Respuesta :
Explanation:
For the given reaction [tex]2NO_{2} \rightarrow 2NO + O_{2}[/tex]
Now, expression for half-life of a second order reaction is as follows.
[tex]t_{1/2} = \frac{1}{[A_{0}]k}[/tex] ....... (1)
Second half life of this reaction will be [tex]t_{1/4}[/tex]. So, expression for this will be as follows.
[tex]t_{1/4}[/tex] = [tex]\frac{1}{k} [\frac{1}{[A]_{f}} - \frac{1}{[A_{0}]}][/tex] ...(2)
where [tex][A]_{f}[/tex] is the final concentration that is, [tex]\frac{[A]_{0}}{4}[/tex] here and [tex][A]_{i}[/tex] is the initial concentration.
Hence, putting these values into equation (2) formula as follows.
[tex]t_{1/4}[/tex] = [tex]\frac{1}{k} [\frac{4}{[A]_{0}} - \frac{1}{[A_{0}]}][/tex]
= [tex]\frac{3}{[A_{0}]k}[/tex] ...... (3)
Now, dividing equation (3) by equation (1) as follows.
[tex]\frac{t_{1/4}}{t_{1/2}}[/tex] = [tex]\frac{3}{[A_{0}]k} \times [A_{0}]k[/tex]
= 3
or, [tex]t_{1/4}[/tex] = 3 [tex]t_{1/2}[/tex]
Thus, we can conclude that one would expect the second half-life of this reaction to be three times the first half-life of this reaction.
