Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Genie123 on January 25, 2018, 05:32:33 PM
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Hey guys, to put it simply i've had issues for the past few days calculating the iodine clock reaction activation energy.
I've tried changing everything, but the iodine clock reaction activation energy just doesn't seem right.
Here's my process, please tell me if there is anything wrong with it, btw it's the persulfate variaton:
2 I- + S2O82- → I2 + 2 SO42-
I2 + 2 S2O32- → 2 I- + S4O62-
Calculating the rate of reaction like this:
Δc/Δt
Δc/Δt=(Δ[I2])/Δt=(Δ S2O32- )/2Δt
I know the concentration of s2o32-, and i know it's volume used, and i know the volume of the final reaction,
so I use Rate of reaction=((C1 V1)/V2)/2Δt formula
Now that I have the rate I can apply it to the rate equation:
k=rate/([I-][S2 O82-])
Then I calculate the c1v1/v2 again to calculate the c2 in the new concentration:
so, (C1 V1)/V2 of S2O82- × (C1 V1)/V2 of I- = x
Then k= rate/x
I do all of these calculations, and plot 1/T vs lnk, and I get the activation energy not at a value close to 50 kj/mol (after multiplying the slope of the graph by -8.31 kj/mol).
Where am I going wrong?
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Perhaps you should show us your data and calculations, so we can see if you have made a mistake there.
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I would love to, but it's for a project, and our schools plagiarism check is high end.
But is there anything wrong with the formulas, thus far?
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It looks OK, subject to the standard assumptions.
Do you know for a fact that the reaction is first order in iodide and persulfate? (I can't remember offhand)
Do you know that the activation energy is supposed to be 50 kJ/mol?
You did plot ln k, not log k, didn't you?
You say you plot 1/T vs lnk. You should be plotting lnk (y-axis) vs 1/T (x-axis). Have you got your slope the wrong way round?
You are using T in K, not °C, aren't you?
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Yeah, i've made sure and double checked all these, they are correct
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Are you making sure that the amounts of both iodide and persulfate are much greater than the amount of thiosulfate you use, so that both the former may be assumed to be constant during the reaction? Otherwise the rate will vary during the reaction and you can't use the equations you use.
Apart from that I can't think of any problems with your procedure; are you sure it's not just a calculating error? (You are using the correct values in your C1V1/V2 expressions - they are not (necessarily) the same for each species!)
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Hi, both persulfate and iodide is greater than the amount of thiosulfate used. (both are double).
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Hi, both persulfate and iodide is greater than the amount of thiosulfate used. (both are double).
I don't think double is sufficiently more to accurately calculate the kinetics, to be honest.
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How do you deduce this, any calculations?
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Are you making sure that the amounts of both iodide and persulfate are much greater than the amount of thiosulfate you use, so that both the former may be assumed to be constant during the reaction? Otherwise the rate will vary during the reaction and you can't use the equations you use.
Apart from that I can't think of any problems with your procedure; are you sure it's not just a calculating error? (You are using the correct values in your C1V1/V2 expressions - they are not (necessarily) the same for each species!)
Hmm, I saw that you said, they are not necessarily the same for each species, maybe that is where the error in calculation comes in?
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Hey, I saw you commented on my post (iodine clock one). Do you mind if I send you some of my data, and calculations applied so you can maybe see where I went wrong?
Sure, post here.
My concern is, that for instance, the concentration of iodide may be dropping from e.g. 2M to 1M over the reaction which is not really constant.
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240 seconds - at 283,15 Kelvin
150 seconds - at 293,15 Kelvin
84 seconds - at 303,15 Kelvin
47 seconds - at 313,15 Kelvin
34 seconds - at 323,15 Kelvin
27 seconds - at 333,15 Kelvin
Concentration of the solutions:
0.500 mol/dm3 KI (I-)
0.010 mol/dm3 K2S2O8 (S2O82-)
0.005 mol/dm3 NaS2O3 (S2O32-)
I used:
10 cm^3 of I-
10 cm^3 of S2O82-
5cm^3 of S2O32-
How I calculated Ea
(Δ S2O32- )/2Δt
C1V1/V2/2Δt
0.005*5/25/240= rate of reaction
did that for all of them.
then for this one i have 2.00 x 10^-6
to find k
k=rate/([I-][S2 O82-])
c1v1/v2 of I- =
0.5*10/25
c1v1/v2 of S2o82-=
0.010*10/25
2.00 x 10^-6/(0.010*10/25*0.5*10/25)
gives me k
ln(k), and then i graph that against their 1/temperatures and get a low reaction rate. Is anything wrong?
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Your procedure looks right, that's what I did and I got an activation energy of 37.36 kJ/mol. You should re-check your calculations. Especially if you're using a calculator and inputting complicated expressions in the denominator.
Here's a quick check. If your concentrations and Δc's are the same in all experiments, and you're only changing the temperature, and you are just interested in Ea, not absolute rate or k values, then you can simplify things by just using the Δt values. In this case, k = const./Δt, so plot ln(1/Δt) vs. 1/T and you should get the same slope as a ln k plot, without the possibility of errors in calculating k.
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Just a thought: did you enter "283,15" into a spreadsheet? It might have interpreted the comma as a thousands separator, not a decimal point. Try it again with a point.
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I got the same value, but the issue is, I did another trial and I added 5 cm^3 of catalyst, so I did everything divided by 30 instead of 25, and I used different times. I got lower reaction times (I.e it took 190 seconds for the 10 °C trial) for every single catalyst trial at the same temperature, but the activation energy came out higher than the uncatalysed reaction. That doesn't make sense, so I must be doing something wrong here?
Or is it the fact that I can't compare 30 cm^3 with 25 cm^3?
Edit: I used the ln/1/t method, and yeah it seems that both the activation energies calculations are correct, for both the catalysed and uncatalysed trials.
So it must be the fact that I can't compare 30 cm^3 with 25 cm^3? Why can't I do that, even though there are lower reaction times, but it still shows up with a higher activation energy?
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I'm sorry, but you're still not giving us enough information to help you sensibly.
I got the same value
Same value as what, for what, doing what differently?
Or is it the fact that I can't compare 30 cm^3 with 25 cm^3?
There's no reason why you can't, as long as you calculate the concentrations correctly in each case. If you use the ln(1/Δt) method, it shouldn't matter, as the difference will show up in the intercept, not the slope.
I used the ln/1/t method, and yeah it seems that both the activation energies calculations are correct
Both of what? Do you mean you now get a "correct" value of Ea, i.e. about 50 kJ/mol? Or do you mean that you get, as expected, the same value of Ea from plots of ln k and ln (1/Δt) (even if this value is wildly off)? Or that you get "correct" values for both catalysed and uncatalysed? How do you know they are correct, especially as you think your result doesn't make sense?
Unless you show full calculations, graphs and results, we can't see where you may have gone wrong.
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Alright sorry, 5 cm^3 of catalyst is added:
These are the catalysed results:
205 - 283.15 K
65 - 293.15 K
46 - 303.15 K
37 -- 313.15 K
17 - 323.15 K
7 - - 333.15 K
I apply the same process but using 30 cm^3 instead of 25cm^3.
I get a higher activation energy for some reason.
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There's clearly something wrong with these data as they give a very poor straight line plot. You can see that just by looking - the rate more than doubles between 10 and 20°, between 40 and 50° and between 50 and 60°, but less than doubles between 20 and 40°. That suggests something amiss with the procedure. What was the catalyst? Could there have been an issue with mixing? Were all the solutions equilibrated at the reaction temperature before mixing? Did you add them all in the same order each time?
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Catalyst was CuSo4 (so copper ions). It was in solution, but there still could be an issue with mixing because a) i only stirred the beaker gently a few times, and I mixed the solutions outside the water bath, i.e not being at it's correct temperature so i'm assuming that's where I went wrong. Is it still possible to have a higher activation energy though, even with faster reaction times?
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Yes, but it's not what you'd expect for a catalysed reaction. You would expect a lower activation energy, and a faster reaction rate than the uncatalysed reaction at all temperatures, though the relative effect (catalysed/uncatalysed rate) would be greater at lower temperatures.
I don't think you can do anything sensible with those data. If possible, do the experiment again, and more carefully.