Kinetics-- a Clock Reaction
Chem. 128
October 4, 2002
web page: October 2, 2002
The reaction we will study involves the oxidation of Iodide Ion,
I-, by peroxydisulfate ion,
S2O82-
2 I- + S2O82- ---> I2 + 2 SO42-
The immediate goal is to determine the rate of the reaction. You will mix the two reagents and time the appearance of a color change.
The next goal is to determine the order of the reaction with respect to Iodide and with respect to peroxydisulfate. You do this by repeating the reaction with different starting concentrations of the two reactants. This allows you to determine how the rate changes with concentration.
The last goal is to see how the rate of the reaction changes with temperature. This is done by repeating an earlier run at several different temperatures. This allows you to compute the Energy of Activation for the reaction.
As mentioned earlier this is a
clock reaction. We deliberately add a small, fixed amount of sodium thiosulfate (S2O32- ) to all of our reaction mixtures. This will rapidly consume any I2 produced and will continue to do so until the thiosulfate is completely consumed. At that point the starch we added will turn dark blue.
- I2 + 2 S2O32- -----> 2 I- + S4O62-
Notice that this reaction has no effect on the reaction we set out to study. It merely provides us a convenient tool for determining when that reaction has reached a specific point.
-
You should also note that the amount of thiosulfate is quite small compared to the other reagents. For example one reaction mixture will be 0.0667 M KI, 0.
033 M peroxydisulfate and 0.
00167 M thiosulfate. When the thiosulfate is used up the peroxidisulfate concentration will still be 0.032M. This justifies saying that we observe the initial rate of this reaction.
- don't confuse the various sulfur compounds
- S2O82- peroxydisulfate ion, S2O32- thiosulfate ion
-
We will also be adding Potassium Chloride or Ammonium Sulfate to some of our reaction mixture. These materials are completely inert in this reaction. Their role is to keep the total concentration of ions the same from run to run. There is a small effect from ionic strength and we are trying to eliminate it as being responsible for any of our observations.
Experimental (Work in pairs on this experiment)
The following reagents are available. These will be provided in bottles with automatic dispensers that deliver a fixed volume of reagent.
- Potassium Iodide, 0.200 M (10 ml Repipet dispenser)
- Ammonium Peroxydisulfate, 0.100M (10 ml Repipet dispenser)
- Sodium Thiosulfate, 0.005 M (10 ml Repipet dispenser)
- Ammonium Sulfate 0.100M (10 ml Repipet dispenser)
- Potassium Chloride 0.100M (10 ml Repipet dispenser)
- Hydrogen Chloride 0.100 M (10 ml Repipet dispenser)
- Copper Sulfate ).1 M
- Starch, 1% in water
Safety and Disposal
These chemicals offer no significant hazards at the concentrations involved.
- Of course, safety goggles are to be worn at all times.
All materials and reaction products can be discarded by rinsing down the sink.
Run #1 --
- Place 20 ml of Ammonium Peroxydisulfate into a clean, dry flask or beaker
- (this is two Repipet portions)
- measure and record the temperature of this solution; the others will be at the same temperature
- Place 20 ml of Potassium Iodide in another beaker
- (this is also two Repipet portions)
- Place 10 ml of Sodium Thiosulfate and 3 drops of starch in a third beaker
-
- Quickly mix all three solutions together and note the exact time of mixing.
- Wait until the solution turns blue and note the time that elapsed
-
- Repeat, varying the concentrations
| run #
| beaker 1
| beaker 2
| beaker 3
|
| 1
(as above)
| 20 ml Peroxydisulfate
| 20 ml Iodide (KI)
| 10 ml thiosulfate
3 drops starch
|
| 2
| 10 ml Peroxydisulfate
10 ml (NH4)2SO4
| 20 ml Iodide (KI)
| 10 ml thiosulfate
3 drops starch
|
| 3
| 20 ml peroxydisulfate
| 10 ml Iodide (KI)
10 ml KCl
| 10 ml thiosulfate
3 drops starch
|
| 4
| 20 ml peroxydisulfate
| 10 ml Iodide (KI)
10 ml HCl
| 10 ml thiosulfate
3 drops starch
|
| 5
| 20 ml peroxydisulfate
| 20 ml Iodide (KI)
1 drop 0.1M CuSO4
| 10 ml thiosulfate
3 drops starch |
- Notice that Runs 1 and 2 differ only in the concentration of Peroxydisulfate.
- Runs 1 and 3 differ only in the concentration of Iodide Ion.
- Runs 3 and 4 have identical concentrations of both reactants; the added HCl will let us see if H+ acts as a catalyst in this reaction. (H+ is a common catalyst.)
- Runs 1 and 5 differ only in the addition of Cu2+, a known catalyst
Repeat Run #1 two more times.
In the first case, place all three beakers into a bath of water 10-15 degrees above room temperature. Wait about 5 minutes for the solutions to reach temperature. Measure and record the temperature in any of the solutions. Mix and time the reaction. (Keep the mixed solution in your water bath to keep the temperature constant.)
In the second case, place all three beakers into a bath cooled about 10 degrees below room temperature. Proceed as above.
(It would be simpler to cool the solutions with ice, but at that temperature the reaction is too slow to be convenient.)
Calculations:
Compute the starting concentrations of Iodide, Thiosulfate and Peroxydisulfate for each reaction mixture.
- Remember the reaction volume is 60 ml, so the starting materials are diluted.
- for example: we added 20 ml of 0.200M KI
- after mixing, the concentration is 0.0167 M KI
Compute the rate of reaction, based on the amount of Iodide Ion consumed
- thiosulfate concentration: 0.00167 M
- this reacted with 0.000833 M of I2
- that was produced by the reaction of 0.000833 M of Peroxydisulfate
If the reaction occurred in 47.0 seconds,
- the
rate is 0.000833M / 47.0 (moles/liter/second)
Finding the order of the reaction:
- Algebraically, our formula is
rate = k * (I-)x (S2O82-)y
x and y are the order of the reaction with respect to the two reactants
- our goal is to determine x and y, hopefully as a simple integer (0, 1, 2)
This is easiest if you compute a ratio for two of the reactions
(the next expression might get garbled as it moves to the Web site)
rate 1 =
k * (I, run1)x (S2O82-, run1 )y = (2/1)x
rate 2 k * (I,
run2 )x (S2O82-,
run2 )y
ln (rate1/rate2) / ln(2) = x
Once you determine x and y, you can compute k for each run; it is the only unknown in the equation
-
rate = k * (I-)x (S2O82-)y
The Effect of Temperature
- Prepare a graph
- on Y axis plot ln(rate)
or ln (rate constant) or ln (1/time)
-
on the X axis plot 1/ (Temp +273.2) that is, 1/TKelvin
- The Arrhenius equation predicts a straight line
- the slope = - EA/R
- where R= 8.314 Joules/K mol
- Measure the slope and evaluate EA