Nitrogen gas adsorbed on charcoal to the extent of 0.387 cm³/g at a pressure of 1.6 atm and at temperature of 200 K, but at 250 K the same amount of adsorption was achieved only when the pressure was

Question

Nitrogen gas adsorbed on charcoal to the extent of 0.387 cm³/g at a pressure of 1.6 atm and at temperature of 200 K, but at 250 K the same amount of adsorption was achieved only when the pressure was increased to 32 atm. The magnitude of molar enthalpy of adsorption (in kcal/mol) of nitrogen on charcoal is (ln 2 = 0.7)

Prashant Kotian · Accepted Answer

Correct answer: 6. For a constant amount of adsorption, the Clausius-Clapeyron equation gives the isosteric enthalpy of adsorption: ln(P₂/P₁) = (-ΔH_ads / R) (1/T₁ - 1/T₂). Here, P₁ = 1.6 atm at T₁ = 200 K, and P₂ = 32 atm at T₂ = 250 K. ln(32/1.6) = ln(20) = ln(10) + ln(2). Taking ln(10) ≈ 2.3 and given ln(2) = 0.7, ln(20) = 3.0. Substituting into the equation: 3.0 = (-ΔH_ads / 2) × (1/200 - 1/250) = (-ΔH_ads / 2) × (50 / 50000) = -ΔH_ads / 2000. ΔH_ads = -6000 cal/mol = -6 kcal/mol. The magnitude is 6 kcal/mol. Therefore, correct answer is 6.

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