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OFB Chapter 14 111/30/2004
Announcements, Nov. 19th
• PRS Quiz results through July 9are posted on the course website.
• July 19, 21– Review Exam 3 and exam analysis– Chapter 14
• Friday July 23 PRS Course Review– Bring PRS unit and calculator– Wrap-up course
• Monday July 26 Final Exam– 2:50-5:40 (this room)– 4 Crib sheets from Exam 1, 2, 3 and
Chapter 14– Part A (1-6), B (7-9), C (10-13), D (14)
Lecture topic: Chemical Kinetics– Check against Kinetics LabCheck
against Kinetics Lab– On Final Exam, NOT 3rd Exam
WebCT set posted (Homework 11) – Ch. 14.16(b), 24(b)– Due Wed. 24th, 1700
• Monday, Nov 22– 3rd Exam , usual format; topics:
Chs. 9-13 (omit 9.5-6, 12.4-5, 13.6-7)- Practice Exam, posted
• Wednesday, Nov 24– RSVP your attendance
• This announcement to be posted
OFB Chapter 14 211/30/2004
Chapter 14: Chemical Kinetics
• 14-1 Rates of Chemical Reactions
• 14-2 Rates & Concentrations• 14-3 The Dependence of
Concentrations on Time• 14-4 Reaction Mechanisms• 14-5 Reaction Mechanism and
Rate Laws• 14-6 Effect of Temperature on
Reaction Rates• 14-7 Kinetics of Catalysis
OFB Chapter 14 311/30/2004
Chapter 14: Chemical Kinetics
• Thus far, we have been using and talking about chemical reactions– Reactants, products, and how much is involved– Chemical equilibrium is “dynamic”
• Chemical Kinetics* concerns how fast a reaction proceeds– Kinetics = Rates of Chemical Reactions– And how to deduce reaction mechanisms from
observed rates of reactions– Activation Energy (Role of Catalysts)
How to analyze, quantify, and predict observed reaction rates?
*Kinetics, Kinematics, Cinematics, Kino
OFB Chapter 14 411/30/2004
Reaction Mechanisms
• Most reactions, as written, actually proceed through a series of steps
• Each Step is called an elementary reaction
• Classes of elementary reactions:1. Unimolecular (a single reactant)
A → B + C (a decomposition)2. Bimolecular (very common)
A + B → products3. Termolecular (less likely event)
A + B + C → products
OFB Chapter 14 511/30/2004
tX
ttXX
sLmolsL
molsLmol
tX
if
if
∆∆
∆∆
][][][ rate average
/ are units
][ rate reaction average
11
=−−
=
••=•
=
=
−−
To measure rates, monitor disappearance of reactants or appearance of products
e.g., NO2 + CO → NO + CO2
∆t]∆[CO
∆t∆[NO]
∆t∆[CO]
∆t]∆[NOrate rxn
2
2
+=+=
−=−=
OFB Chapter 14 611/30/2004
Role of stoichiometryTo measure rates we could monitor the disappearance of reactants or appearance of products
e.g., 2NO2 + F2 → 2NO2F
tFNO
tF
tNOrate
∆∆
∆∆
∆∆
][21
][][21
2
22
+=
−=
−=
OFB Chapter 14 711/30/2004
For a Generalized Reaction
aA + bB → cC + dD
tD
dtC
c
tB
btA
arate
∆∆
∆∆
∆∆
∆∆
][1][1
][1][1
=
+=
−=
−=
tX
ttXX
if
if
∆∆ ][][][
rate average =−−
=
NO2 + CO → NO + CO2
OFB Chapter 14 811/30/2004
Order of a Reaction
→ 221
252 2NO OkON + decomposition
][ 52ONkrate =Called a Rate Expression
Or Rate Law
Called a Rate constant f(Temperature)
→
k[A]rate
k aA n
productse.g.,
=An “nth order” reaction
OFB Chapter 14 911/30/2004
Summary
aA + bB → cC + dD
tD
dtC
c
tB
btA
arate
∆∆
∆∆
∆∆
∆∆
][1][1
][1][1
=
+=
−=
−=
nk[A]rate
products kaA
=
→ Called a Rate Expression
Or Rate Law
nth order reaction
→nm
k
[B]k[A] rateproducts bB aA
=
+
Overall order = m + n
mth order in [A] and nth order in [B]
OFB Chapter 14 1011/30/2004
Example 14-2At elevated temperatures, HI reacts according to the chemical equation
2HI → H2 + I2at 443°C, the rate of reaction increases with concentration of HI, as shown in this table.
Data [HI] RatePoint (mol L-1) (mol L-1 s-1)
1 0.005 7.5 x 10-4
2 0.01 3.0 x 10-3
3 0.02 1.2 x 10-2
a) Determine the order of the reaction with respect to HI and write the rate expressionb) Calculate the rate constant and give its unitsc) Calculate the instantaneous rate of reaction for a [HI] = 0.0020M
OFB Chapter 14 1111/30/2004
( )( )n
n
HIkrate
HIkrate
22
11
][
][
=
=22
k I H 2HI +→
( )( )
( )
part A toanswer 2
n
n
4
3
n1
n2
1
2
pointsdatatwoanyofratio the take
k[HI] rate
and HI inorder second 2n24
0.00500.010
7.5x103.0x10
[HI]k[HI]k
raterate
=
∴==
=
=
−
−
Data [HI] RatePoint (mol L-1) (mol L-1 s-1)
1 0.005 7.5 x 10-4
2 0.01 3.0 x 10-3
3 0.02 1.2 x 10-2
a) Determine the order of the reaction with respect to HI and write the rate expression
OFB Chapter 14 1211/30/2004
Example 14-2 b) Calculate the rate constant and give its units
part A toanswer 2 k[HI] rate =
B toanswer 1-1-
21-
1- 1-4
2
smol L 30k
)L mol (0.0050s L mol 7.5x10
[HI]ratek
=
==−
Example 14-2c) Calculate the instantaneous rate of
reaction for a [HI] = 0.0020M
( )( )C toanswer
1-1-4-
21-1-1-
2
sLmol 10x 1.2
L mol 0.0020s mol L 30
k[HI]rate
=
=
=
OFB Chapter 14 1311/30/2004
Similarly for two or more concentrations
nmp isorder ][][ +== nm BAkrate→products bB aA k+
1)1()1(arek of units −−−− sLmol pp
C B AExample
→+( )( )n1
n2
1
2
pointsdata two anyofratiothetake
[X]k[X]k
raterate
=
initial Rate
[A] [B] mol L-1 s-1
1.0x10-4 1.0x10 -4 2.8x10 -6
1.0x10-4 3.0x10 -4 8.4x10 -6
2.0x10-4 3.0x10 -4 3.4x10 -5
OFB Chapter 14 1411/30/2004
C B AExampleFor →+ initial Rate
[A] [B] mol L-1 s-1
1.0x10-4 1.0x10 -4 2.8x10 -6
1.0x10-4 3.0x10 -4 8.4x10 -6
2.0x10-4 3.0x10 -4 3.4x10 -5
nm BAkrate ][][= ( )( )n1
n2
1
2
pointsdatatwo any ofratiothetake
[X]k[X]k
raterate
=
1st When [A] is constant (1.0x 10-4),
When [B] increases times 3 and rate increases times 3
1
1
][][
][][
BAkrate
BBn
n
=
=
∴
OFB Chapter 14 1511/30/2004
C B AExampleFor →+ initial Rate
[A] [B] mol L-1 s-1
1.0x10-4 1.0x10 -4 2.8x10 -6
1.0E10-4 3.0x10 -4 8.4x10 -6
2.0x10-4 3.0x10 -4 3.4x10 -51][][ BAk m=rate
2nd When [B] is constant (3.0x 10-4),
[A] increases times 2 & rate increases times 4
12
2
][][
24]2[
4][
BAkrate
m
A m
=∴
==
=( )( )m1
m2
1
2
pointsdata two anyofratiothetake
[X]k[X]k
raterate
=
What is the rxn order with respect to
A? B? Overall?
OFB Chapter 14 1611/30/2004
Can now solve for k
1-2- 26
1424
6
12
12
s molL2.8x10
][1x10][1x102.8x10
[B][A]ratek
[B]k[A]rate
=
=
=
=
−−
−
1)1()1(are k of units −−−− smolL pp
Actually Example 13-4
2 NO + O2 → 2 NO2
][][ 22 ONOkrate =
OFB Chapter 14 1711/30/2004
14-3 The Dependence of Concentrations on Time
First Order Reactions
→
k[A]∆t∆[A]1rate
aA
n
productsk
orderfirst for general, in
=−=a
kt0
A of conc initial theis0if
e[A][A]
[A] −= For now, just
accept this important formula
a term that diminishes [A]0
OFB Chapter 14 1811/30/2004
kt0e[A][A]
ReactionOrder 1st afor LawRateIntegrated
−=If we take the ln of both sides
ktAA −= 0]ln[]ln[
Recall y = mx + b or
y = b + mxA plot of ln [A] vs t will be a straight line with the
Intercept = ln[A]0
Slope = -k
OFB Chapter 14 1911/30/2004
ktAA −= 0]ln[]ln[First Order Reaction
-6
-5
-4-3
-2
-1
00.0E+00 2.0E+04 4.0E+04 6.0E+04
Time (in seconds)
ln [A
} (in
mol
/ L
Slope = -k
k = 1.72 x 10-5 s-1
OFB Chapter 14 2011/30/2004
Second Order Reactions
n
k
k[A]∆t∆[A]
21- rate
products A 2order secondfor General,In
==
→
From calculus
o[A]12kt
[A]1
ReactionOrder 2nd afor Law Rate Integrated
+=
0][1intercept
2slope][
12[A]1
A
kA
kto
=
=
+=
This is also the equation for a straight line
y = mx +b
OFB Chapter 14 2111/30/2004
Slope = 2k
Or
k = ½ slope
OFB Chapter 14 2211/30/2004
oAkt
][12
[A]1
+=
Second Order Reaction
0
50
100
150
200
250
0 500 1000
Time (in seconds)
1/[A
] (L
/mol
)
Slope = 2k
OFB Chapter 14 2311/30/2004
Summary
Integrated Rate Laws
First Order Reactions
ktAA −= 0]ln[]ln[
Second Order Reactions
oAkt
][12
[A]1
+=
OFB Chapter 14 2411/30/2004
In practice, if we don’t know the order of the reaction
[A]n plot both
If a plot of ln [A] vs t is a straight line, then the reaction is 1st order
First Order Reaction
-6-5-4-3-2-10
0.0E+00 2.0E+04 4.0E+04 6.0E+04
Time (in seconds)
ln [A
} (in
mol
/ L
Second Order Reaction
050
100150200250
0 500 1000
Time (in seconds)
1/[A
] (L
/mol
)
If a plot of 1/ [A] vs t is a straight line, then the reaction is 2nd order
OFB Chapter 14 2511/30/2004
In a study of the reaction of pyridine (C5H5N) with methyl iodide (CH3I) in a benzene solution, the following initial reaction rates were measured at 25ºC for the different concentrations of the two reactants.
a) Write the rate law for this reactionb) Calculate the rate constant and give its unitsc) Predict the initial reaction rate that would be seen in a solution in which [C5H5N] = 5.0 x 10-5 M and [CH3I] = 2.0 x 10-5 M
In a study of the reaction of pyridine (C5H5N) with methyl iodide (CH3I) in a benzene solution, the following initial reaction rates were measured at 25ºC for the different concentrations of the two reactants.
b) Calculate the rate constant and give its unitsc) Predict the initial reaction rate that would be seen in a solution in which [C5H5N] = 5.0 x 10-5 M and [CH3I] = 2.0 x 10-5 M
]][[ 553 NHCICHkrate =∴
OFB Chapter 14 2911/30/2004
Chemical EquilibriumA direct connection exists between the equilibrium constant of a reaction that takes place in a sequence of steps and the rate constants in each step.
dDcCrkfk
bBaA +←
→+
a.) at equil: forward reaction rate = reverse reaction rate
b.) Keq = kf / kr
eqba
dc
r
f
dcr
baf
dcr
baf
K[B][A][D][C]
kk 2.)
[D][C]k[B][A]k 1.)
[D][C]kratereaction Reverse
[B][A]kratereaction Forward
==
=
=
=
OFB Chapter 14 3011/30/2004
a.) at equil: forward rate = reverse rateb.) Keq = kf / kr (sometimes kn / k-n)
A general reaction to illustrate this principle.
2 A (g) + B (g) ↔ C (g) + D (g)
Suppose the reaction proceeds by the following two step mechanism
1.) A (g) + A (g) ↔ A2 (g)
2.) A2 (g) + B (g) ↔ C (g) + D (g)
k1
k2
k-1
k-2
Rate-1 = k-1[A2]
Rate-2 = k-2[C][D]
Rate1 = k1[A]2
Rate2 = k2[A2][B]
22
1
11 ][
][AA
kkK ==− ]][[
]][[
22
22 BA
DCkkK ==−
][][]][[
]][[]][[
][][
22
22
2
2
1
121
BADC
BADC
AAK
kk
kkKKK
=
=
==
−−
OFB Chapter 14 3111/30/2004
PRS Quiz, Question 1In a study of the reaction of compound A and compound B in solution, the following initial reaction rates were measured at 25ºC for the different concentrations of the two reactants.
[A] [B] Rate(mol L-1) (mol L-1) (mol L-1 s-1)
1.0 x 10-4 1.0 x 10-4 3.0 x 10-6
1.0 x 10-4 3.0 x 10-4 9.0 x 10-6
2.0 x 10-4 3.0 x 10-4 3.6 x 10-5
Q1 What is the order of the reaction with respect to compound A?
1. 02. 13. 24. 3
OFB Chapter 14 3211/30/2004
OFB Chapter 14 3311/30/2004
PRS Quiz, Solution 1In a study of the reaction of compound A and compound B in solution, the following initial reaction rates were measured at 25ºC for the different concentrations of the two reactants.
[A] [B] Rate(mol L-1) (mol L-1) (mol L-1 s-1)
1.0 x 10-4 1.0 x 10-4 3.0 x 10-6
1.0 x 10-4 3.0 x 10-4 9.0 x 10-6
2.0 x 10-4 3.0 x 10-4 3.6 x 10-5
Q1 What is the order of the reaction with respect to compound A?
1. 02. 13. 24. 3
( )( )
( )( )
224
1x10
2x109x10
36x10
[A]k[A]k
raterate
n4-
n4-
6-
6-
n2
n3
2
3
==
=
=
n
n
OFB Chapter 14 3411/30/2004
PRS Quiz, Question 2In a study of the reaction of compound A and compound B in solution, the following initial reaction rates were measured at 25ºC for the different concentrations of the two reactants.
[A] [B] Rate(mol L-1) (mol L-1) (mol L-1 s-1)
1.0 x 10-4 1.0 x 10-4 3.0 x 10-6
1.0 x 10-4 3.0 x 10-4 9.0 x 10-6
2.0 x 10-4 3.0 x 10-4 3.6 x 10-5
Q2 What is the order of the reaction with respect to compound B?
1. 02. 13. 24. 3
OFB Chapter 14 3511/30/2004
OFB Chapter 14 3611/30/2004
( )( )
( )( )
133
1x10
3x103x109x10
[B]k[B]k
raterate
n4-
n4-
6-
6-
n1
n2
1
2
==
=
=
n
n
PRS Quiz, Solution 2In a study of the reaction of compound A and compound B in solution, the following initial reaction rates were measured at 25ºC for the different concentrations of the two reactants.
[A] [B] Rate(mol L-1) (mol L-1) (mol L-1 s-1)
1.0 x 10-4 1.0 x 10-4 3.0 x 10-6
1.0 x 10-4 3.0 x 10-4 9.0 x 10-6
2.0 x 10-4 3.0 x 10-4 3.6 x 10-5
Q2 What is the order of the reaction with respect to compound B?
1. 02. 13. 24. 3
OFB Chapter 14 3711/30/2004
PRS Quiz, Question 3In a study of the reaction of compound A and compound B in solution, the following initial reaction rates were measured at 25ºC for the different concentrations of the two reactants.
[A] [B] Rate(mol L-1) (mol L-1) (mol L-1 s-1)
1.0 x 10-4 1.0 x 10-4 3.0 x 10-6
1.0 x 10-4 3.0 x 10-4 9.0 x 10-6
2.0 x 10-4 3.0 x 10-4 3.6 x 10-5
Q3 What is the rate expression for the above reaction?
1. Rate = k [A]2. Rate = k [A][B]3. Rate = k [A]2[B]2
4. Rate = k [A]2[B]
OFB Chapter 14 3811/30/2004
OFB Chapter 14 3911/30/2004
PRS Quiz, Solution 3In a study of the reaction of compound A and compound B in solution, the following initial reaction rates were measured at 25ºC for the different concentrations of the two reactants.
[A] [B] Rate(mol L-1) (mol L-1) (mol L-1 s-1)
1.0 x 10-4 1.0 x 10-4 3.0 x 10-6
1.0 x 10-4 3.0 x 10-4 9.0 x 10-6
2.0 x 10-4 3.0 x 10-4 3.6 x 10-5
Q3 What is the rate expression for the above reaction?
1. Rate = k [A]2. Rate = k [A][B]3. Rate = k [A]2[B]2
4. Rate = k [A]2[B]
OFB Chapter 14 4011/30/2004
dDcCrkfk
bBaA +←
→+
tD
dtC
ctB
btA
arate
∆∆
=
∆∆
+=
∆∆
−=
∆∆
−=
][1][1][1][1
nmf [B][A]k rate =
Called a Rate Expression Or Rate Law
kf called rate constant
a.) at equil: forward rate = reverse rateb.) Keq = kf / kr
eqba
dc
r
f
dcr
baf
dcr
baf
K[B][A][D][C]
kk 2.)
[D][C]k[B][A]k 1.)
[D][C]kratereaction Reverse
[B][A]kratereaction Forward
==
=
=
=
OFB Chapter 14 4111/30/2004
14-5 Reaction Mechanism & Rate Laws
Typically with a reaction one of several elementary step reaction is the slowest step. This is called the Rate Determining Step (RDS)
Case #1: When the RDS occurs first, the first step is slow and determines the rate of the overall reaction.
]][F[NOk rateRDS theis 1 Step
(fast) FNO F NO 2.)
(slow) FFNO F NO 1.)
221
2k
2
2k
22
2
1
=
→+
+→+
F22NO 2F 22NO →+
OFB Chapter 14 4211/30/2004
]][F[NOk rateRDS theis 1 Step
(fast) FNO F NO 2.)
(slow) FFNO F NO 1.)
221
2k2
2
2k1
22
=
→+
+→+
Reaction Progress
Energy
F + NO2
NO2+ F2
slow fast
NO2F
F22NO 2F 22NO →+
OFB Chapter 14 4311/30/2004
Case #2: When the RDS occurs after one or more Fast steps, mechanisms are often signaled by a reaction order greater than two. The slow step determines the overall rate of the reaction.
22 2NO O 2NOreactionOverall→+
3 molecule reaction. Is it A Termolecular or Bimolecular reactions? Three way collisions are rare. Try a two step mechanism.
]][OO[Nk rate
(slow) 2NO O ON 2.)
(fast) ON NO NO 1.)
2222
2k2
222
221-k
1k
=
→+
←→
+
But N2O2 is a reactive intermediate
OFB Chapter 14 4411/30/2004
]][OO[Nk rate
(slow) 2NO O ON 2.)
(fast) ON NO NO 1.)
2222
2k2
222
221-k
1k
=
→+
←→
+
Reaction Progress
Energy N2O2 + O2
slow
fast
2NO2NO2
22 2NO O 2NO →+
OFB Chapter 14 4511/30/2004
→←
→
]][OO[Nk rate
(slow) 2NO O ON 2.)
(fast) ON NO NO 1.)
2222
2k2
222
221-k
1k
=
+
+
Need to express [intermediates] in terms of other reactants
]O[Nk rate reverse[NO]k rate forward
221-
21
==
2[NO]
]2O2[N
1-k1k
1K
]2O2[N1-k 2[NO]1k
rate reverse rate forward that recall mequilibriuat
==
=
=
2122 [NO]K ]O[N =∴
OFB Chapter 14 4611/30/2004
2122 [NO]K ]O[N =∴
]][OO[Nk rate
(slow) 2NO O ON 2.)
(fast) ON NO NO 1.)
2222
2k2
222
221-k
1k
=
→+
←→
+
Substituting for [N2O2] in the rate expression above
][O[NO]Kk rate 22
12=
OFB Chapter 14 4711/30/2004
→←
→
]][OO[Nk rate
(slow) 2NO O ON 2.)
(fast) ON NO NO 1.)
2222
2k2
222
221-k
1k
=
+
+
Reaction Progress
Energy N2O2 + O2
slow
fast
2NO2NO2
][O[NO]Kk rate 22
12=
22 2NO O 2NO →+
1
11 k
kK−
=
OFB Chapter 14 4811/30/2004
Reaction Mechanism
• Intermediates
•Transition states
Reaction Progress
Energy
slowfast
fast
intermediates
Transition States
OFB Chapter 14 4911/30/2004
Transition States
OFB Chapter 14 5011/30/2004
14-6 The Effect of Temperature on Reaction Rates
RTE a
eA k
Equation Arrenhius−
=
k of units the has andconstant a isfactor" lexponentia-pre" the is A
andmoleper energy are units
energy n Activatiothe is aE where
OFB Chapter 14 5111/30/2004
Aek RTEa
EquationArrenhius
−=
RTElnA lnk a−=
y = mx + bAn Arrenhius Plot
16
16.5
17
17.5
18
18.5
0.0025 0.0027 0.0029 0.0031 0.0033 0.0035
1/T (K-1)
ln k
Slope = - Ea / R
OFB Chapter 14 5211/30/2004
Reaction Progress
Energy
∆E= minus
exothermic
Ear
Eaf
Transition State
∆E = Eaf - Ea
r
The Activation Energy (Ea) is the minimum collision energy that reactants must have in order to form products
OFB Chapter 14 5311/30/2004
Reaction Progress
Energy
∆E= positive
endothermic
Ear
Eaf
Transition State
∆E = Eaf - Ea
r
The Activation Energy (Ea) is the minimum collision energy that reactants must have in order to form products
OFB Chapter 14 5411/30/2004
Chapter 14Chemical Kinetics
• Catalyst– provides a lower energy path, but
it does not alter the energy of the starting material and product
– rather it changes the energy of the transition (s), in the reaction
– A catalyst has no effect on the thermodynamics of the overall reaction
• Inhibitor– is a negative catalyst. It slows the
rate of a reaction frequently by barring access to path of low Eaand thereby forcing the reaction to process by a path of higher Ea.
OFB Chapter 14 5511/30/2004
OFB Chapter 14 5611/30/2004
Kinetics of Catalysis• A catalyst has no effect on the
thermodynamics of the overall reaction
• It only provides a lower energy path• Examples
– Pt and Pd are typical catalysts for hydrogenation reactions (e.g., ethylene to ethane conversion)
– Enzymes act as catalysts• Phases
– Homogenous catalysis – the reactants and catalyst are in the same catalyst (gas or liquid phase)
– Heterogeneous catalysis – reaction occurs at the boundary of two different phases (a gas or liquid at the surface of a solid)