Question

3. (30 pts. total) We discussed the iodination of acetone in class. An alternative, but equivalent, way of writing the reaction is: acid H3C-C-CH3 + 131 acetone triiodide H3C-C-CH2 + 21 + H* iodoacetone iodide Is provides a more useful way to introduce 12 into the system. Iz is also useful because it absorbs visible light, and the rate of the reaction can be monitored by measuring the decrease in its absorbance at 565 nm. (Any time you can measure a process with visible light, you are extremely fortunate because this is the simplest way to do a measurement.) The initial rates obtained by measuring the absorbance of triiodide decay in the laboratory experiment that we have performed here at ISU are given in the table below, which also gives the experimental conditions are listed below that have been designed in order to keep the total volume of the reaction mixture constant throughout the series of experiments. expt. vol. 0.005-M 13, 0.3-M KI (mL) 2.0 vol. 0.0305-M HCI (mL) initial rate (M/s) x 10-6 2.0 2.0 vol. 4.0-M vol. 0.305-M acetone (mL) KI (mL) 2.0 0.0 1.0 1.0 2.0 1.0 1 2.0 1.0 2.0 (deuterated 10.0 acetone) 3 1 .0 2.0 2.0 2.0 1.0 2.0 4.6 +0.3 2.1 +0.5 4.4 +0.3 2.3 +0.5 0.85 +0.05 deuterated acetone 2.0 a. (10 pts.) Find, to one significant figure, which is an integer, the values of a, b, and c in the rate law below: rate = k [H*]* (acetone] [131]. b. (10 pts.) From the value of c, express the rate law in terms of the triiodide concentration and integrate this rate law to obtain an expression for [13](t). c. (5 pts.) What does a plot of [13](t) vs. t look like? How is this useful? More specifically, if a careless student did this experiment and missed the first 100 seconds of the decay of triiodide, would it have made a difference? (Think about your answer to problem 1 in responding.) d. (5 pts.) How much faster of the rate of reaction using protiated (normal) acetone as opposed to deuterated acetone? What does this result suggest about the mechanism of reaction that we discussed in class?

Answer 1

Answer Page 1 As per the given details sate: { [H+39 [ace tone] º [13]° Compair experiment, and a Por experiment i and rate: k, for experiment a rate = kg no [3]" [3]" [3] appool 2 [within comod limit] 2 = [2] => b=) Similarly com peiring experiment and caperiment we can get ha - [7] 13 [a] Similarly Compeining eap and exp @ 4,6 = [?] => 2 [2]" =) 0:1 So a:1, b=1 c=0 93

Page 2 respect to Is it is a zero ordeo b) with 1. So - *[78]" / negative Sogn K2 - 3 L is for on [It is Consumed] => [13] =+kt so the plot of 9 [13] = kt zero order reaction [ 13 ] vs It will be constant [111] slope :-+ slope --K ၇ 1. -> t (time) th time concentration of [[ ] will change Diffn between 1 to 3 and initial time is, only the concertoation of I d) In Protiated acetone it is much faster than dec terated Acetone 406 = 5.47 times faster 0.85 So, Acetone has a major rate in Rxn Mechanism PLEASE HIT LIKE VOTE