The half life of a first order reaction is `60 min`. How long will it take to consume `90%` of the reactant?

Explanation

For the first order reaction
`k = (0.693)/(t_(1//2)) = (0.693)/(60) = 11.55 xx 10^(-3) min^(-1)`
Applying first order kinetic equation,
`t = (2.303)/(k)log.(a)/((a-x))`
Given: `a = 100, x = 90`, i.e., `(a-x) = (100-90) =10`
Hence, `t = (2.303)/(11.5 xx 10^(-3)) log10 = 200 min`
Alternate methof
`t_(1//2) = 60 min`
`(t_(99.9%))/(t_(1//2)) = (log((100)/(100-90)))/(0.3)`
`t_(90%) = (t_(1//2) xx log 10)/(0.3) = (60 xx 1)/(0.3) = 200 min`