• Chemical Kinetics


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    • Abstract: The activation energy for a reaction, E. a , is the minimum energy the reac- tants must possess before reaction can occur, and we ... 2. The slope of the plot gives the value of the activation energy, E ...

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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 The ethylenediamine mol-
H2
H2
ecule and the chloride ions
N N
H2C CH2 are called “ligands.”
Co
H2C CH2
N
N
H2 H2
Cl
The nitrogen atoms of the ethylenediamine molecule all lie in a plane sur-
rounding 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 Information on rate laws
five-coordinate intermediate. and reaction mechanisms
is found in Chapter 15
of Chemistry & Chemical
Slow: trans–[Co(en)2Cl2]+(aq) → [Co(en)2Cl]2+(aq) + Cl-(aq) Reactivity.
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
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
Be sure to notice this
ln (fraction remaining) = - (rate constant)•(time elapsed) equation uses the natural
logarithm. See page A-2 in
the Appendix of Chemistry
C = concentration after time t has elapsed
& Chemical Reactivity.
C0 = original concentration at t = 0
k = first order rate constant
t = time
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 This is the crucial idea in
reactant to red product has a given value. Or, rewriting the equation above in the experiment. In essence
you are measuring the half-
terms of this experiment,
life of the reaction. For a
first order reaction the half-
C [green ion] life is independent of the
=
C0 [green ion] + [red ion] initial concentration. See
pp. 712-722 of Chemistry
the gray color appears at a particular fraction remaining. You will use this & Chemical Reactivity.
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 reac-
tants must possess before reaction can occur, and we can determine this ener-
gy 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
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
Ea 1
ln k = -( )( ) + 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.
C
ln = -kt t = -(1/k)ln(C/C0) for a 1st
C0 order process. Thus, because
the “gray point” in this exper-
Rearranging, we have
iment always occurs at the
same C/C0 ratio, measuring
⎛ 1⎞ C time is a measure of the value
k = ⎜ ⎟ ln 0
⎝ t⎠ C of 1/k at a given temperature.
If we take the log of both sides, we obtain
1 ⎡ C0 ⎤
lnk = ln + ln ⎢ ln C ⎥
t ⎣ ⎦
Now substitute this equation into the equation connecting ln k and Ea.
1 ⎡ C ⎤ E 1
ln + ln ⎢ ln 0 ⎥ = -( a )( ) + ln A
t ⎣ C⎦ R T
Finally, rearrange the equation just above.
R = 8.31 x 10-3 kJ/K•mol
1 E 1 ⎡ C0 ⎤
ln = -( a )( ) + ln A - ln ⎢ ln C ⎥
t R T ⎣ ⎦
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
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 contain-
ing about 100 mL of distilled water. (You can use two, 600-mL beakers as You may do this experi-
ice baths.) The test tubes and graduated cylinder must be dry on the inside ment with another student.
except for some slight condensation of moisture from the air. Please do the calculations
independently.
Fill a 250-mL beaker with water and begin heating it to prepare your con-
stant temperature bath.
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
You may remove the gradu-
of the green stock solution, and record the volume to the nearest 0.1 mL.
ated cylinder from the ice
Add the pink solution dropwise with thorough stirring until the solution bath periodically to observe
assumes a gray color when viewed against a white background. Under the color.
TRIAL 1 on the Report Form, record the volume when the solution last
appears faintly green, when the appearance is gray, and when the appear-
ance 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 immedi-
ately.
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
Chemistry 112 Laboratory: Kinetics Page 87
• 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 The statement in the paragraph
at the left is important. As
temperature declines slightly.)
explained on page 2, you are
3. Observe the immersed test tubes against a white background, and record essentially measuring the half-
the time when the solutions turn gray. life of the reaction. For a first
order reaction the half-life is
Because the rate of the reaction is first-order in the concentration of
independent of the initial con-
trans–[Co(en)2Cl2]+, the gray color should appear at the same time in centration.
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.
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 50-
75 ˚C range. A good procedure is to bring the bath to the desired tempera-
ture 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 Estimate the times here to
the test tube in the water bath. Stir quickly with a stirring rod and observe the nearest second.
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 experi-
ment.)
December 2005
Chemistry 112 Laboratory: Kinetics Page 88
Calculation of Activation Energy See Example 15.10 in
Chemistry & Chemical
1. Use LeastSquares, Mr. Plot, or MS Excel to plot ln(1/time) versus 1/T for
Reactivity for help on this
the 4-6 determinations you did. Be sure to notice that “time” is in seconds and part of the experiment.
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


Use: 0.0254