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3/6/201611
SPEED: DETERMINING THE
RATE EQUATION OF AN IODINE
CLOCK REACTION
Objectives: Gather sufficient data to write a rate equation for the reaction:
2I-(aq) + H2O2(aq) + 2H3O+(aq)  I2(aq) + 4H2O(l)
H3C
CH 3
C
C
H2C
C
H
H
C12
C
C
C
H2
C11
C
C
H2C
H
CH 3
H
C
H
C
H3C
CH 3
We have spent a great deal of time in this course discussing chemical
equilibrium. When 2 competing reactions (a forward and a reverse
reaction) are proceeding at the same rate, then the system is in a state
of dynamic equilibrium. Equilibrium constants can be measured,
equilibrium concentrations can be calculated, and predictions can be
made about what the system will do if you joggle its elbow and move
it away from equilibrium. What you DON’T know is, how fast will
the system REACH equilibrium. Some reactions are notoriously
fast…an example close to home: The first step in the isomerization of
11-cis-retinal to all trans retinal when a photon of light meets a retinal
cone cell is over in picoseconds. See:
http://wunmr.wustl.edu/EduDev/LabTutorials/Vision/Visi
11-cis retinal
on.html
On the other hand, some chemical reactions go very
H
slowly. For example, the reaction of sugar with
oxygen
C
O
H
CH 3
H
C6H12O6(s) (sugar) + 6O2(g) 6CO2(g) + 6H2O(l)
CH 3
all-trans-retinal
H
H3C
CH 3
C
H2C
C
C
C
C
H2C
H
C
C
H2
.
CH 3
C11
C
C
C12
H
H
H
C
C
O
H
is thermodynamically favorable (∆G is -) but Egyptian
tombs contain sugar (in the form of honey) which have not
reacted appreciably in thousands of years.
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THE REACTION:
In this experiment we will study the rate of an “Iodine clock” reaction:
I. 2I-(aq) + H2O2(aq) + 2H3O+(aq)  I2(aq) + 4H2O(l)
Many factors affect the rate of a reaction: Degree of division, temperature, presence or absence of a catalyst,
and concentration of reactants. Observing the maxim that “the essence of science is to change one variable at
time”, we will limit our study to two variables; the concentration of I- and the concentration of H2O2.
The rate law or rate equation for our iodine clock reaction is, of course,
Rate = ∆[I2]/∆t = k[I-]m[H2O2]n[H3O+]o
We will use an acetic acid/sodium acetate buffer to hold the [H3O+] very nearly constant. And, since [H3O+] is
constant, we can write a simpler rate equation:
Rate = ∆[I2]/∆t = k’[I-]m[H2O2]n k’ is the rate constant (k) times [H3O+].
Before you come to lab: The
acetic acid/acetate buffer we
use is 0.5 M CH3COOH and
0.5 M CH3COO-Na+. The
pKa[CH3COOH] = 4.75.
Calculate the pH and [H3O+]
of the buffer.
Your mission today: Determine k’, m and n.
SOME TECHNICAL DETAILS:
The reaction we are studying is:
I. 2I-(aq) + H2O2(aq) + 2H3O+(aq)  I2(aq) + 4H2O(l)
There are two experimental problems:
1. As soon as the reaction starts, and I2 is made, the reverse reaction will also start:
IR. I2(aq) + 4H2O(l)  2I-(aq) + H2O2(aq) + 2H3O+(aq)
2. As soon as the reaction starts, [I-] and [H2O2] start to go down.
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How can we study the effect of concentration of I and H2O2 on reaction I if: reaction IR is going on and [I-]
and [H2O2] are constantly decreasing? There is a rather clever way to do that, by adding a known amount of
sodium thiosulfate, and some starch:
We can prevent reaction IR from occurring by removing I2 as fast as it is made. We can do that by adding
thiosulfate ion (S2O32-):
II. 2S2O32-(aq) + I2(aq)  2I-(aq) + S2O62-(aq)
This has the additional benefit of regenerating I-, so [I-] remains constant until all the S2O32- is used up. At that
point [I2] goes up. I2 reacts with starch, and the solution turns blue:
III. I2 (aq) + starch (aq)  Starch.I2(deep blue)
The time required for a known, quantitative amount of thiosulfate ion to react is the time it takes for the blue
starch-iodine color to appear. During that same time a quantitative amount of I2 is made (equal to half the
moles of thiosulfate). Thus, the time it takes the blue to appear is a measure of ∆[I2]/∆t, the rate.
THE EXPERIMENT:
You will need:
DI water
1 M acetic acid buffer, pH 4.75
0.3000 M KI
0.02000 M Na2S2O3
0.1000 M H2O2
Starch indicator
1 squirt bottle
15 mL
15 mL
15 mL
25 mL
5 mL
25.00 mL volumetric flasks labeled 1-6, with stoppers or parafilm squares
Volumetric pipette 1.00 mL (labels: buffer, KI, Na2S2O3)
Volumetric pipette 2.00 mL (labels: water, KI)
Volumetric pipette 3.00 mL (labels: water, KI, H2O2)
Volumetric pipette 4.00 mL (labels: water)
Volumetric pipette 5.00 mL (labels: H2O2)
Volumetric pipette 10.00 mL (labels: H2O2)
Blue bulb
Pasteur pipette and bulb
Small beakers (50 mL)
White paper
Watch with second hand or a computer
6
3
2
3
1
1
1
1
1
4
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Before you come to lab: Make an Excel Template of the data chart on page 5.
PUT IT ON A FLOPPY DISK & BRING IT TO LAB.
BRING YOUR KNIGHTCARD TO LAB!!!!
1. Get organized: Line up your volumetric flasks IN ORDER. Line up your pipits so that you won’t
use a pipette labeled “water” to pipette H2O2. Notice that you will need a 5.00 mL pipette labeled
“H2O2” and a 2.00 mL pipette labeled “H2O2” to pipette 7.00 mL of H2O2.
Get appropriate volumes of reagents in small beakers from the stock containers in the hood. Be sure you write
down the EXACT concentration of every reagent in your notebook!
2. Get set:
Kinetic
DI
Buffer
0.3 M
Starch
0.02M
0.1M
Trial
1
2
3
4
5
6
H2O (mL)
4.00
3.00
2.00
2.00
0.00
^
(mL)
1.0
1.0
1.0
1.0
1.0
1.0
KI (mL)
1.0
2.0
3.0
1.0
1.0
^
(drops)
5
5
5
5
5
5
Na2S2O3 (mL)
1.0
1.0
1.0
1.0
1.0
1.0
H2O2 (mL)
3.0
3.0
3.0
5.0
10.0
Using the chart, carefully pipette water, buffer, KI, and Na2S2O3 into each of the 25 mL volumetric flasks.
Be sure you use the correct pipits! DO NOT ADD H2O2 YET! Swirl the solutions to mix them, and place the
flasks on a piece of white paper. Add 5 drops of starch indicator to each flask, and swirl them again to mix
them.
3. Almost ready: Designate either you or your lab partner as timekeeper. The non-timekeeper will be in
charge of H2O2. Have a watch with a seconds function, or check out a laptop and open the clock window.
4. GO! Draw up 3.00 mL of H2O2 in a pipette. As you start to add it to flask 1, start timing. Write the
start time INCLUDING SECONDS down in your notebook. Swirl the flask, and time it until it turns blue.
Write the end time INCLUDING SECONDS in your notebook. Repeat for the other 5 flasks.
1. CLEAN UP! RINSE ALL PIPETS WITH DISTILLED H2O. Be sure to rinse through the tips!
Rinse all glassware with distilled water and return them to your bench area!
ALL WASTE FROM THIS EXPERIMENT GOES IN THE JUG LABELLED “IODINE
WASTE” IN THE HOOD.
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Calculations:
1. Set up a table for your results as shown. Include the total volume of each reaction (10.00 mL)
And the exact concentrations of reagents. (SEE ARROWS). Warning: The concentrations used here are
examples only, and are not the concentrations you used.
2. Plug your TIME data (in seconds) into Column J
Log
[H2O2]
3.
Calculate INITIAL [REACTANTS] in columns B and G.
Example: In Trial 1 (Line 12) you diluted 1.0 mL of 0.50101 M KI to a final volume of 10.00 mL.
[I-]initial = 1.0 mL(0.50101 mmole/mL KI)/10.00 mL = 0.050101 M
Log
[H2O2]
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4.
3/6/201666
Calculate log[I-]initial and log[H2O2]initial in Columns C and H
5. Calculate how many mmoles of S2O3= you started with in Column G.
Example: In Trial 1 (Line 12) you used 1.0 mL of 0.04117 M Na2S2O3 .
mmoles S2O3= = 1.0 mL(0.04117 mmole Na2S2O3/ mL) = 0.0.04117 mmoles.
6. Using equation II, calculate how many mmoles of I2 must have reacted with the S2O3= Put that in
column I.
7. In column K, Calculate the reaction rate as mmoles I2 produced/sec.
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Now you are all set to calculate m, n, and k’ for the rate equation:
Rate = ∆[I2]/∆t = k’[I-] m[H2O2] n
Fisrt look at Trials 1,2 and 3. Notice that [I-]
increases from 0.05 M to 0.1 M to 0.15 M, while
[H2O2] remains constant at 0.06 M.
The rate equation is:
1. Rate = k’[I-] m[H2O2] n Taking logs of both sides:
2. Log rate = logk’ + mlog[I-] + nlog[H2O2];
logk’ is always constant at constant temp, and nlog[H2O2] is constant in trials 1,2 and 3.
For neatness, 3. let logk’ + nlog[H2O2] = B
So equation 2 becomes:
4. Log rate = mlog[I-] + B; AH HA! THE EQUATION FOR THE CELEBRATED, BELOVED AND MUCH
SOUGHT AFTER STRAIGHT LINE, with Y = log rate, slope = m, and [I-] = x. LIFE IS GOOD!
I will leave it to you to convince yourself (And write it out neatly for a bonus) that using the data from trials 1,4
and 5 (where [I-] stays constant and [H2O2] changes) allows you to write:
Log rate = nlog[H2O2] + C
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Having come this far in triumph, separate your trial 1,2, and 3 data into one group and your trial 1, 4, and 5
data into another. All you will need is the data in column B, C, G, H, K and L. Graph Log [I-] on the x axis vs
log rate on the y axis, and get the best fit straight line. The slope is “m” for the rate equation:
Rate = k’[I-]m[H2O2]n .
Do the same for trials 1,4 and 5:
Now that you have m and n, calculate k’ for all 5 trials,
and get the average and standard deviation.