With the obtained from the slope of the plot in question 3? Which one do you To calculate the value of Ea for this reaction. ○ Use your data from experiments 2, 4, and 5 and the Arrhenius equation log kk 21 =2− EaR To calculate the activation energy: -(-5220)*( 8 mol ∙ KJ ¿ * ( 10001 kJJ ) = 43 molkJ ■ From the equation found in the graph above, m = − REa. Slope and use it to calculate the energy of activation, east of a, in units of kJ/mol. ○ Using your linear plot of log(Rate) vs 1/T in Part B, measure the experimental ■ The rate law is, Rate = 2 −¿ S 2 O 8 ❑¿ −¿ ¿ I ¿ 2 ¿ 1Ĭoncentration of KI in final solution = 1 0.200/5 = 0 MĬoncentration of N H 4 ¿¿ 2 S 2 O 8 in final solution = 10.100/5 = 0.įor trial 2, 1/t = 1/70 = k(0)(0) for this question 1/t = k(0) System if 1 ml KI, 1 ml KCl, 1 ml N H 4 ¿¿ 2 S 2 O 8, and 1 ml Na 2 S 2 O 3 ○ From your results in part A, calculate the time for the blue color to appear for this The concentration doubled and the rate increased by 3/2. ○ The order of reaction with respect to − I ¿¿ was calculated by comparingĮxperiments 1 and 2. The concentration halved and the rate also halved. ○ The order of reaction with respect to S 22 O − 8 ¿❑¿ was calculated by comparingĮxperiments 2 and 3. Taken for each experiment ( 5 E 79 − 4 =6 E − 6 Ms ). After that step the Molarity was divided by the time Give moles of I 2 (2-6 mol) which was then divided by 0 of solution to The moles of S 22 O − 8 ¿❑¿ was compared to a 2:1 mole ratio with I 2 to ○ The reaction rate was calculated by finding the moles of S 22 O − 8 ¿❑¿ (0 *Ġ). Therefore the dilution factor is ( 15 ∗0¿=0 M. For example, inĮxperiment 1, 1 of − I ¿¿ was used and diluted to 5 before reaction. ○ Initial Concentrations were determined by the dilution factor. ○ From the data below calculate the rate law expression for the reaction of A with B. ■ It will decrease the rate of reaction as fewer collisions will occur resulting ○ How will a decrease in temperature affect the rate of reaction ( increase, decrease, Rate-Controlling Step is the elementary reaction and therefore its ■ Usually, Molecularity is equal to the order of the rate-determining step. ○ How is the molecularity of the rate-controlling step related to the overall reaction ■ A rate-controlling step in a reaction is a step that exerts a strong effect andĭecides the overall rate of the reaction. ○ What is meant by the term “rate-controlling step” of a chemical reaction? ■ Temperature, Reactant Concentration, and Surface Area. ○ In addition to the effects of catalysts, what other factors can influence the rate of a ■ Trial 3: Milimol of persulfate = 01 = 0. ■ Trial 2: Milimol of persulfate = 0 2 = 0. ■ Trial 1: Milimol of persulfate= molarity× volume(ml) = 0×2 = 0. Reacted by the time the blue color appears? ○ For all experiments in part A how many millimoles of persulfate ion (S 2 O 8 2-) have Water bath filled with ice or heated to a temperature and then dipped in the beaker.ĭifferent timings were recorded for the temperatures. The temperature was varied from 0 ∘C ¿ 35 ∘C using a Investigate the effect of temperature on the rate of this reaction, which will allow you toĭetermine the activation energy. The color of the solution will change giving us the total time elapsed in sec. As thiosulfate is a limiting reagent, after complete consumption Thiosulfate was added to convert the I 2 back to − I ¿¿ so that theĬolor change is delayed. Rate of this reaction is determined by measuring the initial reaction rate at several reactantĬoncentrations. Reaction is determined using the method of initial rates. Reaction is to be determined with a stopwatch. When I 2 reacts with starch to form a dark blue iodine/starch complex. Abstract:- The “clock reaction” is a reaction known for its colorless-to-blue color change,Īnd is used to explore the rate at which the reaction takes place.A Kinetic Study of an Iodine Clock Reaction
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