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The Gibbs energy change of a reaction can be used to assess (a) how much heat is absorbed from the surroundings; (b) how much work the system does on the surroundings; (c) the net direction in which the reaction occurs to reach equilibrium; (d) the proportion of the heat evolved in an exothermic reaction that can be converted to various forms of work.

Short Answer

Expert verified
a) The Gibbs energy change doesn't directly measure the heat absorbed from the surroundings, that's measured by the enthalpy change. b) Gibbs free energy change reflects the work a system can do on its surroundings excluding expansionary work. c) The direction of the reaction towards equilibrium is determined by whether Gibbs energy change is negative (forward reaction) or positive (reverse reaction). d) Gibbs energy change doesn't specifically indicate the proportion of evolved heat that can be converted to work in an exothermic reaction, but it can be inferred from the signs of \( \Delta H \) and \( \Delta S \).

Step by step solution

01

Understanding Gibbs Energy

The Gibbs energy change (\( \Delta G \)) is defined as \( \Delta G = \Delta H - T\Delta S \) where \( \Delta H \) is the change in enthalpy (total energy), \( T \) is the absolute temperature and \( \Delta S \) is change in entropy (disorder). Thus, the change in Gibbs Free Energy accounts for both the energy taken in or given out during a reaction (\( \Delta H \)), and the energy absorbed or released due to changes in disorder (\( T\Delta S \)).
02

Answer to (a)

(a) Gibbs energy change doesn't directly measure the amount of heat absorbed from the surroundings but it relies on the enthalpy change (\( \Delta H \)) which is a measure of the total energy change, part of which relates to heat absorbed or released.
03

Answer to (b)

(b) Gibbs energy change directly shows how much reversible work a system can do on its surroundings, excluding expansion work. If \( \Delta G \) is negative, the system can do work to the surroundings. If \( \Delta G \) is positive, work is done on the system by the surroundings.
04

Answer to (c)

(c) The direction in which the reaction occurs to reach equilibrium is directly related to the Gibbs energy change. If \( \Delta G \) is negative, the reaction will proceed forward spontaneously. If \( \Delta G \) is positive, the reaction will proceed in the reverse direction spontaneously.
05

Answer to (d)

(d) The Gibbs energy change doesn't specifically let us know the proportion of the heat evolved that can be converted to various forms of work. However, in an exothermic reaction (\( \Delta H < 0 \)), if there is also an increase in disorder (\( \Delta S > 0 \)), then a portion of the heat released can be used by the system to do work.

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Key Concepts

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

Enthalpy Change
Enthalpy change, represented by the symbol \( \Delta H \), is a measure of the total heat content change in a chemical reaction. It tells us how much energy, in the form of heat, is either absorbed from or released into the surroundings during the reaction. An exothermic reaction, where \( \Delta H < 0 \), releases heat to the surroundings, making it feel warm to the touch. On the contrary, an endothermic reaction absorbs heat, signified by \( \Delta H > 0 \), making the surroundings feel cooler.

Understanding enthalpy change is crucial because it helps predict whether a reaction will be product-favored under constant pressure. Furthermore, knowing \( \Delta H \) allows us to calculate the energy change in a reaction, which is important from the perspective of energy resources and requirements in chemical processes.
Entropy Change
Entropy change, denoted as \( \Delta S \), is a measure of the disorder or randomness within a system. The second law of thermodynamics states that the total entropy of an isolated system always increases over time. This means that systems naturally progress from a state of order to a state of disorder.

When \( \Delta S > 0 \) there's an increase in entropy, indicating a more disordered system after the reaction. Conversely, \( \Delta S < 0 \) implies a decrease in entropy, representing a more ordered system. Entropy change plays a pivotal role in determining reaction spontaneity through the Gibbs energy equation, as a positive entropy change can drive a reaction to be spontaneous even if it's endothermic.
Reaction Spontaneity
The spontaneity of a chemical reaction refers to whether a reaction can proceed without any external input of energy. Spontaneity is governed by the Gibbs free energy change \( \Delta G \), which incorporates both enthalpy and entropy changes according to the relationship \( \Delta G = \Delta H - T\Delta S \).

A negative \( \Delta G \) indicates a spontaneous reaction, as the process can happen on its own under certain conditions. Positive \( \Delta G \) means the reaction is non-spontaneous and requires external energy to proceed. It's essential to note that spontaneity doesn't imply the rate at which a reaction will occur—some spontaneous reactions can be incredibly slow.
Exothermic Reaction
An exothermic reaction is one where energy, primarily in the form of heat, is released to the surroundings. It is characterized by a negative enthalpy change \( \Delta H < 0 \). These reactions often feel hot as they transfer thermal energy to the environment. Exothermic reactions are common in everyday life, including combustion processes like burning wood or fossil fuels, and metabolic reactions in our bodies.

Moreover, exothermic reactions can be spontaneous when also accompanied by an increase in entropy \( \Delta S > 0 \), leading to \( \Delta G \) being negative. However, the ability to do work, such as electrical or mechanical work, depends on both the heat released and the change in entropy during the reaction.

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Most popular questions from this chapter

From the data given in the following table, determine \(\Delta S^{\circ} \quad\) for the reaction \(\quad \mathrm{NH}_{3}(\mathrm{g})+\mathrm{HCl}(\mathrm{g}) \longrightarrow\) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) .\) All data are at \(298 \mathrm{K}\) $$\begin{array}{lcc} \hline & \Delta H_{f}^{\circ}, \mathrm{kJ} \mathrm{mol}^{-1} & \Delta G_{f,}^{\circ} \mathrm{kJ} \mathrm{mol}^{-1} \\ \hline \mathrm{NH}_{3}(\mathrm{g}) & -46.11 & -16.48 \\ \mathrm{HCl}(\mathrm{g}) & -92.31 & -95.30 \\ \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) & -314.4 & -202.9 \\ \hline \end{array}$$

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?

The Gibbs energy available from the complete combustion of 1 mol of glucose to carbon dioxide and water is $$\begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})+6 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 6 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta G^{\circ}=-2870 \mathrm{kJ} \mathrm{mol}^{-1} \end{array}$$ (a) Under biological standard conditions, compute the maximum number of moles of ATP that could form from ADP and phosphate if all the energy of combustion of 1 mol of glucose could be utilized. (b) The actual number of moles of ATP formed by a cell under aerobic conditions (that is, in the presence of oxygen) is about \(38 .\) Calculate the efficiency of energy conversion of the cell. (c) Consider these typical physiological conditions. $$\begin{array}{l} P_{\mathrm{CO}_{2}}=0.050 \mathrm{bar} ; P_{\mathrm{O}_{2}}=0.132 \mathrm{bar} \\\ {[\mathrm{glucose}]=1.0 \mathrm{mg} / \mathrm{mL} ; \mathrm{pH}=7.0} \\ {[\mathrm{ATP}]=[\mathrm{ADP}]=\left[P_{\mathrm{i}}\right]=0.00010 \mathrm{M}} \end{array}$$ Calculate \(\Delta G\) for the conversion of 1 mol ADP to ATP and \(\Delta G\) for the oxidation of 1 mol glucose under these conditions. (d) Calculate the efficiency of energy conversion for the cell under the conditions given in part (c). Compare this efficiency with that of a diesel engine that attains \(78 \%\) of the theoretical efficiency operating with \(T_{\mathrm{h}}=1923 \mathrm{K}\) and \(T_{1}=873 \mathrm{K} .\) Suggest a reason for your result. [ Hint: See Feature Problem 95.]

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.

The following data are given for the two solid forms of \(\mathrm{HgI}_{2}\) at \(298 \mathrm{K}\). $$\begin{array}{llll} \hline & \Delta H_{f}^{\circ} & \Delta G_{f,}^{\circ} & S^{\circ} \\ & \text { kJ mol }^{-1} & \text {kJ mol }^{-1} & \text {J mol }^{-1} \text {K }^{-1} \\ \hline \mathrm{HgI}_{2} \text { (red) } & -105.4 & -101.7 & 180 \\ \mathrm{Hg} \mathrm{I}_{2} \text { (yellow) } & -102.9 & (?) & (?) \\ \hline \end{array}$$ Estimate values for the two missing entries. To do this, assume that for the transition \(\mathrm{HgI}_{2}(\mathrm{red}) \longrightarrow\) \(\mathrm{HgI}_{2}(\text { yellow }),\) the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) at \(25^{\circ} \mathrm{C}\) have the same values that they do at the equilibrium temperature of \(127^{\circ} \mathrm{C}\).

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