1. 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
rate = -
D[A] Dt
D[A] = change in concentration of A over
time period Dt
rate =
D[B] 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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GENERAL CHEMISTRY(KINETIKS)
13.1

2. 13.1 A B time D[A] rate = - Dt D[B] rate = Dt
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13.1

3. 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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4. slope of tangent slope of tangent slope of tangent 13.1
Br2 (aq) + HCOOH (aq) Br- (aq) + 2H+ (aq) + CO2 (g)
slope of
tangent
slope of
tangent
slope of
tangent
average rate = -
D[Br2] Dt
= -
[Br2]final – [Br2]initial
tfinal - tinitial
instantaneous rate = rate for specific instance in time
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GENERAL CHEMISTRY(KINETIKS)
13.1

5. 2NO22NO+O2
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6. Wednesday, April 26, 2017
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7. 13.3

8. Concentrations & Rates
0.3 M HCl
6 M HCl
Mg(s) HCl(aq) ---> MgCl2(aq) + H2(g)
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9. 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?
= 3.5
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10. 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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11. [S2O82-] [I-] Initial Rate (M/s) 1 0.08 0.034 2.2 x 10-4 2 0.017
Determine the rate law and calculate the rate constant for the following reaction from the following data:
S2O82- (aq) + 3I- (aq) SO42- (aq) + I3- (aq)
Experiment
[S2O82-]
[I-]
Initial Rate (M/s) 1
0.08
0.034
2.2 x 10-4 2
0.017
1.1 x 10-4 3
0.16
rate = k [S2O82-]x[I-]y
y = 1
x = 1
rate = k [S2O82-][I-]
Double [I-], rate doubles (experiment 1 & 2)
Double [S2O82-], rate doubles (experiment 2 & 3)
k =
rate
[S2O82-][I-] =
2.2 x 10-4 M/s
(0.08 M)(0.034 M)
= 0.08/M•s
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13.2

12. Deriving Rate Laws
Derive rate law and k for
CH3CHO(g) --> CH4(g) + CO(g)
from experimental data for rate of disappearance of CH3CHO
Expt. [CH3CHO] Disappear of CH3CHO (mol/L) (mol/L•sec)
Rate of rxn = k [CH3CHO]2
GENERAL CHEMISTRY(KINETIKS)

13. Deriving Rate Laws
Rate of rxn = k [CH3CHO]2 Here the rate goes up by ______ when initial conc. doubles. Therefore, we say this reaction is _________________ order. Now determine the value of k. Use expt. #3 data— 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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14. Expession of R &K
Iodide ion is oxidized in acidic solution to triiodide ion, I3- , by hydrogen peroxide.
A series of four experiments was run at different concentrations, and the initial rates of I3- formation were determined.
From the following data, obtain the reaction orders with respect to H2O2, I-, and H+.
Calculate the numerical value of the rate constant.
Copyright © Houghton Mifflin Company.All rights reserved.

15. A Problem to Consider Initial Concentrations (mol/L) H2O2 I- H+
Initial Rate [mol/(L.s)]
Exp. 1
0.010
1.15 x 10-6
Exp. 2
0.020
2.30 x 10-6
Exp. 3
Exp. 4
Comparing Experiment 1 and Experiment 2, you see that when the
H2O2 concentration doubles (with other concentrations constant),
the rate doubles.
This implies a first-order dependence with respect to H2O2.
Copyright © Houghton Mifflin Company.All rights reserved.

16. A Problem to Consider Initial Concentrations (mol/L) H2O2 I- H+
Initial Rate [mol/(L.s)]
Exp. 1
0.010
1.15 x 10-6
Exp. 2
0.020
2.30 x 10-6
Exp. 3
Exp. 4
Comparing Experiment 1 and Experiment 3, you see that when the
I- concentration doubles (with other concentrations constant),
the rate doubles.
This implies a first-order dependence with respect to I-.
Copyright © Houghton Mifflin Company.All rights reserved.

17. A Problem to Consider Initial Concentrations (mol/L) H2O2 I- H+
Initial Rate [mol/(L.s)]
Exp. 1
0.010
1.15 x 10-6
Exp. 2
0.020
2.30 x 10-6
Exp. 3
Exp. 4
Comparing Experiment 1 and Experiment 4, you see that when
the H+ concentration doubles (with other concentrations constant),
the rate is unchanged.
This implies a zero-order dependence with respect to H+.
Copyright © Houghton Mifflin Company.All rights reserved.

18. A Problem to Consider Initial Concentrations (mol/L) H2O2 I- H+
Initial Rate [mol/(L.s)]
Exp. 1
0.010
1.15 x 10-6
Exp. 2
0.020
2.30 x 10-6
Exp. 3
Exp. 4
Because [H+]0 = 1, the rate law is:
The reaction orders with respect to H2O2, I-, and H+, are 1, 1, and 0, respectively.
Copyright © Houghton Mifflin Company.All rights reserved.

19. A Problem to Consider Initial Concentrations (mol/L) H2O2 I- H+
Initial Rate [mol/(L.s)]
Exp. 1
0.010
1.15 x 10-6
Exp. 2
0.020
2.30 x 10-6
Exp. 3
Exp. 4
You can now calculate the rate constant by substituting values from
any of the experiments. Using Experiment 1 you obtain:
Copyright © Houghton Mifflin Company.All rights reserved.

20. A Problem to Consider Initial Concentrations (mol/L) H2O2 I- H+
Initial Rate [mol/(L.s)]
Exp. 1
0.010
1.15 x 10-6
Exp. 2
0.020
2.30 x 10-6
Exp. 3
Exp. 4
You can now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 you obtain:
Copyright © Houghton Mifflin Company.All rights reserved.

21. Zero-Order Reactions [A] is the concentration of A at any time t
rate = -
D[A] Dt
A product
rate = k [A]0 = k
rate
[A]0
D[A] Dt
= k -
k =
=RATE= M/s
[A] is the concentration of A at any time t
[A] = [A]0 - kt
[A]0 is the concentration of A at time t=0
t½ = t when [A] = [A]0/2
t½ =
[A]0 2k
13.3

22. First-Order Reactions
A product
rate = k [A]
D[A] Dt
= k [A] -
rate
[A]
M/s M =
k =
= 1/s or s-1
[A] is the concentration of A at any time t
t½ =
0.693 k
[A]0 is the concentration of A at time t=0
[A] = [A]0exp(-kt)
ln[A] = ln[A]0 - kt
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GENERAL CHEMISTRY(KINETIKS)
13.3

23. Second-Order Reactions
rate = -
D[A] Dt
A product
rate = k [A]2
rate
[A]2
M/s M2 =
D[A] Dt
= k [A]2 -
k =
= 1/M•s
[A] is the concentration of A at any time t 1
[A] =
[A]0
+ kt
[A]0 is the concentration of A at time t=0
t½ = t when [A] = [A]0/2
t½ = 1
k[A]0
13.3

24. 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
Graphs of different rates of reaction copy the chapter from Langmuir for them!

25. Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order
Rate Law
Concentration-Time Equation
Half-Life
t½ =
[A]0 2k
rate = k
[A] = [A]0 - kt t½
ln2 k = 1
rate = k [A]
ln[A] = ln[A]0 - kt 1
[A] =
[A]0
+ kt
t½ = 1
k[A]0 2
rate = k [A]2
13.3

26. Collision Theory O3(g) + NO(g) ---> O2(g) + NO2(g)
10 31 COLLISINS/LIT.S
Reactions require
(A) geometry
(B) Activation energy
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27. Three possible collision orientations-- a) & b) produce reactions,
while c) does not.

28. Collision Theory TEMPERATURE 10 c0 T  100-300% RATE
25 c0 T  35 c  2%  COLLISIN
EFFECTIVE COLLISIONS
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29. Wednesday, April 26, 2017
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30. ① with sufficient energy
The colliding molecules must have a total kinetic energy equal to or
greater than the activation energy, Ea. The activation energy is the mini-
mum 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(平面) 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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31. 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.
Exothermic. Products are more stable than reactants and after product formation, there is release of heat.
endothermic.
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32. Wednesday, April 26, 2017
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33. 13.4 A + B C + D Exothermic Reaction Endothermic Reaction
The activation energy (Ea) is the minimum amount of energy required to initiate a chemical reaction.
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13.4

34. Temperature Dependence of the Rate Constant
k = A • exp( -Ea/RT )
(Arrhenius equation)
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature
A is the frequency factor
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GENERAL CHEMISTRY(KINETIKS)
13.4

35. Arrhenius Equation k = A • exp( -Ea/RT )
Can be arranged in the form of a straight line
ln k = ln A-Ea/RT
Plot ln k vs. 1/T 
slope = -Ea/R
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36. k = A • e( -Ea/RT ) ln k = ln A-Ea/RT= (-Ea/R)(1/T) + ln A
300400C0 Rate20000 Times
400500C0 Rate400 Times
Ea=60KJ/mol
300310C0  Rate2 Times
Ea=260KJ/mol
300310C0  Rate25 Times
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37. Wednesday, April 26, 2017
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38. K in different T Log(K2/K1) =Ea/2.303R(T2-T1)/T1T2
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
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39. Mechanisms
O3 + NO reaction occurs in a single ELEMENTARY step. Most others involve a sequence of elementary steps.
Adding elementary steps gives NET reaction.
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40. Mechanisms
Most rxns. involve a sequence of elementary steps. 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2] 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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41. Most rxns. involve a sequence of elementary steps.
Mechanisms
Most rxns. involve a sequence of elementary steps.
2 I- + H2O H+ ---> I H2O
Rate = k [I-] [H2O2]
Proposed Mechanism Step 1 — slow HOOH + I- --> HOI + OH- Step 2 — fast HOI + I- --> I2 + OH- Step 3 — fast 2 OH- + 2 H+ --> 2 H2O Rate of the reaction controlled by slow step — RATE DETERMINING STEP, rds. Rate can be no faster than rds! The species HOI and OH- are reaction intermediates.
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42. Mechanisms 2NO+F22NOF R=K[NO][F2] 1=NO+F2NOF+F R1=K1[NO][F2]
Rate limiting step(RDS)
2=NO+F NOF R2=K2[NO][F]
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43. SN1
OH- + (CH3)3CBr(CH3)3COH + Br- R=K[(CH3)3CBr] 1)(CH3)3CBr(CH3)3C+ +Br- R1=K1[(CH3)3CBr ] 2) (CH3)3C+ +OH- (CH3)3COH R2=K2[(CH3)3C+] +[OH-]
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44. SN2 OH- +CH3Br  CH3OH + Br- R=K[CH3Br][OH-] Wednesday, April 26, 2017
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45. Wednesday, April 26, 2017
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46. Rate Laws and Mechanisms
NO2 + CO reaction: Rate = k[NO2]2
Two possible mechanisms
Two steps:
Single step
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47. 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
catalyzed
ratecatalyzed > rateuncatalyzed
Ea < Ea ‘
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13.6

48. For the decomposition of hydrogen peroxide:
A catalyst(Br-) speeds up a reaction by lowering the activation energy.
It does this by providing a different mechanism (机制) by which the reac-
tion can occur. The energy profiles for both the catalyzed and uncatalyzed
reactions of decomposition of hydrogen peroxide are shown in Figure below.
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49. Homogeneous Catalysis
N2O (g) (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
Cl2(g)2Cl(g)
N2O(g)+Cl(g)N2(g)+ClO(g)
2ClOCl2(g)+O2(g)
Ea=140KJ/mol
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50. Heterogeneous Catalysis
C2H4 + H2  C2H6 with a metal catalyst a. reactants b. adsorption c. migration/reaction d. desorption
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51. Enzyme Catalysis 13.6 Wednesday, April 26, 2017
GENERAL CHEMISTRY(KINETIKS)
13.6

52. Catalytic Converters catalytic converter catalytic converter 13.6
CO + Unburned Hydrocarbons + O2
CO2 + H2O
catalytic
converter
2NO + 2NO2
2N2 + 3O2
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13.6

53. The End!!!!!!!
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