Chemistry 112 Laboratory: Kinetics
Page 83
Chemical Kinetics
T
his experiment is a study of the rate of a chemical reaction and the
dependence of that rate on the temperature. You will first verify the
reaction is first-order in the reactant of interest, a metal complex, and then
you will study the temperature dependence of the rate. This latter study will
enable you to calculate the activation energy, Ea, for the reaction.
The compound to be studied is the beautiful green, cobalt-based complex
called trans-dichlorobis(ethylenediamine)cobalt(+) ion.
+
Cl
H2C
H2C
H2
N
H2
N
Co
N
H2
N
H2
CH2
The ethylenediamine molecule and the chloride ions
are called “ligands.”
CH2
Cl
The nitrogen atoms of the ethylenediamine molecule all lie in a plane surrounding the Co3+ ion. The line connecting the two Cl- ions is perpendicular
to this plane and passes through the Co3+ ion. The trans portion of the name
indicates this arrangement—the Cl- ions are across the molecule from one
another.
The reaction you will study is the aquation of the green complex ion
trans–[Co(en)2Cl2]+ to form the red complex ion [Co(en)2(H2O)Cl]2+.
trans–[Co(en)2Cl2]+(aq) + H2O(liq) → [Co(en)2(H2O)Cl]2+(aq) + Cl-(aq)
green
red
That is, a Cl- ion is replaced with a water molecule on the Co3+ ion. Reactions
such as this have been studied extensively, and experiments suggest that the
initial, slow step in the reaction is the breaking of the Co–Cl bond to give a
five-coordinate intermediate.
Slow: trans–[Co(en)2Cl2]+(aq) → [Co(en)2Cl]2+(aq) + Cl-(aq)
Fast:
[Co(en)2Cl]2+(aq) + H2O(aq) → [Co(en)2(H2O)Cl]2+(aq)
This means the rate of the overall reaction is a measure of the rate of the slow,
bond-breaking elementary step. The rate-determining, elementary reaction
is first-order in the reactant trans–[Co(en)2Cl2]+. The rate law for this step is
therefore
Rate = k{trans–[Co(en)2Cl2]+}
December 2005
Information on rate laws
and reaction mechanisms
is found in Chapter 15
of Chemistry & Chemical
Reactivity.
Chemistry 112 Laboratory: Kinetics
Page 84
(where k is the rate constant). This is a first order process whose rate depends
on the concentration of the reactant in solution.
A feature of all first-order reactions is that the logarithm of the fraction
of original reactant remaining [ln(C/Co)] after some time has elapsed (t) is
equal to the negative of the product of the rate constant (k) and the time.
ln C = - kt
C0
ln (fraction remaining) = - (rate constant)•(time elapsed)
C = concentration after time t has elapsed
C0 = original concentration at t = 0
k = first order rate constant
t = time
Be sure to notice this
equation uses the natural
logarithm. See page A-2 in
the Appendix of Chemistry
& Chemical Reactivity.
This equation informs us that the time to reach a given fraction remaining is the
same no matter what the initial concentration (C0).
The conclusion above is crucial to understanding this experiment. During
the reaction there is a moment at which the mixture of green reactant and
red product appears gray. This gray color is reached when the ratio of green
reactant to red product has a given value. Or, rewriting the equation above in
terms of this experiment,
C
[green ion]
=
C0
[green ion] + [red ion]
the gray color appears at a particular fraction remaining. You will use this
feature of the reaction to verify that the reaction is first order in the green
complex ion trans–[Co(en)2Cl2]+.
The activation energy for a reaction, Ea, is the minimum energy the reactants must possess before reaction can occur, and we can determine this energy by studying the temperature dependence of the reaction rate. Specifically,
the temperature dependence of the rate constant for a reaction is given by the
equation
k = Ae-E a /RT
where k = rate constant
A = constant
Ea = activation energy
R = gas constant = 8.31 J/K•mol or 8.31 x 10-3 kJ/K•mol
T = temperature (in kelvins)
How can we interpret this equation? First, as T increases, the quotient Ea/RT
becomes smaller (for a given Ea). But, importantly, e-x becomes larger as
the quantity x gets smaller. Therefore, we conclude that the rate constant
k increases as T increases. This is not a surprising conclusion, because it is
reasonable that the reacting molecules acquire more and more energy as the
temperature increases.
December 2005
This is the crucial idea in
the experiment. In essence
you are measuring the halflife of the reaction. For a
first order reaction the halflife is independent of the
initial concentration. See
pp. 712-722 of Chemistry
& Chemical Reactivity.
Chemistry 112 Laboratory: Kinetics
Page 85
In the aquation of trans–[Co(en)2Cl2]+, Ea is related to the energy of the
Co-Cl bond. It is not the energy required to break the bond but rather it is the
energy required to stretch the bond to the point at which it breaks completely
rather than reforming.
Rearranging the equation for the temperature dependence of k we have
ln k = -(
Ea 1
)( ) + ln A
R T
This is an equation of a straight line of the type y = mx + b where y = ln k, m,
the slope, is (-Ea/R), x is (1/T), and b is ln A. This means that if we plot ln
k versus (1/T) we can obtain the activation energy from the slope of the
plot. However, we do not need to measure k in this experiment to obtain Ea.
Instead, go back to the first-order rate law on page 86.
ln
C
= -kt
C0
t = -(1/k)ln(C/C0) for a 1st
order process. Thus, because
the “gray point” in this experiment always occurs at the
same C/C0 ratio, measuring
time is a measure of the value
of 1/k at a given temperature.
Rearranging, we have
⎛ 1⎞ C
k = ⎜ ⎟ ln 0
⎝ t⎠ C
If we take the log of both sides, we obtain
lnk = ln
1
+ ln
t
⎡ C0 ⎤
⎢⎣ ln C ⎥⎦
Now substitute this equation into the equation connecting ln k and Ea.
ln
1
E 1
⎡ C ⎤
+ ln ⎢ ln 0 ⎥ = -( a )( ) + ln A
t
R T
⎣ C⎦
Finally, rearrange the equation just above.
1
E 1
ln = -( a )( ) + ln A - ln
t
R T
⎡ C0 ⎤
⎢⎣ ln C ⎥⎦
This equation informs us that if we plot ln (1/time) versus 1/T (where T is
the kelvin temperature of the experiment), we will have a straight line plot
with a negative slope of Ea/R.
December 2005
R = 8.31 x 10-3 kJ/K•mol
Chemistry 112 Laboratory: Kinetics
Page 86
Experimental Procedure
Begin by preparing an ice-bath in which you cool a 10-mL graduated cylinder,
a 25 x 150 mm test tube, two 18 x 150 mm test tubes, and a flask containing about 100 mL of distilled water. (You can use two, 600-mL beakers as
ice baths.) The test tubes and graduated cylinder must be dry on the inside
except for some slight condensation of moisture from the air.
Fill a 250-mL beaker with water and begin heating it to prepare your constant temperature bath.
You may do this experiment with another student.
Please do the calculations
independently.
Determining the Ratio That Gives the Gray Color
1. Weight out about 0.3 g of trans-[Co(en)2Cl2]Cl and dissolve it in 35 mL of
cold water in your 25 x 150 mm test tube. Mix the solution thoroughly and
keep it cold in the ice bath to inhibit the aquation reaction.
2. To prepare a sample of the final product of the aquation, [Co(en)2Cl(H2O)]2+,
add about 10 mL of the green solution, which you prepared in Step 1, to
an 18 x 150 mm test tube and heat this in your boiling water bath for about
5 minutes. The solution should now be pink, indicating that the following
reaction has occurred.
trans–[Co(en)2Cl2]+(aq) + H2O(liq) → [Co(en)2(H2O)Cl]2+(aq) + Cl-(aq)
green
red
Cool this pink solution in the ice bath.
3. Stand the 10-mL graduated cylinder in an ice bath, add about 4.5-5.0 mL
of the green stock solution, and record the volume to the nearest 0.1 mL.
Add the pink solution dropwise with thorough stirring until the solution
assumes a gray color when viewed against a white background. Under
TRIAL 1 on the Report Form, record the volume when the solution last
appears faintly green, when the appearance is gray, and when the appearance is faintly pink.
4. Empty the graduated cylinder and add about 2.5 mL of the green stock
solution about about 4.5 mL of cold water. Add pink solution and record
the data as above under TRIAL 2.
5. Calculate the ratio indicated on your Report Form for the two trials.
The values should be identical within the limits of your ability to detect
the colors. This ratio represents the fraction of an originally pure trans–
[Co(en)2Cl2]+ sample that will have become aquated (changed to the aqua
complex trans–[Co(en)2(H2O)Cl]2+) when the gray color appears in your
kinetic runs.
Verification that the Aquation Reaction is First-Order
1. Add 5 mL of the green stock solution to a cold 18 x 150 mm test tube.
To another test tube add 2.5 mL of the green stock solution and 2.5 mL
of cold water. Mix well. Immerse these test tubes in the ice bath immediately.
2. Set up a constant temperature bath as demonstrated by your instructor
and adjust the bath temperature to 55-60 ˚C. Simultaneously transfer the
two cold test tubes made up in Step 1 to the bath and begin timing.
December 2005
You may remove the graduated cylinder from the ice
bath periodically to observe
the color.
Chemistry 112 Laboratory: Kinetics
•
The test tubes should be immersed far enough so the liquid level in the
tubes is below the water bath level.
•
Starting at the same temperature and having the same solution volume,
the solutions should warm at the same rate. (Do not worry if the bath
temperature declines slightly.)
3. Observe the immersed test tubes against a white background, and record
the time when the solutions turn gray.
Because the rate of the reaction is first-order in the concentration of
trans–[Co(en)2Cl2]+, the gray color should appear at the same time in
each solution. If the reaction were second-order in the concentration
of the ion, the less concentrated solution should take about twice as
long for the gray color to appear. If the reaction were zero-order, the
more concentrated solution should take twice as long for the gray color
to appear.
Page 87
The statement in the paragraph
at the left is important. As
explained on page 2, you are
essentially measuring the halflife of the reaction. For a first
order reaction the half-life is
independent of the initial concentration.
Measurement of Activation Energy
1. Prepare a new stock solution of trans–[Co(en)2Cl2]+ by dissolving about
0.3 g of the solid in 5 mL of cold water in a small beaker or test tube. Store
the solution in an ice bath.
2 Add 5 mL of water to an 18 x 150 mm test tube. Clamp the tube so it
is immersed in the warm temperature bath. From this point on do not
remove the tube from the bath.
3. Hold the bath temperature steady at ±1 ˚C at some temperature in the 5075 ˚C range. A good procedure is to bring the bath to the desired temperature and then remove the bunsen flame. If the temperature drops outside
the desired range, heat briefly with the flame and stir well.
4. When the bath temperature has been constant for a few minutes, rapidly
add about 8 drops of the cold, green stock solution to the warm water in
the test tube in the water bath. Stir quickly with a stirring rod and observe
the time.
5. Observe the color of the reaction solution as in the previous part of the
experiment and record the time it takes to turn gray.
6. Repeat Steps 4 and 5 at three other temperatures (at the very least) in the
50-75 ˚C range. (It is best to have 4-6 data points for this part of the experiment.)
December 2005
Estimate the times here to
the nearest second.
Chemistry 112 Laboratory: Kinetics
Page 88
Calculation of Activation Energy
1. Use LeastSquares, Mr. Plot, or MS Excel to plot ln(1/time) versus 1/T for
the 4-6 determinations you did. Be sure to notice that “time” is in seconds and
T is in kelvins.
Slope = –E a/R
Best straight
line through the
data points
1/T (where T is the kelvin temperature
2. The slope of the plot gives the value of the activation energy, Ea, because
slope = -Ea/R.
December 2005
See Example 15.10 in
Chemistry & Chemical
Reactivity for help on this
part of the experiment.