Question

`75%` of a first-order reaction was completed in `32` minutes. When was `50%` of the reaction completed
A. `24` min
B. `4` min
C. `16` min
D. `8` min

Answer 1

Correct Answer - C
Let original amount, `N_(0) = 100`
Since `75%` completed, so final amount `N = 100 - 75 = 25`
As we know
`(N_(0))/(N) = 2^(n)`, where `n =` no. of half lives
or, `(100)/(25) = 2^(n)`
or, `4 = 2^(n)` or, `2^(2)` or, `2^(2) = 2^(n)`
`:. n=2`
Since total time `= n xx t_(1//2)`
`32` minutes `= 2 xx t_(1//2)`
`:. t_(1//2) = 16` minutes