WK2 C14 Kinetics

8/9/2019 WK2 C14 Kinetics

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Chapter 14Chapter 14--

ChemicalChemicalReactionReaction

KineticsKinetics


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14.1 Reactionrates14.1 Reactionrates

14.2 Factorsthataffectreactionrate14.2 Factorsthataffectreactionrate

14.3 Rate laws14.3 Rate laws

14.4 Theoryofchemical kinetics14.4 Theoryofchemical kinetics

14.5 Reactionmechanisms14.5 Reactionmechanisms


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14.1 Reaction rates14.1 Reaction rates

Chemical kinetics is the study of the

rate of a reaction

The reaction rate is the speed withwhich reactants disappear and

products form

Reaction rates are always reported aspositive values

The unit is mol L1 s1


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Figure 14.1 Reactionrates

The progressofthereactionA 2B, showingtheconcentration

changesofA andB overtime


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14.1 Reaction rates14.1 Reaction ratesConsiderthereaction A B


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Calculate the average rate at which

[A] changes in the first 50 seconds

2.4 10-4 mol L-1 s-1

rate (A)t

-[A]

-([A]t=50 [A]t=0)

50 s 0 s

-(0.0629 mol L-1 0.0750 mol L-1)

50 s

-(-2.4 10-4 mol L-1 s-1)

=

=

=

=

=


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The rate at a particular time is called

the instantaneous rate

This can be determined from theslope of a tangent to the curve

For example, figure 14.1 at t = 50 s

rate = -(slope of tangent line)dt

=d[A]

= 2.3 10-4 mol L-1 s-1


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For the general reaction:

where a to dare the stoichiometric coefficients of the reactants A and B and the

products Cand D respectively, we define the rate of reaction as

Therefore, for our reaction A 2B we would write


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Estimatingthe Initial Rateofa Reaction

The following data have been measured at 508 C for the reaction 2HI(g) C H2(g) + I2(g)

What is the initial rate of decay of HI, and what is the initial rate of reaction

at this temperature

Time (s)

[HI] (mol L

1)

Time

(s) [HI] (mol L1)

0 0.100 200 0.0387

50 0.0716 250 0.0336

100 0.0558 300 0.0296

150 0.0457 350 0.0265

The initial rate of decay of HI is the instantaneous rate of decay of HI at time zero

The slope of the line can be calculated from the coordinates of any

two points (x1, y1) and (x2, y2) using the equation:

The initial rate of reaction can be calculated using the stoichiometry of the reaction.


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14.214.2 Factors that affectFactors that affect

reaction ratereaction rate

Five principal factors influence

reaction rates:

1.Ch

emical nature of the reactants

2. Ability of the reactants to come in

contact with each others

3. Concentrations of the reactants

4. Temperature

5. Availability of rate-accelerating agents

called catalysts


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14.2 Factors that affect14.2 Factors that affect


Chemical nature of the reactants

During reactions, bonds break and new

bonds form

Some reactions are fast by nature and

others are slow


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The area of contact between the phases

determines the reaction rate


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Concentrations of the reactants

The rates of bothhomogeneous and

heterogeneous reactions are affected by

the concentrations of the reactants

For example, red-hot steel wool bursts into

flame when thrust into pure oxygen

Temperature of the system

Almost all chemical reactions occur

faster at higher temperatures


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Presence of catalysts

Catalysts are substances that increase

the rate of chemical reactions withoutbeing used up

For example, enzymes that direct our

body chemistry are all catalysts


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A rate law is an equation in which the

rate is given as a function of reactant

concentrations

for example, rate = k[HI]n

kis the rate constant of the reaction

n is the order of the reactant

Both rate constant and order must be

experimentally determined


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The value of the rate constant

depends on the particular reaction

being studied and the temperature at

which the reaction occurs

The order can be positive or negative,

an integer or a fraction The ordercannotbe deduced from

the balanced equation


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Worked example 14.2

2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(g)

If the butane concentration is decreasingat a rate of 0.20 mol L1 s1, at what rate

is the oxygen concentration decreasing?

Solution:

1.3 mol O2rate (O2) =

0.20 mol C4H10

L s 2 mol C4H10

13 mol O2x

L s=

Oxygen reacts at a rate of 1.3 mol L1 s1


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Solution

From our definition of rate of reaction, we can write:

We are told that:

we can therefore use this, together with our expression for the rate of reaction to

determine

Oxygen reacts at a rate of 1.3 mol L1

s1


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Types of rate laws:

A differential rate law expresses the rate

as a function of concentration An integrated rate law expresses the

concentration as a function of time

Rate laws help to identify the steps bywhich a chemical reaction occurs

The sum of the individual reaction steps

is called the reaction mechanism


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The differential rate law

Consider the hypothetical reaction:

A + B

products The rate law for the reaction is:

rate = k[A]n[B]m

The values of n and m can bediscovered by looking for patterns in the

rate data


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Suppose that the data in the table below

has been obtained in a series of five

experiments:

A + B C


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For experiments 1, 2 and 3 [B] is held

constant. Any changes in the rate must

be due to the change in [A]

The rate law says that at constant [B] therate must be proportional to [A]n

rateexp 1

rateexp 2

= [A]exp 1

[A]exp 2n

rateexp 1

rateexp 2

=

0.40 mol L-1 s-1

0.20 mol L-1 s-1 = 2

and

[A]exp 1

[A]exp 2=

0.20 mol L-1

0.10 mol L-1= 2


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In the final three experiments, the

concentration of B changes while the

concentration of A is held constant

This time it is [B] that affects the rate

For experiments 3 and 4, we have:

4.0 = 2.0m

Therefore, m must equal 2

rateexp 3

rateexp 4=

[B]exp 3

[B]exp 4m


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We now know that the rate law for the

reaction must be:

rate = k[A]1

[B]2

The overall order is the sum of the orders

for each reactant in the rate law

Thus, ourhypothetical reaction has an

overall order ofthree


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To calculate the value ofkfor the A + B products reaction, we need only

substitute rate and concentration data into the rate law.

rate = k[A]1[B]2

from experiment 1 in table 14-3:

After cancelling units, the value ofkwith the net units is:

Note that the unit of a third-order rate constant is concentration2 time1.


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First-order rate laws Assume A products

is first order in A

Th

e differential rate law is then:

The integrated rate law is:

rate = d[A]

dt= k[A]

[A]t

[A]0 = ktln [A]0e-kt[A]t =or

ln[A

]t=

kt+ ln[

A]0


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First-orderrate laws

Let's assume that ourhypothetical reaction A products is first order in A.

The differential rate law is then:

Rearrangement gives

the concentration from the beginning of the reaction (at t= 0) until a certain

reaction time t


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The equation ln[A]t= kt+ ln[A]0 is of the form y= m x+ b,

where a plot ofyversus xgives a straight line with the slope m and

the intercept b (the intercept may also be denoted by a c).

For the reaction A products,

the plot of ln[A]tversus tis a straight line, if the reaction is first order.Conversely, if the plot doesn't give a straight line,

then the reaction is not first order.


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Forfirst order reactions a plot of the

natural logarithm of concentration versus

reaction time always gives a straight line


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SolutionFirst we calculate the natural logarithm of [N2O5], which gives the

following dataTime (s) ln[N

2

O5

] Time (s) ln[N2

O5

]

0 1.609 200 2.996

50 1.956 300 3.689

100 2.303 400 4.382

The plot is a straight line, which confirms that the reaction is first order in [N2

O5

]

(we could also use a calculator to do this analysis). In this case, the slope of the

line equals k. This gives us:


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Half-lifeoffirst-orderreactions

The half-life ( ) ofareactantisaconvenientwaytodescribehowfast

itreacts, particularlyforanoverall first-orderprocess. A reactant'shalf-life,

, is the time required for half of the reactant to disappear

For a first-order reaction, the half-life of the reactant can be obtained from:

by setting [A]tequal to half of the initial concentration [A]0.

=


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Second-order rate laws

Assume B products is second order

The differential rate law is:

The integrated rate law would be:

rate = d[B]

dt= k[B]2

[B]0

1= kt +

[B]t

1


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When a reaction is second order, a plot

of 1/[B]t versus t should yield a straight

line with a slope k


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Zero-order rate laws

The differential rate law is:

rate = k[C]0 = k(1) = k

The integrated rate law is:

[C]t = kt+ [C]0

A plot of [C] versus time gives a straight

line of slope k


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14.4 Theory of chemical14.4 Theory of chemical

kineticskinetics

Collision theory

One of the simplest models to explain

reaction rates is collision theory

Collision theory says that the rate of a

reaction is proportional to the number of

effective collisions per second among

the reactant molecules

An effective collision is one that actually

gives product molecules


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kineticskinetics

Concentration can influence the number

of effective collisions per second

As reactant concentrations increase, the

number of effective collisions increases

Not every collision between reactant

molecules actually results in a chemical

change

Only a very small fraction of all the

collisions can really lead to a net change

Why is this so?


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kineticskinetics

Molecular orientation

When two reactant molecules collide they

must be oriented correctly for a reaction to

occur (see next slide)


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kineticskinetics

Activation energy

The activation energy (Ea) is the

minimum energy required for a reactionto occur

The activation energy determines the

rate of a reaction

At a given temperature, only a certain

fraction of the collisions possess enough

energy to be effective and form products


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kineticskinetics


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kineticskinetics

The arrangement found on the top of the

potential energy hill is called the

activated complex or transition state

Once the transition state is reached, the

reaction proceeds to give products, with

the release of energy

A reaction intermediate corresponds toan energy minimum between two

transition states


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kineticskinetics

Temperature effects

The rate of a reaction increases with

increasing temperature


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kineticskinetics

The activation energy is related to the

rate constant by the Arrhenius equation

k= Ae-Ea/RT

k = rate constant

Ea = activation energy

R = universal gas constant

T = temperature (in Kelvin)

A = pre-exponential or frequency factor


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kineticskinetics

The activation energy can also be

obtained from two rate components

measured at different temperatures

Most reactions obey the Arrheniusequation to a good approximation

lnk

2k1

=R

Ea

1

T2

1

T1


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14.5 Reaction mechanisms14.5 Reaction mechanisms

Most reactions do not occur in a

single step

The net overall reaction is the result ofa series of simple reactions

Each of these is called an elementary

process The entire set of elementary

processes is the reaction mechanism


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The rate law of an elementary process

can be written from its chemical

equation

This rule only applies to elementary

processes

The overall rate law derived from t

hemechanism must agree with the

observed rate law for the overall

reaction


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The rate-determining step

Consider the gaseous reaction:

2NO2Cl 2NO2 + Cl2

The actual mechanism of the reaction is

the following two-step sequence of

elementary processes:

NO2Cl NO2 + Cl (slow)

NO2Cl + Cl NO2 + Cl2 (fast)

The Cl atom formed is an intermediate


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In any multistep mechanism, one step

is usually much slower than the others

The slow step in a mechanism iscalled the rate-determining step

The rate law for the rate-determining

step is directly related to the rate lawfor the overall reaction


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The steady-state approximation

In complex reaction mechanisms, the

rate-determining step may vary whenreaction conditions are changed

In such cases, the steady-state

approximation is used

This method is the assumption that the

concentration of an intermediate remains

constant as the reaction proceeds


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Consider the hypothetical reaction:

2AB + C2 A2B + C2B

This reaction may proceed via the

following mechanism:2AB A2B2

A2B2 + C2 A2B + C2B

Th

e intermediate in this mec

hanism isA2B2. Thus:

d[A2B2]

dt= 0


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The steady-state approximation is

applied by setting:

rate of A2B2 production = rate of A2B2 consumption

All steps that are producing and

consuming intermediate A2B2 need to be

identified The rate law for each also has to be

written


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(a) Rate of production of A2B2

In this mechanism, A2B2 is only produced

in the forward reaction

The rate law for this step is:

d[A2B2]

dt =k

1[AB]2


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(b) Rate of consumption of A2B2

A2B2 is consumed in the reverse reaction

of the first step and in the second step

The rate laws for these steps are:

d[A2B2]

dt

= k-1[A2B2]

d[A2B2]

dt= k2[A2B2][C2]


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(c) Application of steady-state condition

Under steady-state conditions, we have

for the intermediate A2B2:

k1[AB]2 = k-1[A2B2] + k2[A2B2][C2]

If we solve for the concentration of A2B2:

Now the rate law for the overall reaction

is written

[A2B2] =

k1[AB]2

k-1 + k2[C2]


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14.6 Catalysts14.6 Catalysts

A catalyst is a substance that

changes the rate of a chemical

reaction without being used up

The action caused by a catalyst is

called catalysis

Positive catalysts speed up reactions

Negative catalysts (inhibitors) slow

reactions down


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The catalyst provides a path to the

products that has a rate-determining

step with a lower activation energy

(see next slide)


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Homogeneous catalysts

Exist in the same phase as the reactants

An example is found in the lead chamber

process for manufacturing sulfuric acid.

The reaction is:

S + O2 SO2

SO2 + O2 SO3

SO3 + H2O H2SO4

Unassisted, the second reaction, oxidation of

SO2 to SO3, occurs slowly.


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The SO2 is combined with a mixture ofNO,

NO2, air and steam

The NO2 serves as a catalyst by being an

oxygen carrier and providing a low-energy

path for the oxidation of SO2 to SO3

The NO is then reoxidised to NO2 by oxygen

The reaction is:

NO2 + SO2 NO +SO3

NO + O2 NO2

The NO2 is regenerated


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Heterogeneous catalysts

Are a separate phase from the reactants

They are commonly solids

Usually functions by promoting a

reaction on its surface

An example is the Haber-Bosch process

3H2 +N

2

2N

H3 The reaction takes place on the surface of an

iron catalyst that contains traces of

aluminium and potassium oxides.


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Enzyme kinetics

Enzymes consist of proteins, and are

some of the most powerful homogenous

catalysts

Enzyme catalysis can be represented by

a series of reactions

There are two different hypotheses forhow a substrate is bound into an enzyme

The lock-and-keyhypothesis assumes that

the substrate simple fits into the active site


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The induced fithypothesis assumes that theenzyme molecule changes shape as the

substrate molecule comes close

The action of an enzyme can be described

by th

e Michaelis-Menten mec

hanism


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SummarySummary

Chemical kinetics is the study of the

rate of a reaction

Reaction rates are controlled by fivefactors

The differential rate law shows how

reaction rate depend on concentration

The integrated rate law shows how

concentration depends on time


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SummarySummary

According to collision theory, the rate

of a reaction depends on the number

of effective collisions per second

Colliding molecules must possess a

minimum kinetic energy, called the

activation energy, Ea

The Arrhenius equation relates how

changes in activation energy and

temperature affect a rate constant


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SummarySummary

In more complicated reaction

mechanisms the steady-state

approximation can be used to derive

the rate law

Catalysts are substances that change

a reaction rate but are not consumed

by the reaction Catalysts in living systems are called

enzymes

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WK2 C14 Kinetics