Transcript Document

Chemistry II
Chapter 13
Chemical Kinetics
The Rate of a Chemical Reaction
• kinetics is the study of the factors that affect the speed of a
reaction and the mechanism by which a reaction proceeds
• Factors that influence the speed of a reaction:




physical state of reactants,
temperature,
catalysts,
concentration
The Rate of a Chemical Reaction
Defining Rate
rate is how much a quantity changes in a given period of time,
e.g.
 distance
Speed 
 time
The Rate of a Chemical Reaction
• Chemical reaction - concentration change with time
 concentrat ion
Rate 
 time
• for reactants, a negative sign is used to show a decrease in
concentration
 [product]
 [reactant]
Rate 

 time
 time
• as time goes on, the rate of a reaction generally slows down and stops
 because the concentration of the reactants decreases
at t = 0
[A] = 8
[B] = 8
[C] = 0
at t = 0
[X] = 8
[Y] = 8
[Z] = 0
at t = 16
[A] = 4
[B] = 4
[C] = 4
at t = 16
[X] = 7
[Y] = 7
[Z] = 1


 C

A 1 
CA  C2A
2 
1
Rate  

t 2 tt21  t1 
tt
44 08 0.025.25
Rate  
1616 00

1 X 1 
 Z
ZX  Z2X
2
Rate  

t 2 tt21  t1 
tt
17 08 0.00625
.0625
Rate  
1616 00
at t = 16
[A] = 4
[B] = 4
[C] = 4
at t = 16
[X] = 7
[Y] = 7
[Z] = 1
at t = 32
[A] = 2
[B] = 2
[C] = 6
at t = 32
[X] = 6
[Y] = 6
[Z] = 2
 2C1 A 1 
CA  C2A
Rate  

t 2 tt21  t1 
tt
62 44 0.0125
Rate  
.125
1616 00

1 X 1 
 Z
ZX  Z2X
2
Rate  

t 2 tt21  t1 
tt
26 17  0.00625
.0625
Rate  
1616 00
at t = 32
[A] = 2
[B] = 2
[C] = 6
at t = 32
[X] = 6
[Y] = 6
[Z] = 2
at t = 48
[A] = 0
[B] = 0
[C] = 8
at t = 48
[X] = 5
[Y] = 5
[Z] = 3
 2C1 A 1 
CA  C2A
Rate  

t 2 tt21  t1 
tt
80 62 0.0125
Rate  
.125
1616 00

1 X 1 
 Z
ZX  Z2X
2
Rate  

t 2 tt21  t1 
tt
35 26 0.00625
.0625
Rate  
1616 00
Rate overall  
A2  A1 
A

t 2  t1 
t

0  8
Rate overall  
 0.174
46  0
Rate overall 
C C2  C1 

t 2  t1 
t

8  0
Rate overall 
 0.174
46  0
8


X
X2  X1 
Rate overall  

t 2  t1 
t
Rate overall  
Rate overall 
5  8  0.065
46  0
Z Z2  Z1 

t 2  t1 
t

3  0
Rate overall 
 0.065
46  0
rate of change reactants = rate of change products
9
The Rate of a Chemical Reaction
Stoichiometry
• e.g.
H2 (g) + I2 (g)  2 HI(g)
• for the above reaction, for every 1 mole of H2 used, 1 mole of I2 will
also be used and 2 moles of HI made
 therefore the rate of change will be different
• in order to be consistent, the change in the concentration of each
substance is multiplied by 1/coefficient
[H 2 ]
[I 2 ]
 1  [HI]
Rate  

  
t
t
 2  t
The Rate of a Chemical Reaction
Stoichiometry
• e.g.
H2 (g) + I2 (g)  2 HI(g)
The Rate of a Chemical Reaction
Average Rate
• the average rate is the change in measured concentrations in any
particular time period
 linear approximation of a curve
• the larger the time interval, the more the average rate deviates from
the instantaneous rate
H2
I2
HI
Avg. Rate, M/s Avg. Rate, M/s
Time (s) [H2], M [HI], M
-[H2]/t
1/2 [HI]/t
Stoichiometry tells us that
every 1rate
mole/L
of change
H
Theforaverage
is the
in
0.000 1.000 0.000
used,
the concentration in a given time
2
10.000
20.000
30.000
0.819
0.670
0.549
0.362
0.660
0.902
40.000
50.000
60.000
70.000
0.449
0.368
0.301
0.247
1.102
1.264
1.398
1.506
Ave. rate slows down as reaction
proceeds
2 moles/L of HI are made.
period.
0.0181
0.0181
0.0149
0.0149
Assuming a 1 L container, at 10 s, we used 0.181 moles
of H . Therefore the amount of HI made is 2(0.181
0.0121
moles) 0.0121
= 0.362 moles
2
In the first 10 s, the Δ[H2] is -0.181
0.0100 M, so the rate0.0100
is
0.0081
0.0081
0.0067
0.0067
At 60 s, we used 0.699 moles of H . Therefore the
amount0.0054
of HI made is 2(0.699 moles)0.0054
= 1.398 moles
2
80.000
0.202
1.596
0.0045
90.000
100.000
0.165
0.135
1.670
1.730
0.0037
0.0030
Rate of loss reactant = Rate gain product
 0.181 M
0.0045

10.000 s
0.0037
M
 0.0181
0.0030 s
2.000
1.800
concentration, (M)
1.600
1.400
Concentration vs. Time for H2 + I2 --> 2HI
average rate in a given time
period =  slope of the line
connecting the [H2] points;
and ½ +slope of the line for
[HI]
1.200
the average rate for the
first 80
10 s is 0.0108
40
0.0181 M/s
0.0150
1.000
[H2], M
[HI], M
0.800
0.600
0.400
0.200
Ave. rate slows down as
0.000
reaction
0.000 proceeds
10.000
20.000
30.000
40.000
50.000
time, (s)
60.000
70.000
80.000
90.000
100.000
The Rate of a Chemical Reaction
Instantaneous Rate
• average rate becomes less accurate over longer time spans
• the instantaneous rate is the change in concentration at any one
particular time
 slope at one point of a curve
• determined by taking the slope of a line tangent to the curve at that
particular point
 first derivative of the function
 for you calculus fans
H2 (g) + I2 (g)  2 HI (g)
Using [H2], the
instantaneous
rate at 50 s is:
Rate  
 0.28 M
40 s
Rate  0.0070
M
s
Using [HI], the
instantaneous
rate at 50 s is:
 1  0.56 M
Rate   
 2  40 s
rate reactants = rate
products
Rate  0.0070
M
s
Generalized rate law:
aA + bB → cC + dD
1 [A]
1 [B]
1 [C]
1 [D]
Rate  



a t
b t
c t
d t
Ex 13.1 - For the reaction given, the [I] changes from 1.000 M to 0.868 M
in the first 10 s. Calculate the average rate in the first 10 s and the Δ[H+].
H2O2 (aq) + 3 I(aq) + 2 H+(aq)  I3(aq) + 2 H2O(l)
Solve the equation for
the Rate (in terms of the
change in concentration
of the Given quantity)

 1  [I ]
 1  0.868 M  1.000 M 
Rate    
  
10 s
 3  t
 3
Rate  4.40 10-3
Solve the equation of the
Rate (in terms of the
change in the
concentration for the
quantity to find) for the
unknown value
M
s

 1  [H ]
Rate    
 2  t
[H  ]
 2Rate 
t


[H  ]
M
M
 2 4.40 10-3
  8.80 10-3
s
s
t
Factors Affecting Reaction Rate
- nature of reactants
• nature of the reactants means what kind of reactant molecules and what physical
condition they are in.
 small molecules tend to react faster than large molecules;
 gases tend to react faster than liquids which react faster than solids;
 powdered solids are more reactive than “blocks”
 more surface area for contact with other reactants
 certain types of chemicals are more reactive than others
 e.g., the activity series of metals
 ions react faster than molecules
 no bonds need to be broken
Factors Affecting Reaction Rate
- Temperature
• increasing temperature increases reaction rate
 chemist’s rule of thumb - for each 10°C rise in temperature, the
speed of the reaction doubles
• there is a mathematical relationship between the absolute
temperature and the speed of a reaction discovered by Svante
Arrhenius which will be examined later
Factors Affecting Reaction Rate
- Catalysts
• catalysts are substances which affect the speed of a reaction without
being consumed
• most catalysts are used to speed up a reaction, these are called positive
catalysts
 catalysts used to slow a reaction are called negative catalysts
• homogeneous = present in same phase
• heterogeneous = present in different phase
• how catalysts work will be examined later
Factors Affecting Reaction Rate
- Reactant Concentration
• generally, the larger the concentration of reactant molecules, the
faster the reaction
 increases the frequency of reactant molecule contact
 concentration of gases depends on the partial pressure of the
gas
 higher pressure = higher concentration
• concentration of solutions depends on the solute to solution ratio
(molarity)
The Rate Law:
Effect of Concentration On Reaction Rate
• Mathematical relationship between the rate of the reaction and the
concentrations of the reactants
• for the reaction aA + bB  products the rate law would have the form
given below
 n and m are called the orders for each reactant
 k is called the rate constant
Rate  k[A] [B]
n
m
The Rate Law:
Effect of Concentration On Reaction Rate
• sum of the exponents is called the order of the reaction
• The rate law for the reaction:
2 NO(g) + O2(g) ⇌ 2 NO2(g)
Rate = k[NO]2[O2]
The reaction is second order with respect to [NO],
first order with respect to [O2],
and third order overall
The Rate Law:
Effect of Concentration On Reaction Rate
Sample Rate Laws
Reaction
Rate Law
CH3CN  CH3NC
Rate = k[CH3CN]
CH3CHO  CH4 + CO
Rate = k[CH3CHO]
2 N2O5  4 NO2 + O2
Rate = k[N2O5]2
H2 + I2  2 HI
Rate = k[H2][I2]
The Rate Law:
Effect of Concentration On Reaction Rate
Example:
A → Products
Rate = k[A]1
[A] (M)
Initial Rate
(M/s)
0.10
0.015
0.20
0.030
0.30
0.060
k = 0.015 / 0.10 = 0.15 s-1
If concentration of A doubles, the new rate,
Rate2 = k[2A]1
= 2 k[A]1 = 2 x Rate
The Rate Law:
Effect of Concentration On Reaction Rate
Example:
Zero order:
Second order:
[A] (M)
Initial Rate
(M/s)
[A] (M)
Initial Rate
(M/s)
0.10
0.015
0.10
0.015
0.20
0.015
0.20
0.060
0.30
0.015
0.30
0.240
Concentration of A doubles, the rate is constant
Rate1 = Rate2 = k
Conc. A doubles, new rate
Rate2 = k[2A]2 = 4 x k[A]2
Rate2 = 4 x Rate
Reactant Concentration vs. Time
A  Products
0: Concentration dec.
linearly with time. Rate Is
constant, reaction does
not slow down as [A] dec.
1 and 2: Rate slows as
reaction proceeds since
[A] dec.
Rate = k[A]2
Rate = k[A]
Rate = k
The Integrated Rate Law
• can only be determined experimentally
• graphically
 rate = slope of curve [A] vs. time
 if graph [A] vs time is straight line, then exponent on A in rate law
is 0, rate constant = -slope
 if graph ln[A] vs time is straight line, then exponent on A in rate
law is 1, rate constant = -slope
 if graph 1/[A] vs time is straight line, exponent on A in rate law is
2, rate constant = slope
The Integrated Rate Law
• the half-life, t1/2, of a reaction is
the length of time it takes for the
concentration of the reactants to
fall to ½ its initial value
• the half-life of the reaction
depends on the order of the
reaction
Zero Order Reactions (n = 0)
Rate = -d[A] = k[A]0 = k
dt
 constant rate reactions
Solution: [A] = -kt + [A]0
y = mx + b
∫ -d[A]/dt = ∫ k
∫ d[A] = -∫ k dt
[A] = -kt + C, where C = [A]0
(= integrated rate law)
graph of [A] vs. time is straight line with
slope = -k and y-intercept = [A]0
[A] = [A0]/2, t ½ = [A0]/2k
Units: when Rate = M/sec, k = M/sec
[A]0
[A]
time
First Order Reactions (n = 1)
∫ -d[A]/dt = ∫ k[A]
∫ 1 d[A] = -∫ k dt
[A]
ln[A] = -kt + C, where C = ln[A]0
Rate = -d[A] = k[A]
dt
Solution: ln[A] = -kt + ln[A]0
graph of ln[A] vs. time gives straight line with
slope = -k and y-intercept = ln[A]0

used to determine the rate constant
[A] = [A0]/2,
t½ = ln 2
k
[lna-lnb = ln(a/b)]
ln[A]0
the half-life of a first order reaction is constant
ln[A]
Units: when Rate = M/sec, k =
sec-1
(dim. Analysis: M/s = k.M)
Rate slows as reaction proceeds since [A] dec.
time
Rate Data for hydrolysis of C4H9Cl
Time (sec) [C4H9Cl], M
0.0
0.1000
Show
reaction is
first-order
and find k
50.0
100.0
150.0
0.0905
0.0820
0.0741
200.0
300.0
400.0
500.0
0.0671
0.0549
0.0448
0.0368
800.0
10000.0
0.0200
0.0000
C4H9Cl + H2O  C4H9OH + 2 HCl
Concentration vs. Time for the Hydrolysis of C 4H9Cl
0.12
concentration, (M)
0.1
0.08
0.06
0.04
0.02
0
0
200
400
600
time, (s)
800
1000
C4H9Cl + H2O  C4H9OH + 2 HCl
Rate vs. Time for Hydrolysis of C 4H9Cl
2.5E-04
Rate, (M/s)
2.0E-04
1.5E-04
1.0E-04
5.0E-05
0.0E+00
0
100
200
300
400
time, (s)
500
600
700
800
C4H9Cl + H2O  C4H9OH + 2 HCl
LN([C4H9Cl]) vs. Time for Hydrolysis of C 4H9Cl
0
slope =
-2.01 x 10-3
-0.5
LN(concentration)
-1
k=
2.01 x 10-3 s-1
-1.5
-2
-2.5
-3
t1 
y = -2.01E-03x - 2.30E+00
2
-3.5

0.693
k
0.693
2.0110 3 s -1
 345 s
-4
-4.5
0
100
200
300
400
time, (s)
500
600
700
800
Second Order Reactions
Rate = -d[A] = k[A] = k[A]2
dt
Solution: 1/[A] = kt + 1/[A]0
y = mx + b
graph 1/[A] vs. time gives straight line with
slope = k and y-intercept = 1/[A]0
 used to determine the rate constant
1/[A]
t½ =
1
k[A0]
when Rate = M/sec, k = M-1∙sec-1
l/[A]0
time
Example Second-Order Reaction
Show that the reaction: NO2 → NO + O is incorrect according to the data
below
PNO2,
(mmHg)
Time (hrs.)
ln(PNO2)
1/(PNO2)
0
100.0
4.605
0.01000
30
60
62.5
45.5
4.135
3.817
0.01600
0.02200
90
35.7
3.576
0.02800
120
150
29.4
25.0
3.381
3.219
0.03400
0.04000
180
21.7
3.079
0.04600
210
19.2
2.957
0.05200
240
17.2
2.847
0.05800
Rate Data Graphs For
NO2 → NO + O
Partial Pressure NO2, mmHg vs. Time
100.0
90.0
Pressure, (mmHg)
80.0
Non-linear so
not zero order
70.0
60.0
50.0
40.0
30.0
20.0
10.0
0.0
0
50
100
150
Time, (hr)
200
250
Rate Data Graphs For
NO2 ® NO + O
ln(PNO2) vs. Time
4.8
4.6
4.4
ln(pressure)
4.2
4
Non-linear so not first
order
3.8
3.6
3.4
3.2
3
2.8
2.6
2.4
0
50
100
150
Time (hr)
200
250
Rate Data Graphs For
NO2 → NO + O
1/(PNO2) vs Time
k = 2 x 10-4 M-1 s-1
0.07000
1/PNO2 = 0.0002(time) + 0.01
-1
Inverse Pressure, (mmHg )
0.06000
0.05000
0.04000
0.03000
0.02000
We can deduce actual reaction should be:
2NO2 → 2NO + O2
0.01000
0.00000
0
50
100
150
Time, (hr)
200
250
Tro, Chemistry: A Molecular Approach
47
Ex. 13.4 – The reaction SO2Cl2(g)  SO2(g) + Cl2(g) is first order with a rate constant of 2.90 x 10-4
s-1 at a given set of conditions. Find the [SO2Cl2] at 865 s when [SO2Cl2]0 = 0.0225 M
Given:
Find:
[SO2Cl2]0 = 0.0225 M, t = 865, k = 2.90 x 10-4 s-1
[SO2Cl2]
Concept Plan:
[SO2Cl2]0, t, k
[SO2Cl2]
Relationships:
for a 1st order process : ln[A]  kt  ln[A] 0
Solution:
ln[SO 2Cl2 ]  kt  ln[SO 2Cl2 ]0


ln[SO 2Cl2 ]  2.90 10- 4 s -1 865 s   ln 0.0225
ln[SO 2Cl2 ]  0.251  3.79  4.04
[SO 2Cl2 ]  e(-4.04)  0.0175 M
Check:
the new concentration is less than the original, as expected
Ex 13.2 – Determine the rate law and rate constant for the
reaction NO2(g) + CO(g)  NO(g) + CO2(g)
given the data below.
Write a general
rate law
including all
reactants
Examine the
data and find
two experiments
in which the
concentration of
one reactant
changes, but the
other
concentrations
are the same
Expt.
Initial
Initial Rate
Initial
Number
(M)
(M/s)
Number [NO22], (M) [CO],
[CO], (M)
(M/s)
1.
0.10
0.10
0.0021
2.
0.20
0.10
0.0082
3.
0.20
0.20
0.0083
0.033
4.
0.40
0.10
Rate  k[NO2 ] [CO]
n
m
Comparing Expt #1 and Expt #2, the
[NO2] changes but the [CO] does not
49
Ex 13.2 – Determine the rate law and rate constant for the
reaction NO2(g) + CO(g)  NO(g) + CO2(g)
given the data below.
Determine by
what factor the
concentrations
and rates change
in these two
experiments.
[NO2 ]expt 2
[NO2 ]expt 1
Expt.
Initial
Initial Rate
Initial
Number [NO2], (M) [CO], (M)
(M/s)
1.
2.
0.10
0.20
0.10
0.10
0.0021
0.0082
3.
0.20
0.20
0.0083
4.
0.40
0.10
0.033
0.20 M

2
0.10 M
Rate expt 2
Rate expt 1
0.0082 M s

4
M
0.0021 s
Ex 13.2 – Determine the rate law and rate constant for the
reaction NO2(g) + CO(g)  NO(g) + CO2(g)
given the data below.
Determine to
what power the
concentration
factor must be
raised to equal
the rate factor.
[NO2 ]expt 2
[NO2 ]expt 1
Expt.
Initial
Initial Rate
Initial
Number [NO2], (M) [CO], (M)
(M/s)
1.
0.10
0.10
0.0021
2.
0.20
0.10
0.0082
3.
0.20
0.20
0.0083
4.
0.40
0.10
0.033
0.20 M
n
M
Rate
0
.
0082
expt
2


2
s
 [NO 2 ]expt 2 
Rate expt 2

4
0.10 M 
 
M
Rate expt 1 0.0021 s
 [NO 2 ]expt 1 
Rate expt 1


2n  4
n2
Ex 13.2 – Determine the rate law and rate constant for the
reaction NO2(g) + CO(g)  NO(g) + CO2(g)
given the data below.
Repeat for the
other reactants
Expt.
Initial
Initial Rate
Initial
Number [NO2], (M) [CO], (M)
(M/s)
1.
0.10
0.10
0.0021
2.
0.20
0.10
0.0082
3.
0.20
0.20
0.0083
4.
0.40
0.10
0.033
m
0[CO]

M
Rate
Rate
0
.
0083
expt
3
expt
3
.
20
M
expt
3
s
  [CO]  2  Rate

1
expt 2 
Rateexpt
0.0082 M s
[CO]expt 2 0.10 M
expt2 2
2m  1
m0
[CO]expt 3
Ex 13.2 – Determine the rate law and rate constant for the
reaction NO2(g) + CO(g)  NO(g) + CO2(g)
given the data below.
Substitute the
exponents into
the general rate
law to get the
rate law for the
reaction
Expt.
Initial
Initial Rate
Initial
Number [NO2], (M) [CO], (M)
(M/s)
1.
0.10
0.10
0.0021
2.
0.20
0.10
0.0082
3.
0.20
0.20
0.0083
4.
0.40
0.10
0.033
n = 2, mRate
=0
Rate
n0
m
2
 2k][NO
Rate
k[NO
[CO]
2 ] [CO]
2
 k[NO2 ]
Ex 13.2 – Determine the rate law and rate constant for the
reaction NO2(g) + CO(g)  NO(g) + CO2(g)
given the data below.
Substitute the
concentrations
and rate for any
experiment into
the rate law and
solve for k
Expt.
Initial
Initial Rate
Initial
Number [NO2], (M) [CO], (M)
(M/s)
1.
0.10
0.10
0.0021
2.
0.20
0.10
0.0082
3.
0.20
0.20
0.0083
4.
0.40
0.10
0.033
Rate  k[NO2 ]2
for expt 1
0.0021 M s  k 0.10 M 2
0.0021 M s
-1
-1
k

0
.
21
M

s
0.01 M 2
The Effect of Temperature on Reaction Rate
• Rate constant k is temperature dependent
• Arrhenius investigated this relationship and showed that:

k  A e

 Ea
RT




where T is the temperature in kelvin
R is the gas constant in energy units, 8.314 J/(mol∙K)
A is a constant called the frequency factor
Ea is the activation energy, the extra energy needed to start the
molecules reacting
As x (temperature) increases e-1/x will increase
up to a maximum value of 1, k increases
10
8
ex
6
e-x
y
e1/x
e-1/x
4
2
0
0
10
x
20
As Ea increases k will decrease (follows e-x graph)
The Effect of Temperature on Reaction Rate
Activation Energy and the Activated Complex
• Ea is an energy barrier to the reaction
• amount of energy needed to convert
reactants into the activated complex
 aka transition state
• the activated complex is a chemical
species with partially broken and
partially formed bonds
 always very high in energy
because partial bonds
The Effect of Temperature on Reaction Rate
The Exponential Factor
• e-Ea/RT is a number between 0 and 1
• it represents the fraction of reactant molecules with sufficient energy to
make it over the energy barrier
• that extra energy comes from converting the KE of motion to PE in the
molecule when the molecules collide
• e-Ea/RT decreases as Ea increases
Reaction rate inc.:
-Increasing T increases the Ave. KE of the
molecules
-Increases no. of molecules with sufficient
energy to overcome the energy barrier
The Effect of Temperature on Reaction Rate
Arrhenius Plots
• the Arrhenius Equation can be algebraically solved:
 Ea
ln( k ) 
R
1
   ln  A
T
y = mx + b
where y = ln(k) and x = (1/T)
a graph of ln(k) vs. (1/T) is a straight line
slope of the line = -Ea/R
so Ea = -mR
ey-intercept = A, (unit is the same as k …why?)
k = Ae-Ea/RT
lnk = ln(Ae-Ea/RT)
lnk = lnA + lne-Ea/RT
lnk = lnA – Ea/RT
Ex. 13.7 Determine the activation energy and frequency factor for the
reaction O3(g)  O2(g) + O(g) given the following data:
Temp, K
600
700
k, M-1∙s-1
3.37 x 103
4.83 x 104
Temp, K
1300
1400
k, M-1∙s-1
7.83 x 107
1.45 x 108
800
900
1000
1100
3.58 x 105
1.70 x 106
5.90 x 106
1.63 x 107
1500
1600
1700
1800
2.46 x 108
3.93 x 108
5.93 x 108
8.55 x 108
1200
3.81 x 107
1900
1.19 x 109
Ex. 13.7 Determine the activation energy and frequency factor for the
reaction O3(g)  O2(g) + O(g) given the following data:
use a
spreadsheet to
graph ln(k) vs.
(1/T)
Ex. 13.7 Determine the activation energy and frequency factor for the
reaction O3(g)  O2(g) + O(g) given the following data:
Ea = m∙(-R)
solve for Ea


J 
4 J
Ea  1.12 104 K  8.314
  9.3110
mol  K 
mol

kJ
Ea  93.1
mol
A = ey-intercept
solve for A
A  e 26 .8  4.36 1011
A  4.36 1011 M -1  s 1
The Effect of Temperature on Reaction Rate
The Collision Model
•
•
for most reactions, in order for a reaction to take place, the reacting
molecules must collide into each other.
once molecules collide they may react together or they may not,
depending on two factors –
1. whether the collision has enough energy to "break the bonds holding
reactant molecules together";
2. whether the reacting molecules collide in the proper orientation for new
bonds to form.
The Effect of Temperature on Reaction Rate
The Collision Model
Effective Collisions
• collisions in which these two conditions are met
(and therefore lead to reaction) are called
effective collisions
•
the higher the A value (frequency of effective
collisions), the higher k value and the faster the
reaction rate
•
when two molecules have an effective collision, a
temporary, high energy (unstable) chemical
species is formed - called an activated complex
or transition state
The Effect of Temperature on Reaction Rate
The Collision Model
Orientation Effect
The Effect of Temperature on Reaction Rate
The Collision Model
• A is the factor called the frequency factor and is the number of
molecules that can approach overcoming the energy barrier
• there are two factors that make up the frequency factor – the
orientation factor (p) and the collision frequency factor (z)
 RTEa
k  A e

 Ea

  pze RT


The Effect of Temperature on Reaction Rate
The Collision Model
Orientation Factor
•
proper orientation is when the atoms are aligned so that old bonds can break and the
new bonds can form
•
the more complex the reactants, the less frequently they will collide with the proper
orientation
 reactions between atoms generally have p = 1
 reactions where symmetry results in multiple orientations leading to reaction have p
slightly less than 1
•
for most reactions, the orientation factor is less than 1
 For many, p << 1
e.g.
H(g) + I(g) → HI(g)
H2(g) + I2(g) → 2HI(g)
HCl(g) + HCl(g) → H2(g) + Cl2(g)
Smallest p
Reaction Mechanisms
• we generally describe chemical reactions with an equation listing all
the reactant molecules and product molecules
• but the probability of more than 3 molecules colliding at the same
instant with the proper orientation and sufficient energy to overcome
the energy barrier is negligible
• most reactions occur in a series of small reactions involving 1, 2, or at
most 3 molecules
• describing the series of steps that occur to produce the overall
observed reaction is called a reaction mechanism
• knowing the rate law of the reaction helps us understand the
sequence of steps in the mechanism
Reaction Mechanisms
•
•
•
Overall reaction:
H2(g) + 2 ICl(g)  2 HCl(g) + I2(g)
Mechanism:
1) H2(g) + ICl(g)  HCl(g) + HI(g)
2) HI(g) + ICl(g)  HCl(g) + I2(g)
the steps in this mechanism are elementary steps, meaning that
they cannot be broken down into simpler steps and that the
molecules actually interact directly in this manner without any other
steps
Reaction Mechanisms
Intermediates
1) H2(g) + ICl(g)  HCl(g) + HI(g)
2) HI(g) + ICl(g)  HCl(g) + I2(g)
H2(g) + 2 ICl(g)  2 HCl(g) + I2(g)
•
•
•
notice that the HI is a product in Step 1, but then a reactant in Step 2
since HI is made but then consumed, HI does not show up in the overall
reaction
materials that are products in an early step, but then a reactant in a later
step are called intermediates
Reaction Mechanisms
Molecularity
• the number of reactant particles in an elementary step is called its
molecularity
• a unimolecular step involves 1 reactant particle
• a bimolecular step involves 2 reactant particles
 though they may be the same kind of particle
• a termolecular step involves 3 reactant particles
 though these are exceedingly rare in elementary steps
Reaction Mechanisms
Rate Laws for Elementary Steps
• each step in the mechanism is like its own little reaction – with its
own activation energy and own rate law
• the rate law for an overall reaction must be determined
experimentally
• but the rate law of an elementary step can be deduced from the
equation of the step
H2(g) + 2 ICl(g)  2 HCl(g) + I2(g)
1) H2(g) + ICl(g)  HCl(g) + HI(g)
2) HI(g) + ICl(g)  HCl(g) + I2(g)
Rate = k1[H2][ICl]
Rate = k2[HI][ICl]
Reaction Mechanisms
Rate Laws for Elementary Steps
Reaction Mechanisms
Rate Determining Step
• in most mechanisms, one step occurs slower than the other steps
• the result is that product production cannot occur any faster than the
slowest step – the step determines the rate of the overall reaction
• we call the slowest step in the mechanism the rate determining step
 the slowest step has the largest activation energy
• the rate law of the rate determining step determines the rate law of the
overall reaction
Another Reaction Mechanism
NO2(g) + NO2(g)  NO3(g) + NO(g)
NO3(g) + CO(g)  NO2(g) + CO2(g)
NO2(g) + CO(g)  NO(g) + CO2(g)
Rate = k1[NO2]2 slow
Rate = k2[NO3][CO] fast
Rateobs = k[NO2]2
The first step is slower than the
second step because its activation
energy is larger.
The first step in this
mechanism is the rate
determining step.
The rate law of the first step
is the same as the rate law of
the overall reaction.
Reaction Mechanisms
Validating a Mechanism
in order to validate (not prove) a mechanism, two conditions must
be met:
1. the elementary steps must sum to the overall reaction
2. the rate law predicted by the mechanism must be consistent with the
experimentally observed rate law
Reaction Mechanisms
Mechanisms with a Fast Initial Step
• when a mechanism contains a slow initial step, the rate law will not contain
•
•
•
•
intermediates
when a mechanism contains a fast initial step, the rate limiting step, and
hence the rate law may contain intermediates
We can express[intermediate] in terms of [reactant]
If first step is fast, intermediate products build up (limited by slower step down
the line)
 as they build up they react to re-form reactants
 reaches equilibrium, the forward and reverse reaction rates are equal – so
the concentrations of reactants and products of the step are related
substituting into the rate law of the RDS will produce a rate law in terms of just
reactants
An Example
2 H2(g) + 2 NO(g)  2 H2O(g) + N2(g)
Experimentally observed Rateobs = k [H2][NO]2
Proposed mechanism:
k1
2 NO(g) ⇌ N2O2(g)
Fast
k-1
H2(g) + N2O2(g)  H2O(g) + N2O(g)
H2(g) + N2O(g)  H2O(g) + N2(g)
Slow
Fast
2 H2(g) + 2 NO(g)  2 H2O(g) + N2(g)
(rate limiting)
Reaction Mechanisms
Mechanisms with a Fast Initial Step
• Is the mechanism valid?
1. steps must sum to the over all reaction
2. rate law predicted by mechanism must be consistent with exp.
observation
Since 2nd step is rate limiting, Rate = k2[H2][N2O2]
BUT! Contains intermediate [N2O2] , not consistent with observation
Since 1st step is in equilibrium we can express [intermediate] in terms
of reactants
An Example
k1
2 NO(g) ⇌ N2O2(g)
k-1
H2(g) + N2O2(g)  H2O(g) + N2O(g)
H2(g) + N2O(g)  H2O(g) + N2(g)
Fast
Slow
Rate = k2[H2][N2O2]
Fast
2 H2(g) + 2 NO(g)  2 H2O(g) + N2(g) Rateobs = k [H2][NO]2
for Step 1 Rateforward = Ratereverse
k1[NO]2  k1[N 2O 2 ]
[N 2O 2 ] 
k1
[NO]2
k1
Rate  k 2 [H 2 ][N 2 O 2 ]
Rate  k 2 [H 2 ]
Rate 
k1
[NO]2
k 1
k 2 k1
[H 2 ][NO]2
k 1
Let k2k1/k-1 = k, we have obs rate law
Ex 13.9 Show that the proposed mechanism for the reaction
2 O3(g)  3 O2(g) matches the observed rate law
Rate = k[O3]2[O2]-1
k1
O3(g) ⇌ O2(g) + O(g)
k-1
O3(g) + O(g)  2 O2(g)
Fast
Slow
Rate = k2[O3][O]
(Slow rate has intermediate)
for Step 1 Rateforward = Ratereverse
k1[O3 ]  k1[O 2 ][O]
k1
[O] 
[O3 ][O 2 ]1
k1
Rate  k2[O3 ][O]
k1
Rate  k2[O3 ] [O3 ][O 2 ]-1
k1
k2k1
Rate 
[O3 ]2[O 2 ]-1
k1
Catalysts
• catalysts are substances that affect the rate of a reaction without being
consumed
• catalysts work by providing an alternative mechanism for the reaction
 with a lower activation energy
• catalysts are consumed in an early mechanism step, then made in a
later step
mechanism without catalyst
mechanism with catalyst
O3(g) + O(g)  2 O2(g)
Cl(g) + O3(g) ⇌ O2(g) + ClO(g)
V. Slow
ClO(g) + O(g)  O2(g) + Cl(g)
O3(g) + O(g)  2 O2(g)
Fast
Slow
Catalysts
Demo
2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g)
http://academics.rmu.edu/~short/chem1215/1215demos/oscillating-methanol-7mins.mov
Ozone Depletion over the Antarctic
Catalysts
polar stratospheric clouds
contain ice crystals that
catalyze reactions that
release Cl from
atmospheric chemicals
Catalysts
Homorgeneous and Heterogeneous Catalysts
• homogeneous catalysts are in the same phase as the reactant
particles
 Cl(g) in the destruction of O3(g)
• heterogeneous catalysts are in a different phase than the reactant
particles
 solid catalytic converter in a car’s exhaust system
Catalysts
Enzymes
• because many of the molecules are large and complex, most
biological reactions require a catalyst to proceed at a reasonable rate
• protein molecules that catalyze biological reactions are called
enzymes
• enzymes work by adsorbing the substrate reactant onto an active site
that orients it for reaction
Enzymatic Hydrolysis of Sucrose