Question

If it takes 56.7 min for the concentration of a reactant to drop to 18.0% of its initial value in a first-order reaction, what is the rate constant for the reaction in the units min-1? You do not need to show your work. You can just write the answer in the space provided below

Answer 1

Answer is 0.00302min-1=k

Answer 2

0.00302min⁻¹ = k

Step-by-step explanation:

In a first order reaction the concentration of the reactant decrease following the equation:

ln[A] = -kt + ln[A]₀

ln [A] / [A]₀ = -kt

Where [A] represents actual and initial concentration of the reactant, k rate constant and t time.

As the reactant drop to 18.0% of its initial concentration [A] / [A]₀ = 0.18

And time = 56.7min:

ln [A] / [A]₀ = -kt

ln 0.18 = -k*56.7min

-1.715 / 56.7min = -k

0.00302min⁻¹ = k