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ChemistryChemical KineticsArrhenius EquationMedium2 minQB

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In a reaction carried out at $400 K, 0.0001%$ of the total number of molecules are in activated state. The energy of activation of reaction is -

Answer:

Solution:

Fraction of active molecule $= e^{- \frac{E _{a}}{RT}} = 1 0^{- 6}$

$- \frac{E _{a}}{RT} = ln 1 0^{- 6} = 2.303 lo g 1 0^{- 6}$

$E_{a} = 6 \times 2.303 \times 2 \times 1 0^{- 3} \times 400 = 11.05 kcal$

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:Arrhenius Equation

⏱ 2mℹ️ Source: QB

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