Transcript Slide 1

Exp. 20 - video
(time: 41:13 minutes)
Exp. 20: Spectrophotometric Analysis:
Determination of the Equilibrium
Constant for a Reaction
Chemical Equilibrium
Previously we have assumed that chemical
reactions results in complete conversion of
reactants to products:
A + B
C + D
No A or B remaining or possibly an excess of A or B but not both and
eventually reaction stops.
Many chemical reactions do not completely convert reactants to
products. Stop somewhere between no rxn and complete rxn.
A + B
some left
C + D
some formed
reversible (both directions)
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A + B C + D
A + B C + D
exchange, constant conc.,
Ratef = Rater “equilibrium”
Chemical Equilibrium
• Therefore, many reactions do not go to
completion but rather form a mixture of
products and unreacted reactants, in a
dynamic equilibrium.
– A dynamic equilibrium consists of a forward reaction, in which
substances react to give products, and a reverse reaction, in which
products react to give the original reactants.
– Chemical equilibrium is the state reached by a reaction mixture
when the rates of the forward and reverse reactions have
become equal.
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Graphically we can represent this
A + B
C +D
The concentrations and reaction rate (less collisions, less component) of A and B
decreases over time as the concentrations and reaction rate of C and D increases
(more collisions, more component) over time until the rates are equal and the
concentrations of each components reaches a constant. This occurs at what we call
equilibrium -- Rf = Rr.
If the rates are equal, then there must be a relationship to show this.
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3H2 + CO
CH4 + H2O
Rf=Rr
3H2 + CO

CH4 + H2O 
CH4 + H2O
3H2 + CO
rate decreases over time
rate increases over time
Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin,
New York, NY, 2005.
If we assume these reactions are elementary rxns (based on collisions), we
can write the rate laws directly from the reaction:
A + B
C+ D
Rf = kf [A][B]
For the reverse reaction we have,
C +D
A+ B
Rr = kr [C][D]
We know at equil that Rf = Rr ;therefore, we can set these two expressions as
equal
kf [A][B] = kr [C][D]
Rearrange to put constants on one side we get
kf
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[C ][ D]
K

k r [ A][ B]
Constant divided by constant just call a new constant K. This ratio is given a
special name and symbol called equilibrium constant K relating to the equilibrium
condition at a certain temperature (temp dependent) for a particular reaction relating
conc of each component. This is basically a comparison between forward and
reverse reaction rates. At equilibrium, the ratio of conc of species must satisfy K.
kf
[C ][ D]
K

k r [ A][ B]
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The Equilibrium Constant
• Every reversible system has its own “position of
equilibrium”- K- under any given set of conditions.
– The ratio of products produced to unreacted reactants
for any given reversible reaction remains constant under
constant conditions of pressure and temperature. If the
system is disturbed, the system will shift and all the
concentrations of the components will change until
equilibrium is re-established which occurs when the ratio of
the new concentrations equals "K". Different constant conc
but ratio same as before.
– The numerical value of this ratio is called the equilibrium
constant for the given reaction, K.
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The Equilibrium Constant
• The equilibrium-constant expression for a reaction is
obtained by multiplying the equil concentrations ( or partial
pressures) of products, dividing by the equil concentrations (or
partial pressures) of reactants, and raising each concentration to
a power equal to its coefficient in the balanced chemical
equation.
aA  bB
c
cC  dD
d
[C ] [ D]
Kc 
a
b
[ A] [ B]
c
d
( PC ) ( PD )
Kp 
( PA ) a ( PB )b
The molar concentration of a substance is denoted by writing its formula
in square brackets for aq solutions. For gases can put Pa - atm. As
long as use M or atm, K is unitless; liquids and solids = 1; setup same
for all K’s (Ka, Kb, etc.)
Temp dependent; any changes, ratio will still equal K when equil
10established
In our experiment, we will determine the K for the following
reaction:
Fe3+ + SCN-
FeSCN2+
K?
2
[ FeSCN ]eq
K 
[ Fe3 ]eq[ SCN  ]eq
Fe3+
Starting, [ ]o
MFe3+
Change, D[ ]
-x
Equilibrium, MFe3+ - x
[ ]eq
+
SCN-
FeSCN2+
MSCN-
0
-x
+x
x
MSCN- - x
[ FeSCN2 ]eq
K 

3

[ Fe ]eq[SCN ]eq [M
3
Fe
[x]
- x][M

- x]
SCN
If we can determine x, we can solve for K for
this reaction.
First part of experiment involves making a
calibration curve for [FeSCN2+]
Calibration Curve
We basically force the [FeSCN2+] to equal the initial
[SCN-] by using a 1:1000 ratio of SCN- to Fe3+
pg 139: 20.00 mL of 2.00 x 10-4 M KSCN
20.00 mL of 0.200 M Fe(NO3)3
Therefore, [SCN-]o diluted within rxn = [FeSCN2+]eq
for calibration only not actual K experiment
Calibration Curve
[SCN-]o = [FeSCN2+]eq within rxn (meaning taking into
account dilution)
No.1 Standard (pg 139)
20.00 mL of 2.00 x 10-4 M KSCN
20.00 mL of 0.200 M Fe(NO3)3
CV = CV
(2.00 x 10-4 M SCN-) (20.00 mL) = CSCN- diluted No. 1 (40.00 mL)
[CSCN-diluted No.1] = 1.00 x 10-4 M = [FeSCN2+]No. 1
10.0 x 10-5
1.00 x 10-4
5.00 x 10-5
2.50 x 10-5
blank:
0.00 M
0.00
No. 2 standard
(1.00 x 10-4 M SCN-) (10.00 mL) = CSCN- diluted No. 2 (20.00 mL)
[CSCN-diluted No. 2] = 5.00 x 10-5 M = [FeSCN2+]No. 2
No. 3 is done same way and then measure the absorbance at 460 nm (watch filter
on spec 20) of each for calibration curve.
Tip for plotting: make the three conc the same power of 10 and label axis conc x
10-5 M.
Preparation of solutions for K experiment
pg 139:
Fe3+ solution – CV = CV
Make solutions No. 1 -3 from stock 0.200 M; skip
No.4
0.100
0.0500
0.0250
Preparation of solutions for K experiment
pg. 140:
Initial [SCN-]o
same dilution across (5 mL SCN- and 5 mL Fe3+)
[SCN-]o = 1.00 x 10-4 M (skip No.4)
Initial [Fe3+]o
same dilution but different stock of Fe3+ (5 mL/5 mL)
[Fe3+]o = ½ [Fe3+ solution pg 139] (skip No. 4)
0.100 M
[SCN-]o
[Fe3+]o
0.0500 M
0.0250 M
1.00 x 10-4 1.00 x 10-4 1.00 x 10-4
0.0500 M 0.0250 M
0.0125 M
Determination of K
-Obtain calibration curve from first three solutions (top
of pg 139); this plot will allow you to determine the
[FeSCN2+]eq for each of the three experimental
mixtures.
Mix solutions for exp 1 (pg 140), obtain absorbance
from spec 20 at 460 nm, read absorbance off
calibration curve for [FeSCN2+]eq; technically the x in
K expression
Determination of K
Bottom table pg 140
[Fe3+]eq = [Fe3+]o – x = [Fe3+]o – [FeSCN2+]eq
[SCN-]eq = [SCN-]o – x = [SCN-]o – [FeSCN2+]eq
-Plug values into K expression and solve for K
-Repeat for exp 2 and 3
exp1
eq
eq
exp 2
exp 3
spec 20
spec 20
spec 20
graph = “x”
graph
graph
[Fe3+]o – [FeSCN2+]eq
= 0.0500 – graphexp1 0.0250 – graphexp2 0.0125 – graphexp3
[SCN-]o – [FeSCN2+]eq
1.00 x 10-4 – graphexp2 1.00 x 10-4 – graphexp3
= 1.00 x 10-4 – graphexp1
eq
[ FeSCN2 ]eq
[ FeSCN2 ]eq
[ FeSCN2 ]eq
K 
[ Fe3 ]eq[SCN  ]eq [ Fe3 ]eq[ SCN  ]eq [ Fe3 ]eq[ SCN  ]eq
Goal:
Average K for reaction and std dev.
Only computer plot required but make sure change
axis to some small increment.
Chemicals:
Take only 45 mL of each component.