The inactivation rate of a viral preparation | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $106 	imes 10^{-3} \ min^{-1}$.

Similar Questions

The rate equation for a first-$order$ reaction is given by $R = R_0 e^{-kt}$. $A$ straight line with positive slope is obtained by plotting which of the following? ($R_0 =$ Initial concentration of the reactant,$R =$ concentration of the reactant at time $t$)

The half-life of a substance in a certain enzyme-catalysed reaction is $138 \; s$. The time required for the concentration of the substance to fall from $1.28 \; mg \; L^{-1}$ to $0.04 \; mg \; L^{-1}$ is ....... $s$.

In a first order reaction,the concentration of the reactant is reduced to $25\%$ in one hour. The half-life period of the reaction is: (in $hr$)

If the rate constant of a reaction is $0.03 \ s^{-1}$,how much time does it take for $7.2 \ mol \ L^{-1}$ concentration of the reactant to get reduced to $0.9 \ mol \ L^{-1}$ (in $s$)? (Given: $\log 2 = 0.301$)

Which equation represents the $1/4^{th}$ life for a first-order reaction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module