Physisorbed particles undergo desorption at 27^ with an activation energy of 16.628 kJ mol⁻¹. Assuming first–order process and a frequency factor of 10¹² Hz, the average residence time (in seconds) of the particles on the surface is

Physisorbed particles undergo desorption at 27^ with an activation energy of 16.628 kJ mol⁻¹. Assuming… · ABC26PH0560

Subject: Physical Chemistry · Chapter: Surface Chemistry · Topic: Surface Chemistry – General · Exam: GATE 2005-2021 · Marks: 2 · Difficulty: Medium

Physisorbed particles undergo desorption at $27^\circ\mathrm{C}$ with an activation energy of $16.628\ \mathrm{kJ\ mol^{-1}}$. Assuming first–order process and a frequency factor of $10^{12}\ \mathrm{Hz}$, the average residence time (in seconds) of the particles on the surface is

(a)$8 \times 10^{-10}$

(b)$8 \times 10^{-11}$

(c)$2 \times 10^{-9}$

(d)$1 \times 10^{-12}$

Answer

Answer (as printed): A

Explanation

No explanation was printed for this question.

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