Chapter 14 Chemical Kinetics Dr. Peter Warburton [email protected] .

146
Chapter 14 Chemical Kinetics Dr. Peter Warburton [email protected] http://www.chem.mun.ca/zcourses/ 1051.php

Transcript of Chapter 14 Chemical Kinetics Dr. Peter Warburton [email protected] .

Page 1: Chapter 14 Chemical Kinetics Dr. Peter Warburton peterw@mun.ca .

Chapter 14Chemical Kinetics

Dr. Peter [email protected]://www.chem.mun.ca/zcourses/1051.php

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Rates

Rates can be measured as the concentration change of a chemical (X)

over a period (change) of time

Rate = [X] / t

Often use molL-1 (or M) as units of concentration

so rate often has units molL-1s-1 (or Ms-1 )

01

tt

tt

XXRate 01

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Rates can be expressed in two different ways. We can look at the

formation of a product

(increase in the concentration over time)

or the

disappearance of a reactant

(decrease in the concentration over time).

Rate

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Rate of formation of a product

2 Fe3+(aq) + Sn2 +(aq) → 2 Fe2+(aq) + Sn4+(aq)

t0 = 0.0 s [Fe2+]t0 = 0.0000 M

t1 = 38.5 s [Fe2+]t1 = 0.0010 M

Δt = 38.5 s Δ[Fe2+] = (0.0010 – 0.0000) M

152

2 sM 10 x 2.6s 38.5

M 0.0010

Δt

FeΔFe offormation of rate

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Rate of disappearance of a reactant

2 Fe3+(aq) + Sn2 +(aq) → 2 Fe2+(aq) + Sn4+(aq)

t0 = 0.0 s [Sn2+]t0 = 0.0015 M

t1 = 38.5 s [Sn2+]t1 = 0.0010 M

Δt = 38.5 s Δ[Sn2+] = (0.0010 – 0.0015) M 15

22 sM 10 x 3.1

s 38.5

M) (-0.0005

Δt

SnΔSn of ncedisappeara of rate

The word “disappearance” implies the negative sign. Because of this, all rates are considered to be

POSITIVE!

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a A + b B c C + d D

Reaction rates are always positive, so we must put negative signs in front of reactant concentration changes!

To account for the stoichiometric relationships and their effect on rate, we must always divide by the stoichiometric coefficient for the chemical.

Δt

d

1

Δt

c

1

Δt

b

1

Δt

a

1 ratereaction - - + +

Reaction rates

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Reaction rate

2 Fe3+(aq) + Sn2 +(aq) → 2 Fe2+(aq) + Sn4+(aq)

15

4

4 slidesM 10 x 2.6

2

5 slidesM 10 x 1.3

23

sM 10 x 1.3ratereaction

Δt

SnΔ

1

1

Δt

FeΔ

2

1

Δt

SnΔ

1

1

Δt

FeΔ

2

1 ratereaction

15

15

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Measuring rates

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Measuring rates

1) Average rates

2) Slopes

3) Time = 0

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Problem

At some point in the reaction

2 A + B 3 C

[A] = 0.3629 M. At a time 8.25 minutes later [A] = 0.3187 M. What is the average rate of the disappearance of A in Ms-1, what is the average rate of the formation of C in Ms-1, and what is the average rate of reaction in Ms-1 over that time interval?

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Problem answers

15

14

15

sM 10 x 464ratereaction average

sM 10 x 341C offormation of rate average

sM 10 x 8.93A of ncedisappeara of rate average

.

.

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Instantaneous reaction rates

What’s happening at “this instant in time”?

The initial rate is the instantaneous reaction rate for a

reaction at time zero.

We can use instantaneous reaction rates.

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Rate laws

The rate of a chemical reaction depends on the concentration of some or all of the reactants.

A reactant might not affect the rate, regardless of its

concentration.

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Rate laws

The rate law for a reaction

is the equation showing the

dependence of the reaction rate on

concentrations

of the reactants.

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a A + b B + … products

rate = k [A]m[B]n…

k is a constant for the reaction

AT A GIVEN TEMPERATURE,

and is called the rate constant. The stoichiometry of the balanced reaction equation IS NOT ALWAYS the source of the

rate equation exponents!

m DOES NOT HAVE TO equal a

n DOES NOT HAVE TO equal b

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Reaction order

Reaction order with respect to a given reactant

is the value of the exponent of the rate law equation for the specific reactant only.

The overall reaction order is the

sum of the reaction orders for

all reactants.

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Reaction order example

rate = k [A]2[B]

The reaction order with respect to A is 2 or the reaction is second order in A

The reaction order with respect to B is 1 or the reaction is first order in B

The overall reaction order is 3 (2 + 1 = 3) or the reaction is third order overall

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“Sensitivity” to concentration change

Rate change of the reaction if [A] is

doubled depends on 2m

no change

ratedoubles

ratequadruples

ratehalves

zeroth order in A

first order in A

second order in A

NOTE: negative or non-integer orders

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Method of initial rates

Reaction rate laws are often determined experimentally!

We most commonly carry out a series of experiments in which the

initial rate of the reaction is measured as a function of

different initial concentrations of reactants

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Method of initial rates

If you see a table like this with chemical concentrations or pressures and rate data, chances are good the

question is a method of initial rates problem.

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Method of initial rates

IGNORE THE REACTION with this type of problem. The chemicals in the TABLE

are the interesting ones.You always require at least one more

experimental reaction than your number of chemicals given in your table!

Sometimes we are given a table with an extra experiment which we can use to

check if we’ve done everything correctly.

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2 NO (g) + O2 (g) NO

2 (g)

Since rate laws are always expressed in terms of reactants (and sometimes catalysts – we’ll see these later), lets create a general form of the rate law for this reaction based on what chemicals the TABLE tells us are involved in the rate of the reaction.

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2 NO (g) + O2 (g) NO

2 (g)

rate = k [NO]m[O2]n

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Method of initial rates

For our initial reactant order determination we need to choose a pair of reactions where only one reactant concentration changes. Experiments #1 and #2 fulfill this condition.

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Method of initial rates

rate = k [NO]m[O2]n

Since k is a constant then

k for experiment 1

IS EQUAL TO

k for experiment 2!

k = rate / [NO]m[O2]n

n22

m2

n

12m1

2

1n

22m2

2n

12m1

1

ONO

ONO

rate

rate so

ONO

rate

ONO

rate

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Reaction order w.r.t. NO

n22

m2

n

12m1

2

1

ONO

ONO

rate

rate

0.50 log m0.25 log

(0.50) log0.25 log

(0.50)0.25

M) (0.015M) (0.030

M) (0.015M) (0.015

sM 0.192

sM 0.048

m

m

nm

nm

1-

-1

2m0.301

0.602m

0.50 log

0.25 logm

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Reaction order w.r.t. O2

n32

23

n

1221

3

1

ONO

ONO

rate

rate

0.50 logn 0.50 log

(0.50) log0.50 log

(0.50)0.50

M) (0.030M) (0.015

M) (0.015M) (0.015

sM 0.096

sM 0.048

n

n

n2

n2

1-

-1

1n0.301

0.301n

0.50 log

0.50 logn

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ALTERNATE reaction order w.r.t. O2

n

n2

n2

n2

1-

-1

(0.50)]25.0[0.125

(0.50)(0.50)0.125

M) (0.030M) (0.030

M) (0.015M) (0.015

sM 0.384

sM 0.048

n42

24

n

1221

4

1

ONO

ONO

rate

rate

1n

(0.50)0.50

(0.50)0.25

0.125

n

n

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Our rate law

rate = k [NO]2[O2]1

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Rate constant using experiment 1

k = rate / [NO]2[O2]1

12-4

2

368

1

2

1

22

sM 10 x 1.4k

M 10 x 3.3

sM 0.048

M 0.015M 0.015

sM 0.048

ONO

ratek

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Rate constant using experiment 2

k = rate / [NO]2[O2]1

12-4

2

355

1

2

1

22

sM 10 x 1.4k

M 10 x 1.3

sM 0.192

M 0.015M 0.030

sM 0.192

ONO

ratek

The rate constant is the same, as it should be!

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Check using extra experiment

The rate is the same as the experimentally observed rate (within rounding errors). We MUST have done

everything right!

rate = (1.42 x 104 M-2s-1) [NO]2[O

2]1

11-3

35-12-42

212-42

221-24

2

sM 10 x 3.8rate

M 10 x 2.7sM 10 x 1.4rate

M 0.030M 0.030sM 10 x 1.4rate

ONOsM 10 x 1.4rate

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Units of rate constants

Rate always has units in terms of concentration per time unit.

Usually it is molL-1·s-1 (or M·s-1)

To ensure we get the right units for rate means the rate constant must have different units depending

on the overall reaction order.

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Problem

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

The rate of formation of the red-coloured triiodide ion

[I3-]/t (and therefore the reaction rate – why?) can be

determined by measuring the rate of appearance of the colour.

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Problem

a) What is the rate law for the formation of I3-?

b) What is the value for the rate constant?c) What is the initial rate of formation of triiodide

when the concentrations are [H2O2] = 0.300 M and [I-] = 0.400 M?

Expt Initial [H2O2] (M) Initial [I-] (M) Initial [I3-]/t (Ms-1)

1 0.100 0.100 1.15 x 10-4

2 0.100 0.200 2.30 x 10-4

3 0.200 0.100 2.30 x 10-4

4 0.200 0.200 4.60 x 10-4

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Problem answers

a) rate = k [H2O2] [I-]

b) k = 1.15 x 10-2 M-1s-1

c) rate = 1.38 x 10-3 Ms-1

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Zero order reactions

In a zero order reaction the rate of the reaction does NOT depend on the concentration of ANY of the reactants.

rate = k [NH3]0 = k (1) = k = constant

(g) H 3 (g) N (g) NH 2 22catalystPt

K 11303

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Integrated rate law – 0th order rxn

b

0xmy

t

0t

A tkA

OR

k tA A

0-tk AA

dt kAd

dtk Ad

kdt

Ad

k rate

0t

t

0

A

A

t

0

Page 39: Chapter 14 Chemical Kinetics Dr. Peter Warburton peterw@mun.ca .

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Features of zero order reactions

Concentration versus time graph gives a

straight line.

Integrated rate law

Rate constant is the negative slope (UNITS!)

b

0xmy

t A tkA

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First order reactions

In a first order reaction the overall order of the reaction is 1. A common type of first order reaction is the decomposition of a chemical.

rate = k [H2O

2]1

(g) O2

1 (l) OH (aq) OH 2222

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Integrated rate law – 1st order rxn

b

0xm

y

t

0t

Aln tkAln

OR

k tAln Aln

0-tk A

Aln

dt kA

Ad

dtk A

Ad

Akdt

Ad

Ak rate

0

t

t

0

A

A

t

0

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Integrated rate law – 1st order rxn

Natural logarithm of concentration versus

time graph gives a straight line.

Rate constant is the negative slope

(UNITS!)

b

0xm

y

t Aln tkAln

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Problem

The reaction

A 2 B + C

is first order. If the initial [A] = 2.80 M and k = 3.02 x 10-3 s-1, what is the value of [A] after 325 s?

Answer: [A] = 1.05 M

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Half-life of 1st order reactions

From the integrated rate law of a first order reaction we can look at what time [A] is one-half of the initial concentration (½ [A]0).

We call this time the half-life (t½).

k

0.693t

k

0.693-

k2

1lnt

k t21ln

k tA

A21

ln

k tA

Aln

21

21

21

21

0

0

0

t

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Half-life of 1st order reactions

The half-life (t½) of a first order

reaction is constant!

k

0.693t

k

0.693-

k2

1lnt

k t21ln

k tA

A21

ln

k tA

Aln

21

21

21

21

0

0

0

t

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Half-life of 1st order reactions

[A]t = 1/2 [A]0 at t½

[A]t = 1/4 [A]0 at 2 x t½

[A]t = 1/8 [A]0 at 3 x t½

and so on

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Problem

Consider the first order reaction

A products

with k = 2.95 x 10-3 s-1. What percent of A remains after 150 s?

Answer: % A remaining = 64.2%

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Problem

At what time (in minutes) after the start of the reaction is a sample of H2O2 (aq) two-thirds decomposed if k = 7.30 x 10-4 s-1?

Answer: time = 25.1 min

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1st order reactions of gases

Amounts of gases are often measured

by pressure instead of concentration

RT

PA so

RT

P

V

nor

nRTPVbut V

n

volume

A molesA

A

A

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Problem

Start with DTBP at a pressure of 800.0

mmHg at 147 C.

What will the pressure be at t = 125 min if

t½ = 8.0 x 10 min?

k tP

Pln

k t

RTP

RTP

ln

k tA

Aln

0A

tA

0

A

t

A

0

t

Answer:

k = 8.66 x 10-3 min-1

pressure = 271 mmHg

Page 51: Chapter 14 Chemical Kinetics Dr. Peter Warburton peterw@mun.ca .

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Radioactive decay as 1st order rxns

Many radioactive decay processes are first order…

half-life of 5730 yrs

half-life of 8.04 days

half-life of 4.51 x 109 yrs

eνNC 147

decay β146

-

eνXeI 13154

decay β13153

-

242

23490

decay α23892 HeThU

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Second order reactions

In a second order reaction the overall order of the reaction is 2. If there is only a single reactant chemical, like in

Then the rate is second order with respect to that chemical

rate = k [NO2]2

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

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Integrated rate law – 2nd order rxn

0-tk A

1-

A

1-

dt kA

Ad

dtk A

Ad

Akdt

Ad

Ak rate

0t

t

0

A

A2

2

2

2

t

0

b

0xm

y

t

0t

A

1 tk

A

1

OR

k tA

1

A

1

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Integrated rate law – 2nd order rxn

Inverse of concentration versus

time graph gives a straight line.

Rate constant is the slope (UNITS!)

b

0xm

y

t A

1 tk

A

1

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Half-life of 2nd order reactions

From the integrated rate law of a second order

reaction we can look at what time [A] is one-half of

the initial concentration

(½ [A]0).

We also call this time the

half-life (t½).

21

0

02

1

02

1

21

00

0t

tAk

1

Ak t12

Ak t1

1

211

k tA

1

A21

1

k tA

1

A

1

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Half-life of 2nd order reactions

The half-life (t½) of a second

order reaction is NOT constant!

21

0

02

1

02

1

21

00

0t

tAk

1

Ak t12

Ak t1

1

211

k tA

1

A21

1

k tA

1

A

1

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Half-life of 2nd order reactions

The greater the initial conc. of A

([A]0),

the smaller the half-life!

21

0

02

1

02

1

21

00

0t

tAk

1

Ak t12

Ak t1

1

211

k tA

1

A21

1

k tA

1

A

1

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Half-life of 2nd order reactions

Since

reactant concentration

is halved

over a half-life, the

next half-life

is twice as long

compared to the previous half-life.

21

0

tAk

1

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Pseudo first-order reactions

If we carry out some second (or higher) order reactions under conditions where some reactant concentrations do not change significantly, then the reaction may APPEAR to act like a lower order reaction (pseudo-first order, …)

OHHC COOHCHOH HCOOCCH 5232523

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Kinetics summary - rates

•You can calculate rate with known rate law•If you don’t know a rate law, the rate can be determined from the tangent of a [A] versus time graph or by –[A]/t over a small T

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Kinetics summary – rxn orders

•Method of initial rates•Plot graphs to find a straight line

•First order reactions have constant half-life•Use data in integrated rate laws – k MUST NOT change

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Kinetics summary – rate constants

•Rate constants are related to slope of appropriate straight line graphs•Use data in integrated rate laws to get k. Once you know k you can use the integrated rate law to solve for concentrations or times.•Use half-life of a first order reaction to determine k

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Problem

Without graphing, find the order and the rate constant for the following reaction using the given data

B productsTime (s) [B] (M) Time (s) [B] (M)

0 0.88 100 0.44

25 0.74 150 0.31

50 0.62 200 0.22

75 0.52 250 0.16

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Problem answer

Reaction is first order.

k = 6.93 x 10-3 s-1

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Bumper cars

A gasp of surprise

could be a “reaction”

when riding in a

bumper car

“Reactions” generally occur when the bumps are very hard and when they

come from behind

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Collision theory

A + BC AB + C

At some point in time, the B-C bond starts to break, while the A-B bond starts to form.

At this point, all three nuclei are weakly linked together.

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Collision theory

Molecules tend to repel each other when they get close.

It takes energy to force the molecules close together! This is like forcing together the north poles of two magnets.

This energy is the kinetic energy of the molecules. It converts to potential energy as the molecules get closer.

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Collision theory

A---B---C

has a higher potential energy than either A + B-C or A-B + C

A---B---C is the transition state

or the activated complex

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The difference in energy between

products and reactants is E

The difference in energy between the transition state and the reactants is

Ea – the activation energy

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Temperature is the average kinetic energy of molecules.

Collisions between molecules at higher temperatures are more likely to have

energy GREATER THAN the activation energy.

Higher temperatures mean higher rates of reaction!

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Collisions

1 billion collisions per molecule per second

Every reaction should be almost instantaneous.

This is not the case.

Not every collision breaks the activation energy barrier!

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Collisions

The fraction of collisions that have enough energy to overcome the activation energy barrier is

f = e-Ea/RT

e is approximately 2.7183,

Ea is the activation energy, T is the temperature in Kelvin,

R is the gas law constant (8.3145 JK-1mol-1)

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Bumper cars and energy

bumper car - a more energetic collision is more likely to make us gasp (our “reaction”)molecular collisions -higher energy collisions are more likely to lead to reaction (by overcoming the activation energy)

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Bumper cars and orientation

You are also more likely to gasp if you are hit from behind by another bumper car.

The orientation of how the collision occurs is also important to get a “reaction.”

The same is true for molecules where the fraction of collisions that have the right orientation is p. We call this fraction p the

steric factor.

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General reaction A + BC AB + C

Atom A MUST collide with the B side of BC to form the transition state A---B---C.

If atom A hits the C side, we get a different transition state A---C---B.

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General reaction A + BC AB + C

About half of our collisions won’t give the right reaction EVEN IF THEY BREAK THE ACTIVATION

ENERGY BARRIER!The steric factor p will be about 0.5.

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Collision rate = Z [A] [BC]

Z is a constant related to the collision frequency.

Recall only a fraction (f) of the collisions have a collision energy greater than or equal to

the activation energy.

Of those collisions, only a fraction (p) have the correct orientation to proceed through

the transition state to the products.

General reaction A + BC AB + C

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Reaction rate = p x f x Collision rate Reaction rate = pfZ [A] [BC]

If for our general reaction

rate = k [A] [BC]

k = pfZ = pZ e-Ea/RT = A e-Ea/RT (where A = pZ)

General reaction A + BC AB + C

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Arrhenius Equation

pZ = AAs T increases

k increases

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Problem

AB + CD AC + BD

What is the value of the activation energy for

this reaction? Is the reaction endothermic or

exothermic?

Suggest a plausible

structure for the transition

state.

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Using the Arrhenius Equation

If we know the rate constants for a reaction at two different temperatures, we can then

calculate the activation energy.

k = A e-Ea/RT

ln k = ln (A e-Ea/RT)

ln k = ln (A) + ln (e-Ea/RT)

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ln k = ln (A) – (Ea/RT)

This is the equation for a straight line!

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slope = -Ea/R so Ea = - slope x R

A graph of the natural logarithm

of the rate constant versus

inverse temperature

will give a straight line

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If we know

the rate constants (k1 and k2)

at two temperatures (in Kelvin!) T1 and T2

we will see the rate constant has changed because the temperature has changed.

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One of the things that

HAS NOT CHANGED

with the temperature is the

ACTIVATION ENERGY

-Ea/R (at T1) = -Ea/R (at T2) It is constant just like slope is constant

for a straight line!

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ln k2 = ln (A) – (Ea/RT2)

ln k1 = ln (A) – (Ea/RT1)

Subtract the bottom from the top on both sides

ln k2 – ln k1 = [-(Ea/RT2)] – [-(Ea/RT1)]

Pull out –Ea/R and put it in front

ln k2 – ln k1 = (–Ea/R) (1/T2 – 1/T1)

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Your textbook says

ln (k2/k1) = (Ea/R) (1/T1 – 1/T2)

This is absolutely correct as well! Use whichever form of the relation that

you feel more comfortable with mathematically. There are more forms

given in the text.

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Problem

Rate constants for the decomposition of gaseous dinitrogen pentaoxide are

3.7 x 10-5 s-1 at 25 °C and 1.7 x 10-3 s-1 at 55 °C

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

What is the activation energy of this reaction in kJmol-1?What is the rate constant at 35 C?

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Problem answer

Ea = 104 kJmol-1

k35C = 1.4 x 10-4 s-1

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Reaction Mechanisms

A reaction mechanism is the sequence of molecular events (elementary steps or elementary processes) that defines the pathway from the reactants to the

products in the overall reaction.

The elementary processes describe the behaviour of individual molecules while

the overall reaction tells us stoichiometry.

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The reaction actually takes place in two elementary processes!

2 NO2 NO and NO3

NO3 + CO NO2 and CO2

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

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

Elementary processes must add together to give the overall

equation!

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

Some of the “crossed-out” chemicals are neither reactants nor products in the overall reaction.

For example, in the above reaction NO3 is formed in one elementary step and

consumed in a later elementary step.

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Reaction intermediate

A reaction intermediate is a species that is formed in an elementary process, that is consumed in a later elementary process.

We never see reaction intermediates in the overall reaction!

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Molecularity

The molecularity of an elementary process is the number of molecules

on the reactant side

of the elementary process reaction.

A one molecule elementary reaction is unimolecular.

A two molecule elementary reaction is bimolecular.

A three molecule elementary reaction is termolecular.

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Molecularity

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Chances for molecularity

The chances of a unimolecular reaction occurring only depends on the one molecule, and are good.

A bimolecular reaction requires that two molecules collide with each other. This isn’t difficult and happens quite often.

A termolecular reaction requires that three molecules collide with each other at the same time. The chances of this happening are not very good.

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Bumper cars

Consider bumper cars. Very often, you will hit one other bumper car. It is a very rare occurrence to have a “bumper car” pile-up where three or more cars hit at exactly the same time.

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Problem

A suggested mechanism for the reaction of nitrogen dioxide and molecular fluorine is shown

a) Give the chemical equation for the overall reaction, and identify any reaction intermediates. b) What is the molecularity of each of the elementary processes?

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Problem answer

2 NO2 (g) + F2 (g) 2 NO2F (g)F is a reaction intermediate (formed in step 1 then consumed in step 2)Step 1 is bimolecular.Step 2 is bimolecular.

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Rate Laws and Reaction Mechanisms

UNLIKE an overall reaction the rate law for an elementary process

follows directly from the molecularity of the elementary

reaction!For a general elementary process

a A + b B products

rate = k [A]a [B]b

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Ozone

Unimolecular decomposition of ozone.

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

The rate law will be first order with respect to ozone

rate = k [O3]

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Bimolecular reaction

A + B productsReaction depends on collisions between

molecules A and B

Increase A, you increase # collisions

Increase B, you increase # collisions

rate = k [A] [B]

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Bimolecular reactions for one reactant

A + A products

Increase A, you increase # collisions for EACH A

rate = k [A] [A] = k [A]2

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Mechanisms and overall rate law

The mechanism of the overall reaction is predicted through the elementary processes therefore

the elementary processes will determine the rate law of the

overall reaction!

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One step mechanisms

If the overall reaction occurs in one elementary process, then

the two reactions are the same.

The rate law for the overall reaction is the same as the

elementary reaction rate law.

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Single step mechanism

rate = k [CH3Br] [OH-]

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Rate-determining step

For reaction mechanisms of more than one elementary step, one of the elementary step reactions MAY HAVE a much slower rate than any of the other steps.

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Rate-determining step

The rate-determining step of an overall reaction is an elementary step reaction

which has a much slower rate than any other step.

The overall rate law will match the rate law of the rate-determining step

in this case.

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Proposing mechanisms

A proposed mechanism must satisfy two criteria to be considered plausible:

1) the elementary processes must add up to give the appropriate overall reaction

2) the rate law that arises from the proposed mechanism must be consistent with the observed rate law

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H2 (g) + 2 ICl (g) I2 (g) + 2 HCl (g)

Observed rate law

rate = k [H2][ICl]

HCl 2 IICl 2 H

HCl IICl HI :FAST

HCl HIICl H :SLOW

22

2

2

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H2 (g) + 2 ICl (g) I2 (g) + 2 HCl (g)

The rate law of the rate-determining step israte = k [H2][ICl]

It turns out this is the same as the experimentally observed rate law, so this

mechanism is plausible.

HCl 2 IICl 2 H

HCl IICl HI :FAST

HCl HIICl H :SLOW

22

2

2

The elementary reactions DO add up to the overall reaction.

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2 NO (g) + O2 (g) 2 NO2 (g)

Observed rate = k [NO]2 [O2]

The rate law is termolecular. The chances of a single slow termolecular

step reaction occurring are very poor!

We need a plausible mechanism where the slow step cannot be the first step.

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Fast reversible step – slow step

A reversible reaction is one wherereactants can become products

and the products can become reactants

AT THE SAME TIME.

Eventually both reactionshave the same rate

(the system is at equilibrium)

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Fast reversible step – slow step

Rate law of slow step

rate3 = k3 [N2O2] [O2]2

k2

2k

222

22k

k

NO 2O NO 2

NO 2O ON :SLOW

ONNO 2 :FAST

3

2

1

BUT N2O2 is a reaction intermediate (concentration is effectively constant!)

We can’t have it in the rate law!

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Fast reversible step – slow step

Rate law of fast forward step

rate1 = k1 [NO]2

2k

2

2k

222

22k

k

NO 2O NO 2

NO 2O ON :SLOW

ONNO 2 :FAST

3

2

1

Rate law of fast reverse step

rate2 = k2 [N2O2]

At equilibrium rate1 = rate2

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Fast reversible step – slow step

2k

2

2k

222

22k

k

NO 2O NO 2

NO 2O ON :SLOW

ONNO 2 :FAST

3

3

2

1

k1 [NO]2 = k2 [N2O2]so [N2O2] = k1/k2 [NO]2

Let’s say that k1/k2 = K

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Fast reversible step – slow step

Rate law of slow step

rate3 = k3 [N2O2] [O2]2

k2

2k

222

22k

k

NO 2O NO 2

NO 2O ON :SLOW

ONNO 2 :FAST

3

2

1

We can substitute in [N2O2] from fast rxn! rate3 = k3 K [NO]2 [O2]

If we let k = k3 K then we see the rate law of the slow reaction does match the

observed rate law!

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Just because a mechanism is plausible doesn’t mean it is right!

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Problem

In a proposed two-step mechanism for CO + NO2 CO2 + NO, the second, fast step is

NO3 + CO NO2 + CO2.

What must be the slow step? What would you expect the rate law of the rxn to be?

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Problem answer

Slow step

NO2 + NO2 NO3 + NO.

Rate law: rate = k [NO2]2

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Problem

Show that the proposed mechanism is plausible for the reaction

2 NO2 (g) + F2 (g) 2 NO2F (g)

if the rate law is rate = k [NO2][F2]

(g) FNONO (g) F :FAST

(g) F (g) FNO(g)FNO :SLOW

(g)FNO(g) F (g) NO :FAST

22

2 22

2222

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Steady-state approximation

More than one elementary process may determine the rate of an overall

reaction.

No rate-determining step!

Can use the steady state approximation in this case.

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2 NO (g) + O2 (g) 2 NO2 (g)

Observed rate = k [NO]2 [O2]

We’ve already seen this as the fast reversible step – slow step mechanism

example.

Let’s propose a mechanism WITHOUT making any initial assumptions about

the step rates!

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2 NO (g) + O2 (g) 2 NO2 (g)

2k

222

k22

22k

NO 2 O ON :3 Process

NO NO ON :2 Process

ONNO NO :1 Process

3

2

1

This is essentially the same as the fast reversible step – slow step mechanism, but we treat the reversible reaction as

two separate forward reactions.

Each step has its own rate law.

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2 NO (g) + O2 (g) 2 NO2 (g)

2k

222

k22

22k

NO 2 O ON :3 Process

NO NO ON :2 Process

ONNO NO :1 Process

3

2

1

We know a reaction intermediate ALWAYS has a near-zero concentration that does not change much with time (otherwise we would see it build up!)

This means [N2O2]/t 0 (steady state)

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2 NO (g) + O2 (g) 2 NO2 (g)

2k

222

k22

22k

NO 2 O ON :3 Process

NO NO ON :2 Process

ONNO NO :1 Process

3

2

1

Choose our starting point as the step which has a rate law closest to the observed one, while including the reaction intermediate.

rate = k3 [N2O2] [O2]

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2 NO (g) + O2 (g) 2 NO2 (g)

[N2O2]/t 0

2222

2222

ON ncedisappeara of rate- ONformation of rate

0 ON ncedisappeara of rate ONformation of rate

We form the intermediate in process 1 and consume it in processes 2 and 3

2223222

nce!disappearafor ratesreaction positive used webecause negative

22

2122

OONkONk

3 process rate 2 process rate- ON ncedisappeara of rate

NOk 1 process rate ONformation of rate

BE CAREFUL! These are two different negative signs!

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2 NO (g) + O2 (g) 2 NO2 (g)

22232222

1

22232222

1

2222

OONkONk NOk

OONkONk- NOk

ON ncedisappeara of rate- ONformation of rate

Solve for [N2O2]

232

21

22

232222

1

Okk

NOkON

Okk ON NOk

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2 NO (g) + O2 (g) 2 NO2 (g)

Substitute into starting point

22

232

31

2232

21

3

2223

ONOOkk

kk rate

OOkk

NOkk rate

OONk rate

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2 NO (g) + O2 (g) 2 NO2 (g)

22

232

31 ONOOkk

kk rate

Very close to the observed rate law! How do we make the first part constant?

Now we assume something about step rates!

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2 NO (g) + O2 (g) 2 NO2 (g)

Assume rate2 >> rate3

so k2 [N2O2] >> k3 [N2O2] [O2]

k2 >> k3 [O2] SO k2 + k3 [O2] k2

22

2

31 ONO

k

k

kk rate

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Catalysis

Reaction rates are not just affected by reactant concentrations and

temperatures.

A catalyst is a substance that increases the rate of a reaction

WITHOUT undergoing permanent change in the reaction.

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How does a catalyst work?

A catalyst makes available a different reaction pathway that

is more efficient than the uncatalyzed mechanism because

this pathway has a lower activation energy.

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Homogeneous catalysts

A homogeneous catalyst exists in the same phase as the reactants.

I- is a homogeneous catalyst for the decomposition of hydrogen peroxide

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Decomposition of H2O2

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Decomposition of H2O2

Catalysts don’t appear in the

overall balanced equation!

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Heterogeneous catalysts

A heterogeneous catalyst exists in a different phase

(usually solid) than the reactants.

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Heterogeneous catalysts

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Heterogeneous catalysts

Most catalysts used in industry

are heterogeneous

It is much easier to separate a solid from a gas or liquid (for example) than two liquids or gases).

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Enzymes are catalysts

In living beings catalysts are usually called enzymes

Carbonic anhydrase catalyzes the reaction of carbon dioxide with water

CO2 (g) + H2O (l) H+ (aq) + HCO3- (aq)

The enzyme increases the rate of this reaction by a factor of 106. Equivalent to

about a 200 K increase …

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Enzymes

Lock-and-key model