Question

The integrated rate law allows chemists to predict the reactant concentration after a certain amount of time, or the time it would take for a certain concentration to be reached. The integrated rate law for a first-order reaction is: [A]=[A]0e−kt Now say we are particularly interested in the time it would take for the concentration to become one-half of its inital value. Then we could substitute [A]02 for [A] and rearrange the equation to: t1/2=0.693k This equation caculates the time required for the reactant concentration to drop to half its initial value. In other words, it calculates the half-life.

a. What is the half-life of a first-order reaction with a rate constant of 3.30×10⁻⁴  s−1?

b. What is the rate constant of a first-order reaction that takes 8.90 minutes for the reactant concentration to drop to half of its initial value?

c.A certain first-order reaction has a rate constant of 7.20×10⁻³ s−1. How long will it take for the reactant concentration to drop to 1/8 of its initial value?

Answer 1

a) For a first order reaction,

half life, t_1/2 = 0.693/k

or, t_1/2 = 0.693/(3.30 x 10^-4 /s) = 2.1 x 10³

therefore, half life is 2.1 x 10³ s

b) For a first order reaction, rate constant can be given by:

k = 0.693/t_1/2

Therefore, rate constant is 0.0013 s^-1

c) half-life for the given reaction is:

t_1/2 = 0.693/k = 0.693/(7.20 x 10^-3 /s) = 96.25 s

So, it takes 96.25 s for the initial concentration to become half.

For the initial concentration to become 1/8 of the initial value, 3 half-lives are needed [because, 1/8 = (1/2)³ ]

3 half-lives = 3 x (96.25 s) = 288.75 s

288.75 s = 288.75/60 min = 4.81 min

Hence, it will take 288.75 s or 4.81 min for the reactant concentration to drop to 1/8 of its initial value.