Show that in case of a first order reaction, the time required for 99.9% of the reaction to take place is about ten times than that required for half the reaction.

Answer

`10`

Explanation

`t_(1//2) = ( 0.693 )/(k )`
` t_(99.9 ) = ( 2.303 )/(k ) log"" (1)/(1-0.999) = ( 2.303 )/(k) log"" (1)/(10^(-3))`
` = ( 2.303)/(k ) xx 3 log 10 = ( 6.909)/( k ) `
Dividing (ii) by (i) `((6.909 )/(k)) // ((0.693)/(k )) ~~10`