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Chapter 17: Problem 42

Predict the sign of \(\Delta S^{\circ}\) for each of the following changes. a. \(\mathrm{K}(s)+\frac{1}{2} \mathrm{Br}_{2}(g) \longrightarrow \mathrm{KBr}(s)\) b. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) c. \(\mathrm{KBr}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{Br}^{-}(a q)\) d. \(\mathrm{KBr}(s) \longrightarrow \mathrm{KBr}(l)\)

Short Answer

Expert verified
a. \(\Delta S^{\circ}<0\) (The system becomes more ordered as a solid and a gas form a solid.) b. \(\Delta S^{\circ}<0\) (The number of gas particles decreases, reducing the disorder in the system.) c. \(\Delta S^{\circ}>0\) (A solid dissociates into aqueous ions, increasing the system's disorder.) d. \(\Delta S^{\circ}>0\) (The system's disorder increases as solid KBr becomes liquid KBr.)

Step by step solution

01

Reaction (a)

For the reaction K(s) + 1/2 Br2(g) → KBr(s), we will compare the disorder between the reactants and products. In this reaction, the solid potassium and gaseous bromine combine to form a solid potassium bromide. As a gaseous substance has more disorder than a solid one and the product is a solid, the overall reaction decreases the disorder in the system. Therefore, the entropy change for this reaction will be negative.
02

Reaction (b)

For the reaction N2(g) + 3 H2(g) → 2 NH3(g), we will compare the disorder of the reactants and products. In this case, all reactants and products are gases. The initial reaction molecules are one molecule of nitrogen and three molecules of hydrogen, which combine to form two molecules of ammonia. When comparing the initial number of gas particles (4) to the final gas particles (2), we can see that the disorder of the system decreases in this reaction. Therefore, the entropy change for this reaction is also negative.
03

Reaction (c)

For the reaction KBr(s) → K+(aq) + Br-(aq), the solid potassium bromide dissociates into potassium and bromide ions in a liquid solution. In this case, the initial state is a solid, and the final state is a mixture of aqueous ions. The disorder in the system increases as aqueous ions have more freedom to move and interact than solid particles. Therefore, the entropy change for this reaction is positive.
04

Reaction (d)

For the reaction KBr(s) → KBr(l), solid potassium bromide turns into a liquid state. The difference in disorder between the solid state and the liquid state is that liquid particles can move more freely than solid particles. In this case, the disorder of the system increases during the reaction, meaning the entropy change for this reaction is positive.

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Most popular questions from this chapter

Some water is placed in a coffee-cup calorimeter. When \(1.0 \mathrm{~g}\) of an ionic solid is added, the temperature of the solution increases from \(21.5^{\circ} \mathrm{C}\) to \(24.2^{\circ} \mathrm{C}\) as the solid dissolves. For the dissolving process, what are the signs for \(\Delta S_{\text {sys }}, \Delta S_{\text {surr }}\), and \(\Delta S_{\text {univ }}\) ?

Predict the sign of \(\Delta S\) for each of the following and explain. a. the evaporation of alcohol b. the freezing of water c. compressing an ideal gas at constant temperature d. dissolving \(\mathrm{NaCl}\) in water

Which of the following reactions (or processes) are expected to have a negative value for \(\Delta S^{\circ}\) ? a. \(\mathrm{SiF}_{6}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)+\mathrm{SiF}_{4}(g)\) b. \(4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)\) c. \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g)\) d. \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) e. \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\)

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(2 \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{~S}_{\text {rhombic }}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\) b. \(2 \mathrm{SO}_{3}(g) \longrightarrow 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\) c. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\)

For rubidium \(\Delta H_{\mathrm{vap}}^{\circ}=69.0 \mathrm{~kJ} / \mathrm{mol}\) at \(686^{\circ} \mathrm{C}\), its boiling point. Calculate \(\Delta S^{\circ}, q, w\), and \(\Delta E\) for the vaporization of \(1.00 \mathrm{~mole}\) of rubidium at \(686^{\circ} \mathrm{C}\) and \(1.00\) atm pressure.
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