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Ch.18 - Free Energy and Thermodynamics
Tro - Chemistry: A Molecular Approach 4th Edition
Tro4th EditionChemistry: A Molecular ApproachISBN: 9780134112831Not the one you use?Change textbook
All textbooksTro 4th EditionCh.18 - Free Energy and ThermodynamicsProblem 99
Chapter 18, Problem 99

Is the sign of ΔSuniv for each process positive or negative? Explain for the following: b. the electrolysis of H2O(l) to H2(g) and O2(g) at 298 K c. the growth of an oak tree from a little acorn.

Verified step by step guidance
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insert step 1> Identify the processes involved in each scenario. For part b, the process is the electrolysis of water, and for part c, it is the growth of an oak tree from an acorn.
insert step 2> Recall the second law of thermodynamics, which states that the entropy of the universe (ΔS_{univ}) increases for spontaneous processes.
insert step 3> For part b, consider the electrolysis of water: 2H_2O(l) \rightarrow 2H_2(g) + O_2(g). This process involves breaking bonds in water molecules to form hydrogen and oxygen gases, which increases the disorder or entropy of the system.
insert step 4> For part c, consider the growth of an oak tree: This process involves organizing simple molecules into complex structures, which decreases the entropy of the system. However, the overall entropy change of the universe must be considered, including the surroundings.
insert step 5> Determine the sign of ΔS_{univ} for each process by considering the changes in entropy of the system and surroundings. For part b, the increase in gas molecules suggests a positive ΔS_{univ}. For part c, the decrease in system entropy is offset by energy exchanges with the surroundings, potentially leading to a positive ΔS_{univ} if the process is spontaneous.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Entropy (ΔS)

Entropy is a measure of the disorder or randomness in a system. In thermodynamics, the change in entropy (ΔS) can indicate whether a process is spontaneous. A positive ΔS suggests an increase in disorder, while a negative ΔS indicates a decrease in disorder. Understanding how entropy changes during a process is crucial for determining the sign of ΔSuniv.
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Second Law of Thermodynamics

The Second Law of Thermodynamics states that the total entropy of an isolated system can never decrease over time. It implies that natural processes tend to move towards a state of maximum disorder or entropy. This law is essential for analyzing processes like electrolysis and biological growth, as it helps predict whether the overall entropy change (ΔSuniv) will be positive or negative.
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Spontaneity of Processes

A process is considered spontaneous if it occurs without external intervention, often associated with a positive change in the universe's entropy (ΔSuniv > 0). For the electrolysis of water, the process is non-spontaneous under standard conditions, while the growth of an oak tree is a spontaneous biological process. Evaluating spontaneity involves analyzing both the system and surroundings to determine the overall entropy change.
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Related Practice
Textbook Question

Indicate and explain the sign of ΔSuniv for each process. a. 2 H2(g) + O2(g) → 2 H2O (l) at 298 K.

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Textbook Question

The values of ΔG°f for the hydrogen halides become less negative with increasing atomic number. The ΔG°f of HI is slightly positive. However, the trend in ΔS°f is to become more positive with increasing atomic number. Explain.

Textbook Question

Dinitrogen tetroxide decomposes to nitrogen dioxide: N2O4(g) → 2 NO2(g) ΔH°rxn = 55.3 kJ At 298 K, a reaction vessel initially contains 0.100 atm of N2O4. When equilibrium is reached, 58% of the N2O4 has decomposed to NO2. What percentage of N2O4 decomposes at 388 K? Assume that the initial pressure of N2O4 is the same (0.100 atm).

Textbook Question

A metal salt with the formula MCl2 crystallizes from water to form a solid with the composition MCl2 • 6 H2O. The equilibrium vapor pressure of water above this solid at 298 K is 18.3 mmHg. What is the value of ΔG for the reaction MCl2 • 6 H2O(s) ⇌ MCl2(s) + 6 H2O(g) when the pressure of water vapor is 18.3 mmHg? When the pressure of water vapor is 760 mmHg?