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

Appendix C 271
Insertion of [B]
and [B]
into Eq. C.45 gives
()
× − 2 − () 2 − 2 + 2 () − 2 = 1 [A] 0
− 1
(C.51)
The terms with ± () 2 − 2 cancel, so that
()
× − 2 = 1 [A]
0
− 1
which we rewrite in order to solve () as
()
= 1 [A] 0
− 1 × + 2
or
()
= 1 [A]
0
( 2− 1 )
This DE is easily integrated to obtain ():
(C.52)
(C.53)
(C.54)
— If 2 6= 1 =( 1 + 1 ) we obtain
() = 1 [A] 0
2 − 1
( 2− 1 )
(C.55)
and thus
[B]
= () − 2
= 1 [A] 0
2 − 1
( 2− 1 ) − 2
= 1 [A] 0
2 − 1
− 1
(C.56)
(C.57)
(C.58)
— The case 2 = 1 does not interest us here, as the ∗ state B should be a
longer-lived one.
• General solution for 2 6= 1 :
y
Initial value condition at =0:
[B] = [B]
+[B]
[B] = × − 2 + 1 [A] 0
2 − 1
− 1
(C.59)
(C.60)
[B ( =0)]=0 (C.61)
y
= − 1 [A] 0
2 − 1
(C.62)