# Chapter 12: Chemical Kinetics

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```					Chapter 12: Chemical Kinetics
1.  Reaction Rates
2.  Rate Laws: Differential and Integrated
3.  Experimental Determination of the Rate Law
4.  Integrated Rate Laws & Concentration/Time Problems
5.  Collision Theory of Reactions
a. Reaction Mechanisms
b. Predicting Rate Laws
c. Temperature Dependence of the Rate Constant
(Arrhenius)
6. Catalysis
As the reaction proceeds, it gets slower because the rate depends on [NO2].
Figure 12.1
Example 1:
Determine the rate law and the value of the rate
constant for the following reaction.
2 NO + Br2  2 NOBr

Experiment     [NO]0         [Br2]0     Initial Rate
(mole/L s)
1       1.1 x 10-5    1.2 x 10-5       0.37
2       1.9 x 10-5    1.3 x 10-5      0.69
3       3.5 x 10-5    1.2 x 10-5       1.2
4       4.0 x 10-5    3.0 x 10-5       3.4
Example 2:
Determine the rate law and the value of the
rate constant for the following reaction in which
OH- is a catalyst.
OCl- + I-  OI- + Cl-
Experiment  [OCl-]0     [I-]0    [OH-]0      Rate
(mole/L)   (mole/L)   (mole/L)   (mole/L s)
1       .0040      .0020       1.00      4.8x10-4
2       .0020      .0040       1.00      5.0x10-4
3       .0020      .0020       1.00      2.4x10-4
4       .0020      .0020       0.50      4.6x10-4
5       .0020      .0020       0.25      9.4x10-4
Example 3:
The concentration of H2O2 was     Time (s)    [H2O2]
(mole/L)
monitored over time for the
0.000      1.000
following reaction at 25°C:
120.000      0.910
300.000      0.780
H2O2 (aq)  H2 (g) + O2 (g)        600.000      0.590
1200.000      0.370
Find the rate law and the value   1800.000      0.220
2400.000      0.130
of the rate constant for this
3000.000      0.082
reaction at 25°C.
3600.000      0.050
Example 4:
The decomposition of N2O5 to NO2 and O2 is
first order with a rate constant of
4.80 x 10-4 /s at 45°C.

(a) If the initial concentration of N2O5 is 1.65 x
10-2 mole/L, what is the concentration after
825 s?
(b) How long would it take for the concentration
of N2O5 to decrease to 1.00 x 10-2 mole/L if
its initial concentration wa that given in (a)?
Example 5:
For the following decomposition reaction,
AB  A + B
Rate = k[AB]2
k = 0.20 L/mole s
How long will it take for [AB] to reach one
third of its original value of 1.50 M? What
is [AB] after 10.0 seconds?
Zero Order Half Life
t1/2 = [A]0 / 2k     Time (s)   [A]

0       1.6
Note that this half life
decreases as the reaction      2.0      .80
proceeds.
3.0      .40

There is less reactant to      3.5      .20
Consume and it runs at the
3.75      .10
same rate so it takes less
time.                         3.88      .050
First Order Half Life
t1/2 = 0.693 / k           Time (s)   [A]

The first order half life is         0       1.6
constant
24.0      .80
for a particular reaction and
temperature.                       48.0      .40

72.0      .20
As the initial concentration
decreases, the decrease in rate    96.0      .10
is compensated for by less
120.0     .050
reactant to consume.
Second Order Half Life
t1/2 = 1 / [A]0 k             Time (s)   [A]

Note that this half life increases      0       1.6
as the reaction proceeds.
10.2      .80

The rate decreases so strongly        30.6      .40
with concentration that it still
71.4      .20
takes longer to use up half of
the reactant even though the           153      .10
amount of reactant consumed is
less for each half life.
Example 6:
A decomposition reaction has a rate
constant of 0.0012 yr-1.

a) What is the half life of this reaction?

b) How long does it take for the
concentration of the reactant to
reach 12.5% of its original value?
Example 7:
It took 143 s for 50.0% of a particular
substance to decompose. If the initial
concentration was 0.060 M and the
decomposition reaction follows second
order kinetics, what is the value for the
rate constant?
Example 8:
For the reaction A  products,
successive half-lives are observed to be
10.0, 20.0, and 40.0 min for an
experiment in which [A]0=0.10M.
Calculate the concentration of
A at the following times.
a) 80.0 min
b) 30.0 min
Example 9:
(Book #34)
O + NO2  NO + O2          Time (s)      [O]
atoms/cm3
[NO2] in large excess,
0        5.0x109
1.0x1013molecules/cm3
a) Find the order of the      1.0x10-2    1.9x109
reaction with respect to
[O].                       2.0x10-2    6.8x108
b) Reaction is known to be
first order in [NO2].      3.0x10-2    2.5x108
Find the value of k.
Example 10:
Consider the following mechanism for the decomposition
of hydrogen peroxide catalyzed by I-.

Step 1: H2O2 + I-    H2O + IO- ; fast equilibrium
Step 2: IO- + H2O2  H2O + O2 + I- ; slow

a)   Write a balanced equation for the overall reaction.
b)   List any catalysts.
c)   List any intermediates.
d)   Write the rate law for the overall reaction.
Example 11:
Consider the following mechanism for the production of
phosgene:
Step 1: Cl2 (g)        2 Cl (g); fast equilibrium
Step 2: Cl (g) + CO (g)          COCl (g); fast equilibrium
Step 3: COCl (g) + Cl2 (g)  COCl2 (g) + Cl (g); slow
Step 4: 2 Cl (g)  Cl2 (g); fast

a)   Write a balanced equation for the overall reaction.
b)   List any catalysts.
c)   List any intermediates.
d)   Write the rate law for the overall reaction.
Figure 12.11: 2BrNO  2 NO + Br2
Figure 12.13:
Example 12:
The activation energy of the following
reaction is 3.5 kJ/mole and the change in
enthalpy is -66.6 kJ/mole. Calculate
the activation energy of the reverse
reaction.
OH + HCl  H2O + Cl

Hint: Sketch the reaction energy diagram.
Example 13:
For the reaction A2 + B2  2 AB, the
activation energy of the forward reaction is
125 kJ/mole and that of the reverse
reaction is 85 kJ/mole. Assuming the
reaction occurs in one step:
a) Draw a reaction energy diagram.
b) Calculate DH for the reaction.
c) Sketch a possible transition state.
Figure 12.2:
Collisions with E > Ea at two temperatures.
Example 14:
The isomerization reaction:
CH3NC  CH3CN
has an activation energy of 161 kJ/mole.
If the rate constant at 600K is 0.41/s,
what is it at 1000K?
Example 15:
The rate constant of a reaction is
4.50x10-5 L/mol•s at 195°C and
3.20 x 10-3 L/mol•s at 258°C.
What is the activation energy of this
reaction?
Catalysts Work by Lowering Ea:
They provide a different mechanism!
More collisions result in reaction
when Ea is lower.
Homogeneous Catalysis
In the stratosphere, NO catalyzes the
destruction of ozone, O3.

NO (g) + O3 (g)  NO2 (g) + O2 (g)
O (g) + NO2 (g)  NO (g) + O2 (g)

Overall:
O3 (g) + O (g)  2 O2 (g)
Heterogeneous
Catalysis

C2H4 + H2  C2H6

with a metal catalyst

a. reactants
c. migration/reaction
d. desorption
Enzymes are Catalysts
Enzymes are Catalysts
Enzymes are Catalysts

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