How much heat is evolved when 269 g of ammonia gas condenses to a liquid at its boiling point?
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To calculate the heat evolved when ammonia gas condenses to a liquid at its boiling point, we need to use the equation:
q = m * ΔHv
Where:
q is the heat evolved (in joules),
m is the mass of the substance (in grams), and
ΔHv is the molar enthalpy of vaporization (in J/g).
First, let's find the molar enthalpy of vaporization of ammonia. The molar enthalpy of vaporization represents the energy required to convert one mole of a substance from a liquid to a gas at its boiling point.
The molar enthalpy of vaporization of ammonia is 23.4 kJ/mol. To convert this value to J/g, we need to divide it by the molar mass of ammonia, which is 17.03 g/mol:
ΔHv = (23.4 kJ/mol) / (17.03 g/mol) = 1371.17 J/g
Now we can calculate the heat evolved:
q = m * ΔHv
= (269 g) * (1371.17 J/g)
≈ 368,974.73 J
Therefore, when 269 g of ammonia gas condenses to a liquid at its boiling point, approximately 368,974.73 J of heat is evolved.