[ruetherford]

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?

[Corribus]

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?

[mjc123]

If you add the equations for d[Br·]/dt and d[R·]/dt, you get that k₁[Br₂] = k₄[R·][Br·].