Home
PYQ PracticeFree
JEE Rank Predictor
Answer Key

Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsRate of Reactions/Rate LawMedium2 minQB

ChemistryMediumsingle choice

Passage / Comprehension

The gaseous reaction: $n A_{(g)} ⟶ \frac{1}{2} n B_{(g)}$ is first order with respect to A. The rate constant of the reaction is $k$ . The reaction is studied at a constant pressure and temperature. Initially, the moles of A were ' $a$ ' and no B was present.

What will be the concentration of A at time t , if $n_{1} = 1$ and $n_{2} = 2$ ?

Options:

Answer:

Solution:

$[A] = \frac{( a - n _{1} x )}{V _{F}} = \frac{ae ^{- n_{1} Kt}}{V _{0} [ \frac{n _{2}}{n _{1}} - ( \frac{n _{2}}{n _{1}} - 1 ) e ^{- n_{1} Kt} ]}$

$[A] = [A]_{0} \frac{e ^{- K t}}{[ 2 - e ^{- K t} ]}$

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

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...