Upload
lamkhue
View
220
Download
3
Embed Size (px)
Citation preview
f e v
v – e + f = 2 (10.1.1)
1
C60
2 n
m
f = n + m (10.1.2)
6n + 5m
2
e = (6n + 5m)/2 (10.1.3)
m
1
2 (CH)4
(CH)6
v = (6n + 5m)/3 (10.1.4)
5 3 m 3
m 6
(10.1.1)
v – e + f = (6n + 5m)/3 – (6n + 5m)/2 + (n + m)
= (2 – 3 + 1)n + [(5/3) – (5/2) + 1]m (10.1.5)
= m/6
2 m = 12
n = 0 m = 12
(CH)20
C60 12
IPR: Isolated
Pentagon Rule
12 (10.1.5)
10.1 3
ABC 10.1 ACB
3! = 6
10.1 3
A B C
x
1 10.2 10.1
0
ABC 10.2
(10.1.6)
ACB C A B
(10.1.7)
(10.1.6) 3 (10.1.7)
(10.1.8)
6
x3 – 2x 3
x3 – 2x 0
x3 – 2x = 0 π
3 2 2 {(a, b), (c, d)}
ad – bc 3 3
π
4
2
C60
12,500
DNA
DNA
DNA
DNA
4
4
10.3
5
6 2
7
10.3
v
s
l p
p = sl/v
p = 1 l
p = 1/2 p = 1/3
1/2
p ≈ 1
1/3 < p <1/2
p < 1/3
p = 1
p < 1
BZ
1950
1
1968
BZ
BZ
2BrO3- + 3CH3(COOH)2 + 2H+ 2BrCH(COOH)2 + 3CO2 + 4H2O (10.1.9)
BrO3- + Br- +2H+ HBrO2 + HOBr (10.1.10)
HBrO2 + Br- + H+ 2HOBr (10.1.11)
BrO3- + HBrO2 + 2Ce3+ + 3H+ 2HBrO2 + 2Ce4+ + H2O (10.1.12)
2HBrO2 BrO3- + HOBr + H+ (10.1.13)
BrCH(COOH)2 + 4Ce4+ + 2H2O
4Ce3+ + Br- + HCOOH + 2CO2 + 5H+ (10.1.14)
5 Br-
(10.1.10) (10.1.11) Br-
Br- (10.1.12) (10.1.13)
Ce3+ Ce4+ Ce4+ (10.1.14)
Br- Ce3+ Br-
(10.1.10) (10.1.11)
8.4
3
1
4 ˚C
8
30 K
9
10
8 1
9
10
N2 O2
11 CO2 4
3 1
CO2 1 SF6 23900 CH4 21
2012
2007
© K. Saito, 06/22/2013