Härledning av hastigheten för bildning av vätebromid ur H2

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.)