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(a) What is the meaning of the standard free-energy change, \(\Delta G^{\circ}\), as compared with \(\Delta G\) ? (b) For any process that occurs at constant temperature and pressure, what is the significance of \(\Delta G=0 ?(c)\) For a certain process, \(\Delta G\) is large and negative. Does this mean that the process necessarily occurs rapidly?

Short Answer

Expert verified
(a) \(\Delta G^{\circ}\) is the standard free-energy change under standard conditions, whereas \(\Delta G\) is the free-energy change under specific conditions. (b) \(\Delta G = 0\) under constant temperature and pressure signifies that the system is at equilibrium, with the forward and reverse reaction rates being equal. (c) A large negative \(\Delta G\) indicates a thermodynamically favorable process, but it does not necessarily imply a rapid process, as reaction kinetics also play a role in determining the rate of the process.

Step by step solution

01

Part (a): Difference between \(\Delta G^{\circ}\) and \(\Delta G\)

The standard free-energy change, denoted by \(\Delta G^{\circ}\), refers to the difference in free-energy between the products and reactants under standard conditions (pressure of 1 atm and room temperature of 298K), while \(\Delta G\) represents the free-energy change under any given set of conditions (not necessarily standard). In summary, \(\Delta G^{\circ}\) is the free-energy change under standard conditions and \(\Delta G\) is the free-energy change under specific conditions.
02

Part (b): Significance of \(\Delta G = 0\) under constant temperature and pressure

When \(\Delta G = 0\) under constant temperature and pressure, it means that the system has reached equilibrium. In other words, the forward reaction rate equals the reverse reaction rate, and there is no net change in the concentrations of both the reactants and products. This indicates that the process is balanced and spontaneous in both directions.
03

Part (c): Relationship between large negative \(\Delta G\) and reaction rate

A large negative \(\Delta G\) indicates that the process is thermodynamically favorable and that it will proceed spontaneously. However, it does not necessarily mean that the process occurs rapidly. The rate of the process is determined by the reaction kinetics, which involve the activation energy and the reaction rate constant. Therefore, a thermodynamically favorable process (large negative \(\Delta G\)) could still have a slow rate if its activation energy is high or the reaction rate constant is low.

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Key Concepts

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

Thermodynamics
When exploring the concept of Gibbs Free Energy in thermodynamics, it's essential to differentiate between the standard free-energy change \( \Delta G^{\circ} \) and the general free-energy change \( \Delta G \). The standard change \( \Delta G^{\circ} \) is relevant under specific conditions, namely a pressure of 1 atmosphere and a temperature of 298K, which is approximately room temperature. This gives us a baseline to compare how free energy varies under these controlled settings.
On the other hand, \( \Delta G \) indicates the change in free energy for any given process outside of these standard conditions. This makes \( \Delta G \) more applicable to real-world scenarios, where temperatures and pressures often differ from the controlled standard. Understanding this distinction is crucial because it helps predict whether a reaction or process will occur spontaneously.
Thermodynamics is concerned with energy differences, not the rate of reactions, which is where reaction kinetics plays a crucial role. Knowing the difference between \( \Delta G^{\circ} \) and \( \Delta G \) helps us understand the potential energy changes involved in chemical reactions, and plays an important role in predicting how systems behave.
Reaction Kinetics
Even if a process has a large negative \( \Delta G \), indicating it is thermodynamically favorable, it doesn't guarantee that it will happen quickly. This is where reaction kinetics comes into play. Reaction kinetics considers how quickly reactants convert into products. Two main aspects that influence this rate are the activation energy required for the reaction to occur and the reaction rate constant.
Activation energy is the initial energy needed to start a reaction. If this energy is high, the reaction rate might be slow, despite a favorable \( \Delta G \). The reaction rate constant, often influenced by temperature and the presence of a catalyst, also determines how fast a reaction occurs.
  • A reaction can be spontaneous due to its negative free energy but still require a lot of time to reach completion because of these kinetic factors.
  • Therefore, those working with reactions must consider both thermodynamics and kinetics to fully understand and predict reaction behavior.
While thermodynamics lets us know if a reaction is possible, reaction kinetics tell us how fast it would proceed.
Equilibrium
Gibbs Free Energy plays a crucial role in understanding chemical equilibrium. When \( \Delta G = 0 \), this indicates that a system has reached equilibrium under given conditions. This equilibrium doesn't imply that chemical processes have ceased, rather it means that the rates of the forward and reverse reactions are balanced.
At equilibrium, no net change is observed in the concentrations of reactants and products, leading to a stable system. It's the point where the forward and reverse reactions occur at the same rate, thus maintaining the system's state over time.
In thermodynamics, equilibrium is described as the state of maximum entropy and minimum free energy. This signifies that the system is in its most stable form and spontaneous changes are no longer favorably directed in any one direction.
  • Therefore, when \( \Delta G = 0 \), it's an indicator of a system's balanced nature.
  • This concept is crucial in predicting how chemical reactions progress and how various conditions impact the position of equilibrium.
Understanding equilibrium through Gibbs Free Energy helps in analyzing the conditions necessary to shift the equilibrium position, which is essential in both industrial and laboratory settings.

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Most popular questions from this chapter

Cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) is a liquid hydrocarbon at room temperature. (a) Write a balanced equation for the combustion of \(\mathrm{C}_{6} \mathrm{H}_{12}(l)\) to form \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). (b) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ}\)

Carbon disulfide \(\left(\mathrm{CS}_{2}\right)\) is a toxic, highly flam mable substance. The following thermodynamic data are available for \(\mathrm{CS}_{2}(l)\) and \(\mathrm{CS}_{2}(g)\) at \(298 \mathrm{~K}\) : \begin{tabular}{lrl} \hline & \(\Delta H_{f}^{\circ}(\mathbf{k J} / \mathrm{mol})\) & \(\Delta G_{f}^{0}(\mathbf{k J} / \mathrm{mol})\) \\ \hline \(\mathrm{CS}_{2}(l)\) & \(89.7\) & \(65.3\) \\ \(\mathrm{CS}_{2}(g)\) & \(117.4\) & \(67.2\) \\ \hline \end{tabular} (a) Draw the Lewis structure of the molecule. What do you predict for the bond order of the \(\mathrm{C}-\mathrm{S}\) bonds? (b) Use the VSEPR method to predict the structure of the \(\mathrm{CS}_{2}\) molecule. (c) Liquid \(\mathrm{CS}_{2}\) bums in \(\mathrm{O}_{2}\) with a blue flame, forming \(\mathrm{CO}_{2}(g)\) and \(\mathrm{SO}_{2}(g)\). Write a balanced equation for this reaction. (d) Using the data in the preceding table and in Appendix \(C\), calculate \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) for the reaction in part (c). Is the reaction exothermic? Is it spontaneous at 298 K? (e) Use the data in the preceding table to calculate \(\Delta S^{\circ}\) at \(298 \mathrm{~K}\) for the vaporization of \(\mathrm{CS}_{2}(l)\). Is the sign of \(\Delta S^{\circ}\) as you would expect for a vaporization? (f) Using data in the preceding table and your answer to part (e), estimate the boiling point of \(\mathrm{CS}_{2}(\mathrm{l})\). Do you predict that the substance will be a liquid or a gas at \(298 \mathrm{~K}\) and \(1 \mathrm{~atm}\) ?

For each of the following processes, indicate whether the signs of \(\Delta S\) and \(\Delta H\) are expected to be positive, negative, or about zero. (a) A solid sublimes. (b) The temperature of a sample of \(\mathrm{Co}(s)\) is lowered from \(60^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\) (c) Ethyl alcohol evaporates from a beaker. (d) \(\mathrm{A}\) diatomic molecule dissociates into atoms. (e) A piece of charcoal is combusted to form \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

(a) Express the second law of thermodynamics in words. (b) If the entropy of the system increases during a reversible process, what can you say about the entropy change of the surroundings? (c) In a certain spontaneous process the system undergoes an entropy change, \(\Delta S=42 \mathrm{~J} / \mathrm{K} .\) What can you conclude about \(\Delta S_{\text {surr }} ?\)

Consider the following equilibrium: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ Thermodynamic data on these gases are given in Appendix C. You may assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not vary with temperature. (a) At what temperature will an equilibrium mixture contain equal amounts of the two gases? (b) At what temperature will an equilibrium mixture of 1 atm total pressure contain twice as much \(\mathrm{NO}_{2}\) as \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (c) At what temperature will an equilibrium mixture of \(10 \mathrm{~atm}\) total pressure contain twice as much \(\mathrm{NO}_{2}\) as \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (d) Rationalize the results from parts (b) and (c) by using Le Châtelier's principle. \(\infty\) (Section 15.7)

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