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**Hint:**The convergences of reactants and the pace of a reaction with these fixations, we will view from the outset and second request reactions just as half-life. For a Pseudo-\[{n^{th}}\]-Order Reaction, the reaction rate consistent \[k\] is supplanted by the apparent reaction rate steady \[k'\].

**Complete step by step answer:**

If the reaction isn't worked out explicitly to show an estimation of \[{\nu _A}\], the worth is thought to be \[1\] and doesn't appear in these conditions.

One approach to determine the request \[n\] and to get an estimated incentive for \[k\] or \[k'\], is with the strategy for half-lives. The half-life \[{t_{1/2}}\]is characterized as the time needed for the initial fixation to be divided: \[{\left[ A \right]_{{t_{1/2}}}}{\text{ }} = {\text{ }}{\left[ A \right]_t} = 0/2\]

This shows that there is an overall relationship for all estimations of \[n\] (counting \[n{\text{ }} = {\text{ }}1\]) between the initial focus and the half-life for a reaction concentrated with various initial fixations at a similar temperature: \[\dfrac{{{t_{1/2}}}}{{{{\left[ A \right]}_t}}} = {0^{n{\text{ }} - {\text{ }}1}}{\text{ }} = {\text{ }}constant\].

On the off chance that the reaction is First-Order, the half-life won't change with fixation.

On the off chance that the request is more noteworthy than one, the half-life will diminish as the initial focus is expanded.

In the event that the request is short of what one, the half-life will increment as the initial fixation is expanded.

The request for the reaction (\[n\]) might be found by assurance of the half-lives for a reaction learned at two initial focuses. On the off chance that the subsequent focus is equivalent to multiple times the first: \[\dfrac{{{{\left( {{t_{1/2}}} \right)}_1}}}{{{{\left( {{t_{1/2}}} \right)}_2}}} = {\text{ }}{10^{n{\text{ }} - {\text{ }}1}}\].

Whenever n has been determined, \[k\] or \[k'\] can be determined from the relationship above,

\[\dfrac{{\left( {{2^{n{\text{ }} - {\text{ }}1}}{\text{ }} - {\text{ }}1} \right)}}{{{{\left[ A \right]}_t}}} = {0^{n{\text{ }} - {\text{ }}1}}{\text{ }} = {\text{ }}{\nu _A}^{\left( {n{\text{ }} - {\text{ }}1} \right)}{\text{ }}k{t_{1/2}}\].

So, the required answer is as follows…

\[{t_{1/2}} \propto \dfrac{{1}}{{{a^{n - 1}}}}\]

\[{t_{1/2}} = k\dfrac{{1}}{{{a^{n - 1}}}}\]

\[ln\;\;{t_{1/2}} = ln\;\;k - (n - 1)lo{g_e}a\]

**Hence, the correct option is (A).**

**Note:**

The best estimations of the reaction rate consistent (\[k\]) can be acquired with information taken in the centre third of the reaction (from\[{\left[ A \right]_t}{\text{ }} = {\text{ }}\left( {\dfrac{2}{3}} \right){\text{ }}{\left[ A \right]_t} = 0{\text{ }}to{\text{ }}{\left[ A \right]_t}{\text{ }} = {\text{ }}\left( {\dfrac{1}{3}} \right){\text{ }}{\left[ A \right]_t} = 0\]). Straight Least Squares relapse with \[Y{\text{ }} = {\text{ }}1/{\left[ A \right]_t}^{n{\text{ }} - {\text{ }}1}\]and \[X{\text{ }} = {\text{ }}t\] gives \[{\nu _A}^{\left( {n{\text{ }} - {\text{ }}1} \right)k}\] or \[{\nu _A}^{\left( {n{\text{ }} - {\text{ }}1} \right)k'}\]as the slant.

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