Question

Please help me find the average k, determining the order with respect to H2O2 (B) and H+ (y), and what is the complete rate law by including the experimentally numerical value of k, a, b and y.

The lodine Clock: Initial Rate Law Kinetics - 71 Calculations Below is a description and outline for the calculations you need to complete for this laboratory. Make sure to include all of your calculations. Use extra paper if you need to Initial Concentrations: Since initial concentrations are need when using the instantons rates to calculate the initial rate law, you will need to calculate the [Al for each compound in the rate equation (Na2S 0, 1, H', and H20,) using the dilution equation (m, m;V). Record your results for each run (3, 7). Make sure to use the actual concentrations given on the reagent bottles (1) Determining the initial rates: You will need to convert the time for each run to second (3, 7) so that you can then calculate the rate of reaction (A[1] At) for each run (3, 7). Recall, A[12] may be approximately as the [Na S,O for every run. Show a sample calculation below for Run I. Respect significant figures based on the metric equipment used and show the units. Determining the rates of each reactant: Runs 1 and 2 vary KI, but have constants for the H' and H.O, therefore you can use those two runs to determine a, the order ofl. Recall you want to round the value for x to the nearest whole number. Use rest of the runs to determine the order with respect to H30, () and H' (). Determining the rates constant: You will then be able to calculate k for each of runs by plugging in all of the values into your rate law, and solving for k (4). Complete Rate Law: Now you can write the complete Rate Law, including the experimentally numerical value ofk, a. B. and y (5).

Answer 1

Solution Iodine clock Reaction kinetics. 4, O2 + 2I + 2ut - I2 + 2H20 (slow) 1. +25,034 => 21+ Sy 06? (fast) sale = k [yoffs Tonay? ---0 where k= sale constant d = oder of reaction with respect to I ß = order of reaction with respect to H2O ♡ = order of reaction with respect to H+ also, delle' = - 0590-) = +6[1.) - from the stoichiometry of reaction (12) = } [S203?.]. therefore, wake = + A[5:) = 4[5.01.] — At At At concentration of He = olm concentrention of kI = 0.05 m concentrellicom 94 M42 5,03 = 0.02 m concentration of H2O2 = 0.04 m Total volume of reaction mixture in each Run= = 13 mL CS Scanned with CamScanner

Cono Determination of initial concentration of H2O2, I , ut and so, ²- for Run-1 for (H2O2) Vilyme of 1202 = 1.00 mL. concentration of H2O₂ = 0.04 m final volume = 13.0 mb. from mix = M2 V2 1. M2 = mivi V2 M2 = 0.04 Xl me 0.04M X 1.ome 13mL 13.0mL Ma = B07230 0.003077 M [120,7 =50250 A 0.003077m my for(u+ ) Miv, = Me = M2 V2 mira V2 о: | 1 xuml 13 ht [**] = 0.0307M for (I) mirs M = = M2 z milli V2 0.05 X 1.0mL 13.0mL = 0.00385 m [1] CS Scanned with CamScanner

For (S203-) in Mik= M2 V 2. M2 .= mir . , V2 = 0.02 mx 2 mL . 13 me (5,03'-) = 0.003076 M Similarly do for other suns. . Determination of Rate For Runal Rate = A[s2O3'-) Rate - Z 4 at then blue color appeared, all siog ² were consumed hence final concentration of S2O3' is gere , A[S, O3) = (520g). - [S207) final = 0.003076 - 0 = 0.003076 M Pleage hote that the value of A[sol] is same in all durs since we have taken 2mL s.og in all runso At = 5:48.45 = 348.45 ser = 1 Al303) 2 at I x 2 0:00 3076 m 348.45 sec stre - Date = 4.41 xromo (m.sec) es Scanned with similarly calculate for all guns, CamScanner

- Table J. 1 (1.0.J. Time I sake . I safe constant Is.n. Truth 1 [[ J. 115. .003077 07230 | 348.45 4.41810-6 1 0.003076 1.00385 Run-1 1.0307 003077 1-01230 309.55 4.97X106 Rum-2 0.0307 1.00769 osot 7:00769 10.003076 100.3077 rol230 0.00307 6 694.22 2.22X10 Ruth-3 10.01538/ 100769 006154 90.50 16.99 x16 Run-4 1.01538 0.01538/0.003076 1.01538 1.01538 0.003076 1.012308146.3:8 110:51x160 Runs Determination of order of Reaction: va k [1] [460.3% (n=3" 2. = K (0.00385) " (,00 3077) (0102073 - to = x (0.00789)9 (0.00307)" (0:0907, ". 19 = K (0-00 769)" (.090277" (0DIS 28) - Yu = 6.01533 )(0.006154)" (0.01533)" - us = k l 001538) (01012308) "70-01518 )" Now, = (0.00385) = (0-500) = (0-5)4 => 0.8873 = (0-5) loy 60-8973) = a log (0-5) To = 0.1725 To.007 69). Scanned with CamScanner

0.0 = 4.978606 2.22 x 10-6 1996) - 1 x 2.2397 = (1996) 19912.2397) = log (1996) = 1.1667 Now, Vu 16.99 x 10-6 = (0.578 161655 = (0-5) Po lag (1616.55) = loy (0-5) B = -0.69) I think there is some mistakes in determination of time of blue color appearance. . ß cannot be negative. however taking to the nearest nearest integer vakee, 0=1. B=1 date = k [15] [11902] [**] CS Scanned with,.. CamScanner

Now, dolermination of rale constant for Runnt 71 - K (1202] [7] [47] 40 xVm= R (0.003027m) ( 0-60385M) (0-0307m) Ky = 12.1258 m2mod żset! ki = 12. 1258 mod2th see! Similarly calculate for suh-2, guh-3, run-u, run- 5 and' take the arenage value Scanned with CamScanner