Question

If the enthalpy of vaporisation of water is 186.5kJ/mol, the entropy of its vaporisation will be:
A. 0.5
B. 1.0
C. 1.5
D. 2.0

Answer 1

Hint: Every physical or chemical process is accompanied by a change in enthalpy (\[\Delta H\]) and a change in entropy (\[\Delta S\]). If the process takes place at a constant temperature, then\[\Delta H\]and\[\Delta S\]follow a simple mathematical relation. Using that relation here will answer the question.

Formula used:
\[{\Delta _{vap}}S = \dfrac{{{\Delta _{vap}}H}}{{{T_b}}}\] … (1)
\[{\Delta _{vap}}S\] = entropy change during vaporisation (Units: \[Jmo{l^{ - 1}}{K^{ - 1}}\] or \[kJmo{l^{ - 1}}{K^{ - 1}}\] )
\[{\Delta _{vap}}H\] = change in enthalpy during vaporisation (Units: \[Jmo{l^{ - 1}}\] or \[kJmo{l^{ - 1}}\] )
\[{T_b}\] = boiling point (Units: K)

Complete Step by Step Solution:

The enthalpy of a system (H) is a measure of the energy content of the system. During a physical or chemical process, as reactants convert into products or the physical state of a substance changes (which occurs during vaporisation), there occurs a change in the enthalpy associated with the process.
This change in the enthalpy (\[\Delta H\]) is given by the difference in the enthalpies of the products and the reactants.
\[\Delta H = {H_{products}} - {H_{reac\tan ts}}\]
In this instance, when water is being vaporised, the change in enthalpy is the difference between enthalpies of water in its liquid state and vapour state.
\[{H_2}O(l) \to {H_2}O(g)\]
\[\Delta H = {H_{{H_2}O(l)}} - {H_{{H_2}O(g)}}\]
The \[\Delta H\] , in this case, is known as the enthalpy of vaporisation (\[{\Delta _{vap}}H\] ) or as the latent heat of vaporisation. \[{\Delta _{vap}}H\] simply indicates the amount of energy that will be required to convert a particular quantity/amount of water into water vapour.
 In the question, we are given \[{\Delta _{vap}}H\]of water = 186.5\[kJ/mol\]
Since the temperature of the vaporisation process is not mentioned explicitly, we assume that the vaporisation is occurring at the boiling point of water which is 100 \[^\circ C\] or 373K. Thus, \[{T_b} = 373K\]
Plugging the values of\[{\Delta _{vap}}H\]and \[{T_b}\]of water in equation (1), we get:
\[{\Delta _{vap}}S = \dfrac{{{\Delta _{vap}}H}}{{{T_b}}}\]
\[ \Rightarrow {\Delta _{vap}}S = \dfrac{{186.5kJ/mol}}{{373K}}\]
\[ \Rightarrow {\Delta _{vap}}S = 0.5\] \[kJmo{l^{ - 1}}{K^{ - 1}}\]
Thus, option A is correct.

Note: At the boiling point of water, liquid water is in equilibrium with water vapour. \[{H_2}O(l) \rightleftharpoons {H_2}O(g)\]. For an equilibrium condition, the change in Gibbs free energy (\[\Delta G\] ) = 0. We know that \[\Delta G\] is related to\[\Delta S\] as
\[\Delta G = \Delta H - T\Delta S\] .
For boiling of water, \[T = {T_b}\],\[\Delta H = {\Delta _{vap}}H\] and\[\Delta G = 0\] . Substituting these values in the above equation, we get:
\[{\Delta _{vap}}H - {T_b}{\Delta _{vap}}S = 0\]
\[ \Rightarrow {T_b}{\Delta _{vap}}S = {\Delta _{vap}}H\]
\[ \Rightarrow {\Delta _{vap}}S = \dfrac{{{\Delta _{vap}}H}}{{{T_b}}}\] . This is how equation (1) was derived.