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Author Topic: Chemistry lab: determining an equilibrium constant Kc  (Read 21887 times)

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Che120

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Chemistry lab: determining an equilibrium constant Kc
« on: January 29, 2012, 03:53:06 PM »

Hello :)

We did a lab in my chem class, and I'm completely lost, so I'd really appreciate any *delete me*!

The chemical reaction is: Fe3+ + SCN- <--> FeSCN2+

We're told that the equilibrium constant is given by the equation: Kc = [FeSCN2+]eq / ( [Fe3+]eq * [SCN-]eq )

We had 5 trials to runs the tests and find the absorbance values. Then we have to calculate the equilibrium constant, at room temperature (24°C):

So we have to find the initial concentrations [Fe3+]i (M) and [SCN-]i (M), as well as the concentrations at equilibrium, [FeSCN2+] eq (M), [Fe3+] eq (M), [SCN-]eq (M), and Kc (M^-1), for all 5 trials.

I understand that [FeSCN2+] standard is equal to the initial SCN-, which was made in our test tube 5 (the [SCN-]i concentration in test tube 5 after the dilution in 9ml Fe(NO3)3 is [FeSCN2+]std since we're assuming that all the SCN- has reacted), but beyond this I really have no idea how to go about calculating these values, so any help at all would be greatly appreciated!!

For example, for the initial concentrations, do I just take the values of the stock solutions? My lab report says that these initial concentrations are from after the mixing of the solution but before any reaction has taken place... that is very confusing for me.

We also have to calculate enthalpy (ΔH), which I'm not sure how to do. The equation is ln Kc = -ΔH / RT + ln A, but I'm not sure where to get these values from.

Here's is some more info that's probably needed to calculate these values:

Stock Solution Concentrations:
-- SCN-: 0.001998 M (mol / L^-1)
-- Fe3+ for test tubes 1-4: 0.002060 M (mol / L^-1)
-- Fe3+ for tube 5: 0.2000 M (mol / L^-1)


Absorbances and volumes to compose the solutions:
-- Tube 1: absorbance = 0.033, SCN- = 2 mL, Fe3+ = 5 mL, H2O = 3 mL
-- Tube 2: absorbance = 0.057, SCN- = 3 mL, Fe3+ = 5 mL, H2O = 2 mL
-- Tube 3: absorbance = 0.064, SCN- =4 mL, Fe3+ = 5 mL, H2O = 1 mL
-- Tube 4: absorbance = 0.083, SCN- = 5 mL, Fe3+ = 5 mL
-- Tube 5 ([FeSCN2+]std): absorbance = 0.090, SCN- = 1 mL, Fe3+ = 9 mL


Temperature: 24°C


Absorbances at various temperatures, Tube # 4:
-- 4.7°C, absorbance = 0.191
-- 16.0°C, absorbance = 0.090
-- 23.0°C, absorbance = 0.076
-- 47.5°C, absorbance = 0.083
-- 59.0°C, absorbance = 0.079

I don't necessarily need all the answers, I want to understand how to do the calculations. If I could have an example of the correct calculations to do with an explanation, that would be very helpful and I'd be very grateful!!!

Also, I know that the absorbance values are unusually low; our TA told us that one of the stock solutions was faulty, but that we should just go ahead and use the absorbances that we got to do the lab report.

Thank you very much!!
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UG

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Re: Chemistry lab: determining an equilibrium constant Kc
« Reply #1 on: January 29, 2012, 05:05:37 PM »

For example, for the initial concentrations, do I just take the values of the stock solutions? My lab report says that these initial concentrations are from after the mixing of the solution but before any reaction has taken place... that is very confusing for me.

Stock Solution Concentrations:
-- SCN-: 0.001998 M (mol / L^-1)
-- Fe3+ for test tubes 1-4: 0.002060 M (mol / L^-1)
-- Fe3+ for tube 5: 0.2000 M (mol / L^-1)
For any ion concentration after dilution, a general formula is concentration x Vold/Vnew
So for tube 5, [SCN-]i would be 0.001998 x 1/10 = 0.0001998 M, and [Fe3+]i would be 0.2000 x 9/10 = 0.18 M.
And then for the equilibrium concentrations, assuming all SCN- reacted, then [FeSCN2+] would be ~ 0.0001998 M, [SCN-] ~ 0 M and [Fe3+] ~ 0.1798 M
See if you can work out the initial concentrations for the rest of the tubes.
For the determination of Kc I THINK they want you to plot a graph of absorbance vs. concentration of FeSCN2+. Keep in mind that this graph will only have two points, one at 0,0 and the other at absorbance 0.090 and concentration 0.0001998 M. Usually we would prepare several stock solutions of FeSCN2+ of different concentrations and then measure the absorbances and plot a graph. We would then measure the absorbances of unknown solutions of FeSCN2+ and interpolate the graph to find the concentration. Anyways, once you have plotted this graph, you interpolate your graph, so for each absorbance value, you would find the concentration of FeSCN2+ by reading off the value on the x-axis. These values would be the concentration of FeSCN2+ at equilibrium. Hope my mumbo jumbo makes sense ;)
Please get back to us if you are still stuck.
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Che120

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Re: Chemistry lab: determining an equilibrium constant Kc
« Reply #2 on: January 29, 2012, 07:40:45 PM »

Hi,

Thank you very much for the explanations!! I understand now how to calculate the initial concentrations :)

I'm still a little confused about the concentrations at equilibrium though ([FeSCN2+]eq, [Fe3+]eq, [SCN-]eq), since we can't assume that all of the SCN- has reacted for test tubes 1-4, and since we were told that we have to calculate these (not plot them on a graph).

My TA said that to find the [FeSCN2+]eq for solutions 1-4, we have to use the absorbance of each equilibrium system (Aeq) and the absorbance of the standard solution (Astd, tube 5). So, I think the equation here is [FeSCN2+]eq = ( Aeq  /  Astd )  * [FeSCN2+]std.

So, for tube 1, for example, [FeSCN2+]eq  =  ( 0.033  /  0.090 )  *  0.0001998 M  =  0.00007326...?

Therefore, since Fe3+ + SCN- <--> FeSCN2+
-- [Fe3+]eq =  [Fe3+]i  -  [FeSCN2+]eq  -->  (0.001998 * 5/10)  -  0.00007326  =  0.0009567
-- [SCN-]eq =  [SCN-]i  -  [FeSCN2+]eq  -->  (0.002060 * 2/10)  -  0.00007326  =  0.0003263

So, since we are told that Kc = [FeSCN2+]eq / ( [Fe3+]eq * [SCN-]eq ),
then Kc for tube #1 = 0.00007326 M / ( 0.0009567 M  *  0.0003263 M )  = 234.81

Do you think that I'm on the right track? I'm really not confident when it comes to chem, so I really want to make sure.
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UG

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Re: Chemistry lab: determining an equilibrium constant Kc
« Reply #3 on: January 30, 2012, 09:47:49 AM »

My TA said that to find the [FeSCN2+]eq for solutions 1-4, we have to use the absorbance of each equilibrium system (Aeq) and the absorbance of the standard solution (Astd, tube 5). So, I think the equation here is [FeSCN2+]eq = ( Aeq  /  Astd )  * [FeSCN2+]std.

So, for tube 1, for example, [FeSCN2+]eq  =  ( 0.033  /  0.090 )  *  0.0001998 M  =  0.00007326...?
Yes that looks ok, this way gives the same results as if you had interpolated from a plotted graph.

Therefore, since Fe3+ + SCN- <--> FeSCN2+
-- [Fe3+]eq =  [Fe3+]i  -  [FeSCN2+]eq  -->  (0.001998 * 5/10)  -  0.00007326  =  0.0009567
-- [SCN-]eq =  [SCN-]i  -  [FeSCN2+]eq  -->  (0.002060 * 2/10)  -  0.00007326  =  0.0003263
Your mass balances looks ok too.

So, since we are told that Kc = [FeSCN2+]eq / ( [Fe3+]eq * [SCN-]eq ),
then Kc for tube #1 = 0.00007326 M / ( 0.0009567 M  *  0.0003263 M )  = 234.81
Again looks ok, they may want you to take the average of all your results to get a final Kc value.
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Che120

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Re: Chemistry lab: determining an equilibrium constant Kc
« Reply #4 on: January 31, 2012, 02:14:27 AM »

Thanks a lot!! :)
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