A first-order reaction has a half-life of 27.4 s . How long does it take for the concentration of the reactant in the reaction to fall to one-fourth of its initial value?

Respuesta :

Answer:

54.8 seconds

Explanation:

Given

Reaction is first order

half life is 27.4 seconds

Half life is defined as the time required to complete the reaction 50% of its initial value or the time in which the concentration of reactant reaches to 50% of its initial value.

It will take one half life to reach concentration half of the initial value.

It will take another half life to reach concentration half of the half of the initial concentration which is one fourth of the initial concentration.

So the total time taken to reach one fourth of the initial value = two half lives = 27.4 + 27.4 = 54.8 seconds.

Another approach to solve the same problem is :

Let us calculate the rate constant of the reaction from its half life:

rate constant = k = 0.693 / half life = 0.693 /27.4 = 0.0253

The rate expression for first order reaction is:

[tex]time=\frac{1}{k}[ln\frac{A_{0}}{A_{t}}] =\frac{1}{0.0253}ln[\frac{100}{25}] =54.8 seconds[/tex]

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