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Chapter 14Chemical Kinetics

Copyright McGraw-Hill 2009

14.1 Reaction Rates

• Kinetics: the study of how fast reactions take place

• Some reactions are fast (photosynthesis)

• Some reactions are slow (conversion of diamond to graphite)


Importance of Studying Reaction Rates

• Speed up desirable reactions• Minimize damage by undesirable

reactions• Useful in drug design, pollution control,

and food processing


Rate of Reaction

• Expressed as either: – Rate of disappearance of reactants

(decrease or negative)OR

– Rate of appearance of products (increase or positive)


Average Reaction Rate


Average Reaction Rate

• Equation A B

• rate =

• Why the negative on [A]?

[A] [B]ort t


Average Reaction Rate Br2(aq) + HCOOH(aq)2Br(aq) + 2H+(aq) + CO2(g)

Note: Br2 disappears over time


Average Reaction Rate Br2(aq) + HCOOH(aq) 2Br(aq) + 2H+(aq) + CO2(g)


Calculate Average Rate

• Avg. rate =

• Avg. rate =

2 2[Br ] [Br ]

final initialt tfinal initial

= 3.80 0.0101 0.0120 50 0 M M

s sM / s


Average Rate• Average rate depends on time interval• Plot of [Br2] vs time = curve

• Plot of Rate vs [Br2] = straight line


Instantaneous Rate

• Instantaneous: rate at a specific instance in time (slope of a tangent to the curve)


Rate Constant

• Using data from Table 14.1 - what can you conclude?

25050 Time (s)

1.75 x 1050.005963.52 x 1050.0101Rate (M/s) [Br2]


Rate Constant Answer: • When the [Br2] is halved; the rate is

halved • Rate is directly proportional to [Br2]

• rate = k [Br2] • k = proportionality constant and is

constant as long as temp remains constant


Rate Constant • Calculate the value of the rate constant for

any set of data and get basically the same answer!

• k = rate / [Br2]

53.52 x 10 /s 33.5x 10 /s0.0101

MkM


Stoichiometry and Reaction Rate

• When stoichiometric ratios are not 1:1 rate of reaction is expressed as follows

General equation: aA + bB cC + dD

rate = 1 [A] 1 [B] 1 [C] 1 [D]a t b t c t d t



• Write the rate expression for the following reaction

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



4PH3(g) P4(g) + 6H2(g)

If molecular hydrogen is formed at a rate of 0.168 M/s, at what rate is P4 being produced?

3 4 21 [PH ] 1 [P ] 1 [H ]4 1 6t t t



4PH3(g) P4(g) + 6H2(g)

21 [H ] 1 (0.168 /s) 0.028 /s6 6

M Mt

41 [P ] 0.028 /s1

Mt


14.2 Dependence of Reaction Rate on Reactant

Concentration• Rate law expression For the general equation: aA + bB cC + dD rate law = k[A]x[B]y

k = proportionality constant x and y = the order of the reaction with respect to each reactant


Order • Exponents represent order• Only determined via experimental data

– 1st order - rate directly proportional to concentration

– 2nd order - exponential relationship– 0 order - no relationship

• Sum of exponents (order) indicates overall reaction order


Experimental Determination of Rate Law

• Method of initial rates - examine instantaneous rate data at beginning of reaction


rate = k[F2]x [ClO2]y

Find order (exponents) by comparing data Exp. 1 and 3: [ClO2] is held constant

1st order with respect to [F2] (rate and M directly related)

2

2

[F ]0.203 2

[F ] 0.101

MM

[rate] 32.4 10 /s3 2

3[rate] 1.2 10 /s1

M

M


rate = k[F2]1[ClO2]y

Find order (exponents) by comparing data Exp. 1 and 2: [F2] is held constant

2

2

[ClO ]0.0402 4

[ClO ] 0.0101

MM

1st order with respect to [ClO2] (rate and M directly related)

[rate] 34.8 10 /s2 43[rate] 1.2 10 /s

1

M

M


rate = k[F2]1[ClO2]1 overall order = 2

Find k (use any set of data)

4[rate] 2.1 10 /s 2 19.3 s2 2[A] [B] (0.10 ) (0.015 )

Mk MM M


Determining Rate Law

8.4 x 1020.0300.103

4.2 x 1040.0150.202

2.1 x 1040.0150.101

Initial Rate (M/s)

[B] (M)[A] (M)Exp.


What is Different?

• In experiment 1 and 2; [B] is constant; [A] doubles and rate doubles - the reaction is 1st order with respect to [A]

• In experiment 1 and 3; [A] is constant; [B] doubles but the rate quadruples! This means that the reaction is 2nd order with respect to [B]


Calculate the Rate Constant• Rate = k[A] [B]2 • The rxn is 1st order w/ respect to [A]• The rxn is 2nd order w/ respect to [B]• The rxn is 3rd order overall (1 + 2)

4[rate] 2.1 10 /s 2 19.3 s2 2[A] [B] (0.10 M) (0.015 )

Mk MM


14.3 Dependence of Reactant Concentration on Time

• First-Order reactions may be expressed in several ways

• Example: A products rate = k[A]

[ ]ratet

A


Integrated Rate Law (First order)

y = mx + b

When the two expressions are set equal to each other,we get an expression that can be rearranged in the form of a straight line.

0ln lnA kt A


Graphical Methods • Given concentration and time data,

graphing can determine order


Integrated Rate Law (First order)

• For a 1st order reaction, a plot of ln [A] vs time yields a straight line

• The slope = k (the rate constant)


Try Graphing

132300

150250

170200

193150

220100

2840P (mmHg)Time (s)


Graphing

• Plot ln [Pressure] on y-axis and time on x-axis

• If the plot is a straight line, then the integrated rate law equation can be used to find the rate constant, k, or the slope of the line can be calculated for the rate constant.


Integrated Rate Law

• The rate constant for the reaction 2A B is 7.5 x 103 s1 at 110C. The reaction is 1st order in A. How long (in seconds) will it take for [A] to decrease from 1.25 M to 0.71 M?


• Another form of the integrated rate law

[A]ln

[A]0

t kt

(0.71 M) 3 1ln 7.5 10 s ( )(1.25 M)

75 s

t

t


Your Turn!

• Consider the same first order reaction 2A B, for which k = 7.5 x 103 s1 at 110C. With a starting concentration of [A] = 2.25 M, what will [A] be after 2.0 minutes?


Half-Life (1st order)

• Half-life: the time that it takes for the reactant concentration to drop to half of its original value.


Calculating First Order Half-life

• Half-life is the time that it takes for the reactant concentration to drop to half of its original value.

• The expression for half-life is simplified as

t1/ 2 .693

k


Half-Life

• The decomposition of ethane (C2H6) to methyl radicals (CH3) is a first order reaction with a rate constant of 5.36 x 104 s1 at 700 C. C2H6 2CH3

Calculate the half-life in minutes.


Half-Life

1/2 4 1

0.693t 1293 s5.36 10 s

1 min1293 s 21.5 min60 s


Dependence of Reactant Concentration on Time

Second-order reactions may be expressed in several ways

• Example: A product rate = k[A]2

[A]ratet


Integrated Rate Law for Second Order Reactions

Again, the relationships can be combined to yield the following relationship in the form of astraight line

0

1 1ktA A


Integrated Rate Law for Second Order Reactions

• For a 2nd order reaction, a plot of 1/[A] vs time yields a straight line

• The slope = k (the rate constant)


Calculating Second-Order Half-life

The expression for half-life is simplified as

Note: half-life for 2nd order is inversely proportional to the initial reaction concentration

1/20

1t[A]k


Calculating Second Order Half-life

I(g) + I(g) I2(g) The reaction is second order and has a rate

constant of 7.0 x 109 M1 s1 at 23C. a) If the initial [I] is 0.086 M, calculate the concentration after 2.0 min. b) Calculate the half-life of the reaction when the initial [I] is 0.60 M and when the [I] is 0.42 M.


a) Use integrated rate equation for 2nd order

9 1 1

11 1

1211 1

1 1(7.0 10 s ) 120sA 0.086

8.4 101[A] 1.2 10

8.4 10

MM

M

MM


b) Use the half-life formula for 2nd order (note: half-life does not remain constant for a

2nd-order reaction!)

101/2 9 1 1

101/2 9 1 1

1 2.4 10 s(7.0 10 s ) (0.60 )

1 3.4 10 s(7.0 10 s ) (0.42 )

tM M

tM M


Zero Order

Zero Order reactions may exist but are relatively rare

• Example: A product rate = k[A]0 = k• Thus, a plot of [A] vs time yields a

straight line.


Summary of Orders


14.4 Dependence of Reaction Rate on Temperature

• Most reactions occur faster at a higher temperature.

• How does temperature alter rate?


Collision Theory• Particles must collide in order to react • The greater frequency of collisions, the

higher the reaction rate• Only two particles may react at one time• Many factors must be met:

– Orientation– Energy needed to break bonds (activation

energy)


Collision Theory


Collision Theory

• Though it seems simple, not all collisions are effective collisions

• Effective collisions: a collision that does result in a reaction

• An activated complex (transition state) forms in an effective collision


Activation Energy


The Arrhenius Equation• The dependence of the rate constant of a

reaction on temperature can be expressed

Ea = activation energy R = universal gas constant A = frequency factor T = Kelvin temp

a /E RTk Ae


Arrhenius Equation

In the form of a straight line…what plot will give a straight line?

a 1ln lnEk AR t


Arrhenius Equation

• A plot of ln k vs 1/T will give a straight line

• The slope of line will equal Ea/R

• The activation energy may be found by multiplying the slope by “R”


Graphing with Arrhenius

• Rate constants for the reaction CO(g) + NO2(g) CO2(g) + NO(g)

Were measured at four different temperatures. Plot the data to determine activation energy in kJ/mol.


Graphing

3180.332

3080.184

2980.101

2880.0521

T (Kelvin) k (M1 s1)


Graphing

Steps: • Make a column of ln k data• Make a column of inverse temp (1/T) • Plot ln k vs 1/T • Calculate the slope • Multiply slope by R


Arrhenius Another Way!

• Another useful arrangement of the Arrhenius equation enables calculation of: 1) Ea (with k at two different temps) 2) the rate constant at a different temperature (with Ea, k and temps)

1 a

2 12

1 1ln k ET Tk R


Arrhenius Another Way!

• Use the data to calculate activation energy of the reaction mathematically

7.0 x 101500

6.1 x 102450

2.9 x 103400

k (s1)T (Kelvin)


Arrhenius 1

2a

2 1

a

ln

1 1

32.9 10ln 26.1 108.3145J/Kmol 1 1450 400

91 kJ/mol

kkE R

T T

E


14.5 Reaction Mechanisms• Most reactions occur in a series of steps • The balanced equation does not tell us

how the reaction occurs! • There are often a series of steps which

add together to give the overall reaction • The series of steps is the reaction

mechanism


Reaction Mechanisms

• Most chemical reactions occur in a series of steps

• Energy of activation must be overcome to form intermediates


Reaction Mechanism

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

The reaction cannot occur in a single step. One proposed mechanism:

Step 1: NO + NO N2O2

Step 2: N2O2 + O2 2NO2


Reaction Mechanism

N2O2 is an intermediate in the reaction mechanism

Intermediate: a substance that is produced in an early step and consumed in a later step

Elementary reaction: one that occurs in a single collision of the reactant molecules


Reaction Mechanism

• Molecularity: the number of reactant molecules involved in the collision

• Unimolecular: one reactant molecule • Bimolecular: two reactant molecules • Termolecular: three reactant molecules

(fairly rare)


Rate Determining Step

• If the elementary reactions are known, the order can be written from the stoichiometric coefficients of the rate-determining step

• Rate-determining step: the slowest step in the mechanism


Rate-Determining Step

• Steps of a mechanism must satisfy two requirements– Sum of elementary steps must equal the

overall balanced equation – The rate law must have same rate law as

determined from experimental data


• The decomposition of hydrogen peroxide (2H2O2 2H2O + O2 )

may occur in the following two steps Step 1: H2O2 + I H2O + IO

Step 2: H2O2 + IO H2O + O2 + I

If step 1 is the rate-determining step, then the rate law is

rate = k1[H2O2] [I]

k1¾ ¾k2¾ ¾


• I does not appear in the overall balanced equation

• I serves as a catalyst in the reaction - it is present at the start of the reaction and is present at the end

• IO is an intermediate


Potential Energy Diagram


Reaction Mechanism • Given overall equation:

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

Step 1: I2 2I (fast) Step 2: H2 + 2I 2HI (slow)

rate = k2[H2] [I]2

This rate expression does not meet the requirement..I is an intermediate and should not appear in the rate expression

k2¾ ¾

k1

k

¾ ¾¬ ¾¾


• Consider the first equilibrium step: the forward rate is equal to the reverse rate

k1[I2] = k-1 [I]2 k1/k-1 [I2] = [I]2

If we substitute for [I]2 , the rate law becomes

rate = k[H2] [I2] This now matches the overall balanced

equation!

(When step 2 is rate-determining this substitution is always possible)


14.6 Catalysis• Catalyst - a substance that increases

the rate of a chemical reaction without being used up itself

• Provides a set of elementary steps with more favorable kinetics than those that exist in its absence

• Many times a catalyst lowers the activation energy


Reaction Pathway with Catalyst


Types of Catalysts

• Heterogeneous catalysts - reactants and catalyst are in different phases

• Homogeneous catalysts - reactants and catalysts are dispersed in single phase

• Enzyme catalysts - biological catalysts


Heterogeneous Catalysts

• Most important in industrial chemistry• Used in catalytic converters in

automobiles – Efficient catalytic converter serves two

purposes; oxidizes CO and unburned hydrocarbons into CO2 and H2O; converts NO and NO2 into N2 and O2


Catalytic Converter


Homogeneous Catalysts• Usually dispersed in liquid phase • Acid and base catalyses are the most

important types of homogeneous catalysis in liquid solution

• Advantages of homogeneous catalysts– Reactions performed at room conditions – Less expensive – Can be designed to function selectively


Biological Catalysts

• Enzymes: large protein molecule that contains one or more active sites where interactions with substrates occur

• Enzymes are highly specific (lock and key)


Reaction Pathway without and with Enzyme-Substrate


Enzyme-Substrate Complex


Biological Molecules (binding of glucose to

hexokinase)


Key Points

• Rate of reaction can be determined in several ways – Instantaneous rate – Average rate – Graphing using integrated rate laws– Mechanisms


Key Points

• Write rate law expressions • Calculate rate constant with proper units • Distinguish orders: 1st, 2nd, 0 order • Calculate half-life • Collision theory and relationship to

Arrhenius equation


Key Points

• Calculate activation energy graphically and mathematically

• Reaction mechanisms – Elementary reactions – Molecularity – Rate law from slow step – Intermediates and catalysts


Key Points

• Catalysis – Heterogeneous– Homogeneous– Enzymes

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