You need from the core lesson

• Rate = change in concentration per unit time, in mol dm⁻³ s⁻¹.
• Particles react only if they collide with E ≥ Ea and the right orientation.
• K = products over reactants, each concentration raised to its coefficient.
• Only temperature changes K.

Mission 1: Why doesn't adding more bromine speed up the bromination of propanone?

Elementary steps

• Most reactions happen in a series of elementary steps: single collisions or break-ups.
• The sequence of steps is the mechanism.
• The steps must add up to the overall equation.
• They must also fit the rate equation found by experiment.
• The slowest step is the rate-determining step (RDS).

Intermediates and transition states

• An intermediate forms in one step and is used up in a later one.
• It sits in an energy minimum, so it can, in principle, be detected.
• A transition state sits at an energy maximum: bonds partly made, partly broken.
• A transition state cannot be isolated.
• Intermediates cancel when the steps are added, so they never appear in the overall equation.

Molecularity

• Molecularity: the number of reacting particles in one elementary step.
• Unimolecular: one particle. (CH₃)₃CBr → (CH₃)₃C⁺ + Br⁻.
• Bimolecular: two. (CH₃)₃C⁺ + OH⁻ → (CH₃)₃COH.
• Termolecular: three at once. Rare, because three particles seldom collide together.
• Count particles, not species: 2X → X₂ is bimolecular. Never apply it to an overall equation.

How the mechanism sets the rate equation

• The rate equation contains the reactants of the RDS and of any fast step before it.
• If the slow step is first, the orders match the particles in that step.
• SN1 example: slow first step, so rate = k[(CH₃)₃CBr]. OH⁻ reacts later: zero order.
• Intermediates never appear: replace one by the reactants that formed it.
• Species that react only after the RDS do not appear.

When the slow step is not first

• Overall: P + 2Q → PQ₂.
• Step 1: P + Q ⇌ PQ (fast). Step 2: PQ + Q → PQ₂ (slow).
• The intermediate PQ is made from P and Q, so [PQ] is proportional to [P][Q].
• Rate = k[PQ][Q] becomes rate = k[P][Q]²: third order overall.
• The slow step is bimolecular, but the reaction is third order. The two need not match.

Energy profile of a multistep reaction

• One maximum (transition state) per step; one minimum per intermediate.
• Each step's Ea runs from the species entering that step up to its own maximum.
• A later step starts from the intermediate's dip, not from the reactants.
• The step with the largest Ea (usually also the highest peak) is rate-determining.
• From kinetic data: decide which step is slow, then draw its climb as the larger one.

Mission 2: Double one reactant and the rate quadruples; double another and nothing happens. What does that tell you?

The rate equation

• rate = k[A]ᵐ[B]ⁿ. m and n are the orders with respect to A and B.
• Overall order = m + n.
• Orders are found by experiment. They need not match the coefficients.
• Usually only reactants appear, plus sometimes a catalyst. Products do not normally appear.

What each order does

• When two concentrations change together, multiply their factors. Never add them.

Orders from initial rates (worked example)

For X + Y → Z at constant temperature:

• Compare experiments where only one concentration changes.
• 1 → 2: [X] × 3, rate × 3. First order in X.
• 2 → 3: [Y] × 2, rate × 2. First order in Y.
• rate = k[X][Y]: second order overall.

Units of k

• k = rate ÷ (concentration terms). Work out units by cancelling, not memorizing.
• Second order: (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)² = mol⁻¹ dm³ s⁻¹.
• Zero: mol dm⁻³ s⁻¹. First: s⁻¹. Third: mol⁻² dm⁶ s⁻¹.
• k depends on temperature, not on concentration.

Mission 3: Why does a medicine's level in the blood halve every few hours, however high it starts?

Concentration–time graphs

• Zero order: straight line sloping down. The rate is constant.
• First order: a curve. [A] falls by the same fraction in equal times.
• Second order: steeper at first, then flattens more. Each halving takes longer than the last.
• The rate at any moment is the gradient of the tangent (made positive).

Rate–concentration graphs

• Zero order: a horizontal line. Rate does not depend on [A].
• First order: a straight line through the origin. Gradient = k.
• Second order: a curve rising more and more steeply from the origin.
• Rate against [A]² is a straight line for second order.
• Don't carry the concentration–time shape across: these are different graphs.

Mission 4: Could you measure an activation energy with a cricket?

k rises with temperature

• k is constant only at one temperature. It rises as T rises.
• Arrhenius equation: k = Ae^(−Ea/RT) (in the data booklet).
• e^(−Ea/RT) is the fraction of collisions with E ≥ Ea.
• Ea in J mol⁻¹, R = 8.31 J K⁻¹ mol⁻¹, T in kelvin.

The straight-line form

• Take natural logs: ln k = −Ea/RT + ln A (in the data booklet).
• That is y = mx + c, with y = ln k and x = 1/T.
• Gradient = −Ea/R, so Ea = −gradient × R.
• y-intercept = ln A.
• The line slopes down: higher T means smaller 1/T and larger ln k.

Ea and A from a graph (worked example)

• Two points on the line (k in s⁻¹): (3.00 × 10⁻³ K⁻¹, −6.00) and (3.40 × 10⁻³ K⁻¹, −9.60).
• Gradient = (−9.60 − (−6.00)) ÷ (0.40 × 10⁻³) = −9.00 × 10³ K.
• Ea = 9.00 × 10³ × 8.31 = 7.48 × 10⁴ J mol⁻¹ = 74.8 kJ mol⁻¹.
• The intercept is far off the graph, so use a point: ln A = −6.00 + 9.00 × 10³ × 3.00 × 10⁻³ = 21.0.
• A = e²¹·⁰ = 1.3 × 10⁹ s⁻¹ (A has the units of k).

Two temperatures, no graph

• Write ln k = −Ea/RT + ln A at T₁ and T₂, then subtract: ln A cancels.
• ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂).
• Use natural logs and kelvin. Ea comes out in J mol⁻¹.
• Convert to kJ mol⁻¹ only at the end.
• Not a booklet equation: write the linear form twice and subtract. With more than two points, use the graph.

What A means

• A, the Arrhenius factor, accounts for the frequency of collisions with the proper orientation.
• A stricter orientation requirement means a smaller A.
• A is not the fraction with enough energy: that is e^(−Ea/RT).
• Find A from the intercept, or from one point: ln A = ln k + Ea/RT.

Mission 5: A sealed flask looks unchanging. Can you tell which way it will react?

The reaction quotient, Q

• Q has the same form as K, but uses concentrations at any moment.
• Q < K: too little product. Net forward reaction until Q = K.
• Q > K: too much product. Net backward reaction.
• Q = K: at equilibrium; no net change.
• Divide amounts by volume first. Moles work only if the volumes cancel.

Q in action (worked example)

• CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O, all in one liquid mixture; K ≈ 4.0.
• One mixture holds 1.0 mol acid, 1.0 mol ethanol, 1.5 mol ester and 1.5 mol water.
• Two terms on top and two below, so the volume cancels: Q = (1.5 × 1.5) ÷ (1.0 × 1.0) = 2.25.
• Q < K: net forward reaction; more ester forms.

ICE tables (worked example)

For A(g) + B(g) ⇌ 2C(g), starting with [A] = [B] = 0.50 mol dm⁻³. At equilibrium, [C] = 0.60 mol dm⁻³.

• Changes follow the coefficients: 0.60 of C needs 0.30 each of A and B.
• K = [C]² ÷ ([A][B]) = 0.36 ÷ 0.040 = 9.0.

When K is very small

• Very little reactant reacts, so [reactant]eqm ≈ [reactant]initial.
• Then solve for the product directly. No quadratic needed.
• Z₂(g) ⇌ 2Z(g), K = 4.0 × 10⁻⁸, [Z₂]initial = 0.10 mol dm⁻³.
• [Z]² = K × [Z₂] = 4.0 × 10⁻⁸ × 0.10 = 4.0 × 10⁻⁹, so [Z] = 6.3 × 10⁻⁵ mol dm⁻³.
• Check: only 3.2 × 10⁻⁵ mol dm⁻³ of the 0.10 mol dm⁻³ Z₂ reacted. The approximation holds.

K and ΔG⦵

• ΔG⦵ = −RT ln K (in the data booklet). ΔG⦵ in J mol⁻¹, T in kelvin.
• ΔG⦵ < 0 → K > 1, products favoured. ΔG⦵ > 0 → K < 1. ΔG⦵ = 0 → K = 1.
• Rearranged: K = e^(−ΔG⦵/RT).
• Worked: K = 1.0 × 10³ at 298 K. ΔG⦵ = −8.31 × 298 × ln(1.0 × 10³) = −1.71 × 10⁴ J mol⁻¹ = −17.1 kJ mol⁻¹.

The big ideas

• The rate equation reflects the slow step and any fast steps before it. Orders come from experiment.
• Intermediates sit in dips; transition states sit at peaks. The largest climb is the slow step.
• Orders 0, 1, 2: doubling gives rate × 1, × 2, × 4. Units of k come from cancelling.
• ln k against 1/T: gradient −Ea/R, intercept ln A.
• Q < K: net forward. Q > K: net backward. ICE rows follow the coefficients.
• ΔG⦵ = −RT ln K: negative ΔG⦵ means K > 1.