Question

Part A

What is the half-life of a first-order reaction with a rate constant of 1.80×10⁻⁴  s−1?

Express your answer with the appropriate units.

Part B

What is the rate constant of a first-order reaction that takes 163 seconds for the reactant concentration to drop to half of its initial value?

Express your answer with the appropriate units.

Part C

A certain first-order reaction has a rate constant of 3.10×10⁻³ s−1. How long will it take for the reactant concentration to drop to 18 of its initial value?

Express your answer with the appropriate units.

Answer 1

Answer – Part A ) We are given first order reaction rate constant k = 1.80 *10^-4 s^-1 and need to calculate the half-life for first order reaction.

We know the relationship between the rate constant and half-life as follow –

Rate constant , k = 0.693 / t ½

So, Half-life, t ½ = 0.693 / k

                         = 0.693 / 1.80 *10^-4 s^-1

                          = 3850 s

So, the half-life of a first-order reaction is 3850 seconds.

Part B) In this part we are given time 163 seconds and both the concentration like initial and final. We are given that the reactant concentration to drop to half of its initial value mean if we take initial concentration 1 M then final will be 0.5 M, since it is getting half to initial value.

So values are given as, t = 163 s , [A_o] = 1 M , [A] = 0.5 M

We know first order integrated equation as follow –

ln [A] / [A_o] = - k x t

ln 0.5 M / 1 M = - k x 163 sec

-0.693 = - k x 163 sec

So, k = 0.693 / 163 sec

        = 0.00425 s^-1

The rate constant of a first-order reaction that takes 163 seconds for the reactant concentration to drop to half of its initial value is 0.00425 s^-1

Part C ) We are given rate constant, k is 3.10 *10^-3 s^-1,

We consider the initial reactant concentration is 1 M, so the final reactant concentration is 1/18 M and we need to calculate time for this conversation.

We know first order integrated equation as follow –

ln [A] / [A_o] = - k x t

ln (1/18 M) / 1 M = - 3.10 *10^-3 s^-1 * t

-2.89 = - 3.10 *10^-3 s^-1 * t

So, time, t = -2.89 / - 3.10 *10^-3 s^-1

               = 932 s

It will take 932 seconds for the reactant concentration to drop to 18 of its initial value