How much heat evolves when 275 g of ammonia gas condenses to a liquid at its boiling point?
To determine the amount of heat evolved when ammonia gas condenses to a liquid at its boiling point, we need to know the heat of vaporization of ammonia. The heat of vaporization of ammonia is given as 23.35 kJ/mol.
To calculate the heat evolved, we first need to convert the mass of ammonia from grams to moles. The molar mass of ammonia (NH3) is 17.03 g/mol.
Number of moles of ammonia = mass / molar mass
Number of moles of ammonia = 275 g / 17.03 g/mol ≈ 16.13 mol
Now, we can calculate the heat evolved using the heat of vaporization:
Heat evolved = number of moles × heat of vaporization
Heat evolved = 16.13 mol × 23.35 kJ/mol ≈ 376.5 kJ
Therefore, approximately 376.5 kJ of heat evolves when 275 g of ammonia gas condenses to a liquid at its boiling point.
To find the amount of heat evolved when ammonia gas condenses to a liquid, we need to use the equation:
q = m × ΔHvap
where:
q is the heat evolved (in Joules),
m is the mass of ammonia gas (in grams), and
ΔHvap is the enthalpy of vaporization for ammonia.
First, let's find the enthalpy of vaporization for ammonia. The enthalpy of vaporization of ammonia is 23.35 kJ/mol.
The molar mass of ammonia (NH3) is:
14.01 g/mol (N) + 3(1.01 g/mol) (H) = 17.03 g/mol
Now, we can calculate the number of moles of ammonia gas (NH3) using its molar mass:
moles of NH3 = mass of NH3 / molar mass of NH3
= 275 g / 17.03 g/mol
Next, we convert the moles of ammonia gas to the heat evolved:
q = moles of NH3 × ΔHvap
q = moles of NH3 × ΔHvap × 1000 (to convert kJ to J)
Plugging in the values, we can calculate the heat evolved:
q = (275 g / 17.03 g/mol) × (23.35 kJ/mol) × 1000
Calculating this, we find:
q = 40,439 J
Therefore, the heat evolved when 275 g of ammonia gas condenses to a liquid at its boiling point is approximately 40,439 Joules.
To determine the amount of heat evolved when ammonia gas condenses to a liquid at its boiling point, we need to use the heat of vaporization of ammonia and the molar mass of ammonia gas.
1. Find the heat of vaporization of ammonia: The heat of vaporization is the amount of heat required to change one mole of a substance from a liquid to a gas at a constant temperature. For ammonia, the heat of vaporization is 23.35 kJ/mol.
2. Calculate the moles of ammonia gas: To do this, we need the molar mass of ammonia, which is 17.03 g/mol. Divide the mass of ammonia gas (275 g) by the molar mass to get the number of moles: 275 g / 17.03 g/mol ≈ 16.13 mol.
3. Determine the amount of heat evolved: Multiply the moles of ammonia gas by the heat of vaporization to find the amount of heat evolved. 16.13 mol * 23.35 kJ/mol ≈ 376 kJ.
Therefore, approximately 376 kJ of heat is evolved when 275 g of ammonia gas condenses to a liquid at its boiling point.