60% of a first order reaction was completed in 60 minutes. When was it half completed ?

Explanation

Step 1: To Calculate k `[R]_0=100(say),[R]=100-60=40`
t=60 min
`k=(2.303)/(t)log([R]_0)/([R])`
`=(2.303)/("60 min")log(100)/(40)`
`=(2.303)/("60 min")[log10-log4]`
`=(2.303)/("60 min")[1-0.6021]`
`=(2.303xx0.03979)/(60)min^(-1)`
`=0.0153" min"^(-1)`
Step II : To Calculate `t_(1//2)`
`t_(1//2)=(0.693)/(k)=(0.693)/(0.0153 min^(-1))`
`=(693)/(15.3)min=45.38` min.