Answer:
The enthalpy of vaporization of water at 273 K and 1 bar = 44.9 KJ/mol
Explanation:
Enthalpy of vaporization of water at 273 K, ΔHvap(T₂) is given as;
ΔHvap(T₂) = ΔHvap(T₁) + ΔCp * (T₂ - T₁)
where ΔCp = molar heat capacity of gas - molar heat capacity of liquid
Therefore, ΔCp = (33.6 - 75.3) = -41.70 J/(mol K) = 0.0417 kJ/(molK)
substituting ΔCp = 0.0417 kJ/(mol K) in the initial formula
;
ΔHvap(T) = ΔHvap(T1) + ΔCp * (T₂ - T₁)
ΔHvap(T₂)= 40.7 kJ/mol + {-0.0417 kJ/(mol K) * (272 - 373 K)}
ΔHvap(T₂) = 44.9 kJ/mol
Therefore, enthalpy of vaporization of water at 273 K and 1 bar = 44.9kJ/mol
The correct is D.
Water is a polar molecule and it has polar bonds, which carry partially positive and partially negative charges. This polar bond increases the attraction between molecules of water and thus it requires a greater energy to break the bond between the molecules of water compare to carbon dioxide, which is a non polar molecule. Thus, water has a higher boiling point than carbon dioxide.
A double bond between carbon atoms is longer than a triple bond between carbon atoms.
4P + 502 -> P4O10 this is the answer
