The enthalpy change for the transition of liquid water to steam,\[{\Delta _{vap}}H = 37.3\]\[kJmo{l^{ - 1}}\]. The entropy change for the process is _________________.
A. 111.9\[Jmo{l^{ - 1}}{K^{ - 1}}\]
B. 37.3\[Jmo{l^{ - 1}}{K^{ - 1}}\]
C. 100\[Jmo{l^{ - 1}}{K^{ - 1}}\]
D. 74.6\[Jmo{l^{ - 1}}{K^{ - 1}}\]

Answer
VerifiedVerified
14.4k+ views
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)
\[{\Delta _{vap}}S\] = change in entropy (Units: \[Jmo{l^{ - 1}}{K^{ - 1}}\] or \[kJmo{l^{ - 1}}{K^{ - 1}}\] )
\[{\Delta _{vap}}H\] = change in enthalpy (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 liquid water is transitioning into steam, the change in enthalpy is the difference between the enthalpies of liquid water and steam as shown below:
\[{H_2}O(l) \to {H_2}O(g)\]
\[\Delta H = {H_{{H_2}O(l)}} - {H_{{H_2}O(g)}}\]

Since the transition of liquid water into steam is called vaporisation, the enthalpy change associated with this process is called the enthalpy of vaporisation (\[{\Delta _{vap}}H\]). \[{\Delta _{vap}}H\] simply indicates the amount of energy that will be required to convert a particular quantity/amount of water into water vapour.

Here, it is given that\[{\Delta _{vap}}H\]=37.3\[kJmo{l^{ - 1}}\] = 37300\[Jmo{l^{ - 1}}\] .
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:
\[{\Delta _{vap}}S = \dfrac{{{\Delta _{vap}}H}}{{{T_b}}}\]
\[ \Rightarrow {\Delta _{vap}}S = \dfrac{{37300Jmo{l^{ - 1}}}}{{373K}}\]
\[ \Rightarrow {\Delta _{vap}}S = 100\] \[Jmo{l^{ - 1}}{K^{ - 1}}\]
Thus, option C 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.