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Lecture 26
KineticsKineticsElementary Reactions
Reaction OrderMulti-step processes IntermediatesMulti step processes, Intermediates
Steady State Approximation
Reaction OrderReaction Order• The power to which a the concentration of aspecies is raised in the rate law is the reaction
d h ll d h f ll f horder. The overall order is the sum of all of thepowers of all reactants. Examples:Examples:1. v = k[NO]2[O2] First order in O2, Second order in NO, Third order overall.Second order in NO, Third order overall.2. v = k[A]1/2[B]2 Half order in A, Second orderin B, 2 1/2 order overall. • The order does not need to be an integer.
R ti d d l l itReaction order and molecularity
Reaction order is an empirical quantity.Molecularity refers to an elementary reactionas a step in a reaction mechanism.The rate law of an elementary stepcan be deduced directly from the reaction.Unimolecular reactions are first order.A → Products d[A]/dt = -k[A]Bimolecular reactions are second order.A + B → Products d[A]/dt = -k[A][B]
A l 2NO O 2NOAn example: 2NO + O2 2NO2
The formation of nitric oxide is a termolecularreaction, i.e. three molecules must collide inorder for the products to be formed. In thisexample the reaction is also the elementarystep and so the rate law is v = k[NO]2[O2] as stated earlier. In this case the reaction is
f fthird order because of the fact that threemolecules must combine simultaneously.
h hHowever, the reaction stoichiometry may notalways indicate the molecularity of the reaction.
Q ti R ti O dQuestion: Reaction Order
What is the order of the reaction:2 H2 + O2 2 H2O
A. First orderB. Second orderC. Third orderD. Unknown
Q ti R ti O dQuestion: Reaction Order
What is the order of the reaction:2 H2 + O2 2 H2O
A. First orderB. Second orderC. Third orderD. Unknown
Q ti 2 R ti O dQuestion2: Reaction Order
The detailed mechanism of many enzymes hasbeen shown to include a rate determiningstep:
E + S ESWhat is the order of the overall reaction?A. First orderB. Second orderC. Third order
kD. Unknown
Q ti 2 R ti O dQuestion2: Reaction Order
The detailed mechanism of many enzymes hasbeen shown to include a rate determiningstep:
E + S ESWhat is the order of the overall reaction?A. First orderB. Second orderC. Third order
kD. Unknown
P ll l Fi O d R iParallel First Order Reactions
d h• A → B and A → C so that:• d[A]/dt = -(k1+k2)[A]
kA
• d[B]/dt = k1[A]• d[C]/dt = k2[A]
k1B
• Solve for [A] first:k2
[A] = [A]0exp{ – (k 1 + k 2)t }
C thend[B]dt = k 1[A]0exp{ – (k 1 + k 2)t }dt
d[C]dt = k 2[A]0exp{ – (k 1 + k 2)t }
P ll l Fi O d R iParallel First Order Reactions
h l• The solutions are:
kA
[B] = k 1[A]0 1 exp{ (k + k )t }k1B
[B] = 1[ ]0
k 1 + k 21 – exp{ – (k 1 + k 2)t }
[C] = k 2[A]0
k + k1 – exp{ – (k 1 + k 2)t }
• The production of B and C
k2[ ]
k 1 + k 2p{ ( 1 2) }
occurs with a constant proportion:
C[B] = k 1[ ][C] = 1
k 2
Sequential first-order reactionsC i l iConsecutive elementary reactions
• A → B → C rate equations are:k1
A
• d[A]/dt = -k1[A]• d[B]/dt = k1[A] - k2[B]
1B
1 2
• d[C]/dt = k2[B]• Either k1 or k2 can be the rate
k2
1 2limiting step.C
Sequential first-order reactionsC i l iConsecutive elementary reactions
• First solve eqn. for Ak1
A[A] = [A]oe– k 1t
• Substitute into eqn. for B1
B[ ] [ ]o
d[B] = k [A] k [B] = k [A] e– k 1t k [B]k2
[ ]dt = k 1[A] – k 2[B] = k 1[A]oe k 1t – k 2[B]
[B] = k 1[A]o
k 2 – k 1e– k 1t – e– k 2t
• Similarly for CC2 1
[C] = [A]o 1 – 1k 2 – k 1
(k 2e– k 1t – k 1e– k 2t)
l f fPopulations as a function of time
• The population as a function of time is given by the
l ti t th ti lsolutions to the sequential first order reactions.
• The case shown is A - GreenB Y ll
[A](t) - Initial[B](t) I t di t intermediate with k1 = 1.5k2.
• The population of B growsand reaches a maximum and
B - YellowC - Red
[B](t) Intermediate
[C](t) - Final
Time
and reaches a maximum and then decays.
Time
Rate determining stepRate determining step
• If k2 >> k1 then the first (A → B) step becomes the rate-limiting step.g p
• If the A → B step is rate limiting then little orno B will be observed even though it is gformed on the reaction path.
• If k1 >> k2 then the second (B → C) step 1 2 ( ) pbecomes 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.
St d t t i tiSteady-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:derivatives of intemediates equal to zero:
d[Intermediate]/dt = 0• The build up of intermediate B shown in the figure• 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 pp ythere.
Application of the t d t t i tisteady-state approximation
• The steady state approximation can be applied to the consecutive reaction scheme
k1 k2
A → B → C. if th t ti f B i f i l t tif the concentration of B is fairly constant.
• The result of setting d[B]/dt = 0 is:• The result of setting d[B]/dt = 0 is:k1[A] - k2[B] = 0 and
d[C]/dt = k1[A]. Since d[A]/dt = - k1[A] we see that d[C]/dt d[A]/dt d [C] (1 { k t})d[C]/dt = - d[A]/dt and [C]=(1 - exp{-k1t})
hQuestion: Rate schemes
Which rate equation describes the rate ofdisappearance of A? k2k1
B A C
D
2
k3
1
k-1
A. d[A]/dt = -(k1+ k2 + k3)[A] + k-1[B]B d[A]/dt = (k1+ k2 + k3)[A] - k 1[B]
D
B. d[A]/dt = (k1+ k2 + k3)[A] k-1[B]C. d[A]/dt = - k1[B] - k2[C] - k3[D] + k-1[A] D. d[A]/dt = k1[B] + k2[C] + k3[D] - k-1[A] [ ]/ 1[ ] 2[ ] 3[ ] 1[ ]
hQuestion: Rate schemes
Which rate equation describes the rate ofdisappearance of A? k2k1
B A C
D
2
k3
1
k-1
A. d[A]/dt = -(k1+ k2 + k3)[A] + k-1[B]B d[A]/dt = (k1+ k2 + k3)[A] - k 1[B]
D
B. d[A]/dt = (k1+ k2 + k3)[A] k-1[B]C. d[A]/dt = - k1[B] - k2[C] - k3[D] + k-1[A] D. d[A]/dt = k1[B] + k2[C] + k3[D] - k-1[A] [ ]/ 1[ ] 2[ ] 3[ ] 1[ ]
Example using the steady state i tiapproximation
For the following folding scheme involving asingle intermediate determine the rate ofsingle intermediate determine the rate ofappearance of the folded form F.
k1 k2k1 k2
U ↔ I → F kk-1
d[F]/dt = k2[I]d[I]/dt = k [U] (k + k )[I]d[I]/dt = k1[U] - (k2 + k-1)[I]d[U]/dt = -k1[U] + k-1[I]
Example using the steady state i tiapproximation
Assuming that you have information that allowsyou to use the steady state approximationdetermine the form of d[F]/dt in terms of [U].
k1 k2
U ↔ I → F k-1
A d[F]/dt [U] k k /(k + k )A. d[F]/dt = [U] k1 k2 /(k2 + k-1)B. d[F]/dt = [U] k1 k2
C d[F]/dt = [U] (k + k )/k kC. d[F]/dt = [U] (k2 + k-1)/k1 k2
D. d[F]/dt = [U] (k2 + k-1)
Example using the steady state i tiapproximation
Assuming that you have information that allowsyou to use the steady state approximationdetermine the form of d[F]/dt in terms of [U].
k1 k2
U I FU ↔ I → F k-1
A d[F]/dt = [U] k k /(k + k )A. d[F]/dt = [U] k1 k2 /(k2 + k-1)B. d[F]/dt = [U] k1 k2
C d[F]/dt = [U] (k2 + k )/k k2C. d[F]/dt = [U] (k2 + k-1)/k1 k2
D. d[F]/dt = [U] (k2 + k-1)
Example using the steady state approximation
The equations below show the reasoning behind the solution:
k1 k2
U ↔ I → F k-1
d[I]/dt = k1[U] - (k2 + k-1)[I] = 0k [U] (k k )[I]k1[U] = (k2 + k-1)[I][I] = [U] k1/(k2 + k-1)d[F]/dt k [I] [U] k k /(k + k )d[F]/dt = k2[I] = [U] k1 k2 /(k2 + k-1)
Biexponential kinetics result when the steady-state approximation fails
• We cannot apply the steady state approximation if the concentration of B ppchanges significantly during the process
k1 k2
A → B → C. • In this case we must solve the rate equationsq• In general, for N processes the result will be
kinetics with N exponential time constants. Here: [B](t) = Φ1e-k1t + Φ2e-k2t
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
i t i / / 2or einsteins/s/m2.• 1 einstein = 1 mole of photons.
T t th i t it i W/ 2 t t• To convert the intensity in W/ m2 to to intensity in number of photon we use the relation: hrelation:
Ephoton = hν = hcλ
Photochemistry
• Beer-Lambert law is:
I = I 10– A A = εlc c = [E]• The concentration is c or [E]• The intensity of absorbed light is I – I
I Io10 , A εlc , c [E]
• The intensity of absorbed light is Io I. • The extinction coefficient ε has units of
M-1cm-1 The exctinction coefficientM 1cm 1. The exctinction coefficient depends on the frequency ε(ω). Often ε is tabulated at the peak of absorption abu a d a p a o abso p ospectrum.
Photochemistry
• Therefore, the rate of excitation of the excited state of molecule E is:
– d[E]dt = φ(Io – I)1000
l = φIo(1 – 10– A)1000l
• The quantum yield is φ.• 1000 converts from cm3 to L1000 converts from cm to L
Photochemistry
• We use the approximation:10– A ≈ 1 – 2.303A
• Thus,10 1 2.303A
d[E] 2303φI ε[E]
• Photoexcitation is a pseudo-first order
– d[E]dt = 2303φIoε[E]
NA
• Photoexcitation is a pseudo-first order process where the excitation rate constant is: 2303φI εs
k = 2303φIoεNA
ll lCompeting or parallel processes
S1• Decay of the singlet S1 statecan occur either radiatively by
hν kf kICfluorescence (kf) or by internal conversion (kIC).• d[S ]/dt (k + k )[S ]
S0
• d[S1]/dt = -(kf + kIC)[S1]• The overall decay rate constantis the sum of the rate constants.0 is the sum of the rate constants. The fluorescence quantum yieldis k
Photoexcitationfollowed by return
Φ =k f
k f + k IC
to the S0 ground state.
ldQuestion: Quantum yield
The observed fluorescence quantum yield of Rhodamine is Φf = 0.7. The observed lifetime is13.0 ns. What is the intrinsic fluorescence lifetime?A. 18.6 nsB 13 0 nsB. 13.0 nsC. 9.1 nsD 0 7 nsD. 0.7 ns