# Chemistry 134 � Lab 7 The Kinetic Study of the Iodination of by akimbo

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```									Chemistry 134 – Lab #7 The Kinetic Study of the Iodination of Acetone Goals • To determine the order of the reaction with respect to acetone, iodine, and hydrochloric acid. • To determine k at an assigned temperature. • To determine the activation energy of the reaction, using data from classmates. Background The Differential Rate Law. For the general reaction (1), the rate is expressed as the change of concentration with respect to time, t, of any reactant or product. (1) aA + bB dD + fF  1  ∆[A]  1  ∆[F]  1  d [A]  1  d [F] Rate = −  =  or Rate = −  =  (2)  a  ∆t  a  dt  f  ∆t  f  dt Since A is consumed and F is produced by (1), their rates (derivatives) have opposite signs. Thus we introduce a minus sign in front of ∆[A]/∆t to ensure a positive rate. Also, unless a=f, -∆[A]/∆t and ∆[F]/∆t are different. In general, the initial rate of reaction is a function of the concentrations of reactants, as indicated in (3).  1  ∆[A]  1  ∆[F] Rate = −  =  = k [ A] m [ B] n (3)  a  ∆t  f  ∆t k is the rate constant of the reaction, and [A] and [B] are the molar concentrations of reactants and products in (1). The exponents m and n appearing in (3) are usually either positive integers or zero. Because (3) is a differential equation, it is called the differential rate law for (1). The goal of a kinetics study is to determine the differential rate law for the reaction (i.e., the values of the exponents m and n) and the numerical value of the rate constant k at each of several temperatures. Suppose that appropriate kinetics experiments carried out on (1) give m=2 and n=1. The differential rate law for the reaction is: (4) Rate = k [ A]2 [ B] The exponents m and n are called orders. We say that the reaction is second order in A and first order in B. It is important to realize that there is no necessary connection between the reaction orders m and n in (3) and the stoichiometric coefficients a and b in (l). The Method of Initial Rates. A straightforward way to determine experimentally the specific form of the rate law (3) is to measure the initial rate, -(∆[A]/∆t)0 , as a function of each of the initial concentrations [A]0 , [B]0 , etc, and from the data deduce the reaction orders. Suppose that we found that -(∆[A]/∆t)0 quadrupled when we doubled the [A]0 while keeping the other concentrations constant. We would conclude that m=2. Similarly, if -(∆[A]/∆t)0 decreased by a factor of 2 when [B]0 was cut in half while keeping other concentrations the same, we would conclude that n=1.

It may be difficult to deduce the reaction order when the value is not integral. The reaction order may be calculated by forming a ratio of the two trials: n (5) rate1 k[ A]m [ B]1 1 = rate2 k[ A]m [ B] n 2 2 If the concentration of B is the same in the two trials, then (5) simplifies to:
m rate1 [ A]1 rate1  [ A]1  = =  m or rate2 [ A] 2 rate2  [ A] 2  Finally, we take the log of both sides and then solve for m:  rate1  log    rate1   [ A]1   rate2  log   = m log   ; m=  rate2   [ A] 2   [A]1  log   [ A] 2  m

(6)

(7)

Calculation of Activation Energies. The relationship between the reaction constant, k, and the temperature is given by the equation: Ea (8) − E 1 k = Ae RT or ln k = − a + ln A R T where Ea is the activation energy, R is the gas constant, T is the temperature, k is the rate constant, and A is a constant that depends on the steric requirements of the reaction. The second form of (8) is a linear equation. The activation energy may be determined by plotting ln k vs. 1/T, which yields a line of slope –Ea/R. Recall that the slope of a line is the change in the y coordinate divided by the change in the x. Substituting in our m, x, and y variables yields: E ln k 1 − ln k 2 − a = 1 1 (9) R T1 − T2

The Experiment. Today we will examine the reaction between acetone (CH3 COCH3 ) and iodine, in the presence of an acid catalyst (HCl). The overall reaction may be written as:

(10)

This overall reaction does not tell us about how the reaction actually proceeds, only what the final products will be. To study this reaction, we will need a way to measure the rate of change of the concentration of a reactant or product. Because I2 (aq) is yellow, it provides a convenient way to monitor the reaction. The rate of this reaction will be ∆[ I 2 ] Rate = − (11) ∆t You will have a chance to verify in this experiment that the rate with respect to the iodine is zero. During this experiment, you will use relatively high concentrations of acetone and

hydrochloric acid, and low concentrations of iodine. The result is that the reaction is pseudozero order, because the amounts of the acetone and acid are relatively constant during each trial. Thus, during the time while the iodine is being consumed, the rate will be constant. You will measure the time that it takes for all of the iodine to be consumed by the reaction, and use that as ∆t in the rate expression. The change in [I2 ] is given by: ∆[I2 ] = 0 - [I2 ]o Materials 1000 mL beaker 3 100 (or 250) mL beakers Two similar test tubes 250 mL Erlenmeyer flask Ring stand and clamps Thermometer Hot plate/magnetic stirrer Magnetic stir bar Ice (if assigned a temperature lower than 25°C) Stopwatch 3 graduated pipets (25 mL) 4.0 M acetone (aq) 1.0 M hydrochloric acid (aq) 0.005 M I2 in 0.05M KI (aq) Procedure Setup. Fill a 1000 mL beaker with water, and adjust the temperature of the water to your assigned temperature by heating on the hot plate or adding ice, as appropriate. Use the magnetic stir bar to keep the temperature uniform throughout the beaker. During this experiment, you will need to monitor the temperature, and make adjustments with heat or ice as necessary to maintain the assigned temperature. Obtain two identical test tubes. Clean and rinse thoroughly, and then fill both tubes with distilled water. Hold the test tubes up to a piece of white paper and compare the color. Both should appear the same. (If they do not, try a different pair of test tubes.) Keep one tube filled with water; this is your reference test tube. Empty the other tube, which you will use for the reactions. Pour approximately 60 mL of the following solutions into clean, labeled beakers: 4.0 M acetone, 1.0 M HCl, and 0.005M I2 . Cover each beaker with a watch glass to reduce evaporation.
5 6 4 7 5 6 4 7 3 2 8 3 2 8 9 9 1 1 1 1 1 0

First set of concentrations. For your first trial, you can use 10.0 mL of each of the three reactants, and 20.0 mL of water. (Students working below 20°C: Use 20.0 mL HCl, 10.0 mL acetone, 10 mL I2 , and 10.0 mL water instead.) Pipet the water, the HCl, and the acetone in a 250 mL Erlenmeyer flask. Place the iodine in one of your test tubes. Put the Erlenmeyer flask and the test tube with the iodine into the water bath, as shown in the diagram. (The test tube can float in the bath, as long as it is big enough not to tip over.) Wait 5 minutes for the glassware

the rounded orders. If the orders you calculate are very far from whole numbers, consult the instructor. Calculate the average k value for your four reactions. Report your average k value and assigned temperature to the rest of the class, and record the class k values and temperatures in a data table. Finally, you will determine the activation energy (Ea) of this reaction. To do so, calculate ln(k) and 1/T for each data point from the class. Record these values in your data table. Make a graph of ln k vs. 1/T. (ln k belongs on the y axis.) Draw the best straight line through the data. Use the graph to calculate Ea, as described in the introduction. Pre-lab preparation. A detailed description of the basic procedure should be written in your lab notebook. Make a data table that lists the volume of each reactant (and water) that you will use for your trials. Leave room to record up to three times for each set of conditions. Make a second data table which shows the molarity of each reactant for each trial, and the rate for each trial. Finally, make a data table for the lnk vs. 1/T plot. Sample data tables acetone (mL) 10 HCl (mL) 10 I2 (mL) 10 H2O (mL) 20 Times (s) (∆t) average time

I II III IV

(choose the volumes that go here)

[acetone] (M) I II III IV

[HCl] (M)

[I2] (M)

rate ([I2]/average ∆t) (M/s)

T (°C) T (K) 1/T (1/K) k (Mx/s)* ln k 10 … … … … 40 *The units for k will depend on the order of the reaction. Put the correct units in your data table after you determine the order of the reaction.

Post-lab 1. Please summarize your results. Report your rate law with the experimentally-determined orders, your value for k, and your value for the activation energy. 2. No one in the class was assigned a temperature below 10°C or above 40°C. Using your data on the relationship between k and T, predict the value of k and ∆t for the reaction at 0°C and 50°C, under the conditions listed below. (Show work.) T (°C) 0 50 acetone
mL M

HCl
mL M

I2 (mL)
mL M

H2O
mL M

expected k

expected time (∆t) (s)

10 10

10 10

10 10

20 20

3. If you wanted to study k below 10°C or above 40°C, what changes would you need to make to the experiment protocol?

Catherine Sarisky, 3/01. Figures created with ACD/Chemsketch.

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