The inactivation rate of a viral preparation | vedclass

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(A) Since the unit of the rate constant is $min^{-1}$,it follows first-order kinetics.The integrated rate equation is $K 	imes t = 2.303 \log(A_0 / A

Vedclass · Accepted Answer

$106$

Vedclass · Answer

(A) Since the unit of the rate constant is $min^{-1}$,it follows first-order kinetics.The integrated rate equation is $K 	imes t = 2.303 \log(A_0 / A_t)$.Given that $10 \%$ of the virus is inactivated in $1 \ min$,we have $A_0 = 100$ and $A_t = 100 - 10 = 90$.Substituting the values: $K 	imes 1 = 2.303 	imes \log(100 / 90)$.$K = 2.303 	imes (\log 10 - \log 9) = 2.303 	imes (1 - 2 \log 3)$.$K = 2.303 	imes (1 - 2 	imes 0.477) = 2.303 	imes (1 - 0.954) = 2.303 	imes 0.046 = 0.105938$.$K = 105.938 	imes 10^{-3} \ min^{-1}$.Rounding to the nearest integer,we get $106 	imes 10^{-3} \ min^{-1}$.

$t/\text{min}$	$[A]/M$
$0$	$0.6500$
$x$	$0.0650$
$20$	$0.00065$

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