Question

How do you calculate the enthalpy of vaporization for 1 mole of water?

Answer 1

Hint: Recall that the enthalpy of vaporization determines the amount of heat that is required to change a substance from its liquid phase to gaseous phase. Since we are required to find the same for 1 mole of water, we find the molar enthalpy of vaporization instead, which is nothing but the enthalpy of vaporization for the entire mass of the water in 1 mole of water. To this end, find the mass of 1 mole of water in grams following which you can arrive at the required solution using the enthalpy of vaporization $H_{vapour} = 2258\;Jg^{-1}$ at $100^{\circ}C$
Formula Used:
Molar enthalpy of vaporization: $Q= H_{vapour} \times m$

Complete answer:
We know that enthalpy is a measure of how much heat (is added or removed) and work (is done on or by) a system. It is given as the sum of the internal energy and the product of the pressure and volume of the system, i.e.,
$H = U+PV$
The enthalpy of vaporization is defined as the amount of energy that must be added to a liquid to transform a quantity of that liquid into a gas. For most liquids, the enthalpy of vaporisation is defined at the normal boiling temperature of the liquid. It is equal to the increased internal energy (energy required to overcome intermolecular interactions in the liquid) of the vapour phase relative to the liquid phase in addition to the work done against ambient pressure under such change of state, i.e.,
$\Delta H_{vapour} = \Delta U_{vapour} + P\Delta V$.
We are required to find the enthalpy of vaporization for 1 mole of water. This is nothing but the molar enthalpy of vaporization, which is the amount of energy needed to change one mole of substance from the liquid phase to the vapour phase. The enthalpy of vaporization of water at $100^{\circ} C$ is known to be $H_{vapour} = 2258\;Jg^{-1} $
$H_{vapour} = U_{vapour}$, where the internal energy can be given as
$U_{vapour} = \dfrac{Q}{m}$, where Q is the enthalpy of vaporization, and m is the mass of water.
Therefore, we can calculate the molar enthalpy of vaporization of water as follows:
$H_{vapour} = \dfrac{Q}{m} \Rightarrow Q= H_{vapour} \times m = 2258 \times m$
Now, we need to convert $n =1\; mole$ of water to grams.
The molecular weight (w) of water ($H_2O$) is: $w=2\times 1 + 1 \times 16 = 18\;gmol^{-1}$, since atomic weight of hydrogen is $1\;g$ and that of oxygen is $16\;g$
The number of moles can be given as: $n = \dfrac{m}{w}$
$\Rightarrow m = n \times w = 1 mol \times 18\;gmol^{-1} = 18\;g $
Plugging this back into our molar enthalpy of vaporization expression:
$Q= 2258\;Jg^{-1} \times 18\;g = 40644\;J = 40.6\;kJ$

Note:
Remember that the change in enthalpy of vaporization $\Delta H_{vapour}$ is extensive, whereas the molar enthalpy of vaporization $Q$ is intensive. This means that the enthalpy of vaporizations depends on the amount of mass/material present whereas the molar enthalpy of vaporization is independent of the amount of mass/ material present in the system since it already takes into account the amount of mass in terms of one mole. Enthalpy in general is an extensive property when defined in units of $J$ but becomes intensive when it is quoted in $Jmol^{-1}$ or $Jg^{-1}$.