A first order reaction is 50% complete in 69.3 minutes. Calculate the time for 80% completion of the reaction.

Explanation

Step I : To Calculate k.
`t_(1//2)`=69.3 min
`k=(0.693)/(t_(1//2))=(0.0693)/(69.3" min")=10^(-2)min^(-1)`
Step II : To Calculate t.
`[R]_0=100(say),[R]=100-80=20`
`k=10^(-2)min^(-1),t=?`
`t=(2.303)/(k)log([R]_0)/([R])`
`=(2.303)/(10^(-2)min^(-1))log(100)/(20)`
`(2.303)/(10^(-2)min^(-1))[log10-log2]`
`=(2.303)/(10^(-2)min^(-1))[1-0.3010]`
`=(2.303xx0.6990)/(10^(-2))min=160.98` min.