1
Reaction Kinetics
An Introduction
2
• A condition of equilibrium is reached in a system when 2 opposing changes occur simultaneously at the same rate.
• The rate of a chemical reaction may be defined as the # of mols of a substance which disappear or are formed by the reaction per unit volume in a unit of time.
3
Example
tI
ttIIRate
CI
forward ΔΔ
=−−
=
→
][][][
12
12
4
• The previous rate is for the DISAPPEARANCE of I, therefore:
""
][
oflossorngdisappearimeanssignnegativethewheretIrate
ΔΔ
−=
5
Backwards Example
IstudyratesreactionbothwhengformeanssignpositivethewheretC
tC
ttCC
Ratebkward
min
][][][][
12
12
ΔΔ
+=ΔΔ
=−−
=
6
More Complex Reactions
tZ
tXRate
ngdisappeariisXasfastasXappearingZtodueisthis
tZ
tXRate
ZX
ΔΔ
=ΔΔ
−=
∴
ΔΔ
≠ΔΔ
−=
→
31
3
3
7
In GeneralFor the Reaction:
pP + qQ → rR + sS
tS
stR
rtQ
qtP
pRate
ΔΔ
=ΔΔ
=ΔΔ
−=ΔΔ
−=1111
8
Reaction Order
9
In GeneralFor the Reaction:
pP + qQ → rR + sS
tconsrateortconsalityproportionkwhere
QPkRateor
QPtoalproportionisand
tS
stR
rtQ
qtP
pRate
mn
mn
tantan
][][
][][
1111
=
=
ΔΔ
=ΔΔ
=ΔΔ
−=ΔΔ
−=
10
k
A reaction with an incredibly large rate constant is faster than a reaction with an
incredibly small rate constant.
11
For the Reaction:pP + qQ → rR + sS
The reaction is “n” order in [P] and “m” order in [Q]
OR
Is OVERALL “(n + m)” order
12
KEY!!!!!!
“n” and “m” DO NOT necessarily equal “p”, “q”,
“r” or “s”
13
Example
][
][][41][
21
)()(4)(2
52
2252
2252
ONkor
tO
tNO
tON
Rate
gOgNOgON
=
ΔΔ
=Δ
Δ=
ΔΔ
−=
+→
14
• This reaction is FIRST order in N2O5, NOT SECOND order as one might intuit from the stoichiometry.
15
The Order of The Reaction• = the specification of the empirical (experimentally-
determined) dependence of the rate of the reaction on CONCENTRATIONS
• The order may = 0, a whole number or a non-whole number, e.g.,– 0– 1– 1½– 2
• We’ll focus on whole numbers and 0 (zero) for reactions of the type:
fF + gG → Products• AND! The order of the reaction is defined in terms of
REACTANTS not the products, therefore, products do not need to be specified
16
Zero-Order Reactions
formbmxylineStraighttatFttimeatF
ChemPherenotIntegrate
tkFarrange
ktF
and
kkftFtcoefficienaonlyisf
GFktF
fRate
o
ktoFF
+====
−−
Δ−=Δ
=ΔΔ
−
==ΔΔ
−∴
=ΔΔ
−=
−=
:0][;][
:)(
][:Re
][
'][
][]['][1
][][
00
17
For Zero-Order Reactions
• These reactions are relatively rare
• Occur on metal surfaces
• Reaction rate is INDEPENDENT of concentration of reactants
18
First Order Reactions• Assume reaction is 1st order in F and zero order
in G:
tko
o
eFFOR
tkFF
egrateandtkFFarrange
FktFkkf
GFktF
fRate
−=
−=
Δ=Δ
=ΔΔ
∴=
=ΔΔ
−=
][][
][][ln
:int:Re
]['
][]['1 01
19
• Many radioactive decays fit 1st order reactions:• 226Ra88 → 222Rn86 + 4He2
• 238U92 → 234Th90 + 4He2
• The rate is proportional to [F]
20
The Rate is Proportional to [F]
Double [F] k [F] k [F]2 k [F]3
Rate Change ⇑ X 2 ⇑ X 4 ⇑ X 8
][FktF
=ΔΔ
21
Second Order Reactions: Type 1
tkFF
egrateandarrangeGtorespectwithorderZero
FtorespectwithorderSecond
FktF
andkkf
GFktF
fRate
o
=−
=ΔΔ
−
=
=ΔΔ
−=
][1
][1
:intRe
][
'
][]['1
2
02
22
Second Order Reactions: Type 2
overallorderSECONDGinANDFinorderFIRSTisaction
tkGFFG
FG
partsbyegrateandarrange
GFktF
henceandkkf
GFktG
gtF
fRate
o
o
oo
∴
=−
=ΔΔ
−
=
=ΔΔ
−=ΔΔ
−=
,Re][][][][
ln][][
1)(intRe
][][
,,'
][]['11
Second order reactions are the most common reactions
The rate is proportional to [G], [G]o, [F], [F]o
23
Pseudo-First Order Reactions
• A special kind of 2d order reaction:• Example• 1 M acetyl chloride (AcCl) reacted with 56
M water (XSSV amount) to for HOAC and HCl
AcCl + H2O → HOAc + HCl
24
OrderFirstPSEUDOHenceorderfirstbetoAPPEARSaction
AcClkRatekOHk
soecttosmalltooisOHinchangeThe
OHAcClktd
AcCldRate
−
=∴=
=−=
:,""Re
][','][
,det][
][][][
2
2
2
25
Empirical Method in Determining Reaction Orders
studiedbeingareAinONLYdifferingreactionsdifferentWhere
AA
tt
X
StepSecondAreactionfororderunknowntheX
AA
tt
raterate
StepFirstX
][2
][][log
log
:""
][][
:
2
1
1
2
2
1
1
2
2
1
ΔΔ
=
=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
ΔΔ
=
26
Example
AinorderondisreactionM
MAAtt
X
tMAandBAaction
tMAandBAaction
sec
2239.0477.0
734.1log3log
173.03.0log
'5'15log
][][log
log
'15;173.0][:2Re
'5;3.0][:1Re
2
1
1
2
2
1
∴
===⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
=⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ΔΔ
=
=Δ=→
=Δ=→
27
Practical Application• Reaction rates are proportional to some power of
[reactant]• Determined by using “initial reaction rate method”
28
Compare Reaction 2 with Reaction 1
29
Compare Reaction 3 with Reaction 1
30
• Fuse both effects and the rate equation for this reaction =
Rate = k [Zn] [HCl]2• REMEMBER:• The exponent in a rate equation generally
does NOT match the chemical equation coefficients.
• The exponent MUST be determined experimentally.
31
Another way to do this
32
Compare Reaction 1 with Reaction 2
33
Compare Reaction 3 with Reaction 1
34
Reaction Order Half-Lives
35
11
21
21
1
21
1
21
][1
:
2ln:
2][
:
−−
−
−
==
==
==
sMkF
t
OrderSecond
IONCONCENTRATofTINDEPENDENistOrderFirst
sk
t
OrderFirst
sMk
Ft
OrderZero
o
o
36
Example
• For the reaction:• C → D + E,• Half of the C is used up in 60 seconds.
Calculate the fraction of C used up after 10 minutes – reaction is first order in C.
37
Solution
UpUsedCorUPUSEDCHenceLEFTCC
CCC
CKey
tkCC
ondskandk
t
o
o
o
%9.99999022.0000978.01:000978.0][
93.6][ln)600()01155.0(][ln][ln
1][:][][ln
sec01155.060693.02ln 1
21
−−=−=
−=−=−
=
−=
=== −
38
Example
• 14C is present at about 1.1*10-13 mol% naturally in living matter. A bone dug up showed 9*10-15 mol% 14C. The half life of 14C is 5720 years. How old is the bone?
39
Solution
oldyears
t
tkCC
yrst
k
o
45.2066210*2115.1
503.2
)10*2115.1(10*1.1
10*9ln
][][ln
10*211.15720
693.02ln
4
413
15
14
14
14
21
=−
−
−=
−=
===
−
−−
−
−−
40
Example
• A decomposition reaction occurs in a fixed-volume container at 460°C. Its rate constant is 4.5*10-3 seconds-1. At t = 0, P = 0.75 atm. What is the pressure (P) after 8 minutes?
41
Solution
atmPP
P
P
tkPP
o
0865.016.2288.0ln
)480()10*5.4(75.0lnln
)480()10*5.4(75.0
ln
ln
3
3
=−=+
−=−
−=
−=
−
−
42
Q10 Effect
43
Example
• If the rate of a chemical reaction doubles for every 10°C rise in temperature, how much faster would the reaction proceed at 55°C than at 25°C?
44
Q 10 EffectQ 10 Effect on Reaction Rate
0, 1 10, 2 20, 4 30, 840, 16
50, 32
60, 64
70, 128
0
20
40
60
80
100
120
0 20 40 60
Temperature (Celsius)
Rel
ativ
e R
eact
ion
Rat
e
45
Solution
• Temperature increased 30°C, therefore, reaction rate increases 8-fold
Example:• What if the temperature was increased to
105°C from 25°C?• Temperature increased 80°C, therefore
reaction rate increases 256-fold
46
Example
• How much faster would a reaction go at 100°C than at 25°C?
• For every 10°C increase in temperature, the reaction rate doubles. The change in temperature is 75°C. This is 7.5 10°C increases.
• Hence 27.5 = 181 times faster
47
Example
• In an experiment, a sample of NaOCl was 85% decomposed in 64 minutes. How long would it have taken if the temperature was 50°C higher?
• For every 10°C increase, the reaction rate doubles. 50°C increase is 5 10°C increases.
• Hence: 25 = 32 times faster• So: (64 minutes)/(32 times faster) = 2 minutes
48
There are 4 types of temperature dependence for reaction rates -- 1
• Rate increases with increasing temperature
• NORMAL
49
There are 4 types of temperature dependence for reaction rates -- 2
• Rate increases to a point, then reduces with increasing temperature
• E.g., enzymes being denatured
50
There are 4 types of temperature dependence for reaction rates -- 3
• Rate decreases with increasing temperature
• VERY RARE• Known only for a
few reactions that are multi-step reactions:– A → B Fast step– B → C Rate limiting step
51
There are 4 types of temperature dependence for reaction rates -- 4
• Rate increases with increasing temperature
• Odd behavior• Explosive reaction
when temperature shoots up
• Gradual rise in temperature due to chain reactions
52
• The Q 10 effect is “not entirely perfect”
• There is another way to study temperature dependence:
• Mathematically with Energy of Activation
53
etemperaturabsoluteTtconsGasR
ActivationofEnergyEmoleculestreacbetweencollisionsof
frequencytotalfactorfrequencyAtconsratek
a
TRaEeAk
==
=
==
−=
tan
)tan(
tan
/
54
Take Natural Log (ln) of Above Equation
TRE
Ak a−= lnln
55
Previous Equation Can Be Manipulated
• If you know the rate constants for reactions at 2 different temperatures you can calculate the Ea:
⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
121
2 11lnTTR
Ekk a
56
Example
• At T1 of 308 K, k1 = 0.326 s-1; at T2 of 318 K, k2 = 1.15 s-1. R = 8.314 J/mol-K. Determine Ea
molkJ
molJE
E
E
TTRE
kk
a
a
a
a
6.1028.601,10210*021.1
)314.8()26.1(
)10*021.1(314.8
26.1
3081
3181
314.8326.015.1ln
11ln
4
4
121
2
⇔==−−
−−=
⎟⎠⎞
⎜⎝⎛ −−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
−
−
57
Ea -- Transition
• Top of Energy curve (“hump”) = transition state
• The smaller the Ea, the easier it is for the reaction “to go”
58
Ea – “thermic”
Ea is related to ΔG
59
Reaction Mechanisms
60
Unimolecular Steps
• A single molecule or atom breaks down or rearranges itself into another species.
61
Bimolecular Steps
• Collision of two molecules or atoms; relatively common
62
Termolecular Steps
• 3 molecules or atoms collide simultaneously; relatively rare and SLOW
63
Elementary Steps
• = each step in a mechanism; a specific occurrence in the reaction sequence– The rate of elementary steps are proportional
to the number of collisions per step– The number of collisions per step are
proportional to every reactant concentration– Hence the rate of elementary steps are
proportional to the concentration of every reactant
64
The exponents of an elementary step rate equation are equal to the coefficients in the step’s equation
65
E.g., Bi Elementary Steps
OrderstSOkStepforequationRate
OrderdOSOkStepforequationRate
StepOSOSOStepSOOSO
RxnOverallOSOOSO
1][:2
2][][:1
21
51
321
235
532
2332
+→→+
+→+
66
E.g., Termolecular Elementary Steps
OrderstClHkStepforequationRate
OrderdHClkStepforequationRate
OrderstClkStepforequationRate
StepHClClHStepClHHCl
StepClCl
RxnOverallHClHCl
1][:3
2][][:2
1][:1
3222
12
2
223
22
2
21
22
222
2
22
=
=
=
→→+
→
→+
−
−
−
67
Reaction MechanismsTWO requirements of postulated mechanisms:
1) MUST account for the postulated mechanism2) MUST explain the experimental rate equation
1) – INHERENT in this is the assumption that2) 1 step is incredibly SLOWER than the others,
hence, it determines the speed of the reaction.3) This step is called the rate limiting step.4) Therefore, the overall rate equation is determined
by rate equations for the rate limiting step.5) ALSO: when a catalyst is present, there MUST BE
at least 2 steps in the mechanism!
68
Example
• The empirically derived rate equation for:2NO2 (g) + Cl2 (g) → 2NO2Cl ↑ is
k [NO2] [Cl2]
• Write a two-step mechanism with a slow first step
69
stepslowtheisStepHenceEquationRateEmpiricalStepforEquationRate
BUTClClNOk
EquationRateStepClNOClClNO
ClNOkEquationRate
StepClClNOClNORxnOverallClNOClNO
1,1
][][
22][][
122
2
22
22
222
222
=
→+
+→+→+
70
Comments
• Any steps after Step 1 will NOT effect the rate because step 1 is the rate limiting step
• Add up the 2 steps (just like in thermodynamics) and you get the overall reaction
71
Illustrative Example
)1,Re(2
2][][
2][][
1][][
22
2
2
2
22
purposesveillustratiforwasthisSTEPisalityInLimitingRateisStep
EquationStepEqualsEquationRateEmpiricalthatNoteHIIClk
EquationRateStepHClIHIICl
HIClkEquationRate
StepHClHIHIClEquationRateDerivedyEmpiricallHIIClkRxnOverallIHClHICl
−
+→+
+→+
+→+
72
Example
• The empirical rate equation for
2ClO2 + F2 → 2FClO2
Is: k [ClO2] [F2]
• Write the reaction mechanism (2-step) consistent with the rate equation.
73
LimitingRateisStepHenceEquationRatedDtnyEmpiricall
MatchesEquationRateStepFClOk
EquationRateStepFClOFClO
FClOkEquationRate
StepFFClOFClORxnOverallFClOFClO
1,'
1][][
2][][
122
2
22
22
222
222
→+
+→+→+
74
Steady State Approximation• The majority of mechanisms are NOT as easy as
previous examples.• Let’s return to our Bimolecular Reaction Step Example:
75
Focus• SO5• This is an “intermediate” – a “transition”• An intermediate (or transition) is a compound
that is neither reactant nor product, rather in between the two.
• Intermediates are very difficult to isolate in many cases.
• They are highly reactive and, more often than not, are used up as quickly as they are formed. It is because of this characteristic that we have the Steady State Approximation
76
Steady State Approximation• The concentration of intermediate is a constant• I.e., the rate of formation of the intermediate is
equal to the rate of disappearance of the intermediate:
tdtermediateind
tdtermediateind ][][
−=
77
Example• 2O3 → 3O2
• Empirically derived rate equation = k [O3]2/[O2]
SLOWStepOOO
clearlyreactionbackwardstheforsitthatyoudsreitaskpreferIkitcallsourcessomereactionbackwardsforkratek
FASTStepcatalystOcatalystOO
reactionforwardforkratekFASTStepcatalystOOcatalystO
MECHANISM
k
k
k
32
'min)(
2
1:
23
1
21
32
1
23
3
1
1
⎯→⎯+
=+⎯→⎯++
=++⎯→⎯+
−
−
−
78
• The constant of a reverse reaction is indicated by (-), e.g., k-1
• The intermediate in this mechanism is the oxygen ATOM, hence:
tdOd
tdOd ][][
−=
79
• Knowing this, we need to show that the empirical rate equation and mechanism are consistent with each other.
80
][][
][][
][][][][
][][][][][
][][][][][][][''
21
31
2131
2131
2131
OOkOk
OforSolveOOkOk
catalystoutCancelcatalystOOkcatalystOk
TwoTheEquateNow
catalystOOktdOdcatalystOk
tdOd
RxndBckwofRateRxndForofRate
=
=
=
=−=
−
−
−
−
81
• Stop temporarily• REMEMBER: the rate limiting step is #3;
the rate equation for this step = k3 [O] [O3]• So, substitute the solution for [O] from the
first for’d and bkw’d reactions in to the rate equation for Step 3:
82
equationratederivedyempiricalltoidenticalIsOO
kequation
substituteandk
kkkLet
OOkOk
k
][][
:
][*][
][*
2
23
1
13
321
313
=
=−
−
83
• This series of kinetic equations fulfills the 2 requirements:
• 1) Accounts for the products, and,• 2) Explains the observed rate law
84
Second Example• For 2NO + O2 → 2NO2, the empirical rate equation is:
k [NO]2 [O2]
• The postulated mechanism follows:
SLOWStepNONONO
FASTStepNOONO
k
k
k
22
1
223
32
2
1
1
⎯→⎯+
⇔+−
85
• Show that the empirical rate equation and mechanism are consistent with each other.
• Use the methodology we used in the previous example.
86
tdNOd
tdNOd
andNONONO
ONONO
NOONO
k
k
k
][][
2
33
23
23
32
2
1
1
−=
⎯→⎯+
+⎯→⎯
⎯→⎯+−
87
ytemporarilhereStop
NOk
ONOk
NOforSolveNOkONOk
twotheEquate
NOktd
NOdONOk
tdNOd
EquationRatedBkwEquationRatedFor
][][][:][
][][][:
][][
][][][
''
31
21
3
3121
313
213
=
=
=−=
−
−
−
88
• The rate limiting step is step 3.• The rate equation for step 3 is: k2 [NO3]
[NO]• Substitute the solution for [NO3] into the
rate equation for step 3.
89
][][][][][ 22
21
21 ONOkNOONOk
kk=⎟⎟
⎠
⎞⎜⎜⎝
⎛
−
90
Apply This to Enzymes• Enzymes are, with a couple of exceptions,
proteins• Enzymes are biological catalysts• Enzymes speed up biological reactions
incredibly• For this discussion:
– E = enzyme, – S = substrate and – P = product
91
tdESd
tdESd
ANDSLOWPEESFASTSEESFASTESSE
k
k
k
][][
2
1
1
−=
+⎯→⎯
+⎯→⎯
⎯→⎯+−
92
ytemporarilStop
ESSEkk
ESforSolveESkSEk
MethodShort
][][][
:][][][][
1
1
11
=
=
−
−
93
• Rate limiting step is step 3• Rate equation is: k2 [ES]• Substitute as before:
RxnUniUniaisThisPEESSE
reactionoveralltheWrite
SEkSEk
kk
−+→⇔+
=⎟⎟⎠
⎞⎜⎜⎝
⎛
−
:
][][][][1
21
94
Uni-Uni Reaction – Cleland Plot
95
Bi-molecular Reactions
• An enzyme catalyzed reaction may utilize 2 substrates.
• This reaction is always SEQUENTIAL, however,
• May be – ORDERED or – RANDOM
96
E.g., Ordered Sequential Reaction
E + X + Y → E + R + SE is still enzyme
X and Y are substratesR and S are products
97
Putative MechanismE + X → EX
EX + Y → EXYEXY → ERS SLOW STEP
ERS → ER + SER → E + S
And:
tdEXYd
tdERSd ][][
−=
98
ytemporarilhereStop
EXYEXYXEkkk
EXYforSolve
EXYkYEXkXEk
Equate
EXYktd
EXYdYEXkXEktd
ERSd
][][][][][
][
][][][][][
][][][][][][][
3
21
321
321
=⎟⎟⎠
⎞⎜⎜⎝
⎛
=
=−=
99
• Rate Limiting Step is: k3 [EXY]• Substitute:
][][][][
][][][][][][][][ 213
213
EXYXEk
EXYXEkkEXYXEkkkk
=
=
100
Kinetic Data Tells us:
• The following sequence MUST be taking place:
E + X + Y → EX (FIRST!) → EXY (SECOND!) → ERS → ES + R → E + S
• And is an Ordered Bi Bi Reaction
101
Ordered Bi Bi Reaction –Cleland Plot
102
• In the case where separate experiments about the same system give 2 different rate equations, e.g.,
k [E] [X] [Y] [EX]And
k [E] [X] [Y] [EY]
103
Mechanism = Random Sequential – Cleland Plot
104
• What, though, if an enzyme catalyzed a reaction that bound one substrate, released its product, then binds a SECOND substrate and releases ITS product?
Overall Reaction is: E + X + Y → E + R + S
105
Mechanism
E + X → EXEX → ER 1st rate limiting step
ER → E + RE + Y → EY
EY → ES 2d rate limiting stepES → E + S
106
• Note: while written unidirectionally, in many cases the reactions are reversible
• With 2 rate limiting steps, this reaction and its kinetics get ugly fast.
• This sort of reaction between 2 substrates and the 1 enzyme act like a ping pong game.
107
Ping Pong Mechanism –Cleland Plot
108
Function of Kinetics
• To Determine Reaction Mechanisms