The rate constant for a first order reaction is `60" s"^(-1)`. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

Explanation

For first order reaction, `t= (2.303)/(k)log"" (a)/((a-x))`
Given, `(a-x) = (a)/(16) , k = 60" s"^(-1)`
On putting the values in Eq. (i), we get
`t= (2.303)/( 60" s"^(-1)) log"" (a xx 16)/( a) = (2.303)/( 60" s"^(-1) ) log 16`
`= (2.303)/( 60" s"^(-1) ) log2^(4) = (2.303)/( 60" s"^(-1) ) xx 4 log 2`
`= (2.303)/( 60" s"^(-1) ) xx 4 xx 0.3010 = 4.62 xx 10^(-2)" s"`
Time `=4.62 xx 10^(-2)" s"`