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.

Explanation

If the reaction is of first order , `k=(2.303)/(t) log .(a)/(a-x)`
By putting the values of a and (a-x) obtained at the different phases of the reaction ,
At `t=10min, k =(2.303)/(10)log.(a)/(a-0.37a)=0.0462"min"^(-1)`
At `t=20min, k =(2.303)/(20)log.(a)/(a-0.6a)=0.0458"min"^(-1)`
At `t=30min, k =(2.303)/(30)log.(a)/(a-0.6a)=0.0462"min"^(-1)`
The values of k so obtained are almost same . So , the reaction is of first order.
Average value of
`k=(1)/(3)(0.0462+0.0458+0.0462)=0.046"min"^(-1)`
`t_(1//2)=(0.693)/(k)=(0.693)/(0.046)=15.06min`