60 percent of first order reaction was complete in 60 minutes. When was it half completed?

Explanation

Calculation of rate constant(k)
According to available data,
`a=100, (a-x)=40, t=60 min`
For the first order reaction,
`k=(2.303)/k log a/(a-x) = (2.303)/(60 "min") log 100/40 = 2.303/(60"min") log 2.5`
`=(2.303)/(60"min") xx 0.3979 = 00153 min^(-1)`
Calculation of half life period `(t_(1//2))`
`t_(1//2) = 0.o693/k = 0.693/(0.0153 min^(-1)) = 46.29 min`