For a first order reaction, calculate the ratio between the time taken to complete `3//4th` of the reaction and time to complete half of the reaction.

Explanation

`t_(1//2) = (0.69)/(k), t_(314) = t_(75%)`
`t_(3//4) = (2.303)/(k) log.(a)/((a-(3a)/(4)))`
`= (2.303)/(k)log4`
`= (2.303)/(k) xx 2 xx 0.3010 = (0.69 xx 2)/(k)`
`(t_(3//4))/(t_(1//2)) = (0.69 xx 2)/(k) xx (k)/(069) rArr t_(3//4) = 2t_(1//2)`
Alternate methof
Use direct relation,
`(t_(75%))/(t_(1//2)) = (log.((100)/(100-75)))/(0.3) = (2 log 2)/(0.3) = (2 xx 0.3)/(0.3) = 2`
`:. t_(75%) = 2t_(1//2)`