Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsFirst Order ReactionsMedium2 minPYQ_2020

ChemistryMediumnumerical

If $75 %$ of a first order reaction was completed in $90$ minutes, $60 %$ of the same reaction would be completed in approximately (in minutes) _____
(Take: $\log 2 = 0.30; \log 2.5 = 0.40$ )

Answer:

60.00

Solution:

$T = \frac{2 .303}{K} \log [\frac{100}{100 - x %}]$

$T_{75 %} = \frac{2 .303}{K} \log [\frac{100}{25}] = 90$

$T_{60 %} = \frac{2 .303}{K} \log [\frac{100}{40}]$

$\frac{T_{75 %}}{T_{60 %}} = \frac{2 \log 2}{\log 2 .5}$ $\Rightarrow$ $\frac{90}{T_{60 %}} = \frac{2 \times 0 .3}{0 .4}$

$T_{60 %} = \frac{90 \times 4}{6} = 60 \min$ .

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

⏱ 2mℹ️ Source: PYQ_2020

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