Chapter 13 Chemical Kinetics

General Chemistry II

Textbook - Principles of Chemistry: A Molecular Approach by Nivaldo Tro

Rates of Reaction | 2

Chapter Outlines

Reaction Rate

How is the rate of reaction measured?

The variables/conditions that will affect the reaction rate

Quantitative Dependence of the Rate of Reaction

on Concentrations of Reactants

Quantitative Dependence of the Rate of Reaction

on Temperature

The Rate Law (Differential and Integrated)

Graphing Kinetic Data

Two Theories – Collision & Transition-State

The Arrhenius Equation

Reaction Mechanism

Catalysis

Example 1: Bombardier Beetle – an insect cannon

hydroquinone p-benzoquinone

+ → irritating

& hot

explosion instantaneously occurs

H2CO

HSO3

an indicator +

~ 35 s

Fact - Chemical reactions require

varying lengths of time for completion

Example 2: the “formaldehyde clock reaction” with bisulfite

H2CO + HSO3

C6H6O2 + H2O2 C6H4O2

13 | 4

What is Chemical kinetics ?

The following questions are explored in this chapter:

• How is the rate of a reaction defined and measured?

• What variables or conditions will affect the rate of a

reaction?

• How do you express the relationship of rate to the

variables affecting the rate? (Rate Law, Arrhenius Eqn. etc.)

• What happens on a molecular level during a

chemical reaction? (Reaction Mechanism)

13 | 5

Definition of Reaction Rate

第一时止第二时起

·Speediness/slowness of changes in concentration

the increase in molar concentration of product of

a reaction per unit time

or the decrease in molar concentration of reactant

per unit time.

·The unit for reaction rate is usually mol/(L•s) or M/s.

For the reaction N2O4(aq) 2NO2(g) which one of the curves below

reflects the concentration change of the product

An example about “change in concentration”

13 | 6

!!! The quantity “X” is always treated as X=XfinalXinitial

!!! Square brackets mean molarity

)(O)(NO4)(ON2 2252 ggg

Notations for Reaction Rate

i) Symbolically RRate justor

ii) Using the definition of rate

t

]ON[ 52

t

]O[ 2

t

]NO[ 2 or or

dt

d ]ON[ 52dt

d ]NO[ 2

dt

d ]O[ 2 or or

!!! The average rate vs. the instantaneous rate

For example

·The average rate notation tells one how to calculate the

rate of reaction:

)(O)(NO4)(ON2 2252 ggg

For decomposition of N2O5 2N2O5(g) 4NO2(g) + O2(g)

It was found that during the time interval from t = 600 s to t = 1200 s,

the concentration of N2O5 are

Calculate the average rate of decomposition of N2O5. sM6105.2:Answer

Iodide ion is oxidized by hypochlorite ion in basic solution

In 1.00 M NaOH solution at 25C, the iodide-ion concentration

(equal to the ClO concentration) at different times was as follows

I(aq) + ClO(aq) Cl(aq) + IO(aq)

Calculate the average rate of I over this time interval. 0.000113 M/s

Dinitrogen pentoxide, N2O5, decomposes to nitrogen dioxide and oxygen

The concentration of N2O5 at the beginning of an experiment was

0.0403 M. After 4.00 min, it was 0.00930 M. The average rate of

reaction of N2O5 in this time interval is qian

a. 0.00775 M/min.

b. 0.00233 M/min.

c. 0.0012 M/min.

d. 0.00101 M/min.

●

2N2O5(g) 4NO2(g) + O2(g)

●

PracT: 1 – 2, 4 - 5

13 | 9

Reactants and products for any given reaction have different

rates which are related via the reaction stoichiometry:

– inversely proportional to their coefficients

t

]A[

t

]B[

For a general chemical reaction

for any given rxn, there

are many different rates t

]E[

t

]D[ or

2N2O5(g) 4NO2(g) + O2(g) For example, for

Its rate of reaction can be written as

tΔ

]Δ[O2

tΔ

]OΔ[N

2

1 52tΔ

]Δ[NO

4

1 2PracT: 6 - 8

However, dividing each rate by its coefficient makes

b/c different stoichiometry coefficients

EDBA edba

or or

d

t

c

t

b

t

a

t

]D[]C[]B[]A[but these different rates

hold the relationship as

Note that tΔ

]Δ[NO2tΔ

]Δ[O2

tΔ

]OΔ[N 52

13 | 10

Reaction rates are determined experimentally in a variety

of ways.

For example, samples can be taken and analyzed from

the reaction at several different intervals.

How is the rate of reaction measured?

Continuously following the

reaction is more convenient.

This can be done by

measuring pressure, as

shown on the right, or by

measuring light absorbance

when a color change is part

of the reaction.

Start timing as Solutions I and

II are combined.

Carry out paired runs simultaneously: “head –to-head” comparisons of reaction

rates for different concentrations.

Solutions I and II for 1st run

Solutions I and II for 2st run

Note time at which each changes color. t

]OH[ 22OH 22

R

Lab 2. Kinetics of Dye Bleaching

Monitor concentration of dye (food coloring) using spectrophotometry

› Spectrophotometers connected to laptops

Collect data transfer to Excel

› Allows continuous monitoring of dye concentration

12

Ab

sorb

an

ce

Time

Blue#1 + OCl− colorless products

t

]1#Blue[1#Blue

R

13 | 13

There are FOUR such variables:

Variables/conditions that will affect the reaction rate

#1 - The concentrations of the reactants

#4 - The concentration of the catalyst

#2 - The temperature at which the reaction occurs

#3 - The surface area of the solid reactant

Often the rate of reaction increases when the concentration of a

reactant is increased. In some reactions, however, the rate is

unaffected by the concentration of a particular reactant, as long

as it is present at some concentration.

Usually reactions speed up when the temperature increases.

empirically rate doubles for every 10 temperature rise

The greater the surface area, the faster the reaction speed

A catalyst speeds up both forward and reverse reactions

PracT: 3 D02-Th 时03-始

13 | 14

Each test tube contains potassium

permanganate, KMnO4, and oxalic acid,

H2C2O4, at the same concentrations.

Permanganate ion oxidizes oxalic acid to

CO2 and H2O.

Top: One test tube was placed

in a beaker of warm water (40°C); the

other was kept at room temperature

(20°C).

Bottom: After 10 minutes, the test tube

at 40°C showed a noticeable reaction,

whereas the other did not.

Chemistry here? Temperature DOES

affect rate of reaction, and the higher

the T, faster the reaction is.

13 | 15

Surface Area of a Solid Reactant or

Catalyst

If a reaction involves a solid with a gas or

liquid, then the surface area of the solid

affects the reaction rate. Because the

reaction occurs at the surface of the solid,

the rate increases with increasing surface

area. For example, a wood fire burns faster if

the logs are chopped into smaller pieces.

Similarly, the surface area of a solid catalyst

is important to the rate of reaction.

Right: The photo shows a powder igniting.

13 | 16

The reaction rate usually depends on the concentration of

one or more reactant. This relationship must be determined

by experiment.

Quantitative dependence of the rate of reaction on [rxt]

For the generic reaction

aA + bB + cC dD + eE

the rate law can always be written as

Rate = k[A]m[B]n[C]p

This information is captured in the rate law

第二时止第三时起

The rate law, a mathematical equation that relates the rate

of a reaction to the concentration of a reactant (and

catalyst) raised to various powers. The proportionality

constant, k, is the rate constant.

Familiarize with Rate Law

(RHS)

·individually the order of rxn in each reactant

[A,], [B], …

·the sum of m, n, p, … the “overall order” of rxn

• (LHS)

k called the “rate constant”

pnm [C][B][A] Rate k

many ways but one unique unit !!

exponents m, n, p, etc.

·each gives the dependence of rate on the concentration of the reactant.

pnm

Ratek

]C[]B[]A[

using terms such as “first order in A” if m = 1

“second order in B” if n = 2, etc.

13 | 18

the rate law is found to be

Rate = k[NO2][F2]

In this case, the exponent for both NO2 and F2 is 1 (which does not need to be written).

We say the reaction is first order in each. The reaction is second order overall.

For example, for the reaction

Hydrogen peroxide oxidizes iodide ion in acidic solution:

H2O2(aq) + 3I(aq) + 2H+(aq) I3(aq) + 2H2O(l)

The rate law for the reaction is

Rate = k[H2O2][I]

a. What is the order of reaction with respect to each

reactant species?

b. What is the overall order?

13 | 19

a. The reaction is first order in H2O2.

The reaction is first order in I.

The reaction is zero order in H+.

b. The reaction is second order overall.

13 | 20

a. The concentration of Q does not affect the rate

b/c it’s zeroth order in Q. The rate will depend

only on [R].

b. Rate = k[R]2, i.e., drop [Q]0 since [Q]0 = 1.

第三D时起

13 | 21

The rate law for a reaction must be determined experimentally.

This method measures the initial rate of reaction, i.e., Rate,

using various starting concentrations [A], [B], [C], all

measured at the same temperature.

pnm [C][B][A] Rate k

By determining the rate law, it means to determine the value of

i) individual orders of reaction: m, n, p, etc,… and

ii) the rate constant, k.

We will study the initial rates method of determining the

rate law

How Is the Rate Law Determined?

13 | 22

Returning to the decomposition of N2O5, we have the

following data:

2N2O5(g) 2N2(g) + 5O2(g)

Initial N2O5

Concentration

Initial Rate of Disappearance

of N2O5

Experiment 1 1.0×10-2 M 4.8×10-6 M/s

Experiment 2 2.0×10-2 M 9.6×10-6 M/s

In this case, when the concentration doubles and the rate

doubles, we can prove that m = 1 (see the example followed)

mk ]O[N Rate 52

]O[N Rate 52k

Note that when the [N2O5] doubles from experiment 1 to 2,

the rate doubles.

First write the rate law symbolically as

But what’s the value of m?

So its rate law really is

For the reaction 2NO(g) + O2(g) 2NO2(g)

Obtain the rate law and the rate constant using the

following data for the initial rates of disappearance of NO:

13 | 23

Exp. [NO]0, M [O2]0, M Initial Rate of Reaction of NO, M/s

1 0.0125 0.0253 0.0281

2 0.0250 0.0253 0.112

3 0.0125 0.0506 0.0561

“Chemistry” - The rate law is Rate = k[NO]m[O2]n.

How to find the value of m and n?

!!! To find the order for a particular reactant, you first

choose TWO runs in which the concentration of only this

reactant changes while all others not. Then, you divide the

rate laws for those two experiments. Finally, you solve the

equation to find the order in this reactant.

13 | 24

The generic rate law is Rate = k[NO]m[O2]n.

nm

nm

MM

MM

M

M

0.0253 0.0125

0.0253 0.0250

s0.0281

s0.112

m2 4

We will divide the rate law for run-2 by that for run-1 because the

rate of run-2 is larger than that of run-1.

To find m, we choose two runs, 1 and 2, that [O2] remains constant.

For n, runs 1 and 3 were chosen b/c [NO] remains constant

n2 2

2m

1n

We will divide the rate law for run-3 by that for run-1 because the

rate of run-3 is larger than that of run-1.

nm

nm

MM

MM

M

M

0.0253 0.0125

0.0506 0.0125

s0.0281

s 0.0561

13 | 25

Alternatively, it is also possible to find the order by

examining the data qualitatively.

Again choosing experiments 1 and 2, we note that the

concentration of NO is doubled and the rate is quadrupled.

That means the reaction is second order in NO.

Exp. [NO]0, M [O2]0, M Initial Rate of Reaction of NO, M/s

1 0.0125 0.0253 0.0281

2 0.0250 0.0253 0.112

3 0.0125 0.0506 0.0561

Choosing experiments 1 and 3, we note that the

concentration of O2 is doubled and the rate is doubled.

That means the reaction is first order in O2.

2m

1n

Rate = k[NO]m[O2]n

13 | 26

The rate law now is Rate = k[NO]2[O2].

Using data from experiment 2: k = 7.08×103/M2 s

Using data from experiment 3: k = 7.10×103/M2 s

These values are the same to 2 significant figures.

s

10 7.11

0.0253 0.0125

s0.0281

1: experiment from data Using

2

3

2

MMM

M

k

2

2ONO

Ratek

k

obtain we, for Solving

13 | 27

Summary of Initial Rates Method - Examining the effect

of doubling or tripling the initial concentration of only one

reactant gives us the order in that reactant -

halved (½) −1

doubles stays the same (1) 0

doubles (2) 1 quadruples (4) 2 increases eight-fold (8) 3

(or) triples stays the same (1) 0 triples (3) 1

increases nine-fold (9) 2

If the concentration and the rate the order of reaction

of a reactant, say A of reaction in A must be

increases 27-fold (27) 3 reduces by two-thirds ( ) −1 3

1

Have you seen the effect of [rxt] on the rate of reaction so far?

pnm [C][B][A] Rate k

Iron(II) is oxidized to iron(III) by chlorine in an acidic

solution:

2Fe2+(aq) + Cl2(aq) 2Fe3+ (aq) + 2Cl(aq)

13 | 28

Exp. [Fe2+], M [Cl2], M [H+], M Rate, M/s

1 0.0020 0.0020 1.0 1.0×10-5

2 0.0040 0.0020 1.0 2.0×10-5

3 0.0020 0.0040 1.0 2.0×10-5

4 0.0040 0.0040 1.0 4.0×10-5

5 0.0020 0.0020 0.5 2.0×10-5

6 0.0020 0.0020 0.1 1.0×10-4

The following data were collected.

Determine the rate law and the value of k.

13 | 29

The generic rate law must be

Rate = k[Fe2+]m[Cl2]n[H+]p

Now can you find that n = 1 and p = 1?

Exp. [Fe2+], M [Cl2], M [H+], M Rate, M/s

1 0.0020 0.0020 1.0 1.0×10-5

2 0.0040 0.0020 1.0 2.0×10-5

3 0.0020 0.0040 1.0 2.0×10-5

4 0.0040 0.0040 1.0 4.0×10-5

5 0.0020 0.0020 0.5 2.0×10-5

6 0.0020 0.0020 0.1 1.0×10-4

To find the order in Fe2+, i.e., the value of m, we choose 1 and 2, or 3 and 4. In each case, [Fe2+] doubles and the rate doubles,

So the rate law is Rate = k[Fe2+][Cl2][H+]1

hence m = 1.

13 | 30

The same value for k is obtained using each experiment.

][

][][

H

ClFe 2

2kRate

Once a rate law was determined, one can calculate the

value of its rate constant as

]Cl[]Fe[

]H[

2

2

Rate k

sM.

M. M.

M. s

M.

50200200020

011001 5

Using data from experiment 1:

第三时止第四时起

13 | 31

Note that the units on the rate constant, k, are specific to the

overall order of the reaction.

• For a zero-order reaction, the unit is M/s or mol/(L.s).

• For a first-order reaction, the unit is 1/s or s-1.

• For a second-order reaction, the unit is 1/(Ms) or L/(mol.s).

In other words, If you know the unit of a rate constant, you

can deduce the order of reaction.

PracT: 9 – 15

13 | 32

The rate law of kind, called the

Differential Rate Law, tells us the relationship between the

rate and the concentrations of reactants and catalysts.

To find concentrations at various times, i.e., the extent of a

reaction, we need to use the integrated rate law. The form

used depends on the order of reaction in that reactant.

pnm [C][B][A] Rate k

d03 TTh

Concentration-Time Equations (Integrated rate laws)

Summary

ktt 0]A[

]A[ln

[A]t = [A]0 – kt

Order Differential rate law

Integrated rate law

Units of the rate constant k

k0

1

= k[A]² 2

1 sM

1s

11 sM

for “ aA products” type reactions !!

or time1

or M 1 ·time1

or M·time1 = k[A]0

0

1 1

[A] [A]t

kt

Here [A]t is the concentration of reactant A at time t, and [A]0 is

the initial concentration of A.

The ratio is the fraction of A remaining at time t. 0]A[

]A[ t

Cyclopropane is used as an anesthetic. The isomerization of cyclopropane to propene is a first-order reaction with a rate constant of 9.2/s. If an initial sample of cyclopropane has a concentration of 6.00 M, what will the cyclopropane concentration be after 1.00 s?

13 | 34

Do “translations”:

Rate = k[C3H6], k = 9.2/s, [A]0 = 6.00 M, t = 1.00 s, [A]t = ?

The reaction is first order,

so the integrated rate law is

Solve for [A]t

1

0

Aln (9.2 )1.00 9.2

A

t s s

9.2 4

0

A1.01 10

A

t e M1006.6]A[ 4t

PracT: 16, 17, 20, 21

Half-Life of a Reaction (t1/2)

ktt 0]A[

]A[ln

[A]t = [A]0 – kt

Order Integrated

rate law

0

1

2

kt

693.021

kt

2

]A[ 021

Half-life

a constant !!!

The half-life of a reaction, t½, is the time it takes for the

reactant concentration to decrease to one-half of its

initial value.

0

1 1

[A] [A]t

kt

By substituting ½[A]0 for [A]t, we solve the integrated

rate law for the special case of t = t½.

Ammonium nitrite is unstable because ammonium ion

reacts with nitrite ion to produce nitrogen:

NH4+(aq) + NO2

(aq) N2(g) + 2H2O(l)

In a solution that is 10.0 M in NH4+, the reaction is first

order in nitrite ion (at low concentrations), and the rate

constant at 25°C is 3.0×103/s.

What is the half-life of the reaction?

13 | 36

Do “translations” – k = 3.0×10-3/s, a 1st-order kinetics; t1/2= ?

The equation needed is kt½ = 0.693

min9.3s103.2s1003

6930 2

13

-.

.

kt

0.6931/2

PracT: 18, 19

13 | 37

Graphs of concentration vs. time can be used to determine the order of reaction in a reactant.

• For zero-order reactions, plot of [A] versus t is linear.

• For first-order reactions, plot of ln[A] versus t is linear.

• For second order reactions, plot of 1/[A] versus t is linear.

PracT: 22

This is illustrated on the next slide for the decomposition of NO2.

y = mx + b

y = mx + b

y = mx + b

linear

y vs. x plot

[A]t vs. t

tt

.vs]A[

1

tt .vs]Aln[

13 | 38

Left: The plot of ln[NO2] versus t is not linear, so the reaction is

not first order.

Right: The plot of 1/[NO2] versus t is linear, so the reaction is

second order in NO2.

13 | 39

t½ increase with the initial

concentration, [A]0, for the zero-

order reactions.

t½ is constant for the first-order

reactions, not depend upon [A]0.

t½ decreases with [A]0 for the

second-order reactions.

The reaction is second order.

第四时止第五时起

0

21

21

021

[A]

1:Order Second

6930:OrderFirst

2

[A] :Order Zero

k t

k

. t

k t

/

/

/

13 | 40

Rate Constant (k) and Temperature (T)

The rate constant depends strongly on temperature. How

can we explain this relationship? – two theories

Quantitative dependence of the rate of reaction on T

This dependence is reflected via the rate constant, k,

rather than the rate of reaction itself

1) Collision theory assumes that reactant molecules must

collide with an energy greater than some minimum value

and with the proper orientation.

The minimum energy is called the activation energy, Ea.

13 | 41

According to the collision theory the effect of a temperature

increase on each of the three factors for a reaction to occur.

The rate constant can be given by the equation

k = Zfp

i) The factor Z – certainly, Z will increase with temperature,

as the average velocity of the molecules increases with

temperature.

However, this factor alone cannot explain the dramatic

effect of temperature changes.

where

Z = collision frequency

f = fraction of collisions with the minimum energy

p = orientation factor

13 | 42

ii) The factor f - The fraction of molecular collisions having

the minimum energy required is given by f :

RT

ERT

E

a

a

e

ef 1

iii) The factor p - The reaction rate also depends on p,

the proper orientation for the collision. This factor is

independent of temperature.

T

Now we can explain the dramatic impact of temperature.

f increases exponentially with temperature:

PracT: 23, 24

)(RT

RT

Ea f

13 | 43

Importance of Molecular Orientation in

the Reaction of NO and Cl2

(A) NO approaches with its N atom toward Cl2, and an

NOCl bond forms. Also, the angle of approach is close to

that in the product NOCl.

(B) NO approaches with its O atom toward Cl2. No NOCl

bond can form, so NO and Cl2 collide and then fly apart.

13 | 44

2) Transition-state theory explains the k-T dependence in

terms of the potential energy diagram and an activated

complex (transition state), an unstable high-energy grouping

of atoms that is resulted from the collision of reacting

molecules and can break up to form products.

The potential energy diagram for a reaction visually illustrates

the changes in energy along with the structural changes in

reacting species.

The next slide shows the potential energy diagram for the

reaction

NO + Cl2 NOCl2‡ NOCl + Cl

where NOCl2‡ indicates the activated complex.

http://www.youtube.com/watch?v=h5xvaP6bIZI&feature=player_detailpage#t=35s

http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/activa2.swf

13 | 45

!!! EAB‡

> EA & EB, b/c outer-e repulsion !!!

!!! Ea (forward) = EAB‡

Ereactants !!! Ea (reverse) = ?

Important Quantities in A Potential Energy Diagram

13 | 46

The reaction of NO with Cl2 is an endothermic reaction, i.e., H > 0.

http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/activa2.swf

Watch a flash video

for the reaction below as well as NO2 + CO NO + CO2

Sketch a potential energy diagram for the

decomposition of nitrous oxide.

N2O(g) N2(g) + O(g)

The activation energy for the forward reaction is

251 kJ; Ho = +167 kJ.

Label your diagram appropriately.

What is the activation energy for the reversed

reaction?

13 | 47

13 | 48

Reactants

Products

Ea= 251 kJ

H = 167 kJ

Ea(reverse)

= 84 kJ

E

n

e

r

g

y

p

e

r

m

o

l

Progress of reaction

N2O

N2 + O

PracT: 25, 26, 31, 32

13 | 49

The k – T dependence for most chemical reactions closely

follows an equation in the form

This is called the Arrhenius equation. A is the frequency

factor, a constant. This equation shows that

RT

E a

e k

A

A more useful form of the Arrhenius equation is

211

2 11ln

TTR

E

k

k a

d04 TTh

Note how k and T are paired in this equation

i) When T changes, the value of k changes as well;

ii) If Ea > 0, then increasing T, the k increases.

!!! be careful with

paired k and T !!!

iii) the greater the Ea, the smaller k and lower the rate.

In a series of experiments on the decomposition of

dinitrogen pentoxide, N2O5, rate constants were

determined at two different temperatures:

• At 35°C, the rate constant was1.4×10-4/s.

• At 45°C, the rate constant was 5.0×10-4/s.

What is the activation energy?

What is the value of the rate constant at 55°C?

13 | 50

Do “translations” -

T1 = 35°C = 308 K T2 = 45°C = 318 K

k1 = 1.4×10-4/s k2 = 5.0×10-4/s

Ea = ? k = ? at T = 55°C = 328 K

211

2 11ln

TTR

E

k

k a

What equation to use?

13 | 51

We calculate the both sides

K 318

1

K 308

1

Kmol

J 8.315

E

/s10 1.4

/s10 5.0ln a

4

4

Set up the Arrhenius equation:

1514 Jmol1023.1K1002.1

Kmol

J315.8

273.1

aa E

E

15

15molJ1003.1

Jmol1023.1

273.1

aE

then solve for Ea

13 | 52

K328

1

K308

1

Kmol

J3158

mol

J10041

1041ln

5

4

2

.

.

/s.

k

47621041

ln4

2 ./s.

k

90111041

476.2

4

2 .e/s.

k-

/s .k 3

2 1071

The setup for solving k at T = 55°C = 328 K is

PracT: 27 - 30

13 | 53

Reaction Mechanism

A balanced chemical equation is a description of the overall

result of a chemical reaction. However, what actually

happens on a molecular level may be more involved than

what is represented by this single overall equation. For

example, the reaction may take place in several steps. That

set of steps, i.e., a collection of several sequentially

occurred molecular events, is called the reaction

mechanism.

13 | 54

Each step in the reaction mechanism is called an

elementary reaction and is a single molecular event.

Because an elementary reaction is an actual molecular

event, the rate of an elementary reaction is proportional to

the concentration of each reactant molecule.

Uni-molecular reaction

Bi-molecular reaction

Ter-molecular reaction

Molecularity – the # of colliding

molecular entities involved in an

elementary reaction

!!! This means the stiochiometry coefficient of a reactant in

an elementary reaction IS the reaction order of the reactant.

!!! This means that we can write without any experiments the

rate law directly for elementary reactions.

13 | 55

Rate Law and Reaction Mechanism

A reaction mechanism cannot be directly observed.

We can, however, determine the rate law by experiment

and decide if the reaction mechanism is consistent with

that rate law.

The rate of an overall reaction is determined completely

by the slowest step, the rate-determining step.

A correct reaction mechanism must meet the following:

i) sum of all elementary steps must give the exact

overall reaction;

ii) the rate law derived from the mechanism must

be consistent with that determined experimentally.

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For example, the reaction of NO2 with F2 is believed to

occur in the following elementary steps:

NO2 + F2 NO2F + F (slow step, RDS)

F + NO2 NO2F (fast step)

The overall reaction is

2NO2 + F2 2NO2F

What is the rate law for this mechanism?

Rate = k[NO2][F2]

So this mechanism is correct. (Use the two criteria to make

this conclusion)

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A reaction intermediate is a species produced during a

reaction that does not appear in the net equation because it

reacts in a subsequent step in the mechanism.

For example, the reaction of NO2 with F2 is believed to

occur in the following elementary steps:

NO2 + F2 NO2F + F (slow step, RDS)

F + NO2 NO2F (fast step)

The overall reaction is

2NO2 + F2 2NO2F

F does not appear in the overall reaction equation. It is

produced in the first step and used in the second step.

Thus F is a reaction intermediate.

Write the rate law for the first step in the mechanism

for the decomposition of ozone.

Cl(g) + O3(g) ClO(g) + O2(g)

ClO(g) + O(g) Cl(g) + O2(g)

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Rate = k[Cl][O3]

The mechanism of the decomposition is

Cl(g) + O3(g) ClO(g) + O2(g) (Slow step, RDS)

ClO(g) + O(g) Cl(g) + O2(g)

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What is the overall reaction equation?

O3(g) + O(g) 2O2(g)

Is there a reaction intermediate? If so, what is it?

Yes. ClO.

Is there a catalyst? If so, what is it?

Yes. Cl.

What is the rate law for the overall reaction?

Rate = k[Cl][O3]

Consider the following reaction:

NO2(g) + CO(g) NO(g) + CO2(g)

At 500 K, the reaction mechanism is believed to involve two elementary reactions:

NO2(g) + NO2(g) NO3(g) + NO(g)

NO3(g) + CO(g) NO2(g) + CO2(g)

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(Slow step, RDS)

PracT: 33 – 38

The overall reaction equation is (obtained by adding…)

NO2(g) + CO(g) NO(g) + CO2(g)

Note additionally that i) NO3 is a reaction intermediate: a species that is produced first and used up later,

The rate law then is (that for the RDS step if it’s the first step!!)

Rate = k[NO2][NO2] = k[NO2]2

and ii) there is no catalyst involved.

Catalysis

Catalysis is an increase in the rate of reaction that results

from the addition of a catalyst.

Enzymes are biological catalysts.

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A catalyst is not consumed in a reaction. Rather, it is present in the beginning, is used in one step, and is produced again in a subsequent step. As a result, the catalyst does not appear in the overall reaction equation. Compare this with a reaction intermediate.

A catalyst increases the reaction rate by providing an alternative reaction path with a lower activation energy. When Ea is reduced, k increases exponentially.

This relationship is illustrated on the potential energy diagram for the decomposition of ozone.

1.

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PracT: 39 – 40

The decomposition of hydrogen peroxide is catalyzed

by iodide ion. The reaction mechanism is thought to be

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H2O2(aq) + I(aq) H2O(l) + IO(aq)

IO(aq) + H2O2(aq) H2O(l) + O2(g) + I(aq)

At 25°C, the first step is slow relative to the second step.

What is the rate law predicted by the reaction

mechanism? Is there a reaction intermediate?

Rate = k[H2O2][I-]

I is a catalyst. IO is an intermediate.

Overall reaction equation:

2H2O2(aq) 2H2O(l) + O2(g)

Chapter Ends