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Physiosorbed particles undergo desorption at 270C with an activation energy of 16.628 kJ/mol. Assuming first order process and a frequency factor of 1012 Hz, the average residence time (in sec) of particles on the surface isa)7.8 × 10–10b)8 × 10–11c)2 × 10–9d)1 × 10–12Correct answer is option 'A'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Physiosorbed particles undergo desorption at 270C with an activation energy of 16.628 kJ/mol. Assuming first order process and a frequency factor of 1012 Hz, the average residence time (in sec) of particles on the surface isa)7.8 × 10–10b)8 × 10–11c)2 × 10–9d)1 × 10–12Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Physiosorbed particles undergo desorption at 270C with an activation energy of 16.628 kJ/mol. Assuming first order process and a frequency factor of 1012 Hz, the average residence time (in sec) of particles on the surface isa)7.8 × 10–10b)8 × 10–11c)2 × 10–9d)1 × 10–12Correct answer is option 'A'. Can you explain this answer?.

Solutions for Physiosorbed particles undergo desorption at 270C with an activation energy of 16.628 kJ/mol. Assuming first order process and a frequency factor of 1012 Hz, the average residence time (in sec) of particles on the surface isa)7.8 × 10–10b)8 × 10–11c)2 × 10–9d)1 × 10–12Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry. Download more important topics, notes, lectures and mock test series for Chemistry Exam by signing up for free.

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