The average time for which an oxygen atom remains adsorbed to a tungsten surface is 0.36 s at 2500 K β Surface Chemistry Chemistry Question
Question
The average time for which an oxygen atom remains adsorbed to a tungsten surface is 0.36 s at 2500 K and 0.72 s at 2000 K. The activation energy for desorption (in kcal/mol) is (ln 2 = 0.7)
π‘ Solution & Explanation
The residence time Ο is inversely proportional to the desorption rate constant k, so Ο = ΟβΒ·e^(Eβ/RT). Taking the ratio at two temperatures: Οβ/Οβ = e^((Eβ/R)Β·(1/Tβ - 1/Tβ)). Here, Οβ = 0.36 s at Tβ = 2500 K, and Οβ = 0.72 s at Tβ = 2000 K. 0.72 / 0.36 = 2 = e^((Eβ/R)Β·(1/2000 - 1/2500)). Taking natural logarithm: ln(2) = (Eβ/R) Γ ((5 - 4) / 10000) = Eβ / (10000 R). Given ln(2) = 0.7 and using R = 2 cal/(molΒ·K): 0.7 = Eβ / (10000 Γ 2) = Eβ / 20000. Eβ = 0.7 Γ 20000 = 14000 cal/mol = 14 kcal/mol. Therefore, correct answer is 0014.