Spectroscopic Determination of an Equilibrium Constant 11... · 2012-07-09 · 1 Spectroscopic...

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1

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

2

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

3

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 ==

=

4

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

5

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)

6

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

7

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*

8

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

9

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

10

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

11

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

12

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

13

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

14

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

15

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?

16

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