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For the reaction $A\rightarrow B$ ,the rate law expression is $-\large\frac{d[A]}{dt}$$=k[A]^{3/2}$. If the initial concentration of $A$ is $[A]_0$, which of the following is true?

$\begin{array}{1 1}(a)\;k=\large\frac{2}{t}\normalsize(A_o^{-1/2}-A^{-1/2})\\(b)\;\;K=\large\frac{2}{t}\normalsize(A_o^{1/2}-A^{1/2})\\(c)\;t_{1/2}=\large\frac{k}{2[A_o]^{1/2}}\\(d)\;t_{1/2}=\large\frac{ln\;2}{k}\end{array}$

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Answer: $ t_{1/2}=\large\frac{2}{k}\bigg(\frac{2^{1/2}-1}{[A_o]^{1/2}}\bigg)$
By differential rate law, $\large\frac{d[A]}{dt}=$$-k[A]^{3/2}$
Integrating, $\large \int_{[A_0]}^{[A_t]}$$\large\frac{d[A]}{[A]^3/2}$$=-\large\int_0^t $$kdt$
$\Rightarrow 2([A_o]^{-1/2}-[A_t]^{1/2})=kt$
Simplifying for $t = t_{1/2}=\large\frac{2}{k}\bigg(\frac{2^{1/2}-1}{[A_o]^{1/2}}\bigg)$
answered Dec 16, 2013 by sreemathi.v
edited Jul 26, 2014 by balaji.thirumalai

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