When the S,0, is completely used up, then 12 begins to accumulate. It combines w

Question

When the S,0, is completely used up, then 12 begins to accumulate. It combines with starch to form a dark blue complex, so the "end-point" of the "titration" occurs when the solution turns blue. The time it takes for the blue color to appear is called the period of the clock, or , the time it takes for Reaction 2 to go to completion. The rate of S,0 disappearance is therefore given by its initial concentration divided by the clock period, or [S,0, At. 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 in[son/At·(Two Sr' ions are consumed for every 12 consumed.) (Actually, the la in solution combines with iodide ion to form triiodide ion, 1, which is the species that forms the blue color with starch. That step is not shown to keep the equations simple, but adding it would not change the conclusions.) The following graphs 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 graphs. Initial conditions: 0om 0a14 IS,0-1-0.055 M [S2032-1-0.0015 M) [So,21 = [12] S,0, Time (eec) 0043 00005 EDI IS,0,1 Tme (sec)

Explanation / Answer

Solution:

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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