Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsZero Order ReactionsMedium2 minPYQ_2020

ChemistryMediumnumerical

A sample of milk splits after $60 m i n$ . at $300 K$ and after $40 m i n$ . at $400 K$ when the population of Iactobacillus acidophilus in it doubles. The activation energy (in $kJ / mol$ ) for this process is closest to ______________.
(Given, $R = 8.3 J m o l^{- 1} K^{- 1}, l n (\frac{2}{3}) = 0.4, e^{- 3} = 4.0$ )

Answer:

3.98

Solution:

$[\ln] \frac{k_{2}}{k_{1}} = \frac{Ea}{R} [\frac{1}{T_{1}} \frac{1}{T_{2}}]$
$\ln (\frac{60}{40}) = \frac{Ea}{8 .3} \times \frac{100}{400 \times 300}$
$\ln (\frac{3}{2}) \times 8 .3 \times 1200 = E_{a}$
$⟹ E_{a} = 0 .4 \times 8 .3 \times 1200$
$⟹ E_{a} = 3984 \frac{J}{mol} .$
$⟹ E_{a} = 3 .984 k \frac{J}{mol} .$

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

⏱ 2mℹ️ Source: PYQ_2020

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