For a reaction, given below is the graph of ln k v s T 1 . The activation energy for the reaction is equal to________ cal mol -1 . (Nearest integer). (Given : R = 2 cal K -1 mol -1 )

k = A e − E a / RT ln k = RT − Ea + ln A Slope = R − E a = 5 − 20 E a = 4 R = 8 Cal / mol

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Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsArrhenius EquationMedium2 minQB

ChemistryMediuminteger

For a reaction, given below is the graph of $ln k v s \frac{1}{T}$ . The activation energy for the reaction is equal to________ cal mol^-1 . (Nearest integer). (Given : R = 2 cal K^-1 mol^-1 )

Answer:

Solution:

$k = A e^{- E a / RT}$

$ln k = \frac{- Ea}{RT} + ln A$

Slope $= \frac{- E a}{R} = \frac{- 20}{5}$

$E_{a} = 4 R = 8 Cal / mol$

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:Arrhenius Equation

⏱ 2mℹ️ Source: QB

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Chemistry - Chemical Kinetics Question with Solution | TestHub

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