For the reaction A → B , the rate constant k (in s - 1 ) is given by log 10 k = 20 . 35 - 2 . 47 × 10 3 T The energy of activation in kJ mol - 1 is ________ . (Nearest integer) [Given : R = 8 . 314 J K - 1 mol - 1 ]

log 10 k = logA - E a 2 . 303 RT log 10 k = 20 . 35 - 2 . 47 × 10 3 T E a 2 . 303 R = 2 . 47 × 10 3 E a = 2 . 47 × 103 × 2 . 303 × 8 . 314 1000 = 47 . 29 KJ / mole

Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsZero Order ReactionsMedium2 minPYQ_2021

ChemistryMediumnumerical

For the reaction $A \to B,$ the rate constant $k$ (in $s^{- 1}$ ) is given by
$\log_{10} k = 20.35 - \frac{(2.47 \times 10^{3})}{T}$
The energy of activation in $kJ {mol}^{- 1}$ is ________ . (Nearest integer)
[Given : $R = 8.314 J K^{- 1} {mol}^{- 1}$ ]

Answer:

47.00

Solution:

$\log_{10} k = logA - \frac{E_{a}}{2.303 RT}$
$\log_{10} k = 20.35 - \frac{(2.47 \times 10^{3})}{T}$
$\frac{E_{a}}{2.303 R} = 2.47 \times 10^{3}$
$E_{a} = \frac{2.47 \times 103 \times 2.303 \times 8.314}{1000} = 47.29 KJ / mole$

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

⏱ 2mℹ️ Source: PYQ_2021

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