Kinetics

1
Kinetics
Lecture 26

• Elementary Reactions
• Reaction Order
• Multi-step processes, Intermediates
• Steady State Approximation

2
Reaction Order

• The power to which a the concentration of a
• species is raised in the rate law is the reaction
• order. The overall order is the sum of all of
  the
• powers of all reactants.
• Examples
• 1. v kNO2O2 First order in O2,
• Second order in NO, Third order overall.
• 2. v kA1/2B2 Half order in A, Second order
• in B, 2 1/2 order overall.
• The order does not need to be an integer.

3
Reaction order and molecularity
Reaction order is an empirical quantity. Molecular
ity refers to an elementary reaction as a step in
a reaction mechanism. The rate law of an
elementary step can be deduced directly from the
reaction. Unimolecular reactions are first
order. A Products dA/dt -kA Bimolecular
reactions are second order. A B Products
dA/dt -kAB
4
An example 2NO O2 2NO2
The formation of nitric oxide is a
termolecular reaction, i.e. three molecules must
collide in order for the products to be formed.
In this example the reaction is also the
elementary step and so the rate law is v
kNO2O2 as stated earlier. In this case the
reaction is third order because of the fact that
three molecules must combine simultaneously.
However, the reaction stoichiometry may
not always indicate the molecularity of the
reaction.
5
Question Reaction Order
What is the order of the reaction
2 H2 O2 2 H2O A. First order B.
Second order C. Third order D. Unknown
6
Question Reaction Order
What is the order of the reaction
2 H2 O2 2 H2O A. First order B.
Second order C. Third order D. Unknown
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Question2 Reaction Order
The detailed mechanism of many enzymes has been
shown to include a rate determining step
E S ES What is the order
of the overall reaction? A. First order B. Second
order C. Third order D. Unknown
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Question2 Reaction Order
The detailed mechanism of many enzymes has been
shown to include a rate determining step
E S ES What is the order
of the overall reaction? A. First order B. Second
order C. Third order D. Unknown
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Parallel First Order Reactions

• A B and A C so that
• dA/dt -(k1k2)A
• dB/dt k1A
• dC/dt k2A
• Solve for A first

10
Parallel First Order Reactions

• The solutions are
• The production of B and C occurs with a constant
  proportion

11
Sequential first-order reactionsConsecutive
elementary reactions

• A B C rate equations are
• dA/dt -k1A
• dB/dt k1A - k2B
• dC/dt k2B
• Either k1 or k2 can be the rate limiting step.

12
Sequential first-order reactionsConsecutive
elementary reactions

• First solve eqn. for A
• Substitute into eqn. for B
• Similarly for C

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Populations as a function of time

• The population as a function
• of time is given by the
• solutions to the sequential
• first order reactions.
• The case shown is
• intermediate with k1 1.5k2.
• The population of B grows
• and reaches a maximum and
• then decays.

A - Green B - Yellow C - Red
A(t) - Initial B(t) Intermediate C(t) -
Final
Time




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Rate determining step

• If k2 gtgt k1 then the first (A B) step becomes
  the rate-limiting step.
• If the A B step is rate limiting then little or
• no B will be observed even though it is formed
  on the reaction path.
• If k1 gtgt k2 then the second (B C) step becomes
  the rate limiting step.
• If the B C step is rate limiting then there
  will be a significant build-up of the
  intermediate state B.

15
Steady-state approximation

• Equations representing kinetic networks of more
  than three states are not soluble analytically.
• One means of pushing the techniques as far as
  possible using analytical solutions is to set the
  derivatives of intemediates equal to zero
  dIntermediate/dt 0
• The build-up of intermediate B shown in the
  figure (two slides back) implies that the steady
  state approximation does not work for the system
  shown there.

16
Application of the steady-state approximation

• The steady state approximation can be applied to
  the consecutive reaction scheme
  
• k1 k2
  
• A B C.
  
• if the concentration of B is fairly constant.
• The result of setting dB/dt 0 is k1A -
  k2B 0 and dC/dt k1A. Since dA/dt
  - k1A we see that dC/dt - dA/dt and
  C(1 - exp-k1t)

17
Question Rate schemes

• Which rate equation describes the rate of
• disappearance of A?
• B
  A C
• A. dA/dt -(k1 k2 k3)A k-1B
• B. dA/dt (k1 k2 k3)A - k-1B
• C. dA/dt - k1B - k2C - k3D k-1A
• D. dA/dt k1B k2C k3D - k-1A

18
Question Rate schemes

• Which rate equation describes the rate of
• disappearance of A?
• B
  A C
• 
  
• A. dA/dt -(k1 k2 k3)A k-1B
• B. dA/dt (k1 k2 k3)A - k-1B
• C. dA/dt - k1B - k2C - k3D k-1A
• D. dA/dt k1B k2C k3D - k-1A

19
Example using the steady state approximation

• For the following folding scheme involving a
• single intermediate determine the rate of
• appearance of the folded form F.
  
• k1 k2
• U I F
  k-1
• dF/dt k2I
• dI/dt k1U - (k2 k-1)I
• dU/dt -k1U k-1I

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Example using the steady state approximation

• Assuming that you have information that allows
• you to use the steady state approximation
• determine the form of dF/dt in terms of U.
  
• k1 k2
• U I F
  k-1
• A. dF/dt U k1 k2 /(k2 k-1)
• B. dF/dt U k1 k2
• C. dF/dt U (k2 k-1)/k1 k2
• D. dF/dt U (k2 k-1)

21
Example using the steady state approximation

• Assuming that you have information that allows
• you to use the steady state approximation
• determine the form of dF/dt in terms of U.
• k1 k2
• U I F
  k-1
• A. dF/dt U k1 k2 /(k2 k-1)
• B. dF/dt U k1 k2
• C. dF/dt U (k2 k-1)/k1 k2
• D. dF/dt U (k2 k-1)

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Example using the steady state approximation

• The equations below show the reasoning behind the
  solution
• k1 k2
• U I F
  k-1
• dI/dt k1U - (k2 k-1)I 0
• k1U (k2 k-1)I
• I U k1/(k2 k-1)
• dF/dt k2I U k1 k2 /(k2 k-1)

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Biexponential kinetics result when the
steady-state approximation fails

• We cannot apply the steady state approximation if
  the concentration of B changes significantly
  during the process
• k1 k2
• A B C.
  
• In this case we must solve the rate equations
• In general, for N processes the result will be
  kinetics with N exponential time constants. Here
  B(t) F1e-k1t F2e-k2t

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Photochemistry

• The intensity, I is measured in power per unit
  area of photons per second per unit area. The
  intensity is given in photons/s/m2 or
  einsteins/s/m2.
• 1 einstein 1 mole of photons.
• To convert the intensity in W/ m2 to to intensity
  in number of photon we use the relation

25
Photochemistry

• Beer-Lambert law is
• The concentration is c or E
• The intensity of absorbed light is Io I.
• The extinction coefficient e has units of
• M-1cm-1. The exctinction coefficient depends
  on the frequency e(w). Often e is tabulated at
  the peak of absorption spectrum.

26
Photochemistry

• Therefore, the rate of excitation of the excited
  state of molecule E is
• The quantum yield is f.
• 1000 converts from cm3 to L

27
Photochemistry

• We use the approximation
• Thus,
• Photoexcitation is a pseudo-first order process
  where the excitation rate constant is

28
Competing or parallel processes

• Decay of the singlet S1 state
• can occur either radiatively by
• fluorescence (kf) or by internal
• conversion (kIC).
• dS1/dt -(kf kIC)S1
• The overall decay rate constant
• is the sum of the rate constants.
• The fluorescence quantum yield
• is

Photoexcitation followed by return to the S0
ground state.
29
Question Quantum yield
The observed fluorescence quantum yield of
Rhodamine is Ff 0.7. The observed lifetime
is 13.0 ns. What is the intrinsic fluorescence
lifetime? A. 18.6 ns B. 13.0 ns C. 9.1 ns D. 0.7
ns
30
Question Quantum yield
The observed fluorescence quantum yield of
Rhodamine is Ff 0.7. The observed lifetime
is 13.0 ns. What is the intrinsic fluorescence
lifetime? A. 18.6 ns B. 13.0 ns C. 9.1 ns D. 0.7
ns