XI. 1st-Order Pre-Equilibrium, 2nd-Order Reaction
A. Given:
!!!
!!A
!!!
k1
k-1
B+ C
!!!
B. Rate Equation:
C
k2
P
d[P]/dt = k2[B][C]
Assuming [C] stays low, applying the steady state
approximation provides...
d[C]/dt = k1[A] - k-1[C] - k2[B][C] ≅ 0
XI. 1st-Order Pre-Equilibrium, 2nd-Order Reaction
Solving for [C]...
[C] =
k1 [A]
k"1 + k2 [B]
Therefore...
!
d[P]/dt =
k1k2 [B][A]
k"1 + k2 [B]
or...
!
d[P]/dt =
k2 Keq [B][A]
1+ (k2 [B])/k"1
such that...
!
Keq = k1/k-1
XI. 1st-Order Pre-Equilibrium, 2nd-Order Reaction
C. Flooding Technique:
1. Given: [B0] >> [A0]
d[P]/dt = kobsd[A]
such that...
kobsd =
!
k2 Keq [B]
1 + (k2 [B])/k"1
!!!
!!A
!!!
k1
C
k-1
B+ C
!!!
k2
P
XI. 1st-Order Pre-Equilibrium, 2nd-Order Reaction
2. Limiting Cases:
!!!
!!A
!!!
a. Case 1: k-1 >> k2[B0]
• This is most likely at low [B0].
kobsd =
• This is a rapid pre-equilibrium with
rate limiting B + C → P.
kobsd =
C
k-1
B+ C
!!!
k1k2
[Bo ]
k"1
!
b. Case 2: k-1 << k2[B0]
k1
k2
P
k2 Keq [B]
1 + (k2 [B])/k"1
!
• Most likely at high [B0]
kobsd = k1
• The rate is independent of [B0]; conversion of C to P is fast, making
conversion of A to C rate limiting.
XI. 1st-Order Pre-Equilibrium, 2nd-Order Reaction
c.
!!!
!!A
!!!
Combined Cases 1 and 2:
case 2
k1
k-1
B+ C
!!!
kobsd
case 1
kobsd =
C
k2
P
k2 Keq [B]
1 + (k2 [B])/k"1
[B]
According to eq (1), the fitting function might be written as...
f(x) =
(ab/c)x
1+ (b/c)x
!
However, adjustable parameters b and c are fully correlated.
Thus, it should be written as...
!
XI. 1st-Order Pre-Equilibrium, 2nd-Order Reaction
f(x) =
ax
1+ bx
!!!
!!A
!!!
such that...
!
a = k2Keq
C
k-1
B+ C
!!!
f(x) = kobsd
x = [B]
k1
k2
b = k2/k-1
• Most statistical packages include algorithms for detecting
correlated adjustable parameters.
P
XI. 1st-Order Pre-Equilibrium, 2nd-Order Reaction
3. Shortcut:
!!!
!!A
!!!
Since...
k2[B0] = constant
we can depict the original scheme as...
!!!
!!A
!!!
k1
k-1
C
k2[B0 ]
k1
C
k-1
B+ C
!!!
k2
P
P
Noting the similarity to the simple steady state approximation in Section VIII.B.2,
we rewrite kobsd by simply adding [B0] every time k2 appears. Given kobsd for the
simple steady state expression...
kk
kobsd = 1 2
k"1 + k2
We express kobsd for the more complex case above as...
!
kobsd =
k1k2 [Bo ]
k"1 + k2 [Bo ]
XI. 1st-Order Pre-Equilibrium, 2nd-Order Reaction
E. Saturation Kinetics, A Caveat:
We find that the saturation kinetics from Sections IX and XI are of the same
mathematical form...
!!!
!!A + B
!!!
!!! k
2
C
!!!
Keq
C
kobsd
P
k 2K eq [B]
=
1+ K eq [B]
f(x) =
ax
1+ bx
!!!
Keq
!!A
C
!!!
k2 Keq [Bo ]
kobsd =
!!!
k2
1+ (k2 [Bo ])/k"1
B + C
P
!!!
The key difference is that in the top example you observe C build up at the
saturation plateau (Section IX), whereas in the bottom case you do not
(Section XI).