Question

What is the half-life of a first-order reaction if it takes 4.4 x 102 seconds for the concentration to decrease from 0.50 M to 0.20 M? 2.5 x 102 s 3.3 x 102s 1.6s 21 s 27 s

Answer 1

Answer: The half-life of a first-order reaction is,
`3.3* 10^2s`

Step-by-step explanation:

All the radioactive reactions follows first order kinetics.

Rate law expression for first order kinetics is given by the equation:

`k=(2.303)/(t)\log([A_o])/([A])`

where,

k = rate constant = ?

t = time taken = 440 s

`[A_o]` = initial amount of the reactant = 0.50 M

[A] = left amount = 0.20 M

Putting values in above equation, we get:

`k=(2.303)/(440s)\log(0.50)/(0.20)`

`k=2.083* 10^(-3)s^(-1)`

The equation used to calculate half life for first order kinetics:

`t_(1/2)=(0.693)/(k)`

Putting values in this equation, we get:

`t_(1/2)=(0.693)/(2.083* 10^(-3)s^(-1))=332.69s=3.3* 10^2s`

Therefore, the half-life of a first-order reaction is,
`3.3* 10^2s`