Chemical KineticsA Study of the Rates of

Reactions

Consider a reaction in which

D A + BHow “fast” does the

reactant, D, disappear?

Concentration of D over Time

Time(s) [D] mol/ L

0 0.2000

10 0.1904

20 0.1812

30 0.1725

40 0.1641

50 0.1562

60 0.1487

70 0.1415

80 0.1347

90 0.1282

100 0.1220

This is the raw data collected during theexperiment

Start your analysis by graphing [D] v. Time

Concentration of D over Time

0.0500

0.0700

0.0900

0.1100

0.1300

0.1500

0.1700

0.1900

0.2100

0 50 100 150 200 250

Time(s)

[D]

mo

l/L

Concentration of D over Time y = -0.0006x + 0.1848

R2 = 0.9737

0.0500

0.0700

0.0900

0.1100

0.1300

0.1500

0.1700

0.1900

0.2100

0 50 100 150 200 250

Time(s)

[D]

mo

l/L

The data do not forma straight line.

The R2 value only hasone “9”.

This reaction is NOT zero order! How do we knowthis? The plot is NOT a straight line!

R2= 0.9737

Time(s) [D] mol/ L Ln[D] mol/ L

0 0.2000 -1.609

10 0.1904 -1.659

20 0.1812 -1.708

30 0.1725 -1.758

40 0.1641 -1.807

50 0.1562 -1.856

60 0.1487 -1.906

70 0.1415 -1.955

80 0.1347 -2.005

90 0.1282 -2.054

100 0.1220 -2.103

Next, convert your raw data to natural logsand graph ln[D] v. time

What is the difference between a natural logand a common log?

•Natural logs use a base value of “e”, 2.71 andis designated with the symbol “ln”

•Common logs use a base value of “10” and isdesignated with the symbol “log”

•The conversion between natural and commonlogs is: log x = 2.303 ln x

Natural Log of [D] over Time

-3.000

-2.800

-2.600

-2.400

-2.200

-2.000

-1.800

-1.600

0 50 100 150 200 250

Time(s)

Ln

[D

] m

ol/L

Natural Log of [D] over Time y = -0.0049x - 1.6094

R2 = 1

-3.000

-2.800

-2.600

-2.400

-2.200

-2.000

-1.800

-1.600

0 50 100 150 200 250

Time(s)

Ln

[D

] m

ol/L

This reaction isfirst order.

Ln [D

]

time

The R2 value is 1.0.

The data DO form a straight line.

Plot of ln [D] vs time

Normally, when you get a straight lineyou stop making graphs.

For teaching purposes, let’s go aheadand look at the third kind of graph:

1/[D] v. time

Inverse [D] over time

4.000

6.000

8.000

10.000

12.000

14.000

16.000

18.000

0 50 100 150 200 250

time(s)

1/[

D]

L/m

ol

Inverse [D] over time y = 0.0477x + 3.9571

R2 = 0.9737

4.000

6.000

8.000

10.000

12.000

14.000

16.000

18.000

0 50 100 150 200 250

time(s)

1/[

D]

L/m

ol

Reaction Rates

average rate

• the rate over a specific time interval

instantaneous rate

• the rate for an infinitely small interval

Rate Law

Reaction rate = k [A]m [B]n

where m => order with respect to A

n => order with respect to B

overall order = m + n

Order of Reaction

• exponent of the concentration for a reactant that implies the number of molecules of that species involved in the rate determining step

• first order, exponent equals one

• second order, exponent equals two

Rates are studied for the first few moments of the chemical reaction.

Method of Initial Rates

0.0000

0.0020

0.0040

0.0060

0.0080

0.0100

0.0120

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Time (s)

Co

nce

ntr

atio

n

Concentrations are changed and the change in rate is calculated.

Method of Initial Rates

0.0000

0.0020

0.0040

0.0060

0.0080

0.0100

0.0120

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Time (s)

Co

nce

ntr

atio

n

The initial rate is found by calculating theAVERAGE RATE over the first few moments of the reaction.

Method of Initial Rates

0.0000

0.0020

0.0040

0.0060

0.0080

0.0100

0.0120

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Time (s)

Co

nce

ntr

atio

n

We know that the average rates will be different because the slope of each line is different.

Method of Initial Rates

0.0000

0.0020

0.0040

0.0060

0.0080

0.0100

0.0120

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Time (s)

Co

nce

ntr

atio

n

When several initial concentrations have been tried, we can generate a table of data that looks like this:

2R P

[R], M

0.1

0.2

0.4 1.1x 10-4

Initial Rate (mol/Ls)

2.7 x 10-5

5.4 x 10-5

What would the basic rate law look like for this reaction?

2R P

Rate = k [R]m

The equation says that the concentration of R is proportional to the rate of reaction.

[R]m : rate

We need to solve for m and k. We start by solving for m.

Basic Rate Law: Rate = k [R]m

Chemical Reaction: 2R P

[R], M

0.1

0.2

0.4 1.1x 10-4

Initial Rate (mol/Ls)

2.7 x 10-5

5.4 x 10-5

Solving for musing the Method of

Initial Rates

Rate = k [R]m

Basic Rate Law: Rate = k [R]m

Chemical Reaction: 2R P

By what factor did the concentration increase from experiment 1 to experiment 2? It doubled.

Experiment 2 = 0.2 = 2Experiment 1 0.1

[R], M

0.1

0.2

0.4 1.1x 10-4

Initial Rate (mol/Ls)

2.7 x 10-5

5.4 x 10-5

Basic Rate Law: Rate = k [R]m

This is a 1:1 relationship between concentrationand rate.

How can we use the Basic Rate Law to guide us tothis relationship?

[R]m : rate The rate goes up when the concentration goes up.

2m = 2 The concentration doubled and the rate doubled.

m = 1 Solve for m

Thus, the rate law is: Rate = k [R]1 or Rate = k [R]

Let’s review some basic math:

20 =

21 =

22 =

23 =

24 =

4

2

1

8

16

3m = 9

3m = 3

4m = 16

5m = 125

456m = 1

m = 2

m = 1

m = 2

m = 3

m = 0

Solving for kusing the Method

of Initial Rates

Rate = k [R]m

The Rate Law is : Rate = k [R]

Now, we can solve for k by inserting the information in the table into the Rate Law.

Rate = k [R]

2.7 x 10-5 M/s = k [0.1 M]

k = 2.7 x 10-4 s-1

[R], M

0.1

0.2

0.4 1.1x 10-4

Initial Rate (mol/Ls)

2.7 x 10-5

5.4 x 10-5

Experiment 1: k = 2.7 x 10-4 s-1

Experiment 2: k = 2.7 x 10-4 s-1

Experiment 3: k = 2.7 x 10-4 s-1

The value for k, the proportionality constant, is the same for all three experiment.

It is a good idea to check at least two experimentsto see that k is the same. If you made a mistakefinding the rate law, the k’s will differ!

[R], M

0.1

0.2

0.4 1.1x 10-4

Initial Rate (mol/Ls)

2.7 x 10-5

5.4 x 10-5

We have found

1. The rate law for the reaction RP:

Rate = k [R]

2. The value of the proportionality constant, k:

k = 2.7 x 10-4

Now we can write the rate law for the reaction

RP

Rate = (2.7 x 10-4)[R]

Final form for the Rate Law for this example:

Average Rates

Chemical Kinetics of NO2

Chemical Kinetics of NO2

Decomposition of NO2

0.0000

0.0020

0.0040

0.0060

0.0080

0.0100

0.0120

0 50 100 150 200 250 300 350 400 450

Time(s)

Con

cent

ratio

n (m

ole/

L)

NO2

NO

O2

Disappearance of NO2

0.0000

0.0020

0.0040

0.0060

0.0080

0.0100

0.0120

0 50 100 150 200 250 300 350 400 450

Time (s)

[NO

2]

Rate = [NO2] = 0.0031 mol/L - 0.0100 mol/L = -1.73 x 10-5 mol/L s time 400 s - 0 s

Why is the rate negative?

How do we fix this problem?

Rate = - [NO2] = - (0.0031 mol/L - 0.0100 mol/L) = 1.73 x 10-5 mol/L s

time 400 s - 0 s

We add a negative sign when a chemical is disappearing!

For all reactants:

Rate = - [reactant] time

For all products:

Rate = [product] D time

Use a negative signfor reactant rates!

Omit the negative signfor product rates!

Disappearance of NO2

0.0000

0.0020

0.0040

0.0060

0.0080

0.0100

0.0120

0 50 100 150 200 250 300 350 400 450

Time (s)

[NO

2]

Advantages of the Methodof Initial Rate:

1. Useful when a reaction is reversible. The reverse reaction won’t significantly contribute to the first few moments of the reaction.

2. Useful for very fast or very slow reactions.

Advantages of the Integrated Rate Method:

1. Useful for moderate length reaction.2. Doesn’t require multiple experiments to determine the order of the reaction.

Common Uses of the Method of Initial Rates:

1. To determine the order of the reaction.2. Find the rate constant, k .

Common Uses of the Method of Integrated Rate Laws:1. To determine the order of the reaction.2. Find the rate constant, k .

3. To determine the concentration at a certain time.4. To determine at what time a certain concentration will be reached.

Integrated Rate Laws

A ---> products

rate = - ([A]/t) = k[A]m

average rate

rate = - (d[A]/dt) = k[A]m

instantaneous rate

Integrated Rate Law:A reaction is followed for an extended period of time.

Method of Initial RatesA reaction is followed for only thefirst few moments of the reaction.