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Kinetics appendix.
Mathematical derivations which are here for interest only.
Second order kinetics
(ii) Second order, irreversible reaction (two reagents, different concentrations)
A + B C
k
Mass balance leads to:.
[A] 0 - [A] = [B] 0 - [B] = [C] (assuming, as is generally the case, that [C] 0 = 0)
Then -[B] = [A] 0 - [A] - [B] 0
[ ] [ ] [ ]
[ ][ ]
d A d B d C k A B dt dt dt
The rest of the derivation is given to show it can be done, but we will not be using this
math any further.
In order to simplify the equation it is easier to eliminate both [A] and [B] in favor of [C],
by applying the mass balance condition
[A] = [A] 0 - [C] (assumes [C] 0 = 0.0)
[B] = [B] 0 - [C]
[ ]
[ ] [ ] [ ] [ ]
d C k A C B C dt
Collecting variables
[ ]
[ ] [ ] [ ] [ ]
d C kdt A C B C
This is not so straightforward: one could look at the table of
integrals in the Handbook, and this leads to a solution. Or, you
could apply the method of partial fractions to separate the lhs into
two more manageable fractions.
Suppose that it can be factored; write in general form as
{^ ( 0 )(^0 )} (^0 )^ (^0 )
[ ] [ ] [ ]
[ ] [ ] [ ] [ ] [^ ]^ [^ ]^ [^ ]^ [^ ]
d C vd C wd C
A C B C A^ C^ B^ C
โ โ โ^ โ
To evaluate v and w, recombine into a single fraction, and see if
it is possible to assign logically consistent values to v and w.
0 0
0 0 0 0
0 0
0 0
[ ] [ ] [^ ]^ [^ ]^ [^ ]^ [^ ]^ [^ ]^ [^ ]
[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]
[ ] [ ] [ ]
[ ]
[ ] [ ] [ ] [ ]
vd C wd C^ v^ B^ C^ d C^ w^ A^ C^ d C
A C B C A C B C
v B w A v w C d C A C B C
comparison with the original form, which contains no term in [C],
and a constant of unity, shows that v and w must satisfy:
v+w = 0 v[B] 0 + w[A] 0 = 1
from which v = -w, and w = 1/([A] 0 - [B] 0 )
Now we can write:
[ ] 1 [ ] [ ]
[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]
d C d C d C
A C B C B A A C B C
This can be integrated easily:
{ ( )( )} (^ )^ (^ )^ (^ )
{ (^ )^ (^ ) }
0 0 0 0 0 0
0 0 0 0 0 0
(^0 )
0 0 0 0
0
0 0 0
[ ] 1 [ ] [ ]
[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]
ln [ ] [ ] ln[ ] ln [ ] [ ] ln[ ] [ ] [ ]
1 [^ ]^ [^ ]^ [ ]
ln ln [ ] [ ] [ ] [ ] [ ]
1 [ ] [ ]
ln ln [ ] [ ] [ ] [ ]
d C d C d C
A C B C B A A C B C
A C A B C B
B A
B C B
B A A C A
B B
B A A A
Which leads to the integrated rate equation:
0 0
0
[ ] [ ] (^0)
ln
[ ]
[ ]
ln
[ ]
B A [ ]
B
A
B
A
kt โ
R
S
T
U
V
W
b g
this form is convenient if both [A] and [B] can be readily measured
(^0 )
0 0 0 0
1 [^ ]^ [^ ]^ [ ]
ln ln [ ] [ ] [ ] [ ] [ ]
B C B
kt B A A C A
๏ฃฒ โ^ ๏ฃฝ=
is convenient if [C] is readily measured
0 0 0
0 0 0
1 [ ] [ ] [ ] [ ]
ln ln [ ] [ ] [ ] [ ]
B A A B
kt B A A A
๏ฃฒ โ^ ๏ฃฝ=
d) Reversible lst order reactions
If [A] =[A]o and [B] = [B]o at t = 0 [A]e is the concentration of A at equilibrium, and [B]e
is the concentration of B at equilibrium one finds at any time t,
[A]o - [A] = [B] - [B]o
(this assumes that [A]o > [A]e and [B]o < [B]e ; the alternative is possible but just means
changing the sign of the definition of the progress variable).
It is convenient to define a progress variable x, with a value at equilibrium xe,
x = [A]o - [A] = [B] - [B]o
xe = [A]o - [A]e = [B]e - [B]o and to use a = [A]o , and b = [B]o
The differential rate law is:
1 1
1 1
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
1 1
1 1 1 1 1 1
[ ]
[ ] [ ]
( )(^ )
d A k A k B dt
d a x k a x k b x dt
dx k a k b k x k x dt
dx k a k b k k x dt
dx dt k a k b k k x
dx dt k k k a^ k b x k k
dz k a k b k k dt where z x z k k
โ
โ
โ โ
โ โ
โ โ
โ โ
โ
โ โ โ
1
1
1 1
1 1
1 1 0 0
[ ]
[ ]
ln ( )
e e
e e
e
t t e
e
at equilibrium
k B x b K k A a x
k a k b x k k
dz x x k k dt z x
โ
โ
โ
โ
To turn an equation in x and x (^) e into one in [A], [A]o and [A]e we replace xe and x, using
the original definitions
A B
k 1
k (^) -
o e
e
o e
o e o
e
e
A A
A A
A A
A A A A
x
x x
[ ] [ ]
[ ] [ ]
ln [ ] [ ]
[ ] [ ] [ ] [ ]
ln ln
i.e. approach to equilibrium is a first order process with the observed rate constant the
sum of the forward and reverse k's; i.e.
Since
1
1
[ ]
[ ] [ ] [ ]
[ ]
[ ]
โ
k
k
A
A B A
A
B
K
e
o o e
e
e
we may readily find k 1 and k-1 from kobs and K, but note: If you dont know K, measuring
kobs will not lead to K but only to k 1 + k-
1 1 0
[ ] [ ]
ln ( ) [ ] [ ]
e
e
A A
k k t A A
โ
kobs = k 1 +k-
