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Author: Subject: Kinetics and negative rate orders?
CaptainOfSmug
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[*] posted on 5-4-2014 at 14:11
Kinetics and negative rate orders?


Hello all, I've recently stumbled across a problem that has stumped me. My instructor for my general chem class gave me a sheet which had initial concentrations for two reactants and the initial rates. At first glance I thought it would be very easy but as I looked closer at the values none of the initial concentrations were held constant. I can't figure out a way to solve for the k value without monotonous trial and error... even more frustrating is the values I plug in do not even come near the initial rate unless I use negative exponents. Is a negative rate order possible? My text had no description of them or examples and I'm wondering if I'm missing something obvious.

The information given was as follows:
*Note please do not tell me the answer, I just need a nudge in the right direction if possible;)

Trial [A] [B] Rate (M/s)
1 .0122 .0189 .122
2 .0334 .0885 4.265
3 .0201 .0301 .525

Any help is greatly appreciated, I've been stuck on this for a good day and am finally biting the bullet and asking for a pointer :)
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DraconicAcid
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[*] posted on 5-4-2014 at 15:29


The exponents can be negative, but since the rates increase when you increase the order, they shouldn't be in this case.

I think if you compare two different sets of rates (as a ratio), and take the logs of those ratios, you should get a two-variable two-equation system which should be soluble.

ETA: I just worked through it that way, and it is soluble. A pain in the neck, but it gives very reasonable values for the exponents.

[Edited on 5-4-2014 by DraconicAcid]




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[*] posted on 5-4-2014 at 17:01


Thanks for the fast response! Well I think I may be setting up my systems wrong... or let's just say I chugged through it in a very non elegant way. I ended up with the overall order of the reaction to be 2.5 which doesn't seem right when I plug it back into an equation.
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[*] posted on 5-4-2014 at 17:51


It's easy to tell if you have the exponents correct- calculate k for each run. If they are the same (within rounding errors), you've done it right. If they're not, then go back and find your math error. I didn't get an overall order of 2.5.

Use two ratios. The math won't be elegant, but it should work.




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[*] posted on 5-4-2014 at 18:21


Hmmm, I think trial and error wouldn't be such a bad idea if there is logic behind it. For example, notice that between trials 1 and 3, the [A] and [B] both about double, while the rate about quadruples.

Now, let's assume K is 1... which it probably isn't but it will at least show us how things interact.

4=1(2)^x(2)^y

That gives us a couple values to work with.

That would lead me to assume orders are _____ and ____ order, or ___ and ____ order. (I left it blank because you said you don't want answers :)

Let's verify.

Solve for K, see if it works in all 3 trials.


[Edited on 6-4-2014 by ScienceHideout]




hey, if you are reading this, I can't U2U, but you are always welcome to send me an email!


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CaptainOfSmug
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[*] posted on 5-4-2014 at 19:15


Well I'm still working away at it, as far as using systems of equations... its become painfully clear to me that my algebra chops have weakened significantly haha.

To ScienceHideout, under those observations the only possible x and y values (assuming that neither has an order of 0) would be 1:1, .5:1.5 and vice versa. Unless I'm missing something which is possible
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[*] posted on 5-4-2014 at 20:04


You should be able to get it so that you have two equations in the form xlogA + ylogB = logC. Rearrange one so that x = stuff, and then rewrite the other equation as (stuff)logA + ylogB = logC. That gives you a single equation in one variable- solve for y.



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[*] posted on 6-4-2014 at 10:09


Okay thanks you, the logC would be the rate right? I keep grinding through this (had to open up a precalc book on logs to help remember them) and I keep getting an answer that doens't make sense. my y value comes out dame near 2 but my x value was ~ -1.3 which didn't work out either rounding up or down.

Here's in a nutshell what I did, again I don't want the answer because I won't learn anything but maybe someone can point out where I went wrong? I really do appreciate the help!

Trial 1: xlog(.0122) + ylog(.0189) = log(.122)

Trial 2: xlog(.0334) + ylog(.0885)= log(4.265)

Solving for y in the first trial I got approximately y=0.530096-1.1103X

Plugging in y into trial 2 equation I got: X~-1.30177

I then plugged that x value into y=0.530096-1.1103(-1.30177)
which was ~2
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[*] posted on 6-4-2014 at 11:16


Quote: Originally posted by CaptainOfSmug  
Okay thanks you, the logC would be the rate right? I keep grinding through this (had to open up a precalc book on logs to help remember them) and I keep getting an answer that doens't make sense. my y value comes out dame near 2 but my x value was ~ -1.3 which didn't work out either rounding up or down.

Here's in a nutshell what I did, again I don't want the answer because I won't learn anything but maybe someone can point out where I went wrong? I really do appreciate the help!

Trial 1: xlog(.0122) + ylog(.0189) = log(.122)

Trial 2: xlog(.0334) + ylog(.0885)= log(4.265)

Solving for y in the first trial I got approximately y=0.530096-1.1103X

Plugging in y into trial 2 equation I got: X~-1.30177

I then plugged that x value into y=0.530096-1.1103(-1.30177)
which was ~2


No, you want to use the ratios. So for trial 2 and trial 1, xlog(0.0334/0.0122) + ylog(0.0885/0.0189) = log(4.265/0.122)




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CaptainOfSmug
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[*] posted on 6-4-2014 at 15:48


Thank you so much! This was way to frustrating of a problem for me, I hope my professor heard my curses haha. Anyways I think I finally got it which the x variable was 2 and the y was 1 with an overall order of 3. My rate constant was 43368.98583. I'm not sure if that's what you got but they all worked minus a few rounding differences.
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