Question

If the half-life for a first-order decomposition of a substance is 2.5 hr, how much time is required for 95% of the substance to decompose?

Answer 1

The integrated first order rate law is as follows:

ln[A] = ln[A]₀ - k∙t

here; [A] = concentration of reactant A at time t

[A]₀= initial reactant concentration

k = rate constant;
or we can write it as follows:

t = (ln[A]₀ - ln[A] ) / k

t= ln( [A]₀/[A] ) / k

we know that for first order reaction rate constant k = 0.693/ t ½

k= 0.693 / 2.5 hr

= 0.2772 / hr

according to the problem, time is required for 95% of the substance to decompose means at that time 5% substance remained at the time t= 5% or 5/100 of its initial value,

Therefore

ln[A] = ln[A]₀ - k*t

ln[0.05] = ln[1.00]₀ - 0.2772 / hr*t

-2.996= 0-0.2772 / hr*t

t =10.81 hr