Question

A certain reaction has a fixed half-life of 125 sec, how much time does it take for 25.0% of the original reactant to be reacted?

Answer 1

Given that reaction has a fixed half-life of 125 sec.

That means, half life is independent on initial concentration.

Hence, it is a first order reaction.

For first order reaction,

half life t_1/2 = 0.693 /k where k is rate constant

k = 0.693/ t_1/2  --- Eq (1)

k = 1/t ln { [A]_o/[A]_t} ----- Eq (2)

From Eqs (1) and (2),

0.693/ t_1/2 = (1/t) ln {[A]_o/ [A]_t} ------Eq (3)

Given that

half life t_1/2 = 125 sec

time t = ?

Initial amount = 100 %

Final amount [A]_t = 100 % - 25 % = 75 %

Substitute all the values in Eq (3),

0.693/ t_1/2 = (1/t) ln {[A]_o/ [A]_t}

0.693/ 125 = (1/t) ln {100 %/ 75%}

t = (125/0.693) x ln {100 %/ 75%}

= 51.9 sec

t = 51.9 sec

Therefore,

it take 51.9 sec for 25.0% of the original reactant to be reacted.