please help solve for the rest, thank you

Post by **answerhappygod** » Wed Mar 09, 2022 9:40 am

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Part (A) Excel a Table 1. Calibration Data for [12] versus absorbance [12] Absorbance Calibration Equation 0.0025 M 1.55 0.0010 M 4= 607.364 0.00025 M R . 53 0.9974 Table 2. Initial Concentrations of Reaction Mixtures Reaction Mixture KI (C = (0.010).) HCI (C = [0.10).) Run Number mol/L mol/L 1 D DOS 0.025 2 0.005 0.025 3 DI 0025 0,095 4 0.0025 0.05 KBro, (C. = [0.040)) mol/L 0.01 0.01 0.02 DO All2 n m P Table 3. Calculated Kinetic Values Reaction Mixture Initial Run Number At mol/L's 1 43.86 x 104 2 18024 x 10-6 3 76 77 70-4 4 11. 2061/05 27.618 27.919 13.758 44.074 Table 4. Results for Determining the Experimental Rate Law Average value of k Units of k Standard Deviation of k The Experimental Rate Law at Room Temperature Table 5. Concentration of [12] (also = A[1:]) for Run 1 at 120 seconds. Compound, mol/L [compound initia mol/L A[compound] mol/L [compound):20 sec mol/L Bros H Iz Br
rate: = k{1+1[Brogi(*): and ratez = k[1]](BrO;11"[*! dividing rate ratei we get In(x) rate - k[1") (BOTHER Inty) rate, RIBOST - ()". written as yx, then n = Use this method to determine all orders, and round the orders to the nearest Integer. Record these values on the data/results sheet. 7. Now we have the initial rates, the reaction orders, and the initial concentrations. Using this data calculate the rate constant, k, for each of the experimental runs and record in Table 3 8. In Table 4 record the average value of k, the units of k, and the standard deviation of k. Also write out the complete experimental rate law for the reaction at room temperature. The experimental rate law gives a mathematical expression for how fast the reaction occurs at a given temperature and concentration set. The stoichiometry determines the reactant and product concentrations at any particular time during the course of the reaction (conservation of mass). Using the relative rates of reaction, which relates the stoichiometry of reaction, we will calculate all concentrations (except water) for reaction mixture 1 at 120 seconds. The relative rates of reaction for reaction (1) is: A[Bros) 1 AH) 1012) [Br] 1 AH,0) (5) 6 At 6 or 3 st AM Since we measured the [1] at 120 seconds, we can use this to adjust all other values. An example for calculating the (Bro,] is: A[Br05) 1 Ala! since At divides out, A/Bro, 1 = 4) and since [lol. - 0, then Alu = 14 AF 3 at [Br0.1 = [BrO,+ A/Bro.] = {Bros 1.-) where (Bros ). is the initial concentration of Bro, (6) Use this approach for a mole map approach) to calculate the values for Table 5. 100) = + Part (B) The variation of reaction rate with respect to temperature data will be obtained by using a chemical "clock approach. We will include the same quantity of thiosulfate, 503", In each reaction mixture. The thiosulfate will react with the Iz produced until the 5,07% is used up. 12100 + 25,02 27+ S.O. (7) At this point, the added drop of starch indicator will visually indicate the presence of Is. The time interval between the reaction start and the appearance of the indicator color provides average rate information. All reaction mixtures will have the same reactant composition, but will now be prepared in 13x100 mm test tubes with caps instead of the spectrophotometer cuvettes. Obtaining (12) "Clock" Rate Data to Determine the Temperature Dependence of the Rate Constant The four temperatures at which the reactions will be performed are around 10°C, room temperature, 30°C, and 40'C. The reactant amounts are given in Table 2 below. Table 2. Reagent Quantities for All Temperature Data Runs 0,010 M KI 0.0010 M Na 5,0 0.10 M HCI 0.040 M KBrO,* Starch indicator ul L ul AL Drops 500 500. 500. 500. 1 These quantities will be doubled for handling purposes in the procedure below. The actual amounts used are listed in the Table above.