Show that the time required for 99% completion of a first order reaction In twice the time required for the completion of 90%.

Explanation

For first order reaction
`t = (2. 303)/( K ) log "" (a -x)/( a ) = ( 464)/(K) to eqn (i)`
Iind case `= t _( 90%) = ( 2. 303)/(K) log "" (100)/( 10 ) = (2. 303)/( K ) to eqn (ii)`
Divide eqn (ii) by eqn (i) `= ( t ( 99%))/( t ( 90 %)) = ( 463)/( K ) xx (K )/( 2. 303) = 3`