Does the latent heat of vaporization of a substance vary with its boiling point?

Question

please explain. thanks

Answer 1

yeThe standard enthalpy change of vaporization, ΔvHo, also (less correctly) known as the heat of vaporization is the energy required to transform a given quantity of a substance into a gas. It is measured at the boiling point of the substance, although tabulated values are usually corrected to 298 K: the correction is small, and is often smaller than the uncertainty in the measured value. Values are usually quoted in kJ/mol, although kJ/kg, kcal/mol, cal/g and Btu/lb (obsolete) are also possible, among others.

The standard enthalpy change of condensation (or heat of condensation) is numerically exactly equal to the standard enthalpy change of vaporization, but has the opposite sign: enthalpy changes of vaporization are always positive (heat is absorbed by the substance), whereas enthalpy changes of condensation are always negative (heat is released by the substance).

The enthalpy change of vaporization can be viewed as the energy required to overcome the intermolecular interactions in the liquid (or solid, in the case of sublimation). Hence helium has a particularly low standard enthalpy change of vaporization, 0.0845 kJ/mol, as the van der Waals forces between helium atoms are particularly weak. On the other hand, the molecules in liquid water are held together by relatively strong hydrogen bonds, and its standard enthalpy change of vaporization, 40.8 kJ/mol, is more than five times the energy required to heat the same quantity of water from 0 °C to 100 °C (cp = 75.3 J K−1 mol−1). Care must be taken, however, when using enthalpy changes of vaporization to measure the strength of intermolecular forces, as these forces may persist in the gas phase (as is the case with water), and so the calculated value of the bond strength will be too low. This is particularly true of metals, which often form covalently bonded molecules in the gas phase: in these cases, the standard enthalpy change of atomization must be used to obtain a true value of the bond energy.

An alternative description is to view the enthalpy change of condensation as the heat which must be released to the surroundings to compensate for the drop in entropy when a gas condenses to a liquid. As the liquid and gas are in equilibrium at the boiling point (Tb), ΔvG = 0, which leads to:

ΔvS = Sgas − Sliquid = ΔvH/Tb
As neither entropy nor enthalpy vary greatly with temperature, it is normal to use the tabulated standard values without any correction for the difference in temperature from 298 K. A correction must be made if the pressure is different from 100 kPa, as the entropy of a gas is proportional to its pressure (or, more precisely, to its fugacity): the entropies of liquids vary little with pressure, as the compressibility of a liquid is small.

These two definitions are equivalent: the boiling point is the temperature at which the increased entropy of the gas phase overcomes the intermolecular forces. As a given quantity of matter always has a higher entropy in the gas phase than in a condensed phase (ΔvS is always positive), and from

ΔG = ΔH − TΔS,
the Gibbs free energy change falls with increasing temperature: gases are favored at higher temperatures, as is observed in practice.

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Answer 2

http://www.infoplease.com/ce6/sci/A0861748.html
this mite help

"The latent heat of evaporation is the energy required to overcome the molecular forces of attraction between the particles of a liquid, and bring them to the vapour state, where such attractions are minimal." http://www.physchem.co.za/Heat/Latent.htm

"It is measured at the boiling point of the substance..."
http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_vaporization

so i guess that's a yes