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Consider a system consisting of an ice cube. (a) Under what conditions can the ice cube melt reversibly? (b) If the ice cube melts reversibly, is \(\Delta E\) zero for the process? Explain.

Short Answer

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(a) For an ice cube to melt reversibly, it must be in a state of thermodynamic equilibrium at each stage while the external pressure and temperature should change slowly or be constant, maintaining uniformity throughout the system. This process should occur at 0 degrees Celsius and 1 atmosphere of pressure. (b) ∆E is not zero for a reversible melting process. The internal energy of the system increases because heat (Q) is absorbed to convert the ice into liquid water, while no work (W) is done on the system due to no change in volume.

Step by step solution

01

(a) Conditions for Reversible Melting Process

To have a reversible melting process, several conditions need to be met. First, the system must be in a state of thermodynamic equilibrium at each stage of the process. Second, the external pressure and temperature should change very slowly or be constant, allowing the system to remain in equilibrium as it evolves. Finally, the pressure and temperature must be uniform throughout the system. In the case of the ice cube, it would need to be placed in an environment with the same temperature and pressure as the solid/liquid phase transition of water, which is 0 degrees Celsius and 1 atmosphere of pressure. The process should be carried out slowly, with controlled temperature and pressure changes, ensuring equilibrium at each stage.
02

(b) Determination of ∆E

When considering the change in internal energy (∆E), we need to recall the First Law of Thermodynamics, which states that: \[∆E = Q + W\] where Q is the heat transferred into the system and W is the work done on the system. For the melting process, the work done on the system (W) is zero because there is no change in volume during the melting process for water. The only energy transferred to the system comes in the form of heat (Q). During the reversible melting process, heat is added to the system to overcome the attractive forces between water molecules and convert the ice to liquid water. Since heat is absorbed by the system in this phase transition, the internal energy of the system increases. In conclusion, the change in internal energy (∆E) for a reversible melting process is not zero because heat (Q) is absorbed by the system to convert the ice into liquid water.

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Most popular questions from this chapter

(a) What is the meaning of the standard free-energy change, \(\Delta G^{\circ}\), as compared with \(\Delta G ?\) (b) For any process that occurs at constant temperature and pressure, what is the significance of \(\Delta G=0 ?(c)\) For a certain process, \(\Delta G\) is large and negative. Does this mean that the process necessarily occurs rapidly?

The value of \(K_{a}\) for nitrous acid \(\left(\mathrm{HNO}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to \(K_{a}\). (b) By using the value of \(K_{a}\) calculate \(\Delta G^{\circ}\) for the dissociation of nitrous acid in aqueous solution. (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when \(\left[\mathrm{H}^{+}\right]=5.0 \times 10^{-2} \mathrm{M}\), \(\left[\mathrm{NO}_{2}^{-}\right]=6.0 \times 10^{-4} M\), and \(\left[\mathrm{HNO}_{2}\right]=0.20 \mathrm{M} ?\)

(a) Express the second law of thermodynamics as a mathematical equation. (b) In a particular spontaneous process the entropy of the system decreases. What can you conclude about the sign and magnitude of \(\Delta S_{\text {surr }}\) ? (c) During a certain reversible process, the surroundings undergo an entropy change, \(\Delta S_{\text {surr }}=-78 \mathrm{~J} / \mathrm{K} .\) What is the entropy change of the system for this process?

Consider the reaction \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) .\) (a) Using data from Appendix \(C\), calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). (b) Calculate \(\Delta G\) at \(298 \mathrm{~K}\) if the partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are \(0.40 \mathrm{~atm}\) and \(1.60 \mathrm{~atm}\), respectively.

Predict the sign of the entropy change of the system for each of the following reactions: (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{Ba}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{BaO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l)\) (d) \(\mathrm{FeCl}_{2}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Fe}(s)+2 \mathrm{HCl}(g)\)

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