Download week-6

Take out a clean sheet of paper and print on it
your name and section number
and an indication it is “pop quiz #1”.
Quiz # 4 (Feb 14-18) will be
on sections 14.5-14.7,
and will be given in recitation
Mon Lab, 2:30-5:18
Wed Lab, 2:30-5:18
Pat Bullinger
Chris Beekman
Edwin Motari
Hai Liu
Lin Sun
Chitanya
Patwardhan
Jeremy White
96
Roxana Sierra 97
Ramesh Sharma 98
Lin Sun
99
Mark Lobas
100
Chitanya
Patwardhan 101
90
91
92
93
94
95
Fri: Patric BenziherÆ 102
Take out a clean sheet of paper and print on it
your name and section number
and an indication it is “pop quiz #1”.
Hai Liu Æ 103
A certain reaction has the form A Æ B. At 400 K and
[A]0 = 2.8 x 10 -3 M, data for a plot of [A] vs t were collected.
It was then found that a plot of
1/[A] vs t yielded a
straight line with a slope of 3.60 x 10-2 L·mol -1 ·s -1 .
Wed Lab, 6:30-9:18
Fri Lab, 11:30-2:18
Jen Kljun
Namrata Singh
Mike Johansen
Ray Chammas
Travis Steinke
Lisa Park
Sara Wicke
Patrick Veres
104
105
106
107
108
109
110
111
a) Write an expression for the rate law.
rate = k[A]2
b) Write an expression for the integrated rate law.
1/[A] = 1/[A]0 + kt
c) What is the rate constant for the reaction?
3.60 x 10-2 L·mol -1 ·s -1
d) What is the half-life for the reaction, as given?
t1 =
2
1
1
10 5
=
=
= 9.9 x103
k [ A]0
(3.60 x10−2 )( 2.8 x10 −3 ) (3.6)( 2.8)
A certain reaction has the form A Æ B. At 400 K and
[A]0 = 5.6 x 10 -3 M, data for a plot of [A] vs t were collected.
Week six, a continuation of
It was then found that a plot of
14.1
14.2
14.3
Factors that Affect Reaction Rates
Reaction Rates
Concentration and Rate
14.4
The Change of Concentration with Time
14.5
Temperature and Rate
The Collision Model
Activation Energy
The Orientation Factor
The Arrhenius Equation and Activation Energies
Reaction Mechanisms
Elementary Steps; Multistep Mechanisms
Rate Laws for Elementary Steps
Rate Laws for Multistep Mechanisms
Catalysis
Homogeneous and Heterogeneous Catalysis
Enzymes
1/[A] vs t yielded a
straight line with a slope of 1.80 x 10-2 L·mol -1 ·s -1 .
a) Write an expression for the rate law.
rate = k[A]2
b) Write an expression for the integrated rate law.
1/[A] = 1/[A]0 + kt
c) What is the rate constant for the reaction?
1.80 x 10-2 L·mol -1 ·s -1
d) What is the half-life for the reaction, as given?
t1 =
2
1
1
10 5
=
=
= 9.9 x103
k [ A]0
(1.80 x10−2 )(5.6 x10 −3 ) (1.8)(5.6)
Chapter 14
14.5
14.7
Chemical Kinetics
1
Note the DRAMATIC effect of temperature on k
Temperature and Rate
The Collision Model eg H2 + I2
The Collision Model
• The more molecules present, the greater the
probability of collision and the faster the rate.
• Complication: not all collisions lead to products. In
fact, only a small fraction of collisions lead to
product.
• The higher the temperature, the more energy
available to the molecules and the faster the rate.
• In order for reaction to occur the reactant molecules
must collide in the correct orientation and with
enough energy to form products.
Activation Energy
Activation Energy
• Arrhenius: molecules must posses a
minimum amount of energy to react. Why?
– In order to form products, bonds must be broken
in the reactants.
– Bond breakage requires energy.
• Activation energy, Ea, is the minimum energy
required to initiate a chemical reaction.
Activation Energy
• Consider the rearrangement of acetonitrile:
H3C N C
H3C
N
C
H3C C N
– In H3C-N≡C, the C-N≡C bond bends until the C-N
bond breaks and the N≡C portion is
perpendicular to the H3C portion. This structure
is called the activated complex or transition state.
– The energy required for the above twist and
break is the activation energy, Ea.
– Once the C-N bond is broken, the N≡C portion
can continue to rotate forming a C-C≡N bond.
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Activation Energy
• The change in energy for the reaction is the
difference in energy between CH3NC and
CH3CN.
• The activation energy is the difference in
energy between reactants, CH3NC and
transition state.
• The rate depends on Ea.
• Notice that if a forward reaction is exothermic
(CH3NC → CH3CN), then the reverse
reaction is endothermic (CH3CN → CH3NC).
Activation Energy
• Consider the reaction between Cl and NOCl:
– If the Cl collides with the Cl of NOCl then the
products are Cl2 and NO.
– If the Cl collided with the O of NOCl then no
products are formed.
• We need to quantify this effect.
The Arrhenius Equation
Activation Energy
• Arrhenius discovered most reaction-rate data
obeyed the Arrhenius equation:
k = Ae
− Ea
RT
– k is the rate constant, Ea is the activation energy,
R is the gas constant (8.314 J/K-mol) and T is the
temperature in K.
– A is called the frequency factor, and is a measure
of the probability of a favorable collision.
– Both A and Ea are specific to a given reaction.
3
Sample Exercise 14.8
The Arrhenius Equation
• If we have a lot of data, we can determine Ea
and A graphically by rearranging the
Arrhenius equation:
E
ln k = − a + ln A
RT
Temp. /oC
k / (s-1 )
189.7
2.52 x 10-5
198.9
5.25 x 10-5
230.3
6.30 x 10-5
251.2
3.16 x10-5
• If we do not have a lot of data, then we can
use
− Ea  1 1 
k
 − 
ln 2 =
k1
R  T2 T1 
Sample Exercise 14.8
Temp. (oC)
T (K)
Sample Exercise 14.8
1/T (K-1)
k (s-1 )
ln k
Temp. (oC)
T (K)
1/T (K-1 x 103)
k (s-1 )
189.7
2.52 x 10-5
189.7
462.9
2.160
2.52 x 10-5
198.9
5.25 x
10-5
198.9
472.1
2.118
5.25 x 10-5
230.3
6.30 x 10-5
230.3
503.5
1.986
6.30 x 10-5
251.2
3.16 x10-5
251.2
524.4
1.907
3.16 x10-5
ln k
Sample Exercise 14.8
Temp. (oC)
T (K)
1/T (K-1 x 103)
k (s-1 )
ln k
189.7
462.9
2.160
2.52 x 10-5
-10.589
10-5
-9.855
198.9
472.1
2.118
5.25 x
230.3
503.5
1.986
6.30 x 10-5
-7.370
251.2
524.4
1.907
3.16 x10-5
-5.757
Fig 14.18 For CH3NC Æ CH3CN reaction
4
From the graph we find the slope = -1.9 x 104 K
We can now use these results to calculate the
rate constant at any temperature.
But this is also equal to - Ea/R
ln
Or Ea = -(slope)(R)
k 2 − Ea
=
k1
R
1 1
 − 
 T2 T1 
= - ( - 1.9x104 )(8.314 J mol-1 K-1)(1 kJ / 1000 J)
= 1.62 x 102 kJ/mol
or
162 kJ/mol
To calculate k1 for a temperature of 430.0 K, make substitutions
for all other parameters:
k2 = 2.52 x 10-5 s-1
T2 = 462.9 K
And T1 = 430.0 K
to obtain k1 = 1.0 x 10-6 s -1
Activation EnergyEnergy-orientation factor
k = Ae
− Ea
RT
Multistep Mechanisms and Rate Laws
Overall:
NO2 (g) + CO (g) Æ NO (g) + CO2 (g)
with an observed rate law of
Rate = k[NO2]2
It appears that at temperatures below 225 oC,
the reaction proceeds via two elementary steps:
Which yields two rate
expressions:
NO2 + NO2 Æ NO3 + NO
NO3 + CO Æ NO2 + CO2
Rate1 = k1 [NO2]2
Rate2 = k2 [NO3][CO]
(1)
(2)
Important points:
(a) Multi steps must add up to yield overall reaction.
(b) may involve reactive intermediates (different from activated complexes).
(c) One of these may be the ‘rate-determining’ step—the slower one!
Note that termolecular reactions are extremely unlikely !
In this case given above,
and
step (1) is the rate-determining step
step (2) is a faster step.
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Another example: 2 NO2 + F2 Æ 2 NO2F
observed rate = k[NO2][F2]
Proposed mechanism:
NO2 + F2 Æ NO2F + F
F + NO2 Æ NO2F
(1) slow
(2) fast
Somewhat more complicated:
obs rate = k[O3]2[O2] -1
Proposed mechanism:
O3 == O2 + O
(1) fast, reversible
O + O3 Æ 2 O2
(2) slow
tells us rate = k2[O][O3]
Sum gives overall reaction, and
so that
[O ]
k1[O3 ]
= k' 3
k −1[O2 ]
[O2 ]
[O ] =
[O ]
k1[O3 ]
= k' 3
[O2 ]
k −1[O2 ]
Another example: overall
with an observed rate law of
!!!
and rate = k2 [O ][O3 ] becomes
rate = k2
now what???
assume k1[O3] = k -1[O2][O]
rate = rate1 = k1[NO2][F2]
[O ] =
2 O3 Æ 3 O 2
k1 [O3 ]
[O ]2
[O3 ] = k 3
k −1 [O2 ]
[O2 ]
!!!
2 NO + Br2 Æ 2 NOBr
rate = k[NO]2[Br2]
The proposed mechanism is:
NO + Br2 == NOBr2
(1) (fast)
NOBr2 + NO Æ NOBr
(2) (slow)
Our earlier guideline would suggest we use the slow step to determine the
rate law, giving
rate = k2[NOBr2][NO]
but this presents an immediate problem, since we don’t know what
experimental quantities to put in for [NOBr2] !
The solution comes from an analysis of the reversible fast reaction (1).
rate forward = k 1[NO][Br2]
and rate reverse = k -1[NOBr2]
but these exist in a fast, dynamic equilibrium where rate forward = rate reverse
And this gives us the relationship
rate forward = k 1[NO][Br2] = k -1[NOBr2] = rate reverse
and
[ NOBr2 ] =
k1
k −1
Quiz #4 Next Week
(Feb 14-17)
Covers sections 14.5-14.7
[ NO ][ Br2 ]
and finally
k 
Rate = k 2 [ NOBr2 ][ NO ] = k2  1 [ NO ][ Br2 ][ NO ]
 k −1 
= k [ NO ]2 [ Br2 ]
During lectures next week, we will cover ALL OF
Chapter 15!
The Second Midquarter Exam
will cover
Chapters 14 and 15.
6
Take out a clean sheet of paper and print on it
your name and section number
and an indication it is “pop quiz #1a”.
A certain reaction has the form A Æ B. At 400 K and
[A]0 = 2.8 x 10 -3 M, data for a plot of [A] vs t were collected.
It was then found that a plot of
1/[A] vs t yielded a
Mon Lab, 2:30-5:18
Wed Lab, 2:30-5:18
straight line with a slope of 3.60 x 10-2 L·mol -1 ·s -1 .
Pat Bullinger
Chris Beekman
Edwin Motari
Hai Liu
Lin Sun
Chitanya
Patwardhan
Jeremy White
96
Roxana Sierra 97
Ramesh Sharma 98
Lin Sun
99
Mark Lobas
100
Chitanya
Patwardhan 101
a) Write an expression for the rate law.
rate = k[A]2
b) Write an expression for the integrated rate law.
1/[A] = 1/[A]0 + kt
c) What is the rate constant for the reaction?
3.60 x 10-2 L·mol -1 ·s -1
d) What is the half-life for the reaction, as given?
90
91
92
93
94
95
Fri: Patric BenziherÆ 102
Cl2 + CHCl3 Æ HCl + CCl4
Hai Liu Æ 103
rate = k[Cl2]1/2[CHCl3]
Proposed mechanism:
t1 =
2
1
1
10 5
=
=
= 9.9 x103
k [ A]0
(3.60 x10−2 )( 2.8 x10 −3 ) (3.6)( 2.8)
Catalysis
A catalyst changes the rate of a chemical reaction
without being consumed.
Cl2 == 2 Cl
(1) fast, reversible
Cl + CHCl3 Æ HCl + CCl3
(2) slow
CCl3 + Cl Æ CCl4
(3) fast
• Homogeneous catalysis: catalyst and reaction are in
the same, single phase.
• Heterogeneous catalysis: catalyst and reaction are in
different phases. Often the catalyst is a solid in contact
with gaseous or liquid reactions.
Evaluate the rate constant using this mechanism.
• Enzymes: In living systems, usually a large molecule
which catalyzes a specific reaction, an enzyme-substrate
complex.
• Homogeneous: catalyst and reaction are in same phase:
• Hydrogen peroxide decomposes very slowly:
2H2O2(aq) → 2H2O(l) + O2(g)
• In the presence of the bromide ion, the decomposition
occurs rapidly:
–
–
–
–
2Br-(aq) + H2O2(aq) + 2H+(aq) → Br2(aq) + 2H2O(l).
Br2(aq) is brown.
Br2(aq) + H2O2(aq) → 2Br-(aq) + 2H+(aq) + O2(g).
Br- is a catalyst because it can be recovered at the end of the
reaction and it makes the reaction rate faster.
• Generally, catalysts operate by lowering the activation
energy for a reaction.
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• Catalysts can operate by increasing the number of
effective collisions.
• That is, from the Arrhenius equation: catalysts
increase k be increasing A or decreasing Ea.
• A catalyst may add intermediates to the reaction.
• Example: In the presence of Br-, Br2(aq) is generated
as an intermediate in the decomposition of H2O2.
• When a catalyst adds an intermediate, the activation
energies for both steps must be lower than the
activation energy for the uncatalyzed reaction.
Heterogeneous Catalysis
• The catalyst is in a different phase than the reactants and
products.
• Typical example: solid catalyst, gaseous reactants and
products (catalytic converters in cars).
• Most industrial catalysts are heterogeneous.
• First step is adsorption (the binding of reactant molecules
to the catalyst surface).
• Adsorbed species (atoms or ions) may have increased
reactivity, but they are always easily available.
• Reactant molecules are also adsorbed onto the catalyst
surface and may migrate to active sites.
– Consider the hydrogenation of ethylene:
C2H4(g) + H2(g) → C2H6(g), ∆H = -136 kJ/mol.
– The reaction is slow in the absence of a catalyst.
– In the presence of a metal catalyst (Ni, Pt or Pd) the reaction
occurs quickly at room temperature.
– First the ethylene and hydrogen molecules are adsorbed onto
active sites on the metal surface.
– The H-H bond breaks and the H atoms migrate about the metal
surface.
– When an H atom collides with an ethylene molecule on the
surface, the C-C π bond breaks and a C-H σ bond forms.
– When C2H6 forms it desorbs from the surface.
– When ethylene and hydrogen are adsorbed onto a surface, less
energy is required to break the bonds and the activation energy
for the reaction is lowered.
ENZYMES
• Enzymes are biological catalysts.
• Most enzymes are protein molecules with large molecular
masses (10,000 to 106 ).
• Enzymes have very specific shapes.
• Most enzymes catalyze very specific reactions.
• Substrates are the reactants that undergo reaction at the
active site of an enzyme.
• A substrate locks into an enzyme and a fast reaction
occurs.
• The products then move away from the enzyme.
8
• Only substrates that fit into the enzyme lock can be
involved in the reaction.
• If a molecule binds tightly to an enzyme so that another
substrate cannot displace it, then the active site is blocked
and the catalyst is inhibited (enzyme inhibitors).
• The number of events (turnover number) catalyzed is
large for enzymes (103 - 107 per second).
Considerable research is currently underway to modify
enzymes to prevent undesirable reactions and/or to
prepare new products.
Consider the reaction
‘Fixation’
of N2’
converts it
to
compounds
useful to
plants
Nitrogenase
in legumes
converts
N2 to NH3.
Sample Problem, page 562-563:
2 N2O5 Æ 4 NO2 + O2
Calculate Ea from the following data:
k/s-1
2.0 x 10-5
7.3 x 10-5
2.7 x 10-4
9.1 x 10-4
2.9 x 10-4
T/oC
20
30
40
50
60
The decomposition of formic acid shown on the previous slide is given by
HCOOH (g ) Æ CO2 (g) + H2 (g)
It has been found to be first order at 838 K.
Formic
acid alone,
in the gas phase.
a) Estimate the half-life and first-order rate constant for the
decomposition of pure formic acid and formic acid in the presence of ZnO.
b) What is the effect of ZnO?
Formic acid
in presence of
ZnO.
c) Suppose we express the concentration of formic acid in mol/L. What
effect would that have on the rate constant?
d) The pressure of formic acid at the beginning of the reaction can be read
from the graph. Assume constant T, ideal gas behavior, and areaction
volume of 436 cm3. How many moles of gas are in the container at the
end of the reaction?
e) The standard heat of formation of formic acid vapor is ∆Hof = -378.6
kJ/mol. Calculate ∆Ho for the overall reaction. Assume the activation
energy, Ea , is 184 kJ/mol, sketch an approximate energy profile for the
reaction, and label Ea, ∆Hof , and the transition state.
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