Learners' AlleyDesigned by Khow Willard

Reaction Kinetics

1

Collision Theory

For a reaction to occur, particles must collide with

sufficient kinetic energy, greater than or equal to the activation energy (E_a)
the correct geometry (orientation)

Such a collision is an effective collision. The greater the frequency of effective collisions, the faster the rate. Usually at most two particles collide in one step.

DefinitionActivation energy: the minimum amount of energy reacting particles must possess for an effective collision to take place for reaction to occur.

Collision	Energy ≥ E_a?	Right orientation?	Result
effective	✓	✓	bonds break and form: products
ineffective	✗	✓	bounce apart
ineffective	✓	✗	bounce apart

2

Factors Affecting Rate

2.1Concentration, pressure and surface area

Concentration or pressure ↑

No. of reactant particles per unit volume ↑
→ frequency of collisions ↑, and hence frequency of effective collisions ↑ → rate ↑

Particle size of solid ↓

Larger surface area of contact between reactant particles
→ frequency of effective collisions ↑ → rate ↑

Rate is fastest at the start, when reactant concentrations are highest.

2.2The Boltzmann distribution curve

Features of the curve

Starts at the origin and never touches the energy axis
Area under the curve = total no. of particles
Shaded area = particles with energy ≥ E_a

How the curve changes when T increases

The peak shifts to the right, as the most probable energy increases
The curve flattens out: a larger spread of energies, with more particles at higher energies and fewer at lower energies
The area under the curve stays constant, as the total number of particles is constant

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2

Factors Affecting Rate (continued)

2.3Effect of temperature on rate

As the temperature increases, the average kinetic energy of the particles increases
→ a larger number of reactant particles have energy ≥ E_a (larger shaded area)
→ frequency of effective collisions ↑ → rate ↑ (k ↑). Rate roughly doubles for every 10 °C rise.

2.4Effect of a catalyst on rate

Provides an alternative reaction pathway with a lower E_a
→ more reactant particles possess energy ≥ E_a (larger shaded area)
→ frequency of effective collisions ↑ → rate ↑ (k ↑)

3

Rate of Reaction

DefinitionRate of reaction: the change in concentration of a reactant or product per unit time.

Measuring rate

rate = change in [reactant] or [product]time taken

Rate is always positive: [reactant] falls with time and [product] rises. Units: mol dm⁻³ s⁻¹.

Rates of different species are related by the mole ratio

BrO₃⁻ + 5Br⁻ + 6H⁺ → 3Br₂ + 3H₂O

Br⁻ is used up 5 × as fast as BrO₃⁻
Br₂ is formed 3 × as fast as BrO₃⁻ is used up

So "the rate" must say which species it refers to.

Instantaneous and initial rate

DefinitionInstantaneous rate: gradient of the tangent to the concentration–time graph at time t.

Initial rate: gradient of the tangent at t = 0, the highest rate, as [reactant] is at its maximum.

Topic 08 · Reaction Kinetics39

Learners' AlleyDesigned by Khow Willard

4

Energy Profile Diagrams

One stepone transition state

Two stepsintermediate in the dip, step 1 is the RDS

Heterogeneous catalystone lower hump

Homogeneous catalysttwo lower humps

Transition state: the unstable arrangement at the energy maximum, equally likely to form reactants or products
Intermediate: short-lived species formed in one step and used up in a later step

The rate-determining step (slow step) has the highest E_a
Higher E_a → slower reaction. Very high E_a: reactant is kinetically stable

5

Catalysis

DefinitionCatalyst: a substance that increases the rate of a chemical reaction by providing an alternative reaction pathway with lower activation energy but remains chemically unchanged at the end of the reaction.

It is specific
It is regenerated at the end of the reaction
It does not affect the overall stoichiometry
It does not affect the position of equilibrium or K

DefinitionHomogeneous catalysis: catalyst and reactants in the same phase.

DefinitionHeterogeneous catalysis: catalyst and reactants in different phases.

5.1Heterogeneous: adsorption theory

Adsorption: reactant particles form weak bonds to active sites on the catalyst surface, weakening the bonds within them
Reaction: reactants are held close together in the correct orientation
Desorption: weak bonds to the products break, products diffuse away and the active sites are freed

Process	Catalyst
Haber: N₂ + 3H₂ ⇌ 2NH₃	Fe
Contact: 2SO₂ + O₂ ⇌ 2SO₃	V₂O₅
Hydrogenation of alkenes	Ni
Catalytic converter: 2CO + 2NO → 2CO₂ + N₂	Pt, Pd, Rh

40Topic 08 · Reaction Kinetics

5

Catalysis (continued)

5.2Homogeneous: Fe²⁺ in the I⁻ / S₂O₈²⁻ reaction

Iodide ions are oxidised by peroxodisulfate ions. Without a catalyst this is very slow, adding a little Fe²⁺ speeds it up. 2I⁻(aq) + S₂O₈²⁻(aq) → I₂(aq) + 2SO₄²⁻(aq)

Why is the reaction slow, and how does Fe²⁺ help?

Without a catalyst, the reaction is slow: it has a high E_a, as a large amount of energy is needed to overcome the repulsion between the two negatively charged reactant ions
Step 1: 2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻
Step 2: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂ (Fe²⁺ regenerated)
Each step is between oppositely charged ions → lower E_a for each step

Oxides of nitrogen in the oxidation of atmospheric SO₂

NO₂ + SO₂ → SO₃ + NO
NO + ½O₂ → NO₂ (NO₂ regenerated)
SO₃ + H₂O → H₂SO₄, forming acid rain

5.3Autocatalysis

DefinitionAutocatalysis: catalysis in which one of the products of the reaction catalyses the reaction.

Slow at first: very little catalyst (product) present
Fast as more product, the catalyst, forms
Slow again as the reactants are used up

Example: MnO₄⁻ with ethanedioate ions

2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

The product Mn²⁺ catalyses the reaction, so the purple colour fades slowly at first, then quickly.

5.4Enzymes

Protein catalysts, highly specific (lock-and-key)
Low [S]: first order. High [S]: zero order, as all active sites are saturated
[S] in excess: rate ∝ [E]
pH or T beyond optimum changes the 3D structure and active site → denatured

rate vs [S]levels off: sites saturated

rate vs Toptimum 20–50 °C

rate vs pHoptimum pH 5–8

Topic 08 · Reaction Kinetics41

Learners' AlleyDesigned by Khow Willard

6

The Rate Equation

DefinitionRate equation: the experimentally determined relationship between rate and the concentrations of reactants.

DefinitionOrder of reaction w.r.t. a reactant: the power of its concentration term in the experimentally determined rate equation.

rate = k[A]^x[B]^y

Overall order = x + y

DefinitionRate constant, k: the proportionality constant in the rate equation.

6.1Meaning of order

Order	Rate	[A] × 2: rate
0	independent of [A]	× 1
1	rate ∝ [A]	× 2
2	rate ∝ [A]²	× 4

6.2What changes k?

Change	Rate	k
[reactant] ↑	↑ (if order ≠ 0)	no change
temperature ↑	↑	↑
add a catalyst	↑	↑

6.3Units of k units of rate ÷ (mol dm⁻³)^{overall order}

Overall order	0	1	2	3
Rate equation	rate = k	rate = k[A]	rate = k[A]²	rate = k[A]²[B]
Units of k	mol dm⁻³ s⁻¹	s⁻¹	mol⁻¹ dm³ s⁻¹	mol⁻² dm⁶ s⁻¹

7

Mechanisms and the Rate Equation

Rules for a reaction mechanism

Each elementary step involves at most two reacting particles
Steps add up to the overall equation
The rate equation contains the reactants in the rate-determining step (slow step), with orders = their coefficients in that step
Intermediates do not appear in the overall equation or the rate equation

7.1Proposing a mechanism

Worked example2NO(g) + F₂(g) → 2ONF(g) has rate = k[NO][F₂]. It takes place in two steps, and one step forms ONF and the radical F• in equimolar amounts. Suggest a mechanism.

Slow step contains one NO and one F₂, from the rate equation
Rate equation ≠ overall equation → not a single step, one more NO still to react

Step 1 (slow)NO + F₂ → ONF + F• Step 2 (fast)F• + NO → ONF

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Mechanisms and the Rate Equation (continued)

7.2Testing a mechanism

Worked exampleFor 2A + B → A₂B, the rate equation is found to be rate = k[A]². Which mechanism is consistent with it?

Mechanism 1	Mechanism 2
A + A → A₂ (slow)	A + B → AB (slow)
A₂ + B → A₂B (fast)	AB + A → A₂B (fast)

Mechanism 1: the slow step involves two A → predicted rate = k[A]², which matches the experimental rate equation
Mechanism 2: the slow step involves one A and one B → predicted rate = k[A][B], which does not match
∴ only mechanism 1 is consistent

8

Collecting Rate Data

	Continuous method	Initial rate method
Experiments	one: [reactant] followed over time	several: one concentration changed at a time
Data	concentration at intervals	time taken for a fixed change
Analysis	concentration–time graph, half-life	rate–concentration graph or the ratio method

8.1Physical measurements: a property that changes as the reaction goes

gas volumeMg + 2HCl, H₂ given off

mass lossCaCO₃ + 2HCl, CO₂ escapes

pHCH₃Br + OH⁻, pH falls

conductivityno. of ions changes

colorimetrybrown I₂ used up

pressure at fixed VN₂O₄ → 2NO₂

8.2Titrimetric method

Withdraw portions at intervals, quench, then titrate. Quenching slows the reaction abruptly so the concentration at that time can be found:

rapid cooling in ice
removing the catalyst (e.g. neutralising H⁺ with NaHCO₃)
adding a large volume of cold water to dilute

8.3Clock reactions

Time taken, t, for a sudden visible change.

Sulfur: S₂O₃²⁻ + 2H⁺ → S + SO₂ + H₂O, time for the X to be hidden
Iodine clock: fixed, small amount of S₂O₃²⁻ removes I₂ as it forms, blue-black with starch appears once all S₂O₃²⁻ is used up

A fixed amount of product forms in every run → rate ∝ 1t

Disappearance of colour: the amount of coloured reactant varies between runs, so rate = [coloured reactant]t, it cannot be simplified to 1/t.

Topic 08 · Reaction Kinetics43

Learners' AlleyDesigned by Khow Willard

9

Initial Rates Method

9.1Initial rates

Worked exampleThe initial rates of 2NO(g) + O₂(g) → 2NO₂(g) were measured at different concentrations. Find the rate equation and the rate constant.

Expt	[NO] / mol dm⁻³	[O₂] / mol dm⁻³	rate / mol dm⁻³ h⁻¹
1	0.001	0.001	7 × 10⁻⁶
2	0.001	0.002	14 × 10⁻⁶
3	0.001	0.003	21 × 10⁻⁶
4	0.002	0.003	84 × 10⁻⁶

Let rate = k[NO]^x[O₂]^y. Choose two experiments where only one concentration changes.
Expts 1 and 2: rate 2rate 1 = k(0.001)^x(0.002)^yk(0.001)^x(0.001)^y → 147 = 2^y → y = 1
Expts 3 and 4: rate 4rate 3 = k(0.002)^x(0.003)^yk(0.001)^x(0.003)^y → 8421 = 2^x → x = 2
rate = k[NO]²[O₂]
Expt 1: 7 × 10⁻⁶ = k(0.001)²(0.001) → k = 7000 mol⁻² dm⁶ h⁻¹

Exam tipAlways deduce orders with the ratio method above, choosing two experiments where only one concentration changes. Describing it in words ("when [O₂] doubles, rate doubles") is easy to phrase wrongly, and fails when more than one concentration changes.

Volumes instead of concentrations: water is added to keep the total volume constant, so [reactant] in the mixture ∝ volume of reactant used. Relative rate = 1/t.

10

Graphical Methods

zero orderfirst ordersecond order rate–conc. conc.–time straight line, constant gradient: rate = k constant half-life half-life increases as [A] falls

From a concentration–time graph: linear → zero order, curve → check the half-lives
Rate–conc. for second order: plot rate against [A]² to get a straight line through the origin

44Topic 08 · Reaction Kinetics

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Graphical Methods (continued)

10.1Half-life

DefinitionHalf-life, t½: the time taken for the concentration of a reactant to fall to half of its initial value.

First order: t½ = ln 2k

Any point may be taken as the "initial" concentration, half-lives need not be consecutive
Half-lives must be read from a plotted graph, not from the data table

time	[reactant]	[product]
0	100%	0%
1 t½	50%	50%
2 t½	25%	75%
3 t½	12.5%	87.5%

Worked exampleThe decomposition of hydrogen peroxide, 2H₂O₂ → 2H₂O + O₂, was followed over time. Use the graph to find the order with respect to H₂O₂ and the rate constant.

Rate decreases with a constant half-life of 33 s ∴ first order w.r.t. H₂O₂

rate = k[H₂O₂], k = ln 233 = 0.0210 s⁻¹

11

Pseudo Order

B in large excess (≥ 10 × [A]): why is rate independent of [B]?

B is in large excess
→ only a small fraction of B is used up
→ [B] remains approximately constant
→ rate appears independent of [B]

rate = k′[A]^m, where k′ = k[B]ⁿ

Also applies to a solvent (e.g. H₂O in hydrolysis) or a catalyst: its order cannot be found.

Worked exampleCH₃Br + OH⁻ → CH₃OH + Br⁻ is run with CH₃Br in large excess (1.5 mol dm⁻³) and [OH⁻] = 0.01 mol dm⁻³. The initial rate is 8.33 × 10⁻⁵ mol dm⁻³ min⁻¹. The [OH⁻]–time graph has a constant half-life, and repeating with [CH₃Br] × 1.5 makes the initial rate 1.5 times faster. Find the rate equation and k.

Continuous: [OH⁻]–time graph has a constant t½ → first order w.r.t. OH⁻
Initial rates: rate 2rate 1 = k(1.5 × 1.5)^mk(1.5)^m → 1.5 = 1.5^m → m = 1: first order w.r.t. CH₃Br
rate = k[CH₃Br][OH⁻], so 8.33 × 10⁻⁵ = k(1.5)(0.01)
k = 5.55 × 10⁻³ mol⁻¹ dm³ min⁻¹

Topic 08 · Reaction Kinetics45

Collision Theory

Factors Affecting Rate

2.1Concentration, pressure and surface area

2.2The Boltzmann distribution curve

Factors Affecting Rate (continued)

2.3Effect of temperature on rate

2.4Effect of a catalyst on rate

Rate of Reaction

Energy Profile Diagrams

Catalysis

5.1Heterogeneous: adsorption theory

Catalysis (continued)

5.2Homogeneous: Fe²⁺ in the I⁻ / S₂O₈²⁻ reaction

5.3Autocatalysis

5.4Enzymes

The Rate Equation

6.1Meaning of order

6.2What changes k?

6.3Units of k units of rate ÷ (mol dm⁻³)overall order

Mechanisms and the Rate Equation

7.1Proposing a mechanism

Mechanisms and the Rate Equation (continued)

7.2Testing a mechanism

Collecting Rate Data

8.1Physical measurements: a property that changes as the reaction goes

8.2Titrimetric method

8.3Clock reactions

Initial Rates Method

9.1Initial rates

Graphical Methods

Graphical Methods (continued)

10.1Half-life

Pseudo Order

6.3Units of k units of rate ÷ (mol dm⁻³)^{overall order}