At 100.°C and 1.00 atm, H ° = 40.6kJ/mol for the vaporization of water. Estimate

Question

At 100.°C and 1.00 atm, H° = 40.6kJ/mol for the vaporization of water. EstimateG° for the vaporization of water at 90.°C and110.°C. Assume H° andS° at 100.°C and 1.00 atm do not depend ontemperature.

90.°C____________ kJ/mol
110.°C____________ kJ/mol At 100.°C and 1.00 atm, H° = 40.6kJ/mol for the vaporization of water. EstimateG° for the vaporization of water at 90.°C and110.°C. Assume H° andS° at 100.°C and 1.00 atm do not depend ontemperature.

90.°C____________ kJ/mol
110.°C____________ kJ/mol

Explanation / Answer

Tb= 100 0C = 373 K Svap = Hvap / Tb             = 40.6 kJ/mole /373 K             = 0.1088 kJ/mol-K = 108.8 J/mol-K
Gvap = Hvap - T Svap T = 110 0C = 383 K Gvap = Hvap - T Svap             =40.6 kJ/mole  - 383 K *0.1088 kJ/mol-K T = 90 0C = 363 K Gvap = Hvap - T Svap             =40.6 kJ/mole  - 363 K *0.1088 kJ/mol-K             = 1.1056 kJ/mol Gvap = Hvap - T Svap             =40.6 kJ/mole  - 363 K *0.1088 kJ/mol-K             = 1.1056 kJ/mol

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