Chapter 15 Chemical Kinetics: The Rates of Chemical Reactions

Jeffrey MackCalifornia State University,

Sacramento

Chapter 15

Chemical Kinetics: The Rates of Chemical Reactions

Chemical Kinetics will now provide information about the arrow!

This gives us information on HOW a reaction occurs!

Reactants Products

Chemical Kinetics: The Rates of Chemical Reactions

Thermodynamics – does a reaction take place?

Kinetics – how fast does a reaction proceed?

Kinetics is the study of how fast (Rates) chemical reactions occur.Important factors that affect the rates of chemical reactions:

• reactant concentration or surface area in solids

• temperature• action of catalysts

Our goal: Use kinetics to understand chemical reactions at the particle or molecular level.

Chemical Kinetics

• Reactants go away with time.• Products appear with time.• The rate of a reaction can be measured by

either. In this example by the loss of color with time.

Rate of Reactions

Blue dye is oxidized with bleach. Its concentration decreases with time.The rate — the change in dye conc. with time — can be determined from the plot.

Dye

Con

c

Time

Determining a Reaction Rate

A B

aA + bB cC + dD

Rate of reaction

In general for the reaction:

t

1c

C

reactants go away with time therefore the negative sign…

A1a t

Bt

1b

t

1d

D

Reaction Rate & Stoichiometry

Reaction rate is the change in the concentration of a reactant or a product with time (M/s).


The rate of appearance or disappearance is measured in units of concentration vs. time.

Rate =moles

L

time

There are three “types” of rates1.initial rate2.average rate3.instantaneous rate

= Ms1 or Mmin1 etc...


Rea

ctan

t con

cent

ratio

n (M

)

The concentration of a reactant decreases with time.

Time

Reaction Rates

Rea

ctan

t con

cent

ratio

n (M

)

Time

Initial rate

Reaction Rates

Rea

ctan

t con

cent

ratio

n (M

)

Time

Y [concentration]Average Rate = X time

Reaction Rates

Rea

ctan

t con

cent

ratio

n (M

)

Time

Instantaneous rate (tangent line)

Reaction Rates

Rea

ctan

t con

cent

ratio

n (M

)

Time


Initial rate

Reaction Rates

During the beginning stages of the reaction, the initial rate is very close to the instantaneous rate of reaction.

Problem:Consider the reaction:

Over a period of 50.0 to 100.0 s, the concentration of NO(g) drops from 0.0250M to 0.0100M.a) What is the average rate of disappearance of NO(g)

during this time?

2 22NO (g) O (g) 2NO (g)

a)

Problem:Consider the reaction:

Over a period of 50.0 to 100.0 s, the concentration of NO(g) drops from 0.0250M to 0.0100M.a) What is the rate of rxn?b) What is the average rate of disappearance of NO(g) during this time?

2 22NO (g) O (g) 2NO (g)

= 1.50104 Ms1

(0.0100M 0.0250M)

100.0 s 50.0 s

1 mol of reaction 2 moles NO (g)

RATE= -[NO]t

12

b) [NO]/t = -0.0150/50.0 M/s = -3.00 X 10-4M/s

Practice example

Write the rate expression for the following reaction:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

rate = -[CH4]t = - [O2]

t12

= [H2O]t

12

= [CO2]t

• There are several important factors that will directly affect the rate of a reaction:

• Temperature• The physical state of the reactants• Addition of a catalyst

All of the above can have a dramatic impact on the rate of a chemical process.

Reaction Conditions & Rate

Bleach at 54 ˚C Bleach at 22 ˚C

Reaction Conditions & Rate: Temperature

Reaction Conditions & Rate: Physical State of Reactants

Catalyzed decomposition of H2O2 2 H2O2 2 H2O + O2

Reaction Conditions & Rate: Catalysts

0.3 M HCl 6 M HCl

Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g)

Effect of Concentration on Reaction Rate: Concentration

The rate of reaction must be a function of concentration:• As concentration increases, so do the number of

collisions…• As the number of collisions increase, so does the

probability of a reaction.• This in turn increases the rate of conversion of

reactants to products. • The relationship between reaction rate and

concentration is given by the reaction Rate Law Expression.

• The reaction rate law expression relates the rate of a reaction to the concentrations of the reactants.

The Rate Law Expression

aA + bB cC + dD

x and y are the reactant orders determined from experiment.

x and y are NOT the stoichiometric coefficients.

x yRate k [A] [B]

Each concentration is expressed with an order (exponent).

The rate constant converts the concentration expression into the correct units of rate (Ms1).

For the general reaction:

Reaction Order

A reaction order can be zero, or positive integer and fractional number.

Order Name Rate Law 0 zeroth rate = k[A]0

= k1 first rate = k[A]2 second rate = k[A]2

0.5 one-half rate = k[A]1/2

1.5 three-half rate = k[A]3/2

0.667 two-thirds rate = k[A]2/3

Reaction Orders

Overall order:

or sevenhalves order

note: when the order of a reaction is 1 (first order) no exponent is written.

1 + ½ + 2 = 3.5 = 7/2

Reaction Order

To find the units of the rate constant, divide the rate units by the Molarity raised to the power of the overall reaction order.

1 x yRate (Ms ) k [A] [B]

1Ms x yk M M

x yk M 1Ms

k =1Ms

x yM

1 x y 1M s

if (x+y) = 1 k has units of s-1

if (x+y) = 2 k has units of M-1s-1

Reaction Constant

• The rate constant, k, is a proportionality constant that relates rate and concentration.

• It is found through experiment that the rate constant is a function temperature.

• Rate constants must therefore be reported at the temperature with which they are measured.

• The rate constant also contains information about the energetics and collision efficiency of the reaction.

Reaction Constant

EXAMPLE: The reaction,

2 NO (g) + 2 H2 (g) N2 (g) + 2 H2O (g)

is experimentally found to be first order in H2 and third order in NO

a) Write the rate law.


2 NO (g) + 2 H2 (g) N2 (g) + 2 H2O (g)



Rate(Ms-1) = k [H2] [NO]3

b) What is the overall order of the reaction?


2 NO (g) + 2 H2 (g) N2 (g) + 2 H2O (g)





= 4 Overall order = 1 + 3 “4th order”

c) What are the units of the rate constant?


2 NO (g) + 2 H2 (g) N2 (g) + 2 H2O (g)





= 4 Overall order = 1 + 3 “4th order”

c) What are the units of the rate constant?

x yRate = k [A] [B] If the rate doubles when [A] doubles and [B] stays constant, the order for [A] is?

Rate Law Expression


one… 1

Rate Law Expression


one… 1

Rate Law Expression


Rea

ctan

t con

cent

ratio

n (M

)

During the beginning stages of the reaction, the initial rate is very close to the instantaneous rate of reaction.

Initial rate

Determining a Rate Equation

The reaction of nitric oxide with hydrogen at 1280 °C is as follows:

2NO (g) + 2H2 (g) N2 (g) + 2H2O (g)

From the following experimental data, determine the rate law and rate constant.

Trial [NO]o (M) [H2]o (M) Initial Rate (Mmin-1)

1 0.0100 0.0100 0.00600

2 0.0200 0.0300 0.144

3 0.0100 0.0200 0.0120

Determining Reaction Order: The Method of Initial Rates


1 0.0100 0.0100 0.00600

2 0.0200 0.0300 0.144

3 0.0100 0.0200 0.0120


Notice that in Trial 1 and 3, the initial concentration of NO is held constant while H2 is changed.


2NO (g) + 2H2 (g) N2 (g) + 2H2O (g)


1 0.0100 0.0100 0.00600

2 0.0200 0.0300 0.144

3 0.0100 0.0200 0.0120



2NO (g) + 2H2 (g) N2 (g) + 2H2O (g) This means that any changes to the rate must be due to the changes in H2 which is related to the concentration of H2 & its order!

Rate(M/min) = k [NO]x [H2]y

The rate law for the reaction is given by:

2NO(g) + 2H2(g) N2(g) + 2H2O(g)

Taking the ratio of the rates of Trials 3 and 1 one finds:

0.0120 M/min0.00600 M/min

x y

x yk [0.0100] [0.0200] k [0.0100] [0.0100]

y

y[0.0200][0.0100]

Plugging in the values from the data:

Rate (Trial 3)

Rate (Trial 1)=


2.00

2.00log

Take the log of both sides of the equation:

log(2.00)

log(2.00)

y = 1 Rate(M/min) = k [NO]x[H2]


Rate(2)Rate(1)

0.1440.00600

x

xk [0.0200] [0.0300] k [0.0100] [0.0100]

24 x2 3

x2 8

xlog 2 log 8 x = 3

Similarly for x: Rate(M/min) = k [NO]x[H2]y

k x(2)[NO] y

2 (1)[H ]

k x(1)[NO] y

2 (2)[H ]


The Rate Law expression is:

The order for NO is 3

The order for H2 is 1

The over all order is 3 + 1 =4

2NO(g) + 2H2(g) N2(g) + 2H2O(g)


The Rate constant

Rate(M/min) = k [NO]3[H2]

To find the rate constant, choose one set of data and solve:

310.0120 Mmin k 0.0100 M 0.0200 M

1

30.0120 Mmink

0.0100 M 0.0200 M

1

3 4

0.0120 Mmin0.0100 0.0200 M

5 3 1k 6.00 10 M min


• It is important know how long a reaction must proceed to reach a predetermined concentration of some reactant or product.

• We need a mathematical equation that relates time and concentration:

• This equation would yield concentration of reactants or products at a given time.

• It will all yield the time required for a given amount of reactant to react.

Concentration–Time Relationships:Integrated Rate Laws

For a zero order process where “A” goes onto products, the rate law can be written:

A products

1Rate(Ms ) [A]t

0k[A] = k

k has units of Ms1

For a zero order process, the rate is the rate constant!

Integrated Rate Laws

This is the “average rate”If one considers the infinitesimal changes in concentration and time the rate law equation becomes:

This is the “instantaneous rate”

Zero order kinetics A products


1Rate(Ms ) [A]t

0k[A] = k

[A]t [A]o = k(t 0) = kt

and [A] = [A] at time t = twhere [A] = [A]o at time t = 0

[A]t [A]o = kt

Zero order kinetics


[A]o [A]t = kt

what’s this look like?

y = mx + b

[A]t = kt + [A]o

rearranging…

a plot of [A]t vs t looks like…

[A] t

(mol

s/L)

t (time)

the y-intercept is [A]o

Conclusion: If a plot of reactantconcentration vs. time yieldsa straight line, then the reactantorder is ZERO!

slope = kk has units of M×(time)1

Zero order kinetics


For a first order process, the rate law can be written:

A products

1 [A]Rate(Ms ) k[A]t

This is the “average rate”

If one considers the infinitesimal changes in concentration and time the rate law equation becomes:

1 d[A]Rate(Ms ) k[A]dt

This is the “instantaneous rate”


t

o

[A] t0[A]

d[A] k dt[A]

t

o

[A]ln kt[A]

ktt

o

[A] e[A]

Taking the exponent to each side of the equation:

or ktt o[A] [A] e

Conclusion: The concentration of a reactant governed by first order kinetics falls off from an initial concentration exponentially with time.

First order kinetics


ktt oln [A] [A] e

tln[A]

Taking the natural log of both sides…

k toln [A] e kt

oln[A] ln e

ln[A]t =

rearranging…

= ln[A]o kt

kt + ln[A]o



y = mx + b so a plot of ln[A]t vs t looks like…

ln[A

] t

t (time)

the y-intercept is ln[A]o

Conclusion: If a plot of natural log of reactant concentration vs. time yields a straight line, then the reactant order is FIRST!

slope = k

ln[A]t = kt + ln[A]o

k has units of (time)1



Problem:The decomposition of N2O5(g) following 1st order kinetics.If 2.56 mg of N2O5 is initially present in a container and 2.50 mg remains after 4.26 min, what is the rate constant in s1?

Problem:The decomposition of N2O5(g) following 1st order kinetics.If 2.56 mg of N2O5 is initially present in a container and 2.50 mg remains after 4.26 min, what is the rate constant in s1? Begin with the integrated rate law for a 1st order process:

ktt

o

[A] e[A]

ktt o[A] [A] e

2 5 t

2 5 o

[N O ][N O ]

Wait… what is the volume of the container???Do we need to convert to moles?


2 5 t

2 5 o

[N O ][N O ]

Check it out!You don’t need the volume of the container!

3

2 5 t

2 5 o3

1 g 1 mol2.50 mg108.0 g10 mg

N O 2.50 mg1 g 1 molN O 2.56 mg2.56 mg

108.0 g10 mg

L

L

x

x


taking the natural log and substituting time in seconds:

2.50 mgln k 256 s2.56 mg

k = 9.3 105 s1

2 5 ktt

2 5 o

N O 2.50 meN O 2.56 m

Rate = k[A]2

21 d ARate(Ms ) k A

dt

Integrating as before we find:

t 0

1 1ktA A

A Productsk has units of M1s1

Second order kinetics


y = mx + b so a plot of 1/[A]t vs t looks like…

1/[A

] t

t (time)

the y-intercept is 1/[A]o

Conclusion: If a plot of one over reactant concentration vs. time yields a straight line, then the reactant order is second!

slope = k

Second order kinetics

t 0

1 1ktA A

k has units of M1s1


Summary of Integrated Rate Laws

Half-life of a reaction is the time taken for the concentration of a reactant to drop to one-half of the original value.

Half-life & First-Order Reactions

0t

[A][A]2

For a first order process the half life (t½ ) is found mathematically from:

t 0(1) ln A kt ln A

t 0(2) ln A ln A kt

t0

A(3) ln kt

[ ]A

Start with the integrated rate law expression for a 1st order process

Bring the concentration terms to one side.

Express the concentration terms as a fraction using the rules of ln.

at time = t½

Reaction Half-Life: 1st Order Kinetics

0A(4) ln kt

[A]

01

0 2

A(5) ln kt

[A]2

exchange [A] with [A]0 to reverse the sign of the ln term and cancel the negative sign in front of k

Substitute the value of [A] at the half-life


12

ln 2 kt

12

ln2 0.693tk k

So, knowing the rate constant for a first order process, one can find the half-life!

The half-life is independent of the initial concentration!


1

2

0

0

2 Aln kt

A

Problem:A certain reaction proceeds through first order kinetics.The half-life of the reaction is 180. s.What percent of the initial concentration remains after 900.s?


Using the integrated rate law, substituting in the value of k and 900.s we find:



k = 0.00385 s-112

ln2tk

112

ln2 ln2kt 180.s



k = 0.00385 s-112

ln2tk

112

ln2 ln2kt 180.s

ktt

0

[A][A]

e10.00385 900.st

0

[A] e[A]

s = 0.0312



k = 0.00385 s-112

ln2tk

112

ln2 ln2kt 180.s

ktt

0

[A][A]

e10.00385 900.st

0

[A] e[A]

s = 0.0312

Since the ratio of [A]t to [A]0 represents the fraction of [A] that remains, the % is given by:

100 0.0312 = 3.12%

Elements that decay via radioactive processes do so according to 1st order kinetics:

Element: Half-life:238U 234Th +

14C 14N +

131I 131Xe +

4.5 109 years

5730 years

8.05 days


Tritium decays to helium by beta () decay:

The half-life of this process is 12.3 years

Starting with 1.50 mg of 3H, what quantity remains after 49.2 years.Solution: Begin with the integrated rate law expression for 1st order kinetics.

ktt

0

[A] e[A]

3 0 31 -1 2H Hee


Recall that that the rate constant for a 1st order process is given by:

12

ln(2)kt

ktt

0

[A] e[A]

kt[A] [A] et 0

[3H]t = 1.50 mg 0.693 49.2 years

12.3e years

= 0.094 mg


Notice that 49.2 years is 4 half-lives…49.2 412.3

After 1 half life:1.50 mg

2 = 0.75 mg remains

After 2 half life's:0.75 mg

2 = 0.38 mg remains


2 = 0.19 mg remains


2 = 0.094 mg remains


Arrhenius: Molecules must posses a minimum amount of energy to react. Why?(1) In order to form products, bonds must be

broken in the reactants.(2) Bond breakage requires energy.(3) Molecules moving too slowly, with too little

kinetic energy, don’t react when they collide.

The Activation energy, Ea, is the minimum energy required to initiate a chemical reaction.

Ea is specific to a particular reaction.

A Microscopic View of Reaction Rates

When one writes a reaction all that is seen are the reactants and products. This details the overall reaction stoichiometry.How a reaction proceeds is given by the reaction mechanism.

A Microscopic View of Reaction Rates

The reaction of NO2 and CO (to give NO and CO2) has an activation energy barrier of 132 kJ/mol-rxn. The reverse reaction (NO + CO2 NO2 + CO) requires 358 kJ/mol-rxn. The net energy change for the reaction of NO2 and CO is 226 kJ/mol-rxn.

Activation Energy

The progress of a chemical reaction as the reactants transform to products can be described graphically by a Reaction Coordinate.

Pot

entia

l Ene

rgy

Reaction Progress

reactants

productsHRXN

Eact

In order for the reaction to proceed, the reactants must posses enough energy to surmount a reaction barrier.

Transition State

Activation Energy

The temperature for a system of particles is described by a distribution of energies.

At higher temps, more particles have enough energy to go over the barrier.

E > Ea

E < Ea

Since the probability of a molecule reacting increases, the rate increases.

Activation Energy

Molecules need a minimum amount of energy to react. Visualized as an energy barrier - activation energy, Ea.

Reaction coordinate diagram

Activation Energy

Orientation factors into the equation

The orientation of a molecule during collision can have a profound effect on whether or not a reaction occurs.

When the green atom collides with the green atom on the molecule, a reactive or effective collision occurs.

The reaction occurs only when the orientation of the molecules is just right…

Activation Energy

Orientation factors into the equation

When the green atom collides with the red atom on the molecule, this leads to a non-reactive or ineffective collision occurs.

In some cases, the reactants must have proper orientation for the collision to yield products.

This reduces the number of collisions that are reactive!

Activation Energy

Arhenius discovered that most reaction-rate data obeyed an equation based on three factors:

(1) The number of collisions per unit time.

(2) The fraction of collisions that occur with the correct orientation.

(3) The fraction of the colliding molecules that have an energy greater than or equal to Ea.

From these observations Arrhenius developed the aptly named Arrhenius equation.

The Arrhenius Equation

Both A and Ea are specific to a given reaction.

k is the rate constant

Ea is the activation energy R is the ideal-gas constant (8.314 J/Kmol)

T is the temperature in K

A is known the frequency or pre–exponential factor

In addition to carrying the units of the rate constant, “A” relates to the frequency of collisions and the orientation of a favorable collision probability


Temperature Dependence of the Rate Constant:Increasing the temperature of a reaction generally speeds up the process (increases the rate) because the rate constant increases according to the Arrhenius Equation.

Rate (Ms-1) = k[A]x[B]y

As T increases, the value of the exponential part of the equation becomes less negative thus increasing the value of k.


Reactions generally occur slower at lower T.

Iodine clock reaction. H2O2 + 2 I- + 2 H+ 2 H2O + I2

Room temperature

In ice at 0 oC

Effect of Temperature

Determining the Activation Energy Ea may be determined experimentally. First take natural log of both sides of the Arrhenius equation:

lnaElnk ln

RTA

ln k 1A plot of ln k vs will have a slope

T

of and a y-intercept of ln A. aER

y = mx + b


One can determine the activation energy of a reaction by measuring the rate constant at two temperatures:Writing the Arrhenius equation for each temperature:

Subtracting k1 from k2 we find that:

Determining the Activation Energy


Knowing the rate constants at two temps yields the activation energy.

orKnowing the Ea and the rate constant at one temp allows one to find k(T2)

Determining the Activation Energy


Problem:The activation energy of a first order reaction is 50.2 kJ/mol at 25 °C. At what temperature will the rate constant double?

(1)

(2)

(3)

Problem:The activation energy of a first order reaction is 50.2 kJ/mol at 25 °C. At what temperature will the rate constant double?

(4)

(5)

T2 = 308 K A 10 °C change of temperature doubles the rate!!

algebra!

Catalysts speed up reactions by altering the mechanism to lower the activation energy barrier.

Dr. James Cusumano, Catalytica Inc.

What is a catalyst?

Catalysts and society

Catalysts and the environment

Catalysis

In auto exhaust systems — Pt, NiO 2 CO + O2 2 CO2

2 NO N2 + O2

Catalysis

2. Polymers: H2C=CH2 polyethylene3. Acetic acid: CH3OH + CO CH3CO2H4. Enzymes — biological catalysts

Catalysis

Catalysis and activation energy

Uncatalyzed reaction

Catalyzed reaction

MnO2 catalyzes decomposition of H2O2

2 H2O2 2 H2O + O2

Catalysis

Iodine-Catalyzed Isomerization of cis-2-Butene

Iodine-Catalyzed Isomerization of cis-2-Butene

The overall stoichiometry of a chemical reaction is most often the sum of several steps:

(1) 2AB A2B2

(2) A2B2 + C2 A2B + C2B(3) A2B + C2 A2 + C2B

2AB + A2B2 + A2B + 2C2 A2B2 + A2B + A2 + 2C2B

Net: 2AB + 2C2 A2 + 2C2B

The sequence of steps (1-3) describes a possible “reaction mechanism”.

Reaction Mechanisms

The overall stoichiometry of a chemical reaction is most often the sum of several steps:

(1) 2AB A2B2

(2) A2B2 + C2 A2B + C2B(3) A2B + C2 A2 + C2B

2AB + A2B2 + A2B + 2C2 A2B2 + A2B + A2 + 2C2B

Net: 2AB + 2C2 A2 + 2C2B

The species that cancel out (not part of the overall reaction) are called “reaction intermediates”.

Reaction Mechanisms

(1) 2AB A2B2

(2) A2B2 + C2 A2B + C2B(3) A2B + C2 A2 + C2B

2AB + 2C2 A2 + 2C2B

Each step in the mechanism is called an “elementary step”.The number of reactants in an elementary step is called the “molecularity”.In this example each step (1-3) is a bimolecular process. (2 reactants)

A2 2A is a unimolecular process (1 reactant)

2A + B is a termolecular process (3 reactants)

Reaction Mechanisms

The rate of the overall reaction can never be faster than the “slowest step” in the mechanism.If reaction (1) is the slowest of the three steps in the mechanism…Then it is known as the “rate determining step”

slowfastfast

(1) 2AB A2B2

(2) A2B2 + C2 A2B + C2B(3) A2B + C2 A2 + C2B

2AB + 2C2 A2 + 2C2B

Reaction Mechanisms

slowfastfast

(1) 2AB A2B2

(2) A2B2 + C2 A2B + C2B(3) A2B + C2 A2 + C2B

2AB + 2C2 A2 + 2C2B

The slowest step controls the rate of the reaction.It determines the rate law!

Rate (Ms-1) = k[AB]2

The rate law is not based on the overall reaction:Rate (Ms-1) = k[AB]2[C2]2

Reaction Mechanisms

Consider the following reaction:

2NO2(g) + F2(g) 2FNO2(g)If the reaction proceeded by the overall reaction, the rate law for the reaction would be 3rd order overall.The actual rate law is found to be:

Rate = k[NO2][F2]

Indicating that the slowest step in the mechanism is a bimolecular reaction between NO2 and F2.

Reaction Mechanisms

2NO2(g) + F2(g) 2FNO2(g)Rate = k[NO2][F2]

To explain the observed kinetics, a possible mechanism is proposed:

Reaction Mechanisms

2NO2(g) + F2(g) 2FNO2(g)

Rate = k[NO2][F2]

Since the 1st step in the reaction is the slow step, it determines the kinetics and rate law for the reaction:

Reaction Mechanisms

Validating a Reaction Mechanism:

A mechanism is a proposal of how the reaction proceeds at the molecular level.

1. The individual elementary steps must sum to yield the overall reaction with correct stoichiometry.

2. The predicted reaction rate law must be in agreement with the experimentally determined rate law.

3. Note that there may be more than one mechanism that is in agreement with the reaction stoichiometry and kinetics.

Reaction Mechanisms

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Chapter 15 Chemical Kinetics: The Rates of Chemical Reactions

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