Question

9.0 gm of\[{H_2}O\] is vaporised at 100\[^\circ C\] and 1 atm pressure. If the latent heat of vaporisation of water is x J/gm, then ΔS is given by:
A. \[\dfrac{x}{{373}}\]
B. \[\dfrac{{18x}}{{100}}\]
C. \[\dfrac{{18x}}{{373}}\]
D. \[\dfrac{1}{2} \times \dfrac{{18x}}{{373}}\]

Answer 1

Hint:The latent heat of vaporisation of water is given in terms of the weight of the water being vaporised (in J/g). It can be converted into J/mol knowing the molar mass of water. \[\Delta S\]can be determined from its relationship with the change in enthalpy of vaporisation (\[{\Delta _{vap}}H\] ).

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

Complete step-by-step answer:
The enthalpy of a system (H) is a thermodynamic parameter that is, in essence, a measure of the energy content of the system. During a chemical reaction, as reactants convert into products, the enthalpy associated with the reaction changes. 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.
Here, it is given the latent heat of vaporisation (\[{\Delta _{vap}}H\])\[ = x\] \[J{g^{ - 1}}\] . This means that x J of energy is required to vaporise 1 g of water. We know that the weight of 1 mole of water is 18 g. Thus \[{\Delta _{vap}}H\] for 1 mole of water would be 18x \[Jmo{l^{ - 1}}\] .
The amount of water being vaporised is 9 g which is \[\dfrac{1}{2}\] of a mole. Thus, \[{\Delta _{vap}}H\] for 9 g of water = \[\dfrac{1}{2} \times 18x\] \[Jmo{l^{ - 1}}\].
We know that the boiling point of water (\[{T_b}\]) = \[100^\circ C = 373K\]

Plugging the values of\[{\Delta _{vap}}H\]and\[{T_b}\]in equation (1) we get:
\[\Delta S = \dfrac{{{\Delta _{vap}}H}}{{{T_b}}}\]
\[ \Rightarrow \Delta S = \dfrac{{\dfrac{1}{2} \times 18xJmo{l^{ - 1}}}}{{373K}}\]
\[ \Rightarrow \Delta S = \dfrac{1}{2} \times \dfrac{{18x}}{{373}}\] \[Jmo{l^{ - 1}}{K^{ - 1}}\]

Thus, option D 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 in the above equation, we get:
\[{\Delta _{vap}}H - {T_b}\Delta S = 0\]
\[ \Rightarrow {T_b}\Delta S = {\Delta _{vap}}H\]
\[ \Rightarrow \Delta S = \dfrac{{{\Delta _{vap}}H}}{{{T_b}}}\] .