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Chapter 19: Problem 3

Use ideas from this chapter to explain this famous remark attributed to Rudolf Clausius (1865)\(:^{\prime \prime} \mathrm{Die}\) Energie der Welt ist konstant; die Entropie der Welt strebt einem Maximum zu." ("The energy of the world is constant; the entropy of the world increases toward a maximum.")

Short Answer

Expert verified
The famous remark by Clausius is a restatement of the First Law of Thermodynamics and the Second Law of Thermodynamics respectively. In essence, energy in the universe remains constant (cannot be created or destroyed), while the universe is gradually moving towards a state of maximum disorder or entropy.

Step by step solution

01

Understanding the Energy of the World is Constant

The first part of Clausius's statement refers to the First Law of Thermodynamics. This law explains that energy cannot be created or destroyed, only transformed or transferred. This means, in a closed system (like the world), the total amount of energy remains constant.
02

Understanding the Entropy of the World Increases Toward a Maximum

The second part of Clausius's statement refers to the Second Law of Thermodynamics. This law states that the entropy (level of disorder) in any closed system always tends to increase or, in the best case, stays constant. This means over time, systems naturally progress towards a state of maximum disorder.
03

The Implication of the Statement

Clausius's statement 'The energy of the world is constant; the entropy of the world increases toward a maximum' summarizes the fundamental laws of thermodynamics. It paints an image of a world powered by the constant conversion and redistribution of energy where disorder naturally and constantly increases.

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Most popular questions from this chapter

The following standard Gibbs energy changes are given for \(25^{\circ} \mathrm{C}\) (1) \(\mathrm{SO}_{2}(\mathrm{g})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow \operatorname{COS}(\mathrm{g})+2 \mathrm{CO}_{2}(\mathrm{g})\) \(\Delta G^{\circ}=-246.4 \mathrm{kJ}\) (2) \(\mathrm{CS}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \operatorname{COS}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) \(\Delta G^{\circ}=-41.5 \mathrm{kJ}\) (3) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) \longrightarrow \operatorname{COS}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\) \(\Delta G^{\circ}=+1.4 \mathrm{kJ}\) (4) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\) \(\Delta G^{\circ}=-28.6 \mathrm{kJ}\) Combine the preceding equations, as necessary, to obtain \(\Delta G^{\circ}\) values for the following reactions. (a) \(\operatorname{COS}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow\) \(\begin{aligned} \mathrm{SO}_{2}(\mathrm{g})+\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) & \Delta G^{\circ}=? \end{aligned}\) (b) \(\cos (g)+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow\) \(\mathrm{SO}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \quad \Delta G^{\circ}=?\) \(\left.+\quad \mathrm{H}_{\mathrm{O}} \mathrm{C}(\mathrm{d})=\mathrm{CO}_{-}^{\circ} \mathrm{G}\right)+\mathrm{H}_{-}^{-} \mathrm{S}(\mathrm{q})\) (c) \(\cos (\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) \(\Delta G^{\circ}=?\) Of reactions (a), (b), and (c), which is spontaneous in the forward direction when reactants and products are present in their standard states?

Briefly describe each of the following ideas, methods, or phenomena: (a) absolute molar entropy; (b) coupled reactions; (c) Trouton's rule; (d) evaluation of an equilibrium constant from tabulated thermodynamic data.

\(\mathrm{H}_{2}(\mathrm{g})\) can be prepared by passing steam over hot iron: \(3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g})\) (a) Write an expression for the thermodynamic equilibrium constant for this reaction. (b) Explain why the partial pressure of \(\mathrm{H}_{2}(\mathrm{g})\) is independent of the amounts of \(\mathrm{Fe}(\mathrm{s})\) and \(\mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})\) present. (c) Can we conclude that the production of \(\mathrm{H}_{2}(\mathrm{g})\) from \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) could be accomplished regardless of the proportions of \(\mathrm{Fe}(\mathrm{s})\) and \(\mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})\) present? Explain.

To establish the law of conservation of mass, Lavoisier carefully studied the decomposition of mercury(II) oxide: $$\mathrm{HgO}(\mathrm{s}) \longrightarrow \mathrm{Hg}(1)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ At \(25^{\circ} \mathrm{C}, \Delta H^{\circ}=+90.83 \mathrm{kJ}\) and \(\Delta G^{\circ}=+58.54 \mathrm{kJ}\) (a) Show that the partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in equilibrium with \(\mathrm{HgO}(\mathrm{s})\) and \(\mathrm{Hg}(\mathrm{l})\) at \(25^{\circ} \mathrm{C}\) is extremely low. (b) What conditions do you suppose Lavoisier used to obtain significant quantities of oxygen?

Sodium carbonate, an important chemical used in the production of glass, is made from sodium hydrogen carbonate by the reaction \(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) Data for the temperature variation of \(K_{\mathrm{p}}\) for this reaction are \(K_{\mathrm{p}}=1.66 \times 10^{-5}\) at \(30^{\circ} \mathrm{C} ; 3.90 \times 10^{-4} \mathrm{at}\) \(50^{\circ} \mathrm{C} ; 6.27 \times 10^{-3}\) at \(70^{\circ} \mathrm{C} ;\) and \(2.31 \times 10^{-1}\) at \(100^{\circ} \mathrm{C}\) (a) Plot a graph similar to Figure \(19-12,\) and determine \(\Delta H^{\circ}\) for the reaction. (b) Calculate the temperature at which the total gas pressure above a mixture of \(\mathrm{NaHCO}_{3}(\mathrm{s})\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})\) is \(2.00 \mathrm{atm}\).
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