For the reaction: 3 A ( g ⁡ ) → 2 B ( g ⁡ ) , the rate of formation of ' B ' at 298 K is represented as ln d B dt = − 4 .606 + 2 ln A . The order of reaction is

ln ⁡ d B d t = - 4.606 + 2 ln ⁡ [ A ] 2.303 log d B dt = − 4.606 + 2 × 2.303 log A log d B dt = − 4.606 2.303 + 2 log A log d B dt = − 2 + 2 log A log d B dt = log 10 − 2 + log A 2 log d B dt = log 10 − 2 A 2 d B d t = 1 0 - 2 A 2 Compare with d B d t = k A m d B d t = 1 0 - 2 A 2 Order = 2

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ChemistryChemical KineticsMiscellaneous/MixedEasy2 minPYQ_2020

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For the reaction: $3 A_{(g)} \to 2 B_{(g)}$ , the rate of formation of ' $B$ ' at $298 K$ is represented as $ln (\frac{d [B]}{dt}) = - 4 .606 + 2 ln [A]$ . The order of reaction is

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Answer:

Solution:

$\ln (\frac{d [B]}{d t}) = - 4.606 + 2 \ln [A]$
$2.303 log (\frac{d [B]}{dt}) = - 4.606 + 2 \times 2.303 log [A]$
$log (\frac{d [B]}{dt}) = \frac{- 4.606}{2.303} + 2 log [A]$
$log (\frac{d [B]}{dt}) = - 2 + 2 log [A]$
$log (\frac{d [B]}{dt}) = log 10^{- 2} + log {[A]}^{2}$
$log (\frac{d [B]}{dt}) = log (10^{- 2} {[A]}^{2})$
$(\frac{d [B]}{d t}) = (1 0^{- 2} {[A]}^{2})$
Compare with $(\frac{d [B]}{d t}) = k {[A]}^{m}$
$(\frac{d [B]}{d t}) = (1 0^{- 2} {[A]}^{2})$
Order = 2

Stream:NTA_ABHYASSubject:ChemistryTopic:Chemical KineticsSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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