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Chapter 17: Problem 29

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. \(I_{2}(g) \longrightarrow I_{2}(s)\)

Short Answer

Expert verified
a. For the process \(H_2O(l) \longrightarrow H_2O(g)\), \(\Delta S_{\text{surr}} < 0\). b. For the process \(I_2(g) \longrightarrow I_2(s)\), \(\Delta S_{\text{surr}} > 0\).

Step by step solution

01

Process a: \(H_2O(l) \longrightarrow H_2O(g)\)

This process involves the conversion of water from liquid state to gaseous state. When water evaporates, it absorbs heat from the surroundings, causing the surroundings to cool down. As heat flows out of the surroundings, the entropy of the surroundings decreases. Therefore, for this process, \(\Delta S_{\text{surr}}\) is negative.
02

Process b: \(I_2(g) \longrightarrow I_2(s)\)

This process involves the conversion of iodine from gaseous state to solid state. When iodine undergoes this change, it releases heat to the surroundings, causing the surroundings to warm up. As heat flows into the surroundings, the entropy of the surroundings increases. Therefore, for this process, \(\Delta S_{\text{surr}}\) is positive. To summarize: a. \(\Delta S_{\text{surr}} < 0\) for the process \(H_2O(l) \longrightarrow H_2O(g)\) b. \(\Delta S_{\text{surr}} > 0\) for the process \(I_2(g) \longrightarrow I_2(s)\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phase Change
A phase change occurs when a substance transitions between different states of matter, such as solid, liquid, and gas. During a phase change, the temperature of the substance remains constant, as all the energy provided or released is used to alter the state, not the temperature.
For instance, when water evaporates from liquid to gas, the process requires energy input, known as the heat of vaporization. This energy is absorbed from the surroundings, resulting in a cooling effect. Conversely, when a substance like iodine transitions from gas to solid, it releases energy, which warms the surroundings. These transformations impact the energy exchange and play a significant role in determining the changes in entropy of the surroundings.
Thermodynamics
Thermodynamics is the branch of science concerned with heat, work, and the forms of energy involved in processes. It provides the foundational principles to understand the behavior of energy during chemical reactions and physical transformations.
The second law of thermodynamics is vital here, as it introduces the concept of entropy. This law states that the total entropy of an isolated system can never decrease over time. It governs the directional flow of heat and predicts the feasibility of a process. During a phase change, such as water evaporating, thermodynamics dictates that energy is absorbed from the surroundings, often leading to a decrease in the surrounding's entropy. Alternatively, when iodine solidifies, the heat released increases the surroundings鈥� entropy.
Surroundings Entropy
Surroundings entropy refers to the change in the disorder of the environment due to a particular process. When a process occurs, like a phase change, energy exchanges happen with the surroundings, intriguingly affecting its entropy.
Several factors determine the sign of the change in surroundings entropy (\(\Delta S_{\text{surr}}\)):
  • If a process absorbs heat from the surroundings, as in the case of water evaporation (\(H_2O(l) \longrightarrow H_2O(g)\)), the surrounding's entropy decreases (\(\Delta S_{\text{surr}} < 0\)).
  • Conversely, if a process releases heat to the surroundings, like when iodine solidifies (\(I_2(g) \longrightarrow I_2(s)\)), the surrounding's entropy increases (\(\Delta S_{\text{surr}} > 0\)).
Understanding how these processes impact surrounding entropy can help predict the feasibility and spontaneity of chemical reactions in environmental systems.

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Most popular questions from this chapter

Two crystalline forms of white phosphorus are known. Both forms contain \(\mathrm{P}_{4}\) molecules, but the molecules are packed together in different ways. The \(\alpha\) form is always obtained when the liquid freezes. However, below \(-76.9^{\circ} \mathrm{C}\), the \(\alpha\) form spontaneously converts to the \(\beta\) form: $$\mathrm{P}_{4}(s, \alpha) \longrightarrow \mathrm{P}_{4}(s, \beta)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\) for this process. b. Predict which form of phosphorus has the more ordered crystalline structure (has the smaller positional probability).

It is quite common for a solid to change from one structure to another at a temperature below its melting point. For example, sulfur undergoes a phase change from the rhombic crystal structure to the monoclinic crystal form at temperatures above \(95^{\circ} \mathrm{C}\). a. Predict the signs of \(\Delta H\) and \(\Delta S\) for the process \(S_{\text {rhcmbic }} \longrightarrow\) \(\mathrm{S}_{\text {monoclinic }}\) b. Which form of sulfur has the more ordered crystalline structure (has the smaller positional probability)?

Using data from Appendix 4, calculate \(\Delta G\) for the reaction $$2 \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{~S}_{\text {mombic }}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ for the following conditions at \(25^{\circ} \mathrm{C}\) : $$\begin{array}{l}P_{\mathrm{H}_{2} \mathrm{~S}}=1.0 \times 10^{-4} \mathrm{~atm} \\\P_{\mathrm{SO}_{2}}=1.0 \times 10^{-2} \mathrm{~atm} \\ P_{\mathrm{H}_{2} \mathrm{O}}=3.0 \times 10^{-2} \mathrm{~atm}\end{array}$$

Consider the reactions $$\begin{aligned}\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) & \longrightarrow \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q) \\ \mathrm{Ni}^{2+}(a q)+3 \mathrm{en}(a q) & \longrightarrow \mathrm{Ni}(\mathrm{en})_{3}^{2+}(a q)\end{aligned}$$ where $$\text { en }=\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$$ The \(\Delta H\) values for the two reactions are quite similar, yet \(K_{\text {reaction } 2}>K_{\text {reaction } 1 .}\) Explain.

For each of the following pairs of substances, which substance has the greater value of \(S^{\circ} ?\) a. \(C_{\text {graphite }}(s)\) or \(C_{\text {diamond }}(s)\) b. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{g})\) c. \(\mathrm{CO}_{2}(s)\) or \(\mathrm{CO}_{2}(g)\)
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