UNIT 5

Kinetics

Study the speed of chemical reactions and examine how concentration, temperature, reaction pathways, and catalysts affect reaction rate.

Topics

Reaction Rates and Rate Laws

Learn how reaction rates are measured and use experimental data to determine rate laws, reaction orders, and rate constants.

Concentration and Time

Analyze zero-, first-, and second-order reactions using integrated rate laws, half-life calculations, and concentration-versus-time graphs.

Collision Theory and Mechanisms

Explore activation energy, molecular collisions, reaction mechanisms, energy diagrams, and catalysts.

Reaction Rates and Rate Laws

Chemical kinetics is the study of how quickly reactions occur. A reaction rate describes how the concentration of a reactant or product changes over a period of time.

Measuring Reaction Rate

As a reaction proceeds, reactant concentrations usually decrease while product concentrations increase.

Average rate = Change in concentration Change in time

The Greek letter delta, Δ, means “change in.” Therefore, Δ[A] represents the change in the concentration of substance A.

Reactants: Their concentrations decrease, so a negative sign is used when expressing their rates.

Products: Their concentrations increase, so their rates are positive.

Example: Calculating an Average Rate

The concentration of reactant A decreases from 0.800 M to 0.620 M over a period of 30.0 seconds. Calculate the average rate at which A disappears.

Rate of disappearance = − Δ[A] Δt

Rate = − 0.620 M − 0.800 M 30.0 s

Rate = 0.00600 M/s

The negative sign in the formula changes the negative concentration change into a positive rate.

Stoichiometry and Reaction Rate

The coefficients in a balanced equation relate the rates at which reactants disappear and products appear.

Consider the general reaction:

aA + bB → cC + dD

Rate = − 1 a Δ[A] Δt = − 1 b Δ[B] Δt = 1 c Δ[C] Δt

Dividing by each substance’s coefficient produces one consistent rate for the entire reaction.

Example: Relating Formation Rates

2N₂O₅(g) → 4NO₂(g) + O₂(g)

Suppose O₂ forms at a rate of 0.015 M/s. At what rate does NO₂ form?

Four moles of NO₂ form for every one mole of O₂ that forms.

Rate of NO₂ formation = 4(0.015 M/s) = 0.060 M/s

Rate Laws

A rate law describes how the reaction rate depends on the concentrations of its reactants.

For a reaction involving reactants A and B, a rate law may be:

Rate = k[A]^m[B]ⁿ

In this expression:

k is the rate constant.
[A] and [B] are molar concentrations.
m is the reaction order with respect to A.
n is the reaction order with respect to B.
m + n is the overall reaction order.

The exponents in a rate law must normally be determined from experimental data. They should not be taken directly from the coefficients of the overall balanced equation.

Understanding Reaction Order

Reaction order tells us how changing a reactant’s concentration changes the reaction rate.

Order in a Reactant	When Its Concentration Doubles	Effect on Rate
Zero order	2⁰	Rate does not change
First order	2¹	Rate doubles
Second order	2²	Rate quadruples

Determining a Rate Law from Data

To determine a reaction order, compare two experiments in which only one reactant concentration changes. Then observe how the initial rate changes.

Example: Method of Initial Rates

The following experimental data were collected for a reaction involving A and B.

Experiment	[A] (M)	[B] (M)	Initial Rate (M/s)
1	0.100	0.100	0.0200
2	0.200	0.100	0.0800
3	0.100	0.200	0.0400

Begin with the general form:

Rate = k[A]^m[B]ⁿ

Determine the Order in A

Compare Experiments 1 and 2. The concentration of B remains constant. The concentration of A doubles, while the rate increases by a factor of four.

2^m = 4

m = 2

The reaction is second order in A.

Determine the Order in B

Compare Experiments 1 and 3. The concentration of A remains constant. The concentration of B doubles, and the rate also doubles.

2ⁿ = 2

n = 1

The reaction is first order in B.

The experimental rate law is therefore:

Rate = k[A]²[B]

The overall reaction order is 2 + 1, so the reaction is third order overall.

Calculating the Rate Constant

Once the rate law is known, data from any experiment can be substituted into it to determine the rate constant, k.

Using Experiment 1:

0.0200 M/s = k(0.100 M)²(0.100 M)

k = 0.0200 M/s (0.100 M)²(0.100 M)

k = 20.0 M⁻²s⁻¹

The units of the rate constant depend on the overall reaction order. The value and units of k should be reported together.

Rate Law Versus Balanced Equation

A balanced equation describes the overall amounts of substances consumed and produced. A rate law describes how the measured reaction rate depends on concentration.

Because an overall reaction may occur through several smaller steps, its balanced coefficients do not necessarily equal the exponents in its rate law.

Use experimental data to determine the rate law for an overall reaction. Coefficients can be used as rate-law exponents only when dealing with a confirmed elementary reaction step.

Practice Questions

Question 1: Calculating Average Rate

The concentration of reactant B decreases from 0.900 M to 0.660 M during a period of 40.0 seconds.

Calculate the average rate at which B disappears.

Show solution

Because B is a reactant, include a negative sign in the rate expression.

Rate = − Δ[B] Δt

Rate = − 0.660 M − 0.900 M 40.0 s

Rate = 0.00600 M/s

B disappears at an average rate of 0.00600 M/s.

Question 2: Relating Reaction Rates

Consider the reaction:

2A → 3B

If A disappears at a rate of 0.020 M/s, at what rate does B form?

Show solution

The balanced equation shows that three moles of B form for every two moles of A consumed.

0.020 M/s × 3 mol B 2 mol A = 0.030 M/s

B forms at a rate of 0.030 M/s.

Question 3: Determining a Rate Law

The following data were collected for a reaction involving reactants A and B.

Experiment	[A] (M)	[B] (M)	Initial Rate (M/s)
1	0.100	0.100	0.00600
2	0.200	0.100	0.0120
3	0.100	0.300	0.0540

Determine the order in A, the order in B, and the complete rate law.

Show solution

Compare Experiments 1 and 2. The concentration of A doubles while B remains constant. The rate also doubles.

2^m = 2

m = 1

The reaction is first order in A.

Compare Experiments 1 and 3. The concentration of B triples while A remains constant. The rate increases by a factor of nine.

3ⁿ = 9

n = 2

The reaction is second order in B.

The complete rate law is:

Rate = k[A][B]²

The overall reaction order is 1 + 2, so the reaction is third order overall.

Question 4: Calculating the Rate Constant

Using the data and rate law from Question 3, calculate the rate constant, k.

Show solution

Use the rate law and the values from Experiment 1.

Rate = k[A][B]²

0.00600 M/s = k(0.100 M)(0.100 M)²

k = 0.00600 M/s (0.100 M)(0.100 M)²

k = 6.00 M⁻²s⁻¹

The rate constant is 6.00 M⁻²s⁻¹.

Question 5: Predicting a Change in Rate

A reaction has the following rate law:

Rate = k[A][B]²

What happens to the reaction rate if [A] is doubled while [B] is reduced to one-half of its original value?

Show solution

Doubling [A] changes the rate by a factor of:

2¹ = 2

Reducing [B] to one-half changes the rate by:

(0.5)² = 0.25

Multiply the two effects:

2 × 0.25 = 0.50

The new rate is one-half of the original rate.

Question 6: Interpreting Zero Order

A reaction is zero order in reactant C. The concentration of C is increased from 0.150 M to 0.600 M while all other conditions remain constant.

How will this change affect the reaction rate?

Show solution

Because the reaction is zero order in C, its rate law contains [C]⁰.

[C]⁰ = 1

Changing the concentration of C therefore has no effect on the reaction rate.

VIDEO EXPLANATION

Watch: Determining a Rate Law

Watch a worked example showing how experimental data can be used to determine reaction orders, the rate law, and the rate constant.

Watch on TikTok →

Concentration and Time

An integrated rate law relates reactant concentration to time. It can be used to calculate a concentration, determine a rate constant, or identify the order of a reaction from a graph.

Each reaction order produces a straight line using a different form of reactant concentration. The graph that is linear identifies the reaction order.

Zero-Order Reactions

A zero-order reaction has a rate that does not depend on the concentration of reactant A.

Rate = k

The integrated zero-order rate law is:

[A]_t = [A]₀ − kt

This equation has the same general form as the equation of a straight line:

y = b + mx

A graph of [A] versus time is linear for a zero-order reaction.

The vertical axis is [A].
The horizontal axis is time.
The slope equals −k.
The vertical intercept equals [A]₀.

Zero-order half-life = [A]₀ 2k

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

First-Order Reactions

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

Rate = k[A]

The integrated first-order rate law is:

ln[A]_t = ln[A]₀ − kt

The abbreviation ln represents the natural logarithm. A graph of ln[A] versus time is linear for a first-order reaction.

The vertical axis is ln[A].
The horizontal axis is time.
The slope equals −k.
The vertical intercept equals ln[A]₀.

First-order half-life = ln(2) k = 0.693 k

The half-life of a first-order reaction does not depend on its initial concentration. The same fraction of reactant disappears during each half-life.

Second-Order Reactions

In a second-order reaction involving one reactant, the rate depends on the square of that reactant’s concentration.

Rate = k[A]²

The integrated second-order rate law is:

1 [A]_t = 1 [A]₀ + kt

A graph of 1/[A] versus time is linear for a second-order reaction.

The vertical axis is 1/[A].
The horizontal axis is time.
The slope equals +k.
The vertical intercept equals 1/[A]₀.

Second-order half-life = 1 k[A]₀

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

Identifying Reaction Order from a Graph

Three graphs showing the linear plots for zero-order, first-order, and second-order reactions — The reaction order is identified by determining which concentration plot produces a straight line.

Reaction Order	Linear Plot	Slope	Rate Law
Zero order	[A] versus time	−k	Rate = k
First order	ln[A] versus time	−k	Rate = k[A]
Second order	1/[A] versus time	+k	Rate = k[A]²

Do not identify the reaction order simply by looking at the shape of one concentration graph. Test [A], ln[A], and 1/[A] versus time. The plot with the best straight line identifies the order.

Using the Slope to Find the Rate Constant

After identifying which graph is linear, the rate constant can be determined from its slope.

For zero- and first-order plots, slope = −k.

For a second-order plot, slope = +k.

Example: Determining Order and Rate Constant

A graph of ln[A] versus time produces a straight line with a slope of −0.0350 s⁻¹.

Because ln[A] versus time is linear, the reaction is first order.

Slope = −k

−0.0350 s⁻¹ = −k

k = 0.0350 s⁻¹

Using a First-Order Half-Life

Example: Calculating Half-Life

A first-order reaction has a rate constant of 0.0231 s⁻¹. Calculate its half-life.

t_1/2 = 0.693 k

t_1/2 = 0.693 0.0231 s⁻¹ = 30.0 s

The concentration of the reactant is reduced to one-half of its previous value every 30.0 seconds.

Practice Questions

Question 1: Identifying Reaction Order

A student creates three graphs using concentration data from a reaction. Only the graph of ln[A] versus time produces a straight line.

What is the reaction order? If the line has a slope of −0.0250 s⁻¹, what is the rate constant?

Show solution

A linear graph of ln[A] versus time identifies a first-order reaction.

For a first-order linear plot:

Slope = −k

−0.0250 s⁻¹ = −k

k = 0.0250 s⁻¹

Question 2: Zero-Order Concentration

A zero-order reaction has an initial reactant concentration of 0.800 M and a rate constant of 0.0150 M/s.

What is the reactant concentration after 20.0 seconds?

Show solution

Use the zero-order integrated rate law:

[A]_t = [A]₀ − kt

[A]_t = 0.800 M − (0.0150 M/s)(20.0 s)

[A]_t = 0.800 M − 0.300 M = 0.500 M

The concentration after 20.0 seconds is 0.500 M.

Question 3: First-Order Half-Life

A first-order reaction has a rate constant of 0.0347 s⁻¹. Calculate the reaction’s half-life.

Show solution

Use the first-order half-life equation:

t_1/2 = 0.693 k

t_1/2 = 0.693 0.0347 s⁻¹ = 20.0 s

The reaction’s half-life is 20.0 seconds.

Question 4: Concentration after Several Half-Lives

A first-order reaction begins with a reactant concentration of 0.640 M. What concentration remains after three half-lives?

Show solution

During each half-life, the concentration is reduced to one-half of its previous value.

0.640 M → 0.320 M → 0.160 M → 0.0800 M

After three half-lives, the concentration is 0.0800 M.

This is also one-eighth of the initial concentration:

[A]_t = 0.640 M 2³ = 0.0800 M

Question 5: Second-Order Concentration

A reaction is second order in A. Its initial concentration is 0.200 M, and its rate constant is 0.500 M⁻¹s⁻¹.

Calculate the concentration of A after 30.0 seconds.

Show solution

Use the second-order integrated rate law:

1 [A]_t = 1 [A]₀ + kt

1 [A]_t = 1 0.200 M + (0.500 M⁻¹s⁻¹)(30.0 s)

1 [A]_t = 5.00 M⁻¹ + 15.0 M⁻¹

1 [A]_t = 20.0 M⁻¹

[A]_t = 0.0500 M

After 30.0 seconds, the concentration is 0.0500 M.

Question 6: Evaluating Linear Plots

A student obtains the following values after creating three linear regressions from experimental data:

Plot	R² Value
[A] versus time	0.914
ln[A] versus time	0.999
1/[A] versus time	0.951

Which reaction order is best supported by the data?

Show solution

An R² value closest to 1 indicates that the data most closely form a straight line.

The ln[A] versus time plot has an R² value of 0.999, which is closest to 1. Therefore, the reaction is best classified as first order.

VIDEO EXPLANATION

Watch: Integrated Rate Laws

Watch a worked example showing how graphs, slopes, integrated rate laws, and half-life can be used to analyze concentration over time.

Watch on TikTok →

Collision Theory and Reaction Mechanisms

For a chemical reaction to occur, reactant particles must collide. However, not every collision successfully produces products.

Collision Theory

According to collision theory, a collision results in a reaction only if the particles collide with enough energy and with an effective orientation.

A successful collision requires:

1. Enough energy to overcome the activation energy

2. An orientation that allows the necessary bonds to break and form

A collision that does not meet both conditions will not produce a reaction.

Factors That Affect Reaction Rate

Change	Why the Reaction Rate Usually Increases
Increasing reactant concentration	More reactant particles occupy the same volume, causing collisions to occur more frequently.
Increasing gas pressure	Gas particles are forced closer together, increasing collision frequency.
Increasing temperature	Particles move faster, collide more frequently, and a greater fraction of collisions have enough energy to react.
Increasing a solid’s surface area	More particles are exposed and available for collisions.
Adding a catalyst	The catalyst provides an alternative reaction pathway with a lower activation energy.

Activation Energy

Activation energy, written as E_a, is the minimum energy that reacting particles must possess for a collision to result in a reaction.

During a successful collision, the particles briefly form a high-energy arrangement called the transition state, or activated complex.

A catalyst lowers the activation energy but does not change the energies of the reactants and products or the overall enthalpy change.

Reading an Energy Profile

An energy-profile diagram shows potential energy as a reaction proceeds from reactants to products.

The difference between the reactant energy and the top of the pathway is the forward activation energy.
The highest point represents the transition state.
The difference between product and reactant energy is ΔH for the reaction.
A lower-energy pathway represents a catalyzed reaction.

ΔH = Energy of products − Energy of reactants

If the products are lower in energy than the reactants, ΔH is negative and the reaction is exothermic.

If the products are higher in energy than the reactants, ΔH is positive and the reaction is endothermic.

Temperature and Reaction Rate

Increasing temperature increases the average kinetic energy of the particles. The particles move faster, so collisions occur more frequently.

More importantly, a greater fraction of the particles have energies equal to or greater than the activation energy. Therefore, a greater percentage of collisions successfully form products.

Increasing temperature does not lower the activation energy. It increases the fraction of particles that have enough energy to overcome the existing activation-energy barrier.

Catalysts

A catalyst increases the rate of a reaction by providing an alternative pathway with a lower activation energy. A catalyst participates in the reaction mechanism but is regenerated and is not consumed by the overall reaction.

A catalyst:

• Lowers the activation energy

• Increases both forward and reverse reaction rates

• Does not change ΔH

• Does not change the equilibrium constant

• Does not change the equilibrium amounts of reactants and products

Reaction Mechanisms

A reaction mechanism is a proposed sequence of elementary steps that explains how an overall reaction occurs.

An elementary step represents a single molecular event. When all elementary steps are added together, they must produce the overall balanced reaction.

A valid reaction mechanism must:

1. Add together to produce the correct overall reaction

2. Be consistent with the experimentally determined rate law

Molecularity of Elementary Steps

Molecularity	Description	General Example
Unimolecular	One reactant particle is involved.	A → products
Bimolecular	Two reactant particles collide.	A + B → products
Termolecular	Three reactant particles collide simultaneously.	A + B + C → products

Termolecular elementary steps are uncommon because three particles must collide at the same moment with the required energy and orientation.

Intermediates

An intermediate is produced during one elementary step and consumed during a later step. It does not appear in the overall reaction.

Example: A Two-Step Reaction Mechanism

Consider the following proposed mechanism:

Step 1: NO₂ + F₂ → NO₂F + F slow

Step 2: NO₂ + F → NO₂F fast

Add the two steps and cancel F because it is produced in Step 1 and consumed in Step 2.

2NO₂ + F₂ → 2NO₂F

F is an intermediate because it is formed and then consumed. It does not appear in the overall reaction.

The Rate-Determining Step

The slowest elementary step in a mechanism is commonly called the rate-determining step because it limits how quickly the overall reaction can proceed.

For an elementary step, the rate-law exponents can be taken directly from the coefficients of the reactants in that step.

For the slow first step in the previous mechanism:

NO₂ + F₂ → NO₂F + F

Rate = k[NO₂][F₂]

Coefficients may be used as rate-law exponents for an elementary step. They should not automatically be used as exponents for the overall balanced reaction.

Energy Profiles for Multistep Mechanisms

A multistep reaction produces an energy profile containing multiple peaks and valleys.

Each peak represents a transition state.
Each valley between peaks represents an intermediate.
Each peak corresponds to one elementary step.
The step with the largest activation-energy barrier is generally the slowest step.

If an energy profile has two peaks, the mechanism contains two elementary steps. The valley between the peaks represents an intermediate.

This two-step mechanism has two transition states and one intermediate. The first step has the larger activation-energy barrier and is therefore slower.

Practice Questions

Question 1: Temperature and Reaction Rate

A reaction occurs much faster at 50°C than at 20°C. Explain this observation using collision theory.

Show solution

At the higher temperature, the reactant particles have greater average kinetic energy and move faster. This causes collisions to occur more frequently.

More importantly, a greater fraction of the collisions have enough energy to overcome the activation-energy barrier. Therefore, more collisions successfully produce products.

Increasing temperature does not reduce the activation energy.

Question 2: Reading an Energy Profile

An uncatalyzed reaction has the following energy values:

Reactants: 40 kJ/mol
Transition state: 110 kJ/mol
Products: 15 kJ/mol

Calculate the forward activation energy, reverse activation energy, and ΔH.

Show solution

The forward activation energy is:

E_{a, forward} = 110 kJ/mol − 40 kJ/mol = 70 kJ/mol

The reverse activation energy is:

E_{a, reverse} = 110 kJ/mol − 15 kJ/mol = 95 kJ/mol

The enthalpy change is:

ΔH = 15 kJ/mol − 40 kJ/mol = −25 kJ/mol

Because ΔH is negative, the reaction is exothermic.

Question 3: Effect of a Catalyst

A catalyst is added to a reaction at constant temperature. State how the catalyst affects each of the following:

Activation energy
Reaction rate
ΔH
Equilibrium constant

Show solution

The catalyst provides a different reaction pathway with a lower activation energy. This causes the reaction rate to increase.

The catalyst does not change ΔH because the reactants and products still have the same energies.

The catalyst also does not change the equilibrium constant. It increases both the forward and reverse rates, allowing equilibrium to be reached more quickly.

Question 4: Analyzing a Mechanism

Consider the proposed mechanism:

Step 1: A + B → C + D slow

Step 2: D + E → F fast

Determine the overall reaction, identify the intermediate, and write the rate law predicted by the slow step.

Show solution

Add the elementary steps and cancel D because it is produced in Step 1 and consumed in Step 2.

A + B + E → C + F

The overall reaction is A + B + E → C + F.

D is the intermediate because it is formed and then consumed.

Because the first step is an elementary slow step, its reactant coefficients may be used in its rate law:

Rate = k[A][B]

Question 5: Catalyst Versus Intermediate

Consider the proposed mechanism:

Step 1: A + X → AX

Step 2: AX + B → AB + X

Identify the catalyst and intermediate, and write the overall reaction.

Show solution

Add the two steps and cancel X and AX.

A + B → AB

X is consumed in Step 1 and regenerated in Step 2. Therefore, X is the catalyst.

AX is formed in Step 1 and consumed in Step 2. Therefore, AX is the intermediate.

Question 6: Multistep Energy Diagram

An energy-profile diagram contains three peaks and two valleys between the reactants and products.

How many elementary steps, transition states, and intermediates are represented?

Show solution

Each peak represents one transition state and one elementary step. Therefore, the mechanism contains:

Three elementary steps
Three transition states
Two intermediates

The two valleys between the three peaks represent the two intermediates.

VIDEO EXPLANATION

Watch: Collision Theory and Mechanisms

Watch an explanation of activation energy, catalysts, energy profiles, and multistep reaction mechanisms.

Watch on TikTok →