BLB 11th Chapter 14
1.
2.
3.
Will the reaction occur? Ch. 5, 19
How fast will the reaction occur? Ch. 14
How far will the reaction proceed? Ch. 15
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Review: concentration units, graphing (line, slope)
Work with reaction rates.
Determine a rate law for a given reaction from
experimental data.
Work with rate laws.
Understand the factors that affect the rate of a
chemical reaction.
Calculate the activation energy.
C(s, diamond)
ΔH °rxn =
Is the reaction favorable?
C(s, graphite)
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studies the rates at which chemical reactions
occur.
gives information about how the reaction
occur, that is, the reaction mechanism
Determining the reaction mechanism is the
overall goal of kinetic studies. (sect. 6)
1.
2.
3.
4.
5.
Physical state of the reactants – states that
promote contact have faster rates;
homogeneous vs. heterogeneous
Concentration of the reactants: conc. ↑,
rate ↑ (or pressure for gases)
Temperature: temp. ↑, rate ↑ due to higher
molecular energy and speed (section 5)
Catalysts: rate ↑ by changing the
mechanism and reaction energy (section 7)
Other physical things like stirring and
grinding solid reactants.

Rate – change in some variable per unit time
1
rate 
time
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
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Reaction rate – change in concentration per
unit time; M/s or mol/L·s
Rates are determined by monitoring
concentration as a function of time.
Rates are positive quantities; for reactant A:
[ A]
rate  
  final  initial
t
In this reaction, the
concentration of
butyl chloride,
C4H9Cl, was
measured at
various times.

Rates change over time:
◦ reactant rates decrease
◦ product rates increase
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
[ A]
rate  
t
Instantaneous rate – rate at a specific time
Average rate – Δ[A] over a specific time
interval
Initial rate – instantaneous rate at t = 0
Note: Rates and rate laws are not based on
stoichiometry!! They must be determined
experimentally.
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The molar ratios between reactants and
products correspond to the relative rates
of the reaction.
Relative rates – relationship between rates
of reactant disappearance and product
appearance at a given time.
2 HI(g) → H2(g) + I2(g)
1 [ HI ] [ H 2 ] [ I 2 ]
rate  


2 t
t
t
2 NO2(g) → 2 NO(g) + O2(g)
Rate = 2.4 x 10-5 M/s
Rate = 8.6 x 10-5 M/s
Rate = 4.3 x 10-5 M/s
1.
Differential rate law or rate law (section 3)
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2.
Shows how the reaction rate changes with
concentration
Must be specific in how defined
Integrated rate law (section 4)
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Shows how concentration changes with time
Graphical determination of the order
aA + bB → cC + dD
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General form of rate law:
[A], [B] – conc. in M or P
rate = k[A]m[B]n k – rate constant; units vary
m, n – reaction orders
Reaction orders and, thus, rate laws must be
determined EXPERIMENTALLY!!!
Note: m ≠ a and n ≠ b
Overall order = sum of individual orders
Rate constant is independent of concentration.
Order (m)
Δ[A] by a factor of:
Effect on rate
Zero (0)
2, 4, 15, ½, etc.
None
2
2X
3
3X
2
4X
3
9X
½
¼X
1st (1)
2nd (2)
1.
2.
3.
4.
5.
6.
What is the order with respect to NO?
What is the order with respect to H2?
What is the overall order?
If [NO] is doubled, what is the effect on the
reaction rate?
If [H2] is halved, what is the effect on the
reaction rate?
What are the units of k?
1.
2.
Calculate the rate of reaction when the
concentration of PtCl2(NH3)2 is 0.020M.
What is the rate of Cl¯ production under these
conditions?
(b) Calculate rate when [NO] = 0.025 M and
[H2] = 0.015 M.
Initial Rates Method
1. Find two experiments in which all but one
reactant’s concentration is constant.
2. Observe the relationship between concentration
change and rate change to determine the order
for that reactant.
3. Repeat for other reactant(s).
Exp. #
1
[NH4+]o
0.100
[NO2¯]o
0.0050
Initial Rate (M·s-1)
1.35 x 10-7
2
0.100
0.0100
2.70 x 10-7
3
0.200
0.0100
5.40 x 10-7
Determine the rate law and calculate k.
Exp. #
1
[NO]o
0.10
[Cl2]o
0.10
Initial Rate (M·min-1)
0.18
2
0.10
0.20
0.36
3
0.20
0.20
1.45
Determine the rate law and calculate k.
Exp. # [BrO3¯]o
1
0.10
2
3
4
0.20
0.20
0.10
[Br¯]o
0.10
[H+]o
0.10
Initial Rate (M·s-1)
8.0 x 10-4
0.10
0.20
0.10
0.10
0.10
0.20
1.6 x 10-3
3.2 x 10-3
3.2 x 10-3
Determine the rate law and calculate k.

1st order
integrate
[ A]
rate  
 k[ A]
t
ln[ A]t  kt  ln[ A]o
y = mx + b
Plot: ln[A] vs. t
slope = − k
 [ A]o 
  kt, for[ A]t  [ A]o
ln 
 [ A]t 
 2nd
[ A]
2
rate  
 k[ A]
t
order
integrate
1
1
 kt 
[ A]t
[ A]o
y = mx + b
Plot: 1/[A] vs. t
slope = k
1
1

 kt, for[ A]t  [ A]o
[ A]t [ A]o

zero order
integrate
[ A]
0
rate  
 k[ A]  k
t
[ A]t  kt  [ A]o
y = mx + b
Plot: [A] vs. t
slope = − k
[ A]o  [ A]t  kt, for[ A]t  [ A]o
Pressure
Time (s)
CH3NC (torr)
0
502
2,000
335
5,000
180
8,000
95.5
12,000
41.7
15,000
22.4
Time (min) [C12H22O11]
0
0.316
39
0.274
80
0.238
140
0.190
210
0.146
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t1/2 – time required for the concentration of
a reactant to decrease by half of its initial
value
[ A]0
[ A]t 1
[ A]t 

 @ t1/ 2
2
[ A]0 2

1st order half-life
◦ All half-lives same length
of time
◦ Independent of initial
concentration
 [ A]0 
  kt
ln 
 [ A]t 
 [ A]0 
  kt1/ 2
ln 
 0.5[ A]0 
0.693  kt1/ 2
Note: Radioactive decay
follows 1st order kinetics.
t1/ 2
0.693

k
1st order reaction

2nd order half-life
◦ All half-lives different
length of time
◦ Dependent on initial
concentration

t1/ 2
1

k [ A]0
t1/ 2
[ A]0

2k
Zero half-life
◦ All half-lives different
length of time
◦ Dependent on initial
concentration
Order
Rate law
zero
1st
2nd
rate = k
rate = k[A]
rate = k[A]2
Integrated
[A]t=−kt+[A]0 ln[A]t=−kt+ln[A]0 1/[A]t=kt+1/[A]0
rate law
Straight[A] vs. t
ln[A] vs. t
1/[A] vs. t
line plot
Slope
−k
−k
k
Half-life
(t1/2)
[A]o/2k
0.693/k
1/k[A]0
BLB 39
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Generally, as
temperature
increases, so
does reaction
rate.
This is because
k is temperature
dependent.
Collision theory
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In a chemical reaction, bonds are broken
and new bonds are formed. In order for
molecules to react, they must collide.
Collisions are either effective or ineffective
due to orientation of molecules.
Collisions must have enough energy to
overcome the barrier to reaction, the
activation energy.
Temperature affects the number of
collisions.
Molecular Collisions
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Energy barrier (hump) that must be overcome
for a chemical reaction to proceed
Activated complex or transition state –
arrangement of atoms at the top of the barrier
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Energy difference between the reactant and the
highest energy along the reaction pathway
Reaction specific
Rate of reaction is dependent upon the
magnitude of Ea; Ea ↓, rate ↑ (generally)
Temperature independent
Maxwell-Boltzman Distribution

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At higher
temperatures,
more molecules
will have adequate
energy to react.
This increases the
reaction rate.

Svante Arrhenius developed an equation for
the mathematical relationship between k
and Ea.
Ea

RT
k  Ae

A is the frequency factor, which represents
the number of effective collisions.
k  Ae

Ea
RT
Ea
ln k  ln A 
RT
 Ea  1 
ln k   
   ln A
 R  T 
y
=
m
x
+
b
Ea
m
 Ea   m  R
R
where :
R  8.3145 molJK
T in K
 k2 
ln  
k1 
y ln k 2  ln k1 

m

or
x
1 1
1 1
  
  
 T2 T1 
 T2 T1 
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Reactions occur in a series of elementary
steps collectively called a mechanism.
Determining the reaction mechanism is the
overall goal of kinetic studies.
One step, the rate-determining step, is
much slower than the other.
Usually, an intermediate (isolable) or a
transition state (non-isolable) is formed at
some point during the reaction.
molecularity – the number of molecules that
participate in a reaction
Molecularity is the number of molecules reacting.
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Catalysts – increase the rate of a reaction
without being consumed or changing
chemically
Accomplished by lowering the activation
energy and changing the reaction mechanism.
Heterogeneous vs. homogeneous catalysis
Examples:
◦ Catalytic converter (p. 608)
◦ Enzymes in the body (pp. 609-611)
◦ Ozone depletion (pp. 774-775)
Heterogeneous catalytic ethylene hydrogenation: C2H4 + H2 → C2H6