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 (s)",0,10,20),("Vol. of "KMnO_4"(ml)",11.4,6.9,4.15):}`
Show that the reaction is of first ordre.

Explanation

If the reaction is of 1st order, the rate constant, k calculated by using the equation,
`k=(2.303)/(t)log([R]_0)/([R])`
should be constant
Here `[H_2O_2]prop` vol of `KMnO_4` used
In 1st Case
t=10 s
`([R]_0)/([R])=(11.4)/(6.9)`
`therefore k=(2.303)/(10s)log(11.4)/(6.9)`
`=(2.303xxlog 1.6522)/(10s)`
`=(2.303xx0.2180)/(10s)=0.5025s^(-1)`
In 2nd Case
`([R]_0)/([R])=(11.4)/(4.15)=2.7470`
`k=(2.303)/(20 s)log2.7470`
`=(2.303xx0.4389)/(20s)`
`=0.05054s^(-1)`
As value of k comes out to be practically constant, it is a Ist order reaction
Average value of
`k=1/2[0.05025+0.05054]s^(-1)`
`=1/2=0.05039 s^(-1)`