Questions and answers in Chem538

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Fall 2009:

Q5 (10/19/09): No, I am not assuming d[C2H6]/dt =0 and have included everything...
I also don`t know why it is mentioned in (d) of question 2 that other >> products should be assumed minor (H2 and CH4) and based on this expression for k simplified because they(these products) do not appear in any rate expression (for C2H6 and intermediates) and consequently, can not appear in final expression for k...
A5 (10/19/09): - There are 3 equations for [CH3], [C2H5], and [H] which should have a nonzero solution. Plugging it into the equation for d[C2H6]/dt provides the answer.
- Since there are other products ([CH4]) in smaller quantities, knowing that the amount of produced [C2H6] >> [CH4] suggest certain relationship for the ki in the equations and thus allows simplification.

Q4 (10/19/09): Sorry for bothering you while you are ill, but I am really confused by problem 2 of exam... I am approaching to this problem as following: 1. wrote rate expressions for C2H6 and for 3 intermediates. 2. Assumed dA/dt for each radical (H, CH3 and C2H5) to be equal to 0 3. Trying express intermediates concentrations via C2H6 concentration and then plug them in rate expression for C2H6 ( I am immitating example with H2 and Br2 from textbook and class discussion) In this way I can not express each of concentrations independently (only 2) and after plugging them in rate expression they calcel out! My feeeling is that smth is missing or,maybe, I am approaching in wrong way... Could you, please, give me tip. thanks in advance
A4 (10/19/09): Obviously you are making a mistake somewhere as having d[C2H6]/dt =0 would mean that [C2H6] = const = 0 (since it was 0 at the beginning). Make sure that you have not missed any routes for each of the species changing, e.g. C2H6 appears in 2a, 2b, 2d, 2e.

Q3 (10/19/09): Hi, Could I get a hint on how to combine steps a and b to do part c in problem 2.  I remember what the steady-state approximation is but I'm not seeing how to use it to get k.
A3 (10/19/09): As you see, eqs. (2a-2e) do not add up just what a) asks. You will use that in d). Question c) is a straight forward application of the dif equations that you should derive from (2a-2e), i.e. in question b).

Q2 (9/25/09): Hi, Could I get a hint as to how to deal with solution Y being a product and a reactant in the first chemical reaction of problem 2.14? The pesky two is throwing me off, of course.
A2 (9/25/09): As we did in class for the reversible A <=> B reaction, add the probabilities for both steps. Be careful in setting up the contributions for both cases. In the first step, Y is reacted but 2Y are produced; in the second step 2Y molecules are consumed. Thus dPn/dt for Y is linked to both Pn and Pn-1 from the first reaction and what Pn and Pn+2 from the second reaction. Treat nx as a constant since [X]>>[Y].

Q1 (9/21/09): Hi, I'm attempting to do the LaPlace transform for problem 2.10. Am I supposed to end up doing the determinate of a 4X4 matrix? Did I miss a simplification somewhere? I've tried taking the determinate using matlab and the program refuses to do so because there isn't enough memory on my computer.
A1 (9/21/09): Hello.., The answer is yes and no. Yes, that the matrix is 4 x 4 but no as you can reduce it. There will be a 2x2 matrix for the (001) and (010) concentrations while the remaining (000) and [NO2] concentrations can be calculated afterwards.


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Last updated on 10/19/09