Chem.414 - Physical Chemistry II
Kinetcs
Mechanisms
Rate Laws
T-Dependence
Quantum Chemistry
Translational
Operator Alegbra
Appplications
Theories of Reaction Rates
Collisions
Boltzmann Distributions
Transiton State
Solutions
Diffusion
Ideal/Dilute
Henry/Raoult
E-Chem (Lab)
Spring 2016
Thermodynamics
Phase diagrams
Chemical Kinetics
Concepts of Rxn Rates
First Order
0th Order
2nd Order
nth Order
Order & Molecularity
Mechanisms
Steady State Approximation
T-Dependence of Rxn Rates
Experimental Techniques
Study of Chemical Kinetics
• Rate of reaction
• Dependence of concentration of species
• Dependence of temp., pressure, catalyst
• Control of reactions
• Mechanisms [Dominating step (fast vs. slow)]
• Guide to chemical intuition
Reaction Rates
Reaction Rate and Stoichiometry
• For the reaction
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
we know
d C4 H 9Cl d C4 H 9OH
Rate  

dt
dt
• In general for
aA + bB  cC + dD
1 d A
1 d B 1 d C 1 d D
Rate  



a dt
b dt
c dt
d dt
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
4. Consider the following N2O5 reaction: 2 N2O5(soln) ----> 4 NO2(soln) + O2(g)
Let: C = [N2O5]
(a) Using a graph of C vs. t, obtain tangential slopes and plot dC/dt vs. C. Calculate k after fitting with linear regression.
(b) Plot ln C vs. t. Calculate k after fitting with linear regression.
(c) Plot C vs. t. Fit the data with an appropriate function. Display the equation in standard IRL form with the appropriate variable names for this reaction.
(d) Calculate half-live (t2) and life-time (t). Compare them to the interpolated values from the plot of C vs. t.
Time / s
[N2O5] / M
1
2
3
4
5
6
7
8
9
0
200
400
600
800
1000
1200
1400
1600
1.00
0.88
0.78
0.69
0.61
0.54
0.48
0.43
0.38
10
1800
0.34
11
2000
0.30
ln [N2O5]
d[N2O5]/dt
(tangential
slope)
EXCEL
Non-Linear IRL
1.00
0.90
[N2O5] = 9.91E-01e-5.98E-04*t
R2 = 1.00E+00
[N2O5]
0.80
0.70
0.60
0.50
0.40
0.30
0
200
400
600
800
Time / seconds
1000
1200
1400
1600
1800
2000
The Change of Concentration with Time
Isomeric Transformation of Methyl Isonitrile to Acetonitrile
First Order Reactions (to one component)
CH 3 NC  CH 3CN
198.9o C
ln C  kt  ln C0
Differential and Integrated Rate Laws
n-th Order to One Component (Generalized Rate Laws)
Let:
DRL:
IRL:
C = concentration of reactant A remaining at time t
Co = initial concentration of reactant A (i.e. t=0)
k = rate constant (units depends on n)
Differential and Integrated Rate Laws
Rate Law: First Order to One Component
The Change of Concentration with Time
1
NO2 ( g )  NO( g )  O2 ( g )
2
300o C
Second Order Reactions
1
1
 kt 
C
C0
Rate Law: Second Order to One Component
Gas-Phase Decomposition of Nitrogen Dioxide
1
NO2 ( g )  NO( g )  O2 ( g )
2
300o C
Time / s
[NO2] / M
0.0
0.01000
50.0
0.00787
100.0
0.00649
200.0
0.00481
300.0
0.00380
Is this reaction first or second order?
k = 0.543 unit?
Half-Lives, Rate Constants and Co
Half-Lives, Rate Constants and Co - II
Zeroth Order to One Component - Catalysis
1. Provide the DRL.
2. Determine the IRL.
3. Sketch the IRL: Co=1.00 mol L-1 , k = 5.00x10-3 mol L-1 s-1 .
4. Use Mathcad (or EXCEL) to generate the IRL graph.
5. Obtain the half-life expression.
6. How many half-lives would it take for the reaction to reach
equilibrium (i.e. completion)? [ Hint: Solve the IRL for time
when C=0. Confirm by graph. ]
Summary of Rate Laws to One-Component
DRL
(-dC/dt)
IRL
Linear
Equation
First-Order
Second-Order
Zeroth-Order
kC
kC2
k
1/C = kt + 1/Co
C = -kt + Co
1/C vs. t
C vs. t
C = Co·e-kt
ln C = -kt + ln Co
ln C vs. t
m = -k
Linear Plot
b = ln Co
m=k
m = -k
b = 1/Co
b = Co
Half-Life
ln(2)/k
1/kCo
Co/2k
Units on k
time-1
M-1 time-1
M time-1
Concentration and Rate
•
•
•
•
Exponents in the Rate Law
For a general reaction with rate law
Rate  k[reactant 1]m[reactant 2]n
we say the reaction is mth order in reactant 1 and nth
order in reactant 2.
The overall order of reaction is m + n + ….
A reaction can be zeroth order if m, n, … are zero.
Note the values of the exponents (orders) have to be
determined experimentally. They are not simply related
to stoichiometry.
Method of Initial/Comparative Rates


NH4 (aq)  NO2 (aq)  N 2 ( g )  2 H 2O()
Expt #
1
2
3
[NH4+]o / M
0.100
0.100
0.200
[NO2-]o / M
0.0050
0.0100
0.0100
(Rate)o / M s-1
1.35x10-7
2.70x10-7
5.40x10-7
Three Component Rate Law

BrO3 (aq)  5 Br  (aq)  6 H  (aq)  3 Br2 ()  3 H 2O()
Expt #
[BrO3-]o / M
[Br-]o / M
[H+]o / M
(Rate)o / M s-1
1
0.10
0.10
0.10
8.0x10-4
2
0.20
0.10
0.10
1.6x10-3
3
0.20
0.20
0.10
3.2x10-3
4
0.10
0.10
0.20
3.2x10-3
Techniques for Multiple Component Rate Laws
1. Integration Approach:
Second Order – First Order to each of two components
2. Flooding Technique:
Rate = k [A]x [B]y [C]z
Second Order:
First order to each of two components
Consider:
A + B ---> Products (D)

rate
At
t=0;
At
some time t;
CA
dx
dt
dCA
dt
Let:
ax

dCB
dCD
dt
dt
CA = a ;
Let:
CB
C
B
x (mol/L) of A and B be reacted
bx
CD
x
dx
k ( a  x)  ( b  x)
k dt
( a  x)  ( b  x)
1
expands in partial fractions to
1
( a  x)  ( b  x)
by integration, yields
( a  x)  ( b  x)
1
( b  a)  ( a  x)

1
b  a
 ln ( a  x) 
by integration, yields
1


ab 

x
1
0
1
1
ab
 
dx
ax

dx
 ln ( a  x) 
1
b  a
 k dt

b  x
t
ab
 ln 
ab
 ln ( b  x)
 1  1  d x k 
 1 dt



 a  x b  x
0
1
Integrating yields:
1
b  a
Re-writing to give the following:
1
( b  a)  ( b  x)
b  a
1
( b  a)  ( b  x)
Z

( b  a)  ( a  x)
1
=b
 ln 
b  ( a  x) 
 k t
  a ( b  x) 
b  ( a  x) 

  a ( b  x) 
Z versus t yields straight line.
 ln ( b  x)
Applications of First-Order Processes
1. Radioactive Decay
2. Bacterial Growth
3. Interest and Exponential Growth
[Credit Card]
4. Loan Balance
Interest and Exponential Growth
P = future value
C = initial deposit
r = interest rate (expressed as a fraction: eg. 0.06)
n = # of times per year interest is compounded
t = number of years invested
C  10000
r  0.06
Continuous Compound Interest
lim
n 
n ---> infinity
PE( t)  C e
P( t)  C  1 
n  1
rate 
1 


n 


n t
e


n
r
n t
rate t
3
P( 30)  57  10
r t
3
P( t )
8 10
4
7.2 10
4
6.4 10
4
5.6 10
4
4.8 10
4
PE( 30)  60  10
PE( 30)  P( 30)  3061.56
4
4 10
PE( t )
4
3.2 10
2.4 10
4
1.6 10
4
8000
0
0
3
6
9
12
15
t
18
21
24
27
30
Loan Balance
A person initially borrows an amount A and in return agrees
to make n repayments per year, each of an amount P.
While the person is repaying the loan, interest is
accumulating at an annual percentage rate of r, and this
interest is compounded n times a year (along with each
payment). Therefore, the person must continue paying
these installments of amount P unitl the original amount
and any accumulated interest is repayed. The equation B(t)
gives the amount B that the person still needs to repay after
t years.
B = balance after t years
A = amount borrowed
n = number of payments per year
P = amount paid per payment
r = annual percentage rate (APR)
n t
r  0.05
n  12 P  800
A  115000
B( t)  A   1 

1.5 10
5
1.1 10
5
7 10
4
3 10
4
r


n t
n
1  r   1


n

 P
1  r   1


n

B( t )
1 10
4
5 10
4
0
2
4
6
8
10
t
12
14
16
18
20
Temperature and Rate
The Arrhenius Equation
• Arrhenius discovered most reaction-rate data obeyed the
Arrhenius equation:
k  Ae
 Ea
RT
– k is the rate constant, Ea is the activation energy, R is the gas
constant (8.3145 J K-1 mol-1) and T is the temperature in K.
– A is called the frequency factor.
– A is a measure of the probability of a favorable collision.
– Both A and Ea are specific to a given reaction.
Temperature and Rate
Reaction Mechanisms
• The balanced chemical equation provides information
about the beginning and end of reaction.
• The reaction mechanism gives the path of the reaction.
• Mechanisms provide a very detailed picture of which
bonds are broken and formed during the course of a
reaction.
Elementary Steps
• Elementary step: any process that occurs in a single step.
Reaction Mechanisms
Elementary Steps
• Molecularity: the number of molecules present in an
elementary step.
– Unimolecular: one molecule in the elementary step,
– Bimolecular: two molecules in the elementary step, and
– Termolecular: three molecules in the elementary step.
• It is not common to see termolecular processes
(statistically improbable).
Reaction Mechanisms
Rate Laws for Elementary Steps
• The rate law of an elementary step is determined by its
molecularity:
– Unimolecular processes are first order,
– Bimolecular processes are second order, and
– Termolecular processes are third order.
Rate Laws for Multistep Mechanisms
• Rate-determining step is the slowest of the elementary
steps. [example]
Reaction Mechanisms
Rate Laws for Elementary Steps
Rate Expressions
1

k
If elementary
steps:
vA  wB
xC  yD


k
1
Rate  
1 d A
1 d B 1 d C 1 d D



v dt
w dt
x dt
y dt

-d[A]/dt = vk1[A]v[B]w – vk-1[C]x[D]y
 -d[B]/dt = wk1[A]v[B]w – wk-1[C]x[D]y
 d[C]/dt = xk1[A]v[B]w – xk-1[C]x[D]y
 d[D]/dt = yk1[A]v[B]w – yk-1[C]x[D]y
Reaction Mechanisms
Mechanisms with an Initial Fast Step
2NO(g) + Br2(g)  2NOBr(g)
• The experimentally determined rate law can be:
d[NOBr]/dt = kobs[NO]2[Br2] (or) = kobs’[NO][Br2]
• Consider the following mechanism
k1
NOBr2(g)
Step 1: NO(g) + Br2(g)
(fast)
k-1
Step 2: NOBr2(g) + NO(g)
k2
2NOBr(g) (slow)
Step 1: NO(g) + Br2(g)
Step 2: NOBr2(g) + NO(g)
k1
k-1
k2
NOBr2(g)
(fast)
2NOBr(g) (slow)
Spring 2014
Step 1: NO(g) + Br2(g)
Step 2: NOBr2(g) + NO(g)
k1
k-1
k2
NOBr2(g)
(fast)
2NOBr(g) (slow)
Spring 2014
General Mechanism
Overall Reaction:
A B  C  D
Proposed Mechanism:
k1

A
M C


k
1
Where: D = observable product
M = intermediate
M  B  D
k2
k1

A
M C


k
1
k2
M  B 
D
k1

A
M C


k
1
k2
M  B 
D
k1

A
M C


k
1
k2
M  B 
D
Spring 2014
k1

A
M C
(
)
(
)


k
1
k2
M  B 
D
Spring 2014
Hydrogen-Iodine Reaction
Overall Reaction:
H 2  I 2  2HI
Proposed Mechanism:
k1

I2
2I

Where: I• = free radical

k
2

H 2  2 I  2 HI
k3
d [ HI ]
?
dt
k1

I2
2I 

k
2
3
H 2  2 I  
2 HI
k
k1

I2
2I 

k
2
3
H 2  2 I  
2 HI
k
k1

I2
2I 

k
2
3
H 2  2 I  
2 HI
k
Spring 2012
k1

I2
2I 

k
2
3
H 2  2 I  
2 HI
k
Spring 2012
Rice-Hertzfeld Free Radical
Chain Reaction Mechanism
Overall Reaction:
CH 3CHO ( g )  CH 4 ( g )  CO( g )
Proposed Mechanism:
1
CH 3CHO 
CH 3  CHO 
k
2
CH 3CHO  CH 3 
CH 4  CO  CH 3
k

2CH 3  C2 H 6
k3
d [CH 4 ]
?
dt
chain initiation
chain propagatio n
chain ter min ation

k1
CH 3CHO 
CH 3  CHO 

k2
CH 3CHO  CH 3 
CH 4  CO  CH 3

k3
2CH 3 
C2 H 6
chain initiation

chain
propagatio n
chain ter min ation

k1
CH 3CHO 
CH 3  CHO 

k2
CH 3CHO  CH 3 
CH 4  CO  CH 3

k3
2CH 3 
C2 H 6
chain initiation

chain
propagatio n
chain ter min ation
Kinetics
Experimental Techniques
Data to Conclusions
Hardware
"Instrumentation"
optial rotation
absorption/emission
dielectric constant
refractive index
dilatometric (Vol.)
pressure jump
temperature jump
electric field
conductivity
Software
"Brainmentation"
reaction rates vs. time
concentrations vs. time
initial rates vs. time
"lives" vs. time
guess type of order
computer fits
flooding/isolation
catalysts
mechanisms
Catalysis
Catalysis
Heterogeneous Catalysis
• 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.
Catalysis
Catalysis
Enzymes
• Enzymes are biological catalysts.
• Most enzymes are protein molecules with large molecular
masses (10,000 to 106 amu).
• Enzymes have very specific shapes.
• Most enzymes catalyze very specific reactions.
• Substrates 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.
Catalysis
Enzymes
• 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).
Catalysis
Rate( Michaelis  Menten) 
k 2  [ Et ]  [ S ]
K M  [S ]
Enzymes
Mechanism: Two Intermediates
Overall Reaction:
Experimentally found:
I   OCl   IO   Cl 
d[OI  ] kobs  [ I  ]  [OCl  ]

dt
[OH  ]
Proposed Mechanism:
OCl   H 2 O
k1

HOCl  OH 
very fast equilibrium



k
1
2
I   HOCl 
HOI  Cl 
k
3
OH   HOI 
H 2 O  OI 
k
fast
slow rds
Show that the proposed mechanism is consistent with the observed RL.
Mechanism
Overall Reaction:
Observed Rate Law:
2 O3 ( g )  3 O2 ( g )
Proposed Mechanism:
k1

O3


k
O2  O
d [O3 ]
[O3 ]2

 kobs 
dt
[O2 ]
fast
1
O  O3  2O2
k2
slow
k1

O3



k
O2  O
fast
1
k2
O  O3 
2O2
slow
Chemical Kinetics
DRL  Rate  
dC
 k Cn
dt
0th Order
ln C  kt  ln C0
Mechanisms
k1

A
Concepts of Rxn Rates
First Order
IRL ' s  C versus t
2nd Order
nth Order
Order & Molecularity
1
1
 kt 
C
C0
Steady State Approximation
Experimental Techniques
M C


k
1
k2
M  B 
D
T-Dependence of Rxn Rates
k  Ae
 Ea
RT