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

The rate constant for the decomposition of hydrocarbons is 2.418 × 10^−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?

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प्रश्न

The rate constant for the decomposition of hydrocarbons is 2.418 × 10−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?

संख्यात्मक
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उत्तर

Given: k = 2.418 × 10−5 s−1

T = 546 K

Ea = 179.9 kJ mol−1 

According to the Arrhenius equation,

log A = `log k + E_a/(2.303 RT)`

= `log 2.418 × 10^-5 + 179.9/(2.303 xx 8.314 xx 10^-3 xx 546)`

= (−5 + 0.3834) + 17.2081

= −4.6165 + 17.2081

log A = 12.5924 s−1

⇒ A = Antilog (12.5924) s−1

= 3.902 × 1012 s−1

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पाठ 3: Chemical Kinetics - Exercises [पृष्ठ ८७]

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एनसीईआरटी Chemistry Part 1 and 2 [English] Class 12
पाठ 3 Chemical Kinetics
Exercises | Q 3.23 | पृष्ठ ८७
नूतन Chemistry [English] Class 12 ISC
पाठ 3 Chemical Kinetics
'NCERT TEXT-BOOK' Exercises | Q 4.23 | पृष्ठ २८१

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