chemical kinetics chapter 17, kinetics fall 2009, chem 1310 1
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CHEMICAL KINETICSCHAPTER 17, KineticsFall 2009, CHEM 1310
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KeKin
Kinetics vs
Thermodynamics
A: ReactantsB: Transition state
C: productsE: Forward Activation
Free EnergyF: Reverse Activation
Free Energy
• Most reactions occur through several steps and are not single step reactions.
• Each step in a multi-step reaction is called an elementary reaction.
Types of elementary reactions1. Unimolecular (a single reactant)
2. Bimolecular
3. Termolecular (very unlikely)
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Reaction Mechanisms
• Each step of this reaction is an “elementary step”. Each elementary step has reactant(s), a transition state, and product(s). Products that are consumed in subsequent elementary reaction are called intermediates.
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Reaction Rates: To measure a reaction rate we could monitor the disappearance of reactants or appearance of products.
e.g., 2NO2 + F2 → 2NO2F
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Gen. Rxn: aA + bB → cC + dD
tD
dtC
c
tB
btA
arate
][1][1
][1][1
tX
tt
XX
if
if
][][][
rate average
NO2 + CO → NO + CO2
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Order of a ReactionThe power (n) to which the concentration of A is raised in the rate expression describes the order of the reaction with respect to A.
k[A]rate
Products k
aA n
e.g.,
Do not confuse the order (n) with the stoichiometric coefficient (a).
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mth order in [A]
nth order in [B]
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aA + bB k→ products
rate = k[A]m[B]n
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as before
rate = k[A] m[B] n
= −1
a ⎛ ⎝
⎞ ⎠d[A]
dt= −
1
b ⎛ ⎝
⎞ ⎠d[B]
dt
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A + B → C
rate = k[A]m[B]n (this is the "rate law"
for this reaction)
If you know rates at various concentrations,
take the ratio of rates, solve for n or m.
rate2
rate1
=k [A]2( )
m[B]2( )
n
k [A]1( )m
[B]2( )n
[A] (mol L-1) [B] (mol L-1) Rate (mol L-1 s-1)
1 1.0x10-4 1.0x10 -4
2.8x10 -6
2 1.0x10-4 3.0x10 -4
8.4x10 -6
3 2.0x10-4 3.0x10 -4
3.4x10 -5
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[A] (mol L-1) [B] (mol L-1) Rate (mol L-1 s-1)
1 1.0x10-4 1.0x10 -4
2.8x10 -6
2 1.0x10-4 3.0x10 -4
8.4x10 -6
3 2.0x10-4 3.0x10 -4
3.4x10 -5
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rate2
rate3
=k [A]2( )
m[B]2( )
n
k [A]3( )m
[B]3( )n
8.4 x 10-6( )
3.4 x 10-5( )=
1
4=
1m
2m
m = 2
Example: At elevated temperatures, HI reacts according to the chemical equation
2HI → H2 + I2The 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 0.010 3.0 x 10-3
3 0.020 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 11
INTEGRATED RATE LAWS
• Single Reactant (three cases)– Zero-Order Rate Law (n = 0)– First-Order Rate Law (n = 1)– Second-Order Rate Law (n = 2)
• More than one Reactant– Must state the order of the reaction with respect
to each reactant (rate = k[A]n[B]m[C]p)
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A → B
rate = k[A]n
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A k ⏐ → ⏐ B
rate = −d[A]
dt= k[A]n
rate =d[A]
[A]nAo
A
∫ = −k dtt=0
t
∫
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INTEGRATED RATE LAWSn=0,1,2
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In the real world, if we do not know the order of the reaction we can use experimental plots
to estimate the order.If a plot of [A] vs t is a straight line, then the reaction is zero order.
If a plot of ln[A] vs t is a straight line, then the reaction is 1st order.
If a plot of 1/ [A] vs t is a straight line, then the reaction is 2nd order.
First Order Reaction
-6
-5
-4
-3
-2
-1
0
0.0E+00 2.0E+04 4.0E+04 6.0E+04
Time (in seconds)
ln [
A}
(in
mo
l /
L
Second Order Reaction
0
50
100
150
200
250
0 500 1000
Time (in seconds)1/
[A] (
L /m
ol)
INTEGRATED RATE LAWS
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for zero order (n = 0)
A k ⏐ → ⏐ products
rate = −d[A]
dt= k[A]0
Zero Order Reactions
[A] - [Ao] = -kt
Graph [A] vs t
Slope = -k, intercept = [Ao]
INTEGRATED RATE LAWS
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for first order (n =1)
A k ⏐ → ⏐ products
rate = −d[A]
dt= k[A]1
First Order Reactions
ln[A]-ln[Ao] = -kt
Graph ln[A] vs t
Slope = -k, intercept = [Ao]
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ktAA 0]ln[]ln[
In[N205] versus time.
Slope = - k
2 N2O5 (g) → 4 NO2 (g) + O2 (g)
This graph gives a straight line, and so is
First order with respect to the decomposition of N2O5
If a plot of ln[A] vs t is not a straight line, the
reaction is not first order!
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Dimerization
Data set provided [C4H6] vs time
2 C4H6 (g) → C8H12 (g) [C4H6]˚ = 0.01M
• Most reactions proceed not through a single step but through a series of steps
• Each Step is called an elementary reaction
Types of elementary reactions1. Unimolecular (a single reactant)
E.g., A → B + C (a decomposition)
2. Bimolecular (most common type)E.g., A + B → products
3. Termolecular (less likely event)E.g., A + B + C → products
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Reaction Mechanisms
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(fast) CO NO CO NO 2.)
(slow) NO NO NO NO 1.)
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Notice that NO3 is formed and consumed. This is called a __________________________________.Notice also that Step 1 is bimolecular and Step 2 is bimolecular
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CO NO NO NO
CO NO 2NO
sum
CO NO CO NO overall 22
CHEMICAL EQUILIBRIUM
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A direct connection exists between the equilibrium constant of a reaction and the rate constants.
a) at equilibrium: forward reaction rate = reverse reaction rate.
b) Keq = kf / kr (same as K = k1/k-1)
A ⇌ B
kf
kr
REACTION MECHANISM & RATE LAWS
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(fast) F2NO k
F 2NO 2.)
(slow) FF2NOk
2F 2NO 1.)
2
1
F2
2NO 2
F 2
2NO
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.
Example 15.6
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Reaction Progress
Energy
F + NO2F
NO2+ F2
slow fast
NO2F
RDS theis 1 Step
(fast) F2NO k2
F 2NO 2.)
(slow) FF2NOk1
2F 2NO 1.)
F2
2NO 2
F 2
2NO
Chem 1310 Spring 2009 stop here
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Reaction Progress
Energy
]][OO[Nk rate
(slow) 2NO O ON 2.)
(fast) ON NO NO 1.)
2222
2k2
222
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1-k
1k
22 2NO O 2NO
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Need to express [intermediates] in terms of other reactants
]][OO[Nk rate
(slow) 2NO O ON 2.)
(fast) ON NO NO 1.)
2222
2k2
222
22
1-k
1 k
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]][OO[Nk rate
(slow) 2NO O ON 2.)
(fast) ON NO NO 1.)
2222
2k2
222
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1-k
1k
Substituting for [N2O2] in the rate expression above
22 2NO O 2NO
reaction Overall
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Reaction Progress
Energy
N2O2 + O2
slow
fast
2NO2NO2
]][OO[Nk rate
(slow) 2NO O ON 2.)
(fast) ON NO NO 1.)
2222
2k2
222
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1-k
1 k
][O[NO]Kk rate 22
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22 2NO O 2NO
1
11 k
kK
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Reaction Progress
Energy
Reaction Mechanism
• Intermediates
• Transition states
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A MODEL FOR CHEMICAL KINETICS
RT
E a
eA k
Equation Arrenhius
k of units the has andconstant a is
factor" lexponentia-pre" the is A
and
moleper energy are units
energy n Activatiothe is a
E
where
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ave
ave
or
PV 2 = RT = (KE)3n
3(KE) = RT2
Chapter 5: The Kinetic Molecular Theory of Gases
M
3RTu2
The Meaning of Temperature: temperature is a measure of the average kinetic energy of the gas particles.
The Kelvin temperature of a gas is a measure of the random motions of the particles of gas. With higher temperature, greater motion.
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Chapter 5: Speed Distribution Curves
Maxwell-Boltzmann speed distribution
Temperature is a measure of the average kinetic energy of molecules when their speeds have Maxwell Boltzmann distribution. i.e., the molecules come to thermal equilibrium.
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Transition State, also called Activated Complex
Two requirements must be satisfied for reactants to collide successfully to rearrange to form products
1.
2.
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Consider two different temperatures.
1.) Collisions must have enough energy to produce a reaction. Not all collisions have enough energy to make product
Ecollision > Eact
RT
E a
eA k
Num
ber o
f col
lisio
ns
Distribution of velocities
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2 BrNO (g) → 2 NO (g) + Br2 (g)2.) Molecular Orientation
Relative orientations of the reactants must allow formation of any new bonds to produce products.
Orientation a or b lead to product,
c does not.
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Find the rate constant k at several temperatures. Plot of In(k) versus 1/T for the reaction
Aek RTEa
Equation Arrenhius
RTE
lnA lnk a
y = mx + b
Slope =
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Reaction Progress
Energy
Ear
Eaf
Transition State
The Activation Energy (Ea) is the minimum collision energy that reactants must have in order to form products
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Reaction Progress
Energy
ΔE = Eaf - Ea
r
Ear
Eaf
Transition State
The Activation Energy (Ea) is the minimum collision energy that reactants must have in order to form products
CHEMICAL KINETICS
• Catalyst
• Inhibitor
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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)
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Effect of a catalyst on the number of reaction-producing collisions.