need help with question 1 and 2. Pre-Lab Questions (Use a separate sheet of pape
ID: 564072 • Letter: N
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need help with question 1 and 2.
Pre-Lab Questions (Use a separate sheet of paper to ansuver the following quertions.) 1. Another version of the iodine clock reaction involves reaction of iodide ions with persulfate ions. The following rate data was collected by measuring the time required for the appearance of the blue colan Reaction Time 270 sec 138 sec 142 sec Trial II-1 0.040 M 0.080 M 0.040 M 0.040 M 0.040 M 0.080 M a. In each trial, the blue color appeared after 0.0020 M iodine (L) had been produced. Calculate the reaction rate for each trial by dividing the concentration of iodine formed by the reaction time. b. Compare trials 1 and 2 to determine the order of reaction with respect to iodide ions. How did the concentration of iodide ions change in these two trials, and how did the rate change accordingly? What is c Which two trials should be compared to determine the order of reaction with respect to persulfate ions? d.Wite the raté law for this version öf thie iodine clock reaction. Coutd the rate law have been prédicted [0 -a the reaction order for iodide? What is the reaction order for persulfate? using the coefficients in the balanced chemical equation? Explain. INO 0.0125 M 0.0250 M 0.0125 M Exp. No. 2 initial 0.0250 M 0.0250 M 0.0500 M 0.0282 M/s 0.112 M/s 0.0560 M's Write the rate law including the value for the specific rate constant, k.Explanation / Answer
For question 1, you need to divide the amount provided in the statement by the reaction time
0.002 / 270 (for trial 1) = 7.4 x 10-6
For trial 2
0.002 / 138 = 1.5 x 10-5
For trial 3
0.002 / 142 = 1.408 x 10-5
part b compare trial 1 and 2
Concentration of S2O8 is the same I-
trial 2 concentration of I- is 0.08 and from trial 1 is 0.04 this is that the concentration of I- has doubled
do this for the rates of reaction
1.5 x 10-5 / 7.4 x 10-6 = 1.95 approximately 2 so
when the concentration of I- doubles the rate of reaction doubles this means that the reaction is first order for I-
c)trial 1 and 3 must be compared in order to find the order of reaction for thiosulfate
divide the concentrations of thiosulfate for trial 1 and trial 3
0.08 / 0.04 = 2
divide the rates of reaction for trial 1 and 3
1.408 x 10-5 / 7.4 x 10-6 = 1.9 approximately 2
so when the concentration of thiosulfate doubles the rate of reaction doubles so it is a first order for thiosulfate
rate law is = k * [I-]*[S2O8-2]
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