Derive order of reaction for the decomposition of `H_(2)O_(2)` from the following data.
`{:("Time (in minutes)",0,10,20,30),("Volume of" KMnO_(4),25,16,10.5,7.09):}`
needed for `H_(2)O_(2)`

To calculate the rate of decomposition of H_(2)O_(2) in an aqueous solution , a certain volume of the aqueous solution of H_(2)O_(2) is pipetted out and titrated against KMnO_(4) solution at different interval of time . Intervals The observations are given below : {:(Time ("in min"),0,5,10,20),("Volume of "KMnO_(4)("inML"),46.2,37.1,29.8,19.6):} Show that the decomposition of H_(2)O_(2) in aqueous solution is a first order reaction and calculate its rate constant .

The following data are obtained for the decomposition of H_(2)O_(2) " at " 25^(@)C : {:("Time (min)",0,10,20,30),(% "decomposed",0,37,60,75):} Calculate the order , rate constant and half -life of the reaction.

It takes 10 min for the decomposition of 50% H_(2)O_(2) . If the reaction is of first order, the rate constant will be

What mass of O_(2) would be formed from the reaction of 34.0 g H_(2)O_(2), 180.0 g KMnO_(4) and excess of H_(2)SO_(4) according to following equation. (mol wt. of H_(2)O_(2)=34 amu, KMnO_(4)=158 amu ) ? H_(2)O_(2)+2KMnO_(4)+3H_(2)SO_(4) to K_(2)SO_(4)+2MnSO_(4)+8H_(2)O+5O_(2)