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Chapter 13 - Chemical Kinetics II
Integrated Rate Laws Reaction Rates and Temperature
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Reaction Order - Graphical Picture
Reactant Concentration vs. Time
A → Products
A—->Products
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Integrated Rate Laws
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Zero Order Reactions
Rate = k[A]0 = k (constant rate reactions) [A] = -kt +[A]0 (integrated rate law)
graph of [A] vs t is straight line (slope = -k and y intercept = [A]0)
t½ = [A]0/2k
• Rate = k[A]0 = k
constant rate reactions
• [A] = -kt + [A]0
• graph of [A] vs. time is straight line with slope = -k and y-intercept = [A]0
• t ½ = [A0]/2k
• when Rate = M/sec, k = M/sec
[A]0
[A]
time
slope = - k
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First Order Reactions
Rate = k[A] ln[A] = -kt +ln[A]0 (integrated rate law)
graph of ln[A] vs t is straight line (slope = -k and y intercept = ln[A]0)
t½ = ln2/k = (0.693)/k (constant half-life)
• Rate = k[A]0 = k
constant rate reactions
• [A] = -kt + [A]0
• graph of [A] vs. time is straight line with slope = -k and y-intercept = [A]0
• t ½ = [A0]/2k
• when Rate = M/sec, k = M/sec
[A]0
[A]
time
slope = - k ln[A]
ln[A]0
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Second Order Reactions
Rate = k[A]2 1/[A] = -kt + 1/[A]0 (integrated rate law)
graph of 1/[A] vs t is straight line (slope = +k and y intercept = 1/[A]0)
t½ = 1/(k[A]0)
'me
slope =
+k
1/[A]
1/[A]
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Half-life, t½
the length of time it takes for the concentration of the reactants to fall to one-half its initial value
depends on the order of the reaction
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Zero Order ReactionsDetermining Rate Graphically
First Order Reactions
Second Order Reactions
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Data for 2NO2 (g) → 2NO (g) + O2 (g)
ln[NO2] vs time 1/[NO2] vs time
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Problems Using Integrated Rate Laws
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1) N2O decomposes at 1050 K through 1st order kinetics:
2 N2O —> 2 N2 + O2 Rate = k[N2O], k = 3.4 sec-1
For an initial concentration of [N2O] of 0.20 M, what is the [N2O] after 100 msec.
kt = ln[A]0 - ln[A] [A]0
[A]kt = ln
(3.4 sec-1)(100 x 10-3sec)= ln [A]0
[A]
e =(3.4 sec-1)(100 x 10-3sec) [A]0
[A] e = (.34) [A]0
[A]
1.40 = 0.20 M
x x = 0.14 M
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tmin 0 200 400 600 800 1000
[N2O5] 0.015 0.0096 0.0062 0.004 0.0025 0.0016
ln[N2O5] -4.20 -4.65 -5.08 -5.52 -5.99 -6.43
0
0.00375
0.00750
0.01125
0.01500
0 250 500 750 1000
[N2O5] vs time
-7.00
-5.25
-3.50
-1.75
0
0 250 500 750 1000
ln[N2O5] vs time
2) N2O5 decomposes at 25 ºC through 1st order kinetics:
2 N2O5 —> 4 NO2 + O2
Given the following concentration vs time data, determine the rate constant for the reaction.
[N2O5]
ln[N2O5]
-( )-slope = -2.23
60,000 sec
slope = 3.7 x 10-5 sec-1
[A]0
slope = -k
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3) Cyclopropane is converted to propene (a rearrangement) at 500ºC with first order kinetics with a rate constant of 6.7 x 10-4 sec-1.
C3H6 —> C3H6
How long does it take for the concentration of cyclopropane to drop to 1/2 of its original concentration ? to 1/4 ?
t ½ = 0.69/(6.7 x 10-4sec-1) = 1.0 x 103 sec
t = 2 x t ½= 2.0 x 103 sec1/4
4) N2O5 decomposes at 25 ºC through 1st order kinetics:
2 N2O5 —> 4 NO2 + O2 k = 5.2 x 10-3 sec-1 at 65º
How long will it take for a sample which is 20 mM N2O5 to decrease to 2 mM?
[A]0
[A]kt = ln [A]0
[A] t = ln 1
k= ln 20mM
2 mM 1
5.2 x 10-3
= (192)ln(10) = 442 sec
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5) HI decomposes at 700 K through 2nd order kinetics:
2 HI —> H2 + I2 k = 1.8 x 10-3 M-1sec-1
How long will it take for 1/2 of a 0.10 M sample to decompose ? for 3/4 ?.
1/(1.8 x 10-3 M-1sec-1)(0.010 M)t½ =
t½ = 5.6 x 104 sec
t = 1/(1.8 x 10-3 M-1sec-1)(0.005 M) 1/41/2 →
t = 11.2 x 104 sec 1/41/2 →
Total time for 3/4 to decompose = 1.7 x 105 sec
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Reaction Rates and Temperature
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Temperature: Changing the temperature changes the rate constant
in the rate law.
These factors are incorporated into rate constant k by the Arrhenius equation:
R = universal gas constant (in J/mol·K) T = temperature (in kelvin)
A is a collisional frequency factor. (Includes frequency of collisions and an orientation factor)
Ea = Activation energy, The minimum energy of molecular collisions required to break bonds in reactants, leading to
formation of products.
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Collision Theory
In order for a chemical reaction to occur, reactant molecules must collide with sufficient energy and in the proper orientation.
Effective collisions involve both of these factors.
When two molecules have an effective collision, a temporary, high-energy (unstable) chemical species, the reactive
intermediate*, is formed.
*also referred to as an activated complex or transition state
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Arrhenius equation:
Frequency Factor
Activation Energy
Exponential Factor
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Reaction Energy Profile
High-energy transition state (“activated
complex”)
Activation energy (Ea )
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Reaction Energy Profile
Partially broken and partially formed bonds
The amount of energy necessary to form the “activated complex”
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The Reaction of CH3Br and OH-
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Reaction Energy Diagram for the Reaction of CH3Br and OH-
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The Arrhenius equation can be rearranged:
by
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Graphical Determination of Ea (Arrhenius Plot)
Slope = –Ea/R Intercept = ln A
According to the rearranged Arrhenius equation, the activation energy for a reaction can be determined by plotting k vs 1/T which yields a line whose slope is -Ea/R and whose intercept is lnA.
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Mathematical Determination of Ea
Consider determination of k two different T’s:
By subtraction,
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k = (A)(e- ) Ea RT
The Exponential Factor
A number between 0 and 1, representing the fraction of reactant molecules with sufficient energy to
make it over the energy barrier
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Ea (J/mol)Temp -Ea/RT -Ea/RTe
10,000523 K -2.30
-11.5 0.00001
0.100
The Effect of Ea on the Fraction of Collisions with Sufficient Energy to Allow Reaction
k = (A)(e- ) Ea RT
523 K
523 K
20,000
50,000
-4.60 0.001
The higher the energy barrier, the fewer molecules that have sufficient energy to overcome the energy barrier.
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The Effect of Ea on the Fraction of Collisions with Sufficient Energy to Allow Reaction
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Ea (J/mol) Temp -Ea/RT -Ea/RTe
10,000
10,000
10,000
298 K
523 K
773 K
-4.04
-2.30
-1.56 0.211
0.100
0.0176
The Effect of Temperature on the Fraction of Collisions with Sufficient Energy to Allow Reaction
k = (A)(e- ) Ea RT
The extra kinetic energy of the reactants is converted into potential energy when molecules collide.
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Increased temperature increases kinetic energy of molecules and molecular
collisions.
The Effect of Temperature on the Fraction of Collisions with Sufficient Energy to Allow Reaction
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k = (A)(e- ) Ea RT
Frequency Factor
A number representing the number of reactant molecules which can approach each other with the correct orientation
and sufficient energy to make it over the energy barrier
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A=(p)(z)Frequency Factor
p zOrientation
FactorCollision Frequency
Factorfor reactions of single atoms, p=1
for most molecules p<<1for a collision to overcome the energy
barrier, the reactants mus have sufficient kinetic energy so that the
collision leads to the formation of an “activated complex”
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c
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Effective Collisions-Kinetic Energy Effect