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Chapter 17: Problem 95

Consider two perfectly insulated vessels. Vessel 1 initially contains an ice cube at 0∘C and water at 0∘C. Vessel 2 initially contains an ice cube at 0∘C and a saltwater solution at 0∘C. Consider the process H2O(s)→H2O(l) a. Determine the sign of ΔS,ΔSsurr , and ΔSuniv  for the process in vessel 1 . b. Determine the sign of ΔS,ΔSsurt , and ΔSuniv  for the process in vessel 2 . (Hint: Think about the effect that a salt has on the freezing point of a solvent.)

Short Answer

Expert verified
For Vessel 1, the signs of the entropy changes are: ∆S > 0, ∆S_surr = 0, and ∆S_univ > 0. For Vessel 2, the signs of the entropy changes are: ∆S > 0, ∆S_surr < 0, and the sign of ∆S_univ cannot be determined without more information about the specific process.

Step by step solution

01

Analyze the system

In vessel 1, there is an ice cube at 0°C and water at 0°C. When the ice cube melts, it undergoes a phase transition from solid to liquid. This process increases the disorder of the molecules since the liquid state has more randomness than the solid state. Therefore, the change in entropy of the system (∆S) during this process is positive (∆S > 0).
02

Analyze the surroundings

Since both the ice cube and the water in vessel 1 are initially at the same temperature of 0°C, there is no heat exchange between the system (ice cube) and the surroundings (water), which means the change in entropy of the surroundings (∆S_surr) during this process is zero (∆S_surr = 0).
03

Calculate the entropy change for the universe

The entropy change of the universe (∆S_univ) is the sum of the entropy changes of the system and the surroundings. In this case, we have: ∆S_univ = ∆S + ∆S_surr Since ∆S > 0 and ∆S_surr = 0, ∆S_univ is positive: ∆S_univ > 0 b. Vessel 2: Ice cube melting in saltwater
04

Analyze the system

In vessel 2, there is an ice cube at 0°C and a saltwater solution at 0°C. When the ice cube melts, it undergoes a phase transition from solid to liquid, just like in the previous case. Therefore, the change in entropy of the system (∆S) during this process is also positive (∆S > 0).
05

Analyze the surroundings

The presence of salt in the saltwater solution lowers its freezing point. Thus, even though the saltwater solution is initially at the same temperature as the ice cube (0°C), the melting of the ice cube requires heat to be absorbed from the surroundings. As heat flows from the surroundings (saltwater solution) to the system (ice cube), the entropy of the surroundings decreases. Therefore, the change in entropy of the surroundings during this process is negative (∆S_surr < 0).
06

Calculate the entropy change for the universe

The entropy change of the universe (∆S_univ) is the sum of the entropy changes of the system and the surroundings. In this case, we have: ∆S_univ = ∆S + ∆S_surr Since ∆S > 0 and ∆S_surr < 0, the sign of ∆S_univ is dependent on the magnitudes of the two changes in entropy. However, we cannot determine the sign of ∆S_univ without more information about the specific process.

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Key Concepts

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

Phase Transitions
Understanding phase transitions is essential for grasping many concepts in thermodynamics, which deal with changes in the state of matter. These transformations between states—such as solid, liquid, and gas—occur when energy in the form of heat is either absorbed or released, impacting the movement and arrangement of particles. For instance, when an ice cube melts, like in the example from the textbook, the solid ice with its orderly structured molecules absorbs heat and transitions into a less organized, liquid state.

During this phase change, the temperature of the substance remains constant while it absorbs heat, signifying a latent heat process. The degree of randomness, or entropy, increases, as seen in our vessel 1 scenario, because the liquid's molecules can move more freely than those in the solid. This increase in entropy is a significant prospect in predicting whether a phase transition will occur spontaneously, aligning with the second law of thermodynamics.
Thermodynamics
Thermodynamics is the branch of physical science concerned with heat and its relation to other forms of energy and work. It describes how thermal energy is converted to and from other forms of energy, and how it affects matter. The four laws of thermodynamics lay the foundation for various concepts like the energy changes during phase transitions, entropy, and the principles governing the direction of a chemical process.

For our ice cube scenario, thermodynamics directs us to examine the transfer of energy as heat during the melting process and the subsequent effects on both the system (ice cube) and the surroundings (water or saltwater solution). The first law of thermodynamics, concerning the conservation of energy, and the second law, centering around the natural increase of entropy, are particularly pertinent to understanding these changes during the ice's phase transition.
Gibbs Free Energy
The Gibbs free energy, denoted as 'G', is an essential concept connecting thermodynamics to chemical reactions and phase changes. It represents the maximum reversible work that may be performed by a thermodynamic system at a constant temperature and pressure. 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This is fundamental when considering why the ice melts in our exercise: at a constant temperature, melting is spontaneous if the process leads to a decrease in G.
Enthalpy
Enthalpy, symbolized as H, is a measure of the total heat content of a thermodynamic system under constant pressure. It reflects the ability of a system to do mechanical work as heat is transferred. Enthalpy is a state function, meaning its change—ΔH—depends only on the start and end states, not the path taken. In phase transitions, such as melting, the enthalpy change is associated with the heat absorbed or released.

The melting of an ice cube, as described in the textbook problem, involves a positive enthalpy change, ΔH > 0, because the system absorbs heat. This heat exchange is crucial in maintaining the temperature phase transition. Furthermore, understanding this enthalpic change allows for the calculation of ΔG and ultimately the assessment of the process' spontaneity. Enthalpy changes also play a role in calculating the system's entropy, thereby affecting the entropy change of the universe ΔS_univ.

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Most popular questions from this chapter

Consider the following reaction: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$Calculate \(\Delta G\) for this reaction under the following conditions (assume an uncertainty of \(\pm 1\) in all quantities): a. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=P_{\mathrm{H}_{2}}=200 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=50 \mathrm{~atm}\) b. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=200 \mathrm{~atm}, P_{\mathrm{H}_{2}}=600 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=200 \mathrm{~atm}\)

Is \(\Delta S_{\text {surr }}\) favorable or unfavorable for exothermic reactions? Endothermic reactions? Explain.

Consider the reaction $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ where \(\Delta H^{\circ}=-103.8 \mathrm{~kJ} / \mathrm{mol} .\) In a particular experiment, equal moles of \(\mathrm{H}_{2}(\mathrm{~g})\) at \(1.00 \mathrm{~atm}\) and \(\mathrm{Br}_{2}(\mathrm{~g})\) at \(1.00 \mathrm{~atm}\) were mixed in a \(1.00\) -L flask at \(25^{\circ} \mathrm{C}\) and allowed to reach equilibrium. Then the molecules of \(\mathrm{H}_{2}\) at equilibrium were counted using a very sensitive technique, and \(1.10 \times 10^{13}\) molecules were found. For this reaction, calculate the values of \(K, \Delta G^{\circ}\), and \(\Delta S^{\circ}\).

Gas \(\mathrm{A}_{2}\) reacts with gas \(\mathrm{B}_{2}\) to form gas \(\mathrm{AB}\) at a constant temperature. The bond energy of \(\mathrm{AB}\) is much greater than that of either reactant. What can be said about the sign of \(\Delta H ? \Delta S_{\text {sumr }} ? \Delta S ?\) Explain how potential energy changes for this process. Explain how random kinetic energy changes during the process.

Consider the following reaction: \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) \quad K_{298}=0.090\) For \(\mathrm{Cl}_{2} \mathrm{O}(g)\), \(\Delta G_{\mathrm{f}}^{\circ}=97.9 \mathrm{~kJ} / \mathrm{mol}\) \(\Delta H_{\mathrm{f}}^{\circ}=80.3 \mathrm{~kJ} / \mathrm{mol}\) \(S^{\circ}=266.1 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) a. Calculate \(\Delta G^{\circ}\) for the reaction using the equation \(\Delta G^{\circ}=\) \(-R T \ln (K)\) b. Use bond energy values (Table 8.4) to estimate \(\Delta H^{\circ}\) for the reaction. c. Use the results from parts a and \(b\) to estimate \(\Delta S^{\circ}\) for the reaction. d. Estimate \(\Delta H_{\mathrm{f}}^{\circ}\) and \(S^{\circ}\) for \(\mathrm{HOCl}(g)\). e. Estimate the value of \(K\) at \(500 . \mathrm{K}\). f. Calculate \(\Delta G\) at \(25^{\circ} \mathrm{C}\) when \(P_{\mathrm{H}_{2} \mathrm{O}}=18\) torr, \(P_{\mathrm{Cl}_{2} \mathrm{O}}=2.0\) torr, and \(P_{\mathrm{HOCl}}=0.10\) torr.
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