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Explain order of reaction of complex reaction by giving examples. |
Answer» Solution :(a)The example of complex reaction involving more than THREE molecules :The complex reaction involving more than three molecules in the stoichiometric equation MUST take place in more than one step e.g  This reaction which apparently seems to be of tenth order is actually a second order reaction. This shows that this reaction takes place in several steps but slowest step determine the rate of reaction ."The overall rate of the reaction is controlled by the slowest step in a reaction called the rate determining step" Example :The decomposition of hydrohen proxide which is catalysed by iodide ion in an alkaline medium. `2H_(2)O_(2)(I^(-))/("alkaline medium ")2H_(2)O+O_(2)` The rate equation for this reaction is found to be Rate=`-(d[H_(2)O_(2)])/(dt)=k[H_(2)O_(2)][I^(-)]` If the order of reaction =`(5)/(2)` then unit of rateconstant k is `L^((+3)/(2)) mol ^((-3)/(2))S^(-1)` Thus,in respect of `H_(2)O_(2)`, the order of reaction =1 and In respect of `I^(-)` ,the order of reaction =1 and overall order of reaction =(1+1)=2 So,this reaction is second order reaction. The decomposition of `H_(2)O_(2)` takes place in two steps. (i)`H_(2)O_(2)+I^(-)toH_(2)O+IO^(-)` (slow step) (II)`underset("molecule")underset("second")(H_(2)O_(2))""underset("species")underset("Intermediate")(+IO^(-)toH_(2)O)+underset(("fast step"))(I^(-)+O_(2))` Overall reaction (i)+(ii):`2H_(2)O_(2)to2H_(2)O+O_(2)` Both the steps are bimolecular elementary reaction .Species `IO^(-)` is called as an intermediate since it is formed during the COURSE of the reaction but not in the overall balance equation. The first tep,being slow is the rate determining step.Thus,the rate of formation of intermediate will determined the rate of this reaction. The order of the slowest step=molecularity of the slowest step. |
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