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Copyright 2011 Pearson Education, Inc.

Chapter 1

Chemical Kinetics


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Kinetics: Rates and Mechanisms of Chemical Reactions

1.1 Expressing Reaction 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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The speed of a chemical reaction is called itsreaction rate.

The rate of a reaction is a measure of how fastthe reaction makes products, or how fast thereactants are consumed.

Chemical kinetics is the study of the factors

that affect the rates of chemical reactions:Such as temperature, reactant concentrations

3Tro: Chemistry: A Molecular Approach, 2/e

Chemical KineticsChemical Kinetics


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The rate of a chemical reaction is generallymeasured in terms of how much the concentrationof a reactant decreases in a given period of timeor product concentration increases

For reactants, a negative sign is placed in front ofthe definition:


Defining Reaction RatesDefining Reaction Rates


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For a reaction A p B

Reaction RatesReaction Rates

t(

(!

!

Bofmoles

in timechange

BofmolesofnumberinchangerateAverage

6


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? A ? A ? A

06250061

87Rate

ttXX

tXRate

12

12

.!

!

!

(

(!? A ? A ? A

250061

84Rate

ttAA

tARate

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

? A ? A ? A

250061

04Rate

ttCC

tCRate

12

12

.!

!

!

(

(! ? A ? A ? A

06250061

01Rate

ttZZ

tZRate

12

12

.!

!

!

(

(!



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As time goes on, the rate of a reaction generallyslows down

because the concentration of the reactants

decreases At some time the reaction stops, either

because the reactants run out orbecause thesystem has reached equilibrium.


ReactionReaction Rate Changes Over TimeRate Changes Over Time


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The average rate is the change in measuredconcentrations in any particulartime period linear approximation of a curve

Change of Rate with Time

Most useful units for rates are to look at molarity.Since volume is constant, molarity and moles are

directly proportional.


Average RateAverage Rate


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In most reactions, the coefficients of the balancedequation are not all the same.H2 (g) + I2 (g) p 2 HI(g)

For these reactions, the change in the number of

molecules of one substance is a multiple of thechange in the number of molecules 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 ofeach substance is multiplied by 1/coefficient


Reaction Rate andReaction Rate and StoichiometryStoichiometry


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H2 I2HI

Stoichiometry tells us that forevery 1 mole/L of H2 used,2 moles/L of HI are made.

Assuming a 1 L container, at10 s, we used 0.181 moles ofH2. Therefore the amount ofHI made is 2(0.181 moles) =0.362 moles.

At 60 s, we used 0.699 molesof H2. Therefore the amountof HI made is 2(0.699 moles)= 1.398 moles.

The average rate isthe change in theconcentration in a

given time period

In the first 10 s, the([H2] is 0.181 M,so the rate is



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average rate in a given

time period = slope ofthe line connecting the[H2] points; and +slopeof the line for [HI]

the average rate for thefirst 10 s is 0.0181 M/s





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The instantaneous rate is the change inconcentration at any one particular time.Determined by taking the slope of a line tangent to theconcentartion vs time curve at any particular point.

The initial rate is the reaction rate at theb

eginningof a reaction. slope of a line tangent to the curve at time t= 0No product formed yet, reversed rate = 0


Instantaneous RateInstantaneous Rate


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H2 (g) + I2 (g) p 2 HI (g)Using [H2], the

instantaneousrate at 50 s is

Using [HI], theinstantaneous

rate at 50 s is



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Problem 1.1: For the reaction given, the [I] changes from1.000 M to 0.868 Min the first 10 s. Calculate the average rate in the first 10 s, and ([H+]/(t.

H2O2 (aq) + 3 I

(aq) + 2 H+

(aq) p I3

(aq) + 2 H2O(l)

15

solve the rateequation for the rate(in terms of the

change inconcentration of thegiven quantity)

solve the rateequation (in terms of

the change in theconcentration for thequantity to find) forthe unknown value

Tro: Chemistry: A Molecular Approach, 2/e


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Practice :If 2.4 x 102 g ofNOBr (MM 109.91 g) decomposes in a 2.0 x102 mL flask in 5.0 minutes, find the average rate of Br2production in M/s.

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



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2 mol NOBr:1 mol Br2, 1 mol = 109.91g, 1 mL=0.001 L

Solve:

Conceptual

Plan:

Relationships:

240.0 g NOBr, 200.0 mL, 5.0 min.

([Br2]/(t, M/s

Given:

Find:

Practice If 2.4 x 102 g ofNOBr decomposes in a 2.0 x 102 mLflask in 5.0 minutes, find the average rate of Br2 production

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

17

( M( mole Br2Rate

( min ( s }

5.45 M Br2, 3.0 x 102 s

([Br2]/(t, M/s

240.0 g NOBr, 0.2000 L, 5.0 min.

([Br2]/(t, M/s



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Practice: Consider the reaction:

8A(g)+ 5B(g)p 8C(g)+6D(g)If [C] is increasing at the rate of4.0 mol L1s1, at what rate is[B] changing?

A. -0.40 mol L1s1

B. -2.5 mol L 1s1

C. -4.0 mol L1s1

D. -6.4 mol L1s1E. None of these choices is correct, since its rate ofchange must be positive.

Tro: Chemistry: A Molecular Approach 18


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In general, the differential rate law for the reaction

aA + bB cC + dD

rate = 1a

- = -(

[A](t

1b

([B](t

1c

([C](t

= + 1d

([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.


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SOLUTION:

Expressing Rate in Terms of Changes in Concentration with Time

PROBLEM 1.2: Because it has a nonpolluting product (water vapor), hydrogengas is used for fuel aboard the space shuttle and may be usedby earthbound 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?

-1

2

([H2]

(t= -

([O2]

(t

= +([H2O]

(t

1

2

0.23 mol/L*s = +([H2O]

(t

1

2

= 0.46 mol/L*s([H2O]

(t

rate =(a)

([O2]

(t- = -(b)


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The rate law of a reaction is the mathematicalrelationship between the rate of the reaction and theconcentrations of the reactantsand homogeneous catalysts as well

The rate lawmu

stbe determined experimentally!! The rate of a reaction is directly proportional to theconcentration of each reactant raised to a power

For the reaction aA + bB p products the rate

law would have the form given belown and m are called the orders for each reactant

k is called the rate constant


The Rate LawThe Rate Law


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The exponent on each reactant in the rate lawis called the orderwith respect to that reactant.

The sum of the exponents on the reactants iscalled the overall order of the reaction.

The rate law for the reaction given:2 NO(g)+ O2(g) p 2 NO2(g)is: Rate = k[NO]2[O2]

The reaction is:

second orderwith respect to [NO],

first orderwith respect to [O2],

and third order overall.22Tro: Chemistry: A Molecular Approach, 2/e

Reaction OrderReaction Order


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Using Initial Rates to Determines Rate Laws

A reaction is zero order in a reactant if the change in

concentration of that reactant produces no effect.

A reaction is first order if doubling the concentration

causes the rate to double.

A reaction is nth order if doubling the concentration

causes an2n

increase in rate. Note that the rate constant does not depend on

concentration.

Concentration and RateConcentration and Rate

23


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SOLUTION:

Determining Reaction Order from Rate Laws

PROBLEM 1.3:

For each of the following reactions, determine thereaction order with respect to each reactant and theoverall order from the given rate law.

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

(b) 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 1st order in H2O2, 1st order in I- and zero order in H+,

while being 2nd order overall.


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Rate = k[NO][O3]

Solve:

Conceptual

Plan:

Relationships:

[NO] = 1.00 x 106 M, [O3] = 3.00 x 106 M,Rate = 6.60 x 106 M/sk, M1ys1

Given:

Find:

Example: The rate equation for the reaction ofNO with ozone isRate = k[NO][O3]. If the rate is 6.60 x 10

5 M/sec when [NO] = 1.00 x 106 Mand [O3] = 3.00 x 10

6 M, calculate the rate constant k

Rate, [NO], [O3] k



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Practice:The rate law for the decomposition of acetaldehyde,CH3CHO, is Rate = k[CH3CHO]2.What is the rate of the reaction when the [CH3CHO]= 1.75 x 103 M and the rate constant is 6.73 x 106

M1s1?



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Rate = k[acetaldehyde]2

Solve:

Conceptual

Plan:

Relationships:

[CH3CHO] = 1.75 x 103 M, k= 6.73 x 106 M1s1

Rate, M/s

Given:

Find:

Practice What is the rate of the reaction when the[CH3CHO] = 1.75 x 103 M and the rate constant is 6.73 x

106 M1s1?

k, [acetaldehyde] Rate



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The rate law mustbe determined experimentally. The rate law shows how the rate of a reaction

depends on the concentration of the reactants.

Changing the initial concentration of a reactant willtherefore affect the initial rate of the reaction.

28

if for the reactionA Products

Rate = k[A]n

then doubling the

initial concentrationofA doubles theinitial reaction rate

then doubling the

initial concentrationofA does notchange the initialreaction rate. Thus,n = 0 (Zero Order)

Rate = k

then doubling the

initial concentrationofA quadruples theinitial reaction rate


Finding the Rate Law:Finding the Rate Law:

The Initial Rate MethodThe Initial Rate Method


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Proposed rate law: Rate = k[A]n

If a reaction is Zero Order, the rate of the reactionis always the same.doubling [A] will have no effect on the reaction rate

If a reaction is First Order, the rate is directlyproportional to the reactant concentrationdoubling [A] will double the rate of the reaction

If a reaction is Second Order, the rate is directlyproportional to the square of the reactantconcentrationdoubling [A] will quadruple the rate of the reaction



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Determining the Rate Law when there are

Multiple Reactants

Changing each reactant will effect the overall rateof the reaction.

By changing the initial concentration of onereactant at a time, the effect of each reactantsconcentration on the rate can be determined.

In examining results, we compare differences in

rate for reactions that only differ in theconcentration of one reactant.



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Example 13.2: Determine the rate law and rate constant for thereaction NO2(g) + CO(g) p NO(g) + CO2(g)

given the data below

Comparing Expt #1 and Expt #2,the [NO2] changes but the [CO]

does not31Tro: Chemistry: A Molecular Approach, 2/e


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Example 13.2: Determine the rate law and rate constantfor the reaction NO2(g) + CO(g) p NO(g) + CO2(g)




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n = 2, m = 0



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Substitute theconcentrationsand rate forany experiment

into the ratelaw and solvefork

Expt.

Number

Initial[NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.00823. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

Rate ! k[NO2]2

for expt 1

0.0021 Ms ! k 0.10 M 2

k!0.0021 Ms0.01 M2

! 0.21 M1 y s1



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Practice Determine the rate law and rate constant for thereaction NH4

+ +NO2 pN2 + 2 H2O




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Practice Determine the rate law and rate constant for thereaction NH4

+ +NO2 pN2 + 2 H2O


Rate = k[NH4+]n[NO2

]m



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For the reaction A p Products,the rate law depends on the concentration of A.

Applying calculus to integrate the rate law givesanother equation showing the relationshipbetween the concentration of A and the time of thereaction this is called the Integrated Rate Law


Integrated Rate LawsIntegrated Rate Laws


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Rate = k[A]1 = k[A] (differential rate law)

ln[A] = (k)t + ln[A]initial (integrated rate law)y = m. x + b

Graph ln[A] vs. time gives straight line with slope= k and yintercept = ln[A]initialused to determine the rate constant

t = 0.693/k. The half-life of a first order reaction

is constant.

When Rate = M/sec, k= s140Tro: Chemistry: A Molecular Approach, 2/e

First Order ReactionFirst Order Reaction

? A ? A? A ? A

initialt

initialt

kt

kt

kt

AlnAln

AlnAln

]A[A][

Rate

!

!

!(

(!


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ln[A]initial

ln[A]

time



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Half-Life

The half-life, t1/2, of a reaction is the time ittakes for the concentration of the reactanttofall to its initial value.

The half-life of the reaction depends on theorder of the reaction.


HalfHalf--LifeLife

kkt

693.0ln 21

21 !!


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The Half-Life of a First-Order Reaction Is Constant



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Rate Data for: C4H9Cl + H2O p C4H9OH + HCl



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C4H9Cl + H2O p C4H9OH + 2 HCl



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slope =2.01 x 103

k=

2.01 x 103 s-1



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Rate = k[A]0

= k (differential rate law)

[A] = kt + [A]initial (integrated rate law)

Graph of [A] vs. time is straight line withslope = k and yintercept = [A]initial

t = [Ainitial]/2k = [A0]/2k

When Rate = M/sec, k= M/sec [A]initial

[A]

time48Tro: Chemistry: A Molecular Approach, 2/e

Zero Order ReactionZero Order Reaction


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Rate = k[A]2

(differential rate law)1/[A] = kt + 1/[A]initial (integrated rate law)

Graph 1/[A] vs. time gives straight line withslope = k and yintercept = 1/[A]

initialused to determine the rate constant

t = 1/(k[A0])

When Rate = M/sec, k= M1ys1

For a second order reaction, half-life depends onthe initial concentration.

The lower the initial concentration the higher is the

half-life. 49Tro: Chemistry: A Molecular Approach, 2/e

Second Order ReactionSecond Order Reaction


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PLAN:

SOLUTION:

Determining the Half-Life of a First-Order Reaction

PROBLEM 1.3: Cyclopropane is the smallest cyclic hydrocarbon. It isthermally unstable and rearranges to propene at 1000oCvia the following first-order reaction:

CH2

H2C CH2(g)

(H3C CH CH2 (g)

The rate constant is 9.2 s-1, (a) What is the half-life of the reaction?(b) How long does it take for the concentration of cyclopropane toreach one-quarter of the initial value?

Use the half-life equation, t1/2 =0.693

k

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

t1/2 = 0.693/9.2 s-1 = 0.075 s(a) 2 t1/2 = 2(0.075 s) = 0.150 s(b)


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Copyright 2011 Pearson Education, Inc.52Tro: Chemistry: A Molecular Approach, 2/e


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Table 1 Units of the Rate Constant k

Overall Reaction Order Units ofk(tin seconds)

0 mol/L*s (or mol L-1 s-1)

1 1/s (or s-1)

2 L/mol*s (or L mol -1 s-1)

3 L2 / mol2 *s (or L2 mol-2 s-1)

E l 13 4 Th ti SO Cl SO Cl i fi t d


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Example 13.4: The reaction SO2Cl2(g) p SO2(g) + Cl2(g) is first orderwith a rate constant of 2.90 x 104 s1 at a given set of conditions.

Find the [SO2Cl2] at 865 s when [SO2Cl2]initial = 0.0225 M

54

the new concentration is less than the original,as expected

[SO2Cl2]init = 0.0225 M, t = 865, k= 2.90 x 10-4 s1

[SO2Cl2]

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

[SO2Cl2][SO2Cl2]init, t, k



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PracticeThe reaction Q p 2 R is second order in Q. If theinitial [Q] = 0.010 M and after 5.0 x 102 seconds the[Q] = 0.0010 M, find the rate constant.



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Practice The reaction Q p 2 R is second order in Q. If theinitial [Q] = 0.010 M and after 5.0 x 102 seconds the [Q] =

0.0010 M, find the rate constant

56

[Q]init = 0.010 M, t = 5.0 x 102 s, [Q]t = 0.0010 M

k

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

k[Q]init, t, [Q]t



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Plots of [A] vs. time, ln[A] vs. time, and 1/[A] vs.time allow determination of whether a reaction is

zero, first, or second order. Whichever plot gives a straight line determines

the order with respect to [A]

if linear is [A] vs. time, Rate = k[A]0

if linear is ln[A] vs. time, Rate = k[A]1

if linear is 1/[A] vs. time, Rate = k[A]2


Graphical Determination of theGraphical Determination of the

Rate Law for A ProductRate Law for A Product


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Practice Complete the Table and Determine theRate Equation for the Reaction A p 2 Product

58

What will the rate be when the [A] = 0.010 M?



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Practice Complete the Table and Determine theRate Equation for the Reaction A p 2 Product



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Conclusion for the determination of the rateequation for the reaction A p 2 Product

Because the graph 1/[A] vs. time is linear,the reaction is second order,

63

Rate ! k[A]2

k! slope of the line ! 0.10 M1y s 1


R l ti hi B t O d dR l ti hi B t O d d


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For a zero orderreaction, the half-life is directlyproportionalto the initial concentration. The lower theinitial concentration of the reactants, the shorter thehalf-life

t1/2 = [A]init/2k For a first orderreaction, the half-life is independent

of the concentration.

(t1/2 = constant) t1/2 = ln(2)/k

For a second orderreaction, the half-life is inverselyproportionalto the initial concentration increasingthe initial concentration shortens the half-life t1/2 = 1/(k[A]init)


Relationship Between Order andRelationship Between Order and

HalfHalf--LifeLife

Example 13.6: Molecular iodine dissociates at 625 K with a first-


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Example 13.6: Molecular iodine dissociates at 625 K with a firstorder rate constant of 0.271 s1. What is the half-life of this

reaction?

65

the new concentration is less than the original,as expected

k= 0.271 s1

t1/2

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:Find:

t1/2k



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PracticeThe reaction Q p 2 R is second order in Q. If the

initial [Q] = 0.010 M and the rate constant is 1.8M1s1 find the length of time for [Q] = [Q]init


Practice : The reaction Q p 2 R is second order in Q If the


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Practice : The reaction Q p 2 R is second order in Q. If theinitial [Q] = 0.010 M and the rate constant is 1.8 M1s1 findthe length of time for [Q] = [Q]init

67

[Q]init = 0.010 M, k= 1.8 M1ys1

t1/2, s

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

t1/2[Q]init, k



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For most reactions, for a reaction to take place, thereactingmolecules must collide with each other. on average about 109 collisions per second

Once molecules collide they may react together or

they may not, depending on two factors1. whether the collision has enough energy to "break

the bonds holding reactant molecules together"; and

2. whether the reacting molecules collide in the properorientation for new bonds to form


Collision TheoryCollision Theory


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The greater the number of collisions the faster the

rate. The more molecules present, the greater the

probability of collision and the faster the rate.

The higher the temperature, the more energyavailable to the molecules (collision) and the fasterthe rate.

(Eg. food spoils when not refrigerated.)

Complication: not all collisions lead to products. Infact, only a small fraction of collisions lead toproduct.


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Molecules must possess a certain minimum kineticenergy called the activation energyfor a reactionto occur.

energy needed to break the bonds of the reactingmolecules.

Reaction with low activation energyis easier to takeplace, and is faster.


Activation EnergyActivation Energy


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Effective CollisionsKinetic Energy Factor

For a collision to beeffective, the reacting

molecules must havesufficient kineticenergy to overcomethe energy barrier.



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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.

In order for reaction to occur the reactantmolecules must collide in the correct orientation

and with enough energyto form products.


Orientation FactorOrientation Factor

Consider:


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Consider:

Cl +NOCl p NO + Cl2

There are two possible ways that Cl atoms and NOClmolecules can collide; one is effective and one is not.

Effecti e Collisions


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Effective CollisionsOrientation Effect


C lli i FC lli i F


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The collision frequency is the number of collisionsthat happen per unit time.

The more collisions per second there are, themore collisions can be effective and lead toproduct formation.


Collision FrequencyCollision Frequency


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Effective Collisions

Collisions in which these two conditions are met(and therefore lead to reaction) are calledeffective collisions.

The higher the frequency of effective collisions,the faster the reaction rate.

Any factor that can increase the frequency ofeffective collisions would increase the rate of areaction.



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Under specific conditions, every reaction has its owncharacteristic rate, which may be controlled by fourfactors:

1. Concentrations of reactants2. Chemical nature of reactants3. Temperature4. The use of a catalyst


Factors Affecting Reaction RateFactors Affecting Reaction Rate

F t Aff ti R ti R t


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Factors Affecting Reaction Rate:

Reactant Concentration

Generally, the larger the concentration of reactantmolecules, the faster the reaction increases the frequency of effective collisions

between reactant molecules

concentration of gases depends on the partialpressure of the gas

higher pressure = higher concentration

Concentrations of solutions depend on the solute-to-solution ratio (molarity)



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Factors Affecting Rate Temperature

Increasing the temperature raises the average kineticenergyof the reactant molecules.

Increasing the temperature increases the number ofmolecules with sufficient kinetic energy to overcome theactivation energy, leads to increase in frequency ofeffective collisions. Hence, increasing the rate ofreaction.

The distribution of kinetic energy is shown by the Boltzmanndistribution curves.



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Sufficient kineticenergy to overcomethe activation energy(T1>T2)

Faster rate

Increased number ofmolecules (T1>T2)

Higher kinetic energy requiredto overcome the activationenergy(T2>T1)

Slower rate


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Factors Affecting Rate Catalyst

Catalysts are substances that affect the rate of areaction without being consumed.

Catalysts generally speed up a reaction.

They give the reactant molecules a different pathto follow with a lower activation energy.



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Catalysts

Homogeneous catalysts are in the same phaseas the reactant particlesCl(g) in the destruction of O3(g)

Heterogeneous catalysts are in a differentphase than the reactant particlessolid catalytic converter in a cars exhaust system



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Types of Catalysts


H C t l iH C t l i


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Generally, catalysts operate by lowering the activation

energy for a reaction.

Chlorine atoms are catalysts for the destruction of ozone.

A single CFC molecules can destroy 100,000 ozone

molecules. The catalyst and reacting molecules are in

one phase.

85

Homogeneous CatalysisHomogeneous Catalysis

Hydrogen peroxide decomposes very slowly:


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

But, in the presence of the bromide ion, the decomposition occurs

rapidly. A catalyst may add intermediates to the reaction.

When a catalyst adds an intermediate, the activation energies for

both steps must be lower than the activation energyfor the

uncatalyzed reaction. Example: In the presence of Br-, Br2(aq) is generated as an

intermediate in the decomposition of H2O2.

Heterogeneo s Catal sisHeterogeneo s Catal sis


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The catalyst is in a different phase of the reactants and

products. Typical example: solid catalyst, gaseous

reactants and products (catalytic converters in cars).

Most industrial catalysts are heterogeneous. First step is adsorption (the bindingof reactant molecules

to the catalyst surface).

Adsorbed species (atoms or ions) are very reactive.

Molecules are adsorbed onto active sites on the catalyst

surface.

Heterogeneous CatalysisHeterogeneous Catalysis

87

Consider the hydrogenation of ethylene:


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C2H4(g) + H2(g)p C2H6(g), (H= -136 kJ/mol.

(Ethylene) (Ethane)

In the presence of a metal catalyst (Ni, Pt or Pd) the reaction occurs quickly at roomtemperature.

Ethylene and hydrogen molecules are adsorbed onto active sites on the metal surface.

The H-H bond breaks and the H atoms migrate about the metal surface.

When an H atom collides with an ethylene molecule on the surface, the C-C Tbondbreaks and a C-H Wbond forms.

When C2H6 forms it desorbs from the surface.

When ethylene and hydrogen are adsorbed onto a surface, less energyis required tobreak the bonds and the activation energy for the reaction is lowered.

Enzymes CatalysisEnzymes Catalysis


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Enzymes are biological catalysts.

Most enzymes are protein molecules with large molecular masses(10,000 to 106 amu).

Substrates undergo reaction at the active site of an enzyme.

A substrate locks into an enzyme and a fast reaction occurs.

The products then move away from the enzyme.

Only substrates that fit into the enzyme lock can be involved in the

reaction.

If a molecule binds tightly to an enzyme so that another substrate

cannot displace it, then the active site is blocked and the catalyst isinhibited (enzyme inhibitors).

The number of events (turnover number) catalyzed is large for

enzymes (103 - 107per second).

Enzymes CatalysisEnzymes Catalysis

89


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Enzymes

90

F t Aff ti R ti R t


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Factors Affecting Reaction Rate:Nature of the Reactants

Nature of the reactants means what kind of reactantmolecules and whatphysical conditions they are insmall molecules tend to react faster than large molecules

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

powdered solids are more reactive than blocks

more surface area for contact with other reactants

certain types of chemicals are more reactive than otherse.g. potassium metal is more reactive than sodium

ions react faster than molecules

no bonds need to be broken91Tro: Chemistry: A Molecular Approach, 2/e

Transition State TheoryTransition State Theory


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There is an energy barrierto almost all reactions. There exists an intermediate stage when reactant

molecules change into products.

The activated complex ortransition state is achemical species with partially broken and partiallyformed bonds

an intermediate formed between reactant and

product moleculeshas highest energyand extremely unstable


Transition State TheoryTransition State Theory

Figure 2


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Figure 2

Reaction energy diagrams and possible transition states.


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Energy Profile of a Catalyzed Reaction

polar stratosphericclouds contain icecrystals that catalyzereactions that releaseCl from atmosphericchemicals



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Reaction progress

Potential

EnergySOLUTION:

Drawing Reaction Energy Diagrams and Transition States

PROBLEM 1.5: A key reaction in the upper atmosphere is

O3(g)+ O(g) 2O2(g)The activation energy, Ea(fwd) is 19 kJ, and the enthalpy of reaction, (Hrxnfor the reaction is -392 kJ. Draw a reaction energy diagram for thisreaction, postulate a transition state, and calculate Ea(rev).

PLAN: Exothermic reaction. The reactants are at a higher energy level than the

products and the transition state is slightly higher than the reactants.

O3+O

2O2

Ea= 19 kJ

(Hrxn = -392 kJ

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

411kJ

OO

O

O

breaking

bond

forming

bond

transition state

Example: Rearrangement of Transition State


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Consider the rearrangement of methyl isonitrile to form

acetonitrile:

In H 3C-N|C, the C-N|C bond bends until the C-N bond breaks

and the N|C portion is perpendicular to the H3C portion. This

structure is called the activatedcomplex or transition state.

The energy required for the above twist and break is theactivation energy, E

a.

Once the C-N bond is broken, the N|C portion can

continue to rotate forming a C-C|N bond.

H3C N CC

NH3C H3C C N

96

Example: Rearrangement of Transition State


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The change in energy for the reaction is the difference in energy between

CH3NC and CH3CN. (()

The activation energy is the difference in energy between reactants,CH3NC and transition state. (Ea)

The rate depends on Ea, not ()the lower Ea, the faster the reaction.

* If a forward reaction is exothermic (CH3NC p CH3CN), then the

reverse reaction is endothermic (CH3CN p CH3NC).

Connection between Activation Energy with


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How methyl isonitrile molecule gain enough energy to ov

ercomethe activation energy barrier?

It does so through collision with other molecules.

From kinetic molecular theory, we know that as temperature

increases, the totalkinetic energyincreases.

We can show the fraction of molecules,f, (with energy equal to or

greater than Ea) is

- whereR is the gas constant

(8.314 J/molK).

RT

Ea

ef

!

98

Temperature and Rate

Arrhenius EquationArrhenius Equation


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Changing the temperature changes the rate constant,k, of the rate law

The Arrhenius Equation shows the relationshipbetween absolute temperature, T, and rate constant, k:

R is the gas constant in energy units, 8.314 J/(molK) where T is the temperature in kelvins

A is called the Arrhenius constant (which includes thefrequency factor and the orientation factor).

Ea is the activation energy, the minimum energyneeded for the molecules to react.


Arrhenius EquationArrhenius Equation

The Arrhenius Equation:


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The Arrhenius Equation:The Exponential Factor

The exponential factor in the Arrhenius equation is anumberbetween 0 and 1

The larger the activation energy, Ea, the fewer moleculesthat 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


Arrhenius Plots


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Arrhenius Plots

The Arrhenius Equation: The integrated form

this equation is in the linear form y= mx+ bwhere y= ln(k) and x= (1/T)

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

Slope of Ea/R, or (R)(slope of the line) = Ea (in Joules) R= 8.314 J/(molK)

eyintercept =A (unit is the same as k)



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102

Example 13.7: Determine the activation energy and


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frequency factor for the reaction O3(g) p O2(g) + O(g)given the following data:


Example 13.7: Determine the activation energy and


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frequency factor for the reaction O3(g) p O2(g) + O(g)given the following data:


Example 13.7: Determine the Ea and the Arrheniusf O O O


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constant,A, for the reaction O3(g) p O2(g) + O(g)given the following data:

105

slope, m = 1.12 x 104Kyintercept, b = 26.8


Arrhenius Equation: TwoArrhenius Equation: Two--point Formpoint Form


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If we do not have a lot of data, then we recognize:

Or if you only have two (T,k) data points:

qq pp

!

!

!!

211

2

12

12

2

2

1

1

11ln

lnlnlnln

lnlnandlnln

TTR

E

k

k

A

RT

EA

RT

Ekk

ART

EkA

RT

Ek

a

aa

aa

106

Example 13.8: The reaction NO2(g)+ CO(g) p CO2(g) +NO(g)has a rate constant of 2 57 M1s1 at 701 K and 567 M1s1


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has a rate constant of 2.57 M s at 701 K and 567 M sat 895 K. Find the activation energy in kJ/mol

107

most activation energies are tens to hundredsof kJ/mol so the answer is reasonable

T1 = 701 K, k1 = 2.57 M1

s1

, T2 = 895 K, k2 = 567 M1

s1

Ea, kJ/mol

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:Find:

EaT1, k1, T2, k2


Practice: It is often said that the rate of a reaction doubles for


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every 10 C rise in temperature. Calculate the activation

energy for such a reaction.

108

Hint:make T1 = 300 K, T2 = 310 Kand k2 = 2k1


Practice Find the activation energy in kJ/mol forincreasing the temperature 10 C doubling the rate


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increasing the temperature 10 C doubling the rate

109

most activation energies are tens to hundredsof kJ/mol so the answer is reasonable

T1 = 300 K, T1 = 310 K, k2 = 2 k1

Ea, kJ/mol

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:Find:

EaT1, k1, T2, k2


Reaction MechanismReaction Mechanism


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We generally describe chemical reactions with anequation listing all the reactant molecules and productmolecules.

But the probability of more than 3 molecules collidingat the same instant with the proper orientation andsufficient energy to overcome the energy barrier isnegligible.

Most reactions occur in a series of small reactionsinvolving 1, 2, or at most 3 molecules.

Describing the series of reactions that occurs toproduce the overall observed reaction is called areaction mechanism.

Knowing the rate law of the reaction helps usunderstand the sequence of reactions in themechanism.


Reaction MechanismReaction Mechanism

An Example of a Reaction Mechanism


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An Example of a Reaction Mechanism

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

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

Step 2: HI(g) + ICl(g) p HCl(g) + I2(g) The reactions (1) and (2) in this mechanism are called

the elementary steps, meaning that they cannot bebroken down into simpler steps, and that the

molecules actually interact directly in this mannerwithout any other steps.

* Elementary step: any process that occurs in asingle step.


Multistep Mechanisms


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Some reaction proceed through more than one step:

NO2(g) + NO2(g)pNO3(g) + NO(g)

NO3(g) + CO(g)pNO2(g) + CO2(g)

Notice that if we add the above steps (usingHess Law),we get the overall reaction:

NO2(g) + CO(g)pNO(g) + CO2(g)

112

Intermediates


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

Step (1): H2(g)+ ICl(g)p HCl(g)+ HI(g)Step (2): HI(g)+ ICl(g)p HCl(g) + I2(g)

Notice that the HI is a product in Step 1, but then a

reactant in Step 2. Because HI is made but then consumed, HIdoesnot showup in the overall reaction.

Materials that areproducts in an early mechanism

step, but then a reactant in a later step, are calledintermediates.

Adding equations (1) and (2), removing HI, will givethe overall equation.


Molecularity


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The number of reactant particles in an elementary stepis called its molecularity.

A unimolecularstep involves one particle A bimolecularstep involves two particles

though they may be the same kind of particle

A termolecularstep involves three particlesthough these are exceedingly rare in elementary

steps


Rate Laws for ElementarySteps


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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 determinedexperimentally.

But the rate law of an elementary step can be deduceddirectly from the equation of the step.

The rate law of an elementary step is determined by its molecularity:

Unimolecular processes are first order,

Bimolecular processes are second order, and

Termolecular processes are third order.

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

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

2) HI(g)+ICl(g)p HCl(g) + I2(g) Rate = k2[HI][ICl]115Tro: Chemistry: A Molecular Approach, 2/e

R t L f El t St


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Rate Laws of Elementary Steps


Rate Determining Step


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In most mechanisms, one step occurs slower than theother steps.

The result is that product formation cannot occur anyfaster than the slowest step the step determines therate of the overall reaction

We call the slowest step in the mechanism the ratedetermining stepthe slowest step has the largest activation energy

The rate law of the rate determining step determines therate law of the overall reaction.


Another Reaction Mechanism


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The first step in thismechanism is the rate

determining step.

The first step is slowerthan

the second step because itsactivation energy is larger.

The rate law of the first stepis the same as the rate law ofthe overall reaction.

118

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

1. NO2(g) +NO2(g) p NO3(g) +NO(g) Rate = k1[NO2]2 Slow2. NO3(g) +CO(g) p NO2(g) + CO2(g) Rate = k2[NO3][CO] Fast


Validating a Mechanism


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To validate (not prove) a mechanism, twoconditions must be met:

1. The elementary steps must sum to the overall

reaction.2. The rate law predicted by the mechanism must

be consistent with the experimentally observedrate law.


Mechanisms with a Fast Initial Step


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When a mechanism contains a fast initial step, the ratelimiting step may contain intermediates.

When a previous step is rapid and reaches equilibrium,

the forward and reverse reaction rates are equal sothe concentrations of reactants and products of thestep are related

and theproduct is an intermediate

Substituting into the rate law of the RDS will produce arate law in terms of just reactants.


Example 1: Mechanisms with a Fast Initial Step


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

The experimentally-determined (observed) rate law is:Rate = k[NO]2[Br2]

Consider the following mechanism:

The rate law is (based on Step 2):Rate = k2[NOBr2][NO]

The rate law shouldnot depend on the concentration of

anintermediate (NOBr2 is intermediates (unstable).

NO(g) + Br2(g) NOBr2(g)

k1

k-1

NOBr2(g) + NO(g) 2NOBr(g)k2

Step 1:

Step 2:

(fast)

(slow)

NOBr2 is the intermediate (unstable), so we express the

concentration of NOBr in terms of NO and Br assuming


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concentration of NOBr2 in terms of NO and Br2 assuming

there is an equilibrium in step 1 we have:

]NO][Br[]NOBr[ 21

12

!k

k

By definition of equilibrium:

Therefore, the overall rate lawbecomes:

Note the final rate law is consistent with the

experimentally-observed rate law, Rate = k[NO]2[Br2]

]NOBr[]NO][Br[ 2121 ! kk

][BrNO][NO][]NO][Br[Rate 22

1

122

1

12

!!

k

k

kk

k

k

k

Example 2: Mechanisms with a Fast Initial Step


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1. 2 NO(g) N2O2(g) Fast

2. H2(g) +N2O2(g)p H2O(g) +N2O(g) Slow Rate =k2[H2][N2O2]

3. H2(g) +N2O(g) p H2O(g) +N2(g) Fast

k1

k1

2 H2(g) + 2 NO(g) p 2 H2O(g) + N2(g) Rateobs = k[H2][NO]2


final rate law is consistent

with the observed rate law

Example 13.9: Show that the proposed mechanism for thereaction 2 O3(g) p 3 O2(g) matches the observed rate law:


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3(g) 2(g)

Rateobs = k[O3]2[O2]

1

1. O3(g) O2(g) + O(g) Fast

2. O3(g) +O(g)p 2 O2(g) Slow Rate = k2[O3][O]

k1

k1


final rate law is consistent

with the observed rate law

Practice Mechanism


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Determine the overall reaction, the rate determining

step, the rate law, and identify allintermediates of the following mechanism

1. A + B2 p AB + B Slow

2. A + B p AB Fast



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PENGUMUMAN


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PENGUMUMAN

1) Sila lengkapkan Entrance Survey di

portal i-Learn sebelum 14/12/2011(Rabu)

2) Taklimat pengambilan dan kemasukan ke

Kyushu University, Japanpada

14/12/20111.00-2.00 petang (Rabu) di

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