what is the molar ratio of g6p to f6p at equilibrium at 298k

Chapter 16. Thermodynamics

16.4 Free Energy

Learning Objectives

By the end of this section, yous will be able to:

• Define Gibbs free energy, and describe its relation to spontaneity
• Calculate free energy alter for a process using free energies of formation for its reactants and products
• Summate free free energy change for a process using enthalpies of formation and the entropies for its reactants and products
• Explain how temperature affects the spontaneity of some processes
• Relate standard gratis energy changes to equilibrium constants

One of the challenges of using the second law of thermodynamics to determine if a process is spontaneous is that nosotros must decide the entropy change for the system and the entropy change for the surroundings. An alternative approach involving a new thermodynamic property defined in terms of system backdrop only was introduced in the late nineteenth century past American mathematician Josiah Willard Gibbs. This new property is chosen the Gibbs gratuitous energy change (G) (or simply the free energy), and it is divers in terms of a system'southward enthalpy and entropy as the post-obit:

[latex]G = H\;-\;TS[/latex]

Free energy is a state part, and at constant temperature and pressure, the standard free free energy alter (ΔG°) may be expressed equally the following:

[latex]{\Delta}G = {\Delta}H\;-\;T{\Delta}S[/latex]

(For simplicity's sake, the subscript "sys" volition be omitted henceforth.)

We can understand the relationship between this system property and the spontaneity of a process past recalling the previously derived second law expression:

[latex]{\Delta}S_{\text{univ}} = {\Delta}S\;+\;\frac{q_{\text{surr}}}{T}[/latex]

The offset law requires that q _surr = −q _sys, and at constant pressure q _sys = ΔH, and so this expression may be rewritten equally the following:

[latex]{\Delta}S_{\text{univ}} = {\Delta}South\;-\;\frac{{\Delta}H}{T}[/latex]

ΔH is the enthalpy change of the system. Multiplying both sides of this equation by −T, and rearranging yields the following:

[latex]-T{\Delta}S_{\text{univ}} = {\Delta}H\;-\;T{\Delta}S[/latex]

Comparing this equation to the previous one for costless energy change shows the following relation:

[latex]{\Delta}1000 = -T{\Delta}S_{\text{univ}}[/latex]

The energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, ΔDue south _univ. Table 3 summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators.

ΔSouthward univ > 0	ΔK < 0	spontaneous
ΔS univ < 0	ΔK > 0	nonspontaneous
ΔS univ = 0	ΔM = 0	reversible (at equilibrium)
Table 3. Relation between Process Spontaneity and Signs of Thermodynamic Properties

Computing Complimentary Energy Change

Complimentary energy is a land office, and so its value depends only on the conditions of the initial and final states of the system that have undergone some change. A convenient and common arroyo to the calculation of free energy changes for physical and chemic reactions is by employ of widely bachelor compilations of standard state thermodynamic information. Ane method involves the use of standard enthalpies and entropies to compute standard free free energy changes according to the following relation as demonstrated in Instance 1.

[latex]{\Delta}Thou^{\circ} = {\Delta}H^{\circ}\;-\;T{\Delta}S^{\circ}[/latex]

Case 1

Evaluation of ΔM° Change from ΔH° and ΔSouthward°
Employ standard enthalpy and entropy data from Appendix G to calculate the standard free energy modify for the vaporization of water at room temperature (298 K). What does the computed value for ΔG° say virtually the spontaneity of this process?

Solution
The process of interest is the following:

[latex]\text{H}_2\text{O}(l)\;{\longrightarrow}\;\text{H}_2\text{O}(g)[/latex]

The standard change in gratis free energy may be calculated using the following equation:

[latex]{\Delta}G_{298}^{\circ} = {\Delta}H^{\circ}\;-\;T{\Delta}Southward^{\circ}[/latex]

From Appendix Thousand, here is the information:

Substance	[latex]{\Delta}H_{\text{f}}^{\circ}(\text{kJ/mol})[/latex]	[latex]S_{298}^{\circ}(\text{J/K}{\cdot}\text{mol})[/latex]
HiiO(50)	−286.83	seventy.0
H2O(yard)	−241.82	188.8
Table 4.

Combining at 298 One thousand:

[latex]{\Delta}H^{\circ} = {\Delta}H_{298}^{\circ} = {\Delta}H_{\text{f}}^{\circ}(\text{H}_2\text{O}(g))\;-\;{\Delta}H_{\text{f}}^{\circ}(\text{H}_2\text{O}(l)) = [-241.82\;\text{kJ}\;-\;(-285.83)]\text{kJ/mol} = 44.01\;\text{kJ/mol}[/latex]

[latex]{\Delta}S^{\circ} = {\Delta}S_{298}^{\circ} = S_{298}^{\circ}(\text{H}_2\text{O}(one thousand))\;-\;S_{298}^{\circ}(\text{H}_2\text{O}(fifty)) = 188.viii\;\text{J/mol}{\cdot}\text{K}\;-\;70.0\text{J/K} = 118.8\;\text{J/mol}{\cdot}\text{Grand}[/latex]

[latex]{\Delta}M^{\circ} = {\Delta}H^{\circ}\;-\;T{\Delta}S^{\circ}[/latex]

Converting everything into kJ and combining at 298 1000:

[latex]\brainstorm{array}{r @{{}={}} fifty} {\Delta}G_{298}^{\circ} & {\Delta}H^{\circ}\;-\;T{\Delta}Southward^{\circ} \\[0.5em] & 44.01\;\text{kJ/mol}\;-\;(298\;\text{K}\;\times\;118.8\;\text{J/mol}{\cdot}\text{K})\;\times\;\frac{1\;\text{kJ}}{k\;\text{J}} \terminate{array}[/latex]

[latex]44.01\;\text{kJ/mol}\;-\;35.4\;\text{kJ/mol} = 8.six\;\text{kJ/mol}[/latex]

At 298 K (25 °C) [latex]{\Delta}G_{298}^{\circ}\;{\textgreater}\;0[/latex], and and so boiling is nonspontaneous (non spontaneous).

Check Your Learning
Utilise standard enthalpy and entropy data from Appendix M to summate the standard free energy change for the reaction shown hither (298 1000). What does the computed value for ΔOne thousand° say most the spontaneity of this procedure?

[latex]\text{C}_2\text{H}_6(g)\;{\longrightarrow}\;\text{H}_2(g)\;+\;\text{C}_2\text{H}_4(m)[/latex]

Answer:

[latex]{\Delta}G_{298}^{\circ} = 102.0\;\text{kJ/mol}[/latex]; the reaction is nonspontaneous (not spontaneous) at 25 °C.

Free energy changes may also employ the standard costless free energy of formation ([latex]{\Delta}G_{\text{f}}^{\circ}[/latex]), for each of the reactants and products involved in the reaction. The standard energy of germination is the free energy modify that accompanies the formation of one mole of a substance from its elements in their standard states. Similar to the standard enthalpies of germination, [latex]{\Delta}G_{\text{f}}^{\circ}[/latex] is by definition nil for elemental substances nether standard country conditions. The approach to computing the gratuitous energy change for a reaction using this approach is the aforementioned as that demonstrated previously for enthalpy and entropy changes. For the reaction

[latex]one thousand\text{A}\;+\;north\text{B}\;{\longrightarrow}\;x\text{C}\;+\;y\text{D}[/latex],

the standard complimentary energy change at room temperature may be calculated as

[latex]{\Delta}G_{298}^{\circ} = {\Delta}G^{\circ} = {\sum}v{\Delta}G_{298}^{\circ}(\text{products})\;-\;{\sum}five{\Delta}G_{298}^{\circ}(\text{reactants}) = [x{\Delta}G_{\text{f}}^{\circ}(\text{C})\;+\;y{\Delta}G_{\text{f}}^{\circ}(\text{D})]\;-\;[thou{\Delta}G_{\text{f}}^{\circ}(\text{A})\;+\;n{\Delta}G_{\text{f}}^{\circ}(\text{B})][/latex].

Instance 2

Adding of [latex]{\Delta}G_{298}^{\circ}[/latex]
Consider the decomposition of xanthous mercury(II) oxide.

[latex]\text{HgO}(southward\text{,\;xanthous})\;{\longrightarrow}\;\text{Hg}(l)\;+\;\frac{1}{two}\text{O}_2(thousand)[/latex]

Summate the standard free energy change at room temperature, [latex]{\Delta}G_{298}^{\circ}[/latex], using (a) standard gratuitous energies of germination and (b) standard enthalpies of formation and standard entropies. Exercise the results indicate the reaction to be spontaneous or nonspontaneous under standard conditions?

Solution
The required information are available in Appendix G and are shown hither.

Compound	[latex]{\Delta}G_{\text{f}}^{\circ}(\text{kJ/mol})[/latex]	[latex]{\Delta}H_{\text{f}}^{\circ}(\text{kJ/mol})[/latex]	[latex]S_{298}^{\circ}(\text{J/K}{\cdot}\text{mol})[/latex]
HgO (s, yellow)	−58.43	−90.46	71.thirteen
Hg(l)	0	0	75.ix
Otwo(g)	0	0	205.ii
Table five.

(a) Using free energies of germination:

[latex]{\Delta}G_{298}^{\circ} = {\sum}vGS_{298}^{\circ}(\text{products})\;-\;{\sum}vGS_{298}^{\circ}(\text{reactants})[/latex]

[latex]= [1{\Delta}G_{298}^{\circ}\text{Hg}(l)\;+\;\frac{ane}{2}{\Delta}G_{298}^{\circ}\text{O}_2(g)]\;-\;ane{\Delta}G_{298}^{\circ}\text{HgO}(south\text{,\;yellow})[/latex]

[latex]= [1\;\text{mol}(0\;\text{kJ/mol})\;+\;\frac{ane}{ii}\text{mol}(0\;\text{kJ/mol})]\;-\;1\;\text{mol}(-58.43\;\text{kJ/mol}) = 58.43\;\text{kJ/mol}[/latex]

(b) Using enthalpies and entropies of formation:

[latex]{\Delta}H_{298}^{\circ} = {\sum}v{\Delta}H_{298}^{\circ}(\text{products})\;-\;{\sum}five{\Delta}H_{298}^{\circ}(\text{reactants})[/latex]

[latex]= [one{\Delta}H_{298}^{\circ}\text{Hg}(l)\;+\;\frac{1}{2}{\Delta}H_{298}^{\circ}\text{O}_2(g)]\;-\;1{\Delta}H_{298}^{\circ}\text{HgO}(s\text{,\;yellowish})[/latex]

[latex]= [i\;\text{mol}(0\;\text{kJ/mol})\;+\;\frac{1}{2}\text{mol}(0\;\text{kJ/mol})]\;-\;1\;\text{mol}(-90.46\;\text{kJ/mol}) = ninety.46\;\text{kJ/mol}[/latex]

[latex]{\Delta}S_{298}^{\circ} = {\sum}five{\Delta}S_{298}^{\circ}(\text{products})\;-\;{\sum}five{\Delta}S_{298}^{\circ}(\text{reactants})[/latex]

[latex]= [1{\Delta}S_{298}^{\circ}\text{Hg}(l)\;+\;\frac{one}{2}{\Delta}S_{298}^{\circ}\text{O}_2(yard)]\;-\;1{\Delta}S_{298}^{\circ}\text{HgO}(s\text{,\;yellow})[/latex]

[latex]= [one\;\text{mol}(75.ix\;\text{J/mol\;G})\;+\;\frac{1}{ii}\text{mol}(205.2\;\text{J/mol\;K})]\;-\;i\;\text{mol}(71.13\;\text{J/mol\;K}) = 107.4\;\text{J/mol\;K}[/latex]

[latex]{\Delta}G^{\circ} = {\Delta}H^{\circ}\;-\;T{\Delta}Due south^{\circ} = xc.46\;\text{kJ}\;-\;298.fifteen\;\text{1000}\;\times\;107.4\;\text{J/K}{\cdot}\text{mol}\;\times\;\frac{1\;\text{kJ}}{1000\;\text{J}}[/latex]

[latex]{\Delta}G^{\circ} = (90.46\;-\;32.01)\;\text{kJ/mol} = 58.45\;\text{kJ/mol}[/latex]

Both ways to calculate the standard free free energy alter at 25 °C give the aforementioned numerical value (to iii significant figures), and both predict that the process is nonspontaneous (non spontaneous) at room temperature.

Cheque Your Learning
Calculate ΔG° using (a) costless energies of formation and (b) enthalpies of formation and entropies (Appendix Yard). Practice the results bespeak the reaction to be spontaneous or nonspontaneous at 25 °C?

[latex]\text{C}_2\text{H}_4(k)\;{\longrightarrow}\;\text{H}_2(yard)\;+\;\text{C}_2\text{H}_2(g)[/latex]

Answer:

−141.v kJ/mol, nonspontaneous

Temperature Dependence of Spontaneity

As was previously demonstrated in this affiliate's section on entropy, the spontaneity of a process may depend upon the temperature of the organisation. Phase transitions, for case, will proceed spontaneously in i direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions tin also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy alter to the enthalpy and entropy changes for the process is considered:

[latex]{\Delta}Thou = {\Delta}H\;-\;T{\Delta}Due south[/latex]

The spontaneity of a process, as reflected in the arithmetics sign of its free free energy change, is then determined past the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (kelvin) temperature, it can simply have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:

1. Both ΔH and ΔS are positive. This status describes an endothermic process that involves an increase in system entropy. In this case, ΔG will exist negative if the magnitude of the TΔS term is greater than ΔH. If the TΔSouthward term is less than ΔH, the gratis energy change will be positive. Such a procedure is spontaneous at loftier temperatures and nonspontaneous at low temperatures.
2. Both ΔH and ΔSouth are negative. This condition describes an exothermic process that involves a subtract in system entropy. In this case, ΔGrand will be negative if the magnitude of the TΔS term is less than ΔH. If the TΔDue south term's magnitude is greater than ΔH, the energy change volition be positive. Such a process is spontaneous at low temperatures and nonspontaneous at high temperatures.
3. ΔH is positive and ΔDue south is negative. This condition describes an endothermic process that involves a decrease in arrangement entropy. In this case, ΔChiliad will be positive regardless of the temperature. Such a process is nonspontaneous at all temperatures.
4. ΔH is negative and ΔS is positive. This status describes an exothermic procedure that involves an increase in system entropy. In this case, ΔOne thousand will be negative regardless of the temperature. Such a process is spontaneous at all temperatures.

These 4 scenarios are summarized in Figure 1.

Figure ane. In that location are iv possibilities regarding the signs of enthalpy and entropy changes.

Instance 3

Predicting the Temperature Dependence of Spontaneity
The incomplete combustion of carbon is described by the following equation:

[latex]2\text{C}(s)\;+\;\text{O}_2(g)\;{\longrightarrow}\;ii\text{CO}(g)[/latex]

How does the spontaneity of this procedure depend upon temperature?

Solution
Combustion processes are exothermic (ΔH < 0). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (cyberspace gain of one mole of gas, ΔDue south > 0). The reaction is therefore spontaneous (ΔM < 0) at all temperatures.

Check Your Learning
Pop chemical hand warmers generate heat past the air-oxidation of iron:

[latex]4\text{Iron}(s)\;+\;iii\text{O}_2(1000)\;{\longrightarrow}\;2\text{Fe}_2\text{O}_3(s)[/latex]

How does the spontaneity of this process depend upon temperature?

Answer:

ΔH and ΔS are negative; the reaction is spontaneous at low temperatures.

When considering the conclusions drawn regarding the temperature dependence of spontaneity, information technology is of import to go along in mind what the terms "high" and "low" mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature. A process that is nonspontaneous at 1 temperature just spontaneous at another volition necessarily undergo a modify in "spontaneity" (every bit reflected past its ΔThou) as temperature varies. This is conspicuously illustrated by a graphical presentation of the free energy alter equation, in which ΔG is plotted on the y axis versus T on the x centrality:

[latex]{\Delta}G = {\Delta}H\;-\;T{\Delta}Due south[/latex]

[latex]y = b\;+\;mx[/latex]

Such a plot is shown in Figure 2. A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependent spontaneity as depicted past the two yellow lines in the plot. Each line crosses from ane spontaneity domain (positive or negative ΔG) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the x-intercept of the line, that is, the value of T for which ΔG is nix:

[latex]{\Delta}G = 0 = {\Delta}H\;-\;T{\Delta}Due south[/latex]

[latex]T = \frac{{\Delta}H}{{\Delta}S}[/latex]

And so, saying a process is spontaneous at "high" or "low" temperatures means the temperature is higher up or beneath, respectively, that temperature at which ΔG for the process is aught. As noted earlier, this condition describes a system at equilibrium.

Figure 2. These plots show the variation in ΔGrand with temperature for the four possible combinations of arithmetic sign for ΔH and ΔS.

Example iv

Equilibrium Temperature for a Phase Transition
As defined in the chapter on liquids and solids, the humid indicate of a liquid is the temperature at which its solid and liquid phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Utilise the information in Appendix M to estimate the humid signal of water.

Solution
The process of interest is the post-obit stage change:

[latex]\text{H}_2\text{O}(l)\;{\longrightarrow}\;\text{H}_2\text{O}(g)[/latex]

When this procedure is at equilibrium, ΔG = 0, so the following is true:

[latex]0 = {\Delta}H^{\circ}\;-\;T{\Delta}South^{\circ}\;\;\;\;\;\;\;\text{or}\;\;\;\;\;\;\;T = \frac{{\Delta}H^{\circ}}{{\Delta}Due south^{\circ}}[/latex]

Using the standard thermodynamic data from Appendix 1000,

[latex]\begin{array}{r @{{}={}} l} {\Delta}H^{\circ} & {\Delta}H_{\text{f}}^{\circ}(\text{H}_2\text{O}(g))\;-\;{\Delta}H_{\text{f}}^{\circ}(\text{H}_2\text{O}(l)) \\[0.5em] & -241.82\;\text{kJ/mol}\;-\;(-285.83\;\text{kJ/mol}) = 44.01\;\text{kJ/mol} \terminate{array}[/latex]

[latex]\begin{array}{r @{{}={}} l} {\Delta}Due south^{\circ} & {\Delta}S_{298}^{\circ}(\text{H}_2\text{O}(thou))\;-\;{\Delta}S_{298}^{\circ}(\text{H}_2\text{O}(50)) \\[0.5em] & 188.8\;\text{J/K}{\cdot}\text{mol}\;-\;70.0\;\text{J/K}{\cdot}\text{mol} = 118.8\;\text{J/Yard}{\cdot}\text{mol} \end{array}[/latex]

[latex]T = \frac{{\Delta}H^{\circ}}{{\Delta}Southward^{\circ}} = \frac{44.01\;\times\;10^3\;\text{J/mol}}{118.8\;\text{J/Thousand}{\cdot}\text{mol}} = 370.v\;\text{K} = 97.3\;^{\circ}\text{C}[/latex]

The accustomed value for h2o'south normal boiling betoken is 373.2 K (100.0 °C), and and so this calculation is in reasonable understanding. Notation that the values for enthalpy and entropy changes data used were derived from standard data at 298 Yard (Appendix G). If desired, yous could obtain more than accurate results by using enthalpy and entropy changes determined at (or at to the lowest degree closer to) the actual boiling point.

Bank check Your Learning
Apply the information in Appendix 1000 to guess the boiling point of CS₂.

Reply:

313 G (accepted value 319 K)

Free Free energy and Equilibrium

The free energy alter for a process may be viewed as a measure out of its driving strength. A negative value for ΔG represents a finite driving force for the process in the forward direction, while a positive value represents a driving force for the process in the reverse direction. When ΔYard is zip, the forward and reverse driving forces are equal, and so the process occurs in both directions at the same rate (the organization is at equilibrium).

In the chapter on equilibrium the reaction quotient, Q, was introduced equally a user-friendly measure of the status of an equilibrium system. Recall that Q is the numerical value of the mass action expression for the system, and that you lot may apply its value to identify the management in which a reaction will proceed in order to achieve equilibrium. When Q is lesser than the equilibrium constant, K, the reaction will keep in the forward direction until equilibrium is reached and Q = Chiliad. Conversely, if Q < K, the process will proceed in the reverse direction until equilibrium is accomplished.

The gratuitous energy change for a process taking place with reactants and products present nether nonstandard conditions, ΔG, is related to the standard gratuitous energy change, ΔG°, according to this equation:

[latex]{\Delta}G = {\Delta}1000^{\circ}\;+\;RT\;\text{ln}\;Q[/latex]

R is the gas constant (8.314 J/K mol), T is the kelvin or absolute temperature, and Q is the reaction quotient. We may utilize this equation to predict the spontaneity for a process under whatever given set of conditions as illustrated in Instance 5.

Example v

Calculating ΔG under Nonstandard Weather condition
What is the free free energy change for the process shown here under the specified conditions?

T = 25 °C, [latex]\text{P}_{\text{N}_2} = 0.870\;\text{atm}[/latex], [latex]\text{P}_{\text{H}_2} = 0.250\;\text{atm}[/latex], and [latex]\text{P}_{\text{NH}_3} = 12.9\;\text{atm}[/latex]

[latex]2\text{NH}_3(g)\;{\longrightarrow}\;3\text{H}_2(g)\;+\;\text{Due north}_2(one thousand)\;\;\;\;\;\;\;{\Delta}G^{\circ} = 33.0\;\text{kJ/mol}[/latex]

Solution
The equation relating gratuitous energy change to standard free energy change and reaction caliber may be used straight:

[latex]{\Delta}One thousand = {\Delta}G^{\circ}\;+\;RT\;\text{ln}\;Q = 33.0\;\frac{\text{kJ}}{\text{mol}}\;+\;(8.314\;\frac{\text{J}}{\text{mol\;K}}\;\times\;298\;\text{1000}\;\times\;\text{ln}\;\frac{(0.250^iii)\;\times\;0.870}{12.ix^2}) = 9680\;\frac{\text{J}}{\text{mol}}\;\text{or}\;9.68\;\text{kJ/mol}[/latex]

Since the computed value for ΔG is positive, the reaction is nonspontaneous under these weather.

Check Your Learning
Summate the free free energy change for this same reaction at 875 °C in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these weather condition?

Answer:

ΔG = −136 kJ; yes

For a system at equilibrium, Q = K and ΔK = 0, and the previous equation may exist written equally

[latex]0 = {\Delta}K^{\circ}\;+\;RT\;\text{ln}\;K\;\;\;\;\;\;\;(\text{at\;equilibrium})[/latex]

[latex]{\Delta}Thou^{\circ} = -RT\;\text{ln}\;K\;\;\;\;\;\;\;\text{or}\;\;\;\;\;\;\;K = e^{-\frac{{\Delta}Thou^{\circ}}{RT}}[/latex]

This form of the equation provides a useful link betwixt these two essential thermodynamic properties, and it can be used to derive equilibrium constants from standard gratis energy changes and vice versa. The relations between standard costless free energy changes and equilibrium constants are summarized in Table 6.

K	ΔYard°	Comments
> one	< 0	Products are more abundant at equilibrium.
< i	> 0	Reactants are more arable at equilibrium.
= i	= 0	Reactants and products are equally abundant at equilibrium.
Table vi. Relations between Standard Free energy Changes and Equilibrium Constants

Case half dozen

Calculating an Equilibrium Constant using Standard Complimentary Free energy Change
Given that the standard free energies of formation of Ag⁺(aq), Cl⁻(aq), and AgCl(s) are 77.1 kJ/mol, −131.2 kJ/mol, and −109.8 kJ/mol, respectively, calculate the solubility product, K _sp, for AgCl.

Solution
The reaction of involvement is the following:

[latex]\text{AgCl}(s)\;{\rightleftharpoons}\;\text{Ag}^{+}(aq)\;+\;\text{Cl}^{-}(aq)\;\;\;\;\;\;\;K_{\text{sp}} = [\text{Ag}^{+}][\text{Cl}^{-}][/latex]

The standard gratuitous energy change for this reaction is first computed using standard free energies of formation for its reactants and products:

[latex]{\Delta}G^{\circ} = {\Delta}G_{298}^{\circ} = [{\Delta}G_{\text{f}}^{\circ}\;(\text{Ag}^{+}(aq))\;+\;{\Delta}G_{\text{f}}^{\circ}\;(\text{Cl}^{-}(aq))]\;-\;[{\Delta}G_{\text{f}}^{\circ}\;(\text{AgCl}(s))] = [77.ane\;\text{kJ/mol}\;-\;131.ii\;\text{kJ/mol}]\;-\;[-109.8\;\text{kJ/mol}] = 55.7\;\text{kJ/mol}[/latex]

The equilibrium constant for the reaction may so be derived from its standard free free energy modify:

[latex]K_{\text{sp}} = e^{-\frac{{\Delta}Thou^{\circ}}{RT}} = \text{exp}\;(-\frac{{\Delta}G^{\circ}}{RT}) = \text{exp}(-\frac{55.7\;\times\;10^3\text{J/mol}}{8.314\;\text{J/mol}{\cdot}\text{M}\;\times\;298.15\;\text{Yard}}) = \text{exp}(-22.470) = eastward^{-22.470} = 1.74\;\times\;10^{-10}[/latex]

This result is in reasonable agreement with the value provided in Appendix J.

Check Your Learning
Use the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C.

[latex]2\text{NO}_2(g)\;{\rightleftharpoons}\;\text{N}_2\text{O}_4(g)[/latex]

To further illustrate the relation between these 2 essential thermodynamic concepts, consider the observation that reactions spontaneously go on in a direction that ultimately establishes equilibrium. As may be shown past plotting the free energy change versus the extent of the reaction (for example, every bit reflected in the value of Q), equilibrium is established when the system'due south gratis free energy is minimized (Figure 3). If a system is present with reactants and products present in nonequilibrium amounts (Q ≠ Thou), the reaction will keep spontaneously in the direction necessary to establish equilibrium.

Figure 3. These plots bear witness the energy versus reaction progress for systems whose standard free changes are (a) negative, (b) positive, and (c) zero. Nonequilibrium systems volition proceed spontaneously in whatever management is necessary to minimize free free energy and institute equilibrium.

Primal Concepts and Summary

Gibbs free free energy (G) is a state function defined with regard to system quantities merely and may be used to predict the spontaneity of a process. A negative value for ΔG indicates a spontaneous process; a positive ΔChiliad indicates a nonspontaneous process; and a Δ1000 of nil indicates that the system is at equilibrium. A number of approaches to the computation of free energy changes are possible.

Primal Equations

• Δ1000 = ΔH − TΔDue south
• ΔG = ΔThousand° + RT ln Q
• ΔG° = −RT ln K

Chemistry End of Chapter Exercises

1. What is the difference between ΔM, ΔG°, and [latex]{\Delta}G_{298}^{\circ}[/latex] for a chemical change?
2. A reactions has [latex]{\Delta}H_{298}^{\circ} = 100\;\text{kJ/mol}[/latex] and [latex]{\Delta}S_{298}^{\circ} = 250\;\text{J/mol}{\cdot}\text{K}[/latex]. Is the reaction spontaneous at room temperature? If not, under what temperature conditions volition it become spontaneous?
3. Explain what happens as a reaction starts with ΔThousand < 0 (negative) and reaches the point where ΔG = 0.
4. Use the standard complimentary free energy of germination data in Appendix 1000 to determine the gratuitous energy modify for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each every bit either spontaneous or nonspontaneous at these weather condition.

   (a) [latex]\text{MnO}_2(s)\;{\longrightarrow}\;\text{Mn}(south)\;+\;\text{O}_2(g)[/latex]

   (b) [latex]\text{H}_2(yard)\;+\;\text{Br}_2(50)\;{\longrightarrow}\;2\text{HBr}(g)[/latex]

   (c) [latex]\text{Cu}(s)\;+\;\text{S}(m)\;{\longrightarrow}\;\text{CuS}(s)[/latex]

   (d) [latex]two\text{LiOH}(due south)\;+\;\text{CO}_2(chiliad)\;{\longrightarrow}\;\text{Li}_2\text{CO}_3(due south)\;+\;\text{H}_2\text{O}(g)[/latex]

   (e) [latex]\text{CH}_4(chiliad)\;+\;\text{O}_2(g)\;{\longrightarrow}\;\text{C}(s\text{,\;graphite})\;+\;two\text{H}_2\text{O}(m)[/latex]

   (f) [latex]\text{CS}_2(g)\;+\;3\text{Cl}_2(thousand)\;{\longrightarrow}\;\text{CCl}_4(g)\;+\;\text{Southward}_2\text{Cl}_2(m)[/latex]

5. Utilize the standard energy data in Appendix G to determine the free free energy alter for each of the following reactions, which are run under standard state conditions and 25 °C. Place each as either spontaneous or nonspontaneous at these conditions.

   (a) [latex]\text{C}(s\text{,\;graphite})\;+\;\text{O}_2(grand)\;{\longrightarrow}\;\text{CO}_2(thousand)[/latex]

   (b) [latex]\text{O}_2(g)\;+\;\text{N}_2(g)\;{\longrightarrow}\;2\text{NO}(g)[/latex]

   (c) [latex]2\text{Cu}(s)\;+\;\text{S}(1000)\;{\longrightarrow}\;\text{Cu}_2\text{S}(southward)[/latex]

   (d) [latex]\text{CaO}(s)\;+\;\text{H}_2\text{O}(fifty)\;{\longrightarrow}\;\text{Ca(OH)}_2(south)[/latex]

   (e) [latex]\text{Fe}_2\text{O}_3(s)\;+\;3\text{CO}(1000)\;{\longrightarrow}\;ii\text{Fe}(south)\;+\;3\text{CO}_2(k)[/latex]

   (f) [latex]\text{CaSO}_4\;{\cdot}\;2\text{H}_2\text{O}(southward)\;{\longrightarrow}\;\text{CaSO}_4(southward)\;+\;two\text{H}_2\text{O}(one thousand)[/latex]

6. Given:
   [latex]\brainstorm{array}{ll} \text{P}_4(s)\;+\;v\text{O}_2(m)\;{\longrightarrow}\;\text{P}_4\text{O}_{ten}(s) & {\Delta}G_{298}^{\circ} = -2697.0\;\text{kJ/mol} \\[0.5em] 2\text{H}_2(g)\;+\;\text{O}_2(1000)\;{\longrightarrow}\;2\text{H}_2\text{O}(one thousand) & {\Delta}G_{298}^{\circ} = -457.18\;\text{kJ/mol} \\[0.5em] 6\text{H}_2\text{O}(g)\;+\;\text{P}_4\text{O}_{10}(g)\;{\longrightarrow}\;4\text{H}_3\text{PO}_4(50) & {\Delta}G_{298}^{\circ} = -428.66\;\text{kJ/mol} \cease{assortment}[/latex]

   (a) Determine the standard gratis energy of formation, [latex]{\Delta}G_{\text{f}}^{\circ}[/latex], for phosphoric acrid.

   (b) How does your calculated result compare to the value in Appendix G? Explain.

7. Is the formation of ozone (O_three(chiliad)) from oxygen (O₂(chiliad)) spontaneous at room temperature nether standard land atmospheric condition?
8. Consider the decomposition of red mercury(2) oxide nether standard country conditions.
   [latex]2\text{HgO}(due south\text{,\;red})\;{\longrightarrow}\;ii\text{Hg}(fifty)\;+\;\text{O}_2(g)[/latex]

   (a) Is the decomposition spontaneous nether standard land conditions?

   (b) To a higher place what temperature does the reaction become spontaneous?

9. Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the advisable catalyst. Evaluate the following substances under standard land conditions as suitable candidates for fuels.

   (a) Ammonia: [latex]two\text{NH}_3(g)\;{\longrightarrow}\;\text{Northward}_2(g)\;+\;3\text{H}_2(g)[/latex]

   (b) Diborane: [latex]\text{B}_2\text{H}_6(g)\;{\longrightarrow}\;2\text{B}(g)\;+\;3\text{H}_2(g)[/latex]

   (c) Hydrazine: [latex]\text{Due north}_2\text{H}_4(g)\;{\longrightarrow}\;\text{N}_2(g)\;+\;ii\text{H}_2(m)[/latex]

   (d) Hydrogen peroxide: [latex]\text{H}_2\text{O}_2(l)\;{\longrightarrow}\;\text{H}_2\text{O}(k)\;+\;\frac{1}{two}\text{O}_2(g)[/latex]

10. Calculate ΔG° for each of the post-obit reactions from the equilibrium constant at the temperature given.

    (a) [latex]\text{N}_2(chiliad)\;+\;\text{O}_2(g)\;{\longrightarrow}\;two\text{NO}(g)\;\;\;\;\;\;\;\text{T} = 2000\;^{\circ}\text{C}\;\;\;\;\;\;\;K_{\text{p}} = 4.1\;\times\;10^{-iv}[/latex]

    (b) [latex]\text{H}_2(g)\;+\;\text{I}_2(g)\;{\longrightarrow}\;two\text{HI}(g)\;\;\;\;\;\;\;\text{T} = 400\;^{\circ}\text{C}\;\;\;\;\;\;\;K_{\text{p}} = 50.0[/latex]

    (c) [latex]\text{CO}_2(grand)\;+\;\text{H}_2(g)\;{\longrightarrow}\;\text{CO}(k)\;+\;\text{H}_2\text{O}(g)\;\;\;\;\;\;\;\text{T} = 980\;^{\circ}\text{C}\;\;\;\;\;\;\;K_{\text{p}} = 1.67[/latex]

    (d) [latex]\text{CaCO}_3(s)\;{\longrightarrow}\;\text{CaO}(s)\;+\;\text{CO}_2(g)\;\;\;\;\;\;\;\text{T} = 900\;^{\circ}\text{C}\;\;\;\;\;\;\;K_{\text{p}} = 1.04[/latex]

    (east) [latex]\text{HF}(aq)\;+\;\text{H}_2\text{O}(l)\;{\longrightarrow}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{F}^{-}(aq)\;\;\;\;\;\;\;\text{T} = 25\;^{\circ}\text{C}\;\;\;\;\;\;\;K_{\text{p}} = seven.ii\;\times\;10^{-four}[/latex]

    (f) [latex]\text{AgBr}(s)\;{\longrightarrow}\;\text{Ag}^{+}(aq)\;+\;\text{Br}^{-}(aq)\;\;\;\;\;\;\;\text{T} = 25\;^{\circ}\text{C}\;\;\;\;\;\;\;K_{\text{p}} = 3.3\;\times\;x^{-xiii}[/latex]

11. Calculate ΔG° for each of the post-obit reactions from the equilibrium abiding at the temperature given.

    (a) [latex]\text{Cl}_2(g)\;+\;\text{Br}_2(k)\;{\longrightarrow}\;2\text{BrCl}(g)\;\;\;\;\;\;\;\text{T} = 25\;^{\circ}\text{C}\;\;\;\;\;\;\;K_{\text{p}} = 4.7\;\times\;x^{-2}[/latex]

    (b) [latex]2\text{SO}_2(thousand)\;+\;\text{O}_2(g)\;{\leftrightharpoons}\;2\text{And so}_3(g)\;\;\;\;\;\;\;\text{T} = 500\;^{\circ}\text{C}\;\;\;\;\;\;\;K_{\text{p}} = 48.ii[/latex]

    (c) [latex]\text{H}_2\text{O}(fifty)\;{\rightleftharpoons}\;\text{H}_2\text{O}(k)\;\;\;\;\;\;\;\text{T} = sixty\;^{\circ}\text{C}\;\;\;\;\;\;\;K_{\text{p}} = 0.196\;\text{atm}[/latex]

    (d) [latex]\text{CoO}(southward)\;+\;\text{CO}(g)\;{\rightleftharpoons}\;\text{Co}(southward)\;+\;\text{CO}_2(g)\;\;\;\;\;\;\;\text{T} = 550\;^{\circ}\text{C}\;\;\;\;\;\;\;K_{\text{p}} = 4.ninety\;\times\;10^2[/latex]

    (e) [latex]\text{CH}_3\text{NH}_2(aq)\;+\;\text{H}_2\text{O}(l)\;{\longrightarrow}\;\text{CH}_3\text{NH}_3^{+}(aq)\;+\;\text{OH}^{-}(aq)\;\;\;\;\;\;\;\text{T} = 25\;^{\circ}\text{C}\;\;\;\;\;\;\;K_{\text{p}} = 4.iv\;\times\;ten^{-4}[/latex]

    (f) [latex]\text{PbI}_2(south)\;{\longrightarrow}\;\text{Lead}^{2+}(aq)\;+\;two\text{I}^{-}(aq)\;\;\;\;\;\;\;\text{T} = 25\;^{\circ}\text{C}\;\;\;\;\;\;\;K_{\text{p}} = viii.7\;\times\;ten^{-9}[/latex]

12. Calculate the equilibrium abiding at 25 °C for each of the following reactions from the value of ΔG° given.

    (a) [latex]\text{O}_2(g)\;+\;2\text{F}_2(g)\;{\longrightarrow}\;2\text{OF}_2(g)\;\;\;\;\;\;\;{\Delta}G^{\circ} = -9.2\;\text{kJ}[/latex]

    (b) [latex]\text{I}_2(south)\;+\;\text{Br}_2(l)\;{\longrightarrow}\;2\text{IBr}(one thousand)\;\;\;\;\;\;\;{\Delta}Thousand^{\circ} = seven.3\;\text{kJI}[/latex]

    (c) [latex]2\text{LiOH}(s)\;+\;\text{CO}_2(g)\;{\longrightarrow}\;\text{Li}_2\text{CO}_3(s)\;+\;\text{H}_2\text{O}(g)\;\;\;\;\;\;\;{\Delta}G^{\circ} = -79\;\text{kJ}[/latex]

    (d) [latex]\text{North}_2\text{O}_3(g)\;{\longrightarrow}\;\text{NO}(g)\;+\;\text{NO}_2(g)\;\;\;\;\;\;\;{\Delta}G^{\circ} = -ane.six\;\text{kJ}[/latex]

    (e) [latex]\text{SnCl}_4(l)\;{\longrightarrow}\;\text{SnCl}_4(l)\;\;\;\;\;\;\;{\Delta}G^{\circ} = 8.0\;\text{kJ}[/latex]

13. Calculate the equilibrium constant at 25 °C for each of the post-obit reactions from the value of ΔG° given.

    (a) [latex]\text{I}_2(southward)\;+\;\text{Cl}_2(1000)\;{\longrightarrow}\;2\text{ICl}(g)\;\;\;\;\;\;\;{\Delta}Grand^{\circ} = -10.88\;\text{kJ}[/latex]

    (b) [latex]\text{H}_2(g)\;+\;\text{I}_2(s)\;{\longrightarrow}\;2\text{Howdy}(k)\;\;\;\;\;\;\;{\Delta}G^{\circ} = three.4\;\text{kJ}[/latex]

    (c) [latex]\text{CS}_2(m)\;+\;3\text{Cl}_2(g)\;{\longrightarrow}\;\text{CCl}_4(g)\;+\;\text{S}_2\text{Cl}_2(k)\;\;\;\;\;\;\;{\Delta}G^{\circ} = -39\;\text{kJ}[/latex]

    (d) [latex]2\text{SO}_2(thousand)\;+\;\text{O}_2(g)\;{\longrightarrow}\;2\text{Then}_3(thousand)\;\;\;\;\;\;\;{\Delta}M^{\circ} = -141.82\;\text{kJ}[/latex]

    (e) [latex]\text{CS}_2(g)\;{\longrightarrow}\;\text{CS}_2(l)\;\;\;\;\;\;\;{\Delta}G^{\circ} = -i.88\;\text{kJ}[/latex]

14. Calculate the equilibrium abiding at the temperature given.

    (a) [latex]\text{O}_2(thou)\;+\;two\text{F}_2(g)\;{\longrightarrow}\;2\text{F}_2\text{O}(grand)\;\;\;\;\;\;\;(\text{T} = 100\;^{\circ}\text{C})[/latex]

    (b) [latex]\text{I}_2(south)\;+\;\text{Br}_2(50)\;{\longrightarrow}\;2\text{IBr}(chiliad)\;\;\;\;\;\;\;(\text{T} = 0.0\;^{\circ}\text{C})[/latex]

    (c) [latex]2\text{LiOH}(south)\;+\;\text{CO}_2(g)\;{\longrightarrow}\;\text{Li}_2\text{CO}_3(southward)\;+\;\text{H}_2\text{O}(g)\;\;\;\;\;\;\;(\text{T} = 575\;^{\circ}\text{C})[/latex]

    (d) [latex]\text{North}_2\text{O}_3(g)\;{\longrightarrow}\;\text{NO}(thou)\;+\;\text{NO}_2(thousand)\;\;\;\;\;\;\;(\text{T} = -x.0\;^{\circ}\text{C})[/latex]

    (e) [latex]\text{SnCl}_4(l)\;{\longrightarrow}\;\text{SnCl}_4(g)\;\;\;\;\;\;\;(\text{T} = 200\;^{\circ}\text{C})[/latex]

15. Calculate the equilibrium constant at the temperature given.

    (a) [latex]\text{I}_2(due south)\;+\;\text{Cl}_2(g)\;{\longrightarrow}\;ii\text{ICl}(m)\;\;\;\;\;\;\;(\text{T} = 100\;^{\circ}\text{C})[/latex]

    (b) [latex]\text{H}_2(g)\;+\;\text{I}_2(due south)\;{\longrightarrow}\;2\text{HI}(g)\;\;\;\;\;\;\;(\text{T} = 0.0\;^{\circ}\text{C})[/latex]

    (c) [latex]\text{CS}_2(g)\;+\;three\text{Cl}_2(g)\;{\longrightarrow}\;\text{CCl}_4(g)\;+\;\text{S}_2\text{Cl}_2(m)\;\;\;\;\;\;\;(\text{T} = 125\;^{\circ}\text{C})[/latex]

    (d) [latex]2\text{Then}_2(g)\;+\;\text{O}_2(g)\;{\longrightarrow}\;ii\text{SO}_3(g)\;\;\;\;\;\;\;(\text{T} = 675\;^{\circ}\text{C})[/latex]

    (e) [latex]\text{CS}_2(g)\;{\longrightarrow}\;\text{CS}_2(l)\;\;\;\;\;\;\;(\text{T} = 90\;^{\circ}\text{C})[/latex]

16. Consider the following reaction at 298 K:
    [latex]\text{N}_2\text{O}_4(g)\;{\rightleftharpoons}\;2\text{NO}_2(g)\;\;\;\;\;\;\;K_{\text{P}} = 0.142[/latex]

    What is the standard complimentary energy change at this temperature? Draw what happens to the initial system, where the reactants and products are in standard states, as information technology approaches equilibrium.

17. Determine the normal humid point (in kelvin) of dichloroethane, CH_iiCl_ii. Find the actual boiling point using the Cyberspace or some other source, and calculate the pct error in the temperature. Explain the differences, if any, betwixt the two values.
18. Under what conditions is [latex]\text{N}_2\text{O}_3(g)\;{\longrightarrow}\;\text{NO}(g)\;+\;\text{NO}_2(g)[/latex] spontaneous?
19. At room temperature, the equilibrium abiding (K_w ) for the cocky-ionization of water is 1.00 × ten⁻¹⁴. Using this information, summate the standard free energy change for the aqueous reaction of hydrogen ion with hydroxide ion to produce water. (Hint: The reaction is the reverse of the self-ionization reaction.)
20. Hydrogen sulfide is a pollutant constitute in natural gas. Following its removal, it is converted to sulfur by the reaction [latex]two\text{H}_2\text{S}(one thousand)\;+\;\text{And so}_2(g)\;{\rightleftharpoons}\;\frac{iii}{viii}\text{S}_8(s\text{,\;rhombic})\;+\;2\text{H}_2\text{O}(l)[/latex]. What is the equilibrium constant for this reaction? Is the reaction endothermic or exothermic?
21. Consider the decomposition of CaCO₃(s) into CaO(due south) and CO₂(g). What is the equilibrium partial pressure of CO₂ at room temperature?
22. In the laboratory, hydrogen chloride (HCl(one thousand)) and ammonia (NH₃(chiliad)) ofttimes escape from bottles of their solutions and react to grade the ammonium chloride (NH₄Cl(south)), the white glaze ofttimes seen on glassware. Assuming that the number of moles of each gas that escapes into the room is the same, what is the maximum partial pressure level of HCl and NH₃ in the laboratory at room temperature? (Hint: The partial pressures will be equal and are at their maximum value when at equilibrium.)
23. Benzene tin be prepared from acetylene. [latex]iii\text{C}_2\text{H}_2(g)\;{\rightleftharpoons}\;\text{C}_6\text{H}_6(chiliad)[/latex]. Determine the equilibrium constant at 25 °C and at 850 °C. Is the reaction spontaneous at either of these temperatures? Why is all acetylene not institute as benzene?
24. Carbon dioxide decomposes into CO and O_two at elevated temperatures. What is the equilibrium partial pressure level of oxygen in a sample at 1000 °C for which the initial pressure of CO₂ was i.xv atm?
25. Carbon tetrachloride, an important industrial solvent, is prepared past the chlorination of methyl hydride at 850 K.
    [latex]\text{CH}_4(chiliad)\;+\;4\text{Cl}_2(k)\;{\longrightarrow}\;\text{CCl}_4(g)\;+\;four\text{HCl}(g)[/latex]

    What is the equilibrium constant for the reaction at 850 K? Would the reaction vessel need to be heated or cooled to proceed the temperature of the reaction constant?

26. Acetic acrid, CH_threeCO₂H, can form a dimer, (CH₃CO₂H)_ii, in the gas phase.
    [latex]2\text{CH}_3\text{CO}_2\text{H}(m)\;{\longrightarrow}\;(\text{CH}_3\text{CO}_2\text{H})_2(g)[/latex]

    The dimer is held together by 2 hydrogen bonds with a full forcefulness of 66.5 kJ per mole of dimer.
    

    At 25 °C, the equilibrium abiding for the dimerization is 1.3 × ten³ (pressure in atm). What is ΔS° for the reaction?

27. Nitric acid, HNO₃, can be prepared by the following sequence of reactions:
    [latex]four\text{NH}_3(g)\;+\;5\text{O}_2(g)\;{\longrightarrow}\;four\text{NO}(g)\;+\;6\text{H}_2\text{O}(m)[/latex]
    [latex]2\text{NO}(chiliad)\;+\;\text{O}_2(g)\;{\longrightarrow}\;2\text{NO}_2(one thousand)[/latex]
    [latex]iii\text{NO}_2(yard)\;+\;\text{H}_2\text{O}(fifty)\;{\longrightarrow}\;ii\text{HNO}_3(50)\;+\;\text{NO}(g)[/latex]

    How much heat is evolved when ane mol of NH₃(one thousand) is converted to HNO₃(l)? Assume standard states at 25 °C.

28. Determine ΔChiliad for the post-obit reactions.

    (a) Antimony pentachloride decomposes at 448 °C. The reaction is:

    [latex]\text{SbCl}_5(g)\;{\longrightarrow}\;\text{SbCl}_3(1000)\;+\;\text{Cl}_2(1000)[/latex]

    An equilibrium mixture in a 5.00 L flask at 448 °C contains three.85 m of SbCl₅, 9.14 g of SbCl₃, and 2.84 1000 of Cl_two.

    (b) Chlorine molecules dissociate according to this reaction:

    [latex]\text{Cl}_2(g)\;{\longrightarrow}\;2\text{Cl}(g)[/latex]

    ane.00% of Cl_ii molecules dissociate at 975 Yard and a pressure of 1.00 atm.

29. Given that the [latex]{\Delta}G_{\text{f}}^{\circ}[/latex] for Pb^two+(aq) and Cl⁻(aq) is −24.iii kJ/mole and −131.two kJ/mole respectively, make up one's mind the solubility product, M _sp, for PbCl₂(due south).
30. Determine the standard costless energy change, [latex]{\Delta}G_{\text{f}}^{\circ}[/latex], for the formation of Due south²⁻(aq) given that the [latex]{\Delta}G_{\text{f}}^{\circ}[/latex] for Ag⁺(aq) and Ag₂S(s) are 77.1 k/mole and −39.five kJ/mole respectively, and the solubility production for Ag₂S(s) is viii × ten⁻⁵¹.
31. Determine the standard enthalpy alter, entropy change, and gratis energy alter for the conversion of diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy changes. Explicate why diamond spontaneously changing into graphite is not observed.
32. The evaporation of 1 mole of water at 298 Yard has a standard free energy change of 8.58 kJ.
    [latex]\text{H}_2\text{O}(l)\;{\leftrightharpoons}\;\text{H}_2\text{O}(g)\;\;\;\;\;\;\;{\Delta}G_{298}^{\circ} = 8.58\;\text{kJ}[/latex]

    (a) Is the evaporation of water nether standard thermodynamic conditions spontaneous?

    (b) Determine the equilibrium constant, K_P , for this physical process.

    (c) By calculating ∆One thousand, decide if the evaporation of water at 298 Thousand is spontaneous when the partial pressure level of water, [latex]\text{P}_{\text{H}_2\text{O}}[/latex], is 0.011 atm.

    (d) If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry out when placed outside. In order for laundry to dry, what must be the value of [latex]\text{P}_{\text{H}_2\text{O}}[/latex] in the air?

33. In glycolysis, the reaction of glucose (Glu) to course glucose-6-phosphate (G6P) requires ATP to exist present as described by the following equation:
    [latex]\text{Glu}\;+\;\text{ATP}\;{\longrightarrow}\;\text{G}6\text{P}\;+\;\text{ADP}\;\;\;\;\;\;\;{\Delta}G_{298}^{\circ} = -17\;\text{kJ}[/latex]

    In this procedure, ATP becomes ADP summarized by the post-obit equation:

    [latex]\text{ATP}\;{\longrightarrow}\;\text{ADP}\;\;\;\;\;\;\;{\Delta}G_{298}^{\circ} = -xxx\;\text{kJ}[/latex]

    Make up one's mind the standard gratuitous energy change for the following reaction, and explain why ATP is necessary to drive this procedure:

    [latex]\text{Glu}\;{\longrightarrow}\;\text{M}half dozen\text{P}\;\;\;\;\;\;\;{\Delta}G_{298}^{\circ} = \text{?}[/latex]

34. One of the of import reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to class fructose-vi-phosphate (F6P):
    [latex]\text{G}6\text{P}\;{\leftrightharpoons}\;\text{F}6\text{P}\;\;\;\;\;\;\;{\Delta}G_{298}^{\circ} = 1.7\;\text{kJ}[/latex]

    (a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic weather condition?

    (b) Standard thermodynamic conditions imply the concentrations of G6P and F6P to be ane M, still, in a typical cell, they are not even close to these values. Summate ΔG when the concentrations of G6P and F6P are 120 μK and 28 μOne thousand respectively, and discuss the spontaneity of the forward reaction nether these conditions. Assume the temperature is 37 °C.

35. Without doing a numerical calculation, determine which of the post-obit will reduce the gratis energy change for the reaction, that is, go far less positive or more than negative, when the temperature is increased. Explicate.

    (a) [latex]\text{Due north}_2(g)\;+\;3\text{H}_2(chiliad)\;{\longrightarrow}\;2\text{NH}_3(thousand)[/latex]

    (b) [latex]\text{HCl}(k)\;+\;\text{NH}_3(m)\;{\longrightarrow}\;\text{NH}_4\text{Cl}(s)[/latex]

    (c) [latex](\text{NH}_4)_2\text{Cr}_2\text{O}_7(southward)\;{\longrightarrow}\;\text{Cr}_2\text{O}_3(southward)\;+\;4\text{H}_2\text{O}(thou)\;+\;\text{North}_2(one thousand)[/latex]

    (d) [latex]2\text{Fe}(s)\;+\;3\text{O}_2(1000)\;{\longrightarrow}\;\text{Iron}_2\text{O}_3(s)[/latex]

36. When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any calculations, deduce the signs of ΔG, ΔH, and ΔS for this procedure, and justify your choices.
37. An important source of copper is from the copper ore, chalcocite, a grade of copper(I) sulfide. When heated, the Cu₂S decomposes to class copper and sulfur described past the post-obit equation:
    [latex]\text{Cu}_2\text{Due south}(south)\;{\longrightarrow}\;\text{Cu}(due south)\;+\;\text{Due south}(southward)[/latex]

    (a) Determine [latex]{\Delta}G_{298}^{\circ}[/latex] for the decomposition of Cu₂S(s).

    (b) The reaction of sulfur with oxygen yields sulfur dioxide equally the merely production. Write an equation that describes this reaction, and decide [latex]{\Delta}G_{298}^{\circ}[/latex] for the process.

    (c) The production of copper from chalcocite is performed by roasting the Cu₂S in air to produce the Cu. By combining the equations from Parts (a) and (b), write the equation that describes the roasting of the chalcocite, and explicate why coupling these reactions together makes for a more efficient procedure for the production of the copper.

38. What happens to [latex]{\Delta}G_{298}^{\circ}[/latex] (becomes more negative or more than positive) for the post-obit chemical reactions when the fractional pressure level of oxygen is increased?

    (a) [latex]\text{Southward}(s)\;+\;\text{O}_2(g)\;{\longrightarrow}\;\text{SO}_2(g)[/latex]

    (b) [latex]2\text{SO}_2(chiliad)\;+\;\text{O}_2(g)\;{\longrightarrow}\;\text{SO}_3(g)[/latex]

    (c) [latex]\text{HgO}(s)\;{\longrightarrow}\;\text{Hg}(50)\;+\;\text{O}_2(g)[/latex]

Glossary

Gibbs gratuitous free energy change (G)

thermodynamic property divers in terms of system enthalpy and entropy; all spontaneous processes involve a decrease in G

standard gratuitous energy change (ΔG°)

change in gratuitous free energy for a procedure occurring under standard conditions (ane bar pressure level for gases, one M concentration for solutions)

standard complimentary energy of formation [latex](\Delta Yard^{^circ}_{\text{f}}[/latex]

change in complimentary energy accompanying the germination of one mole of substance from its elements in their standard states

Solutions

Answers to Chemistry End of Chapter Exercises

2. The reaction is nonspontaneous at room temperature.
Above 400 K, ΔG volition become negative, and the reaction will get spontaneous.

4. (a) 465.1 kJ nonspontaneous; (b) −106.86 kJ spontaneous; (c) −53.6 kJ spontaneous; (d) −83.4 kJ spontaneous; (east) −406.vii kJ spontaneous; (f) −30.0 kJ spontaneous

half-dozen. (a) −1124.iii kJ/mol for the standard free energy change. (b) The calculation agrees with the value in Appendix M because gratis free energy is a state function (just similar the enthalpy and entropy), so its modify depends only on the initial and terminal states, not the path between them.

8. (a) The reaction is nonspontaneous; (b) Above 566 °C the process is spontaneous.

10. (a) 1.v × 10² kJ; (b) −21.9 kJ; (c) −v.34 kJ; (d) −0.383 kJ; (east) 18 kJ; (f) 71 kJ

12. (a) K = 41; (b) K = 0.053; (c) K = vi.ix × 10^thirteen; (d) Chiliad = 1.ix; (e) K = 0.04

14. In each of the post-obit, the value of ΔK is not given at the temperature of the reaction. Therefore, nosotros must calculate ΔG from the values ΔH° and ΔS and and then summate ΔG from the relation ΔG = ΔH° − TΔS°.

(a) Chiliad = i.29;

(b) K = 2.51 × ten⁻³;

(c) K = four.83 × 10³;

(d) K = 0.219;

(e) K = 16.ane

xvi. The standard free free energy change is [latex]{\Delta}G_{298}^{\circ} = -RT\;\text{ln}\;K = 4.84\;\text{kJ/mol}[/latex]. When reactants and products are in their standard states (1 bar or 1 atm), Q = 1. Every bit the reaction proceeds toward equilibrium, the reaction shifts left (the amount of products drops while the amount of reactants increases): Q < 1, and [latex]{\Delta}G_{298}[/latex] becomes less positive as it approaches zero. At equilibrium, Q = K, and ΔG = 0.

18. The reaction volition be spontaneous at temperatures greater than 287 K.

twenty. K = 5.35 × 10¹⁵
The process is exothermic.

22. ane.0 × 10^−viii atm. This is the maximum force per unit area of the gases under the stated weather condition.

24. [latex]10 = 1.29\;\times\;10^{-5}\;\text{atm} = \text{P}_{\text{O}_2}[/latex]

26. −0.16 kJ

28. (a) −22.1 kJ; (b) 61.half dozen kJ/mol

30. 90 kJ/mol

32. (a) Under standard thermodynamic atmospheric condition, the evaporation is nonspontaneous; (b) K_p = 0.031; (c) The evaporation of water is spontaneous; (d) [latex]\text{P}_{\text{H}_2\text{O}}[/latex] must e'er exist less than 1000_p or less than 0.031 atm. 0.031 atm represents air saturated with water vapor at 25 °C, or 100% humidity.

34. (a) Nonspontaneous equally [latex]{\Delta}G_{298}^{\circ}\;{\textgreater}\;0[/latex]; (b) [latex]{\Delta}G_{298}^{\circ} = -RT\;\text{ln}\;One thousand\text{,\;}{\Delta}G = 1.7\;\times\;10^3\;+\;(8.314\;\times\;335\;\times\;\text{ln}\;\frac{28}{128}) = -2.5\;\text{kJ}[/latex]. The forwards reaction to produce F6P is spontaneous under these conditions.

36. ΔThousand is negative as the process is spontaneous. ΔH is positive as with the solution becoming common cold, the dissolving must exist endothermic. ΔDue south must be positive as this drives the process, and it is expected for the dissolution of whatever soluble ionic chemical compound.

38. (a) Increasing [latex]P_{\text{O}_2}[/latex] volition shift the equilibrium toward the products, which increases the value of Yard. [latex]{\Delta}G_{298}^{\circ}[/latex] therefore becomes more negative.
(b) Increasing [latex]P_{\text{O}_2}[/latex] will shift the equilibrium toward the products, which increases the value of One thousand. [latex]{\Delta}G_{298}^{\circ}[/latex] therefore becomes more negative.
(c) Increasing [latex]P_{\text{O}_2}[/latex] will shift the equilibrium the reactants, which decreases the value of K. [latex]{\Delta}G_{298}^{\circ}[/latex] therefore becomes more positive.

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Source: https://opentextbc.ca/chemistry/chapter/16-4-free-energy/

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