The rate of decomposition of hydrogen peroxide at a particular temperature was measured by titrating its solution with acidic `KMnO_4` solution. Following results were obtained.
`{:("time,t (min)",0,10,20),("Vol. of "KMnO_4"(ml.)",22.8,13.8,8.3):}`
Show that the reaction is of first order.

Explanation

Here, `V_(KMnO_4prop[H_2O_2]`
Case I : At t=10 min
Her `([R]_0)/([R])=([H_2O_2])/([H_2O_2])=(22.8)/(13.8)`
For a Ist Order reaction
`k=(2.303)/(t)log([R]_0)/([R])`
`=(2.303)/("10 min")log(22.8)/(13.8)`
`=(2.303)/("10 min")log1.652`
`=(2.303xx0.2180)/("10 min")`
`=5.02xx10^(-2)min^(-1)`
Case II : At t=20 min
Here`([R]_0)/([R])=([H_2O_2]_0)/([H_2O_2])=(22.8)/(8.3)`
`therefore k=(2.303)/(t)log([R]_0)/([R])`
`=(2.303)/("20 min")log2.7470`
`=(2.303xx0.4389)/("20 min")`
`=5.05xx10^(-2)" min"^(-1)`
As the value of k comes out to be practically constant, it is a first order reaction.
Average value of
`k=1/2[5.02xx10^(-2)+5.05xx10^(-2)]min^(-1)`
`=5.035xx10^(-2)" min"^(-1)`