Question

Which of the following is true for the reaction?
\[\;{H_2}O\left( l \right) \rightleftharpoons {H_2}O\left( g \right)\] at 100\[^\circ C\]and 1 atmosphere.
A.\[\Delta S = 0\]
B. \[\Delta H = 0\]
C. \[\Delta H = \Delta E\]
D. \[\Delta H = T\Delta S\]

Answer 1

Hint: Every process, physical and chemical, is accompanied by changes in enthalpy (\[\Delta H\]), entropy (\[\Delta S\]) and Gibbs free energy (\[\Delta G\]). At constant temperature, these three quantities share a mathematical relation. The change in Gibbs free energy is linked to the spontaneity of the process.

Complete Step by Step Solution:
We already know that the net change in entropy of the system and surroundings (\[\Delta {S_{sys}} + \Delta {S_{sur}}\]) can serve as a criterion for the spontaneity of a process. When there is no net change in entropy of the system and surroundings, the process is said to be at equilibrium. To study various processes, we must always measure \[\Delta {S_{sys}}\] and\[\Delta {S_{sur}}\]which is not convenient in all cases.
Therefore, entropy change is considered in terms of other state functions which can be determined more conveniently. One such state function is called the free energy function (G). It is given by \[G = H - TS\] where H is the enthalpy, T is the temperature and S is the entropy of the system. G is also called the Gibbs free energy of the system. It is a measure of the maximum work that the system can perform under isothermal and isobaric conditions.
Let \[{G_1},{S_1}\] and\[{H_1}\]represent the thermodynamic functions at an initial state and\[{G_2},{S_2}\]and\[{H_2}\] in the final state. Assuming temperature (T) to be constant, we can have:
\[{G_2} = {H_2} - T{S_2}\] and \[{G_1} = {H_1} - T{S_1}\]
\[\therefore {G_2} - {G_1} = \left( {{H_2} - {H_1}} \right) - T\left( {{S_2} - {S_1}} \right)\]
\[ \Rightarrow \Delta G = \Delta H - T\Delta S\] … (1)

\[\Delta G\] is, by far, the most useful parameter in deciding the spontaneity of a reaction or process. This is because, for spontaneous processes, \[\Delta G < 0\], for non-spontaneous processes,\[\Delta G > 0\] and for reversible processes (where the system is in an equilibrium) \[\Delta G = 0\] .
Substituting \[\Delta G = 0\] in equation (1), we get:
\[\Delta H - T\Delta S = 0\]
\[ \Rightarrow \Delta H = T\Delta S\]
Thus, option D is correct.

Note: Although, according to the Second Law of Thermodynamics, the net entropy change of system and surroundings is zero for an equilibrium condition (\[\left( {\Delta {S_{sys}} + \Delta {S_{sur}}} \right) = 0\] ), students must not confuse option A to be the correct answer because, in option A the \[\Delta S\] mentioned is the change in entropy of the system only. Even in an equilibrium, the entropy of the system keeps changing.