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Topic: Rate Law for Halogenation of a Hydrocarbon?  (Read 974 times)

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Offline ruetherford

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Rate Law for Halogenation of a Hydrocarbon?
« on: December 06, 2018, 02:57:11 PM »
Hi, I'm working on this problem for PChem and am getting some rather hairy answers!


Work:

The rate of the reaction should be: [tex]R = \frac{-d[RH]}{dt} = \frac{-d[Br_{2}]}{dt} = \frac{d[RBr]}{dt} = \frac{d[HBr]}{dt}[/tex]


I used the steady-state approximation to solve for: [tex]0 = \frac {d(Br)}{dt} = 2k_{1}Br_{2} - k_{2}BrRH + k_{2}RBr_{2} - k_{4}BrR[/tex] and [tex]\frac{d(R)}{dt}  = 0 = k_{2}*Br*RH - k_{3}*R*Br_{2} - k_{4}*Br*R[/tex]


And eventually got: [tex][R]   = \frac{[ B][RH]k_2 - [Br_2]k_1}{[Br2]k_3}[/tex] and [tex]
  = \frac{[Br2]([R]k_3 + k_1)}{[RH]k_2}[/tex]

However, when I go on to plug these back into the SS equations (to get [R] and [ Br] in terms of [RH] and [Br2] only), I get some pretty nasty-looking stuff and end up with a crazy-looking quadratic. Am I approaching the problem wrong?
« Last Edit: December 06, 2018, 05:51:04 PM by Borek »

Offline Corribus

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Re: Rate Law for Halogenation of a Hydrocarbon?
« Reply #1 on: December 07, 2018, 03:53:27 PM »
First, maybe it's just a typo in your syntax here but your first steady state approximation expression is wrong (there should be a k3 term). I'm also not following how you got from your steady state expressions to your "and eventually got" expressions. What happened to your k4 terms?
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline mjc123

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Re: Rate Law for Halogenation of a Hydrocarbon?
« Reply #2 on: December 10, 2018, 03:59:02 AM »
If you add the equations for d[Br·]/dt and d[R·]/dt, you get that k1[Br2] = k4[R·][Br·].

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