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
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.
To calculate the molar enthalpy of condensation (H condensation) for ammonia (NH3), we need to use the given information and the molar mass of NH3.
Step 1: Calculate the number of moles of NH3:
Number of moles = mass / molar mass
The molar mass of NH3 is approximately 17.031 g/mol.
Number of moles = 50.0 g / 17.031 g/mol ≈ 2.934 mol
Step 2: Calculate the molar enthalpy of condensation:
H condensation = Energy released / number of moles
H condensation = 68.500 J / 2.934 mol ≈ 23.386 J/mol
Therefore, the molar enthalpy of condensation for NH3 is approximately 23.386 J/mol.
To calculate the molar enthalpy of condensation (H_condensation) for ammonia (NH3), we need to use the equation:
H_condensation = q / n
Where:
- H_condensation is the molar enthalpy of condensation
- q is the energy released in the process (in Joules)
- n is the number of moles of ammonia gas
To find the number of moles (n) of ammonia gas, we need to use the molar mass of ammonia (17.03 g/mol).
n = mass / molar mass
Given:
- Mass of NH3 gas = 50.0 g
- Energy released (q) = 68.500 J
First, let's find the number of moles (n):
n = 50.0 g / 17.03 g/mol
n ≈ 2.9339 mol
Now, we can calculate the molar enthalpy of condensation (H_condensation):
H_condensation = 68.500 J / 2.9339 mol
H_condensation ≈ 23.356 J/mol
Therefore, the molar enthalpy of condensation for ammonia (NH3) is approximately 23.356 J/mol.