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Chapter 6: Problem 19

For the process \(\mathrm{H}_{2} \mathrm{O}(1)(1 \mathrm{bar}, 373 \mathrm{~K}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})(1 \mathrm{bar}, 373 \mathrm{~K})\), the correct set of thermodynamic parameters is [2007] (a) \(\Delta G=0, \Delta S=+\mathrm{ve}\) (b) \(\Delta G=0, \Delta S=-\mathrm{ve}\) (c) \(\Delta G=+\) ve, \(\Delta S=0\) (d) \(\Delta G=-\mathrm{ve}, \Delta S=+\mathrm{ve}\)

Short Answer

Expert verified
Option (a) \(\Delta G = 0, \Delta S = +\text{ve}\).

Step by step solution

01

Understanding the Process

The given process is the conversion of liquid water \((\mathrm{H}_{2}\mathrm{O}(l))\) to water vapor \((\mathrm{H}_{2}\mathrm{O}(g))\) at 1 bar and 373 K. This is the process of boiling at the boiling point temperature.
02

Analyzing Gibbs Free Energy Change (\(\Delta G\))

At the boiling point, the system is at equilibrium between the liquid and gas phases. Therefore, the change in Gibbs Free Energy \((\Delta G)\) for phase transition at equilibrium is zero: \(\Delta G = 0\).
03

Analyzing Entropy Change (\(\Delta S\))

For the process of phase transition from liquid to gas, the randomness or disorder increases. Therefore, the change in entropy \((\Delta S)\) is positive. Hence, \(\Delta S = +\text{ve}\).
04

Selecting the Correct Option

Based on the analysis, \(\Delta G = 0\) and \(\Delta S = +\text{ve}\). The correct option aligning with these conditions is option (a): \(\Delta G = 0, \Delta S = + \text{ve}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gibbs Free Energy
Gibbs Free Energy, denoted as \(G\), is an important concept in thermodynamics that helps us understand reactions and phase changes at constant temperature and pressure. It's a measure of the maximum reversible work that may be performed by a thermodynamic system. More specifically, it determines the spontaneity of a process.

During a phase transition, such as water boiling to form steam, the system is often at equilibrium. This means there's no net change happening. At this equilibrium state, the change in Gibbs Free Energy, \(\Delta G\), is zero.

This concept is essential because it tells us that at the boiling point temperature, water can convert to steam without any net energy change in the system. This balance is crucial for processes that aim to maintain equilibrium under specific conditions.
Phase Transition
A phase transition occurs when a substance changes from one state of matter to another, such as from liquid to gas. In the given example, the transition from liquid water to vapor at 373 K (100掳C) is a classic case of boiling. Boiling occurs at a specific temperature and pressure where the liquid turns into vapor.

Phase transitions are critical because they involve energy changes. When water boils, it absorbs heat energy, allowing it to break intermolecular bonds and spread out into vapor form. This process ensues until the entire liquid becomes gas.

Understanding phase transitions helps in predicting how substances behave under various conditions and is fundamental in fields like chemical engineering and material sciences.
Entropy Change
Entropy, symbolized as \(S\), measures the disorder or randomness within a system. In thermodynamics, when a system undergoes a process, changes in entropy \((\Delta S)\) can reveal much about the system's evolution.

During a phase transition from liquid to gas, such as water boiling into steam, the system's molecules gain more freedom to move, increasing disorder. Because gases have higher entropy than liquids, \(\Delta S\) is positive, indicating an increase in disorder.
  • The molecules spread out and occupy more space.
  • They move more freely compared to the structured arrangement in the liquid state.
Understanding entropy change is essential because it provides insights into the direction of processes and the energy distribution within a system.

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Most popular questions from this chapter

Two moles of a perfect gas undergo the following processes: (a) a reversible isobaric expansion from \((1.0 \mathrm{~atm}, 20.0 \mathrm{~L})\) to \((1.0 \mathrm{~atm}, 40.0\) L); (b) a reversible isochoric change of state from \((1.0 \mathrm{~atm}, 40.0 \mathrm{~L})\) to \((0.5\) atm, \(40.0 \mathrm{~L}\) ); (c) a reversible isothermal compression from \((0.5 \mathrm{~atm}, 40.0 \mathrm{~L})\) to \((1.0\) atm, \(20.0 \mathrm{~L}\) ). (i) Sketch with labels each of the processes on the same \(P-V\) diagram. (ii) Calculate the total work \((W)\) and the total heat change \((q)\) involved in the above processes. (iii) What will be the values of \(\Delta U, \Delta H\) and \(\Delta S\) for the overall process?

Diborane is a potential rocket fuel which undergoes combustion according to the reaction. [2000 - 2 Marks] \(\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) From the following data, calculate the enthalpy change for the combustion of diborane. \(2 \mathrm{~B}(\mathrm{~s})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s}) \quad \Delta H=-1273 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell) \quad \Delta H=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \quad \Delta H=44 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(2 \mathrm{~B}(\mathrm{~s})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{~g}) \quad \Delta H=36 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

For the reaction ; \(\mathrm{A}(1) \rightarrow 2 \mathrm{~B}(\mathrm{~g})\) \(\Delta \mathrm{U}=2.1 \mathrm{kcal}, \Delta \mathrm{S}=20 \mathrm{cal} \mathrm{K}^{-1}\) at \(300 \mathrm{~K}\) Hence \(\Delta \mathrm{G}\) in kcal is.

The internal energy change (in \(\mathrm{J}\) ) when \(90 \mathrm{~g}\) of water undergoes complete evaporation at \(100^{\circ} \mathrm{C}\) is \(.\) (Given : \(\Delta \mathrm{H}_{\text {vap }}\) for water at \(373 \mathrm{~K}=41 \mathrm{~kJ} / \mathrm{mol}, \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) )

For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant \(K\) in terms of change in entropy is described by [Adv. 2017] (a) With increase in temperature, the value of \(K\) for exothermic reaction decreases because the entropy change of the system is positive (b) With increase in temperature, the value of \(K\) for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases (c) With increase in temperature, the value of \(K\) for endothermic reaction increases because the entropy change of the system is negative (d) With increase in temperature, the value of \(K\) for exothermic reaction decreases because favourable change in entropy of the surroundings decreases
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