Show that the time required for the completion of 75% of a reaction of first order is twice the time required for the completion of 50% of the reaction.

Explanation

`(i) t_(0.5) = ( 2.303)/(k) log ""(1)/(1-0.5) = ( 2.303)/(k) log 2 `
`( ii ) t_( 0.75 ) = ( 2.303)/(k ) log ""(1)/(1-0.75 ) = ( 2.303 )/(k) log 4 `
Dividing (ii) by (i), we get ` (t_( 0.75 ))/( t_(0.5)) = ( log 4)/( log 2) = ( 0.6020)/(0.3010)=2`