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The crystalline hydrate \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s)\) loses water when placed in a large, closed, dry vessel at room temperature: $$ \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ This process is spontaneous and \(\Delta H^{\circ}\) is positive at room temperature. (a) What is the sign of \(\Delta S^{\circ}\) at room temperature? (b) If the hydrated compound is placed in a large, closed vessel that already contains a large amount of water vapor, does \(\Delta S^{\circ}\) change for this reaction at room temperature?

Short Answer

Expert verified
(a) \(\Delta S^{\circ}\) is positive at room temperature. (b) No, placing the hydrated compound in a large, closed vessel already containing a large amount of water vapor does not change the intrinsic entropy change (\(\Delta S^{\circ}\)) of the reaction at room temperature.

Step by step solution

01

Calculate change in entropy \(\Delta S^{\circ}\) and its sign at room temperature

The relationship between Gibbs free energy (\(\Delta G^{\circ}\)), enthalpy (\(\Delta H^{\circ}\)), and entropy (\(\Delta S^{\circ}\)) is given by the equation: \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\] Since the process is spontaneous, the change in Gibbs free energy (\(\Delta G^{\circ}\)) must be negative. We are given that the change in enthalpy (\(\Delta H^{\circ}\)) is positive for this process. The only way to satisfy a negative value for \(\Delta G^{\circ}\) is if the term \(T\Delta S^{\circ}\) is greater in magnitude than \(\Delta H^{\circ}\) and therefore has a positive value: \[\Delta H^{\circ} - T\Delta S^{\circ} < 0 \Rightarrow T\Delta S^{\circ} > \Delta H^{\circ}\] Since room temperature is a positive value, the change in entropy (\(\Delta S^{\circ}\)) must also be positive to satisfy the inequality. Thus, we can conclude that: (a) \(\Delta S^{\circ}\) is positive at room temperature.
02

Effect of additional water vapor on \(\Delta S^{\circ}\) at room temperature

Placing the hydrated compound in a large, closed vessel already containing a large amount of water vapor will not change the intrinsic change in entropy (\(\Delta S^{\circ}\)) as \(\Delta S^{\circ}\) depends on the reaction itself and not on the initial conditions of the system. However, it can potentially affect partial pressures of the water below and above the hydrate, which in turn can affect the spontaneity of the process. Nevertheless, the intrinsic change in entropy at room temperature for the reaction will remain the same. Therefore, we can conclude that: (b) No, placing the hydrated compound in a large, closed vessel already containing a large amount of water vapor does not change the intrinsic entropy change (\(\Delta S^{\circ}\)) of the reaction at room temperature.

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Key Concepts

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

Spontaneity
The spontaneity of a reaction refers to whether a process will occur naturally without any external influence. A spontaneous process is one that can proceed on its own. An important component to determine spontaneity is the Gibbs free energy change, represented as \( \Delta G^{\circ} \).
- If \( \Delta G^{\circ} \) is negative, the reaction is spontaneous.- If \( \Delta G^{\circ} \) is positive, the reaction is non-spontaneous.In the example of the crystalline hydrate losing water, the process is noted as spontaneous, meaning it releases water vapor without requiring additional energy or changes. This natural shift, despite having a positive enthalpy change, is still favorable because the entropy change \( \Delta S^{\circ} \) is significant enough to drive the reaction.forward.
Enthalpy
Enthalpy, represented as \( \Delta H^{\circ} \), is a measure of the total heat content of a system. Changes in enthalpy relate to the heat absorbed or released due to a reaction at constant pressure.
- A positive \( \Delta H^{\circ} \) means the reaction absorbs heat, also known as an endothermic reaction.- A negative \( \Delta H^{\circ} \) indicates the release of heat, known as an exothermic reaction.In the crystalline hydrate example, the enthalpy change is positive, signifying that the process of releasing water vapor absorbs heat. Despite this, because the process is spontaneous, another factor—like a significant increase in entropy—compensates for the positive \( \Delta H^{\circ} \). This dissociation highlights that enthalpy alone doesn't determine spontaneity, but must be considered along with entropy and temperature.
Gibbs Free Energy
Gibbs free energy is a crucial thermodynamic quantity, denoted as \( \Delta G^{\circ} \), that dictates the spontaneity of a reaction at constant temperature and pressure. It combines enthalpy, temperature, and entropy through the formula:\[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \]
The equation shows how temperature and entropy impact the Gibbs free energy and ultimately the spontaneity:- Lower \( \Delta G^{\circ} \) (i.e., negative) suggests that a reaction will occur spontaneously.In the example of the crystalline hydrate, even with a positive \( \Delta H^{\circ} \), the process remains spontaneous. This indicates that the \( T\Delta S^{\circ} \) term is sufficiently large in positive value, making \( \Delta G^{\circ} \) negative. This interplay highlights how reactions that absorb heat can still proceed naturally if they result in increases in disorder or randomness (entropy). Thus, Gibbs free energy effectively predicts the feasibility of reactions in varying conditions.

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Most popular questions from this chapter

(a) What sign for \(\Delta S\) do you expect when the pressure on 0.600 mol of an ideal gas at \(350 \mathrm{~K}\) is increased isothermally from an initial pressure of \(76.0 \mathrm{kPa} ?(\mathbf{b})\) If the final pressure on the gas is \(121.6 \mathrm{kPa}\), calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change?

(a) Can endothermic chemical reactions be spontaneous? (b) Can a process be spontaneous at one temperature and nonspontaneous at a different temperature? (c) Water can be decomposed to form hydrogen and oxygen, and the hydrogen and oxygen can be recombined to form water. Does this mean that the processes are thermodynamically reversible? (d) Does the amount of work that a system can doon its Id on the nath of the nrocese?

Isomersare moleculesthat havethesamechemical formula but different arrangements of atoms, as shown here for two isomers of pentane, \(\mathrm{C}_{5} \mathrm{H}_{12} .\) (a) Do you expect a significant difference in the enthalpy of combustion of the two isomers? Explain. (b) Which isomer do you expect to have the higher standard molar entropy? Explain. \([\) Section 19.4\(]\)

(a) What is the difference between a state and a microstate of a system? (b) As a system goes from state A to state B, its entropy decreases. What can you say about the number of microstates corresponding to each state? (c) In a particular spontaneous process, the number of microstates available to the system decreases. What can you conclude about the sign of \(\Delta S\) surr?

The oxidation of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) in body tissue produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} .\) In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and \(\mathrm{CO}_{2}\). (a) Using data given in Appendix \(\mathrm{C}\), compare the equilibrium constants for the following reactions: $$ \begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \rightleftharpoons 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(I)+2 \mathrm{CO}_{2}(g) \end{array} $$ (b) Compare the maximum work that can be obtained from these processes under standard conditions.

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