Lab 8_Themodynamics
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Lab 8: Thermodynamics
This homework uses the virtual lab. Using a computer that is running Microsoft windows or Macintosh OS 10.1
or higher, go to
http://chemcollective.org/vlab/85
I.
Objective
A spontaneous process is one that takes place without a continuous input of energy.
How can reaction
spontaneity be determined or measured?
Both the enthalpy change (
H
) and the entropy change (
S
) during
the reaction are important factors that relate to spontaneity.
Exothermic reactions, where
H
is negative, are
usually spontaneous in nature, but not always.
Likewise, the entropy change is also an imperfect predictor of
spontaneity.
The most reliable predictor of reaction spontaneity is the change in Gibbs Free Energy (
G
) for
the reaction.
Equation 1 relates the Gibbs Free Energy change to the entropy and enthalpy changes for a
reaction.
G
=
H
-
T
S
(1)
If
G
is positive, the reaction is non-spontaneous; if
G
is negative, the reaction is spontaneous.
This
behavior and equation 1 will explain why being exothermic is not an ironclad approach to predicting
spontaneity.
Four different permutations of
H
and
S
are summarized in Table 1; the third set clearly outlines
that an exothermic reaction may indeed produce a non-spontaneous reaction.
Table 1.
Reaction Spontaneity
H
S
-
T
S
G
Spontaneous?
-
+
-
-
always
+
-
+
+
never
-
-
+
-
at low
T
,
+ at high
T
low
T
yes,
high
T
no
+
+
-
+ at low
T
,
- at high
T
low
T
no,
high
T
yes
If we know all the parameters on the right-hand side of equation 1, we can calculate
G
.
Measuring
temperature is not difficult, and we have effective methods to determine
H
using calorimetry.
However,
experimental measurement of
S
is somewhere between impractical and impossible.
When knowledge of the
entropy is important, it is either found theoretically or experimentally determined in an indirect manner that
involves determination of
G
first.
Since we will be evaluating a chemical reaction at equilibrium, it will be
best for us to alter equation 1 slightly and move the thermodynamic parameters to their standard state
designations, as shown in equation 2.
G
° =
H
° -
T
S
°
(2)
The Gibbs Free Energy change is also difficult to measure, and we will find it in an indirect manner by
first determining the equilibrium constant for the reaction of interest
G
° = - R
T
ln
K
(3)
and using equation 3.
You will recognize the
K
term in equation 3 as the equilibrium constant, and
T
as the
absolute temperature.
Despite our reaction having nothing whatsoever to do with gases, the R term is indeed
the ideal gas constant.
Check in your textbook and select the proper value of R that will yield
G
° in units of
DeGrand, M. J.; Abrams, M. L.; Jenkins, J. L. ; Welch, L. E.
J. Chem. Educ.
2011
, 88
, 5
, 634-636.
NAME: _________________________
SECTION: ____________________
joules.
You will be familiar with finding
K
c
values experimentally from an earlier experiment this term, and thus
now have a pathway to calculate
G
° values.
With this in mind, we are going to take a look at a reaction system where the equilibrium position is
highly sensitive to temperature.
Not all reactions respond in this manner.
If we can see the equilibrium shift
as temperature changes, we can calculate that
G
° is also changing.
For the most part, Δ
H
° and Δ
S
° are
insensitive to temperature changes, and we will assume them to be constants within the range of
temperatures we will study in this experiment.
By doing so, we can measure the response of
G
° as we
change temperature, and then by using equation 2 we can make a graph that will allow us to determine both
S
° and
H
° for this system.
The chemical system we will study involves cobalt chloride in aqueous solution.
Like many transition
metals, cobalt dissolves in water to form a +2 cation.
The Co
+2
is stabilized in water by forming 6 coordinate
covalent bonds to yield the compound Co(H
2
O)
6
2+
.
This compound is bright red in color.
Co
+2
can also form
coordinate covalent bonds with the Cl
-
ion.
As Cl
-
is added to the aqueous Co
+2
, the chloride will begin to
displace the coordinated waters, going through a series of stepwise additions that are detailed by reactions 4a
through 4d shown below.
Co(H
2
O)
6
2+
(aq)
+
Cl
-
(aq)
⇌
CoCl(H
2
O)
5
+
(aq)
+
H
2
O
(l)
(4a)
CoCl(H
2
O)
5
+
(aq)
+
Cl
-
(aq)
⇌
CoCl
2
(H
2
O)
2 (aq)
+
3H
2
O
(l)
(4b)
CoCl
2
(H
2
O)
2 (aq)
+
Cl
-
(aq)
⇌
CoCl
3
(H
2
O)
-
(aq)
+
H
2
O
(l)
(4c)
CoCl
3
(H
2
O)
-
(aq)
+
Cl
-
(aq)
⇌
CoCl
4
2-
(aq)
+
H
2
O
(l)
(4d)
We can write a
K
c
expression for any one of these reactions.
Obviously, things will get complicated if
we have to deal with all 4 reactions at once.
By controlling our reactant mixture, we can produce a situation
where all of the intermediate chlorinated forms of cobalt (i.e. CoCl(H
2
O)
5
+
, CoCl
2
(H
2
O)
2
, CoCl
3
(H
2
O)
-
) can be
considered negligible in concentration.
This allows us to write a single reaction formula, shown below as
reaction 5, which is essentially the sum of the four steps 4a through 4d.
Co(H
2
O)
6
2+
(aq)
+
4Cl
-
(aq)
⇌
CoCl
4
2-
(aq)
+
6H
2
O
(l)
(5)
You may recall the reaction in equation 5 from the lab on LeChâtelier’s Principle, where changing
temperature shifted the equilibrium from the pink color of Co(H
2
O)
6
2+
to the brilliant blue of CoCl
4
2-
. We will
take a more quantitative approach here, measuring the equilibrium constant, K
c,
at several temperatures.
In
the laboratory, the concentration of one or both cobalt complexes would be measured using absorbance
spectroscopy (Beer’s law)
in much the same way we did when determining the K
c
for Fe(SCN)
2+
complex
formation.
In this online simulation, concentrations of each substance in the mixture will be available directly
to you on screen. The measured values of K
c
can be used to determine
G
° at each temperature according to
equation 3. Finally, plotting
G
° vs. T will allow determination of
H
° and
S
° for the reaction as the graph
should fit the linear equation:
G
° =
H
° -
T
S
°.
II.
Procedure
1.
Click on the link to open the Chem Collective virtual lab:
http://chemcollective.org/vlab/85
NAME: _________________________
SECTION: ____________________
2.
Using a 100-mL volumetric flask and the 12M HCl solution provided, prepare 100 mL of 6M HCl. Use
M1V1 = M2V2 to determine volume of stock you need for your dilution.
3.
Mix 40 - 60 mL of 1M CoCl
2
and 100 mL of 6M HCl in a 250-mL Erlenmeyer flask and click on the
flask. Record the temperature and equilibrium values of [Co(H
2
O)
6
2+
], [Cl
-
], and [CoCl
4
2-
] for the
mixture.
4.
To adjust the temperature, right-click on the flask and select “Thermal Properties”. Click on
“Insulated from surroundings” and enter a new temperature between 0 - 65 °C. Record the new
values.
5.
Repeat step 4 with 3 additional temperatures.
III.
Data Collection
a)
Raw Data (30 points)
Volume of 1M CoCl
2
,
(
mL)
Temperature,
(°C)
[Co(H
2
O)
6
2+
],
(M)
[Cl
-
],
(M)
[CoCl
4
2-
],
(M)
b)
Calculated Data (40 points)
To receive credit full for this section, you will need to show sample calculations.
You can type them
out or write by hand and insert below, whichever is easier for you.
a)
For each equilibrium mixture, determine the value of K
c
and ΔG° for the equilibrium shown below.
Show a sample calculation.
Co(H
2
O)
6
2+
(aq)
+
4 Cl
-
(aq)
⇌
CoCl
4
2-
(aq)
+
H
2
O
(l)
Temperature,
(°C)
K
c
ΔG°,
(kJ/mol)
Sample calculations:
K
c
:
Your preview ends here
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NAME: _________________________
SECTION: ____________________
ΔG°:
b)
Plot
ΔG°
vs. Temperature in Excel and include a trendline. Include the chart in your report
submission.
c)
Using the equation for the best-fit line, determine the values for ΔH° (kJ/mol) and ΔS° (J/mol
⋅
K).
Show any work.
ΔH°,
(kJ/mol)
ΔS°,
(J/mol
⋅
K)
Calculations:
IV.
Applied Exercises (30 points)
Answer the following questions in 3-4 complete sentences.
c)
Based on the measured ΔG° values, is this equilibrium spontaneous at room temperature? Which
factor, entropy or enthalpy, has the greater impact on spontaneity in this particular case? Explain
your answers.
d)
You have experimentally measured Δ
S
° for this reaction.
Does the sign of this quantity match what
you would have expected from just examining the chemical reaction set being studied?
Discuss
briefly
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