For a first order reaction show that time required for 99% completion is twise the time required for the complation of 90% of reaction .

Explanation

For a first order reaction
` t=( 2.303 )/(k ) log"" (a)/((a-x))`
Let ` .a. = 100 ` , then
` t_(99%) = ( 2.303 )/(k ) log "" (100)/((100 -99)) = ( 2.303 )/(k ) log ""100`
` t_(90%) = ( 2.303)/(k ) log "" (100) /( (100-90)) = ( 2.303 )/(k) log 10`
` therefore (t_(99%))/( t_(90%)) =((2.303 )/(k ) "" log 100 )/( (2.303)/(k ) "" log 10) = ( log 100)/( log 10) =2`
hence ` t_(99% ) = 2 xx t_(99%)`