IB Chemistry · Unit 4 · Reactivity

How fast,
and why.

Some reactions take microseconds. Some take centuries. The same molecular collisions, with very different consequences.

Lessons HL extension Key terms
9Lessons
3HL extensions
42Key terms
SL+HLLevel
Concentration Time rate
Unit 4 · Standard Level

6 lessons to work through.

The required syllabus content for Unit 4, in order. Each card is one lesson-sized checkpoint.

Lesson 1

Rate of Reaction

Unit 1 – Atomic structure and introduction to moles

Lesson 2

Thiosulphate Practical

Unit 1 – Atomic structure and introduction to moles

Lesson 3

Iodine Clock

Unit 1 – Atomic structure and introduction to moles

Lesson 4

Collision Theory

Unit 1 – Atomic structure and introduction to moles

Lesson 5

Maxwell Boltzmann Distribution

Unit 1 – Atomic structure and introduction to moles

Lesson 6

Catalysts and Rate

Unit 1 – Atomic structure and introduction to moles

Lessons in detail

The unit, lesson by lesson.

Each lesson card below mirrors the original teacher deck — syllabus refs, content, worked examples and practice questions in order.

Lesson 1

Rate of reaction

Reactivity 2.2.1

Lesson outcomes

The rate of reaction measures how fast concentration changes over time. Standard units: mol dm⁻³ s⁻¹.

For reactants, [X] decreases — write a negative sign (or take absolute value). For products, [X] increases.

Why divide by stoichiometric coefficient?

If the mole ratio is not 1:1, the rate measured for each species differs. To get a single overall rate, divide each species' rate by its coefficient. For aA + bB → cC + dD:

rate = −(1/a)·d[A]/dt = −(1/b)·d[B]/dt = +(1/c)·d[C]/dt = +(1/d)·d[D]/dt

Following the reaction experimentally

Average vs instantaneous rate

Average rate = Δ[X] / Δt over a time interval — a chord on the concentration-vs-time graph.
Instantaneous rate = gradient of the tangent to the curve at that moment.
Initial rate = instantaneous rate at t = 0, when no products are yet interfering.

Worked example

Rates with non-1:1 stoichiometry

Problem. 2 N₂O(g) → 2 N₂(g) + O₂(g). Over 10 s, [N₂O] decreases by 0.20 mol dm⁻³ and [N₂] increases by 0.20 mol dm⁻³. Find (a) rate with respect to each species, (b) the single overall rate.
Solution. (a) rate w.r.t. N₂O = 0.20/10 = 0.020 mol dm⁻³ s⁻¹. rate w.r.t. N₂ = 0.020. rate w.r.t. O₂ = 0.20/2/10 = 0.010 mol dm⁻³ s⁻¹.
(b) Overall: divide each by its coefficient. (1/2)(0.020) = 0.010, (1/2)(0.020) = 0.010, (1/1)(0.010) = 0.010 — all agree: rate = 0.010 mol dm⁻³ s⁻¹.
Worked example

Rate of H₂ evolution from Mg + HCl

Problem. Mg + 2 HCl → MgCl₂ + H₂. 0.100 dm³ of 0.250 mol dm⁻³ HCl reacts with excess Mg. After 10 s, 24.0 cm³ H₂ collected. After 30 s (complete), 60.0 cm³ collected. Calculate rate of reaction w.r.t. H₂ (a) 0–10 s, (b) full reaction. Assume STP, V unchanged.
Solution. V at STP: 22.7 dm³ mol⁻¹.
(a) 24.0 cm³ = 24.0/22 700 = 1.057 × 10⁻³ mol. [H₂] in 0.100 dm³ = 1.057 × 10⁻² mol dm⁻³. Rate = 1.057 × 10⁻²/10 = 1.06 × 10⁻³ mol dm⁻³ s⁻¹.
(b) 60.0 cm³ = 2.643 × 10⁻³ mol. [H₂] = 2.643 × 10⁻² mol dm⁻³. Rate = 2.643 × 10⁻²/30 = 8.81 × 10⁻⁴ mol dm⁻³ s⁻¹.
The rate decreases because [HCl] decreases as it is consumed.

Try these

  1. Starter: What is a 'rate' in science? Give a mathematical definition. Provide an example.
    Show answer
    Rate = the change of a quantity per unit time. Mathematically: rate = Δ(quantity)/Δt. Examples: speed (rate of change of distance), rate of reaction (change of [reactant or product] per second), heart rate, growth rate.
  2. Why might the rate of a chemical reaction need to be controlled?
    Show answer
    Safety (runaway reactions can cause explosions); product quality (slow reactions allow side reactions); economics (industrial throughput); selectivity (controlling which product forms in a competitive system).
  3. List five experimental methods to follow a reaction rate.
    Show answer
    (1) Mass change (gas evolved, on a balance). (2) Gas volume (gas syringe). (3) Pressure (sealed gas reactions). (4) pH change (acid-base reactions). (5) Colour change / absorbance (colorimeter/spectrophotometer). Plus conductivity for ionic species.
  4. For H₂O₂ → 2 H₂O + O₂, you have [H₂O₂] vs t data. Describe how you would find the instantaneous rate at t = 60 s.
    Show answer
    Plot [H₂O₂] (y) vs time (x). Draw a tangent to the curve at t = 60 s. The gradient (Δy/Δx) is the instantaneous rate of consumption of H₂O₂.
  5. Why is the initial rate (at t = 0) particularly useful for kinetics studies?
    Show answer
    At t = 0 no products exist, so reverse/side reactions cannot interfere. The rate depends only on the starting concentrations — exactly what's needed to determine reaction orders.
  6. For the H₂O₂ → 2 H₂O + O₂ data: at 60 s the gradient is −2.5 × 10⁻⁴ mol dm⁻³ s⁻¹ and at 120 s it is −1.2 × 10⁻⁴ mol dm⁻³ s⁻¹. Explain why the rate decreased over time.
    Show answer
    As the reaction proceeds, [H₂O₂] decreases. Lower [reactant] = fewer particles per unit volume = fewer effective collisions per second = lower rate. By half-way, [H₂O₂] has fallen so the rate is too.
Lesson 2-3

Thiosulphate and iodine clock practicals

Reactivity 2.2.1Tool 1

Lesson outcomes

Thiosulphate · the disappearing cross

S₂O₃²⁻(aq) + 2 H⁺(aq) → S(s) + SO₂(aq) + H₂O(l). Cloudy yellow sulfur precipitates and obscures a cross drawn under the flask. Time how long it takes for the cross to vanish at a fixed observer position. The reciprocal of that time, 1/t, is proportional to the average rate.

Iodine clock

Two competing reactions: H₂O₂ + 2 I⁻ + 2 H⁺ → I₂ + 2 H₂O (slow) and I₂ + 2 S₂O₃²⁻ → 2 I⁻ + S₄O₆²⁻ (fast). Starch indicator turns blue-black the instant all the thiosulphate has been consumed and free I₂ accumulates. Time to colour change vs concentration of one reactant gives the reaction order.

Common improvements

Try these

  1. In the disappearing-cross experiment, why is plotting 1/t vs [S₂O₃²⁻] sensible?
    Show answer
    If rate ∝ [S₂O₃²⁻], then 1/t ∝ rate ∝ [S₂O₃²⁻]. A straight line through the origin confirms first-order behaviour with respect to thiosulphate.
  2. Why is timing the iodine clock with a stopwatch acceptable, even though the disappearing-cross is criticised for stopwatch use?
    Show answer
    The iodine-clock endpoint is sharp (starch turns from colourless to deep blue-black instantly), so reaction time uncertainty is small. The disappearing-cross endpoint is subjective — the cross fades gradually.
Lesson 4

Collision theory · factors affecting rate

Reactivity 2.2.2Reactivity 2.2.3Reactivity 2.2.4

Lesson outcomes

Collision theory says a reaction can only occur if reactant particles:

  1. Collide (they must come into contact).
  2. Have the correct orientation at the moment of collision.
  3. Have combined kinetic energy ≥ activation energy Ea.

The five rate-affecting factors

Reaction profile

A graph of energy (y) against reaction progress (x). Reactants on the left, products on the right, with a "hill" at the transition state. The hill height from reactants is Ea; the net difference between products and reactants is ΔH.

Try these

  1. Starter: state the three requirements for a reaction to occur according to collision theory.
    Show answer
    (1) Particles must physically collide (be in contact). (2) Particles must collide with correct orientation (geometry). (3) Particles must collide with energy ≥ Ea (activation energy).
  2. Are there any exceptions to the orientation requirement?
    Show answer
    Yes — for very small or spherical species (individual atoms, ions, diatomic molecules), orientation matters less because there's no 'wrong way round' to hit. Larger and more asymmetric molecules have a smaller fraction of favourable orientations.
  3. How does collision theory differ from kinetic molecular theory?
    Show answer
    KMT describes general particle behaviour (gas phase, ceaseless random motion, elastic collisions, no IMFs). Collision theory applies the same idea specifically to reactions — particles must collide with enough energy and right orientation to react. Collision theory works in any phase (gas, liquid, solution); KMT is mainly for gases.
  4. How do each of the following affect rate? (a) concentration, (b) pressure, (c) temperature, (d) surface area of a solid.
    Show answer
    (a) ↑ concentration → ↑ rate: more particles per unit volume → more collisions per second. (b) ↑ pressure (gases) → ↑ rate: equivalent to higher concentration. (c) ↑ T → ↑ rate: more molecules with E ≥ Ea (exponentially more, see M-B). (d) ↑ surface area (solid) → ↑ rate: more particles exposed for collision.
  5. Would a heterogeneous reaction (e.g. powder + liquid) or a homogeneous reaction (e.g. solution + solution) react faster?
    Show answer
    Depends — but homogeneous reactions generally proceed faster in solution because all particles are dispersed and can mix freely. Heterogeneous needs the species to find each other at a phase boundary. (Industrially, heterogeneous catalysts compensate by maximising surface area.)
  6. Is a mixture of oil and water homogeneous or heterogeneous?
    Show answer
    Heterogeneous — they form two visible immiscible layers. Mixing energetically forces emulsion but they separate again.
  7. Why is the activation energy needed even when particles collide?
    Show answer
    The collision must break existing bonds before new bonds can form. The kinetic energy of the collision goes into stretching bonds in the transition state — only collisions with Ek ≥ Ea can clear the energy barrier.
  8. Why does heterogeneous catalysis (solid catalyst, gas reactants) generally proceed faster on a powdered catalyst than on a lump?
    Show answer
    Surface area. The reaction occurs at the surface of the solid catalyst (adsorption sites). Powdering exposes many more sites per unit mass.
  9. On a single graph, sketch the curve of V(O₂) vs t for 2 H₂O₂ → 2 H₂O + O₂. Then add three curves showing (a) increased [H₂O₂], (b) lowered T, (c) added MnO₂ catalyst.
    Show answer
    (a) Steeper initial gradient, larger maximum volume (more product overall). (b) Shallower gradient, same maximum. (c) Much steeper gradient, same maximum (catalysts don't change yield — only rate).
  10. Sketch reaction profile diagrams for both an exothermic and endothermic reaction. Label reactants, products, ΔH, and Ea. What is the maximum of the curve called?
    Show answer
    Exo: products below reactants. Endo: products above reactants. Both show a hump (the transition state). The peak is the transition state (or activated complex) — short-lived high-energy arrangement where old bonds are breaking and new ones forming.
  11. Compare the relative stabilities of reactants, transition state and products for an exo- vs endothermic reaction.
    Show answer
    Both: transition state is least stable (highest energy). Exo: products more stable than reactants. Endo: reactants more stable than products. Stability ∝ −energy.
Lesson 5

Maxwell-Boltzmann distribution

Reactivity 2.2.4

Lesson outcomes

Particles in a sample don't all have the same kinetic energy — they have a distribution, called the Maxwell-Boltzmann distribution. Average kinetic energy is proportional to the absolute temperature.

Key features

Effect of raising T

The curve broadens and flattens, with the peak moving right. The area to the right of Ea grows dramatically — even a modest T rise can double the rate. Ea itself does not change with T.

Effect of a catalyst

The catalyst lowers Ea (shifts the dashed vertical line left). More of the existing distribution now lies above Ea — same T, more successful collisions. The M-B curve itself is unchanged.

Try these

  1. What three things does the Maxwell-Boltzmann distribution tell you?
    Show answer
    (1) The total area under the curve = total number of particles. (2) The peak gives the most probable energy. (3) The fraction to the right of Ea gives the proportion of particles with sufficient energy to react.
  2. Sketch how the M-B curve changes when T is raised. What happens to the peak height, peak position, area, and area beyond Ea?
    Show answer
    Peak height: decreases (curve flattens). Peak position: shifts right (higher most-probable energy). Total area: unchanged (same number of particles). Area beyond Ea: increases significantly (more particles have enough energy → higher rate).
  3. Does Ea change with temperature?
    Show answer
    No. Ea is a property of the reaction (the bond rearrangement involved). T changes the distribution of kinetic energies but not the activation energy.
  4. Why does raising T from 20 °C to 30 °C often double the rate of a reaction, even though the average KE only rises by ~3%?
    Show answer
    It's not the average that matters — it's the fraction above Ea. The high-energy tail of the distribution grows exponentially with T (Boltzmann factor e^{−Ea/RT}). A small T rise can substantially shift this tail.
  5. True or false: a catalyst increases the average kinetic energy of the particles.
    Show answer
    False. The catalyst lowers Ea, not the energy of the particles. T is unchanged, so the M-B distribution is unchanged. The vertical Ea line shifts left → more of the existing distribution lies beyond it.
  6. On a single M-B graph, draw two curves for the same gas at T₁ = 300 K and T₂ = 400 K. Mark Ea and shade the regions of 'successful' collisions for each temperature.
    Show answer
    T₂ curve is flatter and shifted right vs T₁. The shaded region (E ≥ Ea) is much larger for T₂ — demonstrating why ↑T → ↑rate.
Lesson 6

Catalysts and rate

Reactivity 2.2.5

Lesson outcomes

A catalyst is a substance that increases the rate of a reaction by providing an alternative pathway with a lower activation energy. It is regenerated in a later step, so it is not consumed overall.

Two flavours

Key facts about catalysts

Try these

  1. What is a catalyst? What are the requirements for a substance to be a catalyst?
    Show answer
    A substance that increases the rate of a reaction by providing an alternative pathway with a lower Ea, without itself being consumed overall. Requirements: must be regenerated by the end of the reaction (used in one step, reformed in another); must lower Ea for the rate-determining step; must not appear in the overall equation.
  2. Sketch a reaction profile for both an exothermic and endothermic reaction, showing curves with and without a catalyst.
    Show answer
    Exothermic: two humps with same start and end heights. Catalysed curve has a lower peak (lower Ea). ΔH unchanged. Endothermic: same logic — both curves have products higher than reactants; catalysed has lower peak.
  3. Using your knowledge of the Maxwell-Boltzmann distribution, explain how a catalyst speeds up a reaction.
    Show answer
    The catalyst lowers Ea — i.e. shifts the vertical Ea line to the left on the M-B graph. The total area is unchanged, but the area beyond the new (lower) Ea is larger → more particles have enough energy to react at the same T → higher rate.
  4. How would you confirm experimentally that a substance is a true catalyst and not a reagent?
    Show answer
    After the reaction, recover the substance unchanged in chemical identity and mass. A catalyst can be used repeatedly without depletion.
  5. If a catalyst lowers Ea, must it lower it for both the forward and reverse reactions equally?
    Show answer
    Yes. The catalyst provides one alternative pathway that both directions use. The forward and reverse Ea values both decrease — by the same factor. This is why catalysts don't shift equilibrium.
  6. Why does a catalyst not change the position of equilibrium?
    Show answer
    It lowers Ea equally for forward and reverse reactions, so both rates increase by the same factor. The ratio of rates (and therefore K) is unchanged.
  7. Distinguish homogeneous and heterogeneous catalysts with examples.
    Show answer
    Homogeneous: same phase as the reactants. E.g. H₂SO₄ (aq) catalysing ester hydrolysis. Heterogeneous: different phase. E.g. Fe(s) catalysing the Haber process (N₂ + H₂ gases) or Pt in catalytic converters. Industrial scale prefers heterogeneous — easier to separate from products.
HL extension

Reaction
mechanisms,
exposed.

HL kinetics: rate laws derived from mechanisms, rate-determining steps, and the Arrhenius equation in its full glory.

HL only

+ 8: Reaction Order- HL ONLY

Unit 1 – Atomic structure and introduction to moles

HL only

Reaction Mechanisms - HL ONLY

Unit 1 – Atomic structure and introduction to moles

HL only

Arrhenius Equation - HL ONLY

Unit 1 – Atomic structure and introduction to moles

Lesson 7-8 HL only

Reaction orders and rate equations

Reactivity 2.2.9 (HL)Reactivity 2.2.10 (HL)Reactivity 2.2.11 (HL)

Lesson outcomes

  • Write a general rate equation rate = k[A]x[B]y.
  • Determine reaction orders experimentally from initial-rate data.
  • Recognise zero, first and second order behaviour from concentration-time and rate-time graphs.
  • Calculate the rate constant k and state its correct units.

The rate equation describes how rate depends on reactant concentrations:

rate = k [A]x [B]y

where k is the rate constant (specific to the reaction at a given T), and x, y are the orders with respect to A and B. The orders are not in general the stoichiometric coefficients — they must be determined experimentally.

Three common orders

OrderBehaviour[X] vs t graphHalf-life
ZeroDoubling [X] → rate unchangedStraight line (decreasing)Halves as [X] decreases
FirstDoubling [X] → rate doublesExponential decayConstant
SecondDoubling [X] → rate × 4Slower-than-exp. decayDoubles each time

Units of k

Depend on the overall order. Rearrange the rate equation: k = rate / ([A]x[B]y). Units = (mol dm⁻³ s⁻¹) / (mol dm⁻³)n where n is overall order:

  • Zero order overall → mol dm⁻³ s⁻¹.
  • First order overall → s⁻¹.
  • Second order overall → mol⁻¹ dm³ s⁻¹.
Worked example

Orders from initial-rate data

Problem. For 2 NO₂ + F₂ → 2 NO₂F, initial-rate data: [NO₂] = 0.10, [F₂] = 0.10, rate = 0.0026; [NO₂] = 0.20, [F₂] = 0.10, rate = 0.0052; [NO₂] = 0.10, [F₂] = 0.20, rate = 0.0052. Find the orders and the rate equation.
Solution. Order w.r.t. NO₂: doubling [NO₂] (Exp 1→2, [F₂] constant) doubles the rate → first order.
Order w.r.t. F₂: doubling [F₂] (Exp 1→3, [NO₂] constant) doubles the rate → first order.
Rate equation: rate = k [NO₂][F₂]. Overall order = 2.
k: from Exp 1, k = 0.0026 / (0.10 × 0.10) = 0.26 mol⁻¹ dm³ s⁻¹.
Worked example

Fe³⁺ + I⁻

Problem. 2 Fe³⁺ + 2 I⁻ → 2 Fe²⁺ + I₂. Initial-rate data give first order in Fe³⁺ and second order in I⁻. (a) Write the rate equation. (b) State overall order. (c) Calculate k if rate = 8.0 × 10⁻⁵ mol dm⁻³ s⁻¹ when [Fe³⁺] = 0.020 and [I⁻] = 0.010 (T = 298 K). State units.
Solution. (a) rate = k [Fe³⁺] [I⁻]². (b) Overall = 1 + 2 = 3.
(c) k = rate / ([Fe³⁺][I⁻]²) = 8.0 × 10⁻⁵ / (0.020 × 0.010²) = 8.0 × 10⁻⁵ / 2.0 × 10⁻⁶ = 40 mol⁻² dm⁶ s⁻¹.

Try these

  1. For 2 Fe³⁺ + 2 I⁻ → 2 Fe²⁺ + I₂: (a) Determine the order with respect to Fe³⁺ and I⁻ from initial-rate data showing first order in Fe³⁺ and second order in I⁻. (b) State the rate equation and overall order. (c) Calculate k if rate = 8.0 × 10⁻⁵ mol dm⁻³ s⁻¹ when [Fe³⁺] = 0.020 and [I⁻] = 0.010 at 298 K. State units.
    Show answer
    (a) 1st in Fe³⁺, 2nd in I⁻. (b) rate = k[Fe³⁺][I⁻]². Overall order = 1 + 2 = 3. (c) k = rate / ([Fe³⁺][I⁻]²) = 8.0 × 10⁻⁵ / (0.020 × 0.0001) = 8.0 × 10⁻⁵ / 2.0 × 10⁻⁶ = 40 mol⁻² dm⁶ s⁻¹.
  2. If a reaction is zero order in reactant X, how does the rate change when [X] is doubled? What graph of [X] vs t would you see?
    Show answer
    Rate unchanged. [X] vs t is a straight downward line (constant slope) until [X] runs out.
  3. What are the units of k for: (a) zero-order overall, (b) first-order overall, (c) second-order overall, (d) third-order overall?
    Show answer
    (a) mol dm⁻³ s⁻¹. (b) s⁻¹. (c) mol⁻¹ dm³ s⁻¹. (d) mol⁻² dm⁶ s⁻¹. Pattern: units = (mol dm⁻³)1−n · s⁻¹ where n is the overall order.
  4. Why must reaction orders be determined experimentally rather than read from the stoichiometric equation?
    Show answer
    The stoichiometric equation describes the overall transformation. The rate equation describes what happens in the rate-determining step, which may involve only some of the species in the overall equation. Orders depend on mechanism, not stoichiometry.
  5. Why does k change with temperature but the orders of reaction do not?
    Show answer
    k contains the Boltzmann factor e^(−Ea/RT) — exponentially sensitive to T. The orders are determined by which species collide in the rate-determining step, which is set by the mechanism, not by T (unless the mechanism itself changes).
Lesson 9 HL only

Reaction mechanisms · the rate-determining step

Reactivity 2.2.6 (HL)Reactivity 2.2.7 (HL)Reactivity 2.2.8 (HL)

Lesson outcomes

  • Define elementary step and molecularity.
  • Identify the rate-determining step (RDS) on a reaction-profile diagram.
  • Predict the rate equation from a proposed mechanism.
  • Identify intermediates (formed in one step, consumed in another).

An elementary step is a single molecular collision in a reaction. Most reactions involve several elementary steps; the sequence is the mechanism. The slowest step controls the overall rate — the rate-determining step (RDS).

Molecularity

Number of particles colliding in the elementary step:

  • Unimolecular — one species (decomposition, isomerisation).
  • Bimolecular — two species. Most common.
  • Termolecular — three species. Very rare (probability of three particles colliding simultaneously is tiny).

Rate equation from mechanism

If the RDS involves species X and Y, then rate ∝ [X][Y]. The orders in the rate equation match the molecularity of the RDS. Species in fast pre-equilibria can appear too (via substitution).

Intermediates are formed in one step and consumed in a later step; they do not appear in the overall equation.

Worked example

Slow first step

Problem. Mechanism for NO₂ + CO → NO + CO₂: Step 1 (slow): NO₂ + NO₂ → NO₃ + NO. Step 2 (fast): NO₃ + CO → NO₂ + CO₂. Deduce the rate equation.
Solution. RDS is step 1: NO₂ + NO₂ → products. Two NO₂ are involved, so rate = k [NO₂]². CO does not appear in the RDS, so it does not appear in the rate equation.
Rate = k [NO₂]² (overall order 2).
Note: NO₃ is an intermediate (made in step 1, consumed in step 2).
Worked example

Slow second step with fast pre-equilibrium

Problem. 2 NO + O₂ → 2 NO₂ via Step 1 (fast equilibrium): NO + NO ⇌ N₂O₂. Step 2 (slow): N₂O₂ + O₂ → 2 NO₂. Deduce the rate equation.
Solution. RDS is step 2: rate = k₂ [N₂O₂][O₂]. But N₂O₂ is an intermediate — we can't measure its concentration directly. Step 1 is at equilibrium, so [N₂O₂] ∝ [NO]². Substituting:
Rate = k [NO]² [O₂] (overall order 3).
The reactant from a fast pre-equilibrium step still appears in the rate equation — even though the rate-determining step is the slow second step.

Try these

  1. What is an elementary step? What is its 'molecularity'?
    Show answer
    An elementary step is a single molecular collision (a single step that proceeds in one go without intermediates). Its molecularity is the number of particles colliding in that step: unimolecular (1), bimolecular (2), termolecular (3 — rare).
  2. What is the probability of a termolecular elementary step occurring? Why is termolecular rare?
    Show answer
    Three-body collisions in the gas phase are very rare — the chance of three particles all being in the same place at the same time with right orientation and enough energy is vanishingly small. Most observed third-order reactions actually proceed via two-step mechanisms with fast pre-equilibrium.
  3. Funnel analogy: in groups, hold three different sized funnels stacked, pour water through. The smallest funnel determines flow. What chemistry concept does this model?
    Show answer
    The smallest funnel = rate-determining step. The flow rate is set by the slowest step, regardless of where it sits in the sequence. Other steps could be made faster but the overall rate won't improve until the bottleneck is widened.
  4. 2 NO₂Cl → 2 NO₂ + Cl₂. Mechanism: Step 1 (slow) NO₂Cl → NO₂ + Cl. Step 2 (fast) NO₂Cl + Cl → NO₂ + Cl₂. Deduce the rate equation and overall order.
    Show answer
    RDS is step 1, unimolecular in NO₂Cl. Rate = k[NO₂Cl]. Overall order = 1.
  5. For 2 ICl + H₂ → 2 HCl + I₂, the experimentally determined rate law is rate = k[ICl][H₂]. Propose a two-step mechanism consistent with this.
    Show answer
    Step 1 (slow): ICl + H₂ → HI + HCl. Step 2 (fast): HI + ICl → I₂ + HCl. Step 1 involves both reactants (consistent with the rate law); step 2 is fast and doesn't appear in the rate equation.
  6. Define an intermediate. Where do intermediates appear on a reaction profile diagram?
    Show answer
    A species formed in one elementary step and consumed in a later step. Doesn't appear in the overall equation. On a reaction profile: at the local minima between two humps (transition states) — they're real species (not transition states) but at higher energy than reactants/products.
  7. For 2 NO + O₂ → 2 NO₂: Step 1 (fast eq) NO + NO ⇌ N₂O₂. Step 2 (slow) N₂O₂ + O₂ → 2 NO₂. Which is the correct rate equation? (a) rate = k[NO]²[O₂], (b) k[NO]², (c) k[N₂O₂][O₂], (d) k[O₂].
    Show answer
    (a) rate = k[NO]²[O₂]. The RDS is step 2, giving rate ∝ [N₂O₂][O₂]. But N₂O₂ is an intermediate; via step 1's fast equilibrium, [N₂O₂] ∝ [NO]². Substituting → rate = k[NO]²[O₂]. The trap is that the intermediate's concentration must be replaced with the species it's derived from.
Lesson 10 HL only

The Arrhenius equation

Reactivity 2.2.12 (HL)

Lesson outcomes

  • State the Arrhenius equation k = A·e−Ea/RT.
  • Linearise: ln k vs 1/T gives a straight line of gradient −Ea/R.
  • Calculate Ea from a graph of ln k vs 1/T.
  • Calculate the activation energy from rate constants at two temperatures.

Empirically the rate constant rises sharply with temperature. The Arrhenius equation captures this:

k = A · e−Ea/RT

where A is the Arrhenius (frequency) factor — related to collision frequency and orientation. Ea is the activation energy. R = 8.31 J K⁻¹ mol⁻¹. T in K.

Linear form

Take the natural log of both sides:

ln k = ln A − Ea / (R T)

Plot ln k (y) against 1/T (x). The line has:

  • Gradient = −Ea / R. So Ea = −R × gradient.
  • y-intercept = ln A.

From two data points

If you have only k₁ at T₁ and k₂ at T₂:

ln(k₂ / k₁) = (Ea / R) × (1/T₁ − 1/T₂)

Worked example

E_a from two rate constants

Problem. k(298 K) = 1.0 × 10⁻⁴ s⁻¹; k(318 K) = 4.0 × 10⁻⁴ s⁻¹. Calculate Ea.
Solution. ln(4.0 × 10⁻⁴ / 1.0 × 10⁻⁴) = ln 4 = 1.386.
1/T₁ − 1/T₂ = 1/298 − 1/318 = 3.356 × 10⁻³ − 3.145 × 10⁻³ = 2.11 × 10⁻⁴ K⁻¹.
Ea = R × ln(k₂/k₁) / (1/T₁ − 1/T₂) = 8.31 × 1.386 / 2.11 × 10⁻⁴ = 54 600 J mol⁻¹ = 54.6 kJ mol⁻¹.

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  1. How does k change with T, using collision theory?
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    k increases sharply with T. Higher T → larger fraction of particles with E ≥ Ea (M-B distribution shifts right and broadens). Ea itself is unchanged by T.
  2. What is the name of the equation that links the activation energy to the rate constant via temperature?
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    The Arrhenius equation: k = A·e^(−Ea/RT). Named after Svante Arrhenius (1889).
  3. What is the significance of each variable in k = A·e^(−Ea/RT)?
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    k: rate constant. A: Arrhenius (frequency) factor, related to collision frequency and orientation; approximately T-independent. Ea: activation energy. R: gas constant (8.31 J K⁻¹ mol⁻¹). T: absolute temperature (K). The exponential factor e^(−Ea/RT) is the fraction of collisions with sufficient energy.
  4. Why does plotting ln k vs 1/T give a straight line?
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    Taking ln of k = A·exp(−Ea/RT) gives ln k = ln A − (Ea/R)(1/T). Setting y = ln k and x = 1/T gives y = c + mx with c = ln A and m = −Ea/R — a straight line.
  5. From an Arrhenius plot with gradient −5500 K, calculate Ea.
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    Gradient = −Ea/R, so Ea = −R × gradient = −8.31 × (−5500) = 45 705 J mol⁻¹ = 45.7 kJ mol⁻¹.
  6. If a reaction has a very high A but a very high Ea, what does this tell you about it?
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    High A means most collisions have correct orientation and frequency. High Ea means very few collisions have enough energy. The reaction is intrinsically slow at low T but accelerates rapidly with T.
  7. For a reaction with Ea = 50 kJ mol⁻¹, calculate the ratio of rate constants at 310 K and 300 K.
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    ln(k₃₁₀/k₃₀₀) = (Ea/R)(1/T₁ − 1/T₂) = (50 000/8.31)(1/300 − 1/310) = 6017 × (1.075 × 10⁻⁴) = 0.647. k₃₁₀/k₃₀₀ = e^0.647 = 1.91. Rate nearly doubles for a 10 K rise — consistent with the 'rate doubles every 10 K' rule of thumb.
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