Calculate the heat (in kJ) required to transform 58.60 g of dinitrogen tetraoxide from a solid at a
temperature of -9.3 °C to a gas at 47 °C. Report your answer to one decimal place. Data:
Molar mass of dinitrogen tetraoxide, N 2 O 4 = 92.011 g/mol Melting point = -9.3 °C Boilin g point = 21°C. Enthalpy of fusion = 14.65 kJ/mol
Enthalpy of vaporization = 38.12 kJ/mol.
Molar heat capacity of the liquid phase = 142.7 J/mol • K Molar heat capacity of the gas phase= 77.3 J/mol • K.
Chemistry assignment help?
2018-01-22 11:31 pm
回答 (2)
2018-01-23 12:06 am
Heat required to raise temperature = (No. of moles) × (Molar heat capacity) × (Temperature raised)
Heat required to change physical state = (No. of moles) × (Enthaly of fusion/vaporization)
No. of moles of N₂O₄
= (58.60 g) / (92.02 g/mol)
= 58.60/92.011 mol
Heat required to change 58.60 g of solid N₂O₄ to liquid N₂O₄ at -9.3°C
= (58.60/92.011 mol) × (14.65 kJ/mol)
= 9.33 kJ
Heat required to change 58.60 g of liquid N₂O₄ from -9.3°C to 21°C
= (58.60/92.011 mol) × (142.7/1000 kJ/mol) × [21 - (-9.3)]°C
= 2.75 kJ
Heat required to change 58.60 g of liquid N₂O₄ to gas N₂O₄ at 21°C
= (58.60/92.011 mol) × (38.12 kJ/mol)
= 24.27 kJ
Heat required to change 58.60 g of N₂O₄ gas from 21°C to 47°C
= (58.60/92.011 mol) × (77.3/1000 kJ/mol) × (47 - 21)°C
= 1.28 kJ
Total heat required
= (9.33 + 2.75 + 24.27 + 1.28) kJ
= 37.6 kJ (to 1 decimal place)
Heat required to change physical state = (No. of moles) × (Enthaly of fusion/vaporization)
No. of moles of N₂O₄
= (58.60 g) / (92.02 g/mol)
= 58.60/92.011 mol
Heat required to change 58.60 g of solid N₂O₄ to liquid N₂O₄ at -9.3°C
= (58.60/92.011 mol) × (14.65 kJ/mol)
= 9.33 kJ
Heat required to change 58.60 g of liquid N₂O₄ from -9.3°C to 21°C
= (58.60/92.011 mol) × (142.7/1000 kJ/mol) × [21 - (-9.3)]°C
= 2.75 kJ
Heat required to change 58.60 g of liquid N₂O₄ to gas N₂O₄ at 21°C
= (58.60/92.011 mol) × (38.12 kJ/mol)
= 24.27 kJ
Heat required to change 58.60 g of N₂O₄ gas from 21°C to 47°C
= (58.60/92.011 mol) × (77.3/1000 kJ/mol) × (47 - 21)°C
= 1.28 kJ
Total heat required
= (9.33 + 2.75 + 24.27 + 1.28) kJ
= 37.6 kJ (to 1 decimal place)
2018-01-22 11:39 pm
Just mechanical...
number of moles is (58.60 g)/(92.011 g/mol) = 0.6369 mol.
Q = (0.6369 mol)*[14.65 kJ/mol + (30.3 K)(0.1427 kJ/molK) +
+ 38.12 kJ/mol + (26 K)(0.0773 kJ/molK) ]
= about 40 kJ but use a calculator.
number of moles is (58.60 g)/(92.011 g/mol) = 0.6369 mol.
Q = (0.6369 mol)*[14.65 kJ/mol + (30.3 K)(0.1427 kJ/molK) +
+ 38.12 kJ/mol + (26 K)(0.0773 kJ/molK) ]
= about 40 kJ but use a calculator.
收錄日期: 2021-04-24 01:02:20
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