Iodination Of Acetone Lab Conclusion Essay


Kinetics in chemistry deals with the rate at which a chemical reaction occurs. This rate, which is referred to as the reaction rate, is defined as the change in concentration of a reactant or product with time, and is measured in M/s. The rate of a reaction is proportional to the concentration of reactants. An equation called the rate law expresses the relationship of the reaction rate to the rate constant, k, and the concentrations of the reactants raised to some powers, x and y, found experimentally. The rate law is expressed as, rate = k [A]x[B]y. The constant k is equal to the rate divided by the concentration of a certain substance. The purpose in this lab was to experimentally determine the rate constant k, as well as the exponential values of x and y in the rate law.


The procedure of this lab was obtained from the student laboratory course website.


Table 1: Reagent Volumes

TrialI2AcetoneH+dH2OTotal Volume
10.5 mL0.8 mL0.8 mL1.9 mL4.0 mL
20.5 mL1.6 mL0.8 mL1.1 mL4.0 mL
30.5 mL0.8 mL1.6 mL1.1 mL4.0 mL


Table 2: Initial Concentrations, Times, and Rate for Each Trial

Trial[I2][Acetone][H+]Time (sec)Rate [I2]k
16.25e-2 M0.68 M0.2 M184 sec3.40e-6 M/sec2.50e-5 M-1s-1
26.25e-2 M1.36 M0.2 M125 sec5.00e-6 M/sec1.84e-5 M-1s-1
36.25e-2 M0.68 M0.4 M88 sec7.10e-6 M/sec2.62e-5 M-1s-1


Sample Calculations:


            To conduct this experiment, the groups placed 1.9 mL of distilled water, 0.8 mL H+ (HCl), and 0.8 mL of an acetone solution into a 4.0 mL cuvette. 0.5 mL of I2 was then added to the cuvette, and the group immediately started timing the reaction. They mixed the contents of the solution by inverting the cuvette several times before placing it into a calibrated spectrometer. The absorbance rate was monitored at 400 nm until it reached a nominal zero value. The time was then stopped, and the groups were able to determine the rate. Two more trials were conducted, first doubling the original volume of the acetone and keeping the H+ at a constant, then in trial 3, doubling the volume of the H+ and using the original volume of acetone – 0.8 mL. Constantly keeping the volume of I2 at 0.5 mL, and doubling one solution while keeping the other constant made it possible to later calculate the value of the rate constant, k. The chemical reaction being studied was chemical kinetics—the rate at which I2 disappeared. To determine the rate of disappearance of I2 in the reaction, the equation M1V1=M2V2was used to find the concentration of I2. Then, that value was divided by the time elapsed to result in the rate. When I2 was first added to the cuvette, it was dark red in color. As the reaction progressed, the solution lost its color and became clear, consuming the I2 completely. At this point, the spectrophotometer displayed that at 400 nm, zero light was being absorbed in the solution. The starting concentrations were varied according to the experiment design in order to calculate the rate law exponents. The rate law does not include [I2] because I2 does not impact the rate of chemical reaction under the conditions selected. It isn’t included because the rate of disappearance of I2 was what was being solved for. The rate law determined from this lab is rate= k[Acetone][H+]. By using the values of [Acetone], [H+], and the rate, and taking the average of the three trials, the value of k was found to be 2.32e-5 M-1s-1.

The goal of determining the values of k, the exponents x and y, and the rate of disappearance of I2 were successfully met. The expected results matched up with the obtained results—the concentration of acetone and H+ are directly related to the rate of reaction. Both [Acetone] and [H+] are first order reactions, resulting in an overall second order reaction. The main source of error in lab came from not correctly measuring out the substances, resulting in a very askew time and rate of reaction. Having incorrect amounts of each solution in the cuvette directly affected the rate at which I2 disappeared, which in turn made the results not as clear or concise.


            The goal of this lab was to experimentally determine how the concentrations of Acetone and H+ affect the rate at which I2 disappears in a reaction by calculating the values of k, the exponents x and y, and putting those values into the rate law equation. Those numbers were found by altering the amounts of either acetone or H+ used in each trial, which made it possible to use an equation provided above to solve for the value of k, and the exponents, x and y, in the rate law equation. It was found that both Acetone and H+ have a direct effect on the reaction rate of I2. The rate law for acetone iodination is rate= k[Acetone][H+]. The average value of k calculated from the three trials was found to be about 2.32e-5 M-1s-1.

Kinetics of the Reaction Between Acetone and Iodine Essay

652 WordsMar 8th, 20133 Pages

Experiment A1:
Kinetics of the Reaction between Acetone and Iodine

The key aim of this experiment was to determine the rate equation for the acid-catalysed iodination of acetone and to hence consider the insinuations of the mechanism of the rate equation obtained.

The stoichiometric equation for the reaction between iodine and acetone is below, followed by the rate equation (where x,y,z and k are the values to be obtained):
-d[I2]/dt = k [I2]x [CH3COCH3]y [H+]z

The procedure was performed as follows: For run 1, 20cm3 of acetone, 10cm3 of sulphuric acid and 145cm3 of water was added to a conical flask. 25cm3 of iodine was then added to this solution which started the reaction and immediately, 20cm3…show more content…

For run 1: (5-5.5)/(7.9-6) = -0.263
For run 2: (7-9)/(14-9) = -0.4
For run 3: (5.5-10)/(17-9) = -0.563

The equation of the line in run 1 between the two points is y – y1 = m(x – x1) y-4.4=-0.263(x-10) therefore y = 0.263x + 1.77

The ratio between the gradients for run 1 and run 2 is 1:2 as 0.4/0.263 is approximately half, whereas the gradient for runs 2 and 3 is 3:2 as 0.563/0.4 is approximately 1.5.

Initial concentration of iodine in run 1:
Using M1V1=M2V2
Initial undiluted conc of iodine = 0.05M and volume of undiluted

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