The time of completion of 90% of a first order reaction is approximately

Options

A. 1.1 times that of half-life
B. 2.2 times that of half-life
C. 3.3 times that of half-life
D. 4.4 times that of half-life

Answer

C

Explanation

`t_(1//2) = ( 2.303 )/( k) log "" (a)/(( a-a //2) = ( 2.303 )/(k ) log"" 2 , t_(0.9 ) = ( 2.303)/(k) log (a)/(a-0.9 a) = ( 2.303 )/(k) log 10`
` (t_(0.9 ))/( t_(1//2)) = ( log 10)/( log 2) = ( 1)/(0.3010 ) = 3.32 ~~ 3.3`