Reaction
Kinetics

H2 Chemistry (9476), JC1

By the end of this topic, you can answer

1

Why does food go off more slowly in a fridge, and why does powdered sugar burn faster than a sugar cube?

2

How do you measure how fast a reaction is going at one exact moment?

3

How does a catalytic converter speed up reactions without being used up, and how is an enzyme different?

4

Why can doubling the concentration of one reactant double the rate, while doubling another does nothing at all?

5

Why does the amount of a drug in the blood halve every few hours, however much was taken?

Recall: what you know from O levels

1Name four factors that change the rate of a reaction.

Concentration (or pressure), particle size, temperature, catalyst.[1]

2Why does powdered CaCO₃ react faster with acid than lumps?

Larger surface area of contact → increases the frequency of effective collisions → increases the rate.[1]

3What does a catalyst do?

Speeds up a reaction without being used up.[1]
1

Collision Theory

For a reaction to occur, particles must collide with

  1. sufficient kinetic energy, greater than or equal to the activation energy (Ea)
  2. 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.

1

Collision Theory

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

AnalogyA key opens a lock only if it is pushed in hard enough (energy) and the right way round (orientation).

Quick checkTwo particles with energy greater than Ea collide. Must they react?

AYes, always
BOnly with the correct orientation
COnly if a catalyst is present
DOnly at high pressure
NoteB: an effective collision needs both energy ≥ Ea and the correct geometry.
2

Factors Affecting Rate

2.1Concentration and pressure

Concentration or pressure ↑

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

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

2

Factors Affecting Rate

2.1Surface area

Particle size of solid ↓

  1. Larger surface area of contact between reactant particles
  2. → frequency of effective collisions ↑ → rate ↑
2

Factors Affecting Rate

2.2The Boltzmann distribution curve

  • Starts at the origin and never touches the energy axis
  • Area under the curve = total no. of particles
  • Shaded area = particles with energy ≥ Ea
2

Factors Affecting Rate

2.3Effect of temperature on rate

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
  • The area under the curve stays constant
  1. As the temperature increases, the average kinetic energy of the particles increases
  2. → a larger number of reactant particles have energy ≥ Ea (larger shaded area)
  3. → frequency of effective collisions ↑ → rate ↑ (k ↑). Rate roughly doubles for every 10 °C rise.
2

Factors Affecting Rate

2.4Effect of a catalyst on rate

  1. Provides an alternative reaction pathway with a lower Ea
  2. → more reactant particles possess energy ≥ Ea (larger shaded area)
  3. → frequency of effective collisions ↑ → rate ↑ (k ↑)
Exam tipThe curve itself does not change with a catalyst: only the Ea line moves. With temperature, the curve changes and Ea stays.
2

Factors Affecting Rate

Quick checkWhy does a 10 °C rise roughly double the rate?

AThe collision frequency doubles.
BThe activation energy halves.
CMany more particles have energy ≥ Ea.
DThe number of particles doubles.
NoteC: the collision frequency rises only a little. The big effect is the larger shaded area above Ea, so effective collisions become much more frequent.
3

Rate of Reaction

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

rate = change in concentration of reactant or producttime 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.

3

Rate of Reaction

  • Initial rate: gradient of the tangent at t = 0, the highest rate, as [reactant] is at its maximum
  • DefinitionInstantaneous rate: gradient of the tangent to the concentration–time graph at time t

Your turnIn 2N₂O₅ → 4NO₂ + O₂, N₂O₅ is used up at 2.0 × 10⁻⁴ mol dm⁻³ s⁻¹. At what rate does NO₂ form?

42 × 2.0 × 10⁻⁴ = 4.0 × 10⁻⁴ mol dm⁻³ s⁻¹[1]
4

Energy Profile Diagrams

One stepone transition state
Two stepsintermediate in the dip, step 1 is the RDS
  • 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
4

Energy Profile Diagrams

Heterogeneous catalystone lower hump
Homogeneous catalysttwo lower humps
  • The rate-determining step (slow step) has the highest Ea
  • Higher Ea → slower reaction. Very high Ea: 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.
  1. It is specific
  2. It is regenerated at the end of the reaction
  3. It does not affect the overall stoichiometry
  4. It does not affect the position of equilibrium or K
  • DefinitionHomogeneous: catalyst and reactants in the same phase
  • DefinitionHeterogeneous: catalyst and reactants in different phases
5

Catalysis

5.1Heterogeneous: adsorption theory (Haber process on Fe)

  1. Adsorption: reactant particles form weak bonds to active sites on the catalyst surface, weakening the bonds within them
  2. Reaction: reactants are held close together in the correct orientation
  3. Desorption: weak bonds to the products break, products diffuse away and the active sites are freed
5

Catalysis

ProcessCatalyst
Haber: N₂ + 3H₂ ⇌ 2NH₃Fe
Contact: 2SO₂ + O₂ ⇌ 2SO₃V₂O₅
Hydrogenation of alkenesNi
Catalytic converter: 2CO + 2NO → 2CO₂ + N₂Pt, Pd, Rh

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

Catalysis

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

Iodide ions are oxidised by peroxodisulfate ions. Without a catalyst this is very slow, but adding a little Fe²⁺ speeds it up.

2I⁻(aq) + S₂O₈²⁻(aq) → I₂(aq) + 2SO₄²⁻(aq)

  1. Without a catalyst, the reaction is slow: it has a high Ea, as a large amount of energy is needed to overcome the repulsion between the two negatively charged reactant ions
  2. Step 1: 2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻
  3. Step 2: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂ (Fe²⁺ regenerated)
  4. Each step is between oppositely charged ions → lower Ea for each step
5

Catalysis

5.3Autocatalysis

DefinitionAutocatalysis: catalysis in which one of the products of the reaction catalyses the reaction.
  1. Slow at first: very little catalyst (product) present
  2. Fast as more product, the catalyst, forms
  3. Slow again as the reactants are used up
Example2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
The product Mn²⁺ catalyses the reaction, so the purple colour fades slowly at first, then quickly.
5

Catalysis

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
5

Catalysis

Quick checkWhich statement about a catalyst is correct?

AIt increases the yield at equilibrium.
BIt is used up in the reaction.
CIt increases the rate constant, k.
DIt increases the energy of the particles.
NoteC: a lower Ea means more effective collisions at the same concentrations, so k increases. It does not change K or the position of equilibrium.
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
  • Overall order = x + y
  • DefinitionRate constant, k: the proportionality constant in the rate equation

rate = k[A]x[B]y

NoteOrders come from experiments, not from the coefficients in the balanced equation.
6

The Rate Equation

6.1Meaning of order

OrderRateIf [A] × 2, rate
0independent of [A]× 1
1rate ∝ [A]× 2
2rate ∝ [A]²× 4

6.2What changes k?

ChangeRatek
[reactant] ↑↑ (if order ≠ 0)no change
temperature ↑↑↑
add a catalyst↑↑
6

The Rate Equation

6.3Units of k

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

Overall orderUnits of k
0 (rate = k)mol dm⁻³ s⁻¹
1 (rate = k[A])s⁻¹
2 (rate = k[A]²)mol⁻¹ dm³ s⁻¹
3 (rate = k[A]²[B])mol⁻² dm⁶ s⁻¹

Your turnrate = k[NO]²[H₂], with rate in mol dm⁻³ min⁻¹. Find the units of k.

Overall order 3: k = mol dm⁻³ min⁻¹(mol dm⁻³)³ = mol⁻² dm⁶ min⁻¹[1]
7

Mechanisms and the Rate Equation

  • Each elementary step involves at most two reacting particles
  • The steps must 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

AnalogyA queue at a single slow ticket counter: however fast the other counters are, the slow one sets the pace.

7

Mechanisms and 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.

  1. Slow step contains one NO and one F₂, from the rate equation
  2. 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
7

Mechanisms and the Rate Equation

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 1Mechanism 2
A + A → A₂ (slow)A + B → AB (slow)
A₂ + B → A₂B (fast)AB + A → A₂B (fast)
  1. Mechanism 1: the slow step involves two A → predicted rate = k[A]², which matches
  2. Mechanism 2: the slow step involves one A and one B → predicted rate = k[A][B], which does not match
  3. ∴ only mechanism 1 is consistent
8

Collecting Rate Data

Continuous methodInitial rate method
Experimentsone: [reactant] followed over timeseveral: one concentration changed at a time
Dataconcentration at intervalstime taken for a fixed change
Analysisconcentration–time graph, half-liferate–concentration graph or the ratio method

8.1Physical measurements

  • gas volume: Mg + 2HCl, H₂ given off
  • mass loss: CaCO₃ + 2HCl, CO₂ escapes
  • pH: CH₃Br + OH⁻, pH falls
  • conductivity: no. of ions changes
  • colorimetry: brown I₂ used up
  • pressure at fixed V: N₂O₄ → 2NO₂
8

Collecting Rate Data

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

Collecting Rate Data

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.

9

Initial Rates Method

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⁻³rate / mol dm⁻³ h⁻¹
10.0010.0017 × 10⁻⁶
20.0010.00214 × 10⁻⁶
30.0010.00321 × 10⁻⁶
40.0020.00384 × 10⁻⁶

Let rate = k[NO]x[O₂]y

Expts 1, 2: 147 = k(0.001)x(0.002)yk(0.001)x(0.001)y → 2 = 2y → y = 1

Expts 3, 4: 8421 = k(0.002)x(0.003)yk(0.001)x(0.003)y → 4 = 2x → 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

9

Initial Rates Method

Exam tipAlways deduce orders with the ratio method, 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.
NoteVolumes 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.
9

Initial Rates Method

Your turnFor A + B → C, find the order with respect to A and B, and the rate equation.

Expt[A][B]initial rate
10.100.102.0 × 10⁻³
20.200.102.0 × 10⁻³
30.100.301.8 × 10⁻²
Let rate = k[A]a[B]b. Experiments 1 and 2: 2.0 × 10⁻³2.0 × 10⁻³ = k(0.20)a(0.10)bk(0.10)a(0.10)b, so 1 = 2a and a = 0. The reaction is zero order with respect to A.[1]
Experiments 1 and 3: 1.8 × 10⁻²2.0 × 10⁻³ = k(0.10)a(0.30)bk(0.10)a(0.10)b, so 9 = 3b and b = 2. The reaction is second order with respect to B.[1]
or Comparing experiments 1 and 2: when [A] is doubled and [B] is kept constant, the initial rate is unchanged, so the reaction is zero order with respect to A.
Comparing experiments 1 and 3: when [B] is tripled and [A] is kept constant, the initial rate increases 9 times, so the reaction is second order with respect to B.
Rate equation: rate = k[B]²[1]
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
10

Graphical Methods

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
time01 t½2 t½3 t½
[reactant]100%50%25%12.5%
[product]0%50%75%87.5%
10

Graphical Methods

10.1Half-life

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]?

  1. B is in large excess
  2. → only a small fraction of B is used up
  3. → [B] remains approximately constant
  4. → 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.

11

Pseudo Order

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.

  1. Continuous: [OH⁻]–time graph has a constant t½ → first order w.r.t. OH⁻
  2. Initial rates: [CH₃Br] × 1.5 → initial rate × 1.5 → first order w.r.t. CH₃Br
  3. rate = k[CH₃Br][OH⁻], so 8.33 × 10⁻⁵ = k(1.5)(0.01)
  4. k = 5.55 × 10⁻³ mol⁻¹ dm³ min⁻¹
11

Pseudo Order

Quick checkThe concentration–time graph of a reactant has a constant half-life. What is the order w.r.t. that reactant?

Azero
Bfirst
Csecond
Dcannot be found
NoteB: only first order has a constant half-life, t½ = ln 2 / k, independent of concentration.

Exit ticket

1Explain why increasing temperature increases rate.

As temperature increases, the average kinetic energy of the reactant particles increases, so a greater proportion of reactant particles have energy greater than or equal to the activation energy, Eₐ.[1]
Hence, the frequency of effective collisions increases, and the rate of reaction increases.[1]

2Define order of reaction with respect to a reactant.

The power of its concentration term in the experimentally determined rate equation.[1]

3A first-order reaction has t½ = 20 s. Find k.

k = ln 2 ÷ 20 = 0.0347 s⁻¹[1]

4What is a homogeneous catalyst?

A catalyst in the same phase as the reactants.[1]