CHEM 3310 k9 deriveRateLawFromMechanism whiteBkgd

CHEM 3310

Chemical Kinetics

Deri ed Rate La s fromDerived Rate Laws from Reaction Mechanisms

Reaction Mechanism

Determine the rate law by experiment

Devise a reaction mechanism

Predict the rate lawfor the mechanism

If the predicted and experimental rate laws

agree

If the predicted and experimental rate laws do

not agree

Look for additional supporting evidence

2CHEM 3310

Reaction Mechanism

• A sequence of one or more elementary reaction steps together forms a reaction mechanismreaction mechanism.

• In a mechanism, elementary steps proceed at various speeds (governedby different rate constants k)by different rate constants, k).

• Elementary reaction steps must be balanced (as do all chemical reactions).

• The slowest step is the rate-determining step. It is the “bottleneck” in the formationof products. p

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Reaction Mechanism

• A rate law derived from a set of mechanisms should only consist of concentrationsof reactants and/or products, no intermediates.

• In predicting the rate law for an elementary step, the exponents for the concentrationterms are the same as the stoichiometric coefficients.

• To propose a mechanism requires the knowledge of chemistry to give plausible elementary processes. In this course, you will not be asked to propose mechanisms but you will be asked to derive the rate laws from given mechanismsmechanisms, but you will be asked to derive the rate laws from given mechanisms.

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Reaction Mechanism - Terminologies

Molecularity is the number of reacting species (i e atoms ions or• Molecularity – is the number of reacting species (i.e. atoms, ions or molecules) in an elementary reaction, which must collidesimultaneously in order to bring about a chemical reaction.

Uni-, Bi-, Termolecular

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Involving 1 species

Involving 2 species

Involving 3 species

Reaction Mechanism

1. Unimolecular Elementary Step

BA 1k CBA 1korBA

There is only one molecule reacting namely species "A" is reacting

CBA or

There is only one molecule reacting, namely species A is reacting. This unimolecular reaction step implies the rate law,

]B[d]A[d Involving a ]A[

dt]B[d

dt]A[d

1k-

Examples:

Involving a single species.

Examples:

Decomposition of hydrogen peroxide

H O (aq) H O (l) + ½ O (g)

Decomposition of ammonium nitrite

NH NO (g) N (g) + 2 H O (g)

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H2O2 (aq) H2O (l) + ½ O2 (g) NH4NO2 (g) N2 (g) + 2 H2O (g)

Recall, this is a 1st

order reaction.

Reaction Mechanism

A B 2. Reversible Unimolecular Elementary Step

At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.

Rforward = Rreverse

k [A] k [B]k1[A] = k-1[B]

Rearrange,

]A[]B[

1

kk

Rearrange,

This is the definition of Keq.

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]A[1k

Reaction Mechanism

A B

2. Reversible Unimolecular Elementary Step (cont’d)

A B

• Species "A" is in equilibrium with species “B"Species A is in equilibrium with species B .

• The forward reaction is governed by k1.

• The reverse reaction is governed by k-1.

The rate, R, is equal to the rate of the forward step minus the rate of the reverse step. This implies the rate law,

]B[]A[dt

]B[ddt

]A[d11 kk R

8CHEM 3310[A ] decreasing

in forward reaction[A ] increasing

in reverse reaction

Reaction Mechanism

3. Bimolecular Elementary Step

k Implies this Implies this

• requires two molecules coming together at the same time.

CBA 2k Implies this rate law.

Implies this rate law.

]B][A[dt

]C[ddt

]B[ddt

]A[d2k

or

DCBA 2k Implies this rate law.


]B][A[dt

]D[ddt

]C[ddt

]B[ddt

]A[d2k

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dtdtdtdt

Reaction Mechanism

3. Bimolecular Elementary Step

DCAA 2kor Implies this t l

Implies this rate lawDCAA

21 ]D[d]C[d]A[d

rate law.rate law.

2221 ]A[

dt]D[d

dt]C[d

dt]A[d k

Examples of reactions involving a bimolecular elementary step.

R t k[NO][O ]NO + O NO + O Involves collisions of two species.

Rate = k[NO][O3]

Rate = k[HI]2

NO + O3 NO2 + O2

2 HI H2 + I2

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Reaction Mechanism

4. Reversible Bimolecular Elementary Step

Implies this l

Implies this lA + B C rate law.rate law.A + B C

]C[d]B[d]A[d

or

]C[]B][A[dt

]C[ddt

]B[ddt

]A[d22 kk-R


Implies this rate law.A + B C + D

]D][C[]B][A[dt

]D[ddt

]C[ddt

]B[ddt

]A[d22 kk- R

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dtdtdtdt

Reaction Mechanism

4. Reversible Bimolecular Elementary Step (cont’d)

or Implies this rate law

Implies this te l2 A C + D rate law.rate law.2 A C + D

]D][C[]A[dt

]D[ddt

]C[ddt

]A[d2

222

1 kk- R

At equilibrium, Rforward = Rreverse where Rforward = k2[A]2

R = k 2[C][D]

2 ]D][C[kKl b

Rreverse k-2[C][D]k2[A]2 = k-2[C][D]

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22

2

]A[]][C[

eq k

KEquilibrium constant

Reaction Mechanism

2 HI H2 (g) + I2 (g)

At equilibrium, the rate of the forward reaction equals the rate of the q qreverse reaction.

Rforward = Rreverse

Rf = k1[HI]2

Rr = k 1[H2] [I2]

k1[HI]2 = k-1[H2] [I2]

r -1[ 2] [ 2]

2221 ]I][H[

eq kK

At the start At equilibrium

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21 ]HI[eqk

K

Reaction Mechanism5 Termolecular Elementary Step5. Termolecular Elementary Step

Termolecular reaction steps require three molecules coming together at the same time.

productsBAAA 3k Implies this rate law.


333

1 ]A[dt

]B[ddt

]A[d kor3 dtdt

productsCBAA 3k Implies this rate law.

Implies this rate law.p

]B[]A[]C[d]B[d]A[d 23

1 k

rate law.rate law.

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]B[]A[dtdtdt 32

k

Reaction Mechanism5 Termolecular Elementary Step

productsDCBA 3kImplies this

rate lawImplies this

rate lawor

5. Termolecular Elementary Step

]C][B][A[]D[d]C[d]B[d]A[d k

rate law.rate law.

]C][B][A[dt

]D[ddt

]C[ddt

]B[ddt

]A[d3k

Example:Example:

The reaction mechanism for

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

involves a 3-body collision one-step mechanism.

2

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Rate = k [NO]2 [O2]

Reaction Mechanism

Recall , the equation in an elementary step represents the reaction at the molecular level, not the overall reaction.

What about higher orders such as 4th or 5th orders?

Simultaneous collision of 3 molecules is rare. In nature, we observe lots of 2-

body collisions, very few 3-body collisions and not much else.

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Derive the rate law of a reaction mechanism

Most balanced equations do not literally describe how a reaction occurs in terms of the collisions made or the actual sequence of events.

The combustion of hexane illustrates this point:

If it were to take things literally as written, the reaction is saying 2 hexane molecules and 19 oxygen molecules somehow collide simultaneously and fight among themselves until 12 molecules of carbon dioxides and 14 molecules

2 C6H14 (g) + 19 O2 (g) → 12 CO2 (g) + 14 H2O (g)

somehow collide simultaneously and fight among themselves until 12 molecules of carbon dioxides and 14 molecules of waters form.

In nature, we observe lots of 2-body collisions, f 3 b d lli i d h lvery few 3-body collisions and not much else.

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A t h i i l i th lli i f NO d O


A one-step mechanism involving the collision of NO and O3.

NO + O3 NO2 + O21 1

Since this is the only step in the reaction mechanism, then the rate law can be written directly from the stoichiometry of the step.

dt]O[d

dt]NO[d

dt]O[d

dt]NO[dRate 223 --

k-- ]O[]NO[d

]O[dd

]NO[dd

]O[dd

]NO[dRate 3223

dtdtdtdt

1 1][][dtdtdtdt 3

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Example:

The overall reaction of

N2O5(g) + NO(g) 3 NO2(g)

occurs in a one-step mechanism where the two reactants collide toform the product.

The derived rate law can be determined to beThe derived rate law can be determined to be

Rate = k [N2O5][NO]

Suggestion of a bimolecular single step mechanism.

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Mechanism: A sequence of one or more elementary reaction steps that


q y pproceed at various speeds.

BA:Step 11 kEach step is governed

CB:Stepp

22 k ac step s go e edby its rate constant.

What would the energy profile diagram look like?

If k1 >> k2, If k1 << k2,

What would the energy profile diagram look like?

Ea1 Ea2Ea1 Ea2

• Step 2 is the rate determining step (RDS) • Step1 is the rate determining step (RDS)

20CHEM 3310

• Step 2 is the rate determining step (RDS). This implies that isolation of B is good.

• Step1 is the rate determining step (RDS).This implies that isolation of B is not easy.



yproceed at various speeds.

(slow)CBA:Step k 11Consider the reaction mechanism

(fast)EDC:Step( )p

k 22

1. Identify A, B, C, D, and E.

for an overall exothermic reaction:

A and B are reactantsC is the intermediateD and E are the products

2. What is the overall reaction?

A + B D + E

In this mechanism the rate law can be written directly from the slowest step

3. What is the rate law?

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In this mechanism, the rate law can be written directly from the slowest step.

Rate = k1 [A] [B]



(slow)CBA:Step k 11Consider the reaction mechanism

yproceed at various speeds.

(fast)EDC:Step(slow)CBA:Step

k

22

1

4 Sketch the reaction coordinate of the reaction

Consider the reaction mechanismfor an overall exothermic reaction:

Since step 1 is the rate determining step, k1 << k2.

4. Sketch the reaction coordinate of the reaction.

Ea1

Ea2

Ea1Ea1 > Ea2

22CHEM 3310

Example:


CO (g) + NO2 (g) NO (g) + CO2 (g)

This reaction proceeds via two reaction mechanisms.

Above 600K a one step mechanismAbove 600K, a one-step mechanism.

Below 600K, a two-step mechanism., p

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Experimentally determinedRate = k[CO][NO2]Example:


CO (g) + NO2 (g) NO (g) + CO2 (g)(balanced reaction)

Above 600K, the reaction mechanism involves the collision between CO and NO2. A one-step elementary step

which describes the collision CO + NO2 NO + CO2 of CO and NO2.

Reasonable - a single lli i f l l

1 1

collision of two molecules (with correct orientation and minimum energy) would lead to the exchange of an oxygen between CO and NO2.2

d]CO[d

d]NO[d

d]CO[d

d]NO[d 22 -- k ]NO[]CO[ 2 1 1

24CHEM 3310

dtdtdtdt k ]NO[]CO[ 2

Derived rate law is consistent with the experimental rate law.

Example:


CO (g) + NO2 (g) NO (g) + CO2 (g)(balanced reaction)

Proposed mechanism is a 2 step mechanism:

Below 600K, the experimentally determined rate law is

Rate = k [NO2] 2

Proposed mechanism is a 2-step mechanism:

25CHEM 3310

Below 600K, experimentallydetermined rate law isRate = k[NO2]2

Example:


CO (g) + NO2 (g) NO (g) + CO2 (g)

k

(fast)CONOCONO:Step

(slow)NONONONO:Step

k

k

223

322

2

1

2

1

1. Identify the intermediate, if any.

2 What is the overall reaction?

NO3 is the intermediate.

2. What is the overall reaction?

NO2 + NO2 + NO3 + CO NO + + NO2 + NO3 +CO2

Y th l t t NO2 + CO CO2 + NO

3. Which is the rate determining step? Step 1

Yes, the elementary steps add to give the overall reaction.

26CHEM 3310

Derive the rate law of a reaction mechanismExample:

Below 600K, experimentallydetermined rate law isRate = k[NO2]2

CO (g) + NO2 (g) NO (g) + CO2 (g)

Proposed mechanism:NO3 is very reactive; consistent with step 1

being the RDS.

(fast)CONOCONO:Step

(slow)NONONONO:Step k

k322

2

1

2

1

4. What is the rate law of the proposed mechanism? Is it consistent ?

(fast)CONOCONO:Step 2232

with the experimentally determined rate law?

Since step 1 is the RDS, use step 1, the bimolecular step involving the collision of two NO2, to determine the rate law.

The mechanism’s rate law is consistent with the experimentally determined rate law.

Rate = rate of the slow step = k1 [NO2]2

27CHEM 3310

The mechanism s rate law is consistent with the experimentally determined rate law. • Confirmed by reacting two NO2 molecules and look for yielding NO3 as a product.• NO3 is highly reactive and is capable of transferring an oxygen atom to CO to give CO2.

Consider the following reaction mechanisms proposed for the thermal


Consider the following reaction mechanisms proposed for the thermal decomposition of NO2.

2 NO2 2 NO + O2

Experimental rate law is

Two possible mechanisms:

Experimental rate law is

Rate = k [NO2] 2

p

)f(

)slow(ONONO:Step k

k 21

21Mechanism 1

)fast(NOONOO:Step k 22 22

Mechanism 2 )slow(NONONO:Step k 32 121 Mechanism 2

)fast(ONONO:Step

)(p

k 2332

22

28CHEM 3310

Which mechanism is consistent with the observed rate law?

Consider the following reaction mechanism proposed for the overall reaction



2 NO2 2 NO + O2

Experimental rate law isExperimental rate law is

Rate = k [NO2] 2

)f(

)slow(ONONO:Step k

k 22

1

21Mechanism 1

)fast(NOONOO:Step k 22 22

Derived rate law from Mechanism 1 isDerived rate law from Mechanism 1 is

Rate = k1 [NO2]

29CHEM 3310




2 NO2 2 NO + O2

Experimental rate law isExperimental rate law is

Rate = k [NO2] 2

Mechanism 2

)fast(ONONO:Step

)slow(NONONO:Step k

k

2332

21

221

)fast(ONONO:Step 232

Derived rate law from Mechanism 2 isDerived rate law from Mechanism 2 is

Rate = k1 [NO2]2

30CHEM 3310



2 NO + O2 2 NO2

Experimentally determined rate law is

Rate = k [NO]2[O2]

Consider a single step mechanism:

22 22 1 NOONO kD i d t l t b22 Derived rate law appears to be

consistent with the experimental rate law. Derived rate law is

R t k [NO]2[O ]Rate = k1 [NO]2[O2]

The mechanism invokes a termolecular step, which is very unlikely since three way collisions are less likely to take place. A mechanism Involving two species

31CHEM 3310

collision would be more probable.




2 NO + O2 2 NO2Consider a two-step mechanism:

(slow)NOOON:Step

)fast(ONNONO:Stepk

2 222

22

221

2

Need to remove N2O2 from the rate law.

Derived rate law is

Rate = rate of the slow step = k2[N2O2][O2]• N2O2 is an intermediate. The rate law cannot contain intermediates.

• Use the fast equilibrium step to find an expression for N2O2.

O 2 Ok1[NO]2 = k-1[N2O2]

2

1

122 ]NO[]ON[

kk Substitute this back into

the above rate equation to remove N O

32CHEM 3310

to remove N2O2.




2 NO + O2 2 NO2

Consider a two-step mechanism:

Experimentally determined rate lawRate = k [NO]2[O2]

(slow)NOOON:Step

)fast(ONNONO:Stepk

2 222

22

221

2

Derived rate law is

Rate = rate of the slow step = k2[N2O2][O2]

2

1

122 ]NO[]ON[

kk

SinceNeed to remove dinitrogen

dioxide, N2O2, from the rate law. Need to remove dinitrogen

dioxide, N2O2, from the rate law.

]O[]NO[Rate

k ; Let]O[]NO[kRate

21

122

2

1

12

k'kk k'

kk

Derived rate law is consistent with the experimental rate law.

This two step mechanism each involving

33CHEM 3310

]O[]NO[Rate 2ka bimolecular step is more plausible.


Ozone depletionExample: Experimental rate law:

122

3 ][][k][OkR t OO

Proposed mechanism:

2 O3 (g) 3 O2 (g)12

32

3 ][][k][O][kRate 2OO

)fast(OOO:Step1

D i d t l

(slow)OOO:Step

)fast(OOO:Step

k 232322

12

]O][O[dt

]O[d32

2 k

Derived rate law:

Need to remove ][O

; Let]O[]O[ ]O[Rate

21

12

21

3132 k

kk k' kkk

Need to removeNeed to remove O from the rate law as O is an intermediate.

From step 1, k1[O3] = k-1[O2][O]

]O[]O[]O[ 31

kk

][O][ORate2

3k'Need to remove O from the rate law as O is an intermediate.

Derived rate law is consistent

34CHEM 3310

]O[][

21k with the experimental rate law.

Steady State Approximation


• The steady-state approximation is a method used to derive a ratelaw


law.

• The method is based on the assumption that one intermediate in the reactionmechanism is consumed as quickly as it is generated. Its concentration remains unchanged for most of the reaction.

• The system reaches a steady-state, When steady-state is reached, there is no change observed in the hence the name of the technique is

called steady state approximation.

is no change observed in the concentration of the intermediate.

35CHEM 3310



Steady State ApproximationExample:

H ( ) 2 ICl ( ) I ( ) 2 HCl ( )

Experimental rate law:

Rate = k[H ][ICl]

Proposed mechanism:

H2 (g) + 2 ICl (g) I2 (g) + 2 HCl (g) Rate = k[H2][ICl]

k

(fast) HCl (g) (g) I)g( ICl HI (g) :Step

)slow( HCl (g) HI (g) )g(ICl (g)H:Stepk

2

2

2

1

21

Since step 1 is the slow step, the derived rate law is consistent with the experimental rate lawis consistent with the experimental rate law.

Rate = k[H2][ICl]

36CHEM 3310

Steady State ApproximationH2 (g) + 2 ICl (g) I2 (g) + 2 HCl (g)

Experimental rate law:


Steady State ApproximationProposed mechanism:

p

Rate = k[H2][ICl]

k )fast(HCl (g)HI (g))g(ICl(g)H:Step 211

Derived rate law: Need to remove HI from the rate

Steady State ApproximationNeed to remove

HI from the rate

(slow) HCl (g) (g) I)g( ICl HI (g) :Step

)fast(HCl (g) HI (g) )g(ICl (g)H:Stepk

22

221

]ICl][HI[dt

]I[d22 k

HI is being produced in step 1 and quickly removed in step 2 [HI] is pretty much constant throughout the reaction (with the

0dt

]HI[d

HI from the rate law as HI is an intermediate. ]ICl][HI[

dt]I[d

22 k

HI from the rate law as HI is an intermediate.

[HI] is pretty much constant throughout the reaction (with the exception of the beginning and the end of the reaction).

dt

]ICl][HI[]ICl][H[dt

]HI[d2210 kk

dtProduction of

HI (step 1)Removal of HI (step 2)

]ICl][H[]ICl][HI[]ICl][H[ 221

kkk

]ICl[]ICl[

]ICl][H[dt

]I[d

2

212

2

kkk

]I[dSubstitute [HI]

I t th t lSubstitute [HI]

i t th t l

37CHEM 3310

]ICl[]ICl][H[]HI[

2

21

kk

]ICl][H[dt

]I[d21

2 kDerived rate law is consistent with the

experimental rate law.

Into the rate law. into the rate law.

Chain Reactions


• Chain reactions are complex reactions that involve chain carriers, reactiveintermediates which react to produce more intermediates.

• The elementary steps in a chain reaction may be classified into:initiation, propagation, inhibition, and termination steps.

Example:

Chlorofluorocarbons (CFCs) destruction of the ozone layer Initiation: Thermally or photochemically

produces Cl radicalsproduces Cl radicals

Propagation: Regenerates more

Cl radicals

Termination: Cl radicals deactivates by reacting

to form an inactive product.Inhibition: A step involving product

38CHEM 3310

pmolecules being destroyed.

Reaction Mechanism Experiment 4: Does this mean that 5+1+6=12 particles must

come together and collide?What about this reaction in Experiment 4?

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

come together and collide?

Experimentally determined rate law is

Rate = k [Br- ][ BrO-3 ][H+ ]2Rate k [Br ][ BrO 3 ][H ]

• First order with respect to Br- and BrO3- ions

• Second order with respect to H+ ions

• Overall reaction is of 1+1+2 = 4.

We observe that the reaction is found to be quite fast. It means that even though the balanced q gequation involves a large number of molecules, the reaction does not proceed by simultaneous collision of all these reacting particles. The mechanism can involve two or maximum three collisions simultaneous.

39CHEM 3310

Reaction Mechanism

What about this reaction in Experiment 4?

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

This is a rather complex mechanism. According to Field et al., 1972; Pelle et al., 2004, the reaction is thought to occur by the following collection of bimolecular elementary reactions.elementary reactions.

The derived rate law for this mechanism is outside the scope of this course It is scope of this course. It is

found to be consistent with the experimental rate law.

• First order with respect to Br- and BrO3

- ions3

• Second order with respect to H+ ions

40CHEM 3310

Reaction Mechanism

Determine the rate law by experiment

Devise a reaction mechanism

Predict the rate lawfor the mechanism

If the predicted and experimental rate laws

agree

If the predicted and experimental rate laws do

not agree

Look for additional supporting evidence

Rate laws can prove a mechanism is wrong, but can’t prove one right!

41CHEM 3310

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CHEM 3310 k9 deriveRateLawFromMechanism whiteBkgd