The half-life (t1/2) for a chemical reaction is the time it takes for the concentration of reactant to decrease to one-half its initial concentration. For a first-order reaction, the half-life is a constant at a given temperature. So the half-life is also the amount of time it takes for the concentration to change from one-half to one-fourth, and from one-fourth to one-eighth, etc. If the half-life for a first-order reaction is 12.0 minutes, what percentage of reactant remains unreacted after 48.0 minutes?

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mauricioalessandrell mauricioalessandrell · Answer 1 · 2019-09-12T08:50:37+02:00

Answer:

The percentage of reactant that remains unreacted is 6.25%

Explanation:

Let´s call [A]₀ the initial concentration of reactant A

In the first 12.0 minutes the concentration of A will drop to [A]₀/2

In 24.0 minutes the concentration of A will be reduced by half again

( [A]₀/2 / 2) = [A]₀ / 4

In 36.0 min the concentration of A will be ([A]₀/4 / 2) = [A]₀ / 8

And after 48.0 min the concentration of A will be ([A]₀/8 /2) = [A]₀/16

If [A]₀ is the 100%, [A]₀ / 16 will be (100 / 16) = 6.25%

Mathematically, you can express the concentration of A in this way:

[A] = [A]₀ / 2ⁿ

where:

[A] = concentration of A

[A]₀ = initial concentration of A

n = t / t(1/2), where t = time and t(1/2) = half-life

In this case:

[A] = [A]₀ / 2^(48 min / 12 min)

[A] = [A]₀ / 2⁴

[A] = [A]₀ / 16