The rate of a reaction quadruples when temperature changes from 2 7 ∘ C to 5 7 ∘ C . Calculate the energy o activation. Given R = 8.314 J K − 1 mol − 1 , lo g 4 = 0.6021 NEET 2024

lo g ( k 1 k 2 ) = 2.303 R E a ( T 1 1 − T 2 1 ) lo g ( 1 4 ) = 2.303 R E a ( 300 1 − 330 1 ) E a = 30 ( lo g ( 4 )) × 2.303 × 8.314 × 300 × 330 = 3.804 × 1 0 4 J / mol = 38.04 kJ / mol

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Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsArrhenius EquationEasy2 minPYQ_2024

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The rate of a reaction quadruples when temperature changes from $2 7^{\circ} C$ to $5 7^{\circ} C$ . Calculate the energy o activation. Given $R = 8.314 J K^{- 1} mol^{- 1}, lo g 4 = 0.6021$ NEET 2024

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Solution:

$lo g (\frac{k _{2}}{k _{1}}) = \frac{E _{a}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}}) lo g (\frac{4}{1}) = \frac{E _{a}}{2.303 R} (\frac{1}{300} - \frac{1}{330}) E_{a} = \frac{( lo g ( 4 )) \times 2.303 \times 8.314 \times 300 \times 330}{30} = 3.804 \times 1 0^{4} J / mol = 38.04 kJ / mol$

Stream:NEETSubject:ChemistryTopic:Chemical KineticsSubtopic:Arrhenius Equation

⏱ 2mℹ️ Source: PYQ_2024

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