₂. [1] ΔH = −394 + 111 = −283 kJ/mol. [1] This is the enthalpy of combustion of carbon monoxide — a toxic gas that burns in air to give CO₂. The reaction is exothermic (negative ΔH).
Q4 [5 marks]

Calculate ΔH for the reaction: CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l) using bond enthalpies. (C–H = 413, O–H = 464, C–O = 358, O=O = 498, C=O = 805 kJ/mol). Structural formula: CH₃OH has 3 C–H, 1 C–O, 1 O–H bonds. [5]

Bonds broken (reactants): 3 × C–H = 3 × 413 = 1239; 1 × C–O = 358; 1 × O–H = 464; 3/2 × O=O = 1.5 × 498 = 747. Total broken = 1239 + 358 + 464 + 747 = 2808 kJ. [2] Bonds formed (products): CO₂ has 2 × C=O = 2 × 805 = 1610; 2H₂O has 4 × O–H = 4 × 464 = 1856. Total formed = 1610 + 1856 = 3466 kJ. [2] ΔH = bonds broken − bonds formed = 2808 − 3466 = −658 kJ/mol. [1] (Experimental value ≈ −726 kJ/mol. Difference because bond enthalpies are averages and assume gaseous species; methanol is liquid.)
Q5 [5 marks]

In an experiment, 50 cm³ of 1.0 mol/dm³ HCl is added to 50 cm³ of 1.0 mol/dm³ NH₃ solution. The temperature rises by 5.2°C. (a) Calculate q and ΔH for this neutralisation. (b) The value is less exothermic than for HCl + NaOH (−57 kJ/mol). Explain why. [5]

(a) q = mcΔT = 100 × 4.18 × 5.2 = 2173.6 J = 2.17 kJ. [1] n(H₂O) = 0.050 × 1.0 = 0.050 mol. ΔH = −2.17/0.050 = −43.5 kJ/mol. [1] (b) HCl + NH₃: strong acid + weak base. NH₃ is only partially ionised — it must FIRST ionise (NH₃ + H₂O ⇌ NH₄⁺ + OH⁻) before neutralisation can occur. The ionisation of NH₃ is endothermic (absorbs energy). This endothermic step partially cancels the exothermic H⁺ + OH⁻ → H₂O step, giving a less negative overall ΔH. [2] For HCl + NaOH (strong + strong): both fully ionised, only H⁺ + OH⁻ → H₂O. No endothermic ionisation step → more exothermic (−57 kJ/mol). [1]
Q6 [5 marks]

Draw energy profile diagrams for (a) an exothermic reaction and (b) an endothermic reaction. For each diagram, mark and define: Ea(forward), Ea(reverse), ΔH, transition state, reactants, and products. Show the effect of a catalyst on diagram (a). [5]

(a) Exothermic profile: x-axis = reaction progress; y-axis = enthalpy. Reactants at higher plateau → curve rises to peak (transition state) → curve falls to products at LOWER plateau. Labels: Reactants (left high), Transition state (peak), Products (right low), Ea(fwd) = peak − reactants (large), Ea(rev) = peak − products (small), ΔH = products − reactants (negative arrow downward). [2] (b) Endothermic: same axes. Reactants at LOWER plateau → peak → products at HIGHER plateau. ΔH positive (upward arrow). Ea(fwd) = large; Ea(rev) = smaller (since products are higher). [1.5] Catalyst effect on (a): draw a second dashed curve with a LOWER peak. Reactants and products at SAME levels (ΔH unchanged). Label new lower Ea(fwd) and Ea(rev). State: catalyst lowers Ea for BOTH directions; does NOT change ΔH or equilibrium position. [1.5]

Section B — Extended Answer

20 marks
Q7 [10 marks]

(a) Explain Hess's Law and why it is valid. Describe the two methods (enthalpy cycle and algebraic) for using Hess's Law with one example of each. [5]
(b) The enthalpy of combustion of carbon (graphite) is −394 kJ/mol and of hydrogen is −286 kJ/mol. The enthalpy of combustion of ethanol (C₂H₅OH) is −1367 kJ/mol. Using these data, calculate the standard enthalpy of formation of ethanol. Show a clear Hess's law cycle. [5]

(a) Hess's Law: the total enthalpy change for a reaction is INDEPENDENT of the route taken between initial and final states. Valid because enthalpy is a STATE FUNCTION — depends only on the state of the system, not the pathway. Consequence of the First Law of Thermodynamics (energy conservation). [1] Method 1 — Enthalpy cycle: draw two routes from elements to products. Route A (direct) = ΔH°f(compound). Route B (via oxides) = uses ΔH°c values. ΔH°f = ΔH(route A) = ΔH(route B). Example: ΔH°f(CH₄) = [ΔH°c(C) + 2ΔH°c(H₂)] − ΔH°c(CH₄) = [−394 + 2(−286)] − (−890) = −76 kJ/mol. [2] Method 2 — Algebraic: reverse and/or scale equations; add to give target equation; sum ΔH values. Example: target CO + ½O₂ → CO₂. Use: (1) C + O₂ → CO₂, ΔH = −394; (2) C + ½O₂ → CO, ΔH = −111. Reverse (2): CO → C + ½O₂, ΔH = +111. Add: CO + ½O₂ → CO₂, ΔH = −394 + 111 = −283 kJ/mol. [2] (b) Target: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l), ΔH°f = ? [0.5] Route via combustion products: elements → CO₂ + H₂O and ethanol → CO₂ + H₂O. ΔH°c(C) × 2 = 2(−394) = −788 kJ; ΔH°c(H₂) × 3 = 3(−286) = −858 kJ; sum = −1646 kJ (elements→CO₂+H₂O). ΔH°c(C₂H₅OH) = −1367 kJ (ethanol→CO₂+H₂O). [2] Hess's Law: ΔH°f + ΔH°c(ethanol) = ΔH°c(elements). ΔH°f = ΔH°c(elements) − ΔH°c(ethanol) = −1646 − (−1367) = −1646 + 1367 = −279 kJ/mol. [2.5] Negative value — formation of ethanol from elements is exothermic.
Q8 [10 marks]

(a) Describe, with full experimental details, how you would use a polystyrene cup calorimeter to determine the enthalpy of neutralisation of dilute hydrochloric acid with dilute sodium hydroxide. Include: apparatus, procedure, measurements, calculation, and sources of error. [6]
(b) Discuss the differences in enthalpy of neutralisation between: (i) strong acid + strong base; (ii) weak acid + strong base; (iii) strong acid + weak base. Explain the differences in terms of ionisation enthalpies. [4]

(a) Apparatus: polystyrene cup + lid, thermometer (±0.1°C), measuring cylinders (25 cm³), balance. [0.5] Procedure: measure 25 cm³ of 1.0 mol/dm³ HCl into polystyrene cup. Record temperature every 30 s for 2 minutes to establish baseline (T₁). Separately measure 25 cm³ of 1.0 mol/dm³ NaOH, record its temperature. Quickly pour NaOH into HCl, stir gently, record temperature every 30 s for 5 minutes. [2] Measurements: initial temperature T₁ (average of acid and base temperatures), maximum temperature T₂ (extrapolate temperature-time graph if needed). ΔT = T₂ − T₁. [1] Calculation: q = mcΔT = (25 + 25) × 4.18 × ΔT (J); n(H₂O) = 0.025 × 1.0 = 0.025 mol; ΔH = −q/n (kJ/mol). [1.5] Sources of error: (1) Heat loss to surroundings → ΔT lower than true → |ΔH| underestimated. Reduce by: lid, insulation, extrapolating T-time graph. (2) Specific heat capacity of NaCl solution ≠ 4.18 → small error. (3) Heat absorbed by polystyrene cup and thermometer ignored. (4) Imprecise measurement of volumes. [1] (b)(i) Strong + Strong (e.g. HCl + NaOH): both fully ionised. Net: H⁺(aq) + OH⁻(aq) → H₂O(l), ΔH = −57.1 kJ/mol. Constant regardless of which strong acid or base. [1] (ii) Weak acid + Strong base (e.g. CH₃COOH + NaOH): CH₃COOH must first ionise: CH₃COOH → H⁺ + CH₃COO⁻, endothermic (+ve). This endothermic step partially cancels the H⁺ + OH⁻ → H₂O exothermic step. Result: ΔH ≈ −55 kJ/mol (less exothermic). [1.5] (iii) Strong acid + Weak base (e.g. HCl + NH₃): NH₃ must first accept H⁺ (or ionise). NH₄⁺ formation from NH₃ + H⁺: the weak base ionisation/protonation is less than fully exothermic. Overall ΔH ≈ −52 kJ/mol (less exothermic than strong+strong). [1.5]
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