Le Chatlier’s principle Lab
Introduction
This experiment will give you an opportunity to determine the equilibrium constant for
the formation of Fe(SCN)2+. Moreover, the experiment will require you to use Le
Chatelier’s principle.
Concept of the experiment:
When the reaction between Fe3+ and SCN- (thiocyanate) ions in aqueous solution comes
to equilibrium, the system consists of the reactants and Fe(SCN)2+.
Fe3+(aq) + SCN-(aq) ↔ Fe(SCN)2+(aq)
The product is a complex ion that has a coordinate covalent bond between the iron atom
and an atom (probably the S atom) from the thiocyanate anion. The color of this complex
ion (redish-orange in the visible spectrum of light) is so intense that thiocyanate ions can
be used to detect very small quantities of Fe3+. Interestingly, Fe(SCN)2+ appears to exist
solely in solution. Solid compounds containing this ion have never been isolated.
The objective of this experiment is to determine the equilibrium constant for this reaction.
The equilibrium constant is given by the expression:
[Fe(SCN)2+]
K= ----------------------[Fe3+][SCN]where the concentrations of the substances are those at equilibrium. If these
concentrations are measured or inferred, K can be calculated easily.
Because the reactants are essentially colorless, whereas the complex ion is deeply colored
(redish-orange), you will use a spectrophotometer to monitor the # transmittance, %T,
due to the complex ion. The absorbance, A, (calculated from: 2-log%T ) is proportional
to the concentration, c of the species that absorbs the light – in this case, Fe(SCN)2+ according to Beer’s law, A = kc. After our previous laboratory – “Spectrophotometer
Lab”, you already know a great deal about using this law. After the concentration of
Fe(SCN)2+ has been measured by way of absorbance, the concentrations of the reactants
can be inferred from their starting concentrations and the concentration of the complex
ion.
There is a problem, however. To determine k, we must determine the absorbance’s of a
series of solutions with known amounts of the complex ion. How can known amounts of
Fe(SCN)2+ be obtained? After all, this substance is an active participant in the
equilibrium with Fe3+ and SCN- ions. Stoichiometric quantities of the reactants will not
yield a known amount of the product.
This difficulty can be avoided. Le Chatelier’s principle indicates that a net reaction from
left to right (that is, in the forward direction) can be achieved when more of a reactant is
added. As more and more of the same reactant is added, more and more of the product
forms. It is possible to add so much of this reactant that essentially all of the other
reactant is converted to the product. You will use limiting quantities of Fe3+ and
overwhelming amounts of SCN- to achieve this result. The amount of Fe(SCN)2+ that is
formed will then be essentially identical to the starting amount of the limiting reactant.
In this manner, you will prepare a series of solutions with known concentrations of
Fe(SCN)2+. You will calculate the absorbance’s of these solutions at 450 nm (blue-green
portion of the visible spectrum), the wavelength of maximum absorbance. When these
absorbances are plotted against the concentrations of Fe(SCN)2+ on a graph, k can be
determined from the slope of the straight line using linear regression. You will do this on
Excel. Once k has been determined, working under these conditions will no longer be
advantageous. In fact, you will determine the equilibrium constant K under conditions in
which substantial amounts of both of the reactants ant the product are present.
Doing the calculations
You will have to account for every dilution in order to do the calculations in this
experiment.
In the determination of k (runs 1-5), you will need to construct a graph in which
absorbance appears on the vertical axis and the concentration of Fe(SCN)2+, in moles per
liter, appears on the horizontal axis. This will be done in Excel. Report k in Table 1.
Print your graph – including the equation for a straight line…..be sure to set the yintercept to 0.
You will you calculated product and reactant concentrations to determine K, the
equilibrium constant, for this reaction. You will then ultimately report the average K for
runs 6-15.
Le Chatlier’s principle Lab
Procedure
**NOTE: Use instructions provide by the instructor – DO NOT USE INSTRUCTIONS
IN CHEMLAB! To remove the instructions on the screen, and free-up more working
area, perform the following operation: click on the OPTIONS tab; then click on LAB
ONLY. The instructions “disappear” and all of the area is now lab space.**
Determining k in Beer’s law
1) Obtain spectrophotometer from equipment menu and select 450nm.
2) Right click on the spectrophotometer and select chemicals.
3) From the chemicals menu add 1mL of 0.0001 M Fe(NO3)3 Note: this chemical
appears as (0.000M) Iron (III) nitrate – Fe(NO3)3 in the chemicals
window………..this is a slight bug in the software where not enough places past
the decimal place are reported. Use 0.0001 M Fe(NO3)3 for your calculations.
4) From the chemicals menu add 5 mL of 1M KSCN.
5) From the chemicals menu add 4 mL of 0.1M HNO3
6) Right click on the spectrophotometer, select spectrophotometer procedures –
insert sample.
7) Record % Transmission from the spectrometer.
8) Right click on the spectrophotometer, select spectrophotometer procedures –
remove sample.
9) Right click on the spectrophotometer and select empty to clear the contents of the
cuvet.
10) Repeat above procedure for remaining samples 2-5 as follows:
Run #
1
2
3
4
5
0.0000001 M
Fe(NO3)3 (mL)
1.0
2.0
3.0
4.0
5.0
1 M KSCN (mL)
0.1 M HNO3 (mL)
5.0
5.0
5.0
5.0
5.0
4.0
3.0
2.0
1.0
0
Determining the equilibrium constant
Do not use the 0.0001 M solution of Fe(NO3)3 in this part of the experiment. Use the
0.0025 M solution of this substance instead. Note: this chemical appears as
(0.003M) Iron (III) nitrate – Fe(NO3)3 in the chemicals window………..this is a
slight bug in the software where not enough places past the decimal place are
reported. Use 0.0025 M Fe(NO3)3 for your calculations.
Do not use the 1 M KSCN solution in this part of the experiment. Use the 0.0025 M
solution instead. Note: this chemical appears as (0.003M) Potassium Thiocyanate –
KSCN in the chemicals window………..this is a slight bug in the software where not
enough places past the decimal place are reported. Use 0.0025 M KSCN for your
calculations.
Repeat the above spectroscopic procedures above for concentrations of reactants as
follows: (record % transmittance for each run).
Note: Add HNO3 first, then the other two reactants – otherwise the software will
not give the correct transmittance data!
Run #
6
7
8
9
10
11
12
13
14
15
0.0025 M Fe(NO3)3
(mL)
1.0
1.0
1.0
1.0
1.0
2.0
2.0
2.0
2.0
2.0
0.0025 M KSCN
(mL)
1.0
1.5
2.0
2.5
3.0
1.0
1.5
2.0
2.5
3.0
0.1 M HNO3 (mL)
5.0
4.5
4.0
3.5
3.0
4.0
3.5
3.0
2.5
2.0
Le Chatlier’s principle Lab
Name:__________
Section:_________
Observations
Data: Determining k in Beer’s law
Table 1
Run
1
2
3
4
5
[Fe(SCN)2+]
(mol/L)
%T
A
(2-log(%T))
1.0 x 105
83.22
0.080
k – from
regression
(L/mol)
Representative calculation for run 1:
[Fe(SCN)2+] = 0.0001 M x (1.0ml / 10mL) = 1.0 x 105
According to Beer’s law, A = kc. When your absorbances are plotted against the
concentrations of Fe(SCN)2+ on a graph, k can be determined from the slope of the
straight line using linear regression. You will do this on Excel. Report k in Table 1.
Print your graph – including the equation for a straight line…..be sure to set the yintercept to 0.
Table 2
Run
%T
A
(2-log(%T))
6
7
8
9
10
11
12
13
14
15
75.31
0.123
Starting
[Fe3+]
(mol/L)
3.6 x 10-4
Starting
[SCN-]
(mol/L)
3.6 x 10-4
[Fe(SCN)2+]
(mol/L)
Equilibrium
[Fe3+] (mol/L)
1.5 x 10-5
3.4 x 10-4
Equilibriu
m [SCN-]
(mol/L)
3.4 x 10-4
Representative calculation for run 6:
[Fe3+] = [SCN-] = 0.0025 M x (1.0mL / 7.0mL) = 3.6 x 10-4 M (starting)
[FeSCN2+] = A/k = 0.123 / (8.1 x 103 L/mol) = 1.5 x 10-5 M
Equilibrium [Fe3+] = [SCN-] = 3.6 x 10-4 M – 1.5 x 10-5 M = 3.4 x 10-4
K = [Fe(SCN)2+]/([Fe3+][SCN-]) = 1.5 x 10-5 / ((3.4 x 10-4)(3.4 x 10-4)) = 1.3 x 102
K
1.3 x 102
Representative calculation for run 7:
[Fe3+] = 0.0025 M x (1.0mL / 7.0mL) = 3.6 x 10-4 M (starting)
[SCN-] = 0.0025M x (1.5mL/7.0mL) = 5.4 x 10-4 M (starting)
[FeSCN2+] = A/k = 0.181 / (7.98 x 103 L/mol) = 2.26 x 10-5 M
Equilibrium [Fe3+] = 3.6 x 10-4 M – 2.26 x 10-5 M = 3.37 x 10-4
Equilibrium [SCN-] = 5.4 x 10-4 M – 2.26 x 10-5 M = 5.17 x 10-4
K = [Fe(SCN)2+]/([Fe3+][SCN-]) = 2.26 x 10-5 / ((3.37 x 10-4)( 5.17 x 10-4)) = 1.3 x 102
Report average K for your set of data:

Average K for runs 6-15 = _____________
Discussion