Question
calcute the molar enthalpy of condensation (H condensation) for ammonia when 50.0g of NH3 gas turn into a liquid at its bioling piont. 68.500J of energy are releasd in the process
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GPT 3.5
To calculate the molar enthalpy of condensation (ΔH condensation), we first need to calculate the moles of ammonia (NH3) gas using its molar mass.
The molar mass of NH3 is:
Nitrogen (N): 14.01 g/mol
Hydrogen (H): 1.01 g/mol
3 Hydrogens in NH3
Molar mass of NH3 = (14.01 g/mol) + (3(1.01 g/mol))
Molar mass of NH3 = 17.03 g/mol
Now, we can calculate the moles of NH3 gas:
moles of NH3 = mass / molar mass
moles of NH3 = 50.0 g / 17.03 g/mol
moles of NH3 = 2.93 mol
Since the reaction is the condensation of NH3 gas to a liquid, we know that the molar enthalpy of condensation (∆H condensation) is given by:
ΔH condensation = -68.500 J / moles of NH3
Substituting the values:
ΔH condensation = -68.500 J / 2.93 mol
ΔH condensation = -23,400 J/mol
Hence, the molar enthalpy of condensation for ammonia is -23,400 J/mol.
The molar mass of NH3 is:
Nitrogen (N): 14.01 g/mol
Hydrogen (H): 1.01 g/mol
3 Hydrogens in NH3
Molar mass of NH3 = (14.01 g/mol) + (3(1.01 g/mol))
Molar mass of NH3 = 17.03 g/mol
Now, we can calculate the moles of NH3 gas:
moles of NH3 = mass / molar mass
moles of NH3 = 50.0 g / 17.03 g/mol
moles of NH3 = 2.93 mol
Since the reaction is the condensation of NH3 gas to a liquid, we know that the molar enthalpy of condensation (∆H condensation) is given by:
ΔH condensation = -68.500 J / moles of NH3
Substituting the values:
ΔH condensation = -68.500 J / 2.93 mol
ΔH condensation = -23,400 J/mol
Hence, the molar enthalpy of condensation for ammonia is -23,400 J/mol.
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