What is the time taken for the 3^{rd} half-li | vedclass

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(C) For a reaction of order $n$,the half-life is given by $t_{1/2} \propto \frac{1}{a^{n-1}}$,where $a$ is the initial concentration. For a second-ord

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$80$

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(C) For a reaction of order $n$,the half-life is given by $t_{1/2} \propto \frac{1}{a^{n-1}}$,where $a$ is the initial concentration. For a second-order reaction $(n=2)$,$t_{1/2} \propto \frac{1}{a}$. Let the initial concentration be $a_0$. After the $1^{st}$ half-life,the concentration becomes $a_1 = \frac{a_0}{2}$. After the $2^{nd}$ half-life,the concentration becomes $a_2 = \frac{a_1}{2} = \frac{a_0}{4}$. After the $3^{rd}$ half-life,the concentration becomes $a_3 = \frac{a_2}{2} = \frac{a_0}{8}$. Since $t_{1/2} = \frac{1}{k \cdot a}$,the $n^{th}$ half-life is $t_{n} = \frac{1}{k \cdot a_{n-1}}$. Given $t_1 = 20 \ s = \frac{1}{k \cdot a_0}$,then $k = \frac{1}{20 \cdot a_0}$. The $3^{rd}$ half-life is $t_3 = \frac{1}{k \cdot a_2} = \frac{1}{k \cdot (a_0/4)} = \frac{4}{k \cdot a_0} = 4 	imes 20 \ s = 80 \ s$.

Rate $(mol \ L^{-1} \ s^{-1})$	$[A]$ $(mol \ L^{-1})$	$[B]$ $(mol \ L^{-1})$
$0.10$	$20$	$0.5$
$0.40$	$x$	$0.5$
$0.80$	$40$	$y$

What is the time taken for the $3^{rd}$ half-life of a second-order decomposition reaction,given that its first half-life is $20 \ s$ (in $s$)?

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What is the time taken for the $3^{rd}$ half-life of a second-order decomposition reaction,given that its first half-life is $20 \ s$ (in $s$)?

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For a chemical reaction $A + B \rightarrow \text{Product}$,the order is $1$ with respect to $A$ and $B$.Rate $(mol \ L^{-1} \ s^{-1})$$[A]$ $(mol \ L^{-1})$$[B]$ $(mol \ L^{-1})$$0.10$$20$$0.5$$0.40$$x$$0.5$$0.80$$40$$y$What is the value of $x$ and $y$?

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