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Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsArrhenius EquationEasy2 minPYQ_2019

ChemistryEasysingle choice

What is the activation energy $(kJ / mol)$ for a reaction if its rate constant doubles when the temperature is raised from $300 K$ to $400 K$ ? $(R = 8.314 J mol^{- 1} K^{- 1})$

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Answer:

Solution:

$lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}}) lo g \frac{2}{1} = \frac{E _{a}}{2.303 \times 8.314} (\frac{1}{300} - \frac{1}{400}) 0.3010 = \frac{E _{a}}{2.303 \times 8.314} (\frac{100}{300 \times 400}) \Rightarrow E_{a} = 0.3010 \times 2.303 \times 8.314 \times 3 \times 400 = 6.916 kJ mol^{- 1}$

Stream:AIIMSSubject:ChemistryTopic:Chemical KineticsSubtopic:Arrhenius Equation

⏱ 2mℹ️ Source: PYQ_2019

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