amazon.co.ukEnergetics and Kinetics
A* revision for enthalpy changes, Hess's law, Born-Haber cycles, collision theory, Maxwell-Boltzmann, rate equations and the Arrhenius equation, with required practicals and exam questions for all boards.
🎯 Learning Objectives
- Define standard enthalpy changes and perform calorimetry calculations using q = mcΔT
- Apply Hess's law using formation and combustion cycles, and calculate ΔH from mean bond enthalpies with limitations
- Construct Born-Haber cycles to find lattice enthalpies and explain differences between theoretical and experimental values
- Calculate ΔG = ΔH − TΔS, determine feasibility and the temperature threshold
- Use collision theory and Maxwell-Boltzmann distributions to explain the effects of temperature, concentration and catalysts
- Determine rate equations and orders by the initial-rates method and calculate k with units
- Use the Arrhenius equation, half-life relationships and ln k vs 1/T plots to find Ea
📋 Key Definitions and Core Concepts
Enthalpy change (ΔH): The heat energy change of a reaction at constant pressure, in kJ mol⁻¹.
Standard enthalpy of formation: The enthalpy change when one mole of a compound forms from its elements in their standard states under standard conditions (100 kPa, 298 K).
Standard enthalpy of combustion: The enthalpy change when one mole of a substance burns completely in oxygen under standard conditions; always exothermic.
Hess's law: The total enthalpy change of a reaction is independent of the route taken.
Lattice enthalpy of formation: The enthalpy change when one mole of solid ionic lattice forms from gaseous ions; always exothermic.
Activation energy (Ea): The minimum energy colliding molecules need for a collision to be effective and lead to reaction.
Order of reaction: The power to which a concentration is raised in the rate equation, determined experimentally.
Half-life (t½): The time for the concentration of a reactant to halve; constant for first-order reactions, t½ = ln2/k.
Gibbs free energy (ΔG): ΔG = ΔH − TΔS; a reaction is thermodynamically feasible when ΔG ≤ 0.
Enthalpy Changes, Calorimetry and Energy Level Diagrams
Enthalpy change (ΔH) is the heat energy change at constant pressure. Exothermic reactions release heat to the surroundings (ΔH negative; products lower than reactants; temperature rises) — combustion, oxidation, neutralisation. Endothermic reactions absorb heat (ΔH positive; products higher; temperature falls) — thermal decomposition, dissolving some salts. Standard conditions for all standard enthalpies: 100 kPa, 298 K, and 1 mol dm⁻³ for solutions; ΔHf° of elements in their standard states is 0.
Definitions to memorise exactly (each says 'one mole'): enthalpy of formation — one mole of a compound forms from its elements in their standard states; enthalpy of combustion — one mole of a substance burns completely in oxygen, always exothermic; enthalpy of neutralisation — one mole of water forms from an acid and alkali, −57.1 kJ/mol for strong acid–strong base; enthalpy of atomisation — one mole of gaseous atoms forms from an element in its standard state, always endothermic.
Calorimetry (required practical: measuring enthalpy of combustion or neutralisation): q = mcΔT, where m is the mass of solution heated (assume density 1 g cm⁻³) and c is specific heat capacity (4.18 J g⁻¹ K⁻¹ for water). Convert to ΔH by dividing by moles: ΔH = −q/n (negative for exothermic). Sources of error for combustion: heat loss to surroundings and apparatus, incomplete combustion (soot), evaporation of fuel, heat capacity of the container not accounted for — all make the measured value less exothermic than data-book values. Improvements: lid, draught shields, flame close to can, or a bomb calorimeter.
Energy level diagrams: y-axis enthalpy, reactants left, products right; exothermic products lower, endothermic products higher. An energy profile additionally shows the activation energy hump and, with a catalyst, a lower hump via an alternative route (two smaller humps). ΔH is the difference in levels; Ea is the hump height from reactants. Label both correctly — confusing them is a classic lost mark.
Hess's Law and Enthalpy Cycles
Hess's law: the total enthalpy change for a reaction is independent of the route taken, because enthalpy is a state function. Two standard cycles are examined. (1) Formation cycle — used when combustion data are given: ΔHr = ΣΔHc(reactants) − ΣΔHc(products). Watch the direction: combustion arrows point away from the reactants/products down to common combustion products (CO₂ + H₂O). (2) Combustion-cycle alternative / formation-data cycle — when ΔHf° data are given: ΔHr = ΣΔHf°(products) − ΣΔHf°(reactants).
Worked example (from formation data): for C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), given ΔHf°(C₂H₅OH) = −277, ΔHf°(CO₂) = −394, ΔHf°(H₂O) = −286 kJ/mol: ΔHc = [(2 × −394) + (3 × −286)] − [−277] = (−788 − 858) + 277 = −1369 kJ/mol. Method marks: correct cycle, correct direction, correct multiples, final value with sign and units.
Bond enthalpy method: ΔH = Σ(bonds broken) − Σ(bonds formed) — breaking bonds is endothermic, forming is exothermic. Example: H₂(g) + Cl₂(g) → 2HCl(g) with H–H = 436, Cl–Cl = 243, H–Cl = 431 kJ/mol: ΔH = (436 + 243) − (2 × 431) = −183 kJ/mol. Limitations (standard explain question): mean bond enthalpies are averages over many different compounds containing that bond, and apply strictly only to gaseous molecules, so they ignore the specific molecular environment — calculations from ΔHf° data are more accurate because they are measured for the exact substance under standard conditions. If any reactant or product is liquid/solid, you cannot use mean bond enthalpies without an extra atomisation step.
Born-Haber Cycles, Lattice Enthalpy and Entropy
Lattice enthalpy of formation: the enthalpy change when one mole of solid ionic lattice forms from its gaseous ions under standard conditions — always exothermic (e.g. Na⁺(g) + Cl⁻(g) → NaCl(s), −787 kJ/mol). The lattice dissociation enthalpy is the reverse, always endothermic. It cannot be measured directly, so a Born-Haber cycle is used: ΔHf° = ΔHat(metal) + ΣIE(metal) + ΔHat(non-metal) + ΣEA(non-metal) + ΔHlatt. Each stage: atomise the metal (endothermic), ionise it (IEs endothermic, larger for each successive electron), atomise the non-metal, add electrons (first EA usually exothermic; second EA always endothermic — energy needed to force an electron onto an already negative ion), then form the lattice.
Worked example: calculate lattice enthalpy of NaCl given ΔHf°(NaCl) = −411, ΔHat(Na) = +107, ΔHat(Cl) = +122, IE₁(Na) = +496, EA₁(Cl) = −349 kJ/mol: −411 = 107 + 122 + 496 − 349 + ΔHlatt, so ΔHlatt = −411 − 376 = −787 kJ/mol. Sign discipline is everything — state whether the answer is formation (negative) or dissociation (positive) enthalpy.
Comparing lattice enthalpies: more exothermic when ions are smaller and more highly charged (MgO >> NaCl). A gap between the theoretical lattice enthalpy (calculated assuming a perfect ionic model) and the experimental Born-Haber value indicates covalent character caused by polarisation of the anion — larger, more highly charged anions (e.g. AgI) polarise more, showing the purely ionic model fails.
Enthalpy of solution: ΔHsol = −ΔHlatt(dissociation) + ΣΔHhyd — equivalently ΔHlatt(formation) + ΣΔHhyd. Hydration enthalpy is always exothermic and more exothermic for smaller, higher-charge ions (Li⁺ most exothermic in Group 1, decreasing down the group). Dissolving is feasible when ΔHsol is exothermic, or when TΔS outweighs an endothermic ΔHsol.
Gibbs free energy: ΔG = ΔH − TΔS (T in kelvin, ΔS converted to kJ K⁻¹ mol⁻¹). Reaction thermodynamically feasible when ΔG ≤ 0. Example: CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ/mol, ΔS = +161 J K⁻¹ mol⁻¹: feasible when T ≥ ΔH/ΔS = 178000/161 = 1106 K (833 °C). Entropy increases on: melting, boiling (large), dissolving gases out of solution, more gas moles produced. Feasible ≠ spontaneous — rate and activation energy still matter (kinetic control).
Rates: Collision Theory and Maxwell-Boltzmann
Collision theory: reactions occur when molecules collide with at least the activation energy (Ea) and with correct orientation — only then is a collision effective. Factors increasing rate: concentration/pressure — more molecules per unit volume, more frequent collisions; temperature — molecules move faster, more frequent collisions and (much more importantly) a far higher proportion exceed Ea; surface area (solids) — more exposed particles; catalyst — alternative route with lower Ea.
The Maxwell-Boltzmann distribution shows the number of molecules with a given energy: starts at the origin (no molecules with zero energy), peaks at the most probable energy, and never touches the x-axis at high energy. When temperature rises, the peak shifts right and flattens — the total area (number of molecules) is unchanged, but the area beyond Ea increases disproportionately, giving an exponentially faster reaction. Drawing marks: label axes, draw the Ea line, shade/hatch the extra area beyond Ea at the higher temperature. With a catalyst, draw a lower Ea line — a much larger area of the unchanged distribution now exceeds it.
Required practical (rates): follow reaction rate by measuring gas volume (gas syringe), mass loss (balance) or colour change/turbidity (disappearing cross), then vary concentration or temperature while keeping everything else constant. Classic experiment: Mg + HCl, or Na₂S₂O₃ + HCl with the disappearing cross. Control variables, plot rate against concentration, repeat and average; state the main error sources (reaction starts before sealing/inserting, timing of the endpoint is subjective) and improvements (bung ready, data-logger, fixed cross distance).
Rate Equations, Orders and the Arrhenius Equation
Rate equations must be determined experimentally — orders cannot be deduced from the equation: rate = k[A]m[B]n; overall order = m + n. Zero order: rate independent of [A] (horizontal concentration–time graph). First order: rate ∝ [A]; concentration–time graph shows constant half-life — t½ = ln2/k; ln[A] = −kt + ln[A]₀ gives a straight line when ln[A] is plotted against t. Second order: rate ∝ [A]²; 1/[A] vs t is linear; half-life doubles as concentration halves.
Determining orders: the initial-rates method — vary one concentration while holding others constant and compare initial rates (e.g. from tangents at t = 0 or the disappearing-cross method). If doubling [A] doubles the rate, order 1 in A; quadruples, order 2; no change, order 0. Once orders are known, substitute concentrations and rates into the rate equation to calculate k with units: s⁻¹ for first order, mol⁻¹ dm³ s⁻¹ for second, mol⁻² dm⁶ s⁻¹ for third overall. Clock reactions approximate initial rate using the time for a fixed small amount of product to appear (rate ∝ 1/t).
The Arrhenius equation: k = Ae−Ea/RT. Rearranged: ln k = −Ea/RT + ln A, so plotting ln k against 1/T gives a straight line with gradient −Ea/R and intercept ln A. Exam calculations: given two rate constants at two temperatures, use ln(k₂/k₁) = −Ea/R × (1/T₂ − 1/T₁). A typical answer: Ea = R × gradient × (−1), with R = 8.31 J K⁻¹ mol⁻¹ and temperatures in kelvin. Increasing temperature raises k because a greater proportion of molecules exceed Ea — the exponential term is the key explanation.
Mechanisms and rate equations connect at A2: in a multi-step mechanism, the rate-determining (slowest) step controls the rate equation; species in the rate-determining step (and species feeding into it via fast pre-equilibrium) appear in the rate equation. If rate = k[(CH₃)₃CBr], the rate-determining step involves only the halogenoalkane — evidence for an SN1 mechanism (see Organic Chemistry: Mechanisms).
💡 Worked Example
Exam-style: For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹. Predict and explain the effect on the equilibrium yield of ammonia of (a) increasing pressure, (b) increasing temperature, (c) adding an iron catalyst.
Mark-scheme logic: (a) 4 mol gas on the left → 2 mol on the right; increasing pressure shifts equilibrium to the side with fewer gas moles, so yield increases. (b) The forward reaction is exothermic, so heating shifts equilibrium in the endothermic (reverse) direction — yield falls. (c) A catalyst lowers Ea and speeds forward and reverse reactions equally, so equilibrium is reached faster but the yield/position is unchanged. Mark points: mole count comparison; Le Chatelier reasoning; exothermic direction logic; catalyst has no effect on position. (AQA 3.1.6.1-3.1.6.2; OCR 5.1.2)
❓ Practice Questions & Mark Schemes
Calculate ΔH for H₂(g) + Cl₂(g) → 2HCl(g) given mean bond enthalpies H–H = 436, Cl–Cl = 243, H–Cl = 431 kJ/mol.
Answer: ΔH = Σ(bonds broken) − Σ(bonds formed) = (436 + 243) − (2 × 431) = 679 − 862 = −183 kJ/mol. Exothermic: more energy released forming the two H–Cl bonds than needed to break H–H and Cl–Cl. (Standard enthalpy of combustion is defined in the definitions section: one mole burns completely in oxygen under standard conditions, always exothermic.)
Why are calculations from mean bond enthalpies less accurate than those from enthalpy of formation data?
Answer: Mean bond enthalpies are averages over many compounds containing the same bond and apply strictly to gaseous molecules, so they ignore the specific molecular environment. ΔH_f° values are experimental measurements for the exact compound under standard conditions, so more accurate.
Explain the trend in hydration enthalpy down Group 1.
Answer: Hydration enthalpy becomes less exothermic down the group: ionic radius increases and charge stays +1, so charge density falls and the attraction between the ion and polar water molecules is weaker.
Rate = k[A][B]². If [A] is doubled and [B] is tripled, by what factor does the rate increase?
Answer: 2 × 3² = 18. First order in A, second order in B, overall third order.
The half-life of a first-order reaction is 10 minutes. Calculate k.
Answer: t½ = ln2/k → k = 0.693/10 = 0.0693 min⁻¹ (= 1.16 × 10⁻³ s⁻¹). Constant half-life independent of initial concentration is the hallmark of first order.
For CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ/mol and ΔS = +161 J K⁻¹ mol⁻¹. Calculate the temperature above which decomposition is feasible.
Answer: Feasible when ΔG ≤ 0 → T ≥ ΔH/ΔS = 178000/161 = 1106 K (833 °C). Convert ΔS to kJ before dividing.