{"id":869,"date":"2021-07-23T09:20:57","date_gmt":"2021-07-23T13:20:57","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/free-energy\/"},"modified":"2022-06-23T09:24:01","modified_gmt":"2022-06-23T13:24:01","slug":"free-energy","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/free-energy\/","title":{"raw":"16.4 Free Energy","rendered":"16.4 Free Energy"},"content":{"raw":"&nbsp;\r\n<div class=\"textbox textbox--learning-objectives\">\r\n<h3><strong>Learning Objectives<\/strong><\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Define Gibbs free energy, and describe its relation to spontaneity<\/li>\r\n \t<li>Calculate free energy change for a process using free energies of formation for its reactants and products<\/li>\r\n \t<li>Calculate free energy change for a process using enthalpies of formation and the entropies for its reactants and products<\/li>\r\n \t<li>Explain how temperature affects the spontaneity of some processes<\/li>\r\n \t<li>Relate standard free energy changes to equilibrium constants<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idm21370272\">One of the challenges of using the second law of thermodynamics to determine if a process is spontaneous is that it requires measurements of the entropy change for the system <em data-effect=\"italics\">and<\/em> 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 <span class=\"no-emphasis\" data-type=\"term\">Gibbs<\/span>. This new property is called the <span data-type=\"term\"><strong>Gibbs free energy<\/strong> <strong>(<em data-effect=\"italics\">G<\/em>)<\/strong><\/span> (or simply the <em data-effect=\"italics\">free energy<\/em>), and it is defined in terms of a system\u2019s enthalpy and entropy as the following:<\/p>\r\n\r\n<div id=\"fs-idm129849344\" style=\"padding-left: 40px\" data-type=\"equation\"><em>G<\/em> = <em>H<\/em> -<em> TS<\/em><\/div>\r\n<p id=\"fs-idm247387040\">Free energy is a state function, and at constant temperature and pressure, the <span data-type=\"term\">free energy change (\u0394<em data-effect=\"italics\">G<\/em>)<\/span> may be expressed as the following:<\/p>\r\n\r\n<div id=\"fs-idm279983248\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em> = \u0394<em>H<\/em> - <em>T<\/em>\u0394<em>S<\/em><\/div>\r\n<p id=\"fs-idm111872960\">(For simplicity\u2019s sake, the subscript \u201csys\u201d will be omitted henceforth.)<\/p>\r\n<p id=\"fs-idm275842432\">The relationship between this system property and the spontaneity of a process may be understood by recalling the previously derived second law expression:<\/p>\r\n\r\n<div id=\"fs-idm33912192\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1996\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4a-300x77.png\" alt=\"\" width=\"160\" height=\"41\" \/><\/div>\r\n<p id=\"fs-idm170940880\">The first law requires that <em data-effect=\"italics\">q<\/em><sub>surr<\/sub> = \u2212<em data-effect=\"italics\">q<\/em><sub>sys<\/sub>, and at constant pressure <em data-effect=\"italics\">q<\/em><sub>sys<\/sub> = \u0394<em data-effect=\"italics\">H<\/em>, so this expression may be rewritten as:<\/p>\r\n\r\n<div id=\"fs-idm195954976\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1997\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4b-300x75.png\" alt=\"\" width=\"148\" height=\"37\" \/><\/div>\r\n<p id=\"fs-idm169794560\">Multiplying both sides of this equation by \u2212<em data-effect=\"italics\">T<\/em>, and rearranging yields the following:<\/p>\r\n\r\n<div id=\"fs-idm248567088\" style=\"padding-left: 40px\" data-type=\"equation\">\u2212<em>T<\/em>\u0394<em>S<\/em><sub>univ<\/sub> = \u0394<em>H<\/em> - <em>T<\/em> \u0394<em>S<\/em><\/div>\r\n<p id=\"fs-idm229909296\">Comparing this equation to the previous one for free energy change shows the following relation:<\/p>\r\n\r\n<div id=\"fs-idm117423648\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em> = \u2212<em>T<\/em>\u0394<em>S<\/em><sub>univ<\/sub><\/div>\r\n<p id=\"fs-idm228637552\">The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, \u0394<em data-effect=\"italics\">S<\/em><sub>univ<\/sub>. <a class=\"autogenerated-content\" href=\"#fs-idm211518768\">(Figure)<\/a> summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators.<\/p>\r\n\r\n<table id=\"fs-idm211518768\" class=\"top-titled\" summary=\"This table has three columns and three rows. The first column has the following: \u201ccapital delta S subscript univ is greater than 0,\u201d \u201ccapital delta S subscript univ is less than 0,\u201d and, \u201ccapital delta S subscript univ equals 0.\u201d The second column contains the following: \u201ccapital delta G is less than 0,\u201d \u201ccapital delta G is greater than 0,\u201d and, \u201ccapital delta G equals 0.\u201d The third column contains the following: \u201cSpontaneous,\u201d \u201cnonspontaneous ( spontaneous in opposite direction ),\u201d and, \u201creversible ( system is at equilibrium ).\u201d\">\r\n<thead>\r\n<tr>\r\n<th colspan=\"3\" data-align=\"center\">Relation between Process Spontaneity and Signs of Thermodynamic Properties<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td data-align=\"center\">\u0394<em data-effect=\"italics\">S<\/em><sub>univ<\/sub> &gt; 0<\/td>\r\n<td data-align=\"center\">\u0394<em data-effect=\"italics\">G<\/em> &lt; 0<\/td>\r\n<td data-align=\"center\">spontaneous<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"center\">\u0394<em data-effect=\"italics\">S<\/em><sub>univ<\/sub> &lt; 0<\/td>\r\n<td data-align=\"center\">\u0394<em data-effect=\"italics\">G<\/em> &gt; 0<\/td>\r\n<td data-align=\"center\">nonspontaneous<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"center\">\u0394<em data-effect=\"italics\">S<\/em><sub>univ<\/sub> = 0<\/td>\r\n<td data-align=\"center\">\u0394<em data-effect=\"italics\">G<\/em> = 0<\/td>\r\n<td data-align=\"center\">at equilibrium<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-idm348811808\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>What\u2019s \u201cFree\u201d about \u0394<em data-effect=\"italics\">G<\/em>?<\/strong><\/h3>\r\n<p id=\"fs-idm361428560\">In addition to indicating spontaneity, the free energy change also provides information regarding the amount of useful work (<em data-effect=\"italics\">w<\/em>) that may be accomplished by a spontaneous process. Although a rigorous treatment of this subject is beyond the scope of an introductory chemistry text, a brief discussion is helpful for gaining a better perspective on this important thermodynamic property.<\/p>\r\n<p id=\"fs-idm351078736\">For this purpose, consider a spontaneous, exothermic process that involves a decrease in entropy. The free energy, as defined by<\/p>\r\n\r\n<div id=\"fs-idm343621088\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em> = \u0394<em>H<\/em> - <em>T<\/em>\u0394<em>S<\/em><\/div>\r\n<p id=\"fs-idm347838176\">may be interpreted as representing the difference between the energy produced by the process, \u0394<em data-effect=\"italics\">H<\/em>, and the energy lost to the surroundings, <em data-effect=\"italics\">T<\/em>\u0394<em data-effect=\"italics\">S<\/em>. The difference between the energy produced and the energy lost is the energy available (or \u201cfree\u201d) to do useful work by the process, \u0394<em data-effect=\"italics\">G<\/em>. If the process somehow could be made to take place under conditions of thermodynamic reversibility, the amount of work that could be done would be maximal:<\/p>\r\n\r\n<div id=\"fs-idm353619776\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em> = <em>w<\/em><sub>max<\/sub><\/div>\r\n<p id=\"fs-idm353511696\">However, as noted previously in this chapter, such conditions are not realistic. In addition, the technologies used to extract work from a spontaneous process (e.g., automobile engine, steam turbine) are never 100% efficient, and so the work done by these processes is always less than the theoretical maximum. Similar reasoning may be applied to a nonspontaneous process, for which the free energy change represents the <em data-effect=\"italics\">minimum<\/em> amount of work that must be done <em data-effect=\"italics\">on<\/em> the system to carry out the process.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idm305910384\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Calculating Free Energy Change<\/strong><\/h3>\r\n<p id=\"fs-idm233417152\">Free energy is a state function, so its value depends only on the conditions of the initial and final states of the system. 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<strong> standard free energy changes, \u0394<em data-effect=\"italics\">G<\/em>\u00b0<\/strong>, according to the following relation.<\/p>\r\n\r\n<div id=\"fs-idm182227520\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<span data-type=\"term\">\u00b0<\/span><\/em> = \u0394<em>H<span data-type=\"term\">\u00b0<\/span><\/em> - <em>T<\/em>\u0394<em>S<span data-type=\"term\">\u00b0<\/span><\/em><\/div>\r\n<div id=\"fs-idm117444432\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm267921040\"><strong>Using Standard Enthalpy and Entropy Changes to Calculate \u0394<em data-effect=\"italics\">G<\/em>\u00b0<\/strong><\/p>\r\nUse standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the vaporization of water at room temperature (298 K). What does the computed value for \u0394<em data-effect=\"italics\">G<\/em>\u00b0 say about the spontaneity of this process?\r\n<p id=\"fs-idm222992784\"><strong>Solution:<\/strong><\/p>\r\nThe process of interest is the following:\r\n<div id=\"fs-idm59589120\" style=\"padding-left: 40px\" data-type=\"equation\">H<sub>2<\/sub>O(<em>l<\/em>) \u27f6 H<sub>2<\/sub>O(<em>g<\/em>)<\/div>\r\n<p id=\"fs-idm23409248\">The standard change in free energy may be calculated using the following equation:<\/p>\r\n\r\n<div id=\"fs-idm165858560\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<span data-type=\"term\">\u00b0<\/span><\/em> = \u0394<em>H<span data-type=\"term\">\u00b0<\/span><\/em> - <em>T<\/em>\u0394<em>S<span data-type=\"term\">\u00b0<\/span><\/em><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm272232416\">From Appendix G:<\/p>\r\n\r\n<table id=\"fs-idm230329280\" class=\"medium unnumbered\" summary=\"This table has three columns and three rows. The first row is a header row and it labels each column: \u201cSubstance,\u201d \u201ccapital delta H subscript f superscript degree symbol ( k J \/ mol),\u201d and \u201cS subscript 298 superscript degree symbol ( J \/ K dot mol ).\u201d Under the \u201cSubstance\u201d column are H subscript 2 O ( l ) and H subscript 2 O ( g ). Under the \u201ccapital delta H subscript f superscript degree symbol ( k J \/ mol)\u201d column are the values negative 286.83 and negative 241.82. Under the \u201cS subscript 298 superscript degree symbol ( J \/ K dot mol )\u201d column are the values 70.0 and 188.8.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th data-align=\"center\">Substance<\/th>\r\n<th data-align=\"center\">\u0394H<sub>f<\/sub>\u00b0(kJ\/mol)<\/th>\r\n<th data-align=\"center\">S\u00b0(J\/K\u00b7mol)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td data-align=\"center\">H<sub>2<\/sub>O(<em data-effect=\"italics\">l<\/em>)<\/td>\r\n<td data-align=\"center\">\u2212285.83<\/td>\r\n<td data-align=\"center\">70.0<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"center\">H<sub>2<\/sub>O(<em data-effect=\"italics\">g<\/em>)<\/td>\r\n<td data-align=\"center\">\u2212241.82<\/td>\r\n<td data-align=\"center\">188.8<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idm123711584\">Using the appendix data to calculate the standard enthalpy and entropy changes yields:<\/p>\r\n\r\n<div id=\"fs-idm164142400\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394H\u00b0 = \u0394H<sub>f<\/sub>\u00b0(H<sub>2<\/sub>O(<em>g<\/em>)) - \u0394H<sub>f<\/sub>\u00b0(H<sub>2<\/sub>O(<em>l<\/em>)) = [\u2212241.82 kJ\/mol - (\u2212285.83 kJ\/mol)] = 44.01 kJ\/mol<\/div>\r\n<div id=\"fs-idm171591584\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394S\u00b0 = S\u00b0(H<sub>2<\/sub>O(<em>g<\/em>)) - S\u00b0(H<sub>2<\/sub>O(<em>l<\/em>)) = 188.8 J\/mol\u00b7K - 70.0 J\/mol\u00b7K =118.8 J\/mol\u00b7K<\/div>\r\n<div id=\"fs-idm129668064\" data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp11689536\">Substitution into the standard free energy equation yields:<\/p>\r\n\r\n<div id=\"fs-idm246654432\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<span data-type=\"term\">\u00b0<\/span><\/em> = \u0394<em>H<span data-type=\"term\">\u00b0<\/span><\/em> - <em>T<\/em>\u0394<em>S<span data-type=\"term\">\u00b0<\/span><\/em> = 44.01 kJ\/mol - (298 K)(0.1188 J\/mol\u00b7K)<\/div>\r\n<div id=\"fs-idm135070448\" style=\"padding-left: 40px\" data-type=\"equation\">\u00a0 \u00a0 \u00a0 \u00a0 = 44.01 kJ\/mol - 35.4 kJ\/mol = 8.6 kJ\/mol<\/div>\r\n<p id=\"fs-idm276144320\">At 298 K (25 \u00b0C), \u0394G\u00b0 &gt; 0, so boiling is nonspontaneous (<em data-effect=\"italics\">not<\/em> spontaneous).<\/p>\r\n<p id=\"fs-idm248587360\"><strong>Check Your Learning:<\/strong><\/p>\r\nUse standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the reaction shown here (298 K). What does the computed value for \u0394<em data-effect=\"italics\">G<\/em>\u00b0 say about the spontaneity of this process?\r\n<div id=\"fs-idm276750416\" style=\"padding-left: 40px\" data-type=\"equation\">C<sub>2<\/sub>H<sub>6<\/sub>(<em>g<\/em>) \u27f6 H<sub>2<\/sub>(<em>g<\/em>) + C<sub>2<\/sub>H<sub>4<\/sub>(<em>g<\/em>)<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div id=\"fs-idm97664544\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm157144416\">\u0394G\u00b0 = 102.0 kJ\/mol; the reaction is nonspontaneous (<em data-effect=\"italics\">not<\/em> spontaneous) at 25 \u00b0C.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idm43672192\">The standard free energy change for a reaction may also be calculated from <span data-type=\"term\"><strong>standard free energy of formation, \u0394<em data-effect=\"italics\">G<\/em><sub>f<\/sub>\u00b0<\/strong>,<\/span> values 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 enthalpy of formation, \u0394G<sub>f<\/sub>\u00b0 is by definition zero for elemental substances under standard state conditions. The approach used to calculate \u0394G\u00b0 for a reaction from \u0394G<sub>f<\/sub>\u00b0 values is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction<\/p>\r\n\r\n<div id=\"fs-idm250621408\" style=\"padding-left: 40px\" data-type=\"equation\"><em>m<\/em>A + <em>n<\/em>B \u27f6 <em>x<\/em>C + <em>y<\/em>D,<\/div>\r\n<p id=\"fs-idm124979680\">the standard free energy change at room temperature may be calculated as<\/p>\r\n<p style=\"padding-left: 80px\"><img class=\"wp-image-2022 alignleft\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l-300x32.png\" alt=\"\" width=\"328\" height=\"35\" \/><\/p>\r\n<p style=\"padding-left: 80px\">\u00a0 \u00a0 \u00a0 <img class=\"alignnone wp-image-2024\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4m-300x26.png\" alt=\"\" width=\"335\" height=\"29\" \/><\/p>\r\n&nbsp;\r\n<div id=\"fs-idm202485584\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm232036496\"><strong>Using Standard Free Energies of Formation to Calculate \u0394<em data-effect=\"italics\">G<\/em>\u00b0<\/strong><\/p>\r\nConsider the decomposition of yellow mercury(II) oxide.\r\n<div id=\"fs-idm217690608\" style=\"padding-left: 40px\" data-type=\"equation\">HgO(<em>s<\/em>) \u27f6 Hg(<em>l<\/em>) + \u00bdO<sub>2<\/sub>(<em>g<\/em>)<\/div>\r\n<p id=\"fs-idm274288288\">Calculate the standard free energy change at room temperature, \u0394G\u00b0, 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?<\/p>\r\n<p id=\"fs-idm70679856\"><strong>Solution:<\/strong><\/p>\r\nThe required data are available in Appendix G and are shown here.\r\n<table id=\"fs-idm232730384\" class=\"medium unnumbered\" summary=\"This table has four columns and four rows. The first row is a header row, and it labels each column: \u201cCompound,\u201d \u201ccapital delta G subscript f superscript degree symbol ( k J \/ mol ),\u201d \u201ccapital delta H subscript f superscript degree symbol ( k J \/ mol ),\u201d and \u201cS subscript 298 superscript degree symbol ( J \/ K dot mol ).\u201d Under the \u201cCompound\u201d column are the following: \u201cH g O ( s, yellow ),\u201d \u201cH g ( l ),\u201d and \u201cO subscript 2 ( g ).\u201d Under the \u201ccapital delta G subscript f superscript degree symbol ( k J \/ mol )\u201d column are the following values: negative 58.43, 0, and 0. Under the \u201ccapital delta H subscript f superscript degree symbol ( k J \/ mol )\u201d column are the values: negative 90.46, 0, and 0. Under the \u201cS subscript 298 superscript degree symbol ( J \/ K dot mol )\u201d column are the values: 71.13, 75.9, and 205.2.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th style=\"width: 74.6771px\" data-align=\"center\">Compound<\/th>\r\n<th style=\"width: 82.4792px\" data-align=\"center\">\u0394G<sub>f<\/sub>\u00b0(kJ\/mol)<\/th>\r\n<th style=\"width: 291.333px\" data-align=\"center\">\u0394H<sub>f<\/sub>\u00b0(kJ\/mol)<\/th>\r\n<th style=\"width: 230.781px\" data-align=\"center\">S\u00b0(J\/K\u00b7mol)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td style=\"width: 75.1771px\" data-align=\"center\">HgO (<em data-effect=\"italics\">s<\/em>, yellow)<\/td>\r\n<td style=\"width: 83.4792px\" data-align=\"center\">\u221258.43<\/td>\r\n<td style=\"width: 292.333px\" data-align=\"center\">\u221290.46<\/td>\r\n<td style=\"width: 231.281px\" data-align=\"center\">71.13<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td style=\"width: 75.1771px\" data-align=\"center\">Hg(<em data-effect=\"italics\">l<\/em>)<\/td>\r\n<td style=\"width: 83.4792px\" data-align=\"center\">0<\/td>\r\n<td style=\"width: 292.333px\" data-align=\"center\">0<\/td>\r\n<td style=\"width: 231.281px\" data-align=\"center\">75.9<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td style=\"width: 75.1771px\" data-align=\"center\">O<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>)<\/td>\r\n<td style=\"width: 83.4792px\" data-align=\"center\">0<\/td>\r\n<td style=\"width: 292.333px\" data-align=\"center\">0<\/td>\r\n<td style=\"width: 231.281px\" data-align=\"center\">205.2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idm192751264\">(a) Using free energies of formation:<\/p>\r\n\r\n<div id=\"fs-idm211589632\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-2022\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l-300x32.png\" alt=\"\" width=\"328\" height=\"35\" \/><\/div>\r\n<div id=\"fs-idm178475312\" style=\"padding-left: 40px\" data-type=\"equation\">= [\u0394G<sub>f<\/sub>\u00b0(Hg(<em>l<\/em>)) + \u00bd\u0394G<sub>f<\/sub>\u00b0(O<sub>2<\/sub>(<em>g<\/em>))] - \u0394G<sub>f<\/sub>\u00b0(HgO(<em>s<\/em>, yellow))<\/div>\r\n<div id=\"fs-idm164697728\" style=\"padding-left: 40px\" data-type=\"equation\">= [0 kJ\/mol + \u00bd(0 kJ\/mol)] - (\u221258.43 kJ\/mol) =58.43 kJ\/mol<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm112946608\">(b) Using enthalpies and entropies of formation:<\/p>\r\n\r\n<div id=\"fs-idm147522592\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-2007\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4j-300x24.png\" alt=\"\" width=\"388\" height=\"31\" \/><\/div>\r\n<div id=\"fs-idm39030688\" style=\"padding-left: 40px\" data-type=\"equation\">= [\u0394H<sub>f<\/sub>\u00b0(Hg(<em>l<\/em>)) + \u00bd\u0394H<sub>f<\/sub>\u00b0(O<sub>2<\/sub>(<em>g<\/em>))] - \u0394H<sub>f<\/sub>\u00b0(HgO(<em>s<\/em>, yellow))<\/div>\r\n<div id=\"fs-idm261408512\" style=\"padding-left: 40px\" data-type=\"equation\">=[0 kJ\/mol + \u00bd(0 kJ\/mol)] - (\u221290.46 kJ\/mol) = 90.46 kJ\/mol<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div id=\"fs-idp6944752\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-1990\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.3h-300x31.png\" alt=\"\" width=\"319\" height=\"33\" \/><\/div>\r\n<div id=\"fs-idm61090368\" style=\"padding-left: 40px\" data-type=\"equation\">= [<em>S<\/em>\u00b0(Hg(<em>l<\/em>) + \u00bd<em>S<\/em>\u00b0(O<sub>2<\/sub>(<em>g<\/em>))] - <em>S<\/em>\u00b0(HgO(s, yellow)<\/div>\r\n<div id=\"fs-idm131288608\" style=\"padding-left: 40px\" data-type=\"equation\">= [75.9 J\/mol\u00b7K + \u00bd(205.2 J\/mol\u00b7K)] - 71.13 J\/mol\u00b7K = 107.4 J\/mol\u00b7K<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div id=\"fs-idm153411216\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<span data-type=\"term\">\u00b0<\/span><\/em> = \u0394<em>H<span data-type=\"term\">\u00b0<\/span><\/em> - <em>T<\/em>\u0394<em>S<span data-type=\"term\">\u00b0 <\/span><\/em>=90.46 kJ\/mol - (298.15 K)(0.1074 J\/mol\u00b7K<\/div>\r\n<div id=\"fs-idm190648624\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394G\u00b0 = 90.46 kJ\/mol - 32.01 kJ\/mol = 58.45 kJ\/mol<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm269603872\">Both ways to calculate the standard free energy change at 25 \u00b0C give the same numerical value (to three significant figures), and both predict that the process is nonspontaneous (<em data-effect=\"italics\">not<\/em> spontaneous) at room temperature.<\/p>\r\n<p id=\"fs-idm59738256\"><strong>Check Your Learning:<\/strong><\/p>\r\nCalculate \u0394<em data-effect=\"italics\">G<\/em>\u00b0 using (a) free energies of formation and (b) enthalpies of formation and entropies (Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 \u00b0C?\r\n<div id=\"fs-idm213619408\" style=\"padding-left: 40px\" data-type=\"equation\">C<sub>2<\/sub>H<sub>4<\/sub>(<em>g<\/em>) \u27f6 H<sub>2<\/sub>(<em>g<\/em>) + C<sub>2<\/sub>H<sub>2<\/sub>(<em>g<\/em>)<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div id=\"fs-idm254989456a\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp2244016\">(a) 140.8 kJ\/mol, nonspontaneous<\/p>\r\n<p id=\"eip-687\">(b) 141.5 kJ\/mol, nonspontaneous<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm353615984\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Free Energy Changes for Coupled Reactions<\/strong><\/h3>\r\n<p id=\"fs-idm350537712\">The use of free energies of formation to compute free energy changes for reactions as described above is possible because \u0394<em data-effect=\"italics\">G<\/em> is a state function, and the approach is analogous to the use of Hess\u2019 Law in computing enthalpy changes (see the chapter on thermochemistry). Consider the vaporization of water as an example:<\/p>\r\n\r\n<div id=\"fs-idm359616576\" style=\"padding-left: 40px\" data-type=\"equation\">H<sub>2<\/sub>O(<em>l<\/em>) \u27f6 H<sub>2<\/sub>O(<em>g<\/em>)<\/div>\r\n<p id=\"fs-idm354717680\">An equation representing this process may be derived by adding the formation reactions for the two phases of water (necessarily reversing the reaction for the liquid phase). The free energy change for the sum reaction is the sum of free energy changes for the two added reactions:<\/p>\r\n\r\n<div id=\"fs-idm360093056\" style=\"padding-left: 40px\" data-type=\"equation\">H<sub>2<\/sub>(<em>g<\/em>) + \u00bdO<sub>2<\/sub>(<em>g<\/em>) \u2192 H<sub>2<\/sub>O(<em>g<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u0394G<sub>f<\/sub>\u00b0(H<sub>2<\/sub>O(<em>g<\/em>))<\/div>\r\n<div style=\"padding-left: 40px\" data-type=\"equation\">H<sub>2<\/sub>O(<em>l<\/em>) \u2192 H<sub>2<\/sub>(<em>g<\/em>) + \u00bdO<sub>2<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0-\u0394G<sub>f<\/sub>\u00b0(H<sub>2<\/sub>O(<em>l<\/em>))<\/div>\r\n<div style=\"padding-left: 40px\" data-type=\"equation\">----------------------------------------<\/div>\r\n<div style=\"padding-left: 40px\" data-type=\"equation\">H<sub>2<\/sub>O(<em>l<\/em>) \u2192 H<sub>2<\/sub>O(<em>g<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u0394G\u00b0=\u0394G<sub>f<\/sub>\u00b0(H<sub>2<\/sub>O(<em>g<\/em>)) - \u0394G<sub>f<\/sub>\u00b0(H<sub>2<\/sub>O(<em>l<\/em>))<\/div>\r\n<p id=\"fs-idm331265792\">This approach may also be used in cases where a nonspontaneous reaction is enabled by coupling it to a spontaneous reaction. For example, the production of elemental zinc from zinc sulfide is thermodynamically unfavorable, as indicated by a positive value for \u0394<em data-effect=\"italics\">G<\/em>\u00b0:<\/p>\r\n\r\n<div id=\"fs-idm347055872\" style=\"padding-left: 40px\" data-type=\"equation\">ZnS(<em>s<\/em>) \u2192\u00a0 Zn(<em>s<\/em>) + S(<em>s<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u0394<em>G<\/em><sub>1<\/sub>\u00b0 = 201.3 kJ<\/div>\r\n<p id=\"fs-idm350667184\">The industrial process for production of zinc from sulfidic ores involves coupling this decomposition reaction to the thermodynamically favorable oxidation of sulfur:<\/p>\r\n\r\n<div id=\"fs-idm356637216\" style=\"padding-left: 40px\" data-type=\"equation\">S(<em>s<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u2192 SO<sub>2<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u0394<em>G<sub>2<\/sub><\/em>\u00b0 = \u2212300.1 kJ<\/div>\r\n<p id=\"fs-idm356330656\">The coupled reaction exhibits a negative free energy change and is spontaneous:<\/p>\r\n\r\n<div id=\"fs-idm343614160\" style=\"padding-left: 40px\" data-type=\"equation\">ZnS(<em>s<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u2192 Zn(<em>s<\/em>) + SO<sub>2<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u0394G\u00b0 = 201.3 kJ + (-300.1 kJ) = 98.8 kJ<\/div>\r\n<p id=\"fs-idm348648656\">This process is typically carried out at elevated temperatures, so this result obtained using standard free energy values is just an estimate. The gist of the calculation, however, holds true.<\/p>\r\n\r\n<div id=\"fs-idm348331936\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm348331680\"><strong>Calculating Free Energy Change for a Coupled Reaction <\/strong><\/p>\r\nIs a reaction coupling the decomposition of ZnS to the formation of H<sub>2<\/sub>S expected to be spontaneous under standard conditions?\r\n<p id=\"fs-idm116939264a\"><strong>Solution:<\/strong><\/p>\r\nFollowing the approach outlined above and using free energy values from Appendix G:\r\n<div id=\"fs-idm343484288\" data-type=\"equation\">Decomposition of zinc sulfide:\u00a0 \u00a0 \u00a0ZnS(<em>s<\/em>) \u2192 Zn(s) + S(<em>s<\/em>)\u00a0 \u00a0 \u00a0\u0394<em>G<\/em><sub>1<\/sub>\u00b0 = 201.3 kJ<\/div>\r\n<div data-type=\"equation\">Formation of hydrogen sulfide:\u00a0 \u00a0 \u00a0S(<em>s<\/em>) + H<sub>2<\/sub>(<em>g<\/em>) \u2192 H<sub>2<\/sub>S(<em>g<\/em>)\u00a0 \u00a0 \u00a0\u0394G<sub>2<\/sub>\u00b0 =- 33.4 kJ<\/div>\r\n<div data-type=\"equation\">Coupled reaction:\u00a0 \u00a0 \u00a0ZnS(<em>s<\/em>) + H<sub>2<\/sub>(<em>g<\/em>) \u2192 Zn(<em>s<\/em>) + H<sub>2<\/sub>S(<em>g<\/em>)\u00a0 \u00a0 \u00a0\u0394G\u00b0 = 201.3 kJ + (-33.4 kJ) = 167.9 kJ<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm323451264\">The coupled reaction exhibits a positive free energy change and is thus nonspontaneous.<\/p>\r\n<p id=\"fs-idm328777088\"><strong>Check Your Learning:<\/strong><\/p>\r\nWhat is the standard free energy change for the reaction below? Is the reaction expected to be spontaneous under standard conditions?\r\n<div id=\"fs-idm360883312\" style=\"padding-left: 40px\" data-type=\"equation\">FeS(<em>s<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u2192 Fe(<em>s<\/em>) + SO<sub>2<\/sub>(<em>g<\/em>)<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div id=\"fs-idm254989456\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm360208448\">\u2212199.7 kJ; spontaneous<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm157251776\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Temperature Dependence of Spontaneity<\/strong><\/h3>\r\n<p id=\"fs-idm158748720\">As was previously demonstrated in this chapter\u2019s 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:<\/p>\r\n\r\n<div id=\"fs-idm182396560\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em>\u00a0= \u0394<em>H<\/em>\u00a0- <em>T<\/em>\u0394<em>S<\/em><\/div>\r\n<p id=\"fs-idm176739120\">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 <em data-effect=\"italics\">T<\/em> 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:<\/p>\r\n\r\n<ol id=\"fs-idm124567280\" type=\"1\">\r\n \t<li><strong data-effect=\"bold\">Both \u0394<em data-effect=\"italics\">H<\/em> and \u0394<em data-effect=\"italics\">S<\/em> are positive.<\/strong> This condition describes an endothermic process that involves an increase in system entropy. In this case, \u0394<em data-effect=\"italics\">G<\/em> will be negative if the magnitude of the <em data-effect=\"italics\">T<\/em>\u0394<em data-effect=\"italics\">S<\/em> term is greater than \u0394<em data-effect=\"italics\">H<\/em>. If the <em data-effect=\"italics\">T<\/em>\u0394<em data-effect=\"italics\">S<\/em> term is less than \u0394<em data-effect=\"italics\">H<\/em>, the free energy change will be positive. Such a process is <em data-effect=\"italics\">spontaneous at high temperatures and nonspontaneous at low temperatures.<\/em><\/li>\r\n \t<li><strong data-effect=\"bold\">Both \u0394<em data-effect=\"italics\">H<\/em> and \u0394<em data-effect=\"italics\">S<\/em> are negative.<\/strong> This condition describes an exothermic process that involves a decrease in system entropy. In this case, \u0394<em data-effect=\"italics\">G<\/em> will be negative if the magnitude of the <em data-effect=\"italics\">T<\/em>\u0394<em data-effect=\"italics\">S<\/em> term is less than \u0394<em data-effect=\"italics\">H<\/em>. If the <em data-effect=\"italics\">T<\/em>\u0394<em data-effect=\"italics\">S<\/em> term\u2019s magnitude is greater than \u0394<em data-effect=\"italics\">H<\/em>, the free energy change will be positive. Such a process is <em data-effect=\"italics\">spontaneous at low temperatures and nonspontaneous at high temperatures.<\/em><\/li>\r\n \t<li><strong data-effect=\"bold\">\u0394<em data-effect=\"italics\">H<\/em> is positive and \u0394<em data-effect=\"italics\">S<\/em> is negative.<\/strong> This condition describes an endothermic process that involves a decrease in system entropy. In this case, \u0394<em data-effect=\"italics\">G<\/em> will be positive regardless of the temperature. Such a process is <em data-effect=\"italics\">nonspontaneous at all temperatures.<\/em><\/li>\r\n \t<li><strong data-effect=\"bold\">\u0394<em data-effect=\"italics\">H<\/em> is negative and \u0394<em data-effect=\"italics\">S<\/em> is positive.<\/strong> This condition describes an exothermic process that involves an increase in system entropy. In this case, \u0394<em data-effect=\"italics\">G<\/em> will be negative regardless of the temperature. Such a process is <em data-effect=\"italics\">spontaneous at all temperatures.<\/em><\/li>\r\n<\/ol>\r\n<p id=\"fs-idm39683376\">These four scenarios are summarized in <a class=\"autogenerated-content\" href=\"#CNX_Chem_16_04_Scenarios\">(Figure)<\/a>.<\/p>\r\n\r\n<div id=\"CNX_Chem_16_04_Scenarios\" class=\"bc-figure figure\">\r\n<div class=\"bc-figcaption figcaption\">There are four possibilities regarding the signs of enthalpy and entropy changes.<\/div>\r\n<span id=\"fs-idm123583264\" data-type=\"media\" data-alt=\"A table with three columns and four rows is shown. The first column has the phrase, \u201cDelta S greater than zero ( increase in entropy ),\u201d in the third row and the phrase, \u201cDelta S less than zero ( decrease in entropy),\u201d in the fourth row. The second and third columns have the phrase, \u201cSummary of the Four Scenarios for Enthalpy and Entropy Changes,\u201d written above them. The second column has, \u201cdelta H greater than zero ( endothermic ),\u201d in the second row, \u201cdelta G less than zero at high temperature, delta G greater than zero at low temperature, Process is spontaneous at high temperature,\u201d in the third row, and \u201cdelta G greater than zero at any temperature, Process is nonspontaneous at any temperature,\u201d in the fourth row. The third column has, \u201cdelta H less than zero ( exothermic ),\u201d in the second row, \u201cdelta G less than zero at any temperature, Process is spontaneous at any temperature,\u201d in the third row, and \u201cdelta G less than zero at low temperature, delta G greater than zero at high temperature, Process is spontaneous at low temperature.\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_16_04_Scenarios-1.jpg\" alt=\"A table with three columns and four rows is shown. The first column has the phrase, \u201cDelta S greater than zero ( increase in entropy ),\u201d in the third row and the phrase, \u201cDelta S less than zero ( decrease in entropy),\u201d in the fourth row. The second and third columns have the phrase, \u201cSummary of the Four Scenarios for Enthalpy and Entropy Changes,\u201d written above them. The second column has, \u201cdelta H greater than zero ( endothermic ),\u201d in the second row, \u201cdelta G less than zero at high temperature, delta G greater than zero at low temperature, Process is spontaneous at high temperature,\u201d in the third row, and \u201cdelta G greater than zero at any temperature, Process is nonspontaneous at any temperature,\u201d in the fourth row. The third column has, \u201cdelta H less than zero ( exothermic ),\u201d in the second row, \u201cdelta G less than zero at any temperature, Process is spontaneous at any temperature,\u201d in the third row, and \u201cdelta G less than zero at low temperature, delta G greater than zero at high temperature, Process is spontaneous at low temperature.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<div id=\"fs-idm299243200\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm221990400\"><strong>Predicting the Temperature Dependence of Spontaneity <\/strong><\/p>\r\nThe incomplete combustion of carbon is described by the following equation:\r\n<div id=\"fs-idm124549040\" style=\"padding-left: 40px\" data-type=\"equation\">2C(<em>s<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u27f6 2CO(<em>g<\/em>)<\/div>\r\n<p id=\"fs-idm291788752\">How does the spontaneity of this process depend upon temperature?<\/p>\r\n<p id=\"fs-idm116939264\"><strong>Solution:<\/strong><\/p>\r\nCombustion processes are exothermic (\u0394<em data-effect=\"italics\">H<\/em> &lt; 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, \u0394<em data-effect=\"italics\">S<\/em> &gt; 0). The reaction is therefore spontaneous (\u0394<em data-effect=\"italics\">G<\/em> &lt; 0) at all temperatures.\r\n<p id=\"fs-idm189692848\"><strong>Check Your Learning:<\/strong><\/p>\r\nPopular chemical hand warmers generate heat by the air-oxidation of iron:\r\n<div id=\"fs-idm247358864\" style=\"padding-left: 40px\" data-type=\"equation\">4Fe(<em>s<\/em>) + 3O<sub>2<\/sub>(<em>g<\/em>) \u27f6 2Fe<sub>2<\/sub>O<sub>3<\/sub>(<em>s<\/em>)<\/div>\r\n<p id=\"fs-idm255092672\">How does the spontaneity of this process depend upon temperature?<\/p>\r\n&nbsp;\r\n<div id=\"fs-idm137038288\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm233909808\">\u0394<em data-effect=\"italics\">H<\/em> and \u0394<em data-effect=\"italics\">S<\/em> are negative; the reaction is spontaneous at low temperatures.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idp2328928\">When considering the conclusions drawn regarding the temperature dependence of spontaneity, it is important to keep in mind what the terms \u201chigh\u201d and \u201clow\u201d 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 \u201cspontaneity\u201d (as reflected by its \u0394<em data-effect=\"italics\">G<\/em>) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which \u0394<em data-effect=\"italics\">G<\/em> is plotted on the <em data-effect=\"italics\">y<\/em> axis versus <em data-effect=\"italics\">T<\/em> on the <em data-effect=\"italics\">x<\/em> axis:<\/p>\r\n\r\n<div id=\"fs-idm161001120\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G <\/em>= \u0394<em>H <\/em>- <em>T<\/em>\u0394<em>S<\/em><\/div>\r\n<div id=\"fs-idm285695760\" style=\"padding-left: 40px\" data-type=\"equation\"><em>y<\/em> = b + m<em>x<\/em><\/div>\r\n<p id=\"fs-idm206759776\">Such a plot is shown in <a class=\"autogenerated-content\" href=\"#CNX_Chem_16_05_TempSpont\">(Figure)<\/a>. 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 \u0394<em data-effect=\"italics\">G<\/em>) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the <em data-effect=\"italics\">x<\/em>-intercept of the line, that is, the value of <em data-effect=\"italics\">T<\/em> for which \u0394<em data-effect=\"italics\">G<\/em> is zero:<\/p>\r\n\r\n<div id=\"fs-idm187292464\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em> = 0 = \u0394<em>H<\/em> - <em>T<\/em>\u0394<em>S<\/em><\/div>\r\n<div id=\"fs-idm280698832\" style=\"padding-left: 40px\" data-type=\"equation\"><em>T<\/em> = \u0394<em>H<\/em>\/\u0394<em>S<\/em><\/div>\r\n<p id=\"fs-idm44045216\">So, saying a process is spontaneous at \u201chigh\u201d or \u201clow\u201d temperatures means the temperature is above or below, respectively, that temperature at which \u0394<em data-effect=\"italics\">G<\/em> for the process is zero. As noted earlier, the condition of \u0394G = 0 describes a system at equilibrium.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_16_05_TempSpont\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">These plots show the variation in \u0394<em data-effect=\"italics\">G<\/em> with temperature for the four possible combinations of arithmetic sign for \u0394<em data-effect=\"italics\">H<\/em> and \u0394<em data-effect=\"italics\">S<\/em>.<\/div>\r\n<span id=\"fs-idm275434144\" data-type=\"media\" data-alt=\"A graph is shown where the y-axis is labeled, \u201cFree energy,\u201d and the x-axis is labeled, \u201cIncreasing temperature ( K ).\u201d The value of zero is written midway up the y-axis with the label, \u201cdelta G greater than 0,\u201d written above this line and, \u201cdelta G less than 0,\u201d written below it. The bottom half of the graph is labeled on the right as, \u201cSpontaneous,\u201d and the top half is labeled on the right as, \u201cNonspontaneous.\u201d A green line labeled, \u201cdelta H less than 0, delta S greater than 0,\u201d extends from a quarter of the way up the y-axis to the bottom right of the graph. A yellow line labeled, \u201cdelta H less than 0, delta S less than 0,\u201d extends from a quarter of the way up the y-axis to the middle right of the graph. A second yellow line labeled, \u201cdelta H greater than 0, delta S greater than 0,\u201d extends from three quarters of the way up the y-axis to the middle right of the graph. A red line labeled, \u201cdelta H greater than 0, delta S less than 0,\u201d extends from three quarters of the way up the y-axis to the top right of the graph.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_16_04_TempSpont-1.jpg\" alt=\"A graph is shown where the y-axis is labeled, \u201cFree energy,\u201d and the x-axis is labeled, \u201cIncreasing temperature ( K ).\u201d The value of zero is written midway up the y-axis with the label, \u201cdelta G greater than 0,\u201d written above this line and, \u201cdelta G less than 0,\u201d written below it. The bottom half of the graph is labeled on the right as, \u201cSpontaneous,\u201d and the top half is labeled on the right as, \u201cNonspontaneous.\u201d A green line labeled, \u201cdelta H less than 0, delta S greater than 0,\u201d extends from a quarter of the way up the y-axis to the bottom right of the graph. A yellow line labeled, \u201cdelta H less than 0, delta S less than 0,\u201d extends from a quarter of the way up the y-axis to the middle right of the graph. A second yellow line labeled, \u201cdelta H greater than 0, delta S greater than 0,\u201d extends from three quarters of the way up the y-axis to the middle right of the graph. A red line labeled, \u201cdelta H greater than 0, delta S less than 0,\u201d extends from three quarters of the way up the y-axis to the top right of the graph.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<div id=\"fs-idm232248480\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm39866480\"><strong>Equilibrium Temperature for a Phase Transition<\/strong><\/p>\r\nAs defined in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its liquid and gaseous phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in Appendix G to estimate the boiling point of water.\r\n<p id=\"fs-idm280706032\"><strong>Solution:<\/strong><\/p>\r\nThe process of interest is the following phase change:\r\n<div id=\"fs-idm181862672\" style=\"padding-left: 40px\" data-type=\"equation\">H<sub>2<\/sub>O(<em>l<\/em>) \u27f6 H<sub>2<\/sub>O(<em>g<\/em>)<\/div>\r\n<p id=\"fs-idm40643152\">When this process is at equilibrium, \u0394<em data-effect=\"italics\">G<\/em> = 0, so the following is true:<\/p>\r\n\r\n<div id=\"fs-idm275207392\" style=\"padding-left: 40px\" data-type=\"equation\">0 = \u0394<em>H<\/em>\u00b0 - <em>T<\/em>\u0394<em>S<\/em>\u00b0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 or\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <em>T<\/em> = \u0394<em>H<\/em>\u00b0\/\u0394<em>S<\/em>\u00b0<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm203234768\">Using the standard thermodynamic data from Appendix G,<\/p>\r\n\r\n<div id=\"fs-idm159129872\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>H<\/em>\u00b0 = \u0394<em>H<\/em><sub>f<\/sub>\u00b0(H<sub>2<\/sub>O(<em>g<\/em>)) - \u0394<em>H<\/em><sub>f<\/sub>\u00b0(H<sub>2<\/sub>O(<em>l<\/em>)) = 241.82 kJ\/mol - (-285.83 kJ\/mol) = 44.01 kJ\/mol<\/div>\r\n<div style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>S<\/em>\u00b0 = <em>S<\/em>\u00b0(H<sub>2<\/sub>O(<em>g<\/em>)) - <em>S<\/em>\u00b0(H<sub>2<\/sub>O(<em>l<\/em>)) = 188.8 J\/K\u00b7mol - 70.0 J\/K\u00b7mol = 118.8 J\/K\u00b7mol<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div id=\"fs-idm250608432\" style=\"padding-left: 40px\" data-type=\"equation\"><em>T<\/em> = \u0394<em>H<\/em>\u00b0\/\u0394<em>S<\/em>\u00b0 = (44.01 \u00d7 10<sup>3<\/sup> J\/mol)\/(118.8 J\/K\u00b7mol) = 370.5 K = 97.3 \u00b0C<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm289288576\">The accepted value for water\u2019s normal boiling point is 373.2 K (100.0 \u00b0C), and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (Appendix G). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point.<\/p>\r\n<p id=\"fs-idm252873760\"><strong>Check Your Learning:<\/strong><\/p>\r\nUse the information in Appendix G to estimate the boiling point of CS<sub>2<\/sub>.\r\n<div id=\"fs-idm230494144\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm181951344\">313 K (accepted value 319 K)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm159117504\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Free Energy and Equilibrium<\/strong><\/h3>\r\n<p id=\"fs-idm166739792\">The free energy change for a process may be viewed as a measure of its driving force. A negative value for \u0394<em data-effect=\"italics\">G<\/em> represents a 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 \u0394<em data-effect=\"italics\">G<\/em> is zero, the forward and reverse driving forces are equal, and the process occurs in both directions at the same rate (the system is at equilibrium).<\/p>\r\n<p id=\"fs-idm189708256\">In the chapter on equilibrium the <em data-effect=\"italics\">reaction quotient<\/em>, <em data-effect=\"italics\">Q<\/em>, was introduced as a convenient measure of the status of an equilibrium system. Recall that you may use <em>Q<\/em> to identify the direction in which a reaction will proceed in order to achieve equilibrium. When <em data-effect=\"italics\">Q<\/em> is lesser than the equilibrium constant, <em data-effect=\"italics\">K<\/em>, the reaction will proceed in the forward direction until equilibrium is reached and <em data-effect=\"italics\">Q<\/em> = <em data-effect=\"italics\">K<\/em>. Conversely, if <em data-effect=\"italics\">Q<\/em> &gt; <em data-effect=\"italics\">K<\/em>, the process will proceed in the reverse direction until equilibrium is achieved.<\/p>\r\n<p id=\"fs-idm39755632\">The free energy change for a process taking place with reactants and products present under <em data-effect=\"italics\">nonstandard conditions<\/em> (pressures other than 1 bar; concentrations other than 1 M) is related to the standard free energy change according to this equation:<\/p>\r\n\r\n<div id=\"fs-idm55093920\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em> = \u0394<em>G<\/em>\u00b0 + <em>RT <\/em>ln <em>Q<\/em><\/div>\r\n<p id=\"fs-idm192589168\"><em data-effect=\"italics\">R<\/em> is the gas constant (8.314 J\/K mol), <em data-effect=\"italics\">T<\/em> is the kelvin or absolute temperature, and <em data-effect=\"italics\">Q<\/em> is the reaction quotient. This equation may be used to predict the spontaneity for a process under any given set of conditions as illustrated in <a class=\"autogenerated-content\" href=\"#fs-idm192477856\">(Figure)<\/a>.<\/p>\r\n\r\n<div id=\"fs-idm192477856\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm158842672\"><strong>Calculating \u0394<em data-effect=\"italics\">G<\/em> under Nonstandard Conditions <\/strong><\/p>\r\nWhat is the free energy change for the process shown here under the specified conditions?\r\n<p id=\"fs-idm178094656\"><em data-effect=\"italics\">T<\/em> = 25 \u00b0C, <em>P<\/em><sub>N2<\/sub> = 0.870 atm, <em>P<\/em><sub>H2<\/sub> = 0.250 atm, and<em> P<\/em><sub>NH3<\/sub> = 12.9 atm<\/p>\r\n\r\n<div id=\"fs-idm8287616\" style=\"padding-left: 40px\" data-type=\"equation\">2NH<sub>3<\/sub>(<em>g<\/em>) \u27f6 3H<sub>2<\/sub>(<em>g<\/em>) + N<sub>2<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u0394<em>G<\/em>\u00b0 = 33.0 kJ\/mol<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm218730976\"><strong>Solution:<\/strong><\/p>\r\nThe equation relating free energy change to standard free energy change and reaction quotient may be used directly:\r\n<div id=\"fs-idm150025600\" style=\"padding-left: 40px\" data-type=\"equation\"><img class=\"alignnone wp-image-2015\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4k-300x55.png\" alt=\"\" width=\"531\" height=\"97\" \/><\/div>\r\n<p id=\"fs-idm159629248\">Since the computed value for \u0394<em data-effect=\"italics\">G<\/em> is positive, the reaction is nonspontaneous under these conditions.<\/p>\r\n<p id=\"fs-idm150306224\"><strong>Check Your Learning:<\/strong><\/p>\r\nCalculate the free energy change for this same reaction at 875 \u00b0C in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?\r\n<div id=\"fs-idm329579920\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm247369440\">\u0394<em data-effect=\"italics\">G<\/em> = \u221247 kJ; yes<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idm247911728\">For a system at equilibrium, <em data-effect=\"italics\">Q<\/em> = <em data-effect=\"italics\">K<\/em> and \u0394<em data-effect=\"italics\">G<\/em> = 0, and the previous equation may be written as<\/p>\r\n\r\n<div id=\"fs-idm228614064\" style=\"padding-left: 40px\" data-type=\"equation\">0 = \u0394<em>G<\/em>\u00b0 + <em>RT <\/em>ln <em>K<\/em><\/div>\r\n<div id=\"fs-idm55951760\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em>\u00b0 = \u2212<em>RT<\/em> ln K<\/div>\r\n<p id=\"fs-idm206226416\">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 <a class=\"autogenerated-content\" href=\"#fs-idm159425040\">(Figure)<\/a>.<\/p>\r\n\r\n<table id=\"fs-idm159425040\" class=\"top-titled\" summary=\"This table has three columns and four rows. The first row is a header row, and it labels each column, \u201cK,\u201d \u201ccapital delta G superscript degree symbol,\u201d and \u201cComments.\u201d Under the \u201cK\u201d column are the following: \u201cgreater than 1,\u201d \u201cless than 1,\u201d and \u201cequal to 1.\u201d Under the \u201ccapital delta G superscript degree symbol\u201d column are the following: \u201cless than 0,\u201d \u201cgreater than 0,\u201d and \u201cequal to 0.\u201d Under the \u201cComments\u201d column are the following: \u201cProducts are more abundant at equilibrium,\u201d \u201cReactants are more abundant at equilibrium,\u201d and \u201cReactants and products are equally abundant at equilibrium.\u201d\">\r\n<thead>\r\n<tr>\r\n<th colspan=\"3\" data-align=\"center\">Relations between Standard Free Energy Changes and Equilibrium Constants<\/th>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<th data-align=\"center\"><em data-effect=\"italics\">K<\/em><\/th>\r\n<th data-align=\"center\">\u0394<em data-effect=\"italics\">G<\/em>\u00b0<\/th>\r\n<th data-align=\"left\">Composition of an Equilibrium Mixture<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td data-align=\"center\">&gt; 1<\/td>\r\n<td data-align=\"center\">&lt; 0<\/td>\r\n<td data-align=\"left\">Products are more abundant<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"center\">&lt; 1<\/td>\r\n<td data-align=\"center\">&gt; 0<\/td>\r\n<td data-align=\"left\">Reactants are more abundant<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"center\">= 1<\/td>\r\n<td data-align=\"center\">= 0<\/td>\r\n<td data-align=\"left\">Reactants and products are comparably abundant<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-idm300346368\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm249874240\"><strong>Calculating an Equilibrium Constant using Standard Free Energy Change <\/strong><\/p>\r\nGiven that the standard free energies of formation of Ag<sup>+<\/sup>(<em data-effect=\"italics\">aq<\/em>), Cl<sup>\u2212<\/sup>(<em data-effect=\"italics\">aq<\/em>), and AgCl(<em data-effect=\"italics\">s<\/em>) are 77.1 kJ\/mol, \u2212131.2 kJ\/mol, and \u2212109.8 kJ\/mol, respectively, calculate the solubility product, <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>, for AgCl.\r\n<p id=\"fs-idm192205040\"><strong>Solution:<\/strong><\/p>\r\nThe reaction of interest is the following:\r\n<div id=\"fs-idm266185728\" style=\"padding-left: 40px\" data-type=\"equation\">AgCl(<em>s<\/em>) \u21cc Ag<sup>+<\/sup>(<em>aq<\/em>) + Cl<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Ag+][Cl<sup>\u2212<\/sup>]<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm150004464\">The standard free energy change for this reaction is first computed using standard free energies of formation for its reactants and products:<\/p>\r\n\r\n<div id=\"fs-idm111103296\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em>\u00b0 = [\u0394<em>G<\/em><sub>f<\/sub>\u00b0(Ag<sup>+<\/sup>(<em>aq<\/em>)) + \u0394<em>G<\/em><sub>f<\/sub>\u00b0(Cl<sup>\u2212<\/sup>(<em>aq<\/em>))] - \u0394<em>G<\/em><sub>f<\/sub>\u00b0(AgCl(<em>s<\/em>)) = [77.1 kJ\/mol - 131.2 kJ\/mol] - (-109.8 kJ\/mol) = 55.7 kJ\/mol<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm180460784\">The equilibrium constant for the reaction may then be derived from its standard free energy change:<\/p>\r\n\r\n<div id=\"fs-idm267274240\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em>\u00b0 = \u2212<em>RT<\/em> ln K<\/div>\r\n<div style=\"padding-left: 40px\" data-type=\"equation\">55.7 \u00d7 10<sup>3<\/sup> J\/mol = -(8.314 J\/K\u00b7mol)(298.15 K) ln K<\/div>\r\n<div style=\"padding-left: 40px\" data-type=\"equation\">ln <em>K<\/em> = -22.5<\/div>\r\n<div style=\"padding-left: 40px\" data-type=\"equation\"><em>K<\/em> = 1.7 \u00d7 10<sup>-10<\/sup><\/div>\r\n<p id=\"fs-idm126979616\">This result is in\u00a0reasonable agreement with the value provided in Appendix J.<\/p>\r\n<p id=\"fs-idm284201456\"><strong>Check Your Learning:<\/strong><\/p>\r\nUse the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 \u00b0C.\r\n<div id=\"fs-idm193826960\" style=\"padding-left: 40px\" data-type=\"equation\">2NO<sub>2<\/sub>(<em>g<\/em>) \u21cc N<sub>2<\/sub>O<sub>4<\/sub>(<em>g<\/em>)<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div id=\"fs-idm197604384\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm135514416\"><em data-effect=\"italics\">K<\/em> = 6.9<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idm66775936\">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 <em data-effect=\"italics\">Q<\/em>), equilibrium is established when the system\u2019s free energy is minimized (<a class=\"autogenerated-content\" href=\"#CNX_Chem_16_04_Gibbs\">(Figure)<\/a>). If a system consists of reactants and products in nonequilibrium amounts (<em data-effect=\"italics\">Q<\/em> \u2260 <em data-effect=\"italics\">K<\/em>), the reaction will proceed spontaneously in the direction necessary to establish equilibrium.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_16_04_Gibbs\" class=\"bc-figure figure\">\r\n<div class=\"bc-figcaption figcaption\">These plots show the free energy versus reaction progress for systems whose standard free energy changes are (a) negative, (b) positive, and (c) zero. Nonequilibrium systems will proceed spontaneously in whatever direction is necessary to minimize free energy and establish equilibrium.<\/div>\r\n<span id=\"fs-idm247894944\" data-type=\"media\" data-alt=\"Three graphs, labeled, \u201ca,\u201d \u201cb,\u201d and \u201cc\u201d are shown where the y-axis is labeled, \u201cGibbs free energy ( G ),\u201d and, \u201cG superscript degree sign ( reactants ),\u201d while the x-axis is labeled, \u201cReaction progress,\u201d and \u201cReactants,\u201d on the left and, \u201cProducts,\u201d on the right. In graph a, a line begins at the upper left side and goes steadily down to a point about halfway up the y-axis and two thirds of the way on the x-axis, then rises again to a point labeled, \u201cG superscript degree sign ( products ),\u201d that is slightly higher than halfway up the y-axis. The distance between the beginning and ending points of the graph is labeled as, \u201cdelta G less than 0,\u201d while the lowest point on the graph is labeled, \u201cQ equals K greater than 1.\u201d In graph b, a line begins at the middle left side and goes steadily down to a point about two fifths up the y-axis and one third of the way on the x-axis, then rises again to a point labeled, \u201cG superscript degree sign ( products ),\u201d that is near the top of the y-axis. The distance between the beginning and ending points of the graph is labeled as, \u201cdelta G greater than 0,\u201d while the lowest point on the graph is labeled, \u201cQ equals K less than 1.\u201d In graph c, a line begins at the upper left side and goes steadily down to a point near the bottom of the y-axis and half way on the x-axis, then rises again to a point labeled, \u201cG superscript degree sign ( products ),\u201d that is equal to the starting point on the y-axis which is labeled, \u201cG superscript degree sign ( reactants ).\u201d The lowest point on the graph is labeled, \u201cQ equals K equals 1.\u201d At the top of the graph is the label, \u201cDelta G superscript degree sign equals 0.\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_16_04_Gibbs-1.jpg\" alt=\"Three graphs, labeled, \u201ca,\u201d \u201cb,\u201d and \u201cc\u201d are shown where the y-axis is labeled, \u201cGibbs free energy ( G ),\u201d and, \u201cG superscript degree sign ( reactants ),\u201d while the x-axis is labeled, \u201cReaction progress,\u201d and \u201cReactants,\u201d on the left and, \u201cProducts,\u201d on the right. In graph a, a line begins at the upper left side and goes steadily down to a point about halfway up the y-axis and two thirds of the way on the x-axis, then rises again to a point labeled, \u201cG superscript degree sign ( products ),\u201d that is slightly higher than halfway up the y-axis. The distance between the beginning and ending points of the graph is labeled as, \u201cdelta G less than 0,\u201d while the lowest point on the graph is labeled, \u201cQ equals K greater than 1.\u201d In graph b, a line begins at the middle left side and goes steadily down to a point about two fifths up the y-axis and one third of the way on the x-axis, then rises again to a point labeled, \u201cG superscript degree sign ( products ),\u201d that is near the top of the y-axis. The distance between the beginning and ending points of the graph is labeled as, \u201cdelta G greater than 0,\u201d while the lowest point on the graph is labeled, \u201cQ equals K less than 1.\u201d In graph c, a line begins at the upper left side and goes steadily down to a point near the bottom of the y-axis and half way on the x-axis, then rises again to a point labeled, \u201cG superscript degree sign ( products ),\u201d that is equal to the starting point on the y-axis which is labeled, \u201cG superscript degree sign ( reactants ).\u201d The lowest point on the graph is labeled, \u201cQ equals K equals 1.\u201d At the top of the graph is the label, \u201cDelta G superscript degree sign equals 0.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm220227936\" class=\"summary\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\r\n<p id=\"fs-idm197297904\">Gibbs free energy (<em data-effect=\"italics\">G<\/em>) 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 \u0394<em data-effect=\"italics\">G<\/em> indicates a spontaneous process; a positive \u0394<em data-effect=\"italics\">G<\/em> indicates a nonspontaneous process; and a \u0394<em data-effect=\"italics\">G<\/em> of zero indicates that the system is at equilibrium. A number of approaches to the computation of free energy changes are possible.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idm248576000\" class=\"key-equations\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Key Equations<\/strong><\/h3>\r\n<ul id=\"fs-idm60433792\" data-bullet-style=\"bullet\">\r\n \t<li>\u0394<em data-effect=\"italics\">G<\/em> = \u0394<em data-effect=\"italics\">H<\/em> \u2212 <em data-effect=\"italics\">T<\/em>\u0394<em data-effect=\"italics\">S<\/em><\/li>\r\n \t<li>\u0394<em data-effect=\"italics\">G<\/em> = \u0394<em data-effect=\"italics\">G<\/em>\u00b0 + <em data-effect=\"italics\">RT<\/em> ln <em data-effect=\"italics\">Q<\/em><\/li>\r\n \t<li>\u0394<em data-effect=\"italics\">G<\/em>\u00b0 = \u2212<em data-effect=\"italics\">RT<\/em> ln <em data-effect=\"italics\">K<\/em><\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-idm233991504\" class=\"exercises\" data-depth=\"1\">\r\n<div id=\"fs-idm153799616\" data-type=\"exercise\">\r\n<div id=\"fs-idm71264384\" data-type=\"solution\">\r\n<p id=\"eip-idm1161148928\"><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\" data-type=\"glossary\">\r\n<h3 data-type=\"glossary-title\"><strong>Glossary<\/strong><\/h3>\r\n<dl id=\"fs-idm119810368\">\r\n \t<dt>Gibbs free energy change (<em data-effect=\"italics\">G<\/em>)<\/dt>\r\n \t<dd id=\"fs-idm119809104\">thermodynamic property defined in terms of system enthalpy and entropy; all spontaneous processes involve a decrease in <em data-effect=\"italics\">G<\/em><\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm119807952\">\r\n \t<dt>standard free energy change (\u0394<em data-effect=\"italics\">G<\/em>\u00b0)<\/dt>\r\n \t<dd id=\"fs-idm119806688\">change in free energy for a process occurring under standard conditions (1 bar pressure for gases, 1 M concentration for solutions)<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm119806032\">\r\n \t<dt>standard free energy of formation (\u0394G<sub>f<\/sub>\u00b0)<\/dt>\r\n \t<dd id=\"fs-idm185413520\">change in free energy accompanying the formation of one mole of substance from its elements in their standard states<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<p>&nbsp;<\/p>\n<div class=\"textbox textbox--learning-objectives\">\n<h3><strong>Learning Objectives<\/strong><\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Define Gibbs free energy, and describe its relation to spontaneity<\/li>\n<li>Calculate free energy change for a process using free energies of formation for its reactants and products<\/li>\n<li>Calculate free energy change for a process using enthalpies of formation and the entropies for its reactants and products<\/li>\n<li>Explain how temperature affects the spontaneity of some processes<\/li>\n<li>Relate standard free energy changes to equilibrium constants<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idm21370272\">One of the challenges of using the second law of thermodynamics to determine if a process is spontaneous is that it requires measurements of the entropy change for the system <em data-effect=\"italics\">and<\/em> 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 <span class=\"no-emphasis\" data-type=\"term\">Gibbs<\/span>. This new property is called the <span data-type=\"term\"><strong>Gibbs free energy<\/strong> <strong>(<em data-effect=\"italics\">G<\/em>)<\/strong><\/span> (or simply the <em data-effect=\"italics\">free energy<\/em>), and it is defined in terms of a system\u2019s enthalpy and entropy as the following:<\/p>\n<div id=\"fs-idm129849344\" style=\"padding-left: 40px\" data-type=\"equation\"><em>G<\/em> = <em>H<\/em> &#8211;<em> TS<\/em><\/div>\n<p id=\"fs-idm247387040\">Free energy is a state function, and at constant temperature and pressure, the <span data-type=\"term\">free energy change (\u0394<em data-effect=\"italics\">G<\/em>)<\/span> may be expressed as the following:<\/p>\n<div id=\"fs-idm279983248\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em> = \u0394<em>H<\/em> &#8211; <em>T<\/em>\u0394<em>S<\/em><\/div>\n<p id=\"fs-idm111872960\">(For simplicity\u2019s sake, the subscript \u201csys\u201d will be omitted henceforth.)<\/p>\n<p id=\"fs-idm275842432\">The relationship between this system property and the spontaneity of a process may be understood by recalling the previously derived second law expression:<\/p>\n<div id=\"fs-idm33912192\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1996\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4a-300x77.png\" alt=\"\" width=\"160\" height=\"41\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4a-300x77.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4a-65x17.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4a-225x58.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4a-350x90.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4a.png 546w\" sizes=\"auto, (max-width: 160px) 100vw, 160px\" \/><\/div>\n<p id=\"fs-idm170940880\">The first law requires that <em data-effect=\"italics\">q<\/em><sub>surr<\/sub> = \u2212<em data-effect=\"italics\">q<\/em><sub>sys<\/sub>, and at constant pressure <em data-effect=\"italics\">q<\/em><sub>sys<\/sub> = \u0394<em data-effect=\"italics\">H<\/em>, so this expression may be rewritten as:<\/p>\n<div id=\"fs-idm195954976\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1997\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4b-300x75.png\" alt=\"\" width=\"148\" height=\"37\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4b-300x75.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4b-65x16.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4b-225x57.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4b-350x88.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4b.png 537w\" sizes=\"auto, (max-width: 148px) 100vw, 148px\" \/><\/div>\n<p id=\"fs-idm169794560\">Multiplying both sides of this equation by \u2212<em data-effect=\"italics\">T<\/em>, and rearranging yields the following:<\/p>\n<div id=\"fs-idm248567088\" style=\"padding-left: 40px\" data-type=\"equation\">\u2212<em>T<\/em>\u0394<em>S<\/em><sub>univ<\/sub> = \u0394<em>H<\/em> &#8211; <em>T<\/em> \u0394<em>S<\/em><\/div>\n<p id=\"fs-idm229909296\">Comparing this equation to the previous one for free energy change shows the following relation:<\/p>\n<div id=\"fs-idm117423648\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em> = \u2212<em>T<\/em>\u0394<em>S<\/em><sub>univ<\/sub><\/div>\n<p id=\"fs-idm228637552\">The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, \u0394<em data-effect=\"italics\">S<\/em><sub>univ<\/sub>. <a class=\"autogenerated-content\" href=\"#fs-idm211518768\">(Figure)<\/a> summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators.<\/p>\n<table id=\"fs-idm211518768\" class=\"top-titled\" summary=\"This table has three columns and three rows. The first column has the following: \u201ccapital delta S subscript univ is greater than 0,\u201d \u201ccapital delta S subscript univ is less than 0,\u201d and, \u201ccapital delta S subscript univ equals 0.\u201d The second column contains the following: \u201ccapital delta G is less than 0,\u201d \u201ccapital delta G is greater than 0,\u201d and, \u201ccapital delta G equals 0.\u201d The third column contains the following: \u201cSpontaneous,\u201d \u201cnonspontaneous ( spontaneous in opposite direction ),\u201d and, \u201creversible ( system is at equilibrium ).\u201d\">\n<thead>\n<tr>\n<th colspan=\"3\" data-align=\"center\">Relation between Process Spontaneity and Signs of Thermodynamic Properties<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td data-align=\"center\">\u0394<em data-effect=\"italics\">S<\/em><sub>univ<\/sub> &gt; 0<\/td>\n<td data-align=\"center\">\u0394<em data-effect=\"italics\">G<\/em> &lt; 0<\/td>\n<td data-align=\"center\">spontaneous<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"center\">\u0394<em data-effect=\"italics\">S<\/em><sub>univ<\/sub> &lt; 0<\/td>\n<td data-align=\"center\">\u0394<em data-effect=\"italics\">G<\/em> &gt; 0<\/td>\n<td data-align=\"center\">nonspontaneous<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"center\">\u0394<em data-effect=\"italics\">S<\/em><sub>univ<\/sub> = 0<\/td>\n<td data-align=\"center\">\u0394<em data-effect=\"italics\">G<\/em> = 0<\/td>\n<td data-align=\"center\">at equilibrium<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-idm348811808\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>What\u2019s \u201cFree\u201d about \u0394<em data-effect=\"italics\">G<\/em>?<\/strong><\/h3>\n<p id=\"fs-idm361428560\">In addition to indicating spontaneity, the free energy change also provides information regarding the amount of useful work (<em data-effect=\"italics\">w<\/em>) that may be accomplished by a spontaneous process. Although a rigorous treatment of this subject is beyond the scope of an introductory chemistry text, a brief discussion is helpful for gaining a better perspective on this important thermodynamic property.<\/p>\n<p id=\"fs-idm351078736\">For this purpose, consider a spontaneous, exothermic process that involves a decrease in entropy. The free energy, as defined by<\/p>\n<div id=\"fs-idm343621088\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em> = \u0394<em>H<\/em> &#8211; <em>T<\/em>\u0394<em>S<\/em><\/div>\n<p id=\"fs-idm347838176\">may be interpreted as representing the difference between the energy produced by the process, \u0394<em data-effect=\"italics\">H<\/em>, and the energy lost to the surroundings, <em data-effect=\"italics\">T<\/em>\u0394<em data-effect=\"italics\">S<\/em>. The difference between the energy produced and the energy lost is the energy available (or \u201cfree\u201d) to do useful work by the process, \u0394<em data-effect=\"italics\">G<\/em>. If the process somehow could be made to take place under conditions of thermodynamic reversibility, the amount of work that could be done would be maximal:<\/p>\n<div id=\"fs-idm353619776\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em> = <em>w<\/em><sub>max<\/sub><\/div>\n<p id=\"fs-idm353511696\">However, as noted previously in this chapter, such conditions are not realistic. In addition, the technologies used to extract work from a spontaneous process (e.g., automobile engine, steam turbine) are never 100% efficient, and so the work done by these processes is always less than the theoretical maximum. Similar reasoning may be applied to a nonspontaneous process, for which the free energy change represents the <em data-effect=\"italics\">minimum<\/em> amount of work that must be done <em data-effect=\"italics\">on<\/em> the system to carry out the process.<\/p>\n<\/div>\n<div id=\"fs-idm305910384\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Calculating Free Energy Change<\/strong><\/h3>\n<p id=\"fs-idm233417152\">Free energy is a state function, so its value depends only on the conditions of the initial and final states of the system. 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<strong> standard free energy changes, \u0394<em data-effect=\"italics\">G<\/em>\u00b0<\/strong>, according to the following relation.<\/p>\n<div id=\"fs-idm182227520\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<span data-type=\"term\">\u00b0<\/span><\/em> = \u0394<em>H<span data-type=\"term\">\u00b0<\/span><\/em> &#8211; <em>T<\/em>\u0394<em>S<span data-type=\"term\">\u00b0<\/span><\/em><\/div>\n<div id=\"fs-idm117444432\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm267921040\"><strong>Using Standard Enthalpy and Entropy Changes to Calculate \u0394<em data-effect=\"italics\">G<\/em>\u00b0<\/strong><\/p>\n<p>Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the vaporization of water at room temperature (298 K). What does the computed value for \u0394<em data-effect=\"italics\">G<\/em>\u00b0 say about the spontaneity of this process?<\/p>\n<p id=\"fs-idm222992784\"><strong>Solution:<\/strong><\/p>\n<p>The process of interest is the following:<\/p>\n<div id=\"fs-idm59589120\" style=\"padding-left: 40px\" data-type=\"equation\">H<sub>2<\/sub>O(<em>l<\/em>) \u27f6 H<sub>2<\/sub>O(<em>g<\/em>)<\/div>\n<p id=\"fs-idm23409248\">The standard change in free energy may be calculated using the following equation:<\/p>\n<div id=\"fs-idm165858560\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<span data-type=\"term\">\u00b0<\/span><\/em> = \u0394<em>H<span data-type=\"term\">\u00b0<\/span><\/em> &#8211; <em>T<\/em>\u0394<em>S<span data-type=\"term\">\u00b0<\/span><\/em><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm272232416\">From Appendix G:<\/p>\n<table id=\"fs-idm230329280\" class=\"medium unnumbered\" summary=\"This table has three columns and three rows. The first row is a header row and it labels each column: \u201cSubstance,\u201d \u201ccapital delta H subscript f superscript degree symbol ( k J \/ mol),\u201d and \u201cS subscript 298 superscript degree symbol ( J \/ K dot mol ).\u201d Under the \u201cSubstance\u201d column are H subscript 2 O ( l ) and H subscript 2 O ( g ). Under the \u201ccapital delta H subscript f superscript degree symbol ( k J \/ mol)\u201d column are the values negative 286.83 and negative 241.82. Under the \u201cS subscript 298 superscript degree symbol ( J \/ K dot mol )\u201d column are the values 70.0 and 188.8.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th data-align=\"center\">Substance<\/th>\n<th data-align=\"center\">\u0394H<sub>f<\/sub>\u00b0(kJ\/mol)<\/th>\n<th data-align=\"center\">S\u00b0(J\/K\u00b7mol)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td data-align=\"center\">H<sub>2<\/sub>O(<em data-effect=\"italics\">l<\/em>)<\/td>\n<td data-align=\"center\">\u2212285.83<\/td>\n<td data-align=\"center\">70.0<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"center\">H<sub>2<\/sub>O(<em data-effect=\"italics\">g<\/em>)<\/td>\n<td data-align=\"center\">\u2212241.82<\/td>\n<td data-align=\"center\">188.8<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idm123711584\">Using the appendix data to calculate the standard enthalpy and entropy changes yields:<\/p>\n<div id=\"fs-idm164142400\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394H\u00b0 = \u0394H<sub>f<\/sub>\u00b0(H<sub>2<\/sub>O(<em>g<\/em>)) &#8211; \u0394H<sub>f<\/sub>\u00b0(H<sub>2<\/sub>O(<em>l<\/em>)) = [\u2212241.82 kJ\/mol &#8211; (\u2212285.83 kJ\/mol)] = 44.01 kJ\/mol<\/div>\n<div id=\"fs-idm171591584\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394S\u00b0 = S\u00b0(H<sub>2<\/sub>O(<em>g<\/em>)) &#8211; S\u00b0(H<sub>2<\/sub>O(<em>l<\/em>)) = 188.8 J\/mol\u00b7K &#8211; 70.0 J\/mol\u00b7K =118.8 J\/mol\u00b7K<\/div>\n<div id=\"fs-idm129668064\" data-type=\"equation\"><\/div>\n<p id=\"fs-idp11689536\">Substitution into the standard free energy equation yields:<\/p>\n<div id=\"fs-idm246654432\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<span data-type=\"term\">\u00b0<\/span><\/em> = \u0394<em>H<span data-type=\"term\">\u00b0<\/span><\/em> &#8211; <em>T<\/em>\u0394<em>S<span data-type=\"term\">\u00b0<\/span><\/em> = 44.01 kJ\/mol &#8211; (298 K)(0.1188 J\/mol\u00b7K)<\/div>\n<div id=\"fs-idm135070448\" style=\"padding-left: 40px\" data-type=\"equation\">\u00a0 \u00a0 \u00a0 \u00a0 = 44.01 kJ\/mol &#8211; 35.4 kJ\/mol = 8.6 kJ\/mol<\/div>\n<p id=\"fs-idm276144320\">At 298 K (25 \u00b0C), \u0394G\u00b0 &gt; 0, so boiling is nonspontaneous (<em data-effect=\"italics\">not<\/em> spontaneous).<\/p>\n<p id=\"fs-idm248587360\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the reaction shown here (298 K). What does the computed value for \u0394<em data-effect=\"italics\">G<\/em>\u00b0 say about the spontaneity of this process?<\/p>\n<div id=\"fs-idm276750416\" style=\"padding-left: 40px\" data-type=\"equation\">C<sub>2<\/sub>H<sub>6<\/sub>(<em>g<\/em>) \u27f6 H<sub>2<\/sub>(<em>g<\/em>) + C<sub>2<\/sub>H<sub>4<\/sub>(<em>g<\/em>)<\/div>\n<div data-type=\"equation\"><\/div>\n<div id=\"fs-idm97664544\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm157144416\">\u0394G\u00b0 = 102.0 kJ\/mol; the reaction is nonspontaneous (<em data-effect=\"italics\">not<\/em> spontaneous) at 25 \u00b0C.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idm43672192\">The standard free energy change for a reaction may also be calculated from <span data-type=\"term\"><strong>standard free energy of formation, \u0394<em data-effect=\"italics\">G<\/em><sub>f<\/sub>\u00b0<\/strong>,<\/span> values 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 enthalpy of formation, \u0394G<sub>f<\/sub>\u00b0 is by definition zero for elemental substances under standard state conditions. The approach used to calculate \u0394G\u00b0 for a reaction from \u0394G<sub>f<\/sub>\u00b0 values is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction<\/p>\n<div id=\"fs-idm250621408\" style=\"padding-left: 40px\" data-type=\"equation\"><em>m<\/em>A + <em>n<\/em>B \u27f6 <em>x<\/em>C + <em>y<\/em>D,<\/div>\n<p id=\"fs-idm124979680\">the standard free energy change at room temperature may be calculated as<\/p>\n<p style=\"padding-left: 80px\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2022 alignleft\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l-300x32.png\" alt=\"\" width=\"328\" height=\"35\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l-300x32.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l-1024x110.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l-768x82.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l-65x7.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l-225x24.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l-350x38.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l.png 1035w\" sizes=\"auto, (max-width: 328px) 100vw, 328px\" \/><\/p>\n<p style=\"padding-left: 80px\">\u00a0 \u00a0 \u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-2024\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4m-300x26.png\" alt=\"\" width=\"335\" height=\"29\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4m-300x26.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4m-1024x90.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4m-768x67.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4m-65x6.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4m-225x20.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4m-350x31.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4m.png 1184w\" sizes=\"auto, (max-width: 335px) 100vw, 335px\" \/><\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idm202485584\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm232036496\"><strong>Using Standard Free Energies of Formation to Calculate \u0394<em data-effect=\"italics\">G<\/em>\u00b0<\/strong><\/p>\n<p>Consider the decomposition of yellow mercury(II) oxide.<\/p>\n<div id=\"fs-idm217690608\" style=\"padding-left: 40px\" data-type=\"equation\">HgO(<em>s<\/em>) \u27f6 Hg(<em>l<\/em>) + \u00bdO<sub>2<\/sub>(<em>g<\/em>)<\/div>\n<p id=\"fs-idm274288288\">Calculate the standard free energy change at room temperature, \u0394G\u00b0, 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?<\/p>\n<p id=\"fs-idm70679856\"><strong>Solution:<\/strong><\/p>\n<p>The required data are available in Appendix G and are shown here.<\/p>\n<table id=\"fs-idm232730384\" class=\"medium unnumbered\" summary=\"This table has four columns and four rows. The first row is a header row, and it labels each column: \u201cCompound,\u201d \u201ccapital delta G subscript f superscript degree symbol ( k J \/ mol ),\u201d \u201ccapital delta H subscript f superscript degree symbol ( k J \/ mol ),\u201d and \u201cS subscript 298 superscript degree symbol ( J \/ K dot mol ).\u201d Under the \u201cCompound\u201d column are the following: \u201cH g O ( s, yellow ),\u201d \u201cH g ( l ),\u201d and \u201cO subscript 2 ( g ).\u201d Under the \u201ccapital delta G subscript f superscript degree symbol ( k J \/ mol )\u201d column are the following values: negative 58.43, 0, and 0. Under the \u201ccapital delta H subscript f superscript degree symbol ( k J \/ mol )\u201d column are the values: negative 90.46, 0, and 0. Under the \u201cS subscript 298 superscript degree symbol ( J \/ K dot mol )\u201d column are the values: 71.13, 75.9, and 205.2.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th style=\"width: 74.6771px\" data-align=\"center\">Compound<\/th>\n<th style=\"width: 82.4792px\" data-align=\"center\">\u0394G<sub>f<\/sub>\u00b0(kJ\/mol)<\/th>\n<th style=\"width: 291.333px\" data-align=\"center\">\u0394H<sub>f<\/sub>\u00b0(kJ\/mol)<\/th>\n<th style=\"width: 230.781px\" data-align=\"center\">S\u00b0(J\/K\u00b7mol)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td style=\"width: 75.1771px\" data-align=\"center\">HgO (<em data-effect=\"italics\">s<\/em>, yellow)<\/td>\n<td style=\"width: 83.4792px\" data-align=\"center\">\u221258.43<\/td>\n<td style=\"width: 292.333px\" data-align=\"center\">\u221290.46<\/td>\n<td style=\"width: 231.281px\" data-align=\"center\">71.13<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"width: 75.1771px\" data-align=\"center\">Hg(<em data-effect=\"italics\">l<\/em>)<\/td>\n<td style=\"width: 83.4792px\" data-align=\"center\">0<\/td>\n<td style=\"width: 292.333px\" data-align=\"center\">0<\/td>\n<td style=\"width: 231.281px\" data-align=\"center\">75.9<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"width: 75.1771px\" data-align=\"center\">O<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>)<\/td>\n<td style=\"width: 83.4792px\" data-align=\"center\">0<\/td>\n<td style=\"width: 292.333px\" data-align=\"center\">0<\/td>\n<td style=\"width: 231.281px\" data-align=\"center\">205.2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idm192751264\">(a) Using free energies of formation:<\/p>\n<div id=\"fs-idm211589632\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-2022\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l-300x32.png\" alt=\"\" width=\"328\" height=\"35\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l-300x32.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l-1024x110.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l-768x82.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l-65x7.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l-225x24.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l-350x38.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4l.png 1035w\" sizes=\"auto, (max-width: 328px) 100vw, 328px\" \/><\/div>\n<div id=\"fs-idm178475312\" style=\"padding-left: 40px\" data-type=\"equation\">= [\u0394G<sub>f<\/sub>\u00b0(Hg(<em>l<\/em>)) + \u00bd\u0394G<sub>f<\/sub>\u00b0(O<sub>2<\/sub>(<em>g<\/em>))] &#8211; \u0394G<sub>f<\/sub>\u00b0(HgO(<em>s<\/em>, yellow))<\/div>\n<div id=\"fs-idm164697728\" style=\"padding-left: 40px\" data-type=\"equation\">= [0 kJ\/mol + \u00bd(0 kJ\/mol)] &#8211; (\u221258.43 kJ\/mol) =58.43 kJ\/mol<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm112946608\">(b) Using enthalpies and entropies of formation:<\/p>\n<div id=\"fs-idm147522592\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-2007\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4j-300x24.png\" alt=\"\" width=\"388\" height=\"31\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4j-300x24.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4j-1024x81.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4j-768x61.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4j-65x5.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4j-225x18.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4j-350x28.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4j.png 1325w\" sizes=\"auto, (max-width: 388px) 100vw, 388px\" \/><\/div>\n<div id=\"fs-idm39030688\" style=\"padding-left: 40px\" data-type=\"equation\">= [\u0394H<sub>f<\/sub>\u00b0(Hg(<em>l<\/em>)) + \u00bd\u0394H<sub>f<\/sub>\u00b0(O<sub>2<\/sub>(<em>g<\/em>))] &#8211; \u0394H<sub>f<\/sub>\u00b0(HgO(<em>s<\/em>, yellow))<\/div>\n<div id=\"fs-idm261408512\" style=\"padding-left: 40px\" data-type=\"equation\">=[0 kJ\/mol + \u00bd(0 kJ\/mol)] &#8211; (\u221290.46 kJ\/mol) = 90.46 kJ\/mol<\/div>\n<div data-type=\"equation\"><\/div>\n<div id=\"fs-idp6944752\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1990\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.3h-300x31.png\" alt=\"\" width=\"319\" height=\"33\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.3h-300x31.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.3h-1024x105.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.3h-768x79.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.3h-65x7.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.3h-225x23.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.3h-350x36.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.3h.png 1190w\" sizes=\"auto, (max-width: 319px) 100vw, 319px\" \/><\/div>\n<div id=\"fs-idm61090368\" style=\"padding-left: 40px\" data-type=\"equation\">= [<em>S<\/em>\u00b0(Hg(<em>l<\/em>) + \u00bd<em>S<\/em>\u00b0(O<sub>2<\/sub>(<em>g<\/em>))] &#8211; <em>S<\/em>\u00b0(HgO(s, yellow)<\/div>\n<div id=\"fs-idm131288608\" style=\"padding-left: 40px\" data-type=\"equation\">= [75.9 J\/mol\u00b7K + \u00bd(205.2 J\/mol\u00b7K)] &#8211; 71.13 J\/mol\u00b7K = 107.4 J\/mol\u00b7K<\/div>\n<div data-type=\"equation\"><\/div>\n<div id=\"fs-idm153411216\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<span data-type=\"term\">\u00b0<\/span><\/em> = \u0394<em>H<span data-type=\"term\">\u00b0<\/span><\/em> &#8211; <em>T<\/em>\u0394<em>S<span data-type=\"term\">\u00b0 <\/span><\/em>=90.46 kJ\/mol &#8211; (298.15 K)(0.1074 J\/mol\u00b7K<\/div>\n<div id=\"fs-idm190648624\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394G\u00b0 = 90.46 kJ\/mol &#8211; 32.01 kJ\/mol = 58.45 kJ\/mol<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm269603872\">Both ways to calculate the standard free energy change at 25 \u00b0C give the same numerical value (to three significant figures), and both predict that the process is nonspontaneous (<em data-effect=\"italics\">not<\/em> spontaneous) at room temperature.<\/p>\n<p id=\"fs-idm59738256\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Calculate \u0394<em data-effect=\"italics\">G<\/em>\u00b0 using (a) free energies of formation and (b) enthalpies of formation and entropies (Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 \u00b0C?<\/p>\n<div id=\"fs-idm213619408\" style=\"padding-left: 40px\" data-type=\"equation\">C<sub>2<\/sub>H<sub>4<\/sub>(<em>g<\/em>) \u27f6 H<sub>2<\/sub>(<em>g<\/em>) + C<sub>2<\/sub>H<sub>2<\/sub>(<em>g<\/em>)<\/div>\n<div data-type=\"equation\"><\/div>\n<div id=\"fs-idm254989456a\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp2244016\">(a) 140.8 kJ\/mol, nonspontaneous<\/p>\n<p id=\"eip-687\">(b) 141.5 kJ\/mol, nonspontaneous<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm353615984\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Free Energy Changes for Coupled Reactions<\/strong><\/h3>\n<p id=\"fs-idm350537712\">The use of free energies of formation to compute free energy changes for reactions as described above is possible because \u0394<em data-effect=\"italics\">G<\/em> is a state function, and the approach is analogous to the use of Hess\u2019 Law in computing enthalpy changes (see the chapter on thermochemistry). Consider the vaporization of water as an example:<\/p>\n<div id=\"fs-idm359616576\" style=\"padding-left: 40px\" data-type=\"equation\">H<sub>2<\/sub>O(<em>l<\/em>) \u27f6 H<sub>2<\/sub>O(<em>g<\/em>)<\/div>\n<p id=\"fs-idm354717680\">An equation representing this process may be derived by adding the formation reactions for the two phases of water (necessarily reversing the reaction for the liquid phase). The free energy change for the sum reaction is the sum of free energy changes for the two added reactions:<\/p>\n<div id=\"fs-idm360093056\" style=\"padding-left: 40px\" data-type=\"equation\">H<sub>2<\/sub>(<em>g<\/em>) + \u00bdO<sub>2<\/sub>(<em>g<\/em>) \u2192 H<sub>2<\/sub>O(<em>g<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u0394G<sub>f<\/sub>\u00b0(H<sub>2<\/sub>O(<em>g<\/em>))<\/div>\n<div style=\"padding-left: 40px\" data-type=\"equation\">H<sub>2<\/sub>O(<em>l<\/em>) \u2192 H<sub>2<\/sub>(<em>g<\/em>) + \u00bdO<sub>2<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0-\u0394G<sub>f<\/sub>\u00b0(H<sub>2<\/sub>O(<em>l<\/em>))<\/div>\n<div style=\"padding-left: 40px\" data-type=\"equation\">&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;-<\/div>\n<div style=\"padding-left: 40px\" data-type=\"equation\">H<sub>2<\/sub>O(<em>l<\/em>) \u2192 H<sub>2<\/sub>O(<em>g<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u0394G\u00b0=\u0394G<sub>f<\/sub>\u00b0(H<sub>2<\/sub>O(<em>g<\/em>)) &#8211; \u0394G<sub>f<\/sub>\u00b0(H<sub>2<\/sub>O(<em>l<\/em>))<\/div>\n<p id=\"fs-idm331265792\">This approach may also be used in cases where a nonspontaneous reaction is enabled by coupling it to a spontaneous reaction. For example, the production of elemental zinc from zinc sulfide is thermodynamically unfavorable, as indicated by a positive value for \u0394<em data-effect=\"italics\">G<\/em>\u00b0:<\/p>\n<div id=\"fs-idm347055872\" style=\"padding-left: 40px\" data-type=\"equation\">ZnS(<em>s<\/em>) \u2192\u00a0 Zn(<em>s<\/em>) + S(<em>s<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u0394<em>G<\/em><sub>1<\/sub>\u00b0 = 201.3 kJ<\/div>\n<p id=\"fs-idm350667184\">The industrial process for production of zinc from sulfidic ores involves coupling this decomposition reaction to the thermodynamically favorable oxidation of sulfur:<\/p>\n<div id=\"fs-idm356637216\" style=\"padding-left: 40px\" data-type=\"equation\">S(<em>s<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u2192 SO<sub>2<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u0394<em>G<sub>2<\/sub><\/em>\u00b0 = \u2212300.1 kJ<\/div>\n<p id=\"fs-idm356330656\">The coupled reaction exhibits a negative free energy change and is spontaneous:<\/p>\n<div id=\"fs-idm343614160\" style=\"padding-left: 40px\" data-type=\"equation\">ZnS(<em>s<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u2192 Zn(<em>s<\/em>) + SO<sub>2<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u0394G\u00b0 = 201.3 kJ + (-300.1 kJ) = 98.8 kJ<\/div>\n<p id=\"fs-idm348648656\">This process is typically carried out at elevated temperatures, so this result obtained using standard free energy values is just an estimate. The gist of the calculation, however, holds true.<\/p>\n<div id=\"fs-idm348331936\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm348331680\"><strong>Calculating Free Energy Change for a Coupled Reaction <\/strong><\/p>\n<p>Is a reaction coupling the decomposition of ZnS to the formation of H<sub>2<\/sub>S expected to be spontaneous under standard conditions?<\/p>\n<p id=\"fs-idm116939264a\"><strong>Solution:<\/strong><\/p>\n<p>Following the approach outlined above and using free energy values from Appendix G:<\/p>\n<div id=\"fs-idm343484288\" data-type=\"equation\">Decomposition of zinc sulfide:\u00a0 \u00a0 \u00a0ZnS(<em>s<\/em>) \u2192 Zn(s) + S(<em>s<\/em>)\u00a0 \u00a0 \u00a0\u0394<em>G<\/em><sub>1<\/sub>\u00b0 = 201.3 kJ<\/div>\n<div data-type=\"equation\">Formation of hydrogen sulfide:\u00a0 \u00a0 \u00a0S(<em>s<\/em>) + H<sub>2<\/sub>(<em>g<\/em>) \u2192 H<sub>2<\/sub>S(<em>g<\/em>)\u00a0 \u00a0 \u00a0\u0394G<sub>2<\/sub>\u00b0 =- 33.4 kJ<\/div>\n<div data-type=\"equation\">Coupled reaction:\u00a0 \u00a0 \u00a0ZnS(<em>s<\/em>) + H<sub>2<\/sub>(<em>g<\/em>) \u2192 Zn(<em>s<\/em>) + H<sub>2<\/sub>S(<em>g<\/em>)\u00a0 \u00a0 \u00a0\u0394G\u00b0 = 201.3 kJ + (-33.4 kJ) = 167.9 kJ<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm323451264\">The coupled reaction exhibits a positive free energy change and is thus nonspontaneous.<\/p>\n<p id=\"fs-idm328777088\"><strong>Check Your Learning:<\/strong><\/p>\n<p>What is the standard free energy change for the reaction below? Is the reaction expected to be spontaneous under standard conditions?<\/p>\n<div id=\"fs-idm360883312\" style=\"padding-left: 40px\" data-type=\"equation\">FeS(<em>s<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u2192 Fe(<em>s<\/em>) + SO<sub>2<\/sub>(<em>g<\/em>)<\/div>\n<div data-type=\"equation\"><\/div>\n<div id=\"fs-idm254989456\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm360208448\">\u2212199.7 kJ; spontaneous<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm157251776\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Temperature Dependence of Spontaneity<\/strong><\/h3>\n<p id=\"fs-idm158748720\">As was previously demonstrated in this chapter\u2019s 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:<\/p>\n<div id=\"fs-idm182396560\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em>\u00a0= \u0394<em>H<\/em>\u00a0&#8211; <em>T<\/em>\u0394<em>S<\/em><\/div>\n<p id=\"fs-idm176739120\">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 <em data-effect=\"italics\">T<\/em> 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:<\/p>\n<ol id=\"fs-idm124567280\" type=\"1\">\n<li><strong data-effect=\"bold\">Both \u0394<em data-effect=\"italics\">H<\/em> and \u0394<em data-effect=\"italics\">S<\/em> are positive.<\/strong> This condition describes an endothermic process that involves an increase in system entropy. In this case, \u0394<em data-effect=\"italics\">G<\/em> will be negative if the magnitude of the <em data-effect=\"italics\">T<\/em>\u0394<em data-effect=\"italics\">S<\/em> term is greater than \u0394<em data-effect=\"italics\">H<\/em>. If the <em data-effect=\"italics\">T<\/em>\u0394<em data-effect=\"italics\">S<\/em> term is less than \u0394<em data-effect=\"italics\">H<\/em>, the free energy change will be positive. Such a process is <em data-effect=\"italics\">spontaneous at high temperatures and nonspontaneous at low temperatures.<\/em><\/li>\n<li><strong data-effect=\"bold\">Both \u0394<em data-effect=\"italics\">H<\/em> and \u0394<em data-effect=\"italics\">S<\/em> are negative.<\/strong> This condition describes an exothermic process that involves a decrease in system entropy. In this case, \u0394<em data-effect=\"italics\">G<\/em> will be negative if the magnitude of the <em data-effect=\"italics\">T<\/em>\u0394<em data-effect=\"italics\">S<\/em> term is less than \u0394<em data-effect=\"italics\">H<\/em>. If the <em data-effect=\"italics\">T<\/em>\u0394<em data-effect=\"italics\">S<\/em> term\u2019s magnitude is greater than \u0394<em data-effect=\"italics\">H<\/em>, the free energy change will be positive. Such a process is <em data-effect=\"italics\">spontaneous at low temperatures and nonspontaneous at high temperatures.<\/em><\/li>\n<li><strong data-effect=\"bold\">\u0394<em data-effect=\"italics\">H<\/em> is positive and \u0394<em data-effect=\"italics\">S<\/em> is negative.<\/strong> This condition describes an endothermic process that involves a decrease in system entropy. In this case, \u0394<em data-effect=\"italics\">G<\/em> will be positive regardless of the temperature. Such a process is <em data-effect=\"italics\">nonspontaneous at all temperatures.<\/em><\/li>\n<li><strong data-effect=\"bold\">\u0394<em data-effect=\"italics\">H<\/em> is negative and \u0394<em data-effect=\"italics\">S<\/em> is positive.<\/strong> This condition describes an exothermic process that involves an increase in system entropy. In this case, \u0394<em data-effect=\"italics\">G<\/em> will be negative regardless of the temperature. Such a process is <em data-effect=\"italics\">spontaneous at all temperatures.<\/em><\/li>\n<\/ol>\n<p id=\"fs-idm39683376\">These four scenarios are summarized in <a class=\"autogenerated-content\" href=\"#CNX_Chem_16_04_Scenarios\">(Figure)<\/a>.<\/p>\n<div id=\"CNX_Chem_16_04_Scenarios\" class=\"bc-figure figure\">\n<div class=\"bc-figcaption figcaption\">There are four possibilities regarding the signs of enthalpy and entropy changes.<\/div>\n<p><span id=\"fs-idm123583264\" data-type=\"media\" data-alt=\"A table with three columns and four rows is shown. The first column has the phrase, \u201cDelta S greater than zero ( increase in entropy ),\u201d in the third row and the phrase, \u201cDelta S less than zero ( decrease in entropy),\u201d in the fourth row. The second and third columns have the phrase, \u201cSummary of the Four Scenarios for Enthalpy and Entropy Changes,\u201d written above them. The second column has, \u201cdelta H greater than zero ( endothermic ),\u201d in the second row, \u201cdelta G less than zero at high temperature, delta G greater than zero at low temperature, Process is spontaneous at high temperature,\u201d in the third row, and \u201cdelta G greater than zero at any temperature, Process is nonspontaneous at any temperature,\u201d in the fourth row. The third column has, \u201cdelta H less than zero ( exothermic ),\u201d in the second row, \u201cdelta G less than zero at any temperature, Process is spontaneous at any temperature,\u201d in the third row, and \u201cdelta G less than zero at low temperature, delta G greater than zero at high temperature, Process is spontaneous at low temperature.\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_16_04_Scenarios-1.jpg\" alt=\"A table with three columns and four rows is shown. The first column has the phrase, \u201cDelta S greater than zero ( increase in entropy ),\u201d in the third row and the phrase, \u201cDelta S less than zero ( decrease in entropy),\u201d in the fourth row. The second and third columns have the phrase, \u201cSummary of the Four Scenarios for Enthalpy and Entropy Changes,\u201d written above them. The second column has, \u201cdelta H greater than zero ( endothermic ),\u201d in the second row, \u201cdelta G less than zero at high temperature, delta G greater than zero at low temperature, Process is spontaneous at high temperature,\u201d in the third row, and \u201cdelta G greater than zero at any temperature, Process is nonspontaneous at any temperature,\u201d in the fourth row. The third column has, \u201cdelta H less than zero ( exothermic ),\u201d in the second row, \u201cdelta G less than zero at any temperature, Process is spontaneous at any temperature,\u201d in the third row, and \u201cdelta G less than zero at low temperature, delta G greater than zero at high temperature, Process is spontaneous at low temperature.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-idm299243200\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm221990400\"><strong>Predicting the Temperature Dependence of Spontaneity <\/strong><\/p>\n<p>The incomplete combustion of carbon is described by the following equation:<\/p>\n<div id=\"fs-idm124549040\" style=\"padding-left: 40px\" data-type=\"equation\">2C(<em>s<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u27f6 2CO(<em>g<\/em>)<\/div>\n<p id=\"fs-idm291788752\">How does the spontaneity of this process depend upon temperature?<\/p>\n<p id=\"fs-idm116939264\"><strong>Solution:<\/strong><\/p>\n<p>Combustion processes are exothermic (\u0394<em data-effect=\"italics\">H<\/em> &lt; 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, \u0394<em data-effect=\"italics\">S<\/em> &gt; 0). The reaction is therefore spontaneous (\u0394<em data-effect=\"italics\">G<\/em> &lt; 0) at all temperatures.<\/p>\n<p id=\"fs-idm189692848\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Popular chemical hand warmers generate heat by the air-oxidation of iron:<\/p>\n<div id=\"fs-idm247358864\" style=\"padding-left: 40px\" data-type=\"equation\">4Fe(<em>s<\/em>) + 3O<sub>2<\/sub>(<em>g<\/em>) \u27f6 2Fe<sub>2<\/sub>O<sub>3<\/sub>(<em>s<\/em>)<\/div>\n<p id=\"fs-idm255092672\">How does the spontaneity of this process depend upon temperature?<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idm137038288\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm233909808\">\u0394<em data-effect=\"italics\">H<\/em> and \u0394<em data-effect=\"italics\">S<\/em> are negative; the reaction is spontaneous at low temperatures.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idp2328928\">When considering the conclusions drawn regarding the temperature dependence of spontaneity, it is important to keep in mind what the terms \u201chigh\u201d and \u201clow\u201d 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 \u201cspontaneity\u201d (as reflected by its \u0394<em data-effect=\"italics\">G<\/em>) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which \u0394<em data-effect=\"italics\">G<\/em> is plotted on the <em data-effect=\"italics\">y<\/em> axis versus <em data-effect=\"italics\">T<\/em> on the <em data-effect=\"italics\">x<\/em> axis:<\/p>\n<div id=\"fs-idm161001120\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G <\/em>= \u0394<em>H <\/em>&#8211; <em>T<\/em>\u0394<em>S<\/em><\/div>\n<div id=\"fs-idm285695760\" style=\"padding-left: 40px\" data-type=\"equation\"><em>y<\/em> = b + m<em>x<\/em><\/div>\n<p id=\"fs-idm206759776\">Such a plot is shown in <a class=\"autogenerated-content\" href=\"#CNX_Chem_16_05_TempSpont\">(Figure)<\/a>. 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 \u0394<em data-effect=\"italics\">G<\/em>) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the <em data-effect=\"italics\">x<\/em>-intercept of the line, that is, the value of <em data-effect=\"italics\">T<\/em> for which \u0394<em data-effect=\"italics\">G<\/em> is zero:<\/p>\n<div id=\"fs-idm187292464\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em> = 0 = \u0394<em>H<\/em> &#8211; <em>T<\/em>\u0394<em>S<\/em><\/div>\n<div id=\"fs-idm280698832\" style=\"padding-left: 40px\" data-type=\"equation\"><em>T<\/em> = \u0394<em>H<\/em>\/\u0394<em>S<\/em><\/div>\n<p id=\"fs-idm44045216\">So, saying a process is spontaneous at \u201chigh\u201d or \u201clow\u201d temperatures means the temperature is above or below, respectively, that temperature at which \u0394<em data-effect=\"italics\">G<\/em> for the process is zero. As noted earlier, the condition of \u0394G = 0 describes a system at equilibrium.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_16_05_TempSpont\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">These plots show the variation in \u0394<em data-effect=\"italics\">G<\/em> with temperature for the four possible combinations of arithmetic sign for \u0394<em data-effect=\"italics\">H<\/em> and \u0394<em data-effect=\"italics\">S<\/em>.<\/div>\n<p><span id=\"fs-idm275434144\" data-type=\"media\" data-alt=\"A graph is shown where the y-axis is labeled, \u201cFree energy,\u201d and the x-axis is labeled, \u201cIncreasing temperature ( K ).\u201d The value of zero is written midway up the y-axis with the label, \u201cdelta G greater than 0,\u201d written above this line and, \u201cdelta G less than 0,\u201d written below it. The bottom half of the graph is labeled on the right as, \u201cSpontaneous,\u201d and the top half is labeled on the right as, \u201cNonspontaneous.\u201d A green line labeled, \u201cdelta H less than 0, delta S greater than 0,\u201d extends from a quarter of the way up the y-axis to the bottom right of the graph. A yellow line labeled, \u201cdelta H less than 0, delta S less than 0,\u201d extends from a quarter of the way up the y-axis to the middle right of the graph. A second yellow line labeled, \u201cdelta H greater than 0, delta S greater than 0,\u201d extends from three quarters of the way up the y-axis to the middle right of the graph. A red line labeled, \u201cdelta H greater than 0, delta S less than 0,\u201d extends from three quarters of the way up the y-axis to the top right of the graph.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_16_04_TempSpont-1.jpg\" alt=\"A graph is shown where the y-axis is labeled, \u201cFree energy,\u201d and the x-axis is labeled, \u201cIncreasing temperature ( K ).\u201d The value of zero is written midway up the y-axis with the label, \u201cdelta G greater than 0,\u201d written above this line and, \u201cdelta G less than 0,\u201d written below it. The bottom half of the graph is labeled on the right as, \u201cSpontaneous,\u201d and the top half is labeled on the right as, \u201cNonspontaneous.\u201d A green line labeled, \u201cdelta H less than 0, delta S greater than 0,\u201d extends from a quarter of the way up the y-axis to the bottom right of the graph. A yellow line labeled, \u201cdelta H less than 0, delta S less than 0,\u201d extends from a quarter of the way up the y-axis to the middle right of the graph. A second yellow line labeled, \u201cdelta H greater than 0, delta S greater than 0,\u201d extends from three quarters of the way up the y-axis to the middle right of the graph. A red line labeled, \u201cdelta H greater than 0, delta S less than 0,\u201d extends from three quarters of the way up the y-axis to the top right of the graph.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-idm232248480\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm39866480\"><strong>Equilibrium Temperature for a Phase Transition<\/strong><\/p>\n<p>As defined in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its liquid and gaseous phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in Appendix G to estimate the boiling point of water.<\/p>\n<p id=\"fs-idm280706032\"><strong>Solution:<\/strong><\/p>\n<p>The process of interest is the following phase change:<\/p>\n<div id=\"fs-idm181862672\" style=\"padding-left: 40px\" data-type=\"equation\">H<sub>2<\/sub>O(<em>l<\/em>) \u27f6 H<sub>2<\/sub>O(<em>g<\/em>)<\/div>\n<p id=\"fs-idm40643152\">When this process is at equilibrium, \u0394<em data-effect=\"italics\">G<\/em> = 0, so the following is true:<\/p>\n<div id=\"fs-idm275207392\" style=\"padding-left: 40px\" data-type=\"equation\">0 = \u0394<em>H<\/em>\u00b0 &#8211; <em>T<\/em>\u0394<em>S<\/em>\u00b0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 or\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <em>T<\/em> = \u0394<em>H<\/em>\u00b0\/\u0394<em>S<\/em>\u00b0<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm203234768\">Using the standard thermodynamic data from Appendix G,<\/p>\n<div id=\"fs-idm159129872\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>H<\/em>\u00b0 = \u0394<em>H<\/em><sub>f<\/sub>\u00b0(H<sub>2<\/sub>O(<em>g<\/em>)) &#8211; \u0394<em>H<\/em><sub>f<\/sub>\u00b0(H<sub>2<\/sub>O(<em>l<\/em>)) = 241.82 kJ\/mol &#8211; (-285.83 kJ\/mol) = 44.01 kJ\/mol<\/div>\n<div style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>S<\/em>\u00b0 = <em>S<\/em>\u00b0(H<sub>2<\/sub>O(<em>g<\/em>)) &#8211; <em>S<\/em>\u00b0(H<sub>2<\/sub>O(<em>l<\/em>)) = 188.8 J\/K\u00b7mol &#8211; 70.0 J\/K\u00b7mol = 118.8 J\/K\u00b7mol<\/div>\n<div data-type=\"equation\"><\/div>\n<div id=\"fs-idm250608432\" style=\"padding-left: 40px\" data-type=\"equation\"><em>T<\/em> = \u0394<em>H<\/em>\u00b0\/\u0394<em>S<\/em>\u00b0 = (44.01 \u00d7 10<sup>3<\/sup> J\/mol)\/(118.8 J\/K\u00b7mol) = 370.5 K = 97.3 \u00b0C<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm289288576\">The accepted value for water\u2019s normal boiling point is 373.2 K (100.0 \u00b0C), and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (Appendix G). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point.<\/p>\n<p id=\"fs-idm252873760\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Use the information in Appendix G to estimate the boiling point of CS<sub>2<\/sub>.<\/p>\n<div id=\"fs-idm230494144\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm181951344\">313 K (accepted value 319 K)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm159117504\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Free Energy and Equilibrium<\/strong><\/h3>\n<p id=\"fs-idm166739792\">The free energy change for a process may be viewed as a measure of its driving force. A negative value for \u0394<em data-effect=\"italics\">G<\/em> represents a 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 \u0394<em data-effect=\"italics\">G<\/em> is zero, the forward and reverse driving forces are equal, and the process occurs in both directions at the same rate (the system is at equilibrium).<\/p>\n<p id=\"fs-idm189708256\">In the chapter on equilibrium the <em data-effect=\"italics\">reaction quotient<\/em>, <em data-effect=\"italics\">Q<\/em>, was introduced as a convenient measure of the status of an equilibrium system. Recall that you may use <em>Q<\/em> to identify the direction in which a reaction will proceed in order to achieve equilibrium. When <em data-effect=\"italics\">Q<\/em> is lesser than the equilibrium constant, <em data-effect=\"italics\">K<\/em>, the reaction will proceed in the forward direction until equilibrium is reached and <em data-effect=\"italics\">Q<\/em> = <em data-effect=\"italics\">K<\/em>. Conversely, if <em data-effect=\"italics\">Q<\/em> &gt; <em data-effect=\"italics\">K<\/em>, the process will proceed in the reverse direction until equilibrium is achieved.<\/p>\n<p id=\"fs-idm39755632\">The free energy change for a process taking place with reactants and products present under <em data-effect=\"italics\">nonstandard conditions<\/em> (pressures other than 1 bar; concentrations other than 1 M) is related to the standard free energy change according to this equation:<\/p>\n<div id=\"fs-idm55093920\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em> = \u0394<em>G<\/em>\u00b0 + <em>RT <\/em>ln <em>Q<\/em><\/div>\n<p id=\"fs-idm192589168\"><em data-effect=\"italics\">R<\/em> is the gas constant (8.314 J\/K mol), <em data-effect=\"italics\">T<\/em> is the kelvin or absolute temperature, and <em data-effect=\"italics\">Q<\/em> is the reaction quotient. This equation may be used to predict the spontaneity for a process under any given set of conditions as illustrated in <a class=\"autogenerated-content\" href=\"#fs-idm192477856\">(Figure)<\/a>.<\/p>\n<div id=\"fs-idm192477856\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm158842672\"><strong>Calculating \u0394<em data-effect=\"italics\">G<\/em> under Nonstandard Conditions <\/strong><\/p>\n<p>What is the free energy change for the process shown here under the specified conditions?<\/p>\n<p id=\"fs-idm178094656\"><em data-effect=\"italics\">T<\/em> = 25 \u00b0C, <em>P<\/em><sub>N2<\/sub> = 0.870 atm, <em>P<\/em><sub>H2<\/sub> = 0.250 atm, and<em> P<\/em><sub>NH3<\/sub> = 12.9 atm<\/p>\n<div id=\"fs-idm8287616\" style=\"padding-left: 40px\" data-type=\"equation\">2NH<sub>3<\/sub>(<em>g<\/em>) \u27f6 3H<sub>2<\/sub>(<em>g<\/em>) + N<sub>2<\/sub>(<em>g<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u0394<em>G<\/em>\u00b0 = 33.0 kJ\/mol<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm218730976\"><strong>Solution:<\/strong><\/p>\n<p>The equation relating free energy change to standard free energy change and reaction quotient may be used directly:<\/p>\n<div id=\"fs-idm150025600\" style=\"padding-left: 40px\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-2015\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4k-300x55.png\" alt=\"\" width=\"531\" height=\"97\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4k-300x55.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4k-1024x188.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4k-768x141.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4k-1536x282.png 1536w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4k-65x12.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4k-225x41.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4k-350x64.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/16.4k.png 2001w\" sizes=\"auto, (max-width: 531px) 100vw, 531px\" \/><\/div>\n<p id=\"fs-idm159629248\">Since the computed value for \u0394<em data-effect=\"italics\">G<\/em> is positive, the reaction is nonspontaneous under these conditions.<\/p>\n<p id=\"fs-idm150306224\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Calculate the free energy change for this same reaction at 875 \u00b0C in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?<\/p>\n<div id=\"fs-idm329579920\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm247369440\">\u0394<em data-effect=\"italics\">G<\/em> = \u221247 kJ; yes<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idm247911728\">For a system at equilibrium, <em data-effect=\"italics\">Q<\/em> = <em data-effect=\"italics\">K<\/em> and \u0394<em data-effect=\"italics\">G<\/em> = 0, and the previous equation may be written as<\/p>\n<div id=\"fs-idm228614064\" style=\"padding-left: 40px\" data-type=\"equation\">0 = \u0394<em>G<\/em>\u00b0 + <em>RT <\/em>ln <em>K<\/em><\/div>\n<div id=\"fs-idm55951760\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em>\u00b0 = \u2212<em>RT<\/em> ln K<\/div>\n<p id=\"fs-idm206226416\">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 <a class=\"autogenerated-content\" href=\"#fs-idm159425040\">(Figure)<\/a>.<\/p>\n<table id=\"fs-idm159425040\" class=\"top-titled\" summary=\"This table has three columns and four rows. The first row is a header row, and it labels each column, \u201cK,\u201d \u201ccapital delta G superscript degree symbol,\u201d and \u201cComments.\u201d Under the \u201cK\u201d column are the following: \u201cgreater than 1,\u201d \u201cless than 1,\u201d and \u201cequal to 1.\u201d Under the \u201ccapital delta G superscript degree symbol\u201d column are the following: \u201cless than 0,\u201d \u201cgreater than 0,\u201d and \u201cequal to 0.\u201d Under the \u201cComments\u201d column are the following: \u201cProducts are more abundant at equilibrium,\u201d \u201cReactants are more abundant at equilibrium,\u201d and \u201cReactants and products are equally abundant at equilibrium.\u201d\">\n<thead>\n<tr>\n<th colspan=\"3\" data-align=\"center\">Relations between Standard Free Energy Changes and Equilibrium Constants<\/th>\n<\/tr>\n<tr valign=\"top\">\n<th data-align=\"center\"><em data-effect=\"italics\">K<\/em><\/th>\n<th data-align=\"center\">\u0394<em data-effect=\"italics\">G<\/em>\u00b0<\/th>\n<th data-align=\"left\">Composition of an Equilibrium Mixture<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td data-align=\"center\">&gt; 1<\/td>\n<td data-align=\"center\">&lt; 0<\/td>\n<td data-align=\"left\">Products are more abundant<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"center\">&lt; 1<\/td>\n<td data-align=\"center\">&gt; 0<\/td>\n<td data-align=\"left\">Reactants are more abundant<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"center\">= 1<\/td>\n<td data-align=\"center\">= 0<\/td>\n<td data-align=\"left\">Reactants and products are comparably abundant<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-idm300346368\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm249874240\"><strong>Calculating an Equilibrium Constant using Standard Free Energy Change <\/strong><\/p>\n<p>Given that the standard free energies of formation of Ag<sup>+<\/sup>(<em data-effect=\"italics\">aq<\/em>), Cl<sup>\u2212<\/sup>(<em data-effect=\"italics\">aq<\/em>), and AgCl(<em data-effect=\"italics\">s<\/em>) are 77.1 kJ\/mol, \u2212131.2 kJ\/mol, and \u2212109.8 kJ\/mol, respectively, calculate the solubility product, <em data-effect=\"italics\">K<\/em><sub>sp<\/sub>, for AgCl.<\/p>\n<p id=\"fs-idm192205040\"><strong>Solution:<\/strong><\/p>\n<p>The reaction of interest is the following:<\/p>\n<div id=\"fs-idm266185728\" style=\"padding-left: 40px\" data-type=\"equation\">AgCl(<em>s<\/em>) \u21cc Ag<sup>+<\/sup>(<em>aq<\/em>) + Cl<sup>\u2212<\/sup>(<em>aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 K<sub>sp<\/sub> = [Ag+][Cl<sup>\u2212<\/sup>]<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm150004464\">The standard free energy change for this reaction is first computed using standard free energies of formation for its reactants and products:<\/p>\n<div id=\"fs-idm111103296\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em>\u00b0 = [\u0394<em>G<\/em><sub>f<\/sub>\u00b0(Ag<sup>+<\/sup>(<em>aq<\/em>)) + \u0394<em>G<\/em><sub>f<\/sub>\u00b0(Cl<sup>\u2212<\/sup>(<em>aq<\/em>))] &#8211; \u0394<em>G<\/em><sub>f<\/sub>\u00b0(AgCl(<em>s<\/em>)) = [77.1 kJ\/mol &#8211; 131.2 kJ\/mol] &#8211; (-109.8 kJ\/mol) = 55.7 kJ\/mol<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm180460784\">The equilibrium constant for the reaction may then be derived from its standard free energy change:<\/p>\n<div id=\"fs-idm267274240\" style=\"padding-left: 40px\" data-type=\"equation\">\u0394<em>G<\/em>\u00b0 = \u2212<em>RT<\/em> ln K<\/div>\n<div style=\"padding-left: 40px\" data-type=\"equation\">55.7 \u00d7 10<sup>3<\/sup> J\/mol = -(8.314 J\/K\u00b7mol)(298.15 K) ln K<\/div>\n<div style=\"padding-left: 40px\" data-type=\"equation\">ln <em>K<\/em> = -22.5<\/div>\n<div style=\"padding-left: 40px\" data-type=\"equation\"><em>K<\/em> = 1.7 \u00d7 10<sup>-10<\/sup><\/div>\n<p id=\"fs-idm126979616\">This result is in\u00a0reasonable agreement with the value provided in Appendix J.<\/p>\n<p id=\"fs-idm284201456\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Use the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 \u00b0C.<\/p>\n<div id=\"fs-idm193826960\" style=\"padding-left: 40px\" data-type=\"equation\">2NO<sub>2<\/sub>(<em>g<\/em>) \u21cc N<sub>2<\/sub>O<sub>4<\/sub>(<em>g<\/em>)<\/div>\n<div data-type=\"equation\"><\/div>\n<div id=\"fs-idm197604384\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm135514416\"><em data-effect=\"italics\">K<\/em> = 6.9<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idm66775936\">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 <em data-effect=\"italics\">Q<\/em>), equilibrium is established when the system\u2019s free energy is minimized (<a class=\"autogenerated-content\" href=\"#CNX_Chem_16_04_Gibbs\">(Figure)<\/a>). If a system consists of reactants and products in nonequilibrium amounts (<em data-effect=\"italics\">Q<\/em> \u2260 <em data-effect=\"italics\">K<\/em>), the reaction will proceed spontaneously in the direction necessary to establish equilibrium.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_16_04_Gibbs\" class=\"bc-figure figure\">\n<div class=\"bc-figcaption figcaption\">These plots show the free energy versus reaction progress for systems whose standard free energy changes are (a) negative, (b) positive, and (c) zero. Nonequilibrium systems will proceed spontaneously in whatever direction is necessary to minimize free energy and establish equilibrium.<\/div>\n<p><span id=\"fs-idm247894944\" data-type=\"media\" data-alt=\"Three graphs, labeled, \u201ca,\u201d \u201cb,\u201d and \u201cc\u201d are shown where the y-axis is labeled, \u201cGibbs free energy ( G ),\u201d and, \u201cG superscript degree sign ( reactants ),\u201d while the x-axis is labeled, \u201cReaction progress,\u201d and \u201cReactants,\u201d on the left and, \u201cProducts,\u201d on the right. In graph a, a line begins at the upper left side and goes steadily down to a point about halfway up the y-axis and two thirds of the way on the x-axis, then rises again to a point labeled, \u201cG superscript degree sign ( products ),\u201d that is slightly higher than halfway up the y-axis. The distance between the beginning and ending points of the graph is labeled as, \u201cdelta G less than 0,\u201d while the lowest point on the graph is labeled, \u201cQ equals K greater than 1.\u201d In graph b, a line begins at the middle left side and goes steadily down to a point about two fifths up the y-axis and one third of the way on the x-axis, then rises again to a point labeled, \u201cG superscript degree sign ( products ),\u201d that is near the top of the y-axis. The distance between the beginning and ending points of the graph is labeled as, \u201cdelta G greater than 0,\u201d while the lowest point on the graph is labeled, \u201cQ equals K less than 1.\u201d In graph c, a line begins at the upper left side and goes steadily down to a point near the bottom of the y-axis and half way on the x-axis, then rises again to a point labeled, \u201cG superscript degree sign ( products ),\u201d that is equal to the starting point on the y-axis which is labeled, \u201cG superscript degree sign ( reactants ).\u201d The lowest point on the graph is labeled, \u201cQ equals K equals 1.\u201d At the top of the graph is the label, \u201cDelta G superscript degree sign equals 0.\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_16_04_Gibbs-1.jpg\" alt=\"Three graphs, labeled, \u201ca,\u201d \u201cb,\u201d and \u201cc\u201d are shown where the y-axis is labeled, \u201cGibbs free energy ( G ),\u201d and, \u201cG superscript degree sign ( reactants ),\u201d while the x-axis is labeled, \u201cReaction progress,\u201d and \u201cReactants,\u201d on the left and, \u201cProducts,\u201d on the right. In graph a, a line begins at the upper left side and goes steadily down to a point about halfway up the y-axis and two thirds of the way on the x-axis, then rises again to a point labeled, \u201cG superscript degree sign ( products ),\u201d that is slightly higher than halfway up the y-axis. The distance between the beginning and ending points of the graph is labeled as, \u201cdelta G less than 0,\u201d while the lowest point on the graph is labeled, \u201cQ equals K greater than 1.\u201d In graph b, a line begins at the middle left side and goes steadily down to a point about two fifths up the y-axis and one third of the way on the x-axis, then rises again to a point labeled, \u201cG superscript degree sign ( products ),\u201d that is near the top of the y-axis. The distance between the beginning and ending points of the graph is labeled as, \u201cdelta G greater than 0,\u201d while the lowest point on the graph is labeled, \u201cQ equals K less than 1.\u201d In graph c, a line begins at the upper left side and goes steadily down to a point near the bottom of the y-axis and half way on the x-axis, then rises again to a point labeled, \u201cG superscript degree sign ( products ),\u201d that is equal to the starting point on the y-axis which is labeled, \u201cG superscript degree sign ( reactants ).\u201d The lowest point on the graph is labeled, \u201cQ equals K equals 1.\u201d At the top of the graph is the label, \u201cDelta G superscript degree sign equals 0.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idm220227936\" class=\"summary\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\n<p id=\"fs-idm197297904\">Gibbs free energy (<em data-effect=\"italics\">G<\/em>) 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 \u0394<em data-effect=\"italics\">G<\/em> indicates a spontaneous process; a positive \u0394<em data-effect=\"italics\">G<\/em> indicates a nonspontaneous process; and a \u0394<em data-effect=\"italics\">G<\/em> of zero indicates that the system is at equilibrium. A number of approaches to the computation of free energy changes are possible.<\/p>\n<\/div>\n<div id=\"fs-idm248576000\" class=\"key-equations\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Key Equations<\/strong><\/h3>\n<ul id=\"fs-idm60433792\" data-bullet-style=\"bullet\">\n<li>\u0394<em data-effect=\"italics\">G<\/em> = \u0394<em data-effect=\"italics\">H<\/em> \u2212 <em data-effect=\"italics\">T<\/em>\u0394<em data-effect=\"italics\">S<\/em><\/li>\n<li>\u0394<em data-effect=\"italics\">G<\/em> = \u0394<em data-effect=\"italics\">G<\/em>\u00b0 + <em data-effect=\"italics\">RT<\/em> ln <em data-effect=\"italics\">Q<\/em><\/li>\n<li>\u0394<em data-effect=\"italics\">G<\/em>\u00b0 = \u2212<em data-effect=\"italics\">RT<\/em> ln <em data-effect=\"italics\">K<\/em><\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-idm233991504\" class=\"exercises\" data-depth=\"1\">\n<div id=\"fs-idm153799616\" data-type=\"exercise\">\n<div id=\"fs-idm71264384\" data-type=\"solution\">\n<p id=\"eip-idm1161148928\">\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\" data-type=\"glossary\">\n<h3 data-type=\"glossary-title\"><strong>Glossary<\/strong><\/h3>\n<dl id=\"fs-idm119810368\">\n<dt>Gibbs free energy change (<em data-effect=\"italics\">G<\/em>)<\/dt>\n<dd id=\"fs-idm119809104\">thermodynamic property defined in terms of system enthalpy and entropy; all spontaneous processes involve a decrease in <em data-effect=\"italics\">G<\/em><\/dd>\n<\/dl>\n<dl id=\"fs-idm119807952\">\n<dt>standard free energy change (\u0394<em data-effect=\"italics\">G<\/em>\u00b0)<\/dt>\n<dd id=\"fs-idm119806688\">change in free energy for a process occurring under standard conditions (1 bar pressure for gases, 1 M concentration for solutions)<\/dd>\n<\/dl>\n<dl id=\"fs-idm119806032\">\n<dt>standard free energy of formation (\u0394G<sub>f<\/sub>\u00b0)<\/dt>\n<dd id=\"fs-idm185413520\">change in free energy accompanying the formation of one mole of substance from its elements in their standard states<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":1392,"menu_order":5,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[48],"contributor":[],"license":[],"class_list":["post-869","chapter","type-chapter","status-publish","hentry","chapter-type-numberless"],"part":848,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/869","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/users\/1392"}],"version-history":[{"count":21,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/869\/revisions"}],"predecessor-version":[{"id":2183,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/869\/revisions\/2183"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/parts\/848"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/869\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/media?parent=869"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapter-type?post=869"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/contributor?post=869"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/license?post=869"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}