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At 1 atm, liquid water is heated above \(100^{\circ} \mathrm{C}\) . For this process, which of the following choices (i-iv) is correct for \(\Delta S_{\text { surr }} ? \Delta S ?\) \(\Delta S_{\text { univ }} ?\) Explain each answer. i. greater than zero ii. less than zero iii. equal to zero iv. cannot be determined

Short Answer

Expert verified
The correct answer for the various entropy changes when liquid water is heated above \(100^{\circ} \mathrm{C}\) at 1 atm is: i. greater than zero, where \(\Delta S\) (the system) is greater than zero, \(\Delta S_{\text{surr}}\) (the surroundings) is less than zero, and \(\Delta S_{\text{univ}}\) (the universe) is greater than zero.

Step by step solution

01

Understand the process

In this exercise, we are given that liquid water is heated above 100°C at 1 atmosphere of pressure. We need to evaluate the entropy changes of the system (∆S), the surroundings (∆Ssurr), and the universe (∆Suniv) and determine which of the given choices is correct. Step 2: Identify the entropy change in the system
02

Entropy change in the system

When we heat liquid water above 100°C (its boiling point at atmospheric pressure), the water turns into steam. During this phase transition, the entropy increases, as the molecules gain more freedom to move in the gaseous state compared to the liquid state. Therefore, the entropy change of the system, \(\Delta S\), is greater than zero. Step 3: Identify the entropy change in the surroundings
03

Entropy change in the surroundings

In order to heat the water and cause the phase transition, heat must be transferred from the surroundings to the water. When heat flows out of the surroundings, the surroundings' temperature decreases, as heat energy is lost. This causes an entropy decrease for the surroundings. Therefore, \(\Delta S_{\text{surr}}\) is less than zero. Step 4: Identify the entropy change in the universe
04

Entropy change in the universe

The second law of thermodynamics states that the entropy of an isolated system (in this case, the universe consisting of the system and the surroundings) cannot decrease, it can only increase or remain constant. Since the entropy change of the system is greater than zero and the entropy change of the surroundings is less than zero, we need to evaluate if their sum is greater than, equal to, or less than zero. In spontaneous processes, the entropy of the universe increases, which means \(\Delta S_{\text{univ}}\) should be greater than zero. Step 5: Determine the correct answer choice based on the entropy changes
05

Choose the correct answer

Based on the conclusions we reached in Steps 2, 3, and 4, we can deduce the following: - ∆S (the system) is greater than zero; - ∆Ssurr (the surroundings) is less than zero; - ∆Suniv (the universe) is greater than zero. Taking all three conclusions into account, we can conclude that the correct answer choice is: i. greater than zero

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Key Concepts

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

Phase Transition
When discussing phase transitions, we're essentially talking about changes between different states of matter: solid, liquid, and gas. A prime example is when water transitions from a liquid to a gas upon heating above its boiling point of 100°C at 1 atm of pressure.

During this process, water molecules move from a tightly packed arrangement in the liquid phase to a more dispersed and energetic arrangement in the gaseous phase. This change is accompanied by an increase in entropy, which is a measure of disorder or randomness in the system.

The energy input required for water to undergo this phase transition is significant. It is why the entropy change, \(\Delta S\), for the system is positive. The more freedom water molecules gain, the higher the entropy. Thus, transitions like evaporation, where materials move to a higher kinetic energy state, tend to increase the system's entropy.
Second Law of Thermodynamics
The second law of thermodynamics is a fundamental principle that guides how energy and entropy behave in the universe. It states that the total entropy of an isolated system can never decrease over time. In other words, natural processes tend to move towards a state of maximum entropy.

This law is crucial in understanding why certain processes happen spontaneously. When we heat liquid water above 100°C, it begins to evaporate, increasing its entropy as it becomes gas. During this, although the system's entropy is increasing, \(\Delta S \, \) the surroundings' entropy, \( \Delta S_{\text{surr}} \, \) might decrease because energy is being used to heat the water. Yet, the second law dictates that the overall entropy of the universe, \( \Delta S_{\text{univ}} \, \) must increase, meaning that the increase in the system's entropy more than compensates for any decrease in the surroundings.
Spontaneous Processes
The concept of spontaneous processes refers to those changes that naturally occur without external intervention. A key indicator of spontaneity in a thermodynamic process is the increase in the universe's total entropy.

In our exercise, the heating of water and its conversion into steam are examples of spontaneous processes. Even though energy is transferred into the system, making the surroundings lose entropy, the overall change results in a positive \( \Delta S_{\text{univ}} \, \) indicating a net increase in entropy.

For a process to be spontaneous, the sum of the entropy changes of the system and its surroundings must be greater than zero. This is aligned with the second law and is often observed in natural phenomena, like water evaporating, ice melting, or even the dissolution of sugar in coffee.

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Most popular questions from this chapter

Consider the following reaction at 298 K: $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ An equilibrium mixture contains \(\mathrm{O}_{2}(g)\) and \(\mathrm{SO}_{3}(g)\) at partial pressures of 0.50 \(\mathrm{atm}\) and 2.0 \(\mathrm{atm}\) , respectively. Using data from Appendix \(4,\) determine the equilibrium partial pressure of \(\mathrm{SO}_{2}\) in the mixture. Will this reaction be most favored at a high or a low temperature, assuming standard conditions?

Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{f} / k_{\mathrm{r}},\) where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation \(\left(k=A e^{-E_{2} / R T}\right)\) b. Why is the following statement false? "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction."

Consider the reaction: $$\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)$$ At \(25^{\circ} \mathrm{C}, \Delta G^{\circ}=-92.50 \mathrm{kJ}\) Which of the following statements is (are) true? a. This is an endothermic reaction. b. \(\Delta S^{\circ}\) for this reaction is negative. c. If the temperature is increased, the ratio \(\frac{\mathrm{PCl}_{5}}{\mathrm{PCl}_{3}}\) will increase. d. \(\Delta G^{\circ}\) for this reaction has to be negative at all temperatures. e. When \(\Delta G^{\circ}\) for this reaction is negative, then \(K\) is greater than 1.00 .

Describe how the following changes affect the positional probability of a substance. a. increase in volume of a gas at constant T b. increase in temperature of a gas at constant V c. increase in pressure of a gas at constant T

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. \(\mathrm{I}_{2}(g) \longrightarrow \mathrm{I}_{2}(s)\)

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