The rate of this reaction is Foand to depend o on the concentration of hydrogen

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The rate of this reaction is Foand to depend o on the concentration of hydrogen ion in the soluthon as well as concentratioes of the to actants. By quation I, the rate law for this reaction is to acetone, iosdine, and hydrogen son, respectively where m. N, and p are the orders of the reaction with respect to acetone and & is the rate comstant for the reaction The rate of this reaction can be expressed as the (small) change in the concentration of All.tht occurs, divide/ by the time interval A, required for the chan ative). Ordinarily, since rate varies with the con- The minus sign is to make the rate positive (as AlL,1 is nega centrations of the reac or indirectly, the concentration of each reactant as a with time, decreasing to very low values as the concentration of at makes reaction rate studies relatively difficult to carry difficult for beginning students to understand tants according to Equation 3, in a rate study it would be necessary to measure, directly function of time; the rate would typically vary markedly least one reactant becomes very low. This out and introduces mathematical complexities that are The iodination of acetone is a rather atypical First of all, iodine has color, so that one can readily follow changes in iodine concentration and very important characteristic of this reaction is that it turns out to be zero order in I, concentrat means (see Equation 3) that the rate of the reaction does not depend on [1l at all: h value of [I] is, as long as it is not itself zero. reaction, in that it can be casily investigated experimentally visually. A second t to be zero order in I, concentration. This t depend on I-] at all: [1P- 1, no matter what the the rate by simply making I, the Because the rate of the reaction does not depend on 111, we can study limiting reagent present in a large excess of acetone and H known initial c concentrations than that of I, their concentrations will not change appreciably during the course of the reac tion, and the rate will remain, by Equation 3, effectively constant until all the iodine is gone, at which time ng reagent present in a large excess of acetone and H ion. We then measure the time required for a If both acetone and H' are present at much higher up completely. I reaction will stop. Under such circumstances, if it takes t seconds for the color of a solution having an initial concentration of I, equal to IIlo to disappear, the rate of the reaction, by Equation 4, would be Although the rate of the reaction is constant during its course under the conditions we have set up, we can vary it by changing the initial concentrations of acetone and H' ion. For example, should we double the initial concentration of acetone over that in Mixture I, keeping [H'1 and [Iyl at the same values they had previously. then the rate of Mixture 2 would, according to Equation 3, differ from that in Mixture 1 (6a) (6b) Dividing the first equation by the second, we see that the k's cancel, as do the terms in the iodine and hydro- rate 2 = k(2[A],-112101 rr gen ion concentrations, since they have the same values in both reactions, and we obtain simply rate 2 (2[A (A rate 1 [A IA) rateT" (TIA)" (TAT) Having measured both rate 2 and rate 1 by Equation 5, we can find their ratio, which must be equal to 2". We can then solve for m either by inspection or using logarithms and so find the order of the reaction with respect to acetone By a similar procedure we can measure the order of the reaction with respect to H ion concentration and also confirm the fact that the reaction is zero order with respect to l2. Having found the order with respect to each reactant, we can then evaluate k, the rate constant for the reaction. The determination of the orders m and p, the confirmation of the fact that n, the order with respect to l, equals zero, and the evaluation of the rate constant k for the reaction at room temperature comprise your

Explanation / Answer

Order of reaction

Let the rate equation be,

rate = k[acetone]^x.[H+]^y.[I2]^z

with, k being the rate constant

and, x, y ad z be order with respect to acetone, H+ and I2

From Mixture results I and II, [H+] and [I2] is same,

rate I/rate II = 3.65 x 10^-6/6.35 x 10^-6 = (0.8/1.6)^x

taking log of both sides and solving for x,

x = 1

From Mixture results II and III, [I2] is same,

rate II/rate III = 6.35 x 10^-6/8.1 x 10^-6 = (1.6/0.8).(0.2/0.4)^y

taking log of both sides and solving for y,

y = 0

From Mixture results III and IV, [acetone] is same,

rate III/rate IV = 8.1 x 10^-6/3.09 x 10^-6 = (0.4/0.2)^0.2.(0.001/0.002)^z

taking log of both sides and solving for z,

z = inverse 1

So the rate equation becomes,

rate = k[acetone]/[I2]

rate constant

                         I                     II                       III                   IV                       Mean              Standard deviation

k            4.56 x 10^-9     3.97 x 10^-9       1.01 x 10^-8     7.72 x 10^-9       6.59 x 10^-9               2.48 x 10^-9

For activation energy one can use the Arrhenius equation,

ln(k2/k1) = Ea/R[1/T1 - 1/T2]

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