Answer to Question #119771 in Chemistry for B

Question

Hi,Please may I ask for help?
What is the rate constant of a first-order reaction that takes 234 seconds for the reactant concentration to drop to half of its initial value?

Answer 1

The time taken for the reactant concentration to drop to half of its initial value is called the half-life of the reaction $t_{1/2}$. For the first order reaction (A$\rightarrow$ products), the rate law is:

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

At half-life:

$\frac{[A]}{[A]_0} = 1/2$ .

Therefore, the rate constant of this reaction is:

$0.5 = e^{-kt_{1/2}}$

$\text{ln}(0.5) = -kt_{1/2}$

$\text{ln}(2) = kt_{1/2}$

$k = \frac{\text{ln}(2)}{t_{1/2}} = \frac{\text{ln}(2)}{234} = 2.96·10^{-3}$ s^-1.

Answer: The rate constant of this first-order reaction is 2.96·10^-3 s^-1.