{"id":1085,"date":"2016-01-11T20:02:54","date_gmt":"2016-01-11T20:02:54","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/gibbs-free-energy-2\/"},"modified":"2020-04-20T16:45:57","modified_gmt":"2020-04-20T16:45:57","slug":"gibbs-free-energy","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/gibbs-free-energy\/","title":{"raw":"Gibbs Free Energy","rendered":"Gibbs Free Energy"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>To gain an understanding of Gibbs free energy.<\/li>\r\n \t<li>To understand the relationship between the sign of Gibbs free energy change and the spontaneity of a process.<\/li>\r\n \t<li>To be able to determine Gibbs free energy using standard free energies of formation.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<span class=\"Apple-style-span\">Figure 18.4. J. Willard Gibbs<\/span>\r\n\r\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/06\/Gibbs_Josiah_Willard.jpg\"><img class=\"alignnone wp-image-1084 size-full\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/291\/2016\/01\/Gibbs_Josiah_Willard-1.jpg\" alt=\"Figure #.#. A portrait of J. Willard Gibbs.\" width=\"362\" height=\"543\" \/><\/a>\r\n\r\n<span class=\"Apple-style-span\">A portrait of J. Willard Gibbs<\/span>[footnote]Gibbs Josiah Willard\/Public Domain[\/footnote]\r\n\r\nJ. Willard Gibbs (1839-1903) proposed a single state function to determine spontaneity:\r\n\r\n<em>G = H - TS<\/em>\r\n\r\nwhere <em>H<\/em> is the enthalpy of the system,\u00a0<em>S<\/em> is the entropy of the system, and <em>G<\/em> is <strong>Gibbs free energy<\/strong>.\r\n\r\nThe change in Gibbs free energy,\u00a0\u0394<em>G<\/em>, is the maximum amount of free energy available to do useful work. For an isothermal process, it can be expressed as:\r\n\r\n\u0394<em>G = <\/em>\u0394<em>H - T<\/em>\u0394<em>S \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/em>or at standard conditions:<em> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/em>\u0394<em>G<\/em>\u2070<em> = <\/em>\u0394<em>H<\/em>\u2070\u00a0<em>-<\/em><em>\u00a0T<\/em>\u0394<em>S<\/em>\u2070\r\n\r\nThis single term, Gibbs free energy (<em>G<\/em>), allows us to avoid calculating the entropy of the surroundings. It is really just a simplification of our previous method of estimating spontaneity:\r\n\r\n\u0394<em>S<\/em><sub>universe<\/sub> = \u0394<em>S<\/em><sub>sys<\/sub> + \u0394<em>S<\/em><sub>surr<\/sub>\r\n\r\n$latex \\Delta\\textit{S}{}_{universe}$ = $latex \\Delta \\textit{S}{}_{sys}$ + ($latex \\frac{{-{\\rm \\ }\\Delta {\\rm H}}_{sys}}{T}$ )\r\n\r\nMultiply both sides of the equation by \u2013<em>T<\/em>:\r\n\r\n-<em>T<\/em>\u0394<em>S<\/em><sub>universe<\/sub>= $latex \\Delta\\textit{S}{}_{sys}$ + ($latex \\frac{{-{\\rm \\ }\\Delta {\\rm H}}_{sys}}{T}$ )\r\n\r\n-<em>T<\/em>\u0394<em>S<\/em><sub>universe<\/sub> = \u0394<em>H<\/em><sub>sys<\/sub> - <em>T\u0394S<\/em><sub>sys<\/sub>\r\n\r\nTherefore \u0394<em>G = <\/em>-<em>T<\/em>\u0394<em>S<\/em><sub>universe<\/sub>\r\n\r\nAs a result of this relationship, the sign of Gibbs free energy provides information on the spontaneity of a given reaction:\r\n\r\nIf \u0394<em>G <\/em>&gt; 0, the reaction is nonspontaneous in the direction written.\r\n\r\nIf \u0394<em>G <\/em>= 0, the reaction is in a state of equilibrium.\r\n\r\nIf \u0394<em>G <\/em>&lt; 0, the reaction is spontaneous in the direction written.\r\n\r\nThe significance of the sign of a change in\u00a0Gibbs free energy parallels the relationship of terms from\u00a0the equilibrium chapter: the reaction quotient, <em>Q<\/em>, and the equilibrium constant, <em>K<\/em>.\r\n\r\nIf <em>Q <\/em>&gt; <em>K<\/em>, the reaction is nonspontaneous in the direction written.\r\n\r\nIf <em>Q <\/em>=<em> K<\/em>, the reaction is in a state of equilibrium.\r\n\r\nIf <em>Q <\/em>&lt; <em>K<\/em>, the reaction is spontaneous in the direction written.\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>Example 4<\/strong>\r\n\r\nCalculate \u0394<em>G<\/em>\u2070 for a reaction where \u0394<em>H<\/em>\u2070\u00a0is equal to 36.2 kJ and\u00a0\u0394<em>S<\/em>\u2070 is equal to 123 J\/K at 298 K. Is this a spontaneous reaction?\r\n\r\nSolution\r\n\r\n\u0394<em>G<\/em>\u2070<em> = <\/em>\u0394<em>H<\/em>\u2070\u00a0<em>- T<\/em>\u0394<em>S<\/em>\u2070\r\n\r\n\u0394<em>G<\/em>\u2070<em> = <\/em>36.2 kJ<em> \u2013<\/em> (298 K x 123 J\/K)\r\n\r\n\u0394<em>G<\/em>\u2070<em> = <\/em>-0.4 kJ\r\n\r\nTherefore the reaction is spontaneous because \u0394<em>G<\/em>\u2070 is negative.\r\n\r\n<\/div>\r\n<h2>Determining \u0394<em>G<\/em>\u2070 from Standard Free Energy of Formation<\/h2>\r\nThe standard Gibbs free energy change, \u0394<em>G<\/em>\u2070, for a reaction can be calculated from the standard free energies of formation, \u0394<em>G<\/em>\u2070<em><sub>f<\/sub>\u00a0<\/em>.\r\n\r\n\u0394<em>G<\/em>\u2070<em><sub>f<\/sub> \u00a0<\/em>= \u2211<em>n<\/em>\u0394<em>G<\/em>\u2070<em><sub>f<\/sub><\/em>(products) - \u2211<em>m<\/em>\u0394<em>G<\/em>\u2070<em><sub>f<\/sub><\/em>(reactants)\r\n\r\nwhere <em>n<\/em> and <em>m<\/em> are the coefficients in the balanced chemical equation of the reaction.\r\n\r\nStandard free energies of formation values are listed in the appendix, \"Standard Thermodynamic Quantities for Chemical Substances at 25\u00b0C.\"\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>Example 5<\/strong>\r\n\r\nCalculate the standard free energy change for the following reaction, using standard free energies of formation:\r\n\r\n5 C(s) + 2 SO<sub>2<\/sub>(g) \u2192 CS<sub>2<\/sub>(g) + 4 CO(g)\r\n\r\nIs this a spontaneous reaction?\r\n\r\nSolution\r\n\r\n\u0394<em>G<\/em>\u2070= \u2211<em>n<\/em>\u0394<em>G<\/em>\u2070<em><sub>f<\/sub><\/em>(products) - \u2211<em>m<\/em>\u0394<em>G<\/em>\u2070<em><sub>f<\/sub><\/em>(reactants)\r\n\r\n\u0394<em>G<\/em>\u2070 = [(4 x \u2212137.2 kJ\/mol) + (67.1 kJ\/mol)] \u2013 [(5 x 0 kJ\/mol) + (2 x \u2212300.1 kJ\/mol)]\r\n\r\n\u0394<em>G<\/em>\u2070= (-481.7 kJ\/mol) \u2013 (-600.2 kJ\/mol)\r\n\r\n\u0394<em>G<\/em>\u2070= 118.5 kJ\/mol\r\n\r\n\u0394<em>G<\/em>\u2070\u00a0has a positive value so this is not a spontaneous process.\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Key Takeaways<\/h3>\r\n<ul>\r\n \t<li>The change in Gibbs free energy (\u0394<em>G<\/em>) \u00a0is the maximum amount of free energy available to do useful work.<\/li>\r\n \t<li>If\u00a0\u0394<em>G\u00a0<\/em>&gt; 0, the reaction is nonspontaneous in the direction written. If\u00a0\u0394<em>G\u00a0<\/em>=0, the reaction is in a state of equilibrium. If\u00a0\u0394<em>G\u00a0<\/em>&lt; 0, the reaction is spontaneous in the direction written.<\/li>\r\n \t<li>The standard Gibbs free energy change,<em>\u00a0<\/em>\u0394<em>G<\/em>\u2070, for a reaction can be calculated from the standard free energies of formation,\u00a0\u0394<em>G<\/em>\u2070<em><sub>f<\/sub><\/em>.<\/li>\r\n<\/ul>\r\n<\/div>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>To gain an understanding of Gibbs free energy.<\/li>\n<li>To understand the relationship between the sign of Gibbs free energy change and the spontaneity of a process.<\/li>\n<li>To be able to determine Gibbs free energy using standard free energies of formation.<\/li>\n<\/ul>\n<\/div>\n<p><span class=\"Apple-style-span\">Figure 18.4. J. Willard Gibbs<\/span><\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/06\/Gibbs_Josiah_Willard.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1084 size-full\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/291\/2016\/01\/Gibbs_Josiah_Willard-1.jpg\" alt=\"Figure #.#. A portrait of J. Willard Gibbs.\" width=\"362\" height=\"543\" srcset=\"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2016\/01\/Gibbs_Josiah_Willard-1.jpg 362w, https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2016\/01\/Gibbs_Josiah_Willard-1-200x300.jpg 200w, https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2016\/01\/Gibbs_Josiah_Willard-1-65x98.jpg 65w, https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2016\/01\/Gibbs_Josiah_Willard-1-225x338.jpg 225w, https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2016\/01\/Gibbs_Josiah_Willard-1-350x525.jpg 350w\" sizes=\"auto, (max-width: 362px) 100vw, 362px\" \/><\/a><\/p>\n<p><span class=\"Apple-style-span\">A portrait of J. Willard Gibbs<\/span><a class=\"footnote\" title=\"Gibbs Josiah Willard\/Public Domain\" id=\"return-footnote-1085-1\" href=\"#footnote-1085-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a><\/p>\n<p>J. Willard Gibbs (1839-1903) proposed a single state function to determine spontaneity:<\/p>\n<p><em>G = H &#8211; TS<\/em><\/p>\n<p>where <em>H<\/em> is the enthalpy of the system,\u00a0<em>S<\/em> is the entropy of the system, and <em>G<\/em> is <strong>Gibbs free energy<\/strong>.<\/p>\n<p>The change in Gibbs free energy,\u00a0\u0394<em>G<\/em>, is the maximum amount of free energy available to do useful work. For an isothermal process, it can be expressed as:<\/p>\n<p>\u0394<em>G = <\/em>\u0394<em>H &#8211; T<\/em>\u0394<em>S \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/em>or at standard conditions:<em> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/em>\u0394<em>G<\/em>\u2070<em> = <\/em>\u0394<em>H<\/em>\u2070\u00a0<em>&#8211;<\/em><em>\u00a0T<\/em>\u0394<em>S<\/em>\u2070<\/p>\n<p>This single term, Gibbs free energy (<em>G<\/em>), allows us to avoid calculating the entropy of the surroundings. It is really just a simplification of our previous method of estimating spontaneity:<\/p>\n<p>\u0394<em>S<\/em><sub>universe<\/sub> = \u0394<em>S<\/em><sub>sys<\/sub> + \u0394<em>S<\/em><sub>surr<\/sub><\/p>\n<p>$latex \\Delta\\textit{S}{}_{universe}$ = $latex \\Delta \\textit{S}{}_{sys}$ + ($latex \\frac{{-{\\rm \\ }\\Delta {\\rm H}}_{sys}}{T}$ )<\/p>\n<p>Multiply both sides of the equation by \u2013<em>T<\/em>:<\/p>\n<p>&#8211;<em>T<\/em>\u0394<em>S<\/em><sub>universe<\/sub>= $latex \\Delta\\textit{S}{}_{sys}$ + ($latex \\frac{{-{\\rm \\ }\\Delta {\\rm H}}_{sys}}{T}$ )<\/p>\n<p>&#8211;<em>T<\/em>\u0394<em>S<\/em><sub>universe<\/sub> = \u0394<em>H<\/em><sub>sys<\/sub> &#8211; <em>T\u0394S<\/em><sub>sys<\/sub><\/p>\n<p>Therefore \u0394<em>G = <\/em>&#8211;<em>T<\/em>\u0394<em>S<\/em><sub>universe<\/sub><\/p>\n<p>As a result of this relationship, the sign of Gibbs free energy provides information on the spontaneity of a given reaction:<\/p>\n<p>If \u0394<em>G <\/em>&gt; 0, the reaction is nonspontaneous in the direction written.<\/p>\n<p>If \u0394<em>G <\/em>= 0, the reaction is in a state of equilibrium.<\/p>\n<p>If \u0394<em>G <\/em>&lt; 0, the reaction is spontaneous in the direction written.<\/p>\n<p>The significance of the sign of a change in\u00a0Gibbs free energy parallels the relationship of terms from\u00a0the equilibrium chapter: the reaction quotient, <em>Q<\/em>, and the equilibrium constant, <em>K<\/em>.<\/p>\n<p>If <em>Q <\/em>&gt; <em>K<\/em>, the reaction is nonspontaneous in the direction written.<\/p>\n<p>If <em>Q <\/em>=<em> K<\/em>, the reaction is in a state of equilibrium.<\/p>\n<p>If <em>Q <\/em>&lt; <em>K<\/em>, the reaction is spontaneous in the direction written.<\/p>\n<div class=\"textbox shaded\">\n<p><strong>Example 4<\/strong><\/p>\n<p>Calculate \u0394<em>G<\/em>\u2070 for a reaction where \u0394<em>H<\/em>\u2070\u00a0is equal to 36.2 kJ and\u00a0\u0394<em>S<\/em>\u2070 is equal to 123 J\/K at 298 K. Is this a spontaneous reaction?<\/p>\n<p>Solution<\/p>\n<p>\u0394<em>G<\/em>\u2070<em> = <\/em>\u0394<em>H<\/em>\u2070\u00a0<em>&#8211; T<\/em>\u0394<em>S<\/em>\u2070<\/p>\n<p>\u0394<em>G<\/em>\u2070<em> = <\/em>36.2 kJ<em> \u2013<\/em> (298 K x 123 J\/K)<\/p>\n<p>\u0394<em>G<\/em>\u2070<em> = <\/em>-0.4 kJ<\/p>\n<p>Therefore the reaction is spontaneous because \u0394<em>G<\/em>\u2070 is negative.<\/p>\n<\/div>\n<h2>Determining \u0394<em>G<\/em>\u2070 from Standard Free Energy of Formation<\/h2>\n<p>The standard Gibbs free energy change, \u0394<em>G<\/em>\u2070, for a reaction can be calculated from the standard free energies of formation, \u0394<em>G<\/em>\u2070<em><sub>f<\/sub>\u00a0<\/em>.<\/p>\n<p>\u0394<em>G<\/em>\u2070<em><sub>f<\/sub> \u00a0<\/em>= \u2211<em>n<\/em>\u0394<em>G<\/em>\u2070<em><sub>f<\/sub><\/em>(products) &#8211; \u2211<em>m<\/em>\u0394<em>G<\/em>\u2070<em><sub>f<\/sub><\/em>(reactants)<\/p>\n<p>where <em>n<\/em> and <em>m<\/em> are the coefficients in the balanced chemical equation of the reaction.<\/p>\n<p>Standard free energies of formation values are listed in the appendix, &#8220;Standard Thermodynamic Quantities for Chemical Substances at 25\u00b0C.&#8221;<\/p>\n<div class=\"textbox shaded\">\n<p><strong>Example 5<\/strong><\/p>\n<p>Calculate the standard free energy change for the following reaction, using standard free energies of formation:<\/p>\n<p>5 C(s) + 2 SO<sub>2<\/sub>(g) \u2192 CS<sub>2<\/sub>(g) + 4 CO(g)<\/p>\n<p>Is this a spontaneous reaction?<\/p>\n<p>Solution<\/p>\n<p>\u0394<em>G<\/em>\u2070= \u2211<em>n<\/em>\u0394<em>G<\/em>\u2070<em><sub>f<\/sub><\/em>(products) &#8211; \u2211<em>m<\/em>\u0394<em>G<\/em>\u2070<em><sub>f<\/sub><\/em>(reactants)<\/p>\n<p>\u0394<em>G<\/em>\u2070 = [(4 x \u2212137.2 kJ\/mol) + (67.1 kJ\/mol)] \u2013 [(5 x 0 kJ\/mol) + (2 x \u2212300.1 kJ\/mol)]<\/p>\n<p>\u0394<em>G<\/em>\u2070= (-481.7 kJ\/mol) \u2013 (-600.2 kJ\/mol)<\/p>\n<p>\u0394<em>G<\/em>\u2070= 118.5 kJ\/mol<\/p>\n<p>\u0394<em>G<\/em>\u2070\u00a0has a positive value so this is not a spontaneous process.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul>\n<li>The change in Gibbs free energy (\u0394<em>G<\/em>) \u00a0is the maximum amount of free energy available to do useful work.<\/li>\n<li>If\u00a0\u0394<em>G\u00a0<\/em>&gt; 0, the reaction is nonspontaneous in the direction written. If\u00a0\u0394<em>G\u00a0<\/em>=0, the reaction is in a state of equilibrium. If\u00a0\u0394<em>G\u00a0<\/em>&lt; 0, the reaction is spontaneous in the direction written.<\/li>\n<li>The standard Gibbs free energy change,<em>\u00a0<\/em>\u0394<em>G<\/em>\u2070, for a reaction can be calculated from the standard free energies of formation,\u00a0\u0394<em>G<\/em>\u2070<em><sub>f<\/sub><\/em>.<\/li>\n<\/ul>\n<\/div>\n<hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-1085-1\">Gibbs Josiah Willard\/Public Domain <a href=\"#return-footnote-1085-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":124,"menu_order":4,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":["jessie-a-key"],"pb_section_license":"cc-by"},"chapter-type":[],"contributor":[61],"license":[52],"class_list":["post-1085","chapter","type-chapter","status-publish","hentry","contributor-jessie-a-key","license-cc-by"],"part":1071,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters\/1085","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/users\/124"}],"version-history":[{"count":2,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters\/1085\/revisions"}],"predecessor-version":[{"id":1459,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters\/1085\/revisions\/1459"}],"part":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/parts\/1071"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters\/1085\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/media?parent=1085"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapter-type?post=1085"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/contributor?post=1085"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/license?post=1085"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}