Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsZero Order ReactionsMedium2 minPYQ_2020

ChemistryMediumnumerical

The rate of a reaction decreased by $3.555$ times when the temperature was changed from $40 ° C to 30 ° C$ . the activation energy (in $kJ {mol}^{- 1}$ ) of the reaction is ________ .

Answer:

100.00

Solution:

$\log (\frac{k_{2}}{k_{1}}) = \frac{Ea}{2 .303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$\log (3 .555) = \frac{Ea}{2 .303 R} [\frac{1}{303} - \frac{1}{313}]$

$1 .268 \times 8 .314 \times 303 = 10 Ea$

So, $Ea = 100 kJ$

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

⏱ 2mℹ️ Source: PYQ_2020

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