Transcript Slide 1

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The End!!!!!!!
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‫منابع‬
‫کتابخانه ‪‬‬
‫شیمی عمومی مورتیمر ‪‬‬
‫کتب ومنابع شیمی ‪‬‬
‫سایت های اینترنتی ‪‬‬
‫‪ www.asadipour.kmu.ac.ir‬‬
‫‪4‬‬
‫‪http:\\aliasadipour.kmu.ac.ir‬‬
‫‪911213.....56 slides‬‬
Chemical Kinetics
Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed?
Reaction rate is the change in the concentration of a
reactant or a product with time (M/s).
A
B
D[A]
rate = Dt
D[A] = change in concentration of A over
time period Dt
D[B]
rate =
Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is
negative.
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A
B
time
rate = -
rate =
D[A]
Dt
D[B]
Dt
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Reaction Rates
 Reaction rate = change in concentration
of a reactant or product with time.
 Three “types” of rates
 initial rate
 average rate
 instantaneous rate
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Br2 (aq) + HCOOH (aq)
2Br- (aq) + 2H+ (aq) + CO2 (g)
slope of
tangent
average rate = -
D[Br2]
Dt
=-
slope of
tangent
slope of
tangent
[Br2]final – [Br2]initial
tfinal - tinitial
instantaneous rate = rate for specific instance in time
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2NO22NO+O2
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Concentrations & Rates
0.3 M HCl
6 M HCl
Mg(s) + 2 HCl(aq) ---> MgCl2(aq) + H2(g)
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Reaction Order
 What is the rate expression for

aA + bB → cC + dD

Rate = k[A]x[B]y

where x=1 and y=2.5?
 Rate = k[A][B]2.5
 What is the reaction order?
 First in A, 2.5 in B
 Overall reaction order?
 2.5 +1 = 3.5
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Interpreting Rate Laws
Rate = k [A]m[B]n[C]p
 If m = 1, rxn. is 1st order in A
Rate = k [A]1
If [A] doubles, then rate goes up by factor of __
 If m = 2, rxn. is 2nd order in A.
Rate = k [A]2
Doubling [A] increases rate by ________
 If m = 0, rxn. is zero order.
Rate = k [A]0
If [A] doubles, rate ________
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Deriving Rate Laws
Derive rate law and k2for
Rate of rxn = k [CH3CHO]
????
CH3CHO(g) → CH
(g) + CO(g)
Therefore, we say this reaction4 is _____ order.
from experimental data for rate of
Here the rate goes
by 3_____
disappearance
ofupCH
CHO when initial conc. doubles.
Expt.
1
2
3
4
[CH3CHO]
(mol/L)
0.10
0.20
0.30
0.40
Disappear of CH3CHO
(mol/L•sec)
0.020
0.081
0.182
0.318
Now determine the value of k. Use expt. #3 data— 1,2,3 or 4
R= k [CH3CHO]2
0.182 mol/L•s = k (0.30 mol/L)2
k = 2.0 (L / mol•s)
Using k you can calc. rate at other values of [CH3CHO] at same T.
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Expression of R &K
A  3B(aq )  2 H  (aq )  X
– A series of four experiments was run at different concentrations, and
the initial rates of X formation were determined.
– From the following data, obtain the reaction orders with respect to A,
B, and H+.
– Calculate the numerical value of the rate constant.
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Presentation of Lecture Outlines, 14–17
Expression of R &K
Exp. 1
Exp. 2
Exp. 3
Exp. 4
Initial Concentrations (mol/L)
A
B
H+
Initial Rate [mol/(L.s)]
0.010
0.010
0.00050
1.15 x 10-6
0.020
0.010
0.00050
2.30 x 10-6
0.010
0.020
0.00050
2.30 x 10-6
0.010
0.010
0.00100
1.15 x 10-6
– Comparing Experiment 1 and Experiment 2, you see that when the
– A concentration doubles (with other concentrations constant),
– the rate doubles.
– This implies a first-order dependence with respect to A.
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Presentation of Lecture Outlines, 14–18
Expression of R &K
Exp. 1
Exp. 2
Exp. 3
Exp. 4
Initial Concentrations (mol/L)
A
B
H+
Initial Rate [mol/(L.s)]
0.010
0.010
0.00050
1.15 x 10-6
0.020
0.010
0.00050
2.30 x 10-6
0.010
0.020
0.00050
4.60 x 10-6
0.010
0.010
0.00100
1.15 x 10-6
– Comparing Experiment 1 and Experiment 3, you see that when the
– B concentration doubles (with other concentrations constant),
– the rate 4 times.
– This implies a second-order dependence with respect to B.
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Presentation of Lecture Outlines, 14–19
Expression of R &K
Exp. 1
Exp. 2
Exp. 3
Exp. 4
Initial Concentrations (mol/L)
A
B
H+
Initial Rate [mol/(L.s)]
0.010
0.010
0.00050
1.15 x 10-6
0.020
0.010
0.00050
2.30 x 10-6
0.010
0.020
0.00050
4.60 x 10-6
0.010
0.010
0.00100
1.15 x 10-6
– Comparing Experiment 1 and Experiment 4, you see that when
– the H+ concentration doubles (with other concentrations constant),
– the rate is the same.
– This implies a zero-order dependence with respect to H+.
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Presentation of Lecture Outlines, 14–20
Expression of R &K
Exp. 1
Exp. 2
Exp. 3
Exp. 4
Initial Concentrations (mol/L)
A
B
H+
Initial Rate [mol/(L.s)]
0.010
0.010
0.00050
1.15 x 10-6
0.020
0.010
0.00050
2.30 x 10-6
0.010
0.020
0.00050
4.60 x 10-6
0.010
0.010
0.00100
1.15 x 10-6
 0
Rate  k[ A][ B] [ H ]  k[ A][ B]
2
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Presentation of Lecture Outlines, 14–21
2
Expression of R &K
Exp. 1
Exp. 2
Exp. 3
Exp. 4
Initial Concentrations (mol/L)
A
B
H+
Initial Rate [mol/(L.s)]
0.010
0.010
0.00050
1.15 x 10-6
0.020
0.010
0.00050
2.30 x 10-6
0.010
0.020
0.00050
4.60 x 10-6
0.010
0.010
0.00100
1.15 x 10-6
– You can now calculate the rate constant by substituting values from
– any of the experiments. Using Experiment 1 you obtain:
Rate  k[ A][ B]
4
2
1.15 10 6
mol
mol
mol 2
 k  (0.010
)  (0.010
)
Ls
L
L
k  1.2 10 L /(mol  s )
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2
Presentation of Lecture Outlines, 14–22
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Zero-Order Reactions
A
k=
product
rate
DA
rate = 
DT
=RATE= M/s
[A]0
rate = k
[A]0 =
k
-
D[A]
Dt
=k
 DA
 KDT
0

[A] = [A]0 - kt
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
ZERO ORDER REACTION
t½ = t when [A] = [A]0/2
12
10
2k
8
[A]
t½ =
[A]0
6
4
2
0
0
5
10
15
TIME
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First-Order Reactions
A
k=
rate
[A]
product
M/s
=
M
rate = -
= 1/s or s-1
t½ = t when [A] = [A]0/2
0.693
t½ =
k
rate = k [A]
Dt
-
D[A]
Dt
= k [A]
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
[A] = [A]0exp(-kt)
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D[A]
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ln[A] = ln[A]0 - kt
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Second-Order Reactions
A
rate
k=
[A]2
1
[A]
rate = -
product
=
M/s
=
M2
1
[A]0
D[A]
rate = k [A]2
Dt
-
= 1/M•s
D[A]
= k [A]2
Dt
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
+ kt
SECOND ORDER REACTION
SECOND ORDER REACTION
0.6
t½ = t when [A] = [A]0/2
0.5
10
1
1/[A]
k[A]0
0.4
8
[A]
t½ =
12
6
4
0.2
2
0.1
0
0
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0.3
1
2
TIME
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3
4
5
0
0
1
2
3
TIME
4
5
13.3
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 First step in
evaluating rate
data is to
graphically
interpret the
order of rxn
 Zeroth order: rate
does not change with
lower concentration
 First, second
orders:
Rate changes as a
function of
concentration
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Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order
0
Rate Law
rate = k
1
rate = k [A]
2
[A]2
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rate = k
Concentration-Time
Equation
[A] = [A]0 - kt
ln[A] = ln[A]0 - kt
1
[A]
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=
1
[A]0
+ kt
Half-Life
t½ =
t½ =
t½ =
[A]0
2k
ln2
k
1
k[A]0
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Collision Theory
O3(g) + NO(g) → O2(g) + NO2(g)
10 31 Collisin/Lit.S
Reactions require
(A) geometry
(B) Activation energy
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Three possible collision orientations-a & b produce reactions,
c does not.
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Collision Theory
 TEMPERATURE
 10 c0 T  100-300% RATE
 25 c0 T  35 c0
 2%  COLLISIN
 EFFECTIVE COLLISIONS
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① with sufficient energy
The colliding molecules must have a total kinetic energy equal to or
greater than the activation energy, Ea.
Ea is the minimum energy of collision required for two molecules to initiate a chemical reaction.
It can be thought of as the hill in below. Regardless of whether the
elevation of the ground on the other side is lower than the original position,
there must be enough energy imparted to the golf ball to get it over
the hill.
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The Collision Theory of Chemical Kinetics
 Activation Energy (Ea)-
the minimum amount of
energy required to
initiate a chemical
reaction.
 Activated Complex
(Transition State)- a
temporary species
formed by the reactant
molecules as a result of
the collision before they
form the product.
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Exothermic Reaction
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Endothermic Reaction
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Temperature Dependence of the Rate Constant
(Arrhenius equation)
k = A • exp( -Ea/RT )
2.45
2.4
2.35
K
2.3
2.25
Ea is the activation energy (J/mol)
2.2
2.15
R is the gas constant (8.314 J/K•mol)
2.1
2.05
0
500
1000
1500
TEMPERATURE
T is the absolute temperature
A is the frequency factor
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Arrhenius Equation
•
k = A • exp( -Ea/RT )
•
Can be arranged in the form of a
0.9
straight line
0.88
0.86
0.84
Ln K
 ln k = ln A-Ea/RT=(-Ea/R)(1/T) + ln A
0.82
0.8
0.78
0.76
 Plot ln k vs. 1/T 
0
0.0005
0.001
0.0015
0.002
0.0025
1/T
 slope = -Ea/R
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•k = A •
300400C0 Rate20000 Times
400500C0 Rate400 Times
e( -Ea/RT )
ln k = ln A-Ea/RT=(-Ea/R)(1/T) + ln A
Ea=60KJ/mol
300310C0  Rate2 Times
Ea=260KJ/mol
300310C0  Rate25 Times
If T
increases
Ea/RT
decreases
REACTION
SPEEDS UP
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k
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-Ea/RT
increases
e-Ea/RT
increases
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K in different T
Ln(K2/K1)=-Ea/R(1/T2-1/T1)
Log(K2/K1)=-Ea/2.303R(1/T2-1/T1)
Log(K2/K1) =Ea/2.303R(T2-T1)/T1T2
Log(K2/K1) =ΔH/2.303R(T2-T1)/T1T2
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Mechanisms
Rate = k [H2O2] [I-]
Most rxns. involve a sequence of elementary steps.
2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O
Proposed Mechanism
Rate = k [H2O2] [I-]
Step 1 — slow HOOH + I- --> HOI + OHStep 2 — fast
HOI + I- --> I2 + OHStep 3 — fast
2 OH- + 2 H+ --> 2 H2O
------------------------------------------------------------------ 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O
Rate can be no faster than slow step
RATE DETERMINING STEP,(
rds.)
The species HOI and OH- are intermediates.
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Mechanisms
NOTE
1. Rate law comes from experiment
2. Order and stoichiometric coefficients not
necessarily the same!
3.
Rate law reflects all chemistry including the
slowest step in multistep reaction.
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Mechanisms

2NO+F22NOF
 1=NO+F2NOF+F
 Rate limiting step(RDS)
 2=NO+F NOF
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R=K[NO][F2]
R1=K1[NO][F2]
R=R1
R2=K2[NO][F]
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SN1
OH- + (CH3)3CBr(CH3)3COH + BrR=K[(CH3)3CBr]
1)(CH3)3CBr(CH3)3C+ +BrR1=K1[(CH3)3CBr ]
2) (CH3)3C+ +OH- (CH3)3COH
R2=K2[(CH3)3C+] +[OH-]
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SN2
 OH- +CH3Br  CH3OH + Br-
 R=K[CH3Br][OH-]
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Rate Laws andMechanisms
Proposed
NO2 + CO →NO(g) + CO2(g)
Rate = k[NO2]2
Experimental
Two possible
mechanisms
Two steps
Two steps
is rational mechanism
Single step
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A catalyst is a substance that increases the rate of a
chemical reaction without itself being consumed.
k = A • exp( -Ea/RT )
Ea
k
uncatalyzed
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ratecatalyzed > rateuncatalyzed
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For the decomposition of hydrogen peroxide:
Br -
A catalyst(Br-) speeds up a reaction by lowering the activation e
It does this by providing a different mechanism by which the react
can occur. The energy profiles for both the catalyzed and uncataly
reactions of decomposition of hydrogen peroxide are shown in Fig
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1)Homogeneous Catalysis
N2O (g)  N2 (g) + O2 (g)
1)
N2O (g)  N2 (g) + O (g)
2)
O (g) + N2O (g)  N2(g)+ O2 (g)
Ea=240KJ/mol
Cl2 Catal.
------------------------------------------------------Cl2(g)2Cl(g)
N2O(g)+Cl(g)N2(g)+ClO(g)
2ClOCl2(g)+O2(g)
Ea=140KJ/mol
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2)Heterogeneous
Catalysis
C2H4 + H2  C2H6
with a metal catalyst
a. reactants
b. adsorption
c. migration/reaction
d. desorption
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Enzyme Catalysis
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Catalytic Converters
CO + Unburned Hydrocarbons + O2
2NO + 2NO2
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catalytic
converter
catalytic
converter
CO2 + H2O
2N2 + 3O2
13.6
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The End!!!!!!!
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