Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsRate of Reactions/Rate LawEasy2 minPYQ_2020

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In the reaction $2 N_{2} O_{5 (g)} ⟶ 4 NO_{2 (g)} + O_{2 (g)}$ the rate of formation of $NO_{2 (g)}$ and $O_{2 (g)}$ are in the ratio of

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Solution:

By Differential Rate law; $\frac{- 1}{2} \frac{d ( N _{2} O _{5} )}{d t} = \frac{1}{4} \frac{d ( N O _{2} )}{d t} = \frac{d ( O _{2} )}{d t}$ d $\frac{Rate of formation of NO _{2}}{Ratc of formation of O _{2}} = \frac{d ( NO _{2} ) / d t}{d ( O _{2} / d t} = 4 : 1$

Stream:MHT_CETSubject:ChemistryTopic:Chemical KineticsSubtopic:Rate of Reactions/Rate Law

⏱ 2mℹ️ Source: PYQ_2020

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