June 19, 2019, 09:51:01 AM
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### Topic: Calculating rate of reaction from a graph (y= n mols consumed, x= time)  (Read 347 times)

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#### Traumatic Acid

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##### Calculating rate of reaction from a graph (y= n mols consumed, x= time)
« on: May 28, 2019, 09:45:39 PM »
Hi all,
Before you tell me that there are other ways to do this, I know, but for some mysterious reason I have to use this method I was taught in class if I'm to get full marks for this report I'm writing. Isn't physical chemistry wonderful?

In the experiment we were testing how temperature effects reaction rate. We used the iodine clock reaction in three different water baths of different temperatures. Reaction is as follows: S2O82- + 2I-  ==> 2SO42- + I2
And the second part of the reaction after the addition of Na2S2O3

I2 + S2O32- ==> 2I- + S4O62-

So that's the important part, obviously there are other steps but it's this aspect of the reaction we are interested in, namely the consumption of S2O8. So basically each reaction was timed, until we had run and reversed the reaction 6-7 times. With some basic stoichiometry it can be concluded that the S2O8 was consumed like so:

 Moles S2O3   Reacted moles S2O8   Total moles S2O8   time (s : ms)0                   0                            0.005                       0   0.0004           0.0002                      0.0048                    196.10.0004           0.0002                      0.0046                    393.70.0004           0.0002                      0.0044                    601.30.0004           0.0002                      0.0042                    880.60.0004           0.0002                      0.004                   1199.30.0004           0.0002                      0.0038                  1587.80.0004           0.0002                      0.0036                  2031.5

Attached is a screenshot of the excell graph created by this data.

We have been given the following equations to use:

(1)

Slope =(y2-y1)/(x2-x1)=  Δy/Δx=  (Δn (S2O8))/Δt.

(2)

Rate = Slope / Volume

I was fortunate enough to have an associate already do the calculations and her answer for R2 is 0.9993.
I can't for the life of me figure out how she came to this conclusion. When I attempt this it goes something like:

S = 0.0036 - 0.005  / 2031.5 - 0 = -0.0000068

R = 0.0000068 / 0.031 (litres)  = -0.000022 (mol / L / s)

Yeah...
Or if I use the actual gradient:

S = 1.4 / 4.1 = 0.341
R = 0.341 / 0.031 =  11.01

Pretty sure I'm wrong, but I found a happy match (or coincidence)

By rearranging the formula for R, using her answer and the gradient (the calculation above) I have concluded that the  volume used in my mate's calculation is 0.31 L. Which is an order of magnitude too high, the volume of the solution was 31 ml (at the end) not 310 ml.
0.9993 = 0.317 / Vol
Vol = 0.317 x 0.9993
Vol = 0.316

If anyone could help me out that would be fantastic!
Cheers!

• Mr. pH