A reactant A forms two products A → k 1 B; Activation energy E a 1 A → k 2 C; Activation energy E a 2 If E a 2 = 2 E a 1 , then k 1 and k 2 will be related as

k 2 = A e - E a 2 R T , k 1 = A e - E a 1 R T So, k 2 k 1 = e − 2E a 1 +E a 1 RT Since, E a 2 = 2 E a 1 k 2 = k 1 e − E a 1 RT

Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasysingle choice

A reactant $(A)$ forms two products
$A \overset{k_{1}}{\to} B;$ Activation energy $E_{a_{1}}$
$A \overset{k_{2}}{\to} C;$ Activation energy $E_{a_{2}}$
If $E_{a_{2}} = 2 E_{a_{1}},$ then $k_{1}$ and $k_{2}$ will be related as

Options:

Answer:

Solution:

$k_{2} = A e^{- \frac{E_{a_{2}}}{R T}}, k_{1} = A e^{- \frac{E_{a_{1}}}{R T}}$
So, $\frac{k_{2}}{k_{1}} = e^{\frac{- {2E}_{a_{1}} {+E}_{a_{1}}}{RT}}$
Since, $E_{a_{2}} = 2 E_{a_{1}}$
$k_{2} = k_{1} e^{\frac{- E_{a_{1}}}{RT}}$

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

⏱ 2mℹ️ Source: PYQ_2020

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