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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

The decomposition of hydrocarbon follows the equation k = (4.5 × 10^11⁢ 𝑠−1)⁢ 𝑒^−28000⁢𝐾/𝑇 Calculate Ea.

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प्रश्न

The decomposition of hydrocarbon follows the equation

k = `(4.5 xx 10^11 s^-1) e^(-28000 K//T)`

Calculate Ea.

संख्यात्मक
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उत्तर

According to the Arrhenius equation, k = `Ae^((-E_a)/(RT))`

∴ `-E_a/(RT) = -(28000 K)/T`

Ea = 28000 K × R

= 28000 K × 8.314 JK−1 mol1

= 232.79 kJ mol−1

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अध्याय 3: Chemical Kinetics - 'NCERT TEXT-BOOK' Exercises [पृष्ठ २८२]

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नूतन Chemistry [English] Class 12 ISC
अध्याय 3 Chemical Kinetics
'NCERT TEXT-BOOK' Exercises | Q 4.26 | पृष्ठ २८२
एनसीईआरटी Chemistry Part 1 and 2 [English] Class 12
अध्याय 3 Chemical Kinetics
Exercises | Q 3.26 | पृष्ठ ८७

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