[Tekker]

Hi guys,

I'm stuck on a problem about integrated rate laws.... Here is the question:

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For the general case of a stoichiometric reaction with a single reactant, i.e.,

            A --> Product

The general form of the integrated rate law is (for all cases except n=1):

([A]0 / [A]t)^(n-1) = (n-1)kt      

Where [A]0 is the initial concentration, [A]t is the concentration after some interval of time, k is the rate constant, and n is the overall order of the reaction.

Note that the above equation applies to any order except the case n=1.  This equation is used by fixing the value of the reaction order (n), and plotting the data.  The value of n which best linearizes the data gives the order of the reaction. 

Start Excel, open the file “203extracredit.xls”. Save the file to your f: drive or to a flashdrive, etc.  This file contains time (first column) and concentration data (second column) for a reaction with a single reactant.  Plot the raw data.

Using the integrated rate law above, determine the order of the reaction, and report the rate constant k.  The possible values of n are n=0, ½, 2, and 3/2.  You’ll need to rearrange the integrated rate law a little to get it in a form where you can plot something vs time.  Show your work!!!!

Turn in your plot of the raw data and whichever plot of the integrated rate law appeared to be the most linear.  Report the value of the rate constant.

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First off, I have rearranged the formula to get it in a form that I can graph, by dividing both sides of the equation by [A]0. This puts (1/[A]t)^(n-1) on the y-axis and t on the x-axis with a slope of (n-1)k and a y-intercept of (1/[A]0)^(n-1). Does this seem reasonable?

Then I plotted all four graphs in excel and unfortunately the last two appear to both be linear. If I was forced to choose then I'd say the 3rd one looks like it is arcing a tiny bit more than. I've uploaded my excel file so you guys can take a look at them.... Would you guys agree that the 4th one appears to be the most linear? Plus the y-intercept is a little closer to the expected value (not by a whole lot, but it's a little closer).

Once I know which graph to use, do I just pick a set of data for a point on the chart (time and final concentrtation) from the chart and plug those into the slope intercept formula and solve for k? Because the value of k could change depending on which point I pick, so is there a better way to do this?

Any help will be greatly appreciated! 

Thanks,
-tkr

[Hunt]

First off, I have rearranged the formula to get it in a form that I can graph, by dividing both sides of the equation by [A]0. This puts (1/[A]t)^(n-1) on the y-axis and t on the x-axis with a slope of (n-1)k and a y-intercept of (1/[A]0)^(n-1). Does this seem reasonable?

Correct

You didn't mention anything about the data given concerning conc v/s time, and the file you uploaded isn't available either.

Once I know which graph to use, do I just pick a set of data for a point on the chart (time and final concentrtation) from the chart and plug those into the slope intercept formula and solve for k? Because the value of k could change depending on which point I pick, so is there a better way to do this?

The slope doesn't change, it's const : (n-1)k. You can choose any 2 points and determine the slope or if [A]_o is given, you can determine it using 1 point.

[Tekker]

Dang, I did forget to post the link... lol Ok, here it is:
http://www.artistcollaboration.com/users/tekker/Physics%20&%20Chemistry/ExtraCredit2.xls

The slope doesn't change, it's const : (n-1)k. You can choose any 2 points and determine the slope or if [A]_o is given, you can determine it using 1 point.

Because this is experimental data, not all of the points fall exacly on the line. So I could pick one of the outliner points and that would change the result slightly. That's why I was wondering if this is the best way to go about finding the value for k.

-tkr

[Donaldson Tan]

Express ln[A]_t in terms of ln(t)

Then plot ln[A]_t against ln(t)

What does the gradient of the graph correspond to?

What does the y-intercept correspond to?

[Tekker]

Express ln[A]_t in terms of ln(t)

Then plot ln[A]_t against ln(t)

This expression is already integrated, so in this case we actually don't need to use the natural log.

What does the gradient of the graph correspond to?

What does the y-intercept correspond to?

I'm not sure what they all mean in any detail other than it's what needs to happen to make the graph linear.... Why do they plot 1/[A]0? I have no idea other than you have to do it to get a linear graph!


I talked with my teacher today, and he said I was leaning towards the right one and the 4th graph is the linear graph.

I also asked him about plugging a point into the equation and he said you can just solve for k using the slope of the graph, where "(n-1)k" is the slope....

For the linear graph,
n = 3/2

Plug the value for n into "n-1".
(3/2)-1 = 1/2.

Therefore, (1/2)k = the slope of the line on the graph. So k is equal to two times the slope of the graph. That's it! Sure is a lot easier than I was making it out to be! LOL  

Thanks for your help guys!
-tkr

[Hunt]

It's good to know you figured it out, I apologize for my late reply, and it seems I also forgot that I don't have microsoft excel installed on the PC I'm currently using ...

I'm not sure what they all mean in any detail other than it's what needs to happen to make the graph linear.... Why do they plot 1/[A]0? I have no idea other than you have to do it to get a linear graph!

Consider the rate for a reactant A , r= -d[A] / dt = k[A]ⁿ

Rearranging : ∫ (d[A] / [A]ⁿ) = - k ∫dt

Assume [A] changes from [A]_o to [A]_t and the time from 0 to t.

Suppose n= 0 , [A]_t = - kt + [A]_o

Plot [A] v/s t , and you'll get a straight line of slope = -k , y-intercept = [A]_o

Take n = 1/2 , [A]^1/2_t = -( k/2 )t + [A]^1/2₀

Plot? [A]^1/2_t v/s t , and you'll get a straight line of slope = -k/2 , y-intercept = [A]^1/2₀

Take n = 1 , Ln[A]_t = - kt + Ln[A]_o

Plot Ln[A] v/s t , and you'll get a straight line of slope = -k , y-intercept = Ln[A]_o

You can also try for order 3/2 or 2 or ... You get to solve a differential equation and then plot the y-variable versus time in such a way that a straight line is obtained. Ofcourse, this is not a must for you can plot for n = 2 , [A] ( instead of 1/[A] ) v/s t and get a hyperbola , which is easier to distinguish from a str line.

I also asked him about plugging a point into the equation and he said you can just solve for k using the slope of the graph, where "(n-1)k" is the slope....

Ofcourse this holds true, also do not forget that with exp data, the points may not be precisely linear as you had mentioned. Computing different values of k ( acc to diff points ) and finding the average of k would get the most probable value of the rate const.