Physical Chemistry 3: — Chemical Kinetics — - Christian-Albrechts ...

3.4 Generalized first-order kinetics* 55
I Solution of differential equations using Laplace transforms: Consider the
Laplace transform of the derivative () of a function (): 27
L [() ] =
Z ∞
We can solve this equation for () via
0
()
= () −¯¯∞
− (3.121)
Z
0 − ∞
= − ( =0)+
0
Z ∞
0
() ¡ −¢ (3.122)
() − (3.123)
= − 0 + () (3.124)
() = L [() ]+ 0
and recover the solution () by inverse Laplace transformation,
() =L −1 ∙ L [() ]+0
¸
(3.125)
(3.126)
I Example: Application to two consecutive first-order reactions. As an example,
we consider the DE’s for two consecutive first-order reactions
A 1
→ B 2
→ C (3.127)
We already solved this problem previously (in Section 2.6.1) using conventional methods,
and repeat the solution now using the Laplace transform method.
Towards these ends, in this example, we shall denote the concentrations [A ()] and
[B ()] as () and () and their Laplace transforms as () and (). Therespective
initial concentrations shall be [A] 0
= 0 and [B] 0
= 0 =0. Thus we have:
• DE for the concentration ():
= 1 0 − 1 − 2 (3.128)
• Laplace transform: () can be found using Eq. 3.124:
L [] =− 0 + () (3.129)
Taking the Laplace transform of the r.h.s. of Eq. 3.128: 28
L [] =L £ 1 0 − 1 ¤ − L [ 2 ()] (3.130)
= 1 0 L £ − 1 ¤ − 2 L [ ()] = 1 0
+ 1
− 2 () (3.131)
= − 0 + () (Eq. 3.129) (3.132)
27 We used in line 2: integration by parts; in line 3: the differivative ( − ) = − − ;inline4:
the definition () =L [ ()] and the abbreviation ( =0)= 0 .
28 We used L [ ]= 1 from Table E.1 and L [ ()] = ().
−