Iodination of Acetone
Written by Isabel
Introduction:
The purpose of this experiment is to determine the rate law for the reaction of iodine with acetone. In order to determine the rate law, we will use initial rates. Since this is an iodination of acetone experiment, the initial rate would be the time it takes for the brownish color of iodine to turn clear. In the end we should expect to see a clear solution and a concentration of zero for iodine. Things that could affect the rate of iodination would be the concentrations of reactants. The formula used to calculate the rate of this reaction was provided by the lab manual and it is equation 1 below:
Rate of Reaction=K [acetone]a[iodine]i[HCl]h
Where ‘a’ is the order with respect to acetone, ‘I’ is the order with respect to iodine and ‘h’ the order with respect to HCl.
After the initial rates are determined, we can then find the value of the rate constant, K.
We now ran an experiment to determine the initial rates of six trials, each time varying the concentration of one reactant at a time because it is known that the rates could depend on the concentrations of the reactants.
Experimental:
We ran our experiment three times but there appeared to be something wrong with the spectrometers at the time so the data used in this lab report was given. Here
is what the experiment was supposed to be like:
A- Setting up the spectrometer.
1- A spectrometer was used to observe the iodine. The spectrometer was set at a wavelength of 410 nm and it was then calibrated using a cuvet filled with water.
2- For water, we set it at 100% transmission and 0% absorbance due to it being colorless, and for the dark colored solution of iodine, it was 100% absorbance, 0% transmission.
B- Preparing the solutions.
1- Six reactions were carried through, each one using a different amount of reactants. The amount of each reactant to be used was given to us in a table for each of the six trials.
2- The table that we were given for the amount of each reactant is below. This is table 1.
| Reaction Number | 4.0 M Acetone(mL) | 1.0M HCl (mL) | Water(mL) | 0.0050M Iodine (mL) |
| 1 | 3 | 3 |
8 |
4 |
| 2 | 6 | 3 |
5 |
4 |
| 3 | 9 | 3 |
2 |
4 |
| 4 | 3 | 6 |
5 |
4 |
| 5 | 3 | 9 |
2 |
4 |
| 6 | 3 | 3 |
4 |
8 |
Table 1. Amount of reactants to be used in each trial
3- After the reactants are mixed, they need to be placed in the cuvet and into the spectrometer as fast as possible. A graduated pipette should be used for the iodine solution and a different one for the rest of the reactants. After pipetting acetone, the pipette should be rinsed with HCl before pipetting the HCl, and the same goes for water.
C- Procedure.
1- For reaction Number One: Pipet into a beaker 3.00 mL acetone, 3.00 mL of HCl and 8.00 mL of water and into another beaker, pipet 4.00 mL of iodine.
2- Pour the iodine solution into the beaker containing the acetone, HCl and water. Mix quickly.
3- Fill the cuvet with the solution and place it in the spectrometer. Observe the %T which increases as iodine reacts. When this goes constant, The reaction is over. Record the time at which this goes constant.
4- Repeat for Reactions 2-6.
5- A seventh reaction is done to confirm the order with respect to iodine
Results and discussion:
Table 2 below describes the time It took for %T to go constant for each one of the reactions. This was all done at room temperature (72oF which is 22.2oC).
| Reaction Number | 4.0 M Acetone(mL) | 1.0M HCl (mL) | Water(mL) | 0.0050M Iodine (mL) | Time (s) |
| 1 |
3 |
3 |
8 |
4 |
266 |
| 2 |
6 |
3 |
5 |
4 |
138 |
| 3 |
9 |
3 |
2 |
4 |
100 |
| 4 |
3 |
6 |
5 |
4 |
133 |
| 5 |
3 |
9 |
2 |
4 |
85 |
| 6 |
3 |
3 |
4 |
8 |
425 |
Table 2. The time taken by each of the reactions to achieve a constant %T
Table 3 below shows the corrected concentrations of each of the reactants, the rate in [I2]/s, and the rate constant K for each one of the reactions.
I will present the data first and then explain and write the calculations done in order to obtain it.
| Reaction Number | Acetone (M) | HCl (M) | Water (mL) | Iodine (M) | Rate ([I2]/s x 105) | Rate Constant (k x 105) | |
| 1 | 6.66×10-1 | 1.66×10-1 | 8.00 | 1.11×10-3 | 4.17×10-1 | 3.76×105 | |
| 2 | 1.33 | 1.66×10-1 | 5.00 | 1.11×10-3 | 8.04×10-1 | 3.64×105 | |
| 3 | 2.0 | 1.66×10-1 | 2.00 | 1.11×10-3 | 1.11 | 3.34×105 | |
| 4 | 6.66×10-1 | 3.33×10-1 | 5.00 | 1.11×10-3 | 8.34×10-1 | 3.76×105 | |
| 5 | 6.66×10-1 | 5.0×10-1 | 2.00 | 1.11×10-3 | 1.30 | 3.9×105 | |
| 6 | 6.66×10-1 | 1.66×10-1 | 4.00 | 2.22×10-3 | 5.22×10-1 | 4.72×105 | |
| 7 | 6.66×10-1 | 1.66×10-1 | 0 | 3.33×10-3 | 4.44×10-1 | 4.01×105 | |
Table 3: Corrected concentrations of reactants, the rate and rate constant.
Calculations are as follows:
A- Calculating the corrected concentrations.
Using the formula below allows us to calculate the corrected concentrations of reactants that should be used.
Equation 2:
M1V1=M2V2
Where M1 is the Molarity of the given substance, V1 is the volume used of that substance, M2 is the total Molarity used and V2 is the total volume used.
They are all calculated in the same way, but I will demonstrate what we did with acetone:
Taking reaction number 1, we used 3mL of a 4.0 M acetone solution. Those are V1 and M1 respectively. M2 is what we need to find and V2 is the total volume of reactants used. Here we used 4.00 mL of iodine, 8.00 mL of water,3.00 mL of HCl and the 3.00 mL of acetone. Adding them all together gives us 18 mL which is V2. Now plugging them into equation 2 gives us:
M1V1= M2V2
(4.0)(3.00)=M2(18)
M2= (4.0 x 3.00)/18
M2= 6.66 X10-1
This is the corrected concentration of acetone in reaction number 1.
Doing the same exact thing for all of them gives us the corrected concentrations for each one.
B- Calculating The Rate.
The rate was calculated using the formula given right below it in table 3.I will also state this here as equation 3.
Rate= [I2]/s x105
Again, I will demonstrate what we did using reaction 1. The concentration of iodine in reaction one came out to be 1.11×10-3 using equation 2. The time it took %T to be constant in reaction one was 266 seconds. Plugging those numbers into equation 3 gives:
Rate=[I2]/s x105
Rate=( (1.11×10-3)/266) x 105
Rate= 4.17×10-1
Doing this for every reaction gave us the corresponding rates.
C- Calculating the rate constant, K.
In order to calculate the rate constant, we need to go back to equation 1. Equation one was the following:
Rate of Reaction=K [acetone]a[iodine]i[HCl]h
We calculated the concentrations of acetone, iodine and HCl in part A, and we calculated the rate of reaction for each one of the reactions in part B. All the values are in table 3. In order to find K, all we are missing here are the orders a, i and h.
The orders can be found using the following equation:
Equation 4.
(Rate 2)/(Rate 1)= (M2/M1)x
Where Rate 2 is the larger of the two rates chosen, Rate 1 is the smaller, M2 is the Corrected concentration corresponding to Rate 2 and M1 is the corrected concentration corresponding to Rate 1. The superscript x is the order with respect to the specific reactant used.
Now the next step is to figure out which equations to use.
To find the order of HCl, I took solutions 1 and 4 because HCl is different while the others are the same.
For the order of acetone, I took solutions 1 and 2 because the concentrations of HCl and I2 are the same.
And for the order of iodine, I took solutions 1 and 6 because the concentrations of HCl and Acetone are the same.
Now starting with HCl, Using equation 4 and plugging in the corresponding values gives:
([8.34×10-1] / [4.17×10-1] ) = ([3.3×10-1] /[1.66×10-1])h
Log(8.34/4.17) = h Log (3.3/1.66)
h= 1.01
The order h with respect to HCl is 1.01.
Second, Acetone: Using equation 4 and plugging in the corresponding values gives:
Log([ 8.04×10-1]/[ 4.17×10-1])= a Log ([1.33]/[6.66×10-1])
=0.285= a(0.947)
a= 0.9
The order a with respect to acetone is 0.9
Third, Iodine: Using equations 4 and plugging in the corresponding values gives:
Log([5.22×10-1]/ [4.17×10-1]) = i Log ([2.22×10-3]/ [1.11×10-3])
0.98= i(0.301)
i= 0.3
The order a with respect to acetone is 0.3
In this lab, we decided to use integers and so we rounded these orders up or down to make a whole number as follows:
a= 0.9 was rounded up to 1
h= 1.01 was rounded down to 1
i= 0.3 was rounded down to 0
Now that we have everything we need to find the rate constant, K, we can now plug everything in equation one which again is the following:
Rate of Reaction=K [acetone]a[iodine]i[HCl]h
The rate constants for all the reactions are in table 3. I will write out the math for one of them because they are all the same thing. Here is how I got the rate constant for reaction one.
Rate of Reaction=K [acetone]a[iodine]i[HCl]h
4.17×10-1= K [ 6.66×10-1]1 [1.66×10-1]1 [1.11×10-3]0
4.17×10-1= K(0.111)
K= (4.17×10-1)/ 0.111
K= 3.756
And since in the table it is given that Rate constant is Kx105 then each K value obtained is multiplied by 105
So the Rate Constant for reaction 1 is: 3.76×105
For the seventh reaction, the order is going to be the same because all the other values are the same so we are going to have to take the same solutions 1 and 6 because those are the same and iodine is different. This confirms the order with respect to iodine and fulfills the purpose of the 7th trial.
Conclusion:
Now that we found the rates and the rate constant, we can tell which reaction went fastest and which concentrations were the best to use according to the reaction rates. Results show that reaction 4 had the highest rate and so the optimal concentrations that should be used for them to react at that rate would be 6.66×10-1 M of acetone, 3.33×10-1 M of HCl, 5.00 mL of water and 1.11×10-3M of iodine. This all reacts at a rate of 8.34×10-1. The rate constant is highest in equation six which means that the reactants are reacting at a constant rate of 4.72×105 however the rate of that reaction is lower than the rate of reaction 4. Temperature here is not a variable because all of these were conducted at room temperature therefore this shows what was stated earlier in the introduction, that the rates vary according to the concentration of the reactants.
