Two kinetics trials are performed at the same temperature with [S_2O^2-_8] held

Question

Two kinetics trials are performed at the same temperature with [S_2O^2-_8] held constant. The first trial has a reaction time of 82 seconds. For the second trial, the [1^-] is four times that of the first trial (the concentration is quadrupled), and the reaction time is 20 seconds. If the change in concentration of S_20^2-_8 is 4.3 times 10^-5 M for both trials, what is the reaction order with respect to iodide ion? How could one determine that the Cu^2+ ion in the catalyst is the active species, rather than the nitrate ion? Give a specific example of a reagent you might use in another reaction trial in order to verify that the Cu^2+ is the active species.

Explanation / Answer

Q1.

[C] = S2O8-2

t1 = 82 s

[C2] = 4(S2O8-2)

t2 = 20 s

dC = 4.3*10^-5

Apply:

Rate = -k*[S2O8-2]^n

rate in t = 82/20 = 4

chang ein concentration = 4x

therefore

4x --> 1/4 time

so

this must be FIRST ORDER

Rate = -k*[S2O8-2]

Q2.

If you make two trials, one with Cu+2 ions and another one with no Cu+2 ions, and you record different times (considerably)

then, you are sure that Cu+2 is acting as a catalyst

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