75% of a first order reaction was completed in 32 min. When was 50% of the reaction completed ?

Options

A. 16 min
B. 8 min
C. 4 min
D. 32 min

Answer

A

Explanation

Given, reaction is 75% completed in 32 min
a=100,X=75
`therefore K=(2.303)/(32)"log"(100)/(100-75) " "cdots (i)`
For 50% completion of reaction,
a=100 ,X=50
`therefore K=(2.303)/(t)"log"(100)/(100-50) " "`(ii)
On equating Eq. (i)and (ii), we get
`therefore (2.303)/(32)"log"(100)/(100-50)=(2.303)/(t)"log"(100)/(100-50)`
or `" " (2.303)(32)"log" 4=(2.303)/(t)` log 2
or `" " (t)/(32)("log2")/("log4") or t=(32xx"log")/(2"log"2)`
` therefore " "` t=16 min