Question

Good day,
What is the half-life of a first-order reaction with a rate constant of 4.30×10−4 s−1?

Answer 1

For the first-order chemical reaction A$\rightarrow$ products, the rate law is:

$[A] = [A]_0e^{-kt}$ ,

where $k$ is the rate constant and $t$ is the time.

The half-life of a chemical reaction is the time that it takes to decrease the concentration of the reagent A by a factor of two:

$\frac{[A]}{[A]_0} = 0.5 = e^{-kt_{1/2}}$ .

Therefore, the half-life of the reaction is:

$t_{1/2} = \frac{\text{ln2}}{k} = \frac{\text{ln2}}{4.30·10^{-4}} = 1612$ s.

Answer: 1612 s.