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(D) For a reaction of order $n 
eq 1$,the integrated rate law is $t = \frac{1}{k(n-1)} [\frac{1}{C_t^{n-1}} - \frac{1}{C_0^{n-1}}]$.For $t_{1/2}$,$C_

Vedclass · Accepted Answer

$t_{7/8} = t_{1/2} \ [2^{3(n-1)} - 1] / [2^{n-1} - 1]$

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(D) For a reaction of order $n 
eq 1$,the integrated rate law is $t = \frac{1}{k(n-1)} [\frac{1}{C_t^{n-1}} - \frac{1}{C_0^{n-1}}]$.For $t_{1/2}$,$C_t = C_0 / 2$:$t_{1/2} = \frac{1}{k(n-1)} [\frac{2^{n-1}}{C_0^{n-1}} - \frac{1}{C_0^{n-1}}] = \frac{2^{n-1} - 1}{k(n-1)C_0^{n-1}}$.For $t_{7/8}$,$C_t = C_0 / 8 = C_0 / 2^3$:$t_{7/8} = \frac{1}{k(n-1)} [\frac{8^{n-1}}{C_0^{n-1}} - \frac{1}{C_0^{n-1}}] = \frac{(2^3)^{n-1} - 1}{k(n-1)C_0^{n-1}} = \frac{2^{3(n-1)} - 1}{k(n-1)C_0^{n-1}}$.Dividing $t_{7/8}$ by $t_{1/2}$:$\frac{t_{7/8}}{t_{1/2}} = \frac{2^{3(n-1)} - 1}{2^{n-1} - 1}$.Therefore,$t_{7/8} = t_{1/2} \ [\frac{2^{3(n-1)} - 1}{2^{n-1} - 1}]$.

Initial $[A_2]$	Initial $[B_2]$	$R.O.R. (r) \ M s^{-1}$
$0.2$	$0.2$	$0.04$
$0.1$	$0.4$	$0.04$
$0.2$	$0.4$	$0.08$

Which of the following options correctly represents the relationship between $t_{7/8}$ and $t_{1/2}$ where $t_{7/8}$ represents the time required for the concentration to become $1/8$ of the original for a reaction of order $n$?

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