Rate constant of a reaction with a virus is 3.3 × 1 0 − 4 s − 1 . Time required for the virus to become 75% inactivated is :

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ChemistryChemical KineticsFirst Order ReactionsMedium2 minQB

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Rate constant of a reaction with a virus is $3.3 \times 1 0^{- 4} s^{- 1}$ . Time required for the virus to become $75%$ inactivated is :

Answer:

Solution:

$T_{50} = \frac{0.693}{k} = 2100 s = 35 min u t e$

Therefore, the time taken for 75% inactivation of a virus = $2 \times T_{50} = 70 min$ .

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:First Order Reactions

⏱ 2mℹ️ Source: QB

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