Use this information to estimate the normal boiling point (in°C) for CCl4. information below. ?
2020-07-14 1:23 am
At 60.0°C the vapor pressure of CCl4 is 60.4 kPa and its enthalpy of vaporization is 29.82 kJ/mol.
回答 (1)
2020-07-14 1:43 am
CCl₄(ℓ) ⇌ CCl₄(g) K = P(vap)
When T₁ = (273.2 + 60.0) K = 333.2 K: K₁ = 60.4 kPa
When T₂ = normal boiling point: K₂ = P(atm) = 101.3 kPa
Enthalpy of vaporization, ΔH = 29.82 kJ/mol = 29820 J/mol
Gas constant, R = 8.314 J/(mol K)
ln(K₁/K₂) = (ΔH/R) [(1/T₂) - (1/T₁)]
ln(60.4/101.3) = (29820/8.314) [(1/T₂) - (1/333.2)]
(1/T₂) - (1/333.2) = -0.00014417
T₂ = 1 / [-0.00014417 + (1/333.2)]
Normal boiling point, T₂ = 350.0 K = 76.8 °C
When T₁ = (273.2 + 60.0) K = 333.2 K: K₁ = 60.4 kPa
When T₂ = normal boiling point: K₂ = P(atm) = 101.3 kPa
Enthalpy of vaporization, ΔH = 29.82 kJ/mol = 29820 J/mol
Gas constant, R = 8.314 J/(mol K)
ln(K₁/K₂) = (ΔH/R) [(1/T₂) - (1/T₁)]
ln(60.4/101.3) = (29820/8.314) [(1/T₂) - (1/333.2)]
(1/T₂) - (1/333.2) = -0.00014417
T₂ = 1 / [-0.00014417 + (1/333.2)]
Normal boiling point, T₂ = 350.0 K = 76.8 °C
收錄日期: 2021-04-24 07:54:44
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