chemical kinetics ppt

7/31/2019 Chemical Kinetics Ppt

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Chemical Kinetics


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We can use thermodynamics to tell if a reaction isproduct- or reactant-favored.

But this gives us no info on HOW FAST reaction goes from

reactants to products. KINETICS the study ofREACTION RATES and their

relation to the way the reaction proceeds, i.e., itsMECHANISM.

Chemical Kinetics

The reactionmechanismis our goal!


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What we have to study ?

Reaction Rates How we measure rates.

Rate LawsHow the rate depends on amounts

of reactants.

Integrated Rate LawsHow to calc amount left or time toreach a given amount.

Half-lifeHow long it takes to react 50% ofreactants.

Arrhenius Equation How rate constant changes with T.

MechanismsLink between rate and molecularscale processes.


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Rate of reaction is typically measured as the change in

concentration with time

This change may be a decrease or an increase

Likewise the concentration change may be of reactants or

products

Rate = ______________ = ______________

change in timechange in time

in [products] in [reactants]

Rate = concentration changetime change

Reaction Rates


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A B

rate = -D[A]

Dt

rate =D[B]

Dt

time


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Note the use of the negative sign

- rate is defined as a positive quantity

- rate of disappearance of a reactant is negative

t

[N2O5]Rate ofreaction=

12 t

[NO2]14 t

[O2]= =-

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


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1. Average reaction rate: a measure of the change in

concentration with time

2. Instantaneous rate: rate of change of concentration at any

particular instant during the reaction

3. Initial rate: instantaneous rate at t = 0- that is, when the reactants are first mixed

Rate may be expressed in

three main ways


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Average reaction rate

In this reaction, theconcentration of butylchloride, C4H9Cl, wasmeasured at various

times, t.

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)Consider the following reaction


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The average rate of thereaction over eachinterval is the change inconcentration divided bythe change in time:

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

Average Rate, M/s


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Note that the average ratedecreases as the reactionproceeds.

This is because as thereaction goes forward,there are fewer collisionsbetween reactant

molecules.



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Instantaneous rate

A plot of concentrationvs. time for this reaction

yields a curve like this. The slope of a line

tangent to the curve atany point is the

instantaneous rate atthat time.


Consider the following reaction


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The reaction slows down

with time because theconcentration of thereactants decreases.



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In this reaction, the ratioof C4H9Cl to C4H9OH is1:1.

Thus, the rate ofdisappearance of C4H9Clis the same as the rate ofappearance of C4H9OH.

Rate =-[C4H9Cl]

t=[C4H9OH]

t



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Write the rate expression for the following reaction:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

rate = -

[CH4]t

= - [O2]t

12

= [H2O]t

12

=

[CO2]t

Example Problem

You should

careful aboutpositive or

negative sign !


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I. Rate is not Constant

Throughout Reaction

II. Reaction Rate isHigher with Higher

Concentration

(-) slope means concentrationis decreasing.

Curved Rather

than Straight LineTell Us That.....


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1. Rate = - A/ t =; + B/ t

Determined from stoichiometry

Uses both reactants & products

2. Rate Law; rate =k[A]x

Determined by experimentaldata- Stoichiometry of equationis irrelevant

Only reactants in rate law

Two MathematicalExpressions toDescribe Reaction

Rate

A B

Consider the reaction


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The Rate Law

The rate law expresses the relationship of the rate of a reactionto the rate constant and the concentrations of the reactantsraised to some powers.

aA + bB cC + dD

Rate = k[A]x[B]y

Where , k is the Rate Law Constant


x and y are determinedexperimentally, and do notdepend on stoichiometriccoefficients from balancedequation


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reaction isxth order in A

reaction isyth order in B

reaction is (x +y)th order overall

Reaction Order

Reaction order tells how quickly rate will

increase when concentration increases

The Rate Law

The rate law expresses the relationship of the rate of a reactionto the rate constant and the concentrations of the reactantsraised to some powers.

aA + bB cC + dD

Rate = k[A]x[B]y



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For a general reaction with rate law

we say the reaction is mth order in reactant 1 and nth order inreactant 2.

The overall order of reaction is m + n + .

A reaction can be zeroth order ifm, n, are zero.

Note the values of the exponents (orders) have to bedetermined experimentally. They are not simply related tostoichiometry.

nm2][reactant1]k[reactantRate

Exponents in the Rate Law


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F2(g) + 2ClO2(g) 2FClO2(g)

rate = k[F2][ClO2]1

For the above

reaction , rate isgiven as



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F2(g) + 2ClO2(g) 2FClO2(g)

rate = k[F2]x[ClO2]

y

Double [F2] with [ClO2] constant

Rate doubles x= 1

Quadruple [ClO2

] with [F2

] constant

Rate quadruples

y= 1

rate = k[F2][ClO2]

Experimental Calculation


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Determine the rate law and calculate the rate constantfor the following reaction from the following data:S2O8

2- (aq) + 3I- (aq) 2SO42- (aq) + I3

- (aq)

Experiment [S2O82-] [I-]

Initial Rate

(M/s)1 0.08 0.034 2.2 x 10-4

2 0.08 0.017 1.1 x 10-4

3 0.16 0.017 2.2 x 10-4

Example Problem


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rate = k[S2O82-]x[I-]y

y = 1

x = 1rate = k[S2O8

2-][I-]

Double [I-], rate doubles (experiment 1 & 2)

Double [S2O82-], rate doubles (experiment 2 & 3)

k=rate

[S2O82-][I-]

=2.2 x 10-4M/s

(0.08M)(0.034M)= 0.08/Ms

Thinkabout thisproblem as

before

Solution


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Molecularity of a reaction

Order of a reaction

Half life

Two Important

terms inkinetics


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Molecularity

Molecularityis the number of molecules coming together toreact in an elementary step.

Elementary reactions are simple reactions (described bymolecularity)

(a) A Products UNI-molecular reaction

e.g. H2C

H2C CH2

CH3 CH

CH2

(b) A + A Products or A + B Products BI-molecular

e.g. CH3I + CH3CH2O- CH3OCH2CH3 + I

-

(c) 2A + B P or A + B + C P Ter-molecular


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Order of a reaction

The order of reaction is determined with respect to eachreactant in the reaction.The order of a reaction is the power to which the

concentration of a reactant is raised.This is the order with respect to one reactant only.

For the reaction

where there are more than one reactants, the order of thereaction is determined with A and then with B.

aA + bB cC + dD


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As mentioned earlier in equation the rate law which is

expressed as rate (r) = k [A]m

[B]n

The exponent 'm' denotes the order of the reaction withrespect to reactant A and the exponent 'n' denotes the orderof the reaction with respect to B.

The overall order of the reaction is then (m + n).

It should be noted that 'm' and 'n' do not necessarily represent

thestoichiometric coefficients 'a' and 'b' of the reaction.

Order of a reaction is experimentally determined.


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Differences between Molecularity and

Order of a Reaction


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Half-Life

Half-life is defined asthe time required forone-half of a reactant

to react.

Because [A] at t1/2 isone-half of the original

[A], [A]t = 0.5 [A]0.


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Types of order of a reaction

0th Order1st Order2nd Order

Pseudo First Order Reaction


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(Order = 0) have a constant rate.

This rate is independent of the conc of the reactants.

The rate law is: k, with k having the units of M/sec.

Zero-Order Reactions

0[A]kt[A]

IntegratedRate Law


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Zero OrderRate Law k

RateConstant

Slope = - kIntegratedRate Law

[A] = -kt + [A]0

Graph [A] versus t Life t =[A]0/2k

Concentration versus time profilefor a zero order reaction

Rate of reaction versus time for azero order reaction


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Ammonia (NH3) gas decomposes over platinum catalyst tonitrogen gas (N2) and hydrogen gas (H2).

The chemical reaction is as follows:

The reaction follows zero order kinetics. Therefore the ratelaw is rate r = k[NH3]

o

For a zero order reaction the concentration versus timeprofile is linear and the rate of reaction versus time hasthe profile.

Example Problem


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k

][][

Ak

dt

AdR

Rearranging gives:

kdtA

Ad

][

][

At time t = 0, [A] = [A]0And when t = t, [A] = [A]t

First-order reactions


Then the rate of disappearance of A is:

A B


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Integrating:

tA

A

dtkA

Adt

0

][

][ 0 ][

][

ktAtA

A ]][ln[

][

][ 0

lnxdxx1

thatRecall

ktAAt

)]ln[](ln[ 0

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

Integrated form of the

1st order rate expression


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ln[A]t

t / s

-slope = -k

Intercept = ln[A]0kt

A

At

0][

][ln

-slope = -k

t / s

ln([A]t/[A]0)

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

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

Antilog gives: [A]t

t / s

Intercept = [A]0


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The first order integrated rate law can be used to determine

the concentration of [A] at any time. It can be determined graphically

Where y = ln[A]

x = time m = -k b = ln[A] 0

Integrated Rate Law: First Order


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First-Order Processes Example

Consider the process in whichmethyl isonitrile is convertedto acetonitrile.

CH3NC CH3CN


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This data was collectedfor this reaction at198.9C.

CH3NC CH3CN



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When ln Pis plotted as a function of time, a

straight line results. Therefore,

The process is first-order.

kis the negative slope: 5.1 10-5 s1.



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Half-Life for first order reaction

For a first-order process, this becomes

0.5 [A]0

[A]0ln = kt1/2

ln 0.5 = kt1/2

0.693 = kt1/2

= t1/20.693

k

NOTE: For a first-order process, the half-life does not dependon [A]0.


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Second-order reactionsTwo possible cases:

Case I : A + A Products

OR 2A Products

Case II : A + B Products

2][

][

2

1Ak

dt

Adr

Rearranging gives:kdt

A

Ad2

][

][2

At time t = 0, [A] = [A]0

And when t = t, [A] = [A]t


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Integrating:

tA

Adtk

A

Adt

0

][

][ 2 2][

][

0

xx

xdxxdx

x

1

12

1 112

2

2

kt

A

tA

A

2

][

1][

][ 0

ORkt

A

tA

A

2

][

1][

][ 0

ktAA

t

2][

1

][

1

0

Integrated form of the

2nd

order rate expressiony = c + mx


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(1/[A]t) / dm3 mol-1

t / s

slope = 2k

Intercept = 1/[A]0

ktAA

t

2][

1

][

1

0

y = c + mx


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Half-Life for second order reaction

For a second-order process,

1

0.5 [A]0= kt1/2 +

1

[A]0

2

[A]0= kt1/2 +

1

[A]0

2

1[A]0

= kt1/21

[A]0=

= t1/21

k[A]0


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15 Chemical Kinetics

Variation of Reaction rates and Order

First order, rate= k [A]

k = rate, 0th

order

[A]

rate

2nd order, rate= k [A]2

The variation ofreaction rates as functions of concentration for

various order is interesting.

Mathematical analysis is an important scientific tool, worth noticing.

[A] = ___?


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For a reaction A + B C + D

The rate law for the reaction is expressed as rate = k [A]m [B]n

If the concentration of B is kept large, then [B] essentiallydoes not change over time and if the rate of reaction isdetermined as rate = k [A] then this reaction is a pseudo firstorder reaction.

Acidic hydrolysis of ethyl acetate is an example of a first orderreaction, where the acid concentration is kept in excess in thesolvent, water.

Pseudo First Order Reaction


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Methods to Determine

Order of a Reaction

For determination of the order of a reaction, following

methods are usually employed.

a) Graphical Methodb) Initial Rate Methodc) Isolation method


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Graphical Method of

Order DeterminationSteps:

First, the data of concentrations versus time is obtained by asuitable method.

Then, the data is plotted as concentration versus time.

From the resulting plot, the instantaneous rates aredetermined by drawing tangents to curve and thencalculating their slopes.

The reaction rate so obtained are plotted againstconcentrations raised to various powers.


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Decomposition rates of N2O5 g at different values of concentrationsof N2O5 as given in the data below:

Example Problem

Solution

These values are plotted as rate versus concentration and

rate versus (concentration)2


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Rate of decomposition of [N2O5]versus concentration of [N2O5]

Rate of decomposition of [N2O5]

versus concentration of [N2O5]2

It is seen that the rate is proportional to the concentration of N2O5 raised

to the power one. Therefore, dinitrogen pentoxide decomposes in a firstorder manner. The value of the rate constant k, as evaluated from thedata, is


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The rate is measured at the beginning of the reaction for severaldifferent initial concentrations of reactants.

[A]t

t / s

Initial rate

Follow reaction to ~ 10%

completion

Initial Rate Method

Important Method inchemical Kinetics


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RecallA + B P, Rate0 = k[A]0

a[B]0b

Taking logs

log Rate0 = log k+ a log [A]0 + b log[B]0

ym xc

** Keep [A]0 constant for varying values of [B]0 to find b

Log Ro

log[B]0

slope = b

Intercept = log k+ a log[A]0


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** Keep [B]0 constant for varying values of [A]0 to find a

from the slope of the graph, log R0 vs log [A]0

** Substitute values ofa, b, [A]0, [B]0 to find k.

However, in some cases, there may be no need to use

the plots as shown previously.

EXAMPLE

R1 = k[A]a[B]b

R2 = k[nA]a[B]b

Dividing R2 by R1

For these experiments, B is kept constant

while A is varied and R1 and R2 are known.


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ba

ba

BAk

BnAk

R

R

][][

][][

1

2

a

a

A

nA

][

][

a

aa

A

An

][

][

a

n

na

R

Rloglog

1

2

n

R

R

a

log

log1

2

(a) IfR2 = 2R1, and n=2, then a = 1, so 1st order with respect to A

(b) IfR2 = 4R1, and n=2, then a = 2, so 2nd order with respect to A


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Concluding:

if n=2,and Rate doubles 1st order

Rate increases by a factor of 4 2nd order

Rate increases by a factor of 9 3rd order


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This technique simplifies the rate law by making all the

reactants except one, in large excess.Therefore,

The dependence of the rate on each reactant can be foundbyisolating each reactant in turn and keeping all other

substances (reactants) in large excess.

Using as example: r= k[A]tm [B]t

n

Make B in excess, so [B]>>[A].

Hence, by the end of the reaction [B] would not have

changed that much, although all of A has been used up

And we can say, [B] [B]0

Isolation Method:


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r= k[A]tm , where k = k[B]0

n

Since A is the reactant that changes, then the rate

becomes dependent on A, and we can say

Created a false first-order (imitating first-order)

PSEUDO-FIRST-ORDER,

Logging both sides gives:

log r= log k + m log [A]ty = c + m x

A plot of log rvs log [A]t gives a straight line with slope = m,and intercept log k

where k is the pseudo-first-order rate constant


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If m = 1, the reaction is said to be pseudo-first-order

With the roles of A and B reversed, n can be found in asimilar manner

kcan then be evaluated using any data set along with the

known values ofm and n

Remember


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Estimate the orders andrate constant k from the results

observed for the reaction? What is the rate when [H2O2] = [I-]= [H+] = 1.0 M?

H2O2 + 3 I- + 2 H+ I3

- + 2 H2O

Exprmt [H2O2] [I-

] [H+

] Initial rate M s-1

1 0.010 0.010 0.0050 1.15e-6

2 0.020 0.010 0.0050 2.30e-6

3 0.010 0.020 0.0050 2.30e-6

4 0.010 0.010 0.0100 1.15e-6

Learn the strategy to determine the rate law from this example.

Figure out the answer without writing down anything.

Example Problem


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Estimate the orders from the results observed for the reaction

H2O2 + 3 I- + 2 H+ I3

- + 2 H2O

Exprmt [H2O2] [I-] [H+] Initial rate M s-1

1 0.010 0.010 0.0050 1.15e-6

2 0.020 0.010 0.0050 2.30e-6 1 for H2O23 0.010 0.020 0.0050 2.30e-6 1 for I-

4 0.010 0.010 0.0100 1.15e-6 0 for H+

1.15e-6 = k[H2O2]x [I-]y [H+]z

1.15e-6 k(0.010)x(0.010)y(0.0050)z exprmt 1 1 1----------- = ------------------------------------- ---- = ---

2.30e-6 k(0.020)x(0.010)y(0.0050)z exprmt 2 2 2

x = 1

( )x

Solution


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Other orders are determined in a similar way as shown

before.Now, lets findk and therateThurs, rate = 1.15e-6 = k(0.010)(0.010) from exprmt 1

k= 1.15e-6 M s-1 / (0.010)(0.010) M3 = 0.0115 M-1 s-1

And the rate law is therefore,d[H2O2] k

rate == 0.0115 [H2O2] [I-]

d t total order 2

Therate when [H2O2] = [I-] = [H+] = 1.0 M

The rate is the same as the rate constant k, when

concentrations of reactants are all unity (exactly 1), doesnt

matter what the orders are.


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N2O5 decomposes according to 1st

order kinetics, and 10% of itdecomposed in 30 s. Estimatek, tand percent decomposed in

500 s.

Example Problem

If the rate-law isknown, what arethe keyparameters?


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15 Chemical Kinetics

N2O5 decomposes according to 1st order kinetics, and 10% of it

decomposed in 30 s. Estimate k, tand percent decomposed in 500 s.

Assume [A]o = [N2O5]o = 1.0, then [A] = 0.9 at t= 30 s or

0.9 = 1.0 e k t apply [A]o = [A] ek t

ln 0.9 = ln 1.0k30 s

0.1054 = 0k* 30k = 0.00351 s 1

t=0.693 /k= 197 s apply k t = ln 2

[A] = 1.0 e 0.00351*500 = 0.173

Percent decomposed: 1.00.173 = 0.827 or 82.7 %

After 2 t (2*197=394 s), [A] = ()2 =, 75% decomposed.

After 3 t (3*197=591 s), [A] = ()3 =1/8, 87.5% decomposed.

Apply integrated rate law to solve problems

Solution


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What is the half-life of N2O5 if it decomposes with a

rate constant of 5.7 x 10-4 s-1?

13.3

tln2

k=

0.693

5.7 x 10-4 s-1= = 1200 s = 20 minutes

How do you know decomposition is first order?

units ofk(s-1)

Example Problem

Solution

Solution


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The reaction 2A B is first order in A with a rate constant

of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decreasefrom 0.88Mto 0.14M?

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

kt = ln[A]0 ln[A]

t =ln[A]0 ln[A]

k= 66 s

[A]0

= 0.88M

[A] = 0.14M

ln[A]0

[A]

k=

ln0.88M

0.14M

2.8 x 10-2 s-1=

Example Problem

Solution


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Solution

[A] / [A]0 = fraction remaining

when t = t1/2 then fraction remaining = _________Therefore, ln (1/2) = - k t1/2

- 0.693 = - k t1/2

t1/2 = 0.693 / kSo, for sugar,

t1/2 = 0.693 / k = 2100 sec = 35 min

Rate = k[sugar] and k = 3.3 x 10

-4

sec

-1

. What is the half-life ofthis reaction?

Example Problem


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Solution

2 hr and 20 min = 4 half-livesHalf-life Time Elapsed Mass Left

1st 35 min 2.50 g

2nd 70 1.25 g3rd 105 0.625 g

4th 140 0.313 g

Rate = k[sugar] and k = 3.3 x 10-4

sec-1

. Half-life is 35min. Start with 5.00 g sugar. How much is left after 2hr and 20 min (140 min)?

Example Problem


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Radioactive decay is a first order process.Tritium f electron + helium3H 0-1e

3He

t1/2 = 12.3 years

If you have 1.50 mg of tritium, how much is left after

49.2 years?

Example Problem


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Solution

ln [A] / [A]0 = -kt

[A] = ? [A]0 = 1.50 mg t = 49.2 y

Need k, so we calc k from: k = 0.693 / t1/2

Obtain k = 0.0564 y-1

Now ln [A] / [A]0 = -kt = - (0.0564 y-1)(49.2 y)

= - 2.77Take antilog: [A] / [A]0 = e

-2.77 = 0.0627

0.0627 = fraction remaining

Start with 1.50 mg of tritium, how much is left after 49.2

years? t1/2 = 12.3 years


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Solution

[A] / [A]0 = 0.0627

0.0627 is the fraction remaining!

Because [A]0 = 1.50 mg, [A] = 0.094 mg

But notice that 49.2 y = 4.00 half-lives

1.50 mg f 0.750 mg after 1 half-life

f0.375 mg after 2f0.188 mg after 3

f0.094 mg after 4

Start with 1.50 mg of tritium, how much is left after 49.2

years? t1/2 = 12.3 years


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Determine the rate law for the following reactiongiven the data below.

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

-(aq) + H2O (l)

[H2O2] [I-] [H+] InitialExpt # (M) (M) (M) Rate (M /s)

1 0.010 0.010 0.00050 1.15 x 10-6

2 0.020 0.010 0.00050 2.30 x 10-6

3 0.010 0.020 0.00050 2.30 x 10-64 0.010 0.010 0.00100 1.15 x 10-6

Example Problem


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Rate = k[H2O2]m[I-]n[H+]pCompare Expts. 1 and 2:

Rate 2 = 2.30 x 10-6 M/s = 2.00

Rate 1 = 1.15 x 10-6 M/s

Rate 2 = k (0.020 M)m(0.010 M)n(.00050 M)p = 2.00

Rate 1 k (0.010 M)

m

(0.010 M)

n

(.00050 M)

p

(0.020)m = 2.0m = 2.00

(0.010)m

2.0m = 2.00 only if m = 1

Rate = k[H2O2]1[I-]n[H+]p

Solution


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Rate = k[H2O2]1[I-]1[H+]pCompare Expts. 1 and 4:

Rate 4 = 1.15 x 10-6 M/s = 1.00

Rate 1 = 1.15 x 10-6 M/s

Rate 4 = k (0.010 M)1(0.010 M)1(.00100 M)p = 1.00

Rate 1 k (0.010 M)

1

(0.010 M)

1

(.00050 M)

p

(0.00100)p = 2.0p = 1.00

(0.00050)p

2p = 1.00 only if p = 0

Rate = k[H2O2]1[I-]1[H+]0

Rate = k[H2

O2

] [I-]


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The following data was used to determine the ratelaw for the reaction:

A + B C

Calculate the value of the rate constant if the rate lawis:

Expt # [A] (M) [B] (M) Initial rate (M /s)

1 0.100 0.100 4.0 x 10-5

2 0.100 0.200 8.0 x 10-5

3 0.200 0.100 16.0 x 10-5

Rate = k [A]2[B]

Example Problem


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Select a set of conditions:Experiment 1: Rate = 4.0 x 10-5 M /s

[A] = 0.100 M

[B] = 0.100 M

Substitute data into the rate law:

4.0 x 10

-5

M = k (0.100 M)

2

(0.100 M)s

Solve for k

k = 4.0 x 10-5 M/s = 0.040 = 0.040 M-2s-1

(0.100 M)2 (0.100M) M2 . s

Rate = k [A]2[B]

Solution

l bl


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A certain pesticide decomposes in water via a first orderreaction with a rate constant of 1.45 yr-1. What will theconcentration of the pesticide be after 0.50 years for asolution whose initial concentration was 5.0 x 10-4 g/mL?

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

0

Given: k = 1.45 yr-1

t = 0.50 yr[A]o = 5.0 x 10

-4 g/mLFind: [A]t=0.5 yr

Example Problem

Solution


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ln[A]0.5 yr = -kt + ln[A]0

ln[A]0.5 yr = - (1.45 x 0.50 yr) + ln (5.0 x 10-4)

yr

ln[A]0.5 yr = - 0.725 + -7.601 = - 8.326

To find the value of [A]0.5 yr

, use the inverse naturallogarithm, ex.

[A]0.5 yr = e-8.326 = 2.42 x 10-4 = 2.4 x 10-4 g/mL

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A certain pesticide has a half life of 0.500 yr. If the initialconcentration of a solution of the pesticide is 2.5 x 10-4g/mL, what will its concentration be after 1.5 years?

Given: [A]0 = 2.5 x 10-4 g/mL

t1/2 = 0.500 yr

t = 1.5 yrFind: [A]1. 5 yr

First, find the

value for k.

Example Problem

Solution


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Find the rate constant, k:k = 0.693/t1/2

k = 0.693 = 1.386 yr-1

0.500 yr

Substitute data into

ln[A]1.5 yr = - (1.386 x 1.5 yr) + ln(2.5 x 10-4)yr

= - 10.373

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


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Solve for [A]1. 5 yr using the inverse natural logarithm:

[A]1. 5 yr = e-10.373 = 3.1 x 10-5 g/mL


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Concentrations

physical state of reactants and

products

Temperature

Catalysts

Factors Affecting Rates


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Factors That Affect Reaction Rates

Physical State of the Reactants

In order to react, molecules must come in contactwith each other.

The more homogeneous the mixture of reactants,the faster the molecules can react.


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Concentration of Reactants

As the concentration of reactants increases, so doesthe likelihood that reactant molecules will collide.



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Temperature

At higher temperatures, reactant molecules havemore kinetic energy, move faster, and collide more

often and with greater energy.



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Presence of a Catalyst

Catalysts speed up reactions by changing themechanism of the reaction.

Catalysts are not consumed during the course ofthe reaction.


T t d R t


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The Collision Model Most reactions speed up as temperature increases. (E.g. food

spoils when not refrigerated.)

When two light sticks are placed in water: one at room temperature

and one in ice, the one at room temperature is brighter than the one

in ice.

The chemical reaction responsible for chemiluminescence is

dependent on temperature: the higher the temperature, the faster

the reaction and the brighter the light.

Temperature and Rate

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The CollisionModel

As temperature

increases, the rateincreases.


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The Collision Model

Since the rate law has no temperature term in it, the rate

constant must depend on temperature.

Consider the first order reaction CH3NC CH3CN. As temperature increases from 190 C to 250 C the rate

constant increases from 2.52 10-5 s-1 to 3.16 10-3 s-1.

The temperature effect is quite dramatic. Why?

Observations: rates of reactions are affected by

concentration and temperature.


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The Collision Model

Goal: develop a model that explains why rates of

reactions increase as concentration and temperature

increases. The collision model: in order for molecules to react they

must collide.

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.


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The Collision Model

The higher the temperature, the more energy available to

the molecules and the faster the rate.

Complication: not all collisions lead to products. In fact,only a small fraction of collisions lead to product.

The Orientation Factor

In order for reaction to occur the reactant molecules must

collide in the correct orientation and with enough energy

to form products.


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Consider:

Cl + NOCl NO + Cl2

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


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Activation Energy Arrhenius: molecules must posses a minimum amount of

energy to react. Why?

In order to form products, bonds must be broken in thereactants.

Bond breakage requires energy.

Activation energy,Ea, is the minimum energy required to

initiate a chemical reaction.


Temperat re and Rate


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Activation Energy

Consider the rearrangement of methyl isonitrile:

In H3C-NC, the C-NC bond bends until the C-N bond breaks

and the NC portion is perpendicular to the H3C portion. This

structure is called the activated complex or transition state.

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

Once the C-N bond is broken, the NC portion can continue to

rotate forming a C-CN bond.

H3C N C

C

NH3C H3C C N



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Activation Energy

The change in energy for the reaction is the difference in

energy between CH3NC and CH3CN.

The activation energy is the difference in energy betweenreactants, CH3NC and transition state.

The rate depends onEa.

Notice that if a forward reaction is exothermic (CH3

NC

CH3CN), then the reverse reaction is endothermic

(CH3CN CH3NC).




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Activation Energy

How does a methyl isonitrile molecule gain enough

energy to overcome the activation energy barrier?

From kinetic molecular theory, we know that astemperature increases, the total kinetic energy increases.

We can show the fraction of molecules,f, with energy

equal to or greater thanEa is

whereR is the gas constant (8.314 J/molK).

RTEa

ef




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Activation Energy


Example Problem


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For each reaction energy diagram below, mark the

location of the reactants, products and transition state.Identify the magnitude of Ea and DHrxn. Is each reactionendothermic or exothermic?

Example Problem



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The Arrhenius Equation

Arrhenius discovered most reaction-rate data obeyed

the Arrhenius equation:

kis the rate constant, Eais the activation energy,R is the gas

constant (8.314 J/K-mol) and Tis the temperature in K.

A is called the frequency factor.

A is a measure of the probability of a favorable collision.

BothA andEa are specific to a given reaction.

RTEa

Aek




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Determining the Activation Energy If we have a lot of data, we can determineEa andA

graphically by rearranging the Arrhenius equation:

From the above equation, a plot of ln kversus 1/Twill

have slope ofEa

/R and intercept of ln A.

ARTEk a lnln



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Can be arranged in the form of a straight line

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

Plot ln k vs. 1/T slope = -Ea/R

Ea/RTdecreases

-Ea/RTincreases

e-Ea/RT

increasesk

increasesREACTIONSPEEDS UP

If Tincreases



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Determining the Activation Energy If we do not have a lot of data, then we recognize

122

1

2121

22

11

11ln

lnlnlnln

lnlnandlnln

TTR

E

k

k

ART

EA

RT

Ekk

A

RT

EkA

RT

Ek

a

aa

aa


Example Problem


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Calculate the activation energy for the rearrangement ofmethyl isonitrile to acetonitrile using the following data.

You can solve this in oneof several ways:

Graph ln k vs. 1/T and

determine the slope;

Find ln k and 1/T and

determine the slope using

two well spaced points

using D(ln k)/D(1/T);

Example Problem

Solution


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Personally, prefer the non-graphical method:

Use two points that are well separated and convert T toKelvin (K = oC + 273.15)

Temp

(K) 1/T (K-1) k

462.9 2.160 x 10-3

2.52 x 10-5

524.4 1.907 x 10-3

3.16 x 10-3

122

1 11lnTTR

E

k

ka


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Plug the values of k and T into the equation. Be careful toput k1 and T1 in the appropriate order:

ln 2.52 x 10-5 = Ea 1 - __1__

3.16 x 10-3 (8.314 J/mol K) 524.4 K 462.9 K

Solve for Ea.

122

1 11lnTTR

E

k

ka


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ln 2.52 x 10-5 = Ea 1 - __1__3.16 x 10-3 (8.314 J/mol K) 524.4 K 462.9 K

Solve for Ea

.

- 4.8315 = (- 3.047 x 10-5 mol/J) Ea

Ea = 1.59 x 105 J/mol

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The balanced chemical equation provides information

about the beginning and end of reaction.

The reaction mechanism gives the path of the reaction. Mechanisms provide a very detailed picture of which

bonds are broken and formed during the course of a

reaction.

Elementary Steps

Elementary step: any process that occurs in a single step.

Reaction Mechanisms

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Elementary Steps Molecularity: the number of molecules present in an

elementary step.

Unimolecular: one molecule in the elementary step, Bimolecular: two molecules in the elementary step, and

Termolecular: three molecules in the elementary step.

It is not common to see termolecular processes

(statistically improbable).

Reaction Mechanisms

R ti M h i
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Multistep Mechanisms Some reaction proceed through more than one step:

NO2(g) + NO2(g) NO3(g) + NO(g)

NO3(g) + CO(g) NO2(g) + CO2(g) Notice that if we add the above steps, we get the overall

reaction:

NO2(g) + CO(g)

NO(g) + CO2(g)

Reaction Mechanisms

Reaction Mechanisms


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Multistep Mechanisms If a reaction proceeds via several elementary steps, then

the elementary steps must add to give the balanced

chemical equation.

Intermediate: a species which appears in an elementary

step which is not a reactant or product.

Reaction Mechanisms

Reaction Mechanisms


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Rate Laws for Elementary Steps 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.

Rate Laws for Multistep Mechanisms

Rate-determining step: is the slowest of the elementary

steps.

Reaction Mechanisms

Reaction Mechanisms


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Rate Laws for Elementary Steps

Reaction Mechanisms

Reaction Mechanisms


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Rate Laws for Multistep Mechanisms Therefore, the rate-determining step governs the overall

rate law for the reaction.

Mechanisms with an Initial Fast Step It is possible for an intermediate to be a reactant.

Consider

2NO(g) + Br2(g)

2NOBr(g)

Reaction Mechanisms

Reaction Mechanisms


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Mechanisms with an Initial Fast Step2NO(g) + Br2(g) 2NOBr(g)

The experimentally determined rate law is

Rate = k[NO]2[Br2] Consider the following mechanism

NO(g) + Br2(g) NOBr2(g)k1

k-1

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

Step 1:

Step 2:

(fast)

(slow)

Reaction Mechanisms

Reaction Mechanisms


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Mechanisms with an Initial Fast Step The rate law is (based on Step 2):

Rate = k2[NOBr2][NO]

The rate law should not depend on the concentration ofan intermediate (intermediates are usually unstable).

Assume NOBr2 is unstable, so we express the

concentration of NOBr2 in terms of NOBr and Br2

assuming there is an equilibrium in step 1 we have

]NO][Br[]NOBr[ 2

1

12

k

k

Reaction Mechanisms

Reaction Mechanisms


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Mechanisms with an Initial Fast Step By definition of equilibrium:

Therefore, the overall rate law becomes

Note the final rate law is consistent with theexperimentally observed rate law.

]NOBr[]NO][Br[ 2121 kk

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

1

122

1

12

k

kk

k

kk

Reaction Mechanisms

Deriving a Rate Law From a Mechanism


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The decomposition of H2O2 in the presence of Ifollow this

mechanism,

i H2O2 + I k1 H2O + IO

slow

ii H2O2 + IO k2 H2O + O2 + I

fast

What is the rate law? Energy

Eai

Eaii

reaction



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The decomposition of H2O2 in the presence of Ifollow this

mechanism,

i H2O2 + I k1 H2O + IO

slow

ii H2O2 + IO k2 H2O + O2 + I

fast

What is the rate law?

Solution

The slow step determines the rate, and the rate law is:

rate = k1 [H2O2] [I]

Since both [H2O2] and [I]are measurable in the system, this

is the rate law.

g

Deriving a rate law from a mechanism - 2


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Derive the rate law for the reaction, H2 + Br2 = 2 HBr,

from the proposed mechanism:

i Br2 2 Br fast equilibrium (k1, k-1)

ii H2 + Br k2 HBr + H slow

iii H + Br k3 HBr fast

Solution:The fast equilibrium condition simply says that

k1 [Br2] = k-1[Br]2

and [Br] = (k1/k-1 [Br2])

The slow step determines the rate law,

rate = k2 [H2] [Br] Br is an intermediate= k2 [H2] (k1/k-1 [Br2])

= k[H2] [Br2]; k = k2 (k1/k-1)

M- s -1

total order 1.5

explain

C t l i


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A catalyst changes the rate of a chemical reaction.

There are two types of catalyst:

homogeneous, and heterogeneous.

Chlorine atoms are catalysts for the destruction of ozone.

Homogeneous Catalysis

The catalyst and reaction is in one phase.

Catalysis

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Homogeneous Catalysis Hydrogen peroxide decomposes very slowly:

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

In the presence of the bromide ion, the decompositionoccurs rapidly:

2Br-(aq) + H2O2(aq) + 2H+(aq) Br2(aq) + 2H2O(l).

Br2(aq) is brown.

Br2(aq) + H2O2(aq) 2Br-(aq) + 2H+(aq) + O2(g).

Br- is a catalyst because it can be recovered at the end of the

reaction.

Catalysis

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Homogeneous Catalysis Generally, catalysts operate by lowering the activation

energy for a reaction.

Catalysis

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Catalysis

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Homogeneous Catalysis Catalysts can operate by increasing the number of

effective collisions.

That is, from the Arrhenius equation: catalysts increase kbe increasingA or decreasingEa.

A catalyst may add intermediates to the reaction.

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

an intermediate in the decomposition of H2O2.

Catalysis

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Homogeneous Catalysis When a catalyst adds an intermediate, the activation

energies for both steps must be lower than the activation

energy for the uncatalyzed reaction. The catalyst is in a

different phase than the reactants and products.

Heterogeneous Catalysis

Typical example: solid catalyst, gaseous reactants and

products (catalytic converters in cars).

Most industrial catalysts are heterogeneous.

Catalysis

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Heterogeneous Catalysis First step is adsorption (the binding of reactant molecules

to the catalyst surface).

Adsorbed species (atoms or ions) are very reactive.

Molecules are adsorbed onto active sites on the catalyst

surface.

Catalysis

Catal sis
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Catalysis

Catalysis


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Heterogeneous Catalysis Consider the hydrogenation of ethylene:

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

The reaction is slow in the absence of a catalyst.

In the presence of a metal catalyst (Ni, Pt or Pd) the reaction

occurs quickly at room temperature.

First the 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.

Catalysis

Catalysis


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Heterogeneous Catalysis When an H atom collides with an ethylene molecule on the

surface, the C-C bond breaks and a C-H bond forms.

When C2H6 forms it desorbs from the surface.

When ethylene and hydrogen are adsorbed onto a surface, less

energy is required to break the bonds and the activation energy

for the reaction is lowered.

Catalysis

chemical kinetics ppt

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