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

3.6 Oscillating reactions* 69
(2) Rate equations:
=+ 1 [A] 0
([X]
+ ) − 2 ([X]
+ )([Y]
+ ) (3.184)
=+ 1 [A] 0
[X]
+ 1 [A] 0
− 2 [X]
[Y] | {z }
= 1 [A] 0
[X]
− 2 [X]
− 2 [Y]
− 2
| {z }
= 1 [A] 0
= − 2 [X]
− 2 (3.185)
=+ 2 ([X]
+ )([Y]
+ ) − 3 ([Y]
+ ) (3.186)
= 2 [X]
[Y]
+ 2 [X]
+ 2 [Y]
+ 2 − |{z} 3 [Y]
− 3 |{z}
= 2 [X]
= 2 [X]
=+ 2 [Y]
+ 2 (3.187)
(3) Simplified DE’s for small displacements: We may neglect the terms containing
. Thus, we obtain two coupled first-order DE’s
= − 2 [X]
(3.188)
=+ 2 [Y]
(3.189)
which can be converted to a single second-order DE by the ansatz
·
= − (3.190)
·
=+ (3.191)
Second-order DE:
··
= − · = − (3.192)
Formal solution of this second-order DE:
() =[X()] − [X]
= 1 + 2 − (3.193)
(4) Eigenvalues: To determine the eigenvalue belonging to the linear DE’s, we have
to solve the determinant
¯
¯0 − −
+ 0 − ¯ − 2 [X]
¯ 2 [Y]
¯ =0 (3.194)
y
2 + 2 [X]
[Y]
=0 (3.195)
y The solutions for are purely imaginary:
12 = ± ¡ ¢
2 2 12
[X]
[Y]
(3.196)
= ± ( 1 3 [A] 0
) 12 (3.197)