Question

Half-life equation for first-order reactions:
t1/2=0.693k  
where t1/2 is the half-life in seconds (s), and kis the rate constant in inverse seconds (s−1).

a) What is the half-life of a first-order reaction with a rate constant of 8.10×10⁻⁴  s^−1? Express your answer with the appropriate units.

b) What is the rate constant of a first-order reaction that takes 151 seconds for the reactant concentration to drop to half of its initial value? Express your answer with the appropriate units.

c) A certain first-order reaction has a rate constant of 3.80×10⁻³ 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

a) Half life = ?

Rate constant = 8.10×10⁻⁴  s^−1

Half life = 0.693 / rate constant = 0.693 / 8.10×10⁻⁴ = 0.08556 X 10^4 = 8.56 X 10^2 second

b) The time required for completion of half reaction = half life = 151 second

So rate constant = 0.693 / 151 = 0.00458 second-1

c) I think you are saying it to drop to 18% of the initial value

Let initial concentration = a

so after a given time the concentration = a-x = 0.18a

Rate constant = 0.0038 s^-1

The rate law for first order reaction is

time = 2.303 / K [ log (a/a-x)]

time = 2.303 / 0.0038 [log 1/0.18)]

time = 451.34 seconds