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Chemical Kinetics

B.Sc. 1st Year Chemistry

Study notes on rate of reaction, rate law, order and molecularity, integrated rate equations, half-life, temperature dependence, Arrhenius equation, activation energy, collision theory and reaction mechanism.

1. Introduction to Chemical Kinetics

Chemical kinetics is the branch of chemistry that deals with the rate of a chemical reaction and the factors that affect the rate. It also helps us understand how a reaction takes place.

Thermodynamics can tell us whether a reaction is possible or favourable under particular conditions, but it does not by itself tell us how fast the reaction will occur.

For example, a reaction may be thermodynamically favourable but extremely slow. Chemical kinetics helps explain this difference.

What does chemical kinetics study?

  • Rate of chemical reaction
  • Factors affecting reaction rate
  • Rate law and rate constant
  • Order of reaction
  • Molecularity
  • Reaction mechanism
  • Activation energy
  • Effect of temperature
  • Effect of catalysts

A simple way to remember the subject is:

Chemical kinetics = How fast? + How does the reaction occur?

2. Rate of a Chemical Reaction

The rate of a reaction tells us how quickly the concentration of reactants decreases or the concentration of products increases with time.

Consider:

A → B

If the concentration of A changes from [A]₁ to [A]₂ during a time interval from t₁ to t₂:

Rate = - Δ[A] / Δt

The negative sign is used because the concentration of a reactant decreases with time.

For product formation:

Rate = + Δ[B] / Δt

For a general reaction

aA + bB → cC + dD

The reaction rate can be written as:

Rate = -1/a Δ[A]/Δt = -1/b Δ[B]/Δt = 1/c Δ[C]/Δt = 1/d Δ[D]/Δt

The coefficients are used so that the rate has the same value regardless of which substance is used for calculation.

Remember: Reactant concentration decreases, so its rate expression has a negative sign. Product concentration increases, so its expression has a positive sign.

Units of rate

If concentration is expressed in mol L⁻¹ and time in seconds:

Unit of rate = mol L⁻¹ s⁻¹

3. Factors Affecting the Rate of Reaction

1. Nature of reactants

Different substances react at different speeds because their bonds, structures and reaction pathways are different.

2. Concentration

Increasing concentration generally increases the number of particles present in a given volume. More frequent effective collisions can occur, so the reaction may become faster.

3. Pressure

For gaseous reactions, increasing pressure generally increases the effective concentration of gases and can increase reaction rate.

4. Temperature

Increasing temperature usually increases reaction rate because particles move faster and a greater fraction of collisions can have enough energy to react.

5. Catalyst

A catalyst provides an alternative reaction pathway with lower activation energy.

6. Surface area

A powdered solid generally reacts faster than a large solid piece because more surface is available for contact with the other reactant.

7. Light

Some photochemical reactions require light to start or accelerate the reaction.

Practical example: A powdered tablet dissolves and reacts faster than the same amount in a large piece because the available surface area is greater.

4. Rate Law

The rate law is an equation that relates the reaction rate to the concentrations of reactants.

For:

A + B → Products

A possible rate law is:

Rate = k[A]^m[B]^n

Here:

  • k = rate constant
  • [A] = concentration of A
  • [B] = concentration of B
  • m = order with respect to A
  • n = order with respect to B

Overall order

Overall order = m + n

The values of m and n are usually determined experimentally.

Important: Do not automatically take the stoichiometric coefficients as the powers in the rate law unless the reaction is an elementary step for which that relationship is justified.

5. Order of Reaction

The order of a reaction is the sum of the powers of concentration terms in the experimentally determined rate equation.

If:

Rate = k[A]^2[B]

Then:

Order with respect to A = 2
Order with respect to B = 1
Overall order = 2 + 1 = 3

Order may be:

  • Zero
  • First
  • Second
  • Third
  • Fractional
  • Sometimes apparent or experimentally determined values

Importance of order

The order helps determine the mathematical form of the rate equation, the units of the rate constant and the way concentration changes with time.

6. Molecularity of a Reaction

Molecularity is the number of reacting species involved in a single elementary reaction step.

Unimolecular reaction

A → Products

Molecularity = 1

Bimolecular reaction

A + B → Products

Molecularity = 2

Termolecular reaction

A + B + C → Products

Molecularity = 3

Order vs molecularity

Order Molecularity
Determined experimentally. Based on an elementary reaction step.
Can be zero or fractional. Usually a positive whole number.
Applies to the overall experimentally observed rate law. Applies to an elementary step.
May be obtained from rate experiments. Obtained from the number of reacting species in an elementary step.
Easy memory: Order comes from the rate law. Molecularity comes from the elementary reaction step.

7. Zero-Order Reaction

For a zero-order reaction, the rate does not depend on the concentration of the reactant.

A → Products
Rate = k[A]^0
Rate = k

Integrated rate equation

[A]t = [A]0 − kt

Where:

  • [A]₀ = initial concentration
  • [A]t = concentration after time t
  • k = rate constant

Half-life

t1/2 = [A]0 / 2k

The half-life of a zero-order reaction depends on the initial concentration.

Unit of k

k = mol L⁻¹ time⁻¹

Graph

For a zero-order reaction, a plot of [A] against time gives a straight line.

Slope = −k
Identification: Straight line for [A] versus t → possible zero-order reaction.

8. First-Order Reaction

In a first-order reaction, the rate is directly proportional to the concentration of one reactant.

A → Products
Rate = k[A]

Since:

Rate = −d[A]/dt

We have:

−d[A]/dt = k[A]

On integration:

ln([A]0/[A]t) = kt

In common logarithm form:

k = 2.303/t × log([A]0/[A]t)

Half-life of first-order reaction

At half-life:

[A]t = [A]0/2

Substituting in the first-order equation:

t1/2 = 0.693/k

Notice that initial concentration does not appear in this expression.

Very important: For a first-order reaction, half-life is independent of the initial concentration.

Unit of k

k = time⁻¹

Graphical test

A plot of ln[A] versus time gives a straight line:

Slope = −k

9. Second-Order Reaction

For a simple second-order reaction involving one reactant:

2A → Products

The rate law can be:

Rate = k[A]^2

Integrated rate equation

1/[A]t − 1/[A]0 = kt

Therefore:

k = 1/t × (1/[A]t − 1/[A]0)

Half-life

t1/2 = 1/(k[A]0)

Thus the half-life depends on the initial concentration.

Unit of k

k = L mol⁻¹ time⁻¹

Graphical test

A plot of 1/[A] against time gives a straight line.

Slope = k
Quick identification:
Zero order → [A] vs t
First order → ln[A] vs t
Second order → 1/[A] vs t

10. Half-Life of a Reaction

Half-life is the time required for the concentration of a reactant to become half of its initial value.

[A]t = [A]0/2

Zero-order

t1/2 = [A]0 / 2k

First-order

t1/2 = 0.693/k

Second-order

t1/2 = 1/(k[A]0)
Order Half-life Depends on [A]₀?
Zero [A]₀ / 2k Yes
First 0.693/k No
Second 1/(k[A]₀) Yes
Most important memory point: First-order half-life does not depend on initial concentration.

11. Determination of Order of Reaction

The order of a reaction can be determined by several experimental methods.

1. Initial rate method

Concentrations of reactants are changed systematically and the initial rates are measured.

Suppose:

Rate = k[A]^m[B]^n

If [A] is doubled while [B] remains constant and the rate becomes four times:

4 = 2^m
m = 2

Therefore the reaction is second order with respect to A.

2. Integrated rate method

Experimental concentration-time data are compared with the integrated equations for zero, first and second order reactions.

3. Half-life method

The relationship between half-life and initial concentration can help identify the order.

4. Graphical method

The graph that gives a straight line indicates the corresponding reaction order.

Plot Order indicated
[A] vs t Zero order
ln[A] vs t First order
1/[A] vs t Second order

12. Pseudo First-Order Reaction

Sometimes a reaction that actually involves two or more reactants behaves experimentally as a first-order reaction because one reactant is present in very large excess.

Consider:

A + B → Products

Suppose:

Rate = k[A][B]

If B is present in a very large excess, its concentration remains almost constant during the reaction.

k[B] = k'

Therefore:

Rate = k'[A]

The reaction now behaves like a first-order reaction.

Example

Hydrolysis of an ester in a large excess of water can behave as pseudo first-order because water concentration changes very little.

Remember: Pseudo first order does not mean the actual reaction molecularity or true order has changed. It is an observed approximation under particular experimental conditions.

13. Effect of Temperature on Reaction Rate

Reaction rate generally increases when temperature increases.

The important reason is not simply that particles move faster. More importantly, a larger fraction of molecules have energy equal to or greater than the activation energy.

Temperature coefficient

For many reactions, the rate may increase substantially for a temperature rise of 10°C, although the exact change depends on the reaction.

A common approximate rule is:

Rate approximately doubles for every 10°C rise

This is only a rough rule and should not be treated as a universal exact law. The Arrhenius equation gives a better quantitative description.

14. Arrhenius Equation

The Arrhenius equation relates the rate constant to temperature and activation energy.

k = A e−Ea/RT

Where:

  • k = rate constant
  • A = frequency or pre-exponential factor
  • Ea = activation energy
  • R = gas constant
  • T = absolute temperature in kelvin

Logarithmic form

ln k = ln A − Ea/RT

In common logarithm form:

log k = log A − Ea/(2.303RT)

Two-temperature form

When rate constants at two temperatures are given:

ln(k₂/k₁) = Ea/R × (1/T₁ − 1/T₂)

Or:

log(k₂/k₁) = Ea/(2.303R) × (1/T₁ − 1/T₂)
Important for numericals: Temperature must be converted to kelvin before using the Arrhenius equation.

15. Activation Energy

Activation energy is the minimum energy barrier that reacting particles must overcome for a reaction to occur.

It is represented by Ea.

Ea = minimum energy required to reach the activated state

Why is activation energy important?

If the activation energy is high, fewer molecules have enough energy to react at a particular temperature, so the reaction may be slow.

If activation energy is lower, a larger fraction of molecules can react.

Effect of catalyst

A catalyst provides a different pathway with lower activation energy.

Uncatalysed pathway → higher Ea
Catalysed pathway → lower Ea

The catalyst does not change the overall thermodynamic energy difference between reactants and products.

16. Collision Theory

Collision theory explains reaction rate in terms of collisions between reacting particles.

Not every collision produces a reaction.

Conditions for an effective collision

  1. The reacting particles must collide.
  2. The collision must have sufficient energy.
  3. The particles must have a suitable orientation.

Energy requirement

The collision must provide enough energy to overcome the activation energy barrier.

Collision energy ≥ Ea → possible successful reaction

If the collision energy is below the activation energy, the particles usually separate without forming products.

Orientation

Even when particles have enough energy, they must collide in a suitable orientation so that the necessary bonds can break and form.

Remember: Effective collision = sufficient energy + suitable orientation.

17. Activated Complex and Transition State

During a chemical reaction, reactants pass through a high-energy arrangement before forming products. This unstable arrangement is called the activated complex or transition state.

A simplified representation is:

Reactants → Activated complex → Products

The energy required to reach this state from the reactants is related to the activation energy.

Energy profile

In an energy profile diagram, the vertical axis usually represents potential energy and the horizontal axis represents reaction progress.

The highest point corresponds approximately to the transition state.

Ea = Energy of transition state − Energy of reactants

A catalyst lowers the height of the energy barrier by providing another pathway.

18. Reaction Mechanism

The reaction mechanism is the sequence of elementary steps through which reactants are converted into products.

A reaction that appears simple overall may actually occur through several smaller steps.

Example of a two-step mechanism

Overall reaction:

A + B → D

Step 1:

A + B → C

Step 2:

C → D

C is formed in one step and consumed in another. Therefore it is an intermediate.

Elementary step

An elementary step represents a single molecular event in the mechanism.

Slow step and rate-determining step

In a simple mechanism, the slowest elementary step can control the overall reaction rate and is often called the rate-determining step.

However, for more complicated mechanisms, deriving the overall rate law requires careful treatment and cannot always be done simply by assuming the slow step.

Important: The experimentally determined rate law is the final test for a proposed mechanism.

19. Catalysis

A catalyst is a substance that changes the rate of a reaction without being consumed permanently in the overall reaction.

The main kinetic role of a catalyst is to provide an alternative pathway with lower activation energy.

What a catalyst does

  • Changes reaction rate.
  • Provides an alternative reaction pathway.
  • Lowers the activation energy of that pathway.

What a catalyst does not do

  • It does not change the overall reaction equation.
  • It does not change the equilibrium constant for a reaction at a fixed temperature.
  • It does not change the final equilibrium composition when equilibrium is reached.
  • It does not make an impossible reaction thermodynamically possible simply by being present.

Homogeneous catalysis

Catalyst and reactants are in the same phase.

Heterogeneous catalysis

Catalyst and reactants are in different phases.

Enzyme catalysis

Enzymes are biological catalysts. They can accelerate biochemical reactions under relatively mild conditions and often show high specificity for their substrates.

Very important: A catalyst changes the speed of reaching equilibrium, not the equilibrium constant itself.

20. Worked Numerical Problems

Numerical 1: Average rate of reaction

The concentration of A decreases from 0.80 mol L⁻¹ to 0.50 mol L⁻¹ in 30 seconds. Calculate the average rate.

Given:

[A]₁ = 0.80 mol L⁻¹
[A]₂ = 0.50 mol L⁻¹
Δt = 30 s

Formula:

Rate = −Δ[A]/Δt

Change in concentration:

Δ[A] = 0.50 − 0.80
Δ[A] = −0.30 mol L⁻¹

Substitution:

Rate = −(−0.30)/30
Rate = 0.010 mol L⁻¹ s⁻¹

Therefore, the average rate is 1.0 × 10⁻² mol L⁻¹ s⁻¹.

Numerical 2: Finding reaction order

For a reaction:

Rate = k[A]^m

When [A] is doubled, the rate becomes four times. Find m.

Given:

Rate₂ / Rate₁ = 4
[A]₂ / [A]₁ = 2

Therefore:

4 = 2^m

Since:

4 = 2²
m = 2

The reaction is second order with respect to A.

Numerical 3: First-order rate constant

A first-order reaction has an initial concentration of 0.80 mol L⁻¹. After 20 minutes the concentration becomes 0.40 mol L⁻¹. Calculate k.

Formula:

k = 2.303/t × log([A]₀/[A]t)

Substitution:

k = 2.303/20 × log(0.80/0.40)
k = 2.303/20 × log 2

Since:

log 2 = 0.3010
k = (2.303 × 0.3010)/20
k ≈ 0.0347 min⁻¹

Therefore:

k ≈ 3.47 × 10⁻² min⁻¹
Numerical 4: Half-life of a first-order reaction

The rate constant of a first-order reaction is 0.020 min⁻¹. Calculate its half-life.

Formula:

t1/2 = 0.693/k

Substitution:

t1/2 = 0.693/0.020
t1/2 = 34.65 min

Therefore, the half-life is approximately 34.7 minutes.

Numerical 5: Zero-order reaction

A zero-order reaction has [A]₀ = 0.50 mol L⁻¹ and k = 0.010 mol L⁻¹ min⁻¹. Find the concentration after 20 minutes.

Formula:

[A]t = [A]₀ − kt

Substitution:

[A]t = 0.50 − (0.010 × 20)
[A]t = 0.50 − 0.20
[A]t = 0.30 mol L⁻¹
Numerical 6: Second-order rate constant

For a second-order reaction, [A]₀ = 0.50 mol L⁻¹ and [A]t = 0.25 mol L⁻¹ after 10 seconds. Calculate k.

Formula:

k = 1/t × (1/[A]t − 1/[A]₀)

Substitution:

k = 1/10 × (1/0.25 − 1/0.50)
k = 0.1 × (4 − 2)
k = 0.20 L mol⁻¹ s⁻¹
Numerical 7: Arrhenius equation

For a reaction, k₁ = 2.0 × 10⁻³ s⁻¹ at 300 K and k₂ = 8.0 × 10⁻³ s⁻¹ at 320 K. Calculate the activation energy.

Use:

ln(k₂/k₁) = Ea/R × (1/T₁ − 1/T₂)

Step 1: Calculate k₂/k₁

k₂/k₁ = (8.0 × 10⁻³)/(2.0 × 10⁻³)
k₂/k₁ = 4

Step 2:

ln 4 = Ea/8.314 × (1/300 − 1/320)

Since:

ln 4 ≈ 1.386

And:

1/300 − 1/320 = 20/(300 × 320) ≈ 2.083 × 10⁻⁴ K⁻¹

Therefore:

1.386 = Ea/8.314 × 2.083 × 10⁻⁴
Ea ≈ 55,300 J mol⁻¹
Ea ≈ 55.3 kJ mol⁻¹

Therefore the activation energy is approximately 55.3 kJ mol⁻¹.

21. Common Mistakes

1. Forgetting the negative sign for reactant rate

Reactant concentration decreases with time, so its rate expression contains a negative sign.

2. Confusing order with molecularity

Order is obtained from the experimentally determined rate law. Molecularity refers to an elementary reaction step.

3. Assuming stoichiometric coefficients always give the rate-law powers

This is not generally correct for an overall complex reaction.

4. Using Celsius in the Arrhenius equation

Always convert:

T(K) = T(°C) + 273.15

5. Forgetting the unit of k

The unit of the rate constant depends on the overall order.

6. Thinking catalyst changes equilibrium constant

A catalyst changes how quickly equilibrium is reached. It does not change K at a fixed temperature.

7. Saying every collision causes reaction

Only effective collisions have suitable energy and orientation.

8. Using the first-order half-life formula for every reaction

t1/2 = 0.693/k

This expression is specifically for a first-order reaction.

22. TU Examination Preparation

Important 2-Mark Questions

  • Define chemical kinetics.
  • Define rate of reaction.
  • What is rate constant?
  • Define order of reaction.
  • Define molecularity.
  • What is activation energy?
  • What is an effective collision?
  • State Arrhenius equation.
  • Define half-life.
  • What is a catalyst?
  • What is a pseudo first-order reaction?

Important 5-Mark Questions

  • Explain the factors affecting reaction rate.
  • Differentiate order and molecularity.
  • Derive the integrated rate equation for a first-order reaction.
  • Explain the half-life of a first-order reaction.
  • Explain collision theory.
  • Explain the Arrhenius equation.
  • Explain pseudo first-order reactions with an example.
  • Explain the effect of temperature on reaction rate.

Important 10-Mark Questions

  • Derive the integrated rate equations for zero-order and first-order reactions and discuss their half-lives.
  • Explain Arrhenius equation and derive the two-temperature form.
  • Explain collision theory and discuss the conditions required for an effective collision.
  • Explain reaction mechanism, elementary reactions, intermediates and rate-determining step.
  • Explain the methods used for determination of order of a chemical reaction.
For numerical questions: Write Given → Formula → Substitution → Calculation → Unit → Final answer. This makes the calculation easy to follow and reduces mistakes.

23. Quick Revision

  • Chemical kinetics studies reaction rate and reaction mechanism.
  • Rate describes concentration change with time.
  • Rate law connects rate with reactant concentrations.
  • Rate = k[A]^m[B]^n.
  • Overall order = m + n.
  • Order is determined experimentally.
  • Molecularity refers to an elementary reaction step.
  • Zero-order equation: [A]t = [A]₀ − kt.
  • First-order equation: k = 2.303/t log([A]₀/[A]t).
  • First-order half-life: t1/2 = 0.693/k.
  • Second-order equation: 1/[A]t − 1/[A]₀ = kt.
  • Second-order half-life: t1/2 = 1/(k[A]₀).
  • Arrhenius equation: k = Ae−Ea/RT.
  • Higher temperature generally increases k.
  • Higher activation energy generally means a slower reaction at the same temperature, all else being comparable.
  • Effective collision requires sufficient energy and proper orientation.
  • Catalyst provides an alternative pathway with lower activation energy.
  • Catalyst does not change the equilibrium constant at a fixed temperature.
  • Reaction mechanism consists of elementary reaction steps.
Formula memory:
Zero order → [A]t = [A]₀ − kt
First order → ln([A]₀/[A]t) = kt
First-order half-life → 0.693/k
Second order → 1/[A]t − 1/[A]₀ = kt
Arrhenius → k = Ae−Ea/RT

24. Quick Check

1. What does chemical kinetics study?
It studies the rate of chemical reactions, factors affecting the rate and reaction mechanisms.
2. What is the overall order of Rate = k[A]²[B]?
Overall order = 2 + 1 = 3.
3. What is the half-life expression for a first-order reaction?
t1/2 = 0.693/k.
4. Does first-order half-life depend on initial concentration?
No. For a first-order reaction, t1/2 = 0.693/k and does not contain the initial concentration.
5. What is the purpose of a catalyst?
A catalyst provides an alternative pathway with lower activation energy and therefore changes the reaction rate.
6. What two conditions are required for an effective collision?
The collision must have sufficient energy and suitable orientation.
7. Which graph gives a straight line for a first-order reaction?
A plot of ln[A] versus time gives a straight line with slope −k.
8. What is the difference between order and molecularity?
Order is obtained from the experimentally determined rate law, while molecularity is the number of reacting species involved in an elementary step.