The time taken for 10 % completion of a first order reaction is 20 minutes. Then for 19 % completion, the reaction will take

Options

A. 40 mins
B. 60 mins
C. 30 mins
D. 50 mins

Answer

A

Explanation

`t = (2*303)/(k) log (a)/(a-x) therefore 20 = (2*303)/(k) log ""(a)/(a-0*10a)`
or `20 = (2*303)/(k)log""(100)/(90)`
In 2nd case, `t= (2*303)/(k)log""(a)/(a-0*19a)`
`= (2*303)/(k) log ""(100)/(81)=(2*303)/(k)log((10)/(9))^(2)" "…(ii)`
Eqn. (ii) Eqn. (i) gives `(t)/(20) =2` or `t = 40 min.`