The average time for which an oxygen atom remains adsorbed to a tungsten surface is 0.36 s at 2500 K — Surface Chemistry Chemistry Question

💡 Solution & Explanation

The residence time τ is inversely proportional to the desorption rate constant k, so τ = τ₀·e^(Eₐ/RT). Taking the ratio at two temperatures: τ₂/τ₁ = e^((Eₐ/R)·(1/T₂ - 1/T₁)). Here, τ₁ = 0.36 s at T₁ = 2500 K, and τ₂ = 0.72 s at T₂ = 2000 K. 0.72 / 0.36 = 2 = e^((Eₐ/R)·(1/2000 - 1/2500)). Taking natural logarithm: ln(2) = (Eₐ/R) × ((5 - 4) / 10000) = Eₐ / (10000 R). Given ln(2) = 0.7 and using R = 2 cal/(mol·K): 0.7 = Eₐ / (10000 × 2) = Eₐ / 20000. Eₐ = 0.7 × 20000 = 14000 cal/mol = 14 kcal/mol. Therefore, correct answer is 0014.