3) In the iodine clock reaction, you are measuring the rate of the reaction of i

Question

3) In the iodine clock reaction, you are measuring the rate of the reaction of iodide ion with peroxydisulfate ion (S20s I (Reaction 1) The rate of the reaction is measured by determining the amount of formed in a particular time, or All2VAr. The is measured by titrating it with a fixed amount of thiosulfate (S20y which rapidly converts the I2 formed in Reaction 1 back to iodide ion: 2szoba 2n (Reaction 2) (While Reaction 2 is said to be "fast" with respect to Reaction 1, it actually can proceed no faster than the rate at which is produced, so its actual rate is controlled by the rate of Reaction 1.) When the S20,2 is completely used up, then l begins to accumulate. It combines with starch to form a dark blue complex, so the end point of the titration occurs when the solution tums blue. The time it takes for the blue color to appear is called the period of the clock, or At, the time it takes for Reaction 2 to go to completion. The rate of S203 disappearance is therefore given by its initial concentration divided by the clock period, or [S203 yar. Because the rate of disappearance of I2 in Reaction 2 is controlled by its rate of appearance in Reaction 1, the rate of both reactions can be determined from the value of t x [S203 yAr. (Two S20,2 lons are consumed for every I2 consumed.) [E] (Actually, the in solution combines with oose iodide ions to form triiodide ion, I which is ooss the species that forms the blue color with starch. [C] That step is not shown to keep the equations simple, but adding it would not change the conclusions.) Time (sec) The graphs to the right illustrate what is happening to the concentration of each of the species involved in these two reactions. Show you understand the reaction by identifying the species associated with each of the lines in the EFJ graphs. [B] [A] Time (sec)

Explanation / Answer

GRAPH-1

[so4]=[c], as it decreases linearly with time.

[I-]=[E], because first its concentration remain constant and after sometime it will changes to I2 thus its concentration decreases as u can see from the graph.

GRAPH-2

[s2o32-]=[B], as it corresponds to given value 0.0015 M and its concentration decreases linearly with time.

[s4o6]2-=[D] as s2o3 changes to s4o62-.

[I2]= [A], as its concentration decreases after reachig a particular point.

[I-]=[F],as it shows a steep increase in the concenration.

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