The decomposition of `H_2 O `, is a first order reaction. When 5 mL portions of `H_2 O_2` are titrated with `KMnO_4` solution at the start of the reaction and 5 minutes later, the volumes of KMnO4 solution required are 37.0 mL and 29.5 mL respectively. Calculate the rate constant of the reaction.

Explanation

Reaction between `H_2 O_2` and acidified `KMnO_4` may be represented as follows:
` 2KMnO_4 +3H_2 SO_4 to K_2 SO_4 +2MnSO_4 +3H_2 O + 5 [O]`
`H_2 O_2 +O to H_2 +O_2 ] xx 5`
Therefore, volume of `KMnO_4` used is proportional to the amount of `H_2 O_2` present. The value of x when t = 0 corresponds to initial concentration .a. and value of x in the numerical problem when t = 5 corresponds to the amount of undecomposed `H_2 O_2`, i.e., (a-x).
If the reaction is of the first order, it must obey the equation,
`k=( 2.303 )/( t) log"" (a)/((a-x)) =( 2.303 )/(t) log"" (V_0)/(V_t)`
In the present case `V_0 = 37.0`
` V_t` (after 5 minutes)= 29.5 mL
` therefore k=( 2.303 )/( 5 ) log "" ( 37.0 )/( 29.5 ) = ( 2.303 )/(5) [ log 37.0 - 29.5 ]`
`= ( 2.303 )/(5 ) [ 1.568 -1.468 ] = (2.303 xx 0.099 )/( 5) = 0.456`