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  • In inactivation rate of a viral prepartion is proportional to the amount of virus. In the first minute after preparation, <img alt="68 \%" src="https://entrancecorner.oncodecogs.com/gif.latex?68%20

In inactivation rate of a viral prepartion is proportional to the amount of virus. In the first minute after preparation, 68 \% of the virus is inactivated. The rate constant for viral inactivation is what \mathrm{min^{-1}}

Option: 1

921.2 \times 10^{-3}


Option: 2

622.6 \times 10^{-1}


Option: 3

725.5 \times 10^{-2}


Option: 4

334.6 \times 10^{-1}


Answers (1)

best_answer

By the unit of rate constant is \mathrm{min ^{-1}} so it must be a first order reaction in 1 Min 68 % in activated so taking

\mathrm{A_0=100 \quad A_t=32} in \mathrm{1 \mathrm{~min}}

So; k \times 1=2.303 \times \log\frac{50}{16}

\mathrm{k=2.303 \times \log 50-\log (2)^4}

\mathrm{\begin{aligned} & k=2.303 \times(\log 50-4 \log 2) \\\\ & k=2.303(1.6-4 \times 0.30) \\\\ & k=2.303(1.6-1.2) \\\\ & k=0.9212 \\\\ & k=921.2 \times 10^{-3} \end{aligned}}

 

Posted by

Divya Prakash Singh

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