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The ratio of $\large\frac{t_{1/2}}{t_{3/4}}$ for $n^{th}$ order is

$\begin{array}{1 1}(a)\;function\; of\;a_o\;alone\\(b)\;function \;of\;n\;alone\\(c)\;function\;of\;a_o\;and\;n\\(d)\;none\end{array}$

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For $n^{th}$ order reaction
$nA\rightarrow$ products
$-\large\frac{d[A]}{dt}=$$k_n[A]^n$
$\int-\large\frac{d[A]}{[A]^n}=$$k_n\int dt$
$[A_o]=a_o$
$t=\large\frac{1}{k_n(n-1)a^{n-1}}$$+c$
$t=\large\frac{1}{k_n(n-1)}$$\bigg[\large\frac{1}{a^{n-1}}-\frac{1}{a_o^{n-1}}\bigg]$
When $t=t_{1/2},$$a=\large\frac{a_o}{2}$
So $k_nt_{1/2}=\large\frac{1}{n-1}\big(\frac{1}{a_o}\big)^{n-1}$$(2^{n-1}-1)$
When $t=t_{3/4},$$a=\large\frac{3a_o}{4}$
So $k_nt_{3/4}=\large\frac{1}{n-1}\big(\frac{1}{a_o}\big)^{n-1}$$(\big(\large\frac{4}{3}\big)^{n-1}$$-1)$
$\large\frac{t_{1/2}}{t_{3/4}}=\frac{2^{n-1}-1}{(4/3)^{n-1}-1}$
Hence (b) is the correct answer.
answered Dec 19, 2013 by sreemathi.v
 

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