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)`

Explanation

The volume of `KMnO_(4)` used at anytime anytime is proportional to conc. `H_(2)O_(2)` at that time.
`{:(,"At t=0",V=25,:.,a,prop,25),((i),"At t=10",V=16,:.,(a-x),prop,16),((ii),"At t=20",V=10.5,:.,(a-x),prop,10.5),((iii),"At t=30",V=7.09,:.,(a-x),prop,7.09):}`
Now use, `K=2.303/t log a/((a-x))`
For (i), `K=2.303/10 log 25/16`
`=4.46xx10^(-2) min^(-1)`
For (ii), `K=2.303/20 log 25/10.5`
`=4.34xx10^(-2) min^(-1)`
For (iii), `K=2.303/30 log 25/7.09`
`4.20xx10^(-2) min^(-1)`
The values of `K` come almost constant and thus confirming first order reaction. For velocity constant take average of all value of `K`.