chem 3310 k9 deriveratelawfrommechanism whitebkgd
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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
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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.
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.
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.
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
Derive the rate law of a reaction mechanism
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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Derive the rate law of a reaction mechanism
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
Derive the rate law of a reaction mechanism
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)
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• 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.
Mechanism: A sequence of one or more elementary reaction steps that
Derive the rate law of a reaction mechanism
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]
Derive the rate law of a reaction mechanism
Mechanism: A sequence of one or more elementary reaction steps that
(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
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Example:
Derive the rate law of a reaction mechanism
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:
Derive the rate law of a reaction mechanism
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
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dtdtdtdt k ]NO[]CO[ 2
Derived rate law is consistent with the experimental rate law.
Example:
Derive the rate law of a reaction mechanism
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:
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Below 600K, experimentallydetermined rate law isRate = k[NO2]2
Example:
Derive the rate law of a reaction mechanism
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.
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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
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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
Derive the rate law of a reaction mechanism
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
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Which mechanism is consistent with the observed rate law?
Consider the following reaction mechanism proposed for the overall reaction
Derive the rate law of a reaction mechanism
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]
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Consider the following reaction mechanism proposed for the overall reaction
Derive the rate law of a reaction mechanism
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
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
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Derive the rate law of a reaction mechanism
Consider the following reaction mechanism proposed for the overall reaction
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
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collision would be more probable.
Consider the following reaction mechanism proposed for the overall reaction
Derive the rate law of a reaction mechanism
Consider the following reaction mechanism proposed for the overall reaction
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.
Consider the following reaction mechanism proposed for the overall reaction
Derive the rate law of a reaction mechanism
Consider the following reaction mechanism proposed for the overall reaction
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
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]O[]NO[Rate 2ka bimolecular step is more plausible.
Derive the rate law of a reaction mechanism
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
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]O[][
21k with the experimental rate law.
Steady State Approximation
Derive the rate law of a reaction mechanism
• The steady-state approximation is a method used to derive a ratelaw
Steady State Approximation
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.
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Steady State Approximation
Derive the rate law of a reaction mechanism
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]
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Steady State ApproximationH2 (g) + 2 ICl (g) I2 (g) + 2 HCl (g)
Experimental rate law:
Derive the rate law of a reaction mechanism
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
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]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
Derive the rate law of a reaction mechanism
• 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.
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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
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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