If 60 % of a first order reaction was completed in 60 minutes, 50 % of the same reaction would be completed in approximately
[log = 4 = 0.60, log 5 = 0.69].

Options

A. 45 minutes
B. 60 minutes
C. 40 minutes
D. 50 minutes.

Answer

A

Explanation

a) For first order reaction
`k=(2.303)/(60 "min") log 100/40`
`=(2.303)/(60"min") log 2.5`
`=(2.303 xx 2.5)/(60 "min") = 0.0153`
`t-(1//2) = 0.693/k = 0.693/(0.0153)=45.3` min.