
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
162.6k+ views
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.
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.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

Instantaneous Velocity - Formula based Examples for JEE

JEE Advanced 2025 Notes
