Zanarkin Posted July 11, 2013 Share Posted July 11, 2013 (edited) According to the mechanism below, what is the correct expression for the rate of consumption of A? In the mechanism, M is a molecule that collides with A and A* is a highly-energized form of A. Step 1 (reversible): A + M <---> A* + M rate constants k1 and k-1 Step 2 (reversible): A* <---> C rate constants k2, k-2 The answer is Rate = (k1 k2[A] [M] - k-1k-2[C] [M] ) / (k-1[M] + k2) My question is, how do you approach these problems? I haven't been able to figure out a way to do these at all. I have been looking at the textbook and the notes but it just seems like a jumbled mess. I need help. Edited July 11, 2013 by SlayerX Quote Link to comment Share on other sites More sharing options...
IceBrand Posted July 11, 2013 Share Posted July 11, 2013 I wish I was able to help you but I haven't a clue where to start. Sorry for being little to no help. Quote Link to comment Share on other sites More sharing options...
Doga Posted July 11, 2013 Share Posted July 11, 2013 Is this chem 2? This looks like some rate law shenanigans. Quote Link to comment Share on other sites More sharing options...
Zanarkin Posted July 11, 2013 Author Share Posted July 11, 2013 Its second term Chemistry. Still considered first year. Quote Link to comment Share on other sites More sharing options...
Doga Posted July 11, 2013 Share Posted July 11, 2013 Well that's what I meant by Chem 2. Second semester General Chemistry. I remember doing rate laws but I don't remember how to go about this. I should find my notes from this course. Quote Link to comment Share on other sites More sharing options...
Zanarkin Posted July 11, 2013 Author Share Posted July 11, 2013 This stuff is like the vain of my excistance. So glad chemistry isn't forced past this course. Quote Link to comment Share on other sites More sharing options...
mewyeon Posted July 11, 2013 Share Posted July 11, 2013 I literally JUST finished doing this kind of stuff last quarter, but I think we had some more information in our questions about which one is the slow/fast step. I actually just tried working at it a bit, and I'm pretty sure there's substitution for the intermediate, A*. Also, I can't be sure, but the fact that there's a subtraction sign in (k1 k2[A] [M] - k-1k-2[C] [M] ) reminds me strongly of Steady State Approximation. If you guys haven't done either of those concepts at all, just ignore me. It's not like my chem grades are good at all, SO. But I tried really hard in that class. Hopefully something in this will give you an idea? :) Quote Link to comment Share on other sites More sharing options...
Zanarkin Posted July 11, 2013 Author Share Posted July 11, 2013 I literally JUST finished doing this kind of stuff last quarter, but I think we had some more information in our questions about which one is the slow/fast step. I actually just tried working at it a bit, and I'm pretty sure there's substitution for the intermediate, A*. Also, I can't be sure, but the fact that there's a subtraction sign in (k1 k2[A] [M] - k-1k-2[C] [M] ) reminds me strongly of Steady State Approximation. If you guys haven't done either of those concepts at all, just ignore me. It's not like my chem grades are good at all, SO. But I tried really hard in that class. Hopefully something in this will give you an idea? :) Yes we did see that. This is for a test tomorrow and I have no idea what to do in these situations... Kinda glad most questions will be about ice tables though (or so i hope). Quote Link to comment Share on other sites More sharing options...
Zanarkin Posted July 11, 2013 Author Share Posted July 11, 2013 (edited) I managed to find the Steady State approximation for [A*] following some of the notes for another example. I got [A*] = (k1[A][M])/(k-1[M] + k2). However i do not understand where that + comes from in that denominator (remember i followed the notes, but they just show steps no explanations). I also have no dea where to go from here. Edited July 11, 2013 by SlayerX Quote Link to comment Share on other sites More sharing options...
deleted35362 Posted July 12, 2013 Share Posted July 12, 2013 my brain exploded Quote Link to comment Share on other sites More sharing options...
Zanarkin Posted July 12, 2013 Author Share Posted July 12, 2013 my brain exploded Mine died long ago. Quote Link to comment Share on other sites More sharing options...
Zanarkin Posted July 12, 2013 Author Share Posted July 12, 2013 Ah, i just figured out where that + comes from. The consumption of [A*] is not just k2[A*], you have to add the reverse from step 1 which is k-1[A*][M]. Well that took a while :/... Now where the second art comes from... I'll update here if i figure it out. Quote Link to comment Share on other sites More sharing options...
Redwall Posted July 12, 2013 Share Posted July 12, 2013 hopefully this gets to you in time for your exam. Um, I don't see the see to make the pseudo steady-state approximation. You don't even have to do any calculus. The three equations of interest are: d[C]/dt + d[A]/dt = 0 (mass conservation) d[A]/dt = k_{-1}[A*][M] - k_{1}[A][M] d[C]/dt = k_{2}[A*] - k_{-2}[C] The strategy is to simply eliminate [A*] from the set of three equations, giving you an expression for d[A]/dt in terms of [A], [C], and [M]. It's just a bunch of algebra. Quote Link to comment Share on other sites More sharing options...
Zanarkin Posted July 12, 2013 Author Share Posted July 12, 2013 Thank you! However, I was taking the test by then ah well. Its still nice to know how its done :D. Quote Link to comment Share on other sites More sharing options...
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