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 = 2.303 xx 0.301`
` 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.