The value of frac{t_{0.875}}{t_{0.50}} for n^ | vedclass

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(C) For an $n^{th}$ order reaction $(n 
eq 1)$,the integrated rate law is given by: $K = \frac{1}{t(n - 1)} \left[ \frac{1}{A_t^{n - 1}} - \frac{1}{A

Vedclass · Accepted Answer

$\frac{8^{n - 1} - 1}{2^{n - 1} - 1}$

Vedclass · Answer

(C) For an $n^{th}$ order reaction $(n 
eq 1)$,the integrated rate law is given by: $K = \frac{1}{t(n - 1)} \left[ \frac{1}{A_t^{n - 1}} - \frac{1}{A_0^{n - 1}} ight]$ For $t_{0.5}$ (half-life),$A_t = \frac{A_0}{2}$,so: $K t_{0.5} = \frac{1}{n - 1} \left[ \frac{2^{n - 1} - 1}{A_0^{n - 1}} ight]$ For $t_{0.875}$,$A_t = A_0 - 0.875 A_0 = 0.125 A_0 = \frac{A_0}{8}$,so: $K t_{0.875} = \frac{1}{n - 1} \left[ \frac{8^{n - 1} - 1}{A_0^{n - 1}} ight]$ Dividing the two expressions: $\frac{t_{0.875}}{t_{0.5}} = \frac{8^{n - 1} - 1}{2^{n - 1} - 1}$

$Observation$	$[A] \ (mol \ L^{-1})$	$[B] \ (mol \ L^{-1})$	$Rate \ (mol \ L^{-1} \ sec^{-1})$
$1$	$0.1$	$0.1$	$2 \times 10^{-3}$
$2$	$0.2$	$0.1$	$4 \times 10^{-3}$
$3$	$0.1$	$0.2$	$8 \times 10^{-3}$

The value of $\frac{t_{0.875}}{t_{0.50}}$ for $n^{th}$ order reaction is

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$Observation$ $[A] \ (mol \ L^{-1})$ $[B] \ (mol \ L^{-1})$ $Rate \ (mol \ L^{-1} \ sec^{-1})$

$1$ $0.1$ $0.1$ $2 \times 10^{-3}$

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$3$ $0.1$ $0.2$ $8 \times 10^{-3}$

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