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Chapter 19: Problem 37

Does the entropy of the system increase, decrease, or stay the same when (a) a solid melts, (b) a gas liquefies, (c) a solid sublimes?

Short Answer

Expert verified
When a solid melts, the entropy of the system increases. When a gas liquefies, the entropy of the system decreases. When a solid sublimes, the entropy of the system increases.

Step by step solution

01

Understand the process of melting

In the process of melting, a solid becomes a liquid. The particles in a solid are tightly packed and have a well-defined structure, while the particles in a liquid have more freedom to move around and are less structured.
02

Analyze the change in entropy

Since the particles in a liquid have more freedom to move around and are less structured, there is an increase in the number of possible energy states. This means that the entropy of the system increases.
03

Answer: Entropy increases during melting

When a solid melts, the entropy of the system increases. (B) A gas liquefies:
04

Understand the process of liquefaction

In the process of liquefaction, a gas becomes a liquid. The particles in a gas are far apart and have a lot of freedom to move around, while the particles in a liquid are relatively closer and have less freedom to move around.
05

Analyze the change in entropy

Since the particles in a liquid have less freedom to move around than in a gas, there is a decrease in the number of possible energy states. This implies that the entropy of the system decreases.
06

Answer: Entropy decreases during liquefaction

When a gas liquefies, the entropy of the system decreases. (C) A solid sublimes:
07

Understand the process of sublimation

In the process of sublimation, a solid becomes a gas without going through the liquid phase. The particles in a solid are tightly packed and have a well-defined structure, while the particles in a gas are far apart and have a lot of freedom to move around.
08

Analyze the change in entropy

Since the particles in a gas have more freedom to move around than in a solid, there is an increase in the number of possible energy states. This results in an increase in the entropy of the system.
09

Answer: Entropy increases during sublimation

When a solid sublimes, the entropy of the system increases.

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Most popular questions from this chapter

A certain reaction has \(\Delta H^{\circ}=+23.7 \mathrm{~kJ}\) and \(\Delta 5^{\circ}=+52.4 \mathrm{~J} / \mathrm{K}\). (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K}\). (d) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

The reaction \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{MgO}(s)\) is highly spontaneous. A classmate calculates the entropy change for this reaction and obtains a large negative value for \(\Delta S^{5}\). Did your classmate make a mistake in the calculation? Explain.

] Consider a system that conststs of two standard playing dice, with the state of the system defined by the sum of the values shown on the top faces. (a) The two arrangements of top faces shown here can be viewed as two possible microstates of the system. Explain. (b) To which state does each microstate correspond? (c) How many possible states are there for the system? (d) Which state or states have the highest entropy? Explain. (e) Which state or states have the lowest entropy? Explain. (f) Calculate the absolute entropy of the two-dice system.

Indicate whether \(\Delta G\) increases, decreases, or does not change when the partial pressure of \(\mathrm{H}_{2}\) is increased in each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(2 \mathrm{HBr}(\mathrm{g}) \longrightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(g)\) (c) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})\)

Using data from Appendix \(C\), write the equilibrium-constant expression and calculate the value of the equilibrium constant and the free-energy change for these reactions at \(298 \mathrm{~K}\) : (a) \(\mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{HBr}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{Br}_{2}(g)\) (c) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\)
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