t 1/4 can be taken as the time taken for the concentration of a reactant to drop to 4 3 of its initial value. If the rate constant for a first order reaction is K , the t 1/4 can be written as

t 1/4 = K 2.303 lo g 1 − 4 1 1 = K 0.29

Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsFirst Order ReactionsMedium2 minPYQ_2005

ChemistryMediumsingle choice

$t_{1/4}$ can be taken as the time taken for the concentration of a reactant to drop to $\frac{3}{4}$ of its initial value. If the rate constant for a first order reaction is $K$ , the $t_{1/4}$ can be written as

Options:

Answer:

Solution:

$t_{1/4} = \frac{2.303}{K} lo g \frac{1}{1 - \frac{1}{4}} = \frac{0.29}{K}$

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

⏱ 2mℹ️ Source: PYQ_2005

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