Question

What is the half-life of a first-order reaction with a rate constant of 4.20×10⁻⁴  s−1? (the answer is 1650s)

What is the rate constant of a first-order reaction that takes 458 seconds for the reactant concentration to drop to half of its initial value?

Answer 1

The relation between half life (t^1/2) and rate constant k can be given as:

t^1/2 = 0.693 / k

k = 4.2 X 10^-4 s^-1 (given)

So, for first question, t^1/2 = 0.693 / 4.2 X 10^-4

                                         t^1/2 = 0.165 X 10⁴ = 1650 s

Similarily,

From above eq. rate constant for 1st order reaction, when t = t^1/2

k =0.693 / t^1/2

and t = t^1/2 = 458 s (given)

So, k =0.693 / 458   = 0.00151 s

k = 1.51 X 10^-3 s