Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsZero Order ReactionsMedium2 minPYQ_2012

ChemistryMediumsingle choice

The activation energy for a reaction which doubles the rate when the temperature is raised from $298 K$ to $308 K$ is

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Answer:

Solution:

Activation energy can be calculated from the equation. $\frac{lo g K _{2}}{lo g K _{1}} = \frac{- E _{a}}{2.303 R} (\frac{1}{T _{2}} - \frac{1}{T _{1}}) Given \frac{lo g K _{2}}{lo g K _{1}} = 2 T_{2} = 308; T_{1} = 298 ∴ lo g 2 = \frac{- E _{a}}{2.303 \times 8.314} (\frac{1}{308} - \frac{1}{298}) E_{a} = 52.9 kJ mol^{- 1}$

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:Zero Order Reactions

⏱ 2mℹ️ Source: PYQ_2012

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