For an n^{th} order reaction where n

Question

(B) For an $n^{th}$ order reaction,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}} \ri

Vedclass · Accepted Answer

$t_{100\%} = \frac{1}{1-n} 	imes \frac{[A]_0^{1-n}}{K}$

Vedclass · Answer

(B) For an $n^{th}$ order reaction,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 $100\%$ completion,the concentration of the reactant $[A]_t$ becomes $0$. Substituting $[A]_t = 0$ into the equation: $K = \frac{1}{t_{100\%}(n-1)} \left[ \frac{1}{0^{n-1}} - \frac{1}{[A]_0^{n-1}} ight]$ Since $n  0$. Then $n-1 = -m$. $K = \frac{1}{t_{100\%}(-m)} \left[ \frac{1}{0^{-m}} - \frac{1}{[A]_0^{-m}} ight]$ $K = \frac{1}{t_{100\%}(-m)} \left[ 0 - [A]_0^m ight]$ $K = \frac{[A]_0^m}{t_{100\%} 	imes m}$ $t_{100\%} = \frac{[A]_0^m}{K 	imes m}$ Substituting $m = 1-n$ back: $t_{100\%} = \frac{[A]_0^{1-n}}{K(1-n)}$

$A / mol \ L^{-1}$	$0.20$	$0.20$	$0.40$
$B / mol \ L^{-1}$	$0.30$	$0.10$	$0.05$
$r_0 / mol \ L^{-1} \ s^{-1}$	$5.07 \times 10^{-5}$	$5.07 \times 10^{-5}$	$1.43 \times 10^{-4}$

Expt	$\frac{-d[NO]}{dt} \ (mol \ L^{-1} \ s^{-1})$	$[NO] \ (mol \ L^{-1})$	$[H_2] \ (mol \ L^{-1})$
$1$	$4.8 \times 10^{-5}$	$1 \times 10^{-2}$	$1 \times 10^{-3}$
$2$	$43.2 \times 10^{-5}$	$3 \times 10^{-2}$	$1 \times 10^{-3}$
$3$	$86.4 \times 10^{-5}$	$3 \times 10^{-2}$	$2 \times 10^{-3}$

For an $n^{th}$ order reaction where $n < 1$,what is the expression for the time required for $100\%$ completion $(t_{100\%})$?

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