Equilibrium Constant
Today’s Experiment:
I.
Fe3+(aq) + HSCN(aq)
FeSCN2+(aq) + H+(aq)
orange
dark red
colorless
colorless
[FeSCN2 ][H  ]
K
 ???
3
[Fe ][HSCN]
1.
2.
3.
4.
Determine [FeSCN2+] using Spec20 and Beer’s Law
Determine the other concentrations from an ICE Table
Calculate K at three different Temperatures
Use the Temperature data to determine DH, DS, and DG for the reaction
II.
Take a look at the Pre-Lab Problems
III.
Beer’s Law and Making a Calibration Curve
1.
2.
Colored compounds absorb light that is shined through them
A = elC
Absorbance = (Extinction Coefficient)(length)(Concentration)
3. We will use test tubes #1-5 to make a calibration curve using Beer’s Law
4. Excess Fe3+ (0.200 M) pushes the reaction to the right: [HSCN]o = [FeSCN2+]
5. We must use Fe3+ solution as a blank to cancel out Fe3+ absorbance
6. Record %T and calculate A for the five different [HSCN] concentrations
I
%T 
Io
 100 
A  log 

 %T 
7. Plot A vs. [FeSCN2+] to give a straight line. Calibration Curve
A
Slope = el
A  εlC  C 
A
A

εl slope
[FeSCN2+]
8. Once you find the Absorbance of any other [FeSCN2+] solution, you can find its
concentration from the calibration curve.
IV.
Procedure for finding K (Tubes #6-9)
1.
2.
3.
4.
Make 4 different solutions of HSCN, Fe3+
K
Find A with the Spec20
Find [FeSCN2+] from the Calibration Curve
Use an ICE Table to find all the other concentrations
Solution
6
Temperature
Room
Initial Molar Concentration
Change in Molar
Concentration
Equilibrium Molar
Concentration
[FeSCN 2 ][H  ]

[Fe3 ][HSCN]
Species
Fe3+
HSCN
FeSCN2+
H+
0.00100
0.000200
0
0.500
-x
-x
+x
+x
x
0.500 + x
0.00100 - x 0.000200 - x
Calculated Values
5.
6.
0.500
All solutions are made with 0.500 M HNO3, so [H+] = 0.500
Other initial concentrations are found using the Dilution Equation
M1V1  M 2 V2


M1 V1 
0.002M 0.001L 
M2 
 [HSCN]o 
 0.00020M
V2 
0.010L
V.
Procedure for finding DH, DS, and DG (Tubes #6-9 at different T’s)
1.
2.
3.
You will use the same tubes (#6-9) at an ice bath (around 5 oC)
Hot tap water bath temperature (around 45 oC) [Not too hot! Boils off HSCN!)
You already have the room temperature data from these tubes (around 25 oC)
ΔG o  RTlnK  ΔH o  TDSo
ΔH o  1  ΔSo
lnK   
R T R
4.
5.
Slope = -DH/R
Intercept = DS/R
lnK
1/T
Plot lnK vs. 1/T for your three different temperatures
Use the following equations to calculate DH, DS, and DG for the reaction
ΔH  (slope)(R)
DS  (intercept )(R)
DG  DH - T D S
DG  -RTlnK (use average K at Room Temp. in Kelvins)
6.
R  8.3145 J/mol  K
Notes:
a.
Use parafilm to cover the test tubes as you mix the solutions thoroughly
b.
Spec20: 0% with nothing in it; 100% with Iron Solution only as Blank
c.
Fill cuvet with most dilute first, rinse with next most dilute, and so on