A first order reaction takes `23.1 ` minutes for `50%` conmpletion. Calculate the time required for 75% completion of this reaction `(log 2 = 0. 301) , ( log 3 = 0. 4771) ( log 4 = 0.6021) `

Explanation

First Order reaction is, `k=(2.303)/(t)log""([A]_(0))/([A])`
`[A]_(0)=100`
`[A]=100-50`
[A]=50
`K=(2.303)/(23.1)log""(100)/(50)`
`K=(2.303)/(23.1)xx log 2`
`K=2.303xx0.301`
`K=0.030" min"^(-1)`
For 75% completion.
`[A]_(0)=100, [A]=100-75=25`
`t=(2.303)/(0.030)log(100)/(25)`
So, `t=(2.303)/(0.030)log 4`
`t=(2.303)/(0.030) xx 0.6621`
= 46.2 min.