| title | layout |
|---|---|
Free Energy |
page |
One of the challenges of using the second law of thermodynamics to determine if a process is spontaneous is that we must determine 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 properties only was introduced in the late nineteenth century by American mathematician Josiah Willard Gibbs{: data-type="term" .no-emphasis}. This new property is called the Gibbs free energy change (G){: data-type="term"} (or simply the free energy), and it is defined in terms of a system’s enthalpy and entropy as the following:
Free energy is a state function, and at constant temperature and pressure, the standard free energy change (ΔG°){: data-type="term"} may be expressed as the following:
(For simplicity’s sake, the subscript “sys” will be omitted henceforth.)
We can understand the relationship between this system property and the spontaneity of a process by recalling the previously derived second law expression:
The first law requires that qsurr = −qsys, and at constant pressure qsys = ΔH, and so this expression may be rewritten as the following:
ΔH is the enthalpy change of the system. Multiplying both sides of this equation by −T, and rearranging yields the following:
Comparing this equation to the previous one for free energy change shows the following relation:
The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, ΔSuniv. [link] summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators.
| Relation between Process Spontaneity and Signs of Thermodynamic Properties | ||
|---|---|---|
| ΔSuniv > 0 | ΔG < 0 | spontaneous |
| ΔSuniv < 0 | ΔG > 0 | nonspontaneous |
| ΔSuniv = 0 | ΔG = 0 | reversible (at equilibrium) |
Free energy is a state function, 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 approach to the calculation of free energy changes for physical and chemical reactions is by use of widely available compilations of standard state thermodynamic data. One method involves the use of standard enthalpies and entropies to compute standard free energy changes according to the following relation as demonstrated in [link].
Solution The process of interest is the following:
| Substance | ΔHf°(kJ/mol)
| S298°(J/K·mol)
| {: valign="top"}|---------- | H2O(l) | −286.83 | 70.0 | {: valign="top"}| H2O(g) | −241.82 | 188.8 | {: valign="top"}{: summary="This table has three columns and three rows. The first row is a header row and it labels each column: “Substance,” “capital delta H subscript f superscript degree symbol ( k J / mol),” and “S subscript 298 superscript degree symbol ( J / K dot mol ).” Under the “Substance” column are H subscript 2 O ( l ) and H subscript 2 O ( g ). Under the “capital delta H subscript f superscript degree symbol ( k J / mol)” column are the values negative 286.83 and negative 241.82. Under the “S subscript 298 superscript degree symbol ( J / K dot mol )” column are the values 70.0 and 188.8." .medium .unnumbered data-label=""}
Combining at 298 K:
and so boiling is nonspontaneous (not spontaneous).
Check Your Learning Use standard enthalpy and entropy data from Appendix G{: .target-chapter} to calculate the standard free energy change for the reaction shown here (298 K). What does the computed value for ΔG° say about the spontaneity of this process?
the reaction is nonspontaneous (not spontaneous) at 25 °C.
Free energy changes may also use the standard free energy of formation{: data-type="term"} (ΔGf°),
for each of the reactants and products involved in the reaction. The standard free energy of formation is the free energy change that accompanies the formation of one mole of a substance from its elements in their standard states. Similar to the standard enthalpies of formation, ΔGf°
is by definition zero for elemental substances under standard state conditions. The approach to computing the free energy change for a reaction using this approach is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction
the standard free energy change at room temperature may be calculated as
using (a) standard free energies of formation and (b) standard enthalpies of formation and standard entropies. Do the results indicate the reaction to be spontaneous or nonspontaneous under standard conditions?
Solution The required data are available in Appendix G{: .target-chapter} and are shown here.
| Compound | ΔGf°(kJ/mol)
| ΔHf°(kJ/mol)
| S298°(J/K·mol)
| {: valign="top"}|---------- | HgO (s, yellow) | −58.43 | −90.46 | 71.13 | {: valign="top"}| Hg(l) | 0 | 0 | 75.9 | {: valign="top"}| O2(g) | 0 | 0 | 205.2 | {: valign="top"}{: summary="This table has four columns and four rows. The first row is a header row, and it labels each column: “Compound,” “capital delta G subscript f superscript degree symbol ( k J / mol ),” “capital delta H subscript f superscript degree symbol ( k J / mol ),” and “S subscript 298 superscript degree symbol ( J / K dot mol ).” Under the “Compound” column are the following: “H g O ( s, yellow ),” “H g ( l ),” and “O subscript 2 ( g ).” Under the “capital delta G subscript f superscript degree symbol ( k J / mol )” column are the following values: negative 58.43, 0, and 0. Under the “capital delta H subscript f superscript degree symbol ( k J / mol )” column are the values: negative 90.46, 0, and 0. Under the “S subscript 298 superscript degree symbol ( J / K dot mol )” column are the values: 71.13, 75.9, and 205.2." .medium .unnumbered data-label=""}
(a) Using free energies of formation:
Check Your Learning Calculate ΔG° using (a) free energies of formation and (b) enthalpies of formation and entropies (Appendix G{: .target-chapter}). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 °C?
(b) 141.5 kJ/mol, nonspontaneous
As was previously demonstrated in this chapter’s section on entropy, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions can also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:
The spontaneity of a process, as reflected in the arithmetic sign of its free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:
- Both ΔH and ΔS are positive. This condition describes an endothermic process that involves an increase in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is greater than ΔH. If the TΔS term is less than ΔH, the free energy change will be positive. Such a process is spontaneous at high temperatures and nonspontaneous at low temperatures.
- Both ΔH and ΔS are negative. This condition describes an exothermic process that involves a decrease in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is less than ΔH. If the TΔS term’s magnitude is greater than ΔH, the free energy change will be positive. Such a process is spontaneous at low temperatures and nonspontaneous at high temperatures.
- ΔH is positive and ΔS is negative. This condition describes an endothermic process that involves a decrease in system entropy. In this case, ΔG will be positive regardless of the temperature. Such a process is nonspontaneous at all temperatures.
- ΔH is negative and ΔS is positive. This condition describes an exothermic process that involves an increase in system entropy. In this case, ΔG will be negative regardless of the temperature. Such a process is spontaneous at all temperatures. {: data-number-style="arabic"}
These four scenarios are summarized in [link].
{: #CNX_Chem_16_04_Scenarios}
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 (net gain of one mole of gas, ΔS > 0). The reaction is therefore spontaneous (ΔG < 0) at all temperatures.
Check Your Learning Popular chemical hand warmers generate heat by the air-oxidation of iron:
When considering the conclusions drawn regarding the temperature dependence of spontaneity, it is important to keep 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 one temperature but spontaneous at another will necessarily undergo a change in “spontaneity” (as reflected by its ΔG) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which ΔG is plotted on the y axis versus T on the x axis:
Such a plot is shown in [link]. A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependent spontaneity as depicted by the two yellow lines in the plot. Each line crosses from one 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 zero:
And so, saying a process is spontaneous at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which ΔG for the process is zero. As noted earlier, this condition describes a system at equilibrium.
{: #CNX_Chem_16_05_TempSpont}
Solution The process of interest is the following phase change:
Check Your Learning Use the information in Appendix G{: .target-chapter} to estimate the boiling point of CS2.
The free energy change for a process may be viewed as a measure of its driving force. 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 ΔG is zero, the forward and reverse driving forces are equal, and so the process occurs in both directions at the same rate (the system is at equilibrium).
In the chapter on equilibrium the reaction quotient, Q, was introduced as a convenient 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 may use its value to identify the direction in which a reaction will proceed in order to achieve equilibrium. When Q is lesser than the equilibrium constant, K, the reaction will proceed in the forward direction until equilibrium is reached and Q = K. Conversely, if Q > K, the process will proceed in the reverse direction until equilibrium is achieved.
The free energy change for a process taking place with reactants and products present under nonstandard conditions, ΔG, is related to the standard free energy change, ΔG°, according to this equation:
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 use this equation to predict the spontaneity for a process under any given set of conditions as illustrated in [link].
T = 25 °C, PN2=0.870 atm,
PH2=0.250 atm,
and PNH3=12.9 atm
Check Your Learning Calculate the 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 conditions?
For a system at equilibrium, Q = K and ΔG = 0, and the previous equation may be written as
This form of the equation provides a useful link between these two essential thermodynamic properties, and it can be used to derive equilibrium constants from standard free energy changes and vice versa. The relations between standard free energy changes and equilibrium constants are summarized in [link].
| Relations between Standard Free Energy Changes and Equilibrium Constants | ||
|---|---|---|
| K | ΔG° | Comments |
| > 1 | < 0 | Products are more abundant at equilibrium. |
| < 1 | > 0 | Reactants are more abundant at equilibrium. |
| = 1 | = 0 | Reactants and products are equally abundant at equilibrium. |
Solution The reaction of interest is the following:
Check Your Learning Use the thermodynamic data provided in Appendix G{: .target-chapter} to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C.
To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of Q), equilibrium is established when the system’s free energy is minimized ([link]). If a system is present with reactants and products present in nonequilibrium amounts (Q ≠ K), the reaction will proceed spontaneously in the direction necessary to establish equilibrium.
{: #CNX_Chem_16_04_Gibbs}
Gibbs free energy (G) is a state function defined with regard to system quantities only and may be used to predict the spontaneity of a process. A negative value for ΔG indicates a spontaneous process; a positive ΔG indicates a nonspontaneous process; and a ΔG of zero indicates that the system is at equilibrium. A number of approaches to the computation of free energy changes are possible.
- ΔG = ΔH − TΔS
- ΔG = ΔG° + RT ln Q
- ΔG° = −RT ln K {: data-bullet-style="bullet"}
for a chemical change?
= 100 kJ/mol and ΔS298°=250 J/mol·K.
Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?
Above 400 K, ΔG will become negative, and the reaction will become spontaneous.
(a) MnO2(s)⟶Mn(s)+O2(g)
(b) H2(g)+Br2(l)⟶2HBr(g)
(c) Cu(s)+S(g)⟶CuS(s)
(d) 2LiOH(s)+CO2(g)⟶Li2CO3(s)+H2O(g)
(e) CH4(g)+O2(g)⟶C(s,graphite)+2H2O(g)
(f) CS2(g)+3Cl2(g)⟶CCl4(g)+S2Cl2(g)
(a) C(s, graphite)+O2(g)⟶CO2(g)
(b) O2(g)+N2(g)⟶2NO(g)
(c) 2Cu(s)+S(g)⟶Cu2S(s)
(d) CaO(s)+H2O(l)⟶Ca(OH)2(s)
(e) Fe2O3(s)+3CO(g)⟶2Fe(s)+3CO2(g)
(f) CaSO4·2H2O(s)⟶CaSO4(s)+2H2O(g)
P4(s)+5O2(g)⟶P4O10(s)ΔG298°=−2697.0 kJ/mol
{: data-type="newline"}
2H2(g)+O2(g)⟶2H2O(g)ΔG298°=−457.18 kJ/mol
{: data-type="newline"}
6H2O(g)+P4O10(s)⟶4H3PO4(l)ΔG298°=−428.66 kJ/mol
(a) Determine the standard free energy of formation, ΔGf°,
for phosphoric acid.
(b) How does your calculated result compare to the value in Appendix G{: .target-chapter}? Explain.
2HgO(s,red)⟶2Hg(l)+O2(g)
(a) Is the decomposition spontaneous under standard state conditions?
(b) Above what temperature does the reaction become spontaneous?
(a) Ammonia: 2NH3(g)⟶N2(g)+3H2(g)
(b) Diborane: B2H6(g)⟶2B(g)+3H2(g)
(c) Hydrazine: N2H4(g)⟶N2(g)+2H2(g)
(d) Hydrogen peroxide: H2O2(l)⟶H2O(g)+12O2(g)
(a) N2(g)+O2(g)⟶2NO(g)T=2000°CKp=4.1×10−4
(b) H2(g)+I2(g)⟶2HI(g)T=400°CKp=50.0
(c) CO2(g)+H2(g)⟶CO(g)+H2O(g)T=980°CKp=1.67
(d) CaCO3(s)⟶CaO(s)+CO2(g)T=900°CKp=1.04
(e) HF(aq)+H2O(l)⟶H3O+(aq)+F−(aq)T=25°CKp=7.2×10−4
(f) AgBr(s)⟶Ag+(aq)+Br−(aq)T=25°CKp=3.3×10−13
102 kJ; (b) −21.9 kJ; (c) −5.34 kJ; (d) −0.383 kJ; (e) 18 kJ; (f) 71 kJ
(a) Cl2(g)+Br2(g)⟶2BrCl(g)T=25°CKp=4.7×10−2
(b) 2SO2(g)+O2(g)⇌2SO3(g)T=500°CKp=48.2
(c) H2O(l)⇌H2O(g)T=60°CKp=0.196 atm
(d) CoO(s)+CO(g)⇌Co(s)+CO2(g)T=550°CKp=4.90×102
(e) CH3NH2(aq)+H2O(l)⟶CH3NH3+(aq)+OH−(aq)T=25°CKp=4.4×10−4
(f) PbI2(s)⟶Pb2+(aq)+2I−(aq)T=25°CKp=8.7×10−9
(a) O2(g)+2F2(g)⟶2OF2(g)ΔG°=−9.2 kJ
(b) I2(s)+Br2(l)⟶2IBr(g)ΔG°=7.3 kJ
(c) 2LiOH(s)+CO2(g)⟶Li2CO3(s)+H2O(g)ΔG°=−79 kJ
(d) N2O3(g)⟶NO(g)+NO2(g)ΔG°=−1.6 kJ
(e) SnCl4(l)⟶SnCl4(l)ΔG°=8.0 kJ
1013; (d) K = 1.9; (e) K = 0.04
(a) I2(s)+Cl2(g)⟶2ICl(g)ΔG°=−10.88 kJ
(b) H2(g)+I2(s)⟶2HI(g)ΔG°=3.4 kJ
(c) CS2(g)+3Cl2(g)⟶CCl4(g)+S2Cl2(g)ΔG°=−39 kJ
(d) 2SO2(g)+O2(g)⟶2SO3(g)ΔG°=−141.82 kJ
(e) CS2(g)⟶CS2(l)ΔG°=−1.88 kJ
(a) O2(g)+2F2(g)⟶2F2O(g)(T=100°C)
(b) I2(s)+Br2(l)⟶2IBr(g)(T=0.0°C)
(c) 2LiOH(s)+CO2(g)⟶Li2CO3(s)+H2O(g)(T=575°C)
(d) N2O3(g)⟶NO(g)+NO2(g)(T=−10.0°C)
(e) SnCl4(l)⟶SnCl4(g)(T=200°C)
(a) K = 1.29;* * * {: data-type="newline"}
(b) K = 2.51 ×
10−3;* * * {: data-type="newline"}
(c) K = 4.83 ×
103;* * * {: data-type="newline"}
(d) K = 0.219;* * * {: data-type="newline"}
(e) K = 16.1
(a) I2(s)+Cl2(g)⟶2ICl(g)(T=100°C)
(b) H2(g)+I2(s)⟶2HI(g)(T=0.0°C)
(c) CS2(g)+3Cl2(g)⟶CCl4(g)+S2Cl2(g)(T=125°C)
(d) 2SO2(g)+O2(g)⟶2SO3(g)(T=675°C)
(e) CS2(g)⟶CS2(l)(T=90°C)
N2O4(g)⇌2NO2(g)KP=0.142
What is the standard free energy change at this temperature? Describe what happens to the initial system, where the reactants and products are in standard states, as it approaches equilibrium.
When reactants and products are in their standard states (1 bar or 1 atm), Q = 1. As the reaction proceeds toward equilibrium, the reaction shifts left (the amount of products drops while the amount of reactants increases): Q < 1, and ΔG298
becomes less positive as it approaches zero. At equilibrium, Q = K, and ΔG = 0.
spontaneous?
10−14. Using this information, calculate 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.)
What is the equilibrium constant for this reaction? Is the reaction endothermic or exothermic?
1015* * * {: data-type="newline"}
The process is exothermic.
10−8 atm. This is the maximum pressure of the gases under the stated conditions.
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 found as benzene?
CH4(g)+4Cl2(g)⟶CCl4(g)+4HCl(g)
What is the equilibrium constant for the reaction at 850 K? Would the reaction vessel need to be heated or cooled to keep the temperature of the reaction constant?
2CH3CO2H(g)⟶(CH3CO2H)2(g)
The dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of dimer.
103 (pressure in atm). What is ΔS° for the reaction?
4NH3(g)+5O2(g)⟶4NO(g)+6H2O(g)
{: data-type="newline"}
2NO(g)+O2(g)⟶2NO2(g)
{: data-type="newline"}
3NO2(g)+H2O(l)⟶2HNO3(l)+NO(g)
How much heat is evolved when 1 mol of NH3(g) is converted to HNO3(l)? Assume standard states at 25 °C.
(a) Antimony pentachloride decomposes at 448 °C. The reaction is:* * * {: data-type="newline"}
SbCl5(g)⟶SbCl3(g)+Cl2(g)
An equilibrium mixture in a 5.00 L flask at 448 °C contains 3.85 g of SbCl5, 9.14 g of SbCl3, and 2.84 g of Cl2.
(b) Chlorine molecules dissociate according to this reaction:* * * {: data-type="newline"}
Cl2(g)⟶2Cl(g)
1.00% of Cl2 molecules dissociate at 975 K and a pressure of 1.00 atm.
(b) 61.6 kJ/mol
for Pb2+(aq) and Cl−(aq) is −24.3 kJ/mole and −131.2 kJ/mole respectively, determine the solubility product, Ksp, for PbCl2(s).
for the formation of S2−(aq) given that the ΔGf°
for Ag+(aq) and Ag2S(s) are 77.1 kJ/mole and −39.5 kJ/mole respectively, and the solubility product for Ag2S(s) is 8 ×
10−51.
H2O(l)⇌H2O(g)ΔG298°=8.58 kJ
(a) Is the evaporation of water under standard thermodynamic conditions spontaneous?
(b) Determine the equilibrium constant, KP, for this physical process.
(c) By calculating ∆G, determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, PH2O,
is 0.011 atm.
(d) If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, what must be the value of PH2O
in the air?
must always be less than Kp or less than 0.031 atm. 0.031 atm represents air saturated with water vapor at 25 °C, or 100% humidity.
Glu+ATP⟶G6P+ADPΔG298°=−17 kJ
In this process, ATP becomes ADP summarized by the following equation:* * * {: data-type="newline"}
ATP⟶ADPΔG298°=−30 kJ
Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:* * * {: data-type="newline"}
Glu⟶G6PΔG298°=?
G6P⇌F6PΔG298°=1.7 kJ
(a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?
(b) Standard thermodynamic conditions imply the concentrations of G6P and F6P to be 1 M, however, in a typical cell, they are not even close to these values. Calculate ΔG when the concentrations of G6P and F6P are 120 μM and 28 μM respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37 °C.
(b) ΔG298°=−RTlnK,
ΔG=1.7×103+(8.314×310×ln28120)=−2.1 kJ.
The forward reaction to produce F6P is spontaneous under these conditions.
(a) N2(g)+3H2(g)⟶2NH3(g)
(b) HCl(g)+NH3(g)⟶NH4Cl(s)
(c) (NH4)2Cr2O7(s)⟶Cr2O3(s)+4H2O(g)+N2(g)
(d) 2Fe(s)+3O2(g)⟶Fe2O3(s)
Cu2S(s)⟶Cu(s)+S(s)
(a) Determine ΔG298°
for the decomposition of Cu2S(s).
(b) The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine ΔG298°
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 explain why coupling these reactions together makes for a more efficient process for the production of the copper.
(becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?
(a) S(s)+O2(g)⟶SO2(g)
(b) 2SO2(g)+O2(g)⟶2SO3(g)
(c) HgO(s)⟶Hg(l)+O2(g)
thus becomes more negative.
(b) Increasing the oxygen partial pressure will yield a decrease in Q and ΔG
thus becomes more negative.
(c) Increasing the oxygen partial pressure will yield an increase in Q and ΔG
thus becomes more positive.
Gibbs free energy change (G) : thermodynamic property defined in terms of system enthalpy and entropy; all spontaneous processes involve a decrease in G ^
standard free energy change (ΔG°) : change in free energy for a process occurring under standard conditions (1 bar pressure for gases, 1 M concentration for solutions) ^
standard free energy of formation (ΔGf°)
: change in free energy accompanying the formation of one mole of substance from its elements in their standard states