 # 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$Jmo{l^{ - 1}}{K^{ - 1}}$ Verified
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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.