wyklad ch.f.9, Galeria, Muzyka, Dokumenty, ka, Materiały od Oli, fizyczna, Egzamin, fizyczna wyklady i zadania

MECHANIZM LINDEMANNA-CHRISTIANSENA
S
→
P
[ ]
∗
d
S
[
2
∗
S
+
S
→
S
+
S
=
k
1
S
dt
[ ]
∗
d
S
[]
[]
∗
∗
S
+
S
→
S
+
S
−
=
k
S
⋅
S
−
1
dt
[ ]
d
P
[]
∗
∗
S
→
P
=
k
S
2
dt
[ ]
∗
d
S
[]
[]
[]
[]
2
∗
∗
=
k
S
−
k
S
⋅
S
−
k
S
1
−
1
2
dt
[ ]
0
∗
d
S
=
dt
[ ]
[]
2
2
[]
k
⋅
S
∗
S
=
1
k
S
k
⋅
+
−
1
[]
[ ]
[]
2
2
d
P
k
k
⋅
S
1
2
=
dt
k
⋅
S
+
k
−
1
1
 []
[ ]
[]
2
2
d
P
k
k
⋅
S
=
1
2
dt
k
⋅
S
+
k
−
1
[]
[ ]
[ ]
∗
∗
k
⋅
S
⋅
S
>>
k
S
−
1
2
[ ]
d
P
k
k
[]
1
S
=
dt
k
−
1
[ ]
d
P
[]
=
k
⋅
S
dt
2
 S
→
P
[ ]
∗
d
S
[][]
∗
S
+
M
→
S
+
M
=
k
S
⋅
M
1
dt
[ ]
∗
d
S
[]
[]
∗
∗
S
+
M
→
S
+
M
−
=
k
S
⋅
M
−
1
dt
[ ]
d
P
[]
∗
∗
S
→
P
=
k
S
2
dt
[ ]
∗
d
S
[][]
[]
[]
[]
∗
∗
=
k
S
⋅
M
−
k
S
⋅
M
−
k
S
1
−
1
2
dt
[ ]
0
∗
d
S
=
dt
[ ]
d
P
[ ]
∗
=
k
S
2
dt
[]
[ ]
d
P
k
k
M
[]
[]
=
1
2
S
dt
k
+
k
M
2
1
−
3
 [
[ ]
[ ]
[ ]
∗
∗
k
M
S
k
S
k
M
k
>>
⇒
>>
−
1
2
−
1
2
[]
d
P
k
k
[]
1
2
=
S
dt
k
−
1
[ ]
d
P
[]
=
k
S
dt
© wgrzybkowski 2003
4
 

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