Two-step mechanisms: parallel
A
P
rate constant = k1
–dcP/dt = (∂cA/∂t)1 = –k1cA
A
Q
rate constant = k2
–dcQ/dt =(∂cA/∂t)2 = –k2cA
dcA/dt = (∂cA/∂t)1 + (∂cA/∂t)2 = –(k1 + k2)cA
CHEM 471: Physical Chemistry
This is now just a standard first order equation; we know the solution
cA = cA,0 exp[–(k1 + k2)t]
dcP/dt = k1cA = k1cA,0 exp[–(k1 + k2)t]
dcQ/dt = k2cA = k2cA,0 exp[–(k1 + k2)t]
Now we can integrate these two equations:
cP =
k1 cA,0 {1
exp [– (k1 + k2 ) t]}
(k1 + k2 )
cQ =
k2 cA,0 {1
The ratio cP/cQ always equals k1/k2
exp [– (k1 + k2 ) t]}
(k1 + k2 )
Parallel mechanism: example
Radon 211 decays 74% of the time by electron capture (giving Francium 211), and 26%
of the time by alpha emission (giving Polonium 207). The half-life of 211Rn is 15 hours.
What are the decay constants? Plot the decay against time.
CHEM 471: Physical Chemistry
kΣ = k1 + k2 = ln 2 /t1/2 = .6931/(15 h × 3600 s h–1) = 1.3 × 10–5 s–1.
k1/kΣ = 0.74
k1 = 0.74 × 1.284 × 10–5 s–1 = 9.5 × 10–6 s–1
k2/kΣ = 0.26
k2 = 0.26 × 1.284 × 10–5 s–1 = 3.3 × 10–6 s–1
Reversible kinetics
CHEM 471: Physical Chemistry
Something deceptively simple
A
P
rate constant = k1
(∂cA/∂t)1 = –k1cA
P
A
rate constant = k–1
(∂cA/∂t)–1 = k–1cP
dcA/dt = (∂cA/∂t)1 + (∂cA/∂t)–1 = –k1cA + k–1cP
cA + cP = cA,0 + cP,0 cP = cA,0 + cP,0 – cA
dcA/dt = –k1cA + k–1(cA,0 + cP,0 –cA)
dcA/dt = –(k1 + k–1)cA + k–1(cA,0 + cP,0)
Change variables: x = –(k1 + k–1)cA + k–1(cA,0 + cP,0)
dx = –(k1 + k–1)dcA
dx/dt = –x(k1 + k–1)
Separation of variables: ln x = –(k1 + k–1)t + C
At t = 0, x0 = –k1cA,0 + k–1 cP,0
ln x = –(k1 + k–1)t + ln[–k1cA,0 + k–1 cP,0]
cA = [x – k–1(cA,0 + cP,0)]/–(k1 + k–1)
cA =
( k1 cA,0 + k
1 cP,0 ) exp [
(k1 + k 1 ) t]
(k1 + k 1 )
k
1
(cA,0 + cP,0 )
Reversible kinetics
cA =
( k1 cA,0 + k
1 cP,0 ) exp [
(k1 + k 1 ) t]
(k1 + k 1 )
k
1
(cA,0 + cP,0 )
CHEM 471: Physical Chemistry
At time t = 0, cA = cA,0
Time variation depends on
exp[–(k1 + k–1)t]
dcA/dt = –k1cA + k–1cP
If dcA/dt = 0
cP/cA = K= k1/k–1
Fig 11.5: Reversible first-order
reaction, with cA,0 = 1.0, cP,0 = 0, k1 =
2.0, k–1 = 0.5.
Two-step mechanisms: series
A
B
rate constant = k1
–(∂cB/∂t)1 = dcA/dt = –k1cA
B
P
rate constant = k2
(∂cB/∂t)2 = –dcP/dt = –k2cB
dcB/dt = (∂cB/∂t)1 + (∂cB/∂t)2 = k1cA – k2cB
CHEM 471: Physical Chemistry
The first equation is just standard first order; we know the solution
cA = cA,0 exp(–k1t)
dcB/dt = k1cA,0 exp(–k1t) – k2cB
We can solve this by Laplace transform methods
k1 cA,0
cB =
[exp ( k1 t)
k
exp ( k2 t)]
kΔ = k2 – k1
Exercise: prove this solves the equation for dcB/dt !
dcP
k1 k2 cA,0
=
[exp ( k1 t)
dt
k
cA,0
cP =
[k1 exp ( k2 t)
k
exp ( k2 t)]
k2 exp ( k1 t)] + C
cP = cA,0 1
[k2 exp ( k1 t) k1 exp ( k2 t)]
k
C = cA,0
⇥
CHEM 471: Physical Chemistry
Two-step mechanisms: series
Fig 11.6: Series first order reaction,
showing the conversion of A (blue), with
an initial concentration cA,0 = 1 mM, to B
(red) and finally to P(green).
Rate coefficients: k1 = 0.1 mM s–1, k2 =
0.01 mM s–1.
Fig 11.7: Series first order reaction, but
now with a fast second step.
Rate coefficients: k1 = 0.01 mM s–1, k2 =
0.25 mM s–1.
Steady state approximation: two step mechanism
A
B
rate constant = k1
–(∂cB/∂t)1 = dcA/dt = –k1cA
B
P
rate constant = k2
(∂cB/∂t)2 = –dcP/dt = –k2cB
CHEM 471: Physical Chemistry
dcB/dt = (∂cB/∂t)1 + (∂cB/∂t)2 = k1cA – k2cB
We’ve already solved this exactly, in lecture 36. Now let’s apply the steady state
approximation
Steady state approximation: the rate of change of an intermediate or
intermediates is zero.
dcB
= k1 cA –k2 cB 0
dt
k1
This gives a simple relation for cA and cB
cB = cA
k2
We already have an exact solution for cA: cA = cA,0 exp(–k1t)
k1
cB = cA,0 exp ( k1 t) dcP = k1 cA,0 exp ( k1 t) cP = cA,0 exp ( k1 t) + C
k2
dt
C = cA,0
cP = cA,0 [1 exp ( k1 t)]
Obviously, if cB does not change, B cannot accumulate, and the rate of
formation of C is the rate of destruction of A.
CHEM 471: Physical Chemistry
Comparison with exact solution
The approximation is pretty close, because [B] is consumed as fast as it is
produced! k2 > k1
CHEM 471: Physical Chemistry
If k2 < k1 it doesn’t work!
The approximation is lousy because [B], the intermediate, builds up!
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