Question

Initial concentration of reactant for $$n^{th}$$order reaction is '$$a$$'. Which of the following relations is correct about $$t_{1/2}$$​ of the reaction?
(A)$$ln{t}_{1/2}=ln\left(constant\right)-\left(n-1\right)lo{g}_{e}a$$
(B)$$ln{t}_{1/2}=lnn+ln\left(constant\right)-lna$$
(C)$$t_{1/2}lnn=ln\left(constant\right)+ln{a}_0$$ ​
(D)$$ln{t}_{1/2}=nln{a}_0$$

Answer 1

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 $$k^{\prime }$$.

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 $$k^{\prime }$$, 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: $${\left[A\right]}_{t_{1/2}}={\left[A\right]}_t=0/2$$
This shows that there is an overall relationship for all estimations of $$n$$ (counting $$n=1$$) between the initial focus and the half-life for a reaction concentrated with various initial fixations at a similar temperature: $$\frac{t_{1/2}}{{\left[A\right]}_t}=0^{n-1}=constant$$.
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: $$\frac{{\left(t_{1/2}\right)}_1}{{\left(t_{1/2}\right)}_2}={10}^{n-1}$$.
Whenever n has been determined, $$k$$ or $$k^{\prime }$$ can be determined from the relationship above,
$$\frac{\left(2^{n-1}-1\right)}{{\left[A\right]}_t}=0^{n-1}={{\nu }_A}^{\left(n-1\right)}k{t}_{1/2}$$.
So, the required answer is as follows…
$$t_{1/2}\propto \frac{1}{a^{n-1}}$$
$$t_{1/2}=k\frac{1}{a^{n-1}}$$
$$ln{t}_{1/2}=lnk-(n-1)lo{g}_{e}a$$
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$${\left[A\right]}_t=\left(\frac{2}{3}\right){\left[A\right]}_t=0to{\left[A\right]}_t=\left(\frac{1}{3}\right){\left[A\right]}_t=0$$). Straight Least Squares relapse with $$Y=1/{{\left[A\right]}_t}^{n-1}$$and $$X=t$$ gives $${{\nu }_A}^{\left(n-1\right)k}$$ or $${{\nu }_A}^{\left(n-1\right)k^{\prime }}$$as the slant.