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Explain the important distinctions between each of the following pairs: (a) spontaneous and nonspontaneous processes; (b) the second and third laws of thermodynamics; (c) \(\Delta G\) and \(\Delta G^{\circ}\).

Short Answer

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In short, (a) spontaneous processes occur naturally while nonspontaneous demand external effort, (b) the second law of thermodynamics indicates heat always flows from hot to cold and entropy doesn't decrease, while the third law points out that at absolute zero, entropy reaches a constant. (c) \(\Delta G\) predicts spontaneity under given conditions, whereas \(\Delta G^{\circ}\) does this under standard conditions.

Step by step solution

01

Distinguishing Spontaneous and Nonspontaneous Processes

Spontaneous processes are those that occur naturally without any external influence. They increase the entropy of the universe. Nonspontaneous processes, on the other hand, need an external influence to occur. They decrease the entropy of the universe.
02

Differentiating the Second and Third Laws of Thermodynamics

The second law of thermodynamics implies that in any cyclic process, the entropy will either increase or remain the same; it never decreases. It also establishes that heat flows from a body at a higher temperature to a body at a lower temperature. The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches absolute zero.
03

Understanding \(\Delta G\) and \(\Delta G^{\circ}\)

\(\Delta G\) (Gibbs Free Energy Change) measures the total energy change of a system during a reaction at constant pressure and temperature. It predicts whether a reaction will be spontaneous. \(\Delta G^{\circ}\) indicates the change in Gibbs free energy under standard conditions (temperature 298K, pressure 1 atm, and all substances in their standard states).

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Key Concepts

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

Spontaneous and Nonspontaneous Processes
Understanding the difference between spontaneous and nonspontaneous processes is fundamental in chemistry and is often the first step in predicting how a reaction will proceed.

Spontaneous processes are phenomena that can occur without any external input of energy. A hallmark of spontaneous processes is that they tend to lead to an increase in the overall entropy of the universe. Entropy can be thought of as the measure of disorder within a system, and spontaneous processes tend to disperse energy and matter, creating more disorder. Examples of spontaneous processes include ice melting at room temperature and iron rusting.

Nonspontaneous processes, in contrast, require an input of energy to proceed because they lead to a decrease in the universe's entropy. An example would be freezing water into ice in a standard freezer. While this may seem counterintuitive because we encounter it so commonly, it's important to remember that the freezer is an external energy source, transferring heat from the water to the surrounding environment, thus making the process nonspontaneous.

In the context of chemical reactions, these concepts help predict whether a reaction can occur on its own, which is essential for understanding reaction possibilities and designing chemical processes.
Second and Third Laws of Thermodynamics
The second and third laws of thermodynamics set foundational rules for energy transitions within physical systems, profoundly impacting fields such as chemistry, physics, and engineering.

The second law is particularly well-known, encapsulating the idea that 'entropy of the universe always increases in a spontaneous process.' This law suggests that energy systems tend to evolve towards a state of maximum entropy or disorder. It also covers the concept that heat naturally flows from a hotter body to a colder one. What this means is that no process is completely efficient, as some energy is always lost to disorder, and hence perpetual motion machines cannot exist.

Meanwhile, the third law of thermodynamics states that as the temperature of a system uniformly approaches absolute zero, the entropy of the system approaches a constant minimum. This makes perfect crystalline substances at absolute zero the reference point for entropy measurements. This law is critical when considering the behaviour of substances near absolute zero and helps in calculating the absolute entropies of substances.
Gibbs Free Energy Change
Gibbs Free Energy Change, represented by the symbol \( \Delta G \), plays a pivotal role in thermodynamics and in predicting reaction spontaneity.

Essentially, \( \Delta G \) indicates whether a process or chemical reaction will occur spontaneously under constant pressure and temperature. A negative \( \Delta G \) value signifies that a process is exergonic (releases energy) and is spontaneous. Conversely, a positive \( \Delta G \) value implies that the process is endergonic (requires energy) and is nonspontaneous, needing an input of energy to proceed.

The concept of \( \Delta G \) is central to the study of chemical reactions, as it incorporates both enthalpy (heat content) and entropy (disorder) in a single value to predict the energy balance of a reaction. \( \Delta G \) is also temperature-dependent because entropy changes with temperature, altering the spontaneity of the reaction. Understanding how to calculate and interpret \( \Delta G \) is an integral skill for understanding reaction dynamics and for the practical design of chemical processes.
Entropy
Entropy is a fundamental concept in thermodynamics, symbolized by \( S \), and is often referred to as the measure of disorder or randomness in a system.

The tendency of systems to move towards greater entropy is a driving force behind many physical and chemical processes. This movement is intuitive if you consider that there are typically more disordered states (high entropy) available to a system than ordered states (low entropy). For instance, there are many more ways to arrange particles in a gas (high entropy) compared to a solid (low entropy).

In a chemical context, reactions resulting in an increase in entropy are usually favored. However, entropy is not the only factor that determines whether a reaction is spontaneous—enthalpy changes also play a significant role as captured by the Gibbs Free Energy formula: \( \Delta G = \Delta H - T\Delta S \). Understanding entropy and its effect on chemical systems is crucial for predicting reaction behavior, especially when combined with other thermodynamic quantities like temperature and enthalpy.

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Most popular questions from this chapter

Sodium carbonate, an important chemical used in the production of glass, is made from sodium hydrogen carbonate by the reaction \(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) Data for the temperature variation of \(K_{\mathrm{p}}\) for this reaction are \(K_{\mathrm{p}}=1.66 \times 10^{-5}\) at \(30^{\circ} \mathrm{C} ; 3.90 \times 10^{-4} \mathrm{at}\) \(50^{\circ} \mathrm{C} ; 6.27 \times 10^{-3}\) at \(70^{\circ} \mathrm{C} ;\) and \(2.31 \times 10^{-1}\) at \(100^{\circ} \mathrm{C}\) (a) Plot a graph similar to Figure \(19-12,\) and determine \(\Delta H^{\circ}\) for the reaction. (b) Calculate the temperature at which the total gas pressure above a mixture of \(\mathrm{NaHCO}_{3}(\mathrm{s})\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})\) is \(2.00 \mathrm{atm}\).

A handbook lists the following standard a handbook lists the following standard enthalpies of formation at \(298 \mathrm{K}\) for cyclopentane, \(\mathrm{C}_{5} \mathrm{H}_{10}: \quad \Delta H_{\mathrm{f}}^{\mathrm{g}}\left[\mathrm{C}_{5} \mathrm{H}_{10}(1)\right]=-105.9 \mathrm{kJ} / \mathrm{mol} \quad\) and \(\Delta H_{\mathrm{f}}^{\mathrm{o}}\left[\mathrm{C}_{5} \mathrm{H}_{10}(\mathrm{g})\right]=-77.2 \mathrm{kJ} / \mathrm{mol}\) (a) Estimate the normal boiling point of cyclopentane. (b) Estimate \(\Delta G^{\circ}\) for the vaporization of cyclopentane at \(298 \mathrm{K}\). (c) Comment on the significance of the sign of \(\Delta G^{\circ}\) at \(298 \mathrm{K}\)

A tabulation of more precise thermodynamic data than are presented in Appendix D lists the following values for \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(298.15 \mathrm{K},\) at a standard state pressure of 1 bar. $$\begin{array}{llll} \hline & \Delta H_{f}^{\circ}, & \Delta G_{f,}^{\circ} & S_{\prime}^{\circ} \\\ & \text { kJ mol }^{-1} & \text {kJ mol }^{-1} & \text {J mol }^{-1} \text {K }^{-1} \\ \hline \mathrm{H}_{2} \mathrm{O}(1) & -285.830 & -237.129 & 69.91 \\ \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) & -241.818 & -228.572 & 188.825 \\ \hline \end{array}$$ (a) Use these data to determine, in two different ways, \(\Delta G^{\circ}\) at \(298.15 \mathrm{K}\) for the vaporization: \(\mathrm{H}_{2} \mathrm{O}(1,1 \mathrm{bar}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g}, 1 \mathrm{bar}) .\) The value you obtain will differ slightly from that on page 838 because here, the standard state pressure is 1 bar, and there, it is 1 atm. (b) Use the result of part (a) to obtain the value of \(K\) for this vaporization and, hence, the vapor pressure of water at \(298.15 \mathrm{K}\) (c) The vapor pressure in part (b) is in the unit bar. Convert the pressure to millimeters of mercury. (d) Start with the value \(\Delta G^{\circ}=8.590 \mathrm{kJ}\), given on page 838 and calculate the vapor pressure of water at 298.15 K in a fashion similar to that in parts (b) and (c). In this way, demonstrate that the results obtained in a thermodynamic calculation do not depend on the convention we choose for the standard state pressure, as long as we use standard state thermodynamic data consistent with that choice.

Following are some standard Gibbs energies of formation, \(\Delta G_{f}^{2},\) per mole of metal oxide at \(1000 \mathrm{K}: \mathrm{NiO},\) \(-115 \mathrm{kJ} ; \mathrm{MnO},-280 \mathrm{kJ} ; \mathrm{TiO}_{2},-630 \mathrm{kJ} .\) The standard Gibbs energy of formation of \(\mathrm{CO}\) at \(1000 \mathrm{K}\) is \(-250 \mathrm{kJ}\) per mol CO. Use the method of coupled reactions (page 851 ) to determine which of these metal oxides can be reduced to the metal by a spontaneous reaction with carbon at \(1000 \mathrm{K}\) and with all reactants and products in their standard states.

To establish the law of conservation of mass, Lavoisier carefully studied the decomposition of mercury(II) oxide: $$\mathrm{HgO}(\mathrm{s}) \longrightarrow \mathrm{Hg}(1)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ At \(25^{\circ} \mathrm{C}, \Delta H^{\circ}=+90.83 \mathrm{kJ}\) and \(\Delta G^{\circ}=+58.54 \mathrm{kJ}\) (a) Show that the partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in equilibrium with \(\mathrm{HgO}(\mathrm{s})\) and \(\mathrm{Hg}(\mathrm{l})\) at \(25^{\circ} \mathrm{C}\) is extremely low. (b) What conditions do you suppose Lavoisier used to obtain significant quantities of oxygen?

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