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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: kelvinLTR on March 13, 2013, 02:20:44 PM

Title: Pseudo First Order
Post by: kelvinLTR on March 13, 2013, 02:20:44 PM

In the reaction 2C :rarrow:D, the following mechanism is suggested:

2C ::equil:: C + C*  (forward rate constant k₁ and backward rate constant k₂) C* is a reactive intermediate

C*  :rarrow: D  (rate constant k₃)

two questions are asked

1. What is the condition for the reaction to be pseudo-first order
2. What is the condition for the reaction to be 2nd order

I found d[D]/dt=k₁k₃[C]/k₂ using the steady state approximation for [C*]
But got stuck at those two questions.

My argument is for the reaction to be pseudo first order, [C*] formed must disappear much faster than it is formed, thus k₂>>k₁

am I right? and can someone help me with the 2nd part

thanks in advance  :)

Title: Re: Pseudo First Order
Post by: Corribus on March 13, 2013, 02:46:17 PM

Quote from: kelvinLTR on March 13, 2013, 02:20:44 PM

I found d[D]/dt=k₁k₃[C]/k₂ using the steady state approximation for [C*]

Granted, it's been quite a while since I solved rate law problems, but I believe your difficulty may be related to this.

Title: Re: Pseudo First Order
Post by: kelvinLTR on March 13, 2013, 03:29:02 PM

in the question, in an earlier part, it is asked to derive the rate of formation of [D] in terms of [C]

Title: Re: Pseudo First Order
Post by: Corribus on March 13, 2013, 03:32:32 PM

And I'm saying I didn't get the same answer you did. ;)

Title: Re: Pseudo First Order
Post by: kelvinLTR on March 13, 2013, 03:51:42 PM

mmm I got :

d[D]/dt = k₃[C*]
d[C*]/dt=k₁[C]²-k₂[C][C*]=0 in steady state
therefore [C*]=k₁[C]/k₂

substituting d[D]/dt=k₁k₃[C]/k₂

Title: Re: Pseudo First Order
Post by: Corribus on March 13, 2013, 04:07:00 PM

Quote from: kelvinLTR on March 13, 2013, 03:51:42 PM

d[C*]/dt=k₁[C]²-k₂[C][C*]=0 in steady state

You are forgetting to include the loss of the intermediate C* to form your final product via rate k₃.

Title: Re: Pseudo First Order
Post by: kelvinLTR on March 13, 2013, 04:09:43 PM

oh silly me  :) thanks a lot. I'll try again

Title: Re: Pseudo First Order
Post by: kelvinLTR on March 13, 2013, 04:15:53 PM

figured it out. Thank you very much  :)

Title: Re: Pseudo First Order
Post by: Corribus on March 13, 2013, 04:22:48 PM

No problem.  Should be easy to find the conditions of your limits now.