chm096 1 chem kinetics rm

7/27/2019 CHM096 1 Chem Kinetics Rm

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Kinetics: Rates and Mechanisms of Chemical Reactions

1.1 ExpressingReaction Rate: Differential rate law

Graphical representation of reaction rate

1.2 The Rate Law : Order of reaction, Initial rate method

Integrated rate law, rate constant, half -life

1.3 Collision Theory: Activation energy, effective collisions

1.6 Arrhenius Equation: Determination of rate constant and

activation energy

1.7 Reaction Mechanism: Intermediates, molecularity,

rate determining step

1.4 Factors affecting reaction rate: Concentration, temperature, catalyst,

nature of reactants

1.5 Transition State Theory: Activated complex


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1.1 Chemical Kinetics

The speed of a chemical reaction is called its

reaction rate

The rate of a reaction is a measure of how fast

the reaction makes products, or how fast the

reactants are consumed. Chemical kinetics is the study of the factors

that affect the rates of chemical reactions

Such as temperature, reactant

concentrations, addition of catalyst etc.

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Figure 1.1 Reaction rate: the central focus of chemical kinetics


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1.2 Expressing the Reaction Rate

Reaction rate is defined as the changes in the concentrations of

reactants or products per unit time

Reactant concentrations decrease while product concentrations

Increase. For the reactionA B

Where [ ] denote the concentration in mol dm-3 and , the change of

Unit for rate is mol dm-3 s-1 ormol/L s-1 or mol/dm3 min-1

Rate of reaction = concentration time

= - [A2] [A1]

t2-t1

= - [Reactant]

time

Or [product]Rate of reaction = +

t=

[B2] [B1]

t2-t

1


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Figure 1.2 The wide range of reaction rates.


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06250061

87Rate

tt

XX

t

XRate

12

12

.

250061

84Rate

tt

AA

t

ARate

12

12

.

at t = 0

[A] = 8[B] = 8

[C] = 0

at t = 0

[X] = 8[Y] = 8

[Z] = 0

at t = 16

[A] = 4

[B] = 4

[C] = 4

at t = 16

[X] = 7

[Y] = 7

[Z] = 1

250061

04Rate

tt

CC

t

CRate

12

12

.

06250061

01Rate

tt

ZZ

t

ZRate

12

12

.

7


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Initial rate is defined as the rate at the start of the reaction when an

infnitesimally small amount of the reactant has been used up and isgiven by the gradient to the curve at time t = 0

Instantaneous rate is the rate at a particular time and is given by the

tangent to the curve at that time in the concentration-time graph.

Average rate is the rate given by the average change in concentrationof a reactant or product per unit time over a certain time interval.

For

Average rate = [ B ] at time t2 [ B ] at time t1

t2 t1= [B]

t

t2 = final time ; t1 = initial time

1.3 Three different types of rates

A B


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Table 1.1 Concentration of O3 at Various Time in its

Reaction with C2H4 at 303K

C2H4(g) + O3(g) C2H4O(g) + O2(g)

Time (s) Concentration of O3 (mol/L)

0.0

20.0

30.0

40.0

50.0

60.0

10.0

3.20x10-5

2.42x10-5

1.95x10-5

1.63x10-5

1.40x10-5

1.23x10-5

1.10x10-5

(conc O3)-t


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Figure 1.3

The concentrations of O3 vs. time during its reaction with C2H4

- [C2H4]

trate =

[O2

]

t

or

(a) Initial rate, t = 0 s

(d) Instantaneous rate at t = 35 s

h

w

(b) Average rate from t = 10s to

t= 30s is concentration = htime w

rate =

C2H4(g) + O3(g) C2H4O(g) + O2 (g)


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Figure 1.4 Plots of [C2H4] and [O2] vs. time.


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In general, for the reaction

aA + bB cC + dD

rate =

1

a- = -

[A]t

1

b

[B]t

1

c

[C]t= +

1

d

[D]t= +

The numerical value of the rate depends upon the substance that

serves as the reference. The rest is relative to the balancedchemical equation.

1.4 Reaction Rate and Stoichiometry


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For the reaction below, the coefficients of the balanced equationare not all the same

H2 (g) + I2 (g) 2 HI(g)

Therefore, the change in the number of molecules of onesubstance is a multiple of the change in the number ofmolecules of another

for the above reaction, for every 1 mole of H2 used, 1 mole of

I2 will also be used and 2 moles of HI made

therefore the rate of change will be different

To be consistent, the change in the concentration of eachsubstance is multiplied by 1/coefficient

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average rate in a given time

period = slope of the line

connecting the [H2] points; and

slope of the line for [HI]

the average rate for the

first 10 s is 0.0181 M/s





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Sample Problem 1.1

PLAN:

SOLUTION:

Expressing Rate in Terms of Changes in

Concentration with Time

PROBLEM: Because it has a nonpolluting product (water vapor), hydrogen

gas is used for fuel aboard the space shuttle and may be usedby Earth-bound engines in the near future.

2H2(g) + O2(g) 2H2O(g)

(a) Express the rate in terms of changes in [H2], [O2], and [H2O] with time.

(b) When [O2] is decreasing at 0.23 mol/L*s, at what rate is [H2O] increasing?

Choose [O2] as a point of reference since its coefficient is 1. For every

molecule of O2 which disappears, 2 molecules of H2 disappear and 2

molecules of H2O appear, so [O2] is disappearing at half the rate of change

of H2 and H2O.

- 12

[H2]t

[O2]

t

= +

[H2O]t

12

0.23mol/L*s= +

[H2O]

t

1

2; = 0.46mol/L*s[H2O]

t

rate =(a)

[O2]

t- = -(b)

= -

[O2]t

D t i R t f ti b d th


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Sample Problem 1.2Determine Rate of reaction based on the

stoichiometry of a reaction

PROBLEM: Dinitrogen pentoxide, N2O5, decomposes to form nitrogen dioxide

and oxygen.2 N2O5(g) 4NO2 (g) + O2(g)

NO2 is produced at a rate of 5.0 x 10-6 Ms-1. What is the

corresponding rate of disappearance of N2O5 and rate of

formation of O2?

(2.5 x 10-6, 1.25x 10-6M/s)

Sample Problem 1.3

PROBLEM: Ammonia,NH3, reacts with oxygen to form nitric oxide, NO and

water vapour.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)At a specific time in the reaction, NH3 is disappearing at a rate of

0.068 Ms-1. What is the corresponding rate of disappearance ofO2and production of NO and H2O.

(0.085 , 0.068 , 0.102 M/s)


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If 2.4 x 102 g of NOBr (MM 109.91 g) decomposes in a

2.0 x 102 mL flask in 5.0 minutes, find the average rate ofBr2 production in M/s.

2 NOBr(g) 2 NO(g) + Br2(l)

Sample Problem 1.4

M mole Br2 Rate min s }

0.018 M/s


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1.5 Rate Laws and order of reaction

A. Differential rate law or rate law is an experimentally

determined equation that expresses how the rate of a reaction

depends on the concentrations of the reactants

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

Rate

[N2O5]x

= k [N2O5]x

where k = rate constant

x = order of reaction determined experimentally, not equivalent to the

stoichiometric coefficient of a balanced equation


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B. The order of reaction is defined as the power to which the

concentration of a reactant is raised in the rate equation or rate law.

It must be determined experimentally.

Overall orderof a reaction = sum of the powers of the concentration

terms in the rate law

The order of each reactant concentration explains how the rate of a

reaction varies with the concentration of a particular reactant. For

exampleRate = k [N2O5]

x, when [N2O5] is doubled,

If x = 0; the rate remains unchanged

If x = 1; the rate doubles

If x = 2; the rate quadruples / increase by a factor of 4

If x = 3; the rate increases eightfold / factor of 8


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For the reaction

A product

The proposed rate law : Rate = k[A]n

If a reaction is Zero Order, the rate of the reaction is

always the same

doubling [A] will have no effect on the reaction rate

If a reaction is First Order, the rate is directly

proportional to the reactant concentration

doubling [A] will double the rate of the reaction

If a reaction is Second Order, the rate is directlyproportional to the square of the reactant concentration

doubling [A] will quadruple the rate of the reaction

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For a general reaction, A B

Order ofreaction

Differential rate law Units of rate constant (k)(t in seconds)

zero Rate = k[A]o = k k = mol dm-3 s-1 or mol L s-1

The rate is independent of the

concentration of the reactant as

long as some is present.

First Rate = k[A] k = rate/[A]= mol dm-3 s-1

mol dm-3

= s-1

second Rate = k[A]2 k = rate/[A]2= mol dm-3 time-1

mol2 dm-6

= dm3mol-1 s-1 or L mol-1s-1

Table 1.2: order of reaction and differential rate law


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Sample Problem 1.5

SOLUTION:

Determining Reaction Order from Rate Laws

PROBLEM: For each of the following reactions, determine the reaction order

with respect to each reactant and the overall order from thegiven rate law.

(a) 2NO(g) + O2(g) 2NO2(g); rate = k[NO]2[O2]

(b) CH3CHO(g) CH4(g) + CO(g); rate = k[CH3CHO]3/2

(c) H2O2(aq) + 3I-

(aq) + 2H+

(aq) I3-

(aq) + 2H2O(l); rate = k[H2O2][I-

]

PLAN: Look at the rate law and not the coefficients of the chemical reaction.

(a) The reaction is 2nd order in NO, 1st order in O2, and 3rd order overall.

(b) The reaction is 3/2 order in CH3CHO and 3/2 order overall.

(c) The reaction is 1st order in H2O2, 1st order in I- and zero order in H+,

while being 2nd order overall.


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In order to determine the order of a reaction, an

experiment has to be carried out to measure thereaction rates. The results obtained from these

experiments are used to determine the order of the

reaction.

The order of reaction and the rate constant can bedetermined by

a) the initial rate method

b) the half-life method

c) Unit of the rate constant

d) linear plots based on the integrated rate law

1.6 To determine the order of a reaction


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1.7 Experimental Determining of Reaction Orders

Initial rate method

Run a series of experiments, each of which starts with a different set

of reactant concentrations, and from each obtain an initial rate.

See Table 1.3 for data on the reaction

O2(g) + 2NO(g) 2NO2(g) rate = k[O2]m[NO]n

Compare 2 experiments in which the concentration of one reactant variesand the concentration of the other reactant(s) remains constant.

k[O2]2m[NO]2

n

k[O2]1m[NO]1

n=

rate2

rate1=

[O2]2m

[O2]1m

=

6.40x10-3mol/L*s

3.21x10-3mol/L*s

[O2]2

[O2]1

m

=1.10x10-2mol/L

2.20x10-2mol/L m

; 2 = 2m m = 1

Do a similar calculation for the other reactant(s).


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Table 1.3 Initial Rates for a Series of Experiments in the

Reaction Between O2 and NO

Experiment

Initial Reactant

Concentrations (mol/L)Initial Rate

(mol/L*s)

1

2

3

4

5

O2 NO

1.10x10-2

1.30x10-2

3.21x10-3

1.10x10-2 3.90x10-2 28.8x10-3

2.20x10-2

1.10x10-2

3.30x10-2

1.30x10-2

2.60x10-2

1.30x10-2

6.40x10-3

12.8x10-3

9.60x10-3

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


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Sample Problem 1.6

PLAN:

SOLUTION:

Determining Reaction Order from Initial Rate Data

PROBLEM: Many gaseous reactions occur in a car engine and exhaust

system. One of these is

NO2(g) + CO(g) NO(g) + CO2(g) rate = k[NO2]m[CO]n

Use the following data to determine the individual and overall reaction orders.

Experiment Initial Rate(mol/L*s) Initial [NO2] (mol/L) Initial [CO] (mol/L)

1

2

3

0.0050

0.080

0.0050

0.10

0.10

0.40

0.10

0.10

0.20

Solve for each reactant using the general rate law using the

method described previously.

rate = k[NO2]m[CO]n

First, choose two experiments in which [CO] remains

constant and the [NO2] varies.


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Sample Problem 1.6 Determining Reaction Order from Initial Rate Data

continued

0.080

0.0050

rate 2

rate 1

[NO2]2

[NO2]1

m=

k[NO2]m

2[CO]n

2

k[NO2]m

1 [CO]n

1

=

0.40

0.10=

m

;16 = 4m and m = 2

k[NO2]m

3[CO]n3

k[NO2]m

1 [CO]n1

[CO]3

[CO]1

n=

rate 3

rate 1=

0.0050

0.0050=

0.20

0.10

n;

1 = 2n and n = 0

The reaction is 2nd

order in NO2.

The reaction is

zero order in CO.

rate = k[NO2]2[CO]0 = k[NO2]

2


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Sample Problem 1.7

PROBLEM: Consider the following reaction and the data given in the

table, determine the rate law for the reaction and also

calculate the rate constant, k.

NH4+(aq) + NO2

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

Expt Initial [NH4+]

Initial [NO2-

] Initial rate (moldm-3

s-1)

1 0.10 0.005 1.35 x 10-7

2 0.10 0.01 2.70 x 10-7

3 0.20 0.01 5.40 x 10-7


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The following table shows the results for the reaction

BrO3-(aq) + 5Br-(aq) + 6H+(aq) 3Br2(l) + 3H2O(l)

Sample Problem 1.8

PROBLEM:

Expt Initial [BrO3- ] Initial [Br-] Initial [H+ ] Initial rate (mol dm-3s-1)

1 0.10 0.10 0.10 8.00 x 10-4

2 0.20 0.10 0.10 1.60 x 10-3

3 0.20 0.20 0.10 3.20 x 10-3

4 0.10 0.10 0.20 3.20 x 10-3

Determine the order of the reaction with respect to each

reactant and the value of the rate constant.


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1.8 Integrated Rate Laws

For the reaction A Products, the rate of the

reaction depends on the concentration of A Applying calculus to integrate the rate law gives

another equation showing the relationship between

the concentration of A and the time of the reaction

this is called the Integrated Rate Law

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The integrated Rate Law expresses how the

concentrations of the reactants depend on time. The

expression of an integrated rate law depends on the

order of a reaction.


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Almost all experimental work in chemical kinetics deals

with integrated rate laws. The integrated rate laws

derived below are for simple reactions with only one

reactant.

(I) For zero-order reaction: A products

By the rate law (differential); Rate = - d[A] = k

dt

Integrating the differential rate law, we obtained the integrated

rate law

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

where [A]o = initial concentration of A

A plot of[A]t versus t gives a straight line, slope = -k and

y-intercept = [A]o


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The half- life of a reaction, t1/2 is the time required for the concentration

of a reactant to decrease to half its original concentration

[A]o = -kt + [A]o

kt = [A]o

t = [A]o ; ** the half lifeof a zero orderreaction is

2k directly proportionalto [A]o

(II) For first- order reaction aA products

By the rate law , Rate = - d[A] = k[A]

dt

Integrated first-order rate law gives

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

ln [A]o = kt

[A]t


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A plot ofln [A]t versus t will give a straight line. slope =k

and y-intercept = ln [A]o.

The half-life is derived by rewriting the integrated rate law as

ln{ [A]o / [A]t } = kt

At t = t , [A]t = [A]o ; ln 2 = kt

0.693 = kt and t = 0.693/ k

** the half lifeof a first order reaction is independent of the initial

concentration of reactant. i.e 1st half-life (t 2nd half-life (t

(III) For second-order reaction aA products

By the rate law, Rate = -d[A] = k[A]2

dt

Integration of the rate law will give the integrated second order rate

law as

1 = kt + 1

[A]t [A]o


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At half-life , t , where [A]t = [A]o

1 = kt + 1

[A]o [A]o

kt = 2 - 1 = 1

[A]o [A]o [A]o

t = 1

k[A]o

A plot of1/[A]t versus t give a straight line with a slope = k

and the y-intercept = 1/[A]o

** The half life of a second order reaction is inversely proportional tothe initial concentration of the reactant

Thus , the second half-life (t ) = 2(t


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Summary of Integrated Rate Laws

rate = -[A]

t

= k[A]

rate = -[A]

t= k[A]0

rate = -

[A]

t= k[A]2

f ir st order rate equation

second order rate equation

zero order rate equati on

ln[A]0

[A]t

= kt ln [A]0 = kt + ln [A]t

1

[A]t

1

[A]0- = kt

1

[A]t

1

[A]0+= kt

[A]t - [A]0 = - kt


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1.8.1 Relationship Between Order and Half-Life

For a zero order reaction, the half-life is directlyproportional to the initial concentration. The lower theinitial concentration of the reactants, the shorter thehalf-life

t1/2= [A]init/2k

For a first order reaction, the half-life is independent ofthe concentration. (t1/2 = constant)

t1/2 = ln(2)/k

For a second order reaction, the half-life is inversely

proportional to the initial concentration increasing theinitial concentration shortens the half-life

t1/2 = 1/(k[A]init)

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Table 1.4 An Overview of Zero-Order, First-Order, and

Simple Second-Order Reactions

Zero OrderFirst Order Second Order

Plot for straight line

Slope, y-intercept

Half-life

Rate law rate = k rate = k[A] rate = k[A]2

Units fork mol/L*s S-1 L/mol*s

Integrated rate

law in straight-

line form

[A]t = -kt + [A]0 ln[A]t =

-kt + ln[A]0

1/[A]t =

kt + 1/[A]0

[A]t vs. t ln[A]t vs. t 1/[A]t vs t

-k, [A]0 -k, ln[A]0 k, 1/[A]0

[A]0/2k ln 2/k1/k[A]0


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Figure 1.5 Integrated rate laws and reaction order

ln[A]t = -kt + ln[A]0

1/[A]t = kt + 1/[A]0

[A]t = -kt + [A]0

f O f f


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Figure 1.6 A plot of [N2O5] vs. time for three half-lives.

For a first-order process, the half-life is a

constant and is independent of initial

concentration ln 2 0.693T =

k k=


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Figure 1.7 Graphical determination of the reaction order for the

decomposition of N2O5.

The decomposition

reaction of N2O5 is afirst order reaction,

only the plot of ln[N2O5]

vs t is linear. Plots of

[N2O5] vs t and

1/[N2O5] vs t not linear.


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Sample Problem 1.9

PLAN:

SOLUTION:

Determining Reaction Concentration at a Given Time

apply integrated rate law

PROBLEM: At 10000C, cyclobutane (C4H8) decomposes in a first-order

reaction, with the very high rate constant of 87s-1, to twomolecules of ethylene (C2H4).

(a) If the initial C4H8 concentration is 2.00M, what is the

concentration after 0.010 s?

(b) What fraction of C4H8 has decomposed in this time?

Find the [C4H8] at time, t, using the integrated rate law for a 1st order

reaction. Once that value is found, divide the amount decomposed by

the initial concentration.

; ln2.00

[C4H8]t

= (87s-1)(0.010s)

[C4H8]t = 0.83mol/L

ln[C4H8]0

[C4H8]t

= kt

(a)

(b) [C4H8]0 - [C4H8]t

[C4H8]0

=2.00M - 0.87M

2.00M= 0.58


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Sample Problem 1.10

PROBLEM: The decomposition of N2O5 is a first-order reaction with a rate

constant of 5.1 x 10-4

s-1

at 45o

C.2N2O5 (g) 4NO (g) + O2(g)

a) If the initial concentration of N2O5 is 0.25 M, what is

the concentration after 3.2 min?

b) How long will it take for the concentration of N2O5 to

decrease from 0.25M to 0.15M?

c) How long will it take to convert 62% of the starting

material?

d) Calculate the half-life of the reaction .

(0.25 M, 17 min, 32 min)


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Sample Problem 1.11

PROBLEM: Butadiene reacts to form its dimer according to the equation

2C4H6(g) C8H12(g)The following data were collected for this reaction at a given

temperature.

[C4H6(g) ] moldm-3 Time (s) [C4H6(g) ] moldm

-3 Time (s)

0.01000 0.000 0.00313 36000.00625 1000 0.00270 4400

0.00476 1800 0.00241 5200

0.00370 2800 0.00208 6200

a) Is the reaction first order or second order?

b) What is the value of the rate constant for the reaction?

c) What is the half-life of the reaction?


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Sample Problem 1.12

PLAN:

SOLUTION:

Determining the Half-Life of a First-Order Reaction

PROBLEM: Cyclopropane is the smallest cyclic hydrocarbon. Because its

600 bond angles allow poor orbital overlap, its bonds are weak.As a result, it is thermally unstable and rearranges to propene at

10000C via the following first-order reaction:

CH2

H2C CH2(g)

H3C CH CH2 (g)

The rate constant is 9.2s-1, (a) What is the half-life of the reaction? (b) How

long does it take for the concentration of cyclopropane to reach one-quarter of

the initial value?

Use the half-life equation, t =0.693

k, to find the half-life.

One-quarter of the initial value means two half-lives have passed.

t = 0.693/9.2s-1 = 0.075s(a) 2 t = 2(0.075s) = 0.150s(b)

S l P bl 1 13


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The reaction Q 2 R is second order in Q. If the initial

[Q] = 0.010 M and after 5.0 x 102

seconds the [Q] =0.0010 M,

(i) find the rate constant.

(ii) what is the length of time for [Q] = [Q]initial

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Sample Problem 1.13


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(ii)

(i)


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1.9 Collision Theory: Basis of the Rate Law

The Collision theoryis developed from the kinetic theory to explain

the effects of concentration and temperature on reaction rates.

The basic concept of collision is that all chemical reactions occur as a

result of collisions between reacting molecules. However, for a

reaction between two particles to occur, an effective collision must

take place.

Two requirements must be satisfied foreffective collisions betweentwo reacting molecules to take place.

1. The reacting molecules must collide with energy which must be

equal to or greater than the activation energy of the reaction (i.e.

the minimum amount of energy required to initiate a chemical

reaction). (Fig 1.8 & 1.9)

2. The molecules must approach each other in the right orientation

before a collision could lead to product formation. (Fig 1.11)

1 9 1 Activation Energy


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1.9.1 Activation Energy

The Activation Energyis defined as the minimum amount of kinetic

energy required for a chemical reaction to take place and form

products.This energy is needed to break chemical bonds of the reactants and to

rearrange atoms and valence electrons to form the products as the

reaction proceeds.

Reaction with low activation energy is easier to take place, and is faster

A reaction profile or energy profile is a plot of energies (potential

energy) versus progress of reaction as shown in Fig 1.9 & 1.14. The

progress of reaction represents the extent of the reaction.

Fig 1.10 (a) and (b) shows the reaction profiles of an exothermic and

endothermic reactions, respectively.

Ea = Activation energy

H = Enthalpy of reaction or Heat of reaction


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1.9.2 Effective Collisions depend on TWO Factors

For a collision to

be effective, the

reacting molecules

must havesufficient kinetic

energy to

overcome the

energy barrier.

49

Figure 1.8I. Kinetic Energy Factor

Fig re 1 9


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Figure 1.9


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heat inH > 0

Ea

Figure 1.10 Reaction profiles of an exothermic and endothermic

reaction

H < 0H < 0H < 0H < 0H < 0H < 0H < 0H < 0

heat outH < 0

Ea

(a) Exothermic reaction (b) Endothermic reaction


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II. Orientation Factor

The proper orientation results when the atomsare aligned in such a way that the old bonds canbreak and the new bonds can form

The more complex the reactant molecules, theless frequently they will collide with the properorientation

52

Fi 1 11


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Figure 1.11

ON

ClO

O

O

Cl

Cl

Cl


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1.10 The Transition State Theory - Molecular nature

of the Activated complex

There is an energy barrier (activation energy) to almost allreactions

There exists an intermediate stage when reactant molecules

change into products.

The activated complex ortransition state is a chemical

species with partially broken and partially formed bonds

an intermediate formed between reactant and product

molecules

has highest energy and extremely unstable

54

Ea(forw) = activation energy for the forward reaction(from reactant to peak)

Ea(rev) = activation energy for the reverse reaction

(from product to peak)


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Figure 1.12 Energy-level diagram for a reaction

A reaction can occur in either direction, so the diagram shows two

activation energies. The forward reaction is exothermic because

the reactants have more energy than the products and the

Ea(forward) < Ea(reverse)

Collision

Energy

ACTIVATEDSTATE

Ea

(reverse)

PRODUCTS

REACTANTS

Ea

(forward)

CollisionEnergy

For molecules to

react, they must

collide with enough

energy to reach an

activated state. Thisminimum collision

energy is the

activation energy,

Ea

A l l di f th f ti f lli i


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Figure 1.13 An energy-level diagram of the fraction of collisions

exceeding Ea.

R ti fil f th ti f CH B d OH


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Figure 1.14 Reaction energy profile for the reaction of CH3Br and OH-.

1 2 3 4 5

R f t th ti fil f th ti b t CH B


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Refer to the reaction energy profile for the reaction between CH3Br

and OH- shown in Fig 1.14:

CH3Br + OH- CH3OH + Br

[1] Two molecules that are far apart but speeding toward each otherhave high kinetic energy and low potential energy. As the

molecules get closer some kinetic energy is converted to potential

energy as the electron clouds repel each other.

[2] When the molecules collide, kinetic energy is changed into potentialenergy because the molecules are distorted during collision to

break bonds and rearrange the atoms ready to form product

molecules. If this potential energy is less than the activation energy

the molecules bounce off without forming product.

[3] A fraction of the molecules which collide with sufficient kineticenergy along the line of approach to carry them over the

activation barrier, the peak in the reaction profile, then reaction

occur. At the peak, the C-Br bond lengthen and weaken while the

C-OH bond began to form.


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At this point during the transformation, the two reacting molecules are

highly distorted, forming the highly energetic and unstable species called

the transition state, oractivated complex as shown in Fig 1.14 (3)

and Fig 1.15. The dotted line in the Fig represents bonds in the processof breaking and making. C is surrounded by 5 atoms

This complex only forms if the molecules collide in an effective orientation

and with energy equal to or greater than the activation energy. Thus the

Ea is the quantity needed to stretch and deform bonds in order to reach

the transition state

[4] The potential energy of the complex starts to decrease as the new

bond C-O bond in the product shorten and strengthen and the C-Br

bond in the reactant molecule weakens.

[5] At separations to the right of the maximum, the potential energyrapidly falls to a low value as the product molecules separate .


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Figure 1.15

Nature of the transition state in the reaction between CH3

Br and OH-.

CH3Br + OH- CH3OH + Br

-

transition state or activated complex

Figure 1 16


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Figure 1.16

Reaction energy diagrams and possible transition states.


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1-62 L.H. SimReaction progress

Potential

Energy

Sample Problem 1.14

SOLUTION:

Drawing Reaction Energy profile

PROBLEM: A key reaction in the upper atmosphere is

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

The Ea(fwd) is 19 kJ, and the Hrxn for the reaction is -392 kJ. Draw a

reaction energy diagram for this reaction, postulate a transition state, and

calculate Ea(rev).

PLAN: Consider the relationships among the reactants, products and

transition state. The reactants are at a higher energy level than theproducts and the transition state is slightly higher than the

reactants.

O3+O

2O2

Ea= 19kJ

Hrxn = -392kJ

Ea(rev)= (392 + 19)kJ =

411kJ

OO

O

O

breakingbond

formingbond

transition state

1 11 Factors Affecting Reaction Rate


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1.11 Factors Affecting Reaction Rate

Any factor that can increase the frequency of effective

collisions would increase the rate of a reaction

Under specific conditions, every reaction has its own

characteristic rate, which may be controlled by four factors:

1. Concentrations of reactants

2. Chemical nature and physical state of reactants

3. Temperature

4. The use of a catalyst

Tro: Chemistry: A Molecular Approach 63

(A) Concentration of reactants


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(A) Concentration of reactants

Rate of reaction number of collision

Second

OR Rate of reaction frequency of collision

According to the collision theory, as the number of reacting

molecules in a given volume increase, the frequency of collisions

(the number of collisions that happen per unit time) will be increasewhich then increase the amount ofeffective collisions (the

probability of collisions with sufficient energies for a reaction to

occur). This lead to an increase in the rate of a reaction. (Fig 1.17)

Concentration of gases depends on the partial pressure of the gas

higher pressure = higher concentration

Concentrations of solutions depend on the solute-to-solution ratio

(molarity)

Fi 1 17


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Figure 1.17

The dependence of possible collisions on the product

of reactant concentrations.

A

A

B

B

A

A

B

B

A

4 collisions

Add another

molecule of A6 collisions

Add another

molecule of B

A

A

B

B

A B

(B) Nature of the Reactants


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(B) Nature of the Reactants

Nature of the reactants means what kind of reactant molecules

and what physical state they are in

small molecules tend to react faster than large molecules

gases tend to react faster than liquids, which react faster than

solids

powdered solids are more reactive than blocks

more surface area for contact with other reactants leading tohigher frequency of collision. The probability of effective

collision will be increased. Surface area can be increased by

grinding the solids into fine powders. (Fig 1.18)

certain types of chemicals are more reactive than others

e.g. potassium metal is more reactive than sodium

ions react faster than molecules

no bonds need to be broken

66


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Figure 1.18 The effect of surface area on reaction rate.

(C) Temperature


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Increasing the temperature raises the average kinetic energy of the

reactant molecules.

Increasing the temperature increases the number of molecules with

sufficient kinetic energy to overcome the activation energy, leads to

an increase in the frequency of effective collisions between

molecules. (Fig 1.19)

The distribution of kinetic energy is shown by the Boltzmanndistribution curves shown in Fig 1.19

Fig 1.19 shows that at both temperatures T1 and T2 (T2 .> T1), a

relatively small fraction of collisions have sufficient kinetic energy

the activation energy to result in a reaction. Moreover, the

fraction ofeffective collisionsor collisions with the requiredactivation energy increases exponentially with an increasein

temperature from T1 to T2 (see Fig 1.20).

( ) p

Figure 1 19


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Figure 1.19

The effect of temperature on the distribution of collision energies

1 12 The Arrhenius Equation The Effect of Temperature


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1.12 The Arrhenius Equation The Effect of Temperature

on Reaction Rate

k AeEaRT

where kis the kinetic rate constant at T

Ea is the activation energy

R is the gas constant = 8.314 J K-1mol-1

T is the Kelvin temperature

A is the collision frequency factor

The exponential factor in the Arrhenius

equation is a number between 0 and 1

The larger the activation energy, Ea, the fewer molecules that have

sufficient energy to overcome the energy barrier, the smaller the rate

constant k.

Increase in temperature, T, will gives a largerkvalue therefore

increasing the temperature will increase the reaction rate

This type of relationship is called an exponential relationship


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yp p p p

which means that a small rise in temperature results in a large

increase in the fraction of molecules collide with energy equal to

or greater than the activation energy (Fig 1.19).

1.13 Arrhenius Plots

The Arrhenius Equation: The integrated form

This equation is in the linear form y= mx+ b

where y= ln(k) andx= (1/T)

A graph ofln(k) vs. (1/T) is a straight line

eyintercept =A (unit is the same as k)


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Figure 1.20 Dependence of the rate constant on temperature


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Figure 1.21 Graphical determination of the activation energy

ln k= -Ea/R (1/T) + ln A

Sample Problem 1.15


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Determine the activation energy and frequency factor for the reaction

O3(g) O2(g) + O(g) given the following data:

74

Plot the graph ln kversus 1/T. From the straight line curve,


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slope = - Ea/R

y-intercept = ln (frequency factor, A). Figure 1.22

slope m = -1 12 x 104 K


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slope, m 1.12 x 10 K

yintercept, b = 26.8

Or ln A = 26.8, anti ln (26.8) = 4.36 x 1011 M-1.s-1


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Table 1.5 The Effect of Ea and T on the Fraction (f) of Collisions

with Sufficient Energy to Allow Reaction

Ea (kJ/mol) f(at T = 298 K)

50 1.70x10-9

75 7.03x10-14

100 2.90x10-18

T f(at Ea = 50 kJ/mol)

250C(298K) 1.70x10-9

350C(308K) 3.29x10-9

450C(318K) 6.12x10-9

1 14 A h i E ti T P i t F


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1.14 Arrhenius Equation: Two-Point Form

If only two data points i.e two T and two kare

given, the following forms of the Arrhenius

Equation can be used:

78

Sample Problem 1.16 Determining the Energy of Activation


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PLAN:

SOLUTION:

PROBLEM: The decomposition of hydrogen iodide,

2HI(g) H2(g) + I2(g)

has rate constants of 9.51x10-9L/mol*s at 500. K and 1.10x10-5 L/mol*s at

600. K. Find Ea.

Use the modification of the Arrhenius equation to find Ea.

ln

k2

k1 =

Ea

R

1

T1

1

T2

-

R= - (8.314J/mol*K)

Ea = 1.76x105 J/mol = 176 kJ/mol

ln1

500K

1

600K-

1.10x10-5L/mol*s

9..51x10-9L/mol*s=

8.314 J/Mol*K

Ea

100500 x 600

ln (0.11567 x 104) =8.314 J/Mol*K

Ea

7.0533 Ea x 4.009 x 10-5

=

Sample Problem 1.17


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Determining ActivationEnergy graphically

PROBLEM:

The 2nd order rate constant for the decomposition of N2O

into N2 and O2 has been measured at a series of

temperatures.

(a) Determine graphically the activation energy for the

reaction.

(b) Refer to the graph, predict the rate constant at 800K

k (1/M) 1.87x 10-3

0.0113 0.0569 0.244

t (oC) 600 650 700 750

Figure 1.23


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Information sequence to determine the kinetic parameters of a reaction.

Series of plots

of concentra-

tion vs. time Initialrates Reaction

ordersRate constant

(k) and actual

rate law

Integrated

rate law(half-life,

t1/2)

Rate constant

and reaction

order

Activation

energy, Ea

Plots of

concentration

vs. time

Find k atvaried T

Determine slope

of tangent at t0 for

each plot

Compare initial

rates when [A]

changes and [B] isheld constant and

vice versa

Substitute initial rates,

orders, and concentrationsinto general rate law:

rate = k[A]m[B]n

Use direct, ln or

inverse plot to

find order

Rearrange to

linear form and

graph

Find k atvaried T

1.15 REACTION MECHANISMS


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The reaction mechanismof a chemical reaction is the sequence of

reaction steps orelementary reaction sthat ultimately lead from the

initial reactants to the final products of a reaction.

Anelementary reaction or elementary steprepresents a single

molecular event, such as dissociation or collision of molecules, in the

progress of the overall reaction.

An Example of a Reaction Mechanism

Overall reaction:

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

Plausible Mechanism:

Step 1: H2(g) + ICl(g) HCl(g) + HI(g)

Step 2: HI(g) + ICl(g)

HCl(g) + I2(g) The reactions (1) and (2) in this mechanism are called the

elementary steps, meaning that they cannot be broken downinto simpler steps, and that the molecules actually interact directlyin this manner without any other steps


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A catalystdoes not appear in the net equation, it is consumed in the

first step and produced in the second or last step of the reaction

mechanism.

Areaction intermediate is a substance produced during a

elementary reaction that does not appear in the net equation because

it reacts in a subsequent step or elementary reaction in the

mechanism. In other words, Reactionintermediates are products inan early mechanism step, but then a reactant in a later step

E.g: Overall reaction: 2H2O2(aq) 2H2O(l) + O2(g)

Plausible mechanism : H2O2 + I- H2O + OI

-

H2O2 + OI- H2O + O2 + I

-

OI- = intermediate; I- = catalyst

1.15.1 Rate Laws for elementary steps


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Each step in the mechanism is like its own little reaction

with its own activation energy and own rate law

The rate law for an overall reaction must be determined

experimentally

But the rate law of an elementary step can be deduced

directly from the equation of the step

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

Elementary steps in the reaction mechanism:

1) H2(g) + ICl(g) HCl(g) + HI(g) Rate = k1[H2][ICl]

2) HI(g) + ICl(g) HCl(g) + I2(g) Rate = k2[HI][ICl]

84


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Molecularity refers to the number of species (free atoms, ions or

molecules) that must collide or dissociates to produce the reaction

indicated by a particular elementary step.

But the probability of more than 3 molecules colliding at the same

instant with the proper orientation and sufficient energy to overcome

the energy barrier is negligible. Most reactions occur in a series of

small reactions involving 1, 2, or at most 3 molecules

Molecularity

Table 1.6 Rate Laws for General Elementary Steps

Rate Law

A + B product

2A + B product

Unimolecular

Bimolecular

Bimolecular

Termolecular

Rate = k[A]

Rate = k[A]2

Rate = k[A][B]

Rate = k[A]2[B]

A product

2A product

Elementary Step

1.15.2 The Rate-Determining Step of a Reaction Mechanism


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The overall rate of a reaction is related to the rate of the

slowest, or rate-determining step. The slowest step or rate

determining step has the largest activation energy

The rate law of a reaction is determined by the rate of the rate -

determining step of the acceptable mechanism of the reaction.

The rate law of the rate determining step determines the rate

law of the overall reaction.

1.15.3 A plausible reaction Mechanism must satisfy two

requirements

a) The elementary steps must be physically reasonable and add

up to the overall equation.

b) The mechanism must correlated with the experimentally

determined rate law.

1.15.4 A mechanism with a slow step followed by a fast step


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A mechanism can never be proved absolutely correct. We can only

say that a mechanism which satisfy the two requirements is

possibly correct or acceptable .

Example :

The balanced equation for the reaction of NO2(g) and F2(g) is

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

The experimentally rate law is Rate = k[NO2][F2]

A suggested mechanism for the reaction is

k1

NO2 + F2 NO2F + F slowk2F + NO2 NO2F fast

Is this an acceptable mechanism for the reaction?

Requirement 1: Add up the two elementary reactions


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Requirement 1: Add up the two elementary reactions

k1NO2 + F2 NO2F + F slow

k2F + NO2 NO2F fast

2NO2(g) + F2(g) 2NO2F(g) (same as the overall reaction

given)

Requirement 2: write the rate law based on the rate determining step

(i.e. the slow step)

Rate = k1[NO2][F2] (same as the experimentally determined

rate law given )

The mechanism is acceptable as it satisfies the 2 requirements.

Another Reaction Mechanism


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The first step in this mechanism

is the rate determining step.

The first step is slower than the

second step because its

activation energy is larger.

The rate law of the first step is

the same as the rate law of theoverall reaction.

89

NO2(g) + CO(g) NO(g) + CO2(g) Rateobs = k[NO2]2

1. NO2(g) + NO2(g) NO3(g) + NO(g) Rate = k1[NO2]2 Slow

2. NO3(g) + CO(g) NO2(g) + CO2(g) Rate = k2[NO3][CO] Fast

Sample Problem 1.18

Determining Molecularity and Rate Laws for


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PLAN:

SOLUTION:

Determining Molecularity and Rate Laws for

Elementary Steps

PROBLEM: The following two reactions are proposed as elementary steps in

the mechanism of an overall reaction:(1) NO2Cl(g) NO2(g) + Cl(g)

(2) NO2Cl(g) + Cl(g) NO2(g) + Cl2(g)

(a) Write the overall balanced equation.

(b) Determine the molecularity of each step.

(a) The overall equation is the sum of the steps.

(b) The molecularity is the sum of the reactant particles in the step.

2NO2Cl(g) 2NO2(g) + Cl2(g)

(c) Write the rate law for each step.

rate2 = k2 [NO2Cl][Cl]

(1) NO2Cl(g) NO2(g) + Cl(g)

(2) NO2Cl(g) + Cl(g) NO2(g) + Cl2(g)

(a)

Step(1) is unimolecular.

Step(2) is bimolecular.

(b)

rate1 = k1 [NO2Cl](c)

Sample Problem 1.19


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PROBLEM: (a) Consider the following two-step reaction mechanism for a

chemical reaction.

Step 1. NO2(g) + NO2(g) NO3(g) + NO(g) ..(slow)

Step 2. NO3(g) + CO(g) NO2(g) + CO2(g).(fast)

(i) Write the chemical equation for the overall reaction.

(ii) Identify the reaction intermediate

(iii) What is the order of the reaction?

(b) The decomposition of hydrogen iodide occurs via the action of the

following elementary steps

I. HI + HI H2I2 .Slow

II. H2I2 H2 + 2I fast

III. 2I I2 .fast

(i) Write the rate law of the reaction and state the order of thereaction

(ii) What is the molecularity of each of the elementary step?

(iii) Write the stoichiometric equation for the overall reaction.

(iv) State the reaction intermediate.

1.15.5 Mechanism which involve a fast reversible step


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Example: The reaction of Cl2 with CHCl3 is given by the equation

Cl2(g) + CHCl3(g) HCl(g) + CCl4(g)

The rate law determined experimentally is Rate = k[Cl2] [CHCl3]

A proposed mechanism for the reaction is

k1Cl2 2Cl fast

k -1

k2Cl + CHCl3 HCl + CCl3 slow

k3

CCl3 + Cl CCl4 fast

Is this an acceptable mechanism for the reaction?

I) Cl2 2ClCl + CHCl3 HCl + CCl3

CCl3 + Cl CCl4

Cl2(g) + CHCl3(g) HCl(g) + CCl4(g)

II) Rate = k2[Cl][CHCl3]

Cl i i t di t hi h i ll d t b i th t l T b tit t


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Cl is an intermediate which is allowed to be in the rate law. To substitute

the intermediate, we make use of the equilibrium of the fast step.

At equilibrium, rate of forward reaction=rate of reverse reaction

k1[Cl2] = k-1[Cl]2

[Cl] = (k1/k-1) [Cl2]

Substitute into the above rate law, Rate = k2(k1/k-1)[Cl2]

[CHCl3]

= k[Cl2][CHCl3]

Since the mechanism satisfy the two requirements,

it is acceptable

Sample Problem 1.20

PROBLEM: Consider the following reaction

2O3(g) 3O2(g)

The experimentally rate law is Rate = k [O3]2

[O2]

The mechanism proposed for the reaction is

k1

O O + O ( fast equilibrium)


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O3 O2 + O ( fast equilibrium)

k -1

k2

O + O3 2O2 (slow)Verify the acceptability of the proposed mechanism for the above reaction.

Sample Problem 1.21

The equation for the reaction is Tl3+

+ Hg22+

Tl+

+ 2Hg2+

The experimental determined rate law is : rate = k [Tl3+] [Hg22+]

[Hg2+]

Verify the acceptability of the proposed mechanisms below :

k1

Hg22+ Hgo + Hg2+ (fast)k -1

k2Tl3+ + Hgo Tl+ + Hg2+ (slow)

1.16 CATALYSTS


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Acatalystis a substance that increases the rate of a chemical

reaction without itself being consumed.

In general a catalyst lowers the activation energy, Ea, of the reaction

by providing a different mechanism or energy pathway for the

reaction concerned. Lowering Ea increases the rate constant, k, and

thereby increases the rate of the reaction

A catalyst increases the rate of the forwardand the reverse

reactions to the same extent.

A catalyzed reaction yields the products more quickly, but does not

yield more product than the uncatalyzed reaction.

Two types of catalysis:

a) Homogenous catalyst is one that is present in the same phaseor physical state as the reactant molecules. Eg enzyme reaction

b) Heterogeneous catalyst is in a different phase from the

reacting molecules, usually in the solid phase .

Figure 1.24


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Reaction energy profile of a catalyzed and an uncatalyzed process.

Examples of reactions catalyzed by homogeneous catalysts


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1. The depletion of ozone in the stratosphere by Cl atoms is an

example ofhomogeneous catalysis.

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

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

The uncatalyzed reaction is O3 + O 2O2 .

2. The decomposition of H2O2 catalyzed by I- ion.H2O2 + I

- H2O + IO-

H2O2 + IO- H2O + O2 + I

-

The uncatalyzed reaction is 2H2O2 2H2O + O2 .

3. The hydrolysis of an organic ester (RCOOR) to form a carboxylicacid (RCOOH), the reaction is catalyzed by a strong acid (H+).

R C O R + H2O R C OH + R OH (uncatalyzed)

OO

Figure 1.25 The metal-catalyzed hydrogenation of ethylene-


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g e eta cata y ed yd oge at o o et y e e

heterogeneous catalysis

H2C CH2 (g) + H2 (g) H3C CH3 (g)Ni


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CH3CH CH2 + H2 CH3CH CH2

HH

Ni

Example 2: Hydrogenation of an alkene catalyzed by Ni

propenepropane

T bl 1 7


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Table 1.7

chm096 1 chem kinetics rm

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