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Chapter 18: Problem 37

Explain the following nursery rhyme in terms of the second law of thermodynamics. Humpty Dumpty sat on a wall; Humpty Dumpty had a great fall. All the King's horses and all the King's men Couldn't put Humpty together again.

Short Answer

Expert verified
The nursery rhyme 'Humpty Dumpty' can be interpreted in terms of the second law of thermodynamics. Humpty Dumpty's fall and subsequent breakage represent an irreversible increase in entropy in a system, consistent with the second law of thermodynamics. The failed attempts to restore Humpty Dumpty to his original state underscore that in a closed system, processes increasing entropy are irreversible.

Step by step solution

01

Identify the Initial State

Initially, Humpty Dumpty is sitting on a wall, which represents a stable, ordered state.
02

Recognize the Change

Humpty Dumpty falls and breaks, which represents a change in state. This change is dramatic and disorderly as he shatters into pieces.
03

Apply the Second Law of Thermodynamics

The change from an ordered state (whole Humpty Dumpty) to a disordered state (broken Humpty Dumpty) is an example of increasing entropy. The second law of thermodynamics suggests that this process is irreversible in an isolated system because the overall entropy cannot decrease.
04

Interpret the Final Passage

All the King's horses and all the King's men couldn't put Humpty together again. This line directly relates to the concept of irreversibility in the second law of thermodynamics. The inability of the King's men and horses to reverse Humpty's broken state represents the fact that processes in an isolated system are irreversible when entropy increases.

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Most popular questions from this chapter

From the values of \(\Delta H\) and \(\Delta S,\) predict which of the following reactions would be spontaneous at \(25^{\circ} \mathrm{C}\) : Reaction \(\mathrm{A}: \Delta H=10.5 \mathrm{~kJ} / \mathrm{mol}, \Delta S=30 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) reaction \(\mathrm{B}: \Delta H=1.8 \mathrm{~kJ} / \mathrm{mol}, \Delta S=-113 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). If either of the reactions is nonspontaneous at \(25^{\circ} \mathrm{C}\), at what temperature might it become spontaneous?

Calculate \(\Delta G^{\circ}\) for the following reactions at \(25^{\circ} \mathrm{C}\) : (a) \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)\) See Appendix 2 for thermodynamic data.

Give a detailed example of each of the following, with an explanation: (a) a thermodynamically spontaneous process; (b) a process that would violate the first law of thermodynamics; (c) a process that would violate the second law of thermodynamics; (d) an irreversible process; (e) an equilibrium process.

In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to produce \(\mathrm{Ni}(\mathrm{CO})_{4},\) which is a gas and can therefore be separated from solid impurities: $$ \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) $$ Given that the standard free energies of formation of \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) are \(-137.3 \mathrm{~kJ} / \mathrm{mol}\) and \(-587.4 \mathrm{~kJ} / \mathrm{mol}\), respectively, calculate the equilibrium constant of the reaction at \(80^{\circ} \mathrm{C}\). Assume that \(\Delta G_{f}^{\circ}\) is temperature independent.

Explain what is meant by a spontaneous process. Give two examples each of spontaneous and nonspontaneous processes.
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