Reaction between S2O82- ions and I- is
2I- + S2O8 2- ---> 2 SO4 2- + I2
the iodine produced here can react with S2O3 2- and the EXCESS
iodine will react with starch solution to form a blue black complex
as shown:
I2 + 2S2O3 2- ---> S2O6 2- + 2I-
In a particular experiment, 20cm3 of 0.300 mol dm-3 of KI
was added. 10cm3 of 0.005 mol dm-3 of S2O3- was added, and also
20cm3 0.002 mol dm-3 of S2O8 2-. 10 cm3 of starch was added.
given that initial rate of reaction is 0.5 X (no. of moles
of S2O3 2-)/time mol dm-3 s-1
1. calculate rate of reaction.
2. determine rate constant k.
for part 1, i calculated rate to be 5.7 X 10^-7 mol dm-3
s-1.
But i have difficulties with part 2. i wrote the rate equation
as
rate = k [S2O8 2-][I-] (btw, do i equate intial rate to this
rate in part 2?)
Do i have the use the concentration of S2O8 2- and I- after
dilution? If so, what is the final volume upon dilution? (it doesnt
include that of starch yes? since starch does not affect the rate
of reaction)
thanks
How did you get 5.7 X 10^-7 mol dm-3 s-1? Show your working.
How did you get rate = k [S2O8 2-][I-]? Why do you
assume the reaction is first order wrt both
peroxodilsulfate(VI) and iodide ions?
This question is problematic, because it tries to combine two
(or three, if you include the starch-iodine complex) separate
reactions into one experiment, and one rate equation.
While the info it provides can arguably be used to form a
rate equation (hint : think of the rate determining step as the
single slowest step of all the elementary steps for both reactions
combined), this is still a poorly set question (I doubt it's a
Cambridge Qn, but even Cambridge Qns can be poorly set, let alone
JC Tutorial/Prelim Qns).
Regarding your question,
Do i have the use the concentration of S2O8 2- and I- after
dilution? If so, what is the final volume upon dilution? (it doesnt
include that of starch yes? since starch does not affect the rate
of reaction)
You have a misconception here. It's not the point whether starch
directly participates in the reaction or not. It's the fact that
adding starch solution increases the total volume of solution,
which affects the molarities of the reactants of the other two
redox reactions, and THAT (ie. change in molarities) affects the
rate of the slowest (ie. rate determining) step of the reaction(s).
If you've understood what I've just said, you should be able to
self-answer your own question now.