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The inactivation of a viral preparation is found to be first order reaction.If in beginning $1.5\%$ of virus is inactivated then calculate time required for 50V inactivation.

$\begin{array}{1 1}(a)\;47min\\(b)\;42.69min\\(c)\;45min\\(d)\;46.17min\end{array}$

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The rate law
$k_1=\large\frac{1}{[A]}\times \frac{d[A]}{dt}$
For a finite,small change
$k_1=-\large\frac{1}{[A]}\times \frac{\Delta [A]}{\Delta t}$
Since in beginning [A] is not changing appreciably ,we have
$\large\frac{\Delta A}{A}=$$1.5%$(per minute)
$k_1=\large\frac{0.015}{606}$$=2.5\times 10^{-4}s^{-1}$
$t_{1/2}=\large\frac{0.693}{k_1}=\frac{0.693}{2.5\times 10^{-4}s^{-1}}$
$\;\;\;\;\;=2.77\times 10^3s$
Hence (d) is the correct answer.
answered Dec 19, 2013 by sreemathi.v

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