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(a) Express the second law of thermodynamics as a mathematical equation. (b) In a particular spontaneous process the entropy of the system decreases. What can you conclude about the sign and magnitude of \(\Delta S_{\text {surr }} ?\) (c) During a certain reversible process, the surroundings undergo an entropy change, \(\Delta S_{\text {surr }}=-78 \mathrm{~J} / \mathrm{K}\). What is the entropy change of the system for this process?

Short Answer

Expert verified
(a) The second law of thermodynamics can be expressed as \(\Delta S_\text{total} = \Delta S_\text{system} + \Delta S_\text{surr} \geq 0\). (b) For a spontaneous process with a decrease in system entropy, \(\Delta S_\text{surr} > 0\) and \(\Delta S_\text{surr} > |\Delta S_\text{system}|\) must be satisfied. (c) Given \(\Delta S_\text{surr} = -78 \mathrm{~J} / \mathrm{K}\) for a reversible process, the entropy change of the system is \(\Delta S_\text{system} = 78 \mathrm{~J} / \mathrm{K}\).

Step by step solution

01

a) Second Law of Thermodynamics

The second law of thermodynamics states that for any spontaneous (natural) process, the total entropy of a system and its surroundings always increases, or in mathematical terms: \(\Delta S_\text{total} = \Delta S_\text{system} + \Delta S_\text{surr} \geq 0\)
02

b) Spontaneous Process with Entropy Decrease

In this case, we are given a spontaneous process where the entropy of the system decreases, meaning \(\Delta S_\text{system} < 0\). Since the process is spontaneous, the total entropy change should be non-negative, or \(\Delta S_\text{total} \geq 0\). From the second law of thermodynamics equation, we can conclude that for the spontaneous process to occur, the entropy change of the surroundings, \(\Delta S_\text{surr}\), must be positive and greater in magnitude compared to the entropy decrease of the system: \(\Delta S_\text{surr} > 0\) and \(\Delta S_\text{surr} > |\Delta S_\text{system}|\)
03

c) Entropy Change of the System

We are given that during a certain reversible process, the surroundings undergo an entropy change of \(\Delta S_\text{surr} = -78 \mathrm{~J} / \mathrm{K}\). For a reversible process, the total entropy change is always zero, or: \(\Delta S_\text{total} = \Delta S_\text{system} + \Delta S_\text{surr} = 0\) Now we can solve for the entropy change of the system, \(\Delta S_\text{system}\), as follows: \(\Delta S_\text{system} = -(\Delta S_\text{surr})\) Plugging in the given value for \(\Delta S_\text{surr}\), we have: \(\Delta S_\text{system} = -(-78 \mathrm{~J} / \mathrm{K}) = 78 \mathrm{~J} / \mathrm{K}\) Thus, the entropy change of the system during this reversible process is \(78 \mathrm{~J} / \mathrm{K}\).

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Key Concepts

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

Entropy Change
Entropy can be thought of as a measure of disorder or randomness in a system. A highly ordered system has low entropy, while a more disordered system has high entropy. The second law of thermodynamics tells us that the total entropy of an isolated system always tends to increase over time.

In mathematical terms, the change in entropy, denoted by \(\Delta S\), is a key concept in understanding how energy is transferred within a system as heat. If a process leads to a more disordered state, \(\Delta S\) is positive, indicating an increase in entropy. Conversely, if the order within a system increases, \(\Delta S\) is negative — but such processes often require an external influence since they don’t happen spontaneously.

The concept of entropy is crucial in determining whether a reaction or process can occur naturally. Intuitively, it's like seeing a room become messier over time — that’s a natural increase in entropy. Cleaning up the room decreases its entropy, but requires energy and effort — a parallel to non-spontaneous processes.
Spontaneous Process
A spontaneous process is a change that occurs naturally without needing to be driven by an external energy source. For a process to be spontaneous, it doesn't need to happen quickly — it simply means that the process can proceed without any outside intervention once it has been initiated.

It's crucial to understand that, according to the second law of thermodynamics, for a spontaneous process to occur, the total entropy change of the system and its surroundings (\(\Delta S_{total}\)) must be equal to or greater than zero.

For instance, when ice melts into water at room temperature, it's a spontaneous process. The system (ice and water) may see an increase in entropy as it becomes less ordered (solid to liquid), while the surrounding environment may slightly decrease in entropy due to heat absorption, but the total entropy of the system and surroundings will increase. This aligns with the principle that the total entropy of the universe is always increasing.
Reversible Process
A reversible process in thermodynamics is an idealized or theoretical process that happens in such a manner that the system and surroundings can be returned to their original states without any net change to the universe. In other words, a reversible process can proceed in both forward and reverse directions without loss of energy and without changing the total entropy.

In practice, all natural processes are irreversible because they produce an increase in the total entropy of the system and surroundings. However, reversible processes are a valuable conceptual tool, allowing us to define a maximum efficiency for thermodynamic systems, such as engines.

An example of an ideal reversible process is the isothermal expansion of an ideal gas. If done infinitesimally slowly, the gas can be compressed back to its original volume without any net entropy change. Reversible processes are a cornerstone in thermodynamics because they represent the best-case scenario, giving us a benchmark for the efficiency of real, irreversible processes.
System and Surroundings
In thermodynamics, any region or quantity of matter under study is called a system. Everything external to the system that has the potential to affect it is called the surroundings. Systems can be isolated, closed, or open depending on whether they exchange energy or matter with their surroundings.

Isolated systems do not exchange heat, work, or matter with the surroundings, whereas closed systems can exchange energy but not matter, and open systems can exchange both. In the context of the second law of thermodynamics, the exchange of heat between a system and its surroundings is what drives the change in entropy.

Understanding the interactions between a system and its surroundings is key to determining how energy transformations will proceed. It isn't enough to just look at what happens within the boundaries of the system; one must also consider the impact on and inputs from the environment around it to fully grasp thermodynamic processes like heat flow, work, and changes in internal energy.

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Most popular questions from this chapter

Aceticacid can be manufactured by combining methanol with carbon monoxide, an example of a carbonylation reaction: $$ \mathrm{CH}_{3} \mathrm{OH}(l)+\mathrm{CO}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{COOH}(l) $$ (a) Calculate the equilibrium constant for the reaction at \(25^{\circ} \mathrm{C}\). (b) Industrially, this reaction is run at temperatures above \(25^{\circ} \mathrm{C}\). Will an increase in temperature produce an increase or decrease in the mole fraction of acetic acid at equilibrium? Why are elevated temperatures used? (c) At what temperature will this reaction have an equilibrium constant equal to \(1 ?\) (You may assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, and you may ignore any phase changes that might occur.)

The value of \(K_{a}\) for nitrous acid \(\left(\mathrm{HNO}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to \(K_{a} .\) (b) By using the value of \(K_{a}\) calculate \(\Delta G^{\circ}\) for the dissociation of nitrous acid in aqueous solution. (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when \(\left[\mathrm{H}^{+}\right]=5.0 \times 10^{-2} \mathrm{M}\), \(\left[\mathrm{NO}_{2}^{-}\right]=6.0 \times 10^{-4} M\), and \(\left[\mathrm{HNO}_{2}\right]=0.20 \mathrm{M?}\)

Propanol \(\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\right)\) melts at \(-126.5^{\circ} \mathrm{C}\) and boils at \(97.4^{\circ} \mathrm{C}\). Draw a qualitative sketch of how the entropy changes as propanol vapor at \(150^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) is cooled to solid propanol at \(-150^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Consider the vaporization of liquid water to steam at a pressure of 1 atm. (a) Is this process endothermic or exothermic? (b) In what temperature range is it a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) At what temperature are the two phases in equilibrium?

Consider the polymerization of ethylene to polyethylene. cos (Section 12.6) (a) What would you predict for the sign of the entropy change during polymerization ( \(\Delta S_{\text {poly }}\) )? Explain your reasoning, (b) The polymerization of ethylene is a spontaneous process at room temperature. What can you conclude about the enthalpy change during polymerization \(\left(\Delta H_{\text {poly }}\right) ?(\mathrm{c})\) Use average bond enthalpies (Table 8.4) to estimate the value of \(\Delta H_{\text {poly }}\) per ethylene monomer added. (d) Polyethylene is an addition polymer. By comparison, Nylon 66 is a condensation polymer. How would you expect \(\Delta S_{\text {poly }}\) for a condensation polymer to compare to that for an addition polymer? Explain.

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