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Spectroscopic Determination of
an Equilibrium Constant
Experiment 11
Experiment 11
Goal:
� To determine the equilibrium constant, Keq, for the Fe3+/SCN-/FeSCN2+ system spectroscopically
Method:
� Using different starting concentrations, measure FeSCN2+ at its λmax
� Determine Keq from equilibrium concentrations
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Review: Molecular Spectra
400 500 600 700
λλλλ (nm)
Absorption EmissionIntensity
Transmittance, T
I0 I1 TI
I
0
1 =
T is ratio of light “in” vs. light “out”
or fraction of light passing through
Depends on:
�# of molecules b & c
�molecules’ identity ε
Pathlength, b
Concentration,c
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Molar Extinction Coefficient, ε
I
Ilogε
0
1M cm, 1
−=
εεεε Probability of absorption
–Specific to molecule
–Function of λ
λλλλmax
–Most efficient absorption/lowest T
–“Best λ” for experiment
T depends on absorbing molecule
ε
0
1M cm, 110
I
I −=
Absorbance
Beer’s Law: Defines absorbance, A
Tlogεbc A −==
A: Proportional to concentration
high T → low A
( ) εbc-bcε-
bc
0
1M cm, 11010
I
IT ==
=
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Beer’s Law
ε,b constant
εbc TlogA =−=
A−− === 1010%100
%TT εbc
c varied
Equilibrium
Reactant and product concentrations remain constant
Molecular level: rapid activity (dynamic)
Macroscopic level:unchanging
At equilibrium: rateforward = ratereverse
B A
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Example N2O4(g) 2NO2(g)
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
At equilibrium, relative [N2O4] and [NO2] remain constant
Equilibrium Constant
Quantitative determination of extent of reaction
Reactants: colorless
Product: colored (monitor by spectroscopy)
]O[N
][NOK
42
2
2=eq
N2O4(g) 2NO2(g)
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K is a number
small K
K < 10-2
favors reactant
large K
K > 102
favors product
intermediate K
10-2 < K < 102
favors neither
Equilibrium Constant
Here: Fe3+ + SCN- FeSCN2+
Reactants: colorless
Product: colored (monitor by spectroscopy)
You determine Keq
]SCN][Fe[
][FeSCNK
3
2
−+
+
=eq
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Fe3+/SCN-/FeSCN2+
Fe3+ SCN- FeSCN2+
Initial
Fe3+ SCN- FeSCN2+
Approaching equil.
Fe3+ SCN- FeSCN2+
Equilibrium
Le Châtelier’s Principle
. . . if a change is imposed on a system at equilibrium, the
position of the equilibrium will shift in a direction that
tends to reduce that change
� Concentration
� Temperature
� Pressure, Volume, Catalysts*
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Le Châtelier’s Principle
Fe3+ + SCN- FeSCN2+
Temperature changes
exothermic or endothermic
heat = “product” or “reactant”
Concentration changes
Part 1 (week 1)
Reaction energetics: exothermic or endothermic?
�Mix: 2mL NaNO3 2 × 10-3 M
+ 8mL NaSCN 2 × 10-3 M color
+ 10mL Fe(NO3)3 2 × 10-3 M
�Three test tubes (+ 1 cuvette 2/3 full)
� Ice bath
� Hot water bath
� Room temperature
�Compare colors after 10 minutes color
�Is the reaction exo- or endothermic? exo/endo
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Example data: Part 1
Conclusion:
Reaction is endothermicReason:
Concentration of colored product increases with addition of heat
Reaction:
Heat + Fe3+ + SCN- FeSCN2+
NaSCN Fe(NO3)3 NaNO3
intial conc. (M) 2x10-3
2x10-3
2x10-3
vol. (mL) 8.00 10.00 2.00
pre-eq. conc. (M) 8x10-4
1x10-3
2x10-4
Bath conditions Equil. Mixture
Initial red-orange
Ice almost tranparent
Hot darker red-orange
Part 2 (week 1)
Spectral profile of FeSCN2+ for λλλλmax
�NaNO3 = blank / 100% T
�Record %T %T
� 370 – 560 nm
� 5 or 10 nm intervals near Tmin (Amax)
� 20 nm intervals elsewhere
�Calculate A A
�Plot A vs. λ plot
�Determine λλλλmax λλλλmax
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Find λmax
370 560
λλλλ (nm)
AbsorptionTransmittance
Intensity
Find λλλλ where
%T is lowest
A is highest
Example data: Part 2
λmax ~460-465 nm (absorbs green-blue/appears red-orange)
wavelength (nm) %T A
370 52.0 0.284
380 52.0 0.284
390 52.5 0.280
400 52.5 0.280
410 51.0 0.292
420 48.0 0.319
430 41.5 0.382
440 37.0 0.432
450 34.5 0.462
455 33.0 0.481
460 32.5 0.488
465 32.5 0.488
470 33.0 0.481
475 34.0 0.469
480 35.0 0.456
490 36.5 0.438
500 40.0 0.398
510 44.0 0.357
520 48.5 0.314
530 55.0 0.260
540 61.0 0.215
550 67.0 0.174
560 73.0 0.137
Absorbance vs. Wavelength for FeSCN2+
0.100
0.150
0.200
0.250
0.300
0.350
0.400
0.450
0.500
0.550
350 375 400 425 450 475 500 525 550 575
wavelength (nm)
absorbance
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Part 3 (week 1)
Make solutions of known [FeSCN-]
� Use large [Fe3+] and small [SCN-]
[Fe3+] in excess
[SCN-] all gone
� initial [SCN-] ≈≈≈≈ equilibrium [FeSCN2+]
� After experiment, verify this assumption
Part 3 (week 1)
Measure A for various concentrations
�Make 100 mL 0.1 M Fe(NO3)3 X
�Make strongest NaSCN/Fe(NO3)3 Y
5 mL 2×10-3 M NaSCN
in 50 mL flask/fill with X
�Dilutions (10 mL total volume)
1 , 3 , 5 , 7 , 9 mL Y + pure Y
Filled to mark with X
MmL
MmLNaSCNM
43
10250
1025 −−
×=×⋅
=
NaSCN Fe(NO3)3 Solutions
2.0x10-4 M 0.1
M 10 mL total
1 mL 9 mL 2.0x10-5 M
3 mL 7 mL 6.0x10-5 M
5 mL 5 mL 1.0x10-4 M
7 mL 3 mL 1.4x10-4 M
9 mL 1 mL 1.8x10-4 M
10 mL 0 mL 2.0x10-4 M
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Data
Measure absorbance as each is prepared A
Plot A vs. conc. → Beer’s Law plot plot
A = εεεεbc b = pathlength
y = mx c = conc.
Slope = m = εεεεb εεεε
Plot Absorbance
vs. FeSCN2+eq
SCN-initial %T FeSCN
2+eq Absorbance
2.0x10-5 M 88.6 2.0x10
-5 M 0.05
6.0x10-5 M 65.2 6.0x10
-5 M 0.19
1.0x10-4 M 48.4 1.0x10
-4 M 0.32
1.4x10-4 M 33.0 1.4x10
-4 M 0.48
1.8x10-4 M 23.5 1.8x10
-4 M 0.63
2.0x10-4 M 19.0 2.0x10
-4 M 0.72
y = 3477.1x
R2 = 0.9923
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.0E+00
2.0E-05
4.0E-05
6.0E-05
8.0E-05
1.0E-04
1.2E-04
1.4E-04
1.6E-04
1.8E-04
2.0E-04
Molarity
Absorbance
Beer’s Law Plot
cεbA ⋅=
xy ⋅= m
y∆
x∆
εb =∆
∆=
x
yslope
For m = εb = 3477
If b = 1.00 cm:
ε = 3477 L.mol-1cm-1
FeSCN2+
eq Absorbance
2.0x10-5 M 0.05
6.0x10-5 M 0.19
1.0x10-4 M 0.32
1.4x10-4 M 0.48
1.8x10-4 M 0.63
2.0x10-4 M 0.72
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Part 4 (week 2)
Determine Keq for [FeSCN-] formation
� Use 2 × 10-3 M NaNO3, Fe(NO3)3, NaSCN
� Take %T and find A
� Use Beer’s Law plot to find [FeSCN2+]eq
NaSCN Fe(NO3)3 NaNO3
0 mL 5 mL 5 mL
1 mL 5 mL 4 mL
2 mL 5 mL 3 mL
3 mL 5 mL 2 mL
4 mL 5 mL 1 mL
5 mL 5 mL 0 mL
←←←←blank
Beer’s Law Plot Results and FeSCN2+eq
εbAc =
NaSCN Fe(NO3)3 NaNO3 %T A FeSCN2+
eq
2.0x10-3 M 2.0x10
-3 M 2.0x10
-3 M % --- 3477
0 mL 5 mL 5 mL 100.0 0.000 0.00E+00
1 mL 5 mL 4 mL 70.8 0.150 4.31E-05
2 mL 5 mL 3 mL 49.0 0.310 8.92E-05
3 mL 5 mL 2 mL 39.8 0.400 1.15E-04
4 mL 5 mL 1 mL 30.2 0.520 1.50E-04
5 mL 5 mL 0 mL 24.0 0.620 1.78E-04
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Data Analysis – 5 ICE tables
ICE Table Fe3+
SCN-
FeSCN2+
Initial (mol/L)
Change (mol/L)
Equilibrium (mol/L)
x: Amount of product created
[Fe3+]init [SCN-]init 0
– x – x + x
[Fe3+]init - x [SCN-]init - x x
ICE Table 1 Fe3+
SCN-
FeSCN2+
Initial (mol/L) 1.0x10-3 M 2.0x10
-4 M 0 M
Change (mol/L) -4.3x10-5 M -4.3x10
-5 M +4.3x10
-5 M
Equilibrium (mol/L) 9.6x10-4 M 1.6x10
-4 M 4.3x10
-5 M
Example ICE Table – solution 1
Data Analysis
For 5 solutions, determine Keq
Find average Keq, σK, and relative error (σ/Kavg)
Compare values
Discuss
)]SCN)([]Fe([]SCN[]Fe[
][FeSCNK
33
2
xx
x
initiniteqeq
eq
eq−−
==−+−+
+
.280)106.1)(106.9(
103.4
]SCN[]Fe[
][FeSCNK
44
5
3
2
1, =××
×==
−−
−
−+
+
eqeq
eq
eq
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Data Analysis
Go back and check assumption in Part 3
[FeSCN2+]eq = [SCN-]init? (sml [SCN-]init & lg [Fe3+]init)
With: [FeSCN2+]eq = Keq. [Fe3+]eq
[SCN-]eq
0.09 < [Fe3+] < 0.10 M and 2 × 10-5 < [SCN-] < 2 × 10-4 M
If [SCN-]init → 0 then [FeSCN2+]eq → large
[SCN-]eq
Data Analysis Example
If Keq = 1000 and [Fe3+] = 0.09 M
eqeq
eq
eq]Fe[K
]SCN[
][FeSCN3
2
+
−
+
⋅=
90)09.0)(1000(]Fe[K 3 ==⋅ +
eqeq
eq
eq
][SCN
][FeSCN
1
9090
−
−
==
~1% SCN- left at equilibrium
90 out of 91 SCN- → FeSCN2+
~1% error
Using:
What effect does this have of Keq?
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Report
Abstract
� Method of determining results
� Summary of results
Results including
� Observations Part 1
� Spectral Profile Part 2
� Beer’s law plot Part 3
� ICE tables, individual and average Keq Part 4
Sample calculations including
� Absorbance from transmittance
� Concentration calculations
� Keq calculation
� Assumption tests
Discussion/review questions