In a first order reaction, the concentration of the reactants is reduced to `25%` in one hour. The half-life periof of the reactions is

Options

A. `2 hr`
B. `4 hr`
C. `1//2 hr`
D. `1//4 hr`

Answer

C

Explanation

Use direct formula:
`(t_(1//2))/(t_(25%("left"))) = (0.3)/(log((100)/(25)))`
`t_(1//2) = t_(25%) xx (0.3)/(log 4)`
`= 1hr xx (0.3)/(0.6) = (1)/(2) hr`
Note: `25%` of the amount is left.