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
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!!