Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsArrhenius EquationEasy2 minPYQ_2024

ChemistryEasysingle choice

Calculate the activation energy of a reaction, whose rate constant doubles on raising the temperature from 300 K to 600 K .

Options:

Answer:

Solution:

$lo g (\frac{k _{2}}{k _{1}}) = \frac{E _{a}}{2.303 R} [\frac{1}{T _{1}} - \frac{1}{T _{2}}]$ or, $lo g (\frac{2 k}{k}) = \frac{E _{a}}{2.303 \times 8.3} \times \frac{[ 600 - 300 ]}{300 \times 600}$ $E_{a} \approx 3.45 kJ / mol$

Stream:BITSATSubject:ChemistryTopic:Chemical KineticsSubtopic:Arrhenius Equation

⏱ 2mℹ️ Source: PYQ_2024

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