1. No category

Patrik Lundström, 150128
Härledning av hastigheten för bildning av vätebromid ur H2 och Br2
med två olika sätt att initiera kedjereaktionen
Reaktion: H2 + Br2 → 2HBr (A = H2, B = Br2, C = HBr. Små bokstäver = koncentrationer)
Mekanism: (X = Br⋅ , Y = H⋅, M = inert kropp, t.ex. kärlvägg. Små bokstäver = koncentrationer)
a) Initiering:
1) B + M → 2X + M
𝑣𝑎 = 𝑘𝑎 ⋅ 𝑏 ⋅ 𝑚
2) B + hν → 2X
𝑣𝑎 = 𝜙 ⋅ 𝐼𝑎𝑎𝑎
eller
(Iabs är intensiteten på absorberad strålning och ϕ är kvantutbytet för reaktionen)
b) Propagering:
X+A→C+Y
Y+B→C+X
c) Retardering:
Y+C→A+X
d) Terminering:
2X+M→B+M
I båda fallen vill vi beräkna
state.
𝑑𝑑
𝑑𝑑
𝑣𝑏 = 𝑘𝑏 ⋅ 𝑎 ⋅ 𝑥
𝑣𝑏′ = 𝑘𝑏′ ⋅ 𝑏 ⋅ 𝑦
𝑣𝑐 = 𝑘𝑐 ⋅ 𝑐 ⋅ 𝑦
𝑣𝑑 = 𝑘𝑑 ⋅ 𝑥 2 ⋅ 𝑚
𝑑𝑑
(eller 𝑣 = ½ 𝑑𝑑 ) under antagandet att alla intermediat är i steady-
1) Initiering genom reaktion med M
𝑑𝑑
= 𝑘𝑏 ⋅ 𝑎 ⋅ 𝑥 + 𝑘𝑏′ ⋅ 𝑏 ⋅ 𝑦 − 𝑘𝑐 ⋅ 𝑐 ⋅ 𝑦
𝑑𝑑
Steady-state på x och y
Patrik Lundström, 150128
𝑑𝑑
𝑑𝑑
𝑑𝑑
𝑑𝑑
= 2 ⋅ 𝑘𝑎 ⋅ 𝑏 ⋅ 𝑚 − 𝑘𝑏 ⋅ 𝑎 ⋅ 𝑥 + 𝑘𝑏′ ⋅ 𝑏 ⋅ 𝑦 + 𝑘𝑐 ⋅ 𝑐 ⋅ 𝑦 − 2 ⋅ 𝑘𝑑 ⋅ 𝑥 2 ⋅ 𝑚 = 0 (1)
= 𝑘𝑏 ⋅ 𝑎 ⋅ 𝑥 − 𝑘𝑏′ ⋅ 𝑏 ⋅ 𝑦 − 𝑘𝑐 ⋅ 𝑐 ⋅ 𝑦 = 0
(2)
(1) + (2) ger
𝑘𝑎 ⋅ 𝑏 ½
2 ⋅ 𝑘𝑎 ⋅ 𝑏 ⋅ 𝑚 − 2 ⋅ 𝑘𝑑 ⋅ 𝑥 ⋅ 𝑚 = 0 ⇔ 𝑥 = �
�
𝑘𝑑
2
Sätt in uttrycket för x i (2):
𝑘𝑏 ⋅ 𝑎 ⋅ �
1
2
𝑘𝑎 ⋅ 𝑏
� − 𝑘𝑏′ ⋅ 𝑏 ⋅ 𝑦 − 𝑘𝑐 ⋅ 𝑐 ⋅ 𝑦 = 0 ⇔ 𝑦 =
𝑘𝑑
Sätt in i uttrycket för den sökta hastigheten
½
1
𝑑𝑑
𝑘𝑎
= 𝑘𝑏 ⋅ � � ⋅ 𝑎 ⋅ 𝑏 2 +
𝑘𝑑
𝑑𝑑
𝑘𝑎 ½
� ⋅ 𝑎 ⋅ 𝑏½
𝑘𝑑
𝑘𝑏′ ⋅ 𝑏 + 𝑘𝑐 ⋅ 𝑐
𝑘𝑏 ⋅ �
3
𝑘𝑎 ½
𝑘 ½
� ⋅ 𝑘𝑏 ⋅ 𝑘𝑏′ ⋅ 𝑎 ⋅ 𝑏 2 − � 𝑎 � ⋅ 𝑘𝑏 ⋅ 𝑘𝑐 ⋅ 𝑎 ⋅ 𝑏 ½ ⋅ 𝑐
𝑘𝑑
𝑘𝑑
′
𝑘𝑏 ⋅ 𝑏 + 𝑘𝑐 ⋅ 𝑐
�
Förlängning med (𝑘𝑏′ ⋅ 𝑏 + 𝑘𝑐 ⋅ 𝑐) ger
𝑑𝑑
=
𝑑𝑑
2⋅�
𝑘𝑎 ½
� ⋅ 𝑘𝑏 ⋅ 𝑎 ⋅ 𝑏 3/2 𝑘 ⋅ 𝑎 ⋅ 𝑏 3/2
𝑘𝑑
=
𝑘𝑐
𝑏 + 𝑘′ ⋅ 𝑐
𝑏+ ′ ⋅𝑐
𝑘𝑏
2) Initiering genom fotolys
𝑑𝑑
= 𝑘𝑏 ⋅ 𝑎 ⋅ 𝑥 + 𝑘𝑏′ ⋅ 𝑏 ⋅ 𝑦 − 𝑘𝑐 ⋅ 𝑐 ⋅ 𝑦
𝑑𝑑
𝑑𝑑
= 2 ⋅ 𝜙 ⋅ 𝐼𝑎𝑎𝑎 − 𝑘𝑏 ⋅ 𝑎 ⋅ 𝑥 + 𝑘𝑏′ ⋅ 𝑏 ⋅ 𝑦 + 𝑘𝑐 ⋅ 𝑐 ⋅ 𝑦 − 2 ⋅ 𝑘𝑑 ⋅ 𝑥 2 ⋅ 𝑚 = 0
𝑑𝑑
𝑑𝑑
= 𝑘𝑏 ⋅ 𝑎 ⋅ 𝑥 − 𝑘𝑏′ ⋅ 𝑏 ⋅ 𝑦 − 𝑘𝑐 ⋅ 𝑐 ⋅ 𝑦 = 0
𝑑𝑑
Patrik Lundström, 150128
(1) + (2) ger
𝜙 ⋅ 𝐼𝑎𝑎𝑎 ½
2 ⋅ 𝜙 ⋅ 𝐼𝑎𝑎𝑎 − 2 ⋅ 𝑘𝑑 ⋅ 𝑥 ⋅ 𝑚 = 0 ⇒ 𝑥 = �
�
𝑚 ⋅ 𝑘𝑑
2
Sätt in i (2)
½
𝑑𝑑
𝜙 ⋅ 𝐼𝑎𝑎𝑎
= 𝑘𝑏 ⋅ 𝑎 ⋅ �
� − 𝑘𝑏′ ⋅ 𝑏 ⋅ 𝑦 − 𝑘𝑐 ⋅ 𝑐 ⋅ 𝑦 = 0 ⇒ 𝑦 =
𝑑𝑑
𝑚 ⋅ 𝑘𝑑
Sätt in i uttrycket för den sökta hastigheten
½
𝑑𝑑
𝜙 ⋅ 𝐼𝑎𝑎𝑎
= 𝑘𝑏 ⋅ 𝑎 ⋅ �
� +
𝑚 ⋅ 𝑘𝑑
𝑑𝑑
𝜙 ⋅ 𝐼𝑎𝑎𝑎 ½
�
𝑚 ⋅ 𝑘𝑑
(𝑘𝑏′ ⋅ 𝑏 − 𝑘𝑐 ⋅ 𝑐)𝑘𝑏 ⋅ 𝑎 ⋅ �
Förlängning med (𝑘𝑏′ ⋅ 𝑏 + 𝑘𝑐 ⋅ 𝑐) ger
𝑑𝑑
=
𝑑𝑑
𝜙 ⋅ 𝐼𝑎𝑎𝑎 ½
�
𝑚 ⋅ 𝑘𝑑
𝑘𝑏′ ⋅ 𝑏 + 𝑘𝑐 ⋅ 𝑐
𝑘𝑏 ⋅ 𝑎 ⋅ �
𝑘𝑏′ ⋅ 𝑏 + 𝑘𝑐 ⋅ 𝑐
𝜙 ⋅ 𝐼𝑎𝑎𝑎 ½
𝜙 ⋅ 𝐼𝑎𝑎𝑎 ½
� + (𝑘𝑏′ ⋅ 𝑏 − 𝑘𝑐 ⋅ 𝑐)𝑘𝑏 ⋅ 𝑎 ⋅ �
�
𝑚 ⋅ 𝑘𝑑
𝑚 ⋅ 𝑘𝑑
𝑘𝑏′ ⋅ 𝑏 + 𝑘𝑐 ⋅ 𝑐
′
𝜙 ⋅ 𝐼𝑎𝑎𝑎 ½ 𝑎 ⋅ 𝑏 2 ⋅ 𝑘𝑏 ⋅ 𝑘𝑏 (𝜙 ⋅ 𝐼 )½ ⋅ 𝑎 ⋅ 𝑏
′
𝑎𝑎𝑎
2 ⋅ 𝑘𝑏 ⋅ 𝑘𝑏 �
� ⋅ ½
𝑚½
𝑘𝑑½
𝑘𝑑
𝑚
=
=
𝑘
𝑘𝑏′ ⋅ 𝑏 + 𝑘𝑐 ⋅ 𝑐
𝑏 + 𝑐′ ⋅ 𝑐
𝑘𝑏
𝑎⋅𝑏
𝑘(𝜙 ⋅ 𝐼𝑎𝑎𝑎 )½ ⋅ ½
𝑚
=
𝑏 + 𝑘′ ⋅ 𝑐
(𝑘𝑏′ ⋅ 𝑏 + 𝑘𝑐 ⋅ 𝑐)𝑘𝑏 ⋅ 𝑎 ⋅ �
(Notera att k inte är definierad på samma sätt i de två fallen medan k´ är det.)