A monolayer of N₂ molecules (effective area 0.15 nm²) is absorbed on the surface of 1.00 g of Fe/Al₂O₃ catalyst at 77 K, the boiling point of liquid nitrogen. Upon warming, the nitrogen occupies 2.24

Question

A monolayer of N₂ molecules (effective area 0.15 nm²) is absorbed on the surface of 1.00 g of Fe/Al₂O₃ catalyst at 77 K, the boiling point of liquid nitrogen. Upon warming, the nitrogen occupies 2.24 cm³ at 0°C and 760 torr. The surface area of catalyst (in m²) is (Avogadro’s number N_A = 6 × 10²³)

Prashant Kotian · Accepted Answer

Correct answer: 9. Volume of desorbed N₂ at STP (0°C and 1 atm) = 2.24 cm³ = 2.24 × 10⁻³ L. Moles of N₂ = Volume at STP / 22.4 L/mol = (2.24 × 10⁻³) / 22.4 = 10⁻⁴ mol. Number of N₂ molecules = 10⁻⁴ mol × 6 × 10²³ molecules/mol = 6 × 10¹⁹ molecules. Total surface area of the catalyst = Number of molecules × effective area per molecule = 6 × 10¹⁹ × 0.15 nm² = 0.9 × 10¹⁹ nm². Converting to m²: 0.9 × 10¹⁹ × 10⁻¹⁸ m² = 0.9 × 10 = 9 m². Therefore, correct answer is 9.

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