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

The decomposition of hydrocarbon follows the equation k = (4.5×1011⁢𝑠−1)⁢𝑒−28000⁢𝐾/𝑇 Calculate Ea. - Chemistry

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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`

E = 28000 K × R

= 28000 K × 8.314 JK−1 mol1

= 232.79 kJ mol−1

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अध्याय 3: Chemical Kinetics - Exercises [पृष्ठ ८७]

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एनसीईआरटी Chemistry Part 1 and 2 [English] Class 12
अध्याय 3 Chemical Kinetics
Exercises | Q 3.26 | पृष्ठ ८७

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