Chapter 13: Chemical Kinetics

Mrs. Brayfield

13.2: The Rate of a Chemical Reaction

When we measure rates (outside of chemistry) we may use something like miles per hour

But in chemistry, we are more interested in knowing the change in concentration of reactants over time

So our units are:

For example we can look at the rate for the following reaction:

Rate of a Reaction

We can also express the rate as:

Or, if we look at the products we can express the rate as:

We use the ½ in front of the rate because for every 1 mol of

reactants we have ½ mol of products

Note that the concentrations of reactants decrease and the

concentration of the product increases

Rate of a Reaction

Expressing Rate

The instantaneous rate of the reaction is the slope of

the curve at any point (on the graph from previous slide)

However, we can generalize the rate by using a generic

chemical equation:

Expressing Rate Example

For the balanced chemical reaction of:

Predict the rate of change in concentration of H2O2

during the time interval [from previous problem: in the

first 10.0 seconds of the reaction the concentration of I-

dropped from 1.000M to 0.868M]

Rates of a Chemical Reaction

Homework Problems: #1, 2, 4, 6, 8

13.3: The Rate Law

The rate of a chemical reaction typically depends on

more than one reactant

A rate law is the relationship between the rate of the

reaction and the concentration of the reactants:

For the reaction

Rate = k[A]n

Where: k = constant; n = a number called the reaction order

The reaction order is a number that determines how

the rate depends on the concentration of the reactants

Rate Law

The value of n determines how the rate depends on

concentration of the reactant:

If n = 0, the reaction is zero order

This means the rate is independent of the concentration of A

If n = 1, the reaction is first order

This means that the rate is directly proportional to the concentration

of A

If n = 2, the reaction is second order

This means that the rate is proportional to the square of the

concentration of A

Rate Law

Rate Law

The order of a reaction can only be determined ONLY by

experiment!

When we have multiple reactants, simply add them to the

rate law:

For the reaction:

Where m is the order with respect to A and n is the order

with respect to B

HOWEVER the overall order is the sum of the exponents

Rate Law Example

Consider the following reaction:

The initial rate of the reaction was measured at several

different concentrations of the reactants with the following

results:

From the data, determine:

a. The rate law for the reaction

b. The rate constant (k) for the reaction

[CHCl3] (M) [Cl2] (M) Initial Rate (M/s)

0.010 0.010 0.0035

0.020 0.010 0.0069

0.020 0.020 0.0098

0.040 0.040 0.027

Rate Law Example, cont.

From 1 to 2: [A] doubles while [B] is constant

The rate also doubles

This tells us that the rate order for [A] is first order

From 2 to 3: [A] is constant while [B] doubles

The rate is 1.5 times the previous rate

This tells us that the rate order for [B] is ½ order

The rate law is: Rate = k[A][B]1/2

[CHCl3] (M) [Cl2] (M) Initial Rate (M/s)

1 0.010 0.010 0.0035

2 0.020 0.010 0.0069

3 0.020 0.020 0.0098

4 0.040 0.040 0.027

Rate Law Example, cont.

To find k, pick one data set to plug into the rate equation:

Set #4:

Homework Problems #9, 10, 11, 12, 14, 16, 18

[CHCl3] (M) [Cl2] (M) Initial Rate (M/s)

1 0.010 0.010 0.0035

2 0.020 0.010 0.0069

3 0.020 0.020 0.0098

4 0.040 0.040 0.027

13.4: The Integrated Rate Law

So far we have looked at the rate of a reaction and the

change in concentration of a reactant

But we really want to know the change in concentration

of a reactant over time

For this we need to find the integrated rate law

There are three integrated rate laws, one for each order

reaction

First-Order Integrated Rate Law

We know that for a single reactant A:

And since

You can use calculus to integrate this rate law to find:

Where:

[A]t is the concentration of A at any time t

k is the rate constant

[A]0 is the initial concentration of A

First-Order Integrated Rate Law

Notice that the equation looks like the equation for a

straight line:

First-Order Integrated Rate Law Example

See page 484 in the book

In these problems you must first graph the data as shown

Excel is good

Use the graph to predict the concentration of SO2Cl2 at

1900s

Second Example

Cyclopropane rearranges to form propene in the gas

phase according to the following reaction:

The reaction is first order in cyclopropane and has a

measured rate constant of 3.36x10-5s-1 at 720K. If the

initial cyclopropane concentration is 0.0445M, what will

be the cyclopropane concentration after 235.0 minutes?

Second-Order Integrated Rate Law

We know that for a single reactant A:

And since

You can use calculus to integrate this rate law to find:

Where:

[A]t is the concentration of A at any time t

k is the rate constant

[A]0 is the initial concentration of A

Second-Order Integrated Rate Law

Notice that the equation looks like the equation for a

straight line:

Zero-Order Integrated Rate Law

We know that for a single reactant A:

And since

You can use calculus to integrate this rate law to find:

Where:

[A]t is the concentration of A at any time t

k is the rate constant

[A]0 is the initial concentration of A

Zero-Order Integrated Rate Law

Notice that the equation looks like the equation for a

straight line:

Second-Order Integrated Rate Law Example

Use the following data to show that the reaction is NOT

first order: [see excel sheet]

Finding the Rate Order Example

Using the following data find the overall rate order for

the reaction of a single reactant:

It is second order!

Time (s) [A] (M)

0 0.1000

5 0.0141

10 0.0078

15 0.0053

20 0.0040

Half-Life

The half-life (t½) of a reaction is the time required for the

concentration of a reactant to fall to one-half of its initial

value

For first-order reactions:

At a time equal to the half-life (t = t½):

Note that for a first-order reaction t½ does not depend on conc.

Half-Life

Half-Life Example

A first-order reaction has a half-life of 26.4 seconds. How

long will it take for the concentration of the reactant to

fall to one-eighth of its initial value?

1st half-life: 1/2 initial

2nd half-life: 1/4 initial

3rd half-life: 1/8 initial

3 half-lives x 26.4 seconds = 79.2 seconds

Half-Life

For a second-order reaction, the half-life equation is:

For a zero-order reaction, the half-life equation is:

See your book for the proof

On page 490 there is a nice summary chart of the three

rate laws

Homework Problems: #20, 22, 24, 26, 28

13.5: The Affect of Temperature on Rate

Changing the temperature affects the rate of a reaction

The rate constant, k, is only constant when the temperature

remains constant

Chemist Svante Arrhenius investigated this relationship:

Where A is the frequency factor

R is the ideal gas constant

T is the temperature

Ea is the activation energy

Activation Energy

The activation energy is the energy barrier or hump that

must be surmounted for the reactants to be turned into

products

Activation Energy

Activation Energy

The higher the Ea, the slower the reaction rate

The higher the hill the harder it is to go over it

Collision Theory

In the collision model a chemical reaction occurs after a

sufficiently energetic collision between two reactant

molecules

http://phet.colorado.edu/en/simulation/reactions-and-

rates

Homework Problems #31, 43

13.6: Reaction Mechanisms

Most chemical reactions do NOT take place in just one step alone (they take multiple steps)

For example:

Does not just take one step to complete

This does not show the intermediate steps that it took to complete

The full reaction is:

This is called a reaction mechanism

Or a series of individual steps by which an overall reaction occurs

Reaction Mechanisms

Each step in a reaction mechanism is called an elementary

step

Which cannot be broken into smaller steps

In any reaction mechanism, the elementary steps must

add to the overall reaction:

In this case HI is the reaction intermediate

This is not found in the overall reaction but plays a key role in

the mechanism

Reaction Mechanisms

If we look at the reaction:

And if the experimentally determined rate is:

We can say the possible mechanism is:

The slow step is called the rate determining step

We can look at the proposed energy diagram

Reaction Mechanisms

Reaction Mechanisms

The mechanism doesn’t have to be proven to be valid

Simply because other mechanism may exist and can be proven

The rate determining step is also the step that the rate

law is written for the overall reaction

So what is the rate law for the following reaction:

Reaction Mechanisms

If the overall reaction is:

And the experimental rate law is:

Given the proposed mechanism, determine if the

mechanism is valid:

Reaction Mechanisms

So the rate law for the rate determining step is:

However N2O2 is an intermediate and rate laws cannot

be expressed in terms of the intermediate

So we need to find a way to substitute the [N2O2]…

Reaction Mechanisms Example

Predict the overall reaction and rate law that would result

from the following two-step reaction:

Homework Problems: #45, 47, 48

13.7 Catalysis

A catalyst is a substance that increases the rate of a

chemical reaction but is not consumed by the reaction

For example:

This reaction by itself is very slow

However when you introduce a Cl atom (from

chlorofluorocarbons) this happens very quickly:

Catalysis

Homework Problem #50

Review Problems #54, 56, 61, 62, 71