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Chapter 16: Problem 33

Predict the sign of \(\Delta S_{\text {surr }}\) for the following processes. a. \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) b. \(\mathrm{I}_{2}(g) \longrightarrow \mathrm{I}_{2}(s)\)

Short Answer

Expert verified
For the processes: a. \(H_2O(l) \longrightarrow H_2O(g)\), the entropy change for the surroundings is positive. Therefore, \(\Delta S_{surr} > 0\). b. \(I_2(g) \longrightarrow I_2(s)\), the entropy change for the surroundings is negative. Therefore, \(\Delta S_{surr} < 0\).

Step by step solution

01

a. H2O(l) 鈫� H2O(g)

For the process: \(H_2O(l) \longrightarrow H_2O(g)\) The phase transition is from a liquid to a gas. Gaseous molecules have greater mobility and more possible arrangements than liquid molecules. As energy is transferred from the system to the surroundings, the disorder and randomness increase in the surroundings. As a result, the entropy change for the surroundings is positive. So, the sign of 螖Ssurr in this case is: \[\Delta S_{surr} > 0\]
02

b. I2(g) 鈫� I2(s)

For the process: \(I_2(g) \longrightarrow I_2(s)\) The phase transition is from a gas to a solid. Solid molecules have less mobility and fewer possible arrangements than gaseous molecules. As energy is transferred from the surroundings to the system, the disorder and randomness decrease in the surroundings. Therefore, the entropy change for the surroundings is negative. So, the sign of 螖Ssurr in this case is: \[\Delta S_{surr} < 0\]

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Most popular questions from this chapter

Consider the dissociation of a weak acid HA \(\left(K_{\mathrm{a}}=4.5 \times 10^{-3}\right)\) in water: $$\mathrm{HA}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{A}^{-}(a q)$$ Calculate \(\Delta G^{\circ}\) for this reaction at \(25^{\circ} \mathrm{C}.\)

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

Calculate \(\Delta G^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\) at \(600 . \mathrm{K}\) using the following data: $$\begin{aligned} &\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6} \text { at } 600 . \mathrm{K}\\\ &2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37} \mathrm{at} 600 . \mathrm{K} \end{aligned}$$

As \(\mathrm{O}_{2}(l)\) is cooled at 1 atm, it freezes at \(54.5 \mathrm{K}\) to form solid I. At a lower temperature, solid I rearranges to solid II, which has a different crystal structure. Thermal measurements show that \(\Delta H\) for the \(\mathrm{I} \rightarrow\) II phase transition is \(-743.1 \mathrm{J} / \mathrm{mol}\), and \(\Delta S\) for the same transition is \(-17.0 \mathrm{J} / \mathrm{K} \cdot\) mol. At what temperature are solids I and II in equilibrium?

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) b. \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{HCl}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)
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