Molecular iodine, i2(g), dissociates into iodine atoms at 625 k with a first-order rate constant of 0.271 s−1.
(a) what is the half-life for this reaction?
(b) if you start with 0.052 m i2 at this temperature, how much will remain after 5.16 s assuming that the iodine atoms do not recombine to form i2?

a) Half life = 2.56 s

b) Remaining concentration after 5.16 s = 0.0128 M

Explanation:

Given,

Rate constant =

a)

Half life of a first order reaction is given by,

Half life =

             =

b)

Integrated rate law for first order reaction is:

Where,

A = Concetration after time 't'

A0 = Initial concentration = 0.052 M

k = rate constant = 0.271 s^-1

t = time = 5.16 s

a electronegativity decreases beacuse the size of nucleus is bigger.