Chapter 14
Chemical Kinetics
Chemical Kinetics CH 14
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Factors affecting chemical reaction
Rate of reaction
Average rate, Instantaneous rate
Rate law
Order of reaction
First order reaction
Second order reaction
Half - life time
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Chemical Kinetics:
How fast is the chemical reaction,
(i.e. studying of rates of chemical
processes).
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Factors That Affect Reaction Rates
1. Reactant concentration: As the concentration
of reactants increases, so does that reactant
molecules will collide and rate of reaction
increases.
2. Temperature: As temperature increases, the
reaction rate increases, reactant molecules have
more kinetic energy, move faster, and collide more
often and with greater energy.
3. Catalysts: catalyst increases chemical reactions
by changing mechanism.
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Reaction Rates
Speed of a reaction is measured by: the change in
concentration with time.
For a reaction A  B
change in number of moles of B
Average rate 
change in time
 moles of B

t
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Reaction Rates
AB
Rates of reactions can be determined by monitoring the change in
concentration of either reactants or products
as a function of time. [A] / t
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Reaction Rates
 At t = 0 (time zero) there is 1.00 mol A (100 red
spheres) and no B present = zero.
 At t = 20 min, there is 0.54 mol A and 0.46 mol B.
 At t = 40 min, there is 0.30 mol A and 0.70 mol B.
 Calculating average rate:
 moles of B
Average rate  
t

moles of B at t  20  moles of B at t  0

20 min  0 min
0.46 mol  0 mol

 0.023 mol/min
20 min  0 min
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Reaction Rates
For the reaction:
AB
There are two ways of measuring rate:
1. The speed at which the products appear (i.e. change in
moles of B per unit time), or
2. The speed at which the reactants disappear (i.e. the
change in moles of A per unit time).
 moles of A 
[ A]
Average rate with respect to A  

t
t
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Example: Reaction of butyl chloride to give butanol.
Average rate
decreases as
the reaction
goes on.
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C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
The average rate of the
reaction over each
interval = the change in
concentration divided
by the change in time:
Average Rate, M/s
0.0905  0.1000
4
[C 4 H 9Cl ]


1
.
9
x
10
M /S
averagerate  

50.0  0.0
t
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Instantaneous Rate & Average Rate
• Instantaneous rate defines as The rate at any
instant in time and it is the slope of the
tangent to the curve.
• Average rate: is the change in reactant or
product concentration to the change of time.
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Instantaneous Rate & Average Rate
Example:
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
– If We plot [C4H9Cl] with respect to t.
– The units for average rate are mol/L·s or M/s.
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Calculate: 1. average rate?
Average rate= Y2-Y1
X2-X1
averagerate  
0.0671  0.1
0.4

 1.6 x10 4 M / S
200  0
200
2. instantaneous rate at Z point
Z
Ins tan tan ousrate  
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0.030
0.4

 6.67 x10 4 M / S
600
200
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Reaction Rates and Stoichiometry
• What if the ratio is not 1:1?
H2(g) + I2(g)  2 HI(g)
• Only 1/2 HI is made for each H2 used.
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Reaction Rates and Stoichiometry
• In General, for the reaction
aA + bB
cC + dD
(-) sign because Reactants (decrease) with time
(+) sign because Products (increase) with time
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• For Example
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
C4H9Cl C4H9OH
Rate  

t
t
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• For Example
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Concentration and Rate
• In general rates increase as concentrations increase.
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
Constant
increases
increases
Constant
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Concentration and Rate
• From previous table, for the reaction
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
we note:
 as [NH4+] doubles with [NO2-] constant the rate doubles,
 as [NO2-] doubles with [NH4+] constant, the rate doubles
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Concentration and Rates
For the reaction
The above equation is called the rate law,
and k is the rate constant.
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Rate Law
• For a general reaction with rate law
m
n
Rate  k[reactant 1] [reactant 2]
m: order in reactant 1 and n: order in reactant 2.
• The total order of reaction = (m + n + ….)
• The total order of reaction = zero, if m = 0, n = 0.
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Concentration and Rate
• This reaction is
First-order in [NH4+]
First-order in [NO2−]
• The overall reaction order: is the sum of the
exponents on the reactants in the rate law.
• The overall order of this reaction= 1+1= 2
( i.e. second-order).
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AB
Differential
Rate Law
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xt

x0
 dx
  ln x  c
x
[ A ]t

[A]0: the initial
concentration at t = 0.
[A]t: the concentration
after time, t >0.
[ A ]0
 d [ A]

[ A]
t

kdt
0
 ln[ A]  c  kt
t  0, [ A]  [ A]0
 ln[ A]t  ln[ A]0  kt
multiply 
ln[ A]t  ln[ A]0   kt
ln
[ A]t
  kt
[ A]0
[ A]t
 e  kt
[ A]0
[ A]t  [ A]0 e  kt
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First Order Reactions
When [A]t is plotted as a function of time, a curve results.
• Slope = - k
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First Order Reactions
Straight Line Equation
Slope= + m
intercept = b
y = mx + b
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First Order Reactions
ln At  kt  ln A0
A plot of ln[A]t vs t
is a straight line.
slope = -k
intercept = ln[A]0
ln[A]t
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First Order Reactions
• Half-life t1/2: is the time taken for the concentration of
a reactant to drop to half its original value.
• For a first order process, when t = t½,
so [A]t = ½[A]0.
[ A]t
ln
 kt
[ A]0
 Half- life time
doesn’t depend on
concentration of
reactant
t1  
2
2   0.693
ln 1
k
k
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Second Order Reactions
• For a second order reaction with just one reactant.
rate  
d [ A]
 k [ A]2
dt
d [ A]
  kdt
2
[ A]
[ A ]t

[ A ]0

Differential Equation
t
d [ A]
   kdt
2
[ A]
0
dx
1

c
2
x
x
[A] = [A]0 , t=0
1
1

 kt
[ A]t
[ A]0
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1
 kt 
[ A]t
[ A]0
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The Change of Concentration with Time
Second Order Reactions
1
1
 kt 
At
A0
y = mx + b
A plot of 1/[A] vs. t is a straight line
with a slope of k.
Intercept= 1/[A]0
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Half-Life of Second Order
For a second-order process, set [A]t=0.5 [A]0 .
1
1
[ A]0
2
t1
2

1
 kt 1
2
[ A]0
1

k A0
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Determining the order of chemical reaction
Example
The decomposition of NO2 at 300°C is described by
the equation
NO2 (g)
NO (g) + 1/2 O2 (g)
and yields these data:
Time (s)
0.0
50.0
100.0
200.0
300.0
[NO2], M
0.01000
0.00787
0.00649
0.00481
0.00380
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Determining the order of chemical reaction
Graphing ln [NO2] vs. t yields:
The plot is not a straight line,
so the process is not firstorder in [A].
Time (s)
0.0
50.0
[NO2], M
0.01000
0.00787
ln [NO2]
-4.610
-4.845
100.0
200.0
300.0
0.00649
0.00481
0.00380
-5.038
-5.337
-5.573
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Does not fit:
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Determining the order of chemical reaction
A graph of 1/[NO2] vs. t
gives this plot.
Time (s)
0.0
50.0
[NO2], M
0.01000
0.00787
1/[NO2]
100
127
100.0
0.00649
154
200.0
300.0
0.00481
0.00380
208
263
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• This is a straight
line. Therefore,
the process is
second-order in
[NO2].
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Practice Problems
CH.14 in the book
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