`75%` of a first-order reaction was completed in `32` minutes, when was `50%` of the reaction completed ?

Options

A. 16 minutes
B. 24 minutes
C. 8 minutes
D. 4 minutes

Answer

A

Explanation

`k=(2.303)/(32)"log"(100)/(25)`
`k=(2.303log4)/(32)=(2.303log2)/(16)`
`t_(1//2)=(2.303)/(k)"log"(100)/(50)`
`=(2.303log2)/(k)=16 min`
By substituting the value of k)