A student conducting the iodine clock experiment accidentally makes an S_2O_8^2-

Question

A student conducting the iodine clock experiment accidentally makes an S_2O_8^2- stock solution that is too concentrated. How will this affect the rate measurement? Reaction 1: 3I^- (aq) + S_2O_8^2- (aq) rightarrow l_3^- (aq) + 2SO_4^2- (aq) slow Reaction 2: l_3^- (aq) + 2S_2O_3^2- (aq) rightarrow 3L^- (aq) + S_4O_6^2 (aq) fast Reaction 3: l_3 (aq) + starch (aq) rightarrow 3I^- ---- starch (bluish black) fast Not all of the 2S_2SO_3^2-, will react, which will increase the calculated rate. It will take less time for the color to change, as rate increases with concentration of reactant. It will take longer for the color to change, as the reaction takes longer when more reactant is present. Not all of the 2S_2O^2 will react, which will decrease the calculated rate.

Explanation / Answer

Option (b)

Rate of a reaction depends on the concentration of reactants.

So, for increased concentration it will take less time for the colour to change.

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