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3. Chemical Kinetics

3. Chemical Kinetics — NEB Chemistry Notes

🎯 Learning Outcomes

Candidates should be able to:

  • 3.1 Define chemical kinetics.
  • 3.2 Explain and use the terms: rate of reaction, rate equation, and rate constant.
  • 3.3 Qualitatively explain the factors affecting the rate of reaction.
  • 3.4 Use collision theory to explain the influence of temperature, concentration, and particle size on chemical reaction rate.
  • 3.5 Define activation energy and activated complex.
  • 3.6 Derive and use the integrated rate equations and half-life for zero- and first-order reactions.
  • 3.7 Construct and use rate equations, calculating an initial rate using concentration data.
  • 3.8 Explain the significance of the Arrhenius equation and solve related problems.
  • 3.9 Explain and use the terms catalyst and catalysis (homogeneous, heterogeneous).
  • 3.10 Describe enzymes as biological catalysts.
  • 3.11 Explain the role of a catalyst in the reaction mechanism.
  • 3.12 Solve numerical problems based on rate, rate constant, and order of zero- and first-order reactions.
1 Introduction

Have you ever wondered why food spoils faster in summer than in winter? Why does a car engine rust over the years, but a sparkler burns in seconds? Or why do doctors use particular drug dosages? All these phenomena involve the speed of chemical reactions — and that is exactly what Chemical Kinetics is about.

Chemical kinetics: The branch of Chemistry that deals with the rates of chemical reactions, the factors that affect them, and the mechanisms (step-by-step pathways) of those reactions.

Some reactions occur very rapidly, such as the explosion of fireworks, while others occur very slowly, such as the rusting of iron. Chemical kinetics helps us understand why these rates differ.

Kinetics tells us how fast a reaction happens — not just whether it can happen (thermodynamics).

⚡ Why Does Kinetics Matter?

  • Industrial chemistry: Controlling reaction speed maximises product yield and reduces cost.
  • Medicine: Drug metabolism rates determine safe dosages and dosing schedules.
  • Environment: Understanding rates of atmospheric reactions helps control pollution.
  • Food science: Preservative chemistry relies on slowing down decomposition reactions.
  • Biology: Every enzyme in your body is a natural kinetics optimiser.
2 Concept of Reaction Rate

The rate of a chemical reaction tells us how quickly reactants are consumed or products are formed over time.

Rate of Reaction: The decrease in concentration of a reactant (or increase in concentration of a product) per unit time.

Average rate of reaction

It is the rate of reaction measured over a long interval of time. Let us consider a general reaction:

A → C

If the initial concentration of reactant at time t₁ is [A₁] and the final concentration at time t₂ is [A₂]. Then, the average rate of reaction in terms of the reactants is:

Average rate = change in concentration / change in time
= − ([A₂] − [A₁]) / (t₂ − t₁) = − ΔA / Δt
💡 Why the Negative Sign?
Reactant concentration decreases over time, so Δ[A] is negative. The negative sign is included to ensure the rate is always a positive quantity. When expressing rate in terms of products, no negative sign is needed because product concentration increases.

Hence, the rate of the reaction in terms of product can be expressed as follows.

Rate (R) = ΔC / Δt   where ΔC = C₂ − C₁ (increase in concentration) and Δt = t₂ − t₁ (time taken)

Instantaneous rate of reaction

Instantaneous rate of reaction: the rate of reaction measured at any instant of time. It is expressed as follows:

dx/dt = limΔt→0 (ΔC/Δt)   where dx = very small change in concentration, dt = very small change in time

On a concentration vs. time graph, this is the slope of the tangent line at that point.

dx dt Reactant Concentration Time
Instantaneous rate as the slope (dx/dt) of the tangent to the concentration–time curve
3 Collision Theory & Activation Energy

Collision theory of reaction rate

Ineffective collision: particles just bounce apart unchanged.
Effective collision: leads to product formation.

Proper orientation — the molecules must align correctly for bonds to break and form.

A A B B Improper collision A A + B B
Improper collision: wrong orientation → no new bonds form
A B A B Proper collision A B + A B
Proper collision: correct orientation → bonds break and re-form into products

Sufficient energy — enough kinetic energy to overcome the activation energy barrier.

Ineffective CollisionEffective Collision
Particles bounce apart unchangedBonds break, and new bonds form → Products are formed
Energy < activation energy, OR wrong orientationEnergy ≥ activation energy AND correct orientation
No reaction occursReaction occurs successfully
▶ How to speed up chemical reactions (and get a date) — Aaron Sams

Concept of activation energy

The Activation Energy (Eₐ) of a reaction is the minimum energy required by colliding particles for a collision to be effective. Reaction pathway diagrams show how the activation energy provides a barrier to reaction.

Reactants Products Energy Progress of Reaction
Energy profile of a reaction: the blue dot must climb the activation-energy barrier before forming products

Activated Complex (Transition State) is the high-energy, unstable arrangement of atoms at the peak of the energy barrier. It exists for an extremely short time before breaking apart into products.

4 Factors Affecting Reaction Rate

Nature of Reactants

  • Ionic reactions are usually fast.
  • Covalent reactions are generally slow because bonds must be broken first.

Surface Area of the reactants

Greater surface area increases the rate of reaction because more particles are exposed for collision. For example, powdered calcium carbonate reacts faster than marble chips.

Concentration

As concentration increases, the frequency of collisions increases, resulting in an increased reaction rate.

Pressure

Similarly, when pressure is increased in a gaseous reaction, the frequency of collisions increases, increasing the reaction rate.

Temperature

At higher temperatures, molecules have more kinetic energy, so a higher percentage of successful collisions occurs between reactant molecules.

Temperature increases the rate of reaction because of two factors:

  1. The frequency of effective collisions is greater because of the greater kinetic energy of the molecules.
  2. A greater proportion of the molecules have kinetic energy greater than the activation energy.

The second reason has a far greater effect.

FactorEffectExplanation (Collision Theory)
Nature of reactantsIonic reactions are fast; covalent bonds require breaking → slowerIonic species are already separated; covalent bonds need energy to break
Surface area (solid)More surface area → fasterMore particles are exposed to collisions
ConcentrationHigher concentration → faster rateMore particles per volume → more frequent collisions
Pressure (gases)Higher pressure → faster rateParticles pushed closer together → collision frequency increases
TemperatureHigher temperature → much faster rate① Molecules move faster (more collisions) ② Far more important: a larger fraction of molecules have energy ≥ Eₐ
5 The Arrhenius Equation

The Arrhenius equation mathematically shows how temperature affects the rate constant.

k = A e−Eₐ/RT
  • k = the rate constant
  • A = pre-exponential factor or Arrhenius factor
  • Eₐ = activation energy
  • R = universal gas constant
  • T = absolute temperature

Even a small increase in temperature causes a large increase in k because of the exponential term.

📝 Worked Example 1

The rate constant for a reaction is 2.0 × 10⁻³ s⁻¹ at 300 K and 8.0 × 10⁻³ s⁻¹ at 320 K. Calculate the activation energy.

Solution:

Using: log(k₂/k₁) = Eₐ / 2.303R × (1/T₁ − 1/T₂)

log(8.0×10⁻³ / 2.0×10⁻³) = Eₐ / (2.303 × 8.314) × (1/300 − 1/320)
log(4) = Eₐ / 19.14 × (2.083 × 10⁻⁴)
0.602 = Eₐ × 1.088 × 10⁻⁵
Eₐ = 0.602 / 1.088 × 10⁻⁵ ≈ 55,330 J mol⁻¹ ≈ 55.3 kJ mol⁻¹
📝 Worked Example 2

The activation energy for a reaction is 80 kJ mol⁻¹ and A = 1.0 × 10¹ s⁻¹. Calculate the rate constant at 27 °C.

Solution:

T = 27 + 273 = 300 K, Eₐ = 80,000 J mol⁻¹, R = 8.314 J mol⁻¹ K⁻¹

Using k = A × e⁻⁽Eₐ/RT⁾:
Eₐ/RT = 80,000 / (8.314 × 300) = 80,000 / 2,494.2 = 32.08
k = 1.0 × 10⁹ × e⁻³²·⁰⁸
k = 1.0 × 10⁹ × 9.16 × 10⁻¹⁴
k ≈ 9.16 × 10⁻⁵ s⁻¹
6 Catalysis
Catalysis: the increase in the rate of a chemical reaction brought about by the addition of particular substances that are not used up in the reaction.

Catalysts provide an alternative reaction pathway with lower activation energy.

Catalyst: a substance that increases the rate of a reaction but remains chemically unchanged itself at the end of the reaction.

Catalysts increase the rate because they make the reaction go by a different reaction pathway (mechanism) that has a lower activation energy than the uncatalysed reaction.

Energy Reaction coordinate Transition state Activation energy of uncatalyzed reaction Activation energy of catalyzed reaction H₂C=CH₂ + H₂ H₃C–CH₃ ΔH < 0 exothermic
Effect of a catalyst on the activation energy of an exothermic reaction

Homogeneous Catalyst

When a catalyst and the reactants are in the same phase.

  • Concentrated H₂SO₄ (liquid) catalysing the esterification of an alcohol with a carboxylic acid (both liquids).
  • Fe²⁺ ions (aqueous) catalysing the reaction between S₂O₈²⁻ and I⁻ in aqueous solution.
  • Acid-catalysed hydrolysis of sucrose in aqueous solution.

Heterogeneous Catalyst

A catalyst that is in a different phase from the reactants.

  • Iron (Fe) catalyst in the Haber process — N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
  • Nickel in the hydrogenation of vegetable oils (margarine production)
  • Vanadium(V) oxide (V₂O₅) in the Contact process for manufacturing H₂SO₄
7 Enzyme Catalysis

Inside every living cell, thousands of chemical reactions are occurring every second — made possible by remarkable protein catalysts called enzymes.

Enzyme: A protein molecule that acts as a biological catalyst, increasing the rate of a specific biochemical reaction in living organisms.

Key Features

  • Highly specific: One enzyme catalyzes one type of reaction.
    Example: Diastase hydrolyzes starch → maltose, but does not hydrolyze cellulose.
  • Optimum temperature: Maximum activity at a specific temperature (~37 °C for human enzymes). Too high → denatures (loses shape). Too low → too slow.
  • Optimum pH: Each enzyme works best at a specific pH.
    Example: Pepsin (stomach) works at pH ~2; trypsin (small intestine) works at pH ~8.
  • Analogy: An enzyme is like a specific key that fits only one lock. Change the lock (different substrate), and the key doesn’t work.
▶ Pre-lab: Liver and Enzyme activity — YouTube

Role of Catalyst in Reaction Mechanism

A catalyst works by providing an alternative mechanism with a lower activation energy for the rate-determining step. This can involve:

  1. The catalyst forms an intermediate with one of the reactants (homogeneous catalysis).
  2. Reactants adsorbing onto the catalyst surface, where bonds are weakened (heterogeneous catalysis).

Either way, the energy barrier is reduced, and the reaction proceeds faster.

8 Rate Law, Rate Constant & Order of Reaction

Experiments show that the reaction rate depends on the concentration of reactants in a specific mathematical way. This relationship is captured in the rate equation (also called the rate law).

Rate equation: an equation showing the relationship between the rate constant and the concentrations of the species that affect the rate of reaction. The rate equals the rate constant (k) multiplied by the concentration of each reactant raised to certain whole-number powers (called orders).

The rate equation for the general reaction P + Q → R + S is:

rate = k[P]ˣ [Q]ʸ

where k = rate constant, x = order of reaction with respect to reactant P, y = order of reaction with respect to reactant Q, overall order = x + y

Rate constant: the proportionality constant, k, in a rate equation.

The overall order of the reaction is the sum of all the orders in the rate equation. For this reaction, the overall order of reaction would be x + y.

Rate has units of mol L⁻¹ s⁻¹. Concentration has units of mol L⁻¹. The units of the rate constant, k, depend on the overall order of the reaction, and the units can be calculated.

Order of reaction: the power to which the concentration of the reactant is raised in the experimentally determined rate equation. If a rate is directly proportional to concentration, it is a first-order reaction; if the rate is directly proportional to the square of the concentration, it is a second-order reaction. The overall order of reaction is the sum of these powers.
⚠️ Critical Point — Order is Experimental, NOT Stoichiometric
The order of a reaction CANNOT be determined simply by looking at the balanced equation. It must be determined experimentally. For example, the reaction 2NO₂ → 2NO + O₂ has a stoichiometric coefficient of 2 for NO₂, but the actual rate law is Rate = k[NO₂]², which happens to match — but this is coincidental, not a rule!
▶ Stop AI Hallucinations | Verified Science with Bohrium
9 Zero-Order & First-Order Reactions

Zero-order reaction

A zero-order reaction proceeds at a constant rate, unaffected by the reactant’s concentration.

e.g., Formation of HCl gas from H₂ and Cl₂ gas in the presence of sunlight on the surface of water.

H₂(g) + Cl₂(g)  —hυ→  2 HCl(g)

For a zero-order reaction, Rate = k [A]⁰ = k, i.e., rate = k. So, units of k = mol L⁻¹ s⁻¹

💡 What Does Zero-Order Mean Physically?
Zero-order behaviour often occurs when a catalyst surface is completely saturated with reactant. Adding more reactant cannot increase the rate because there are no free active sites remaining. The catalyst (not the reactant concentration) is the limiting factor.

Integrated rate law for a zero-order reaction

Consider a general zero-order reaction in which reactant A is converted to a product. Let:

  • a = initial concentration of reactant (mol L⁻¹)
  • x = amount of reactant converted to product after time t (mol L⁻¹)
  • (a − x) = concentration of reactant at time t (mol L⁻¹)
A ⟶ Product    At t = 0: [A] = a  |  At t = t: [A] = (a − x)

For a zero-order reaction, the rate is independent of the concentration of the reactant. The rate depends on the zeroth power of concentration: Rate ∝ [A]⁰ = constant.

Expressing the rate in terms of the change in x with respect to time:

dx/dt = k₀(a − x)⁰  ⟹  dx/dt = k₀ … (i)

where k₀ is the zero-order rate constant. Since (a−x)⁰ = 1, the rate equals the rate constant at all times. Equation (i) is the differential rate equation for the zero-order reaction.

Rearranging equation (i) to separate variables and integrating both sides:

dx = k₀ dt
∫dx = k₀∫dt
x = k₀t + C … (ii)

At t = 0, x = 0. Substituting into equation (ii): 0 = k₀ × 0 + C ⟹ C = 0.

Substituting C = 0 back into equation (ii): x = k₀t … (iii)

Rearranging for the rate constant: k₀ = x / t … (iv)

Equations (iii) and (iv) are the integrated rate equations for a zero-order reaction. Equation (iii) shows that x varies linearly with time, and equation (iv) gives the rate constant directly from the amount of reactant consumed per unit time.

  • The rate is constant and independent of the concentration of the reactant.
  • A plot of concentration [A] vs. time t gives a straight line with slope −k₀.
  • The rate constant k₀ has units of mol L⁻¹ s⁻¹ (concentration per unit time).
  • Common examples include enzyme-catalysed reactions at saturating substrate concentrations, and reactions on solid catalytic surfaces.
[R]₀ [R] time
Zero order: [R] falls linearly with time

Integrated rate law for a first-order reaction

A first-order reaction is a chemical reaction in which the rate depends on the concentration of only one reactant raised to the first power.

Some examples of first-order reactions:

Decomposition of H₂O₂:  2H₂O₂ → 2H₂O + O₂,  dx/dt = k₁[H₂O₂]¹

Decomposition of N₂O₅:  2N₂O₅ → 4NO₂ + O₂,  dx/dt = k₁[N₂O₅]¹

Let us consider a reaction in which reactant A changes into a product. Let a be the initial concentration of reactant A. After a certain time t, x mol of reactant changes into product.

A ⟶ Product    At t = 0: [A] = a  |  At t = t: [A] = (a − x)

For a first-order reaction, the rate depends on only one concentration term: Rate ∝ [A]¹

dx/dt ∝ (a − x)
dx/dt = k₁(a − x) … (i)

where k₁ is the rate constant (also called the velocity constant or specific rate of reaction). Equation (i) is the differential rate equation for the first-order reaction.

Rearranging equation (i) to separate variables and integrating both sides (using the standard result ∫dx/(ax+b) = ln(ax+b)/a + C):

dx / (a − x) = k₁ dt
∫dx / (a − x) = k₁∫dt
−ln(a − x) = k₁t + C … (ii)

At t = 0, x = 0. Substituting into equation (ii): −ln(a − 0) = k₁ × 0 + C ⟹ C = −ln a.

Substituting the value of C back into equation (ii):

−ln(a − x) = k₁t − ln a
ln a − ln(a − x) = k₁t
ln [a / (a − x)] = k₁t … (iii)
k₁ = (1/t) ln [a / (a − x)]  =  (2.303/t) log [a / (a − x)] … (iv)
  • The rate depends linearly on the concentration of a single reactant.
  • The integrated rate equation is logarithmic, giving a straight line when ln[A] is plotted against t.
  • The rate constant k₁ has units of s⁻¹ (or min⁻¹, h⁻¹, etc.).

Half-life reaction

Half-life (t½): time taken for the amount (or concentration) of the limiting reactant in a reaction to decrease to half its initial value.

At t = t½, [A]t = [A]₀/2. Substituting this boundary condition into the zero-order integrated rate law [A]t = [A]₀ − kt:

[A]₀/2 = [A]₀ − k·t½
k·t½ = [A]₀ − [A]₀/2 = [A]₀/2
t½ = [A]₀ / 2k  (zero-order half-life)

For a first-order reaction, substituting [A]t = [A]₀/2 into the natural-log integrated rate law:

ln([A]₀ / 2[A]₀) = −k·t½
ln(1/2) = −k·t½  ⟹  ln(2) = k·t½
t½ = ln(2)/k ≈ 0.693/k  (first-order half-life)

First-order reactions have a fixed half-life time that is independent of the initial concentration — this is their key distinguishing feature.

Kinetic ParameterZero-Order ReactionFirst-Order Reaction
Differential Rate LawRate = kRate = k[A]
Integrated Rate Law[A]t = [A]₀ − kt[A]t = [A]₀ · e−kt
Half-Life Expressiont½ = [A]₀ / (2k)t½ = 0.693 / k
Rate Constant (k) Unitsmol L⁻¹ s⁻¹s⁻¹
Concentration DependenceDirectly proportional to [A]₀Completely independent of [A]₀
10 Pseudo-Order & Second-Order Reactions

Pseudo-order reaction

A pseudo-order reaction is a reaction in which one reactant is present in large excess, so its concentration remains nearly constant during the reaction. As a result, the rate appears to depend only on the concentration of the other reactant.

A pseudo-first-order reaction is actually a second-order reaction that behaves like a first-order reaction.

Example
Hydrolysis of ethyl ethanoate (ethyl acetate) with excess water:
CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH
Rate = k[CH₃COOC₂H₅][H₂O], but since water is in vast excess, [H₂O] ≈ constant.
So Rate ≈ k′[CH₃COOC₂H₅], where k′ = k[H₂O] = pseudo-first-order rate constant.

Second-order reaction

Rate ∝ [reactant]² — if concentration doubles, rate quadruples.

e.g., 2NO₂ → 2NO + O₂

For a second-order reaction, rate = k[A]². So, units of rate constant, k = mol⁻¹ L s⁻¹

[R]₀ [R] time a ½[R]₀ ¼[R]₀ time b ½[R]₀ ¼[R]₀ (t½)₁ (t½)₂ time c
The three most common types of concentration–time graphs: a zero order, b first order, c second order. [R]₀ is the concentration at time t = 0.
Overall OrderUnits of k
Zero ordermol L⁻¹ s⁻¹
First orders⁻¹
Second ordermol⁻¹ L s⁻¹
11 Order and Molecularity of Reaction
FeatureMolecularityOrder
DefinitionNo. of particles colliding simultaneously in the rate-determining stepSum of powers of concentrations in the experimental rate equation
ValuesWhole numbers only: 1, 2, 3Can be 0, 1, 2, or even fractional
How determinedFrom the reaction mechanism (theoretical)From experimental data (empirical)
For complex reactionsRefers to the slowest (rate-determining) stepOverall observed value from the experiment

📌 Chapter Summary

  • Chemical kinetics is the study of reaction rates, the factors that affect them, and reaction mechanisms.
  • Rate of reaction = Δ[concentration] / Δtime. Units: mol L⁻¹ s⁻¹.
  • Collision theory: Effective collisions require (i) energy ≥ Eₐ and (ii) correct orientation.
  • Activation energy (Eₐ): minimum energy needed for a reaction; the activated complex forms at the energy peak.
  • Factors increasing rate: increased surface area, concentration, pressure (gases), temperature; adding a catalyst.
  • Arrhenius equation: k = Ae⁻⁽Eₐ/RT⁾. Temperature has an exponential effect on k.
  • Rate equation: Rate = k[A]ˣ[B]ʸ. Order is determined experimentally, NOT from stoichiometric coefficients.
  • Zero order: Rate = k; t½ = [A]₀/2k (half-life decreases over time).
  • First order: Rate = k[A]; t½ = 0.693/k (constant half-life — key feature).
  • Second order: Rate = k[A]²; k units: mol⁻¹ L s⁻¹.
  • A catalyst lowers Eₐ by providing an alternative reaction pathway; it is regenerated.
  • Homogeneous catalysis: catalyst in the same phase as reactants. Heterogeneous: different phases.
  • Enzymes: biological protein catalysts; highly specific; have optimum temperature and pH; follow the lock-and-key model.

📖 Key Terms Glossary

TermDefinition
Activation Energy (Eₐ)Minimum energy needed for colliding particles to react.
Activated ComplexHigh-energy, unstable arrangement of atoms at the energy peak.
Arrhenius Equationk = Ae⁻⁽Eₐ/RT⁾; mathematically links rate constant to temperature.
CatalystSubstance increasing reaction rate without being consumed; lowers Eₐ.
Collision TheoryA theory that explains reactions occur through effective collisions.
EnzymeBiological protein catalyst; highly specific; has optimum T and pH.
Half-Life (t½)Time for the reactant concentration to fall to half its initial value.
MolecularityNumber of particles colliding in the rate-determining elementary step.
Order of ReactionPower to which the concentration is raised in the experimental rate equation.
Rate Constant (k)Proportionality constant in the rate equation; depends on T and Eₐ.
Rate EquationRate = k[A]ˣ[B]ʸ; experimental relationship between rate and concentrations.
Rate-Determining StepSlowest step in a mechanism; controls the overall reaction rate.
Reaction MechanismA sequence of elementary steps by which reactants form products.

*** This is not a complete note. It is to guide you. It is recommended to study the prescribed textbooks along with this material. ***

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