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Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsSecond & nth Order ReactionsHard2 minPYQ_2007

ChemistryHardsingle choice

Decay constant of a radioactive substance is $69.3 sec^{- 1}$ , find $t_{1/16}$ of the same substance.

Options:

Answer:

Solution:

$1 ⟶ t_{1/2} \frac{1}{2} ⟶ t_{1/2} \frac{1}{4} ⟶ t_{1/2} \frac{1}{8} ⟶ t_{1/2} \frac{1}{16}$ So, $t_{1/16} = 4 t_{1/2}$ $= 4 \times \frac{0.693}{λ} = 4 \times \frac{0.693}{69.3} = 4 \times 1 0^{- 2} sec .$

Stream:AIIMSSubject:ChemistryTopic:Chemical KineticsSubtopic:Second & nth Order Reactions

⏱ 2mℹ️ Source: PYQ_2007

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