{"id":954,"date":"2017-12-14T21:52:01","date_gmt":"2017-12-14T21:52:01","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/gibbs-free-energy\/"},"modified":"2017-12-14T21:52:01","modified_gmt":"2017-12-14T21:52:01","slug":"gibbs-free-energy","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/chapter\/gibbs-free-energy\/","title":{"raw":"Gibbs Free Energy","rendered":"Gibbs Free Energy"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ul><li>To gain an understanding of Gibbs free energy.<\/li>\n\t<li>To understand the relationship between the sign of Gibbs free energy change and the spontaneity of a process.<\/li>\n\t<li>To be able to determine Gibbs free energy using standard free energies of formation.<\/li>\n<\/ul><\/div>\n<span class=\"Apple-style-span\">Figure 18.4. J. Willard Gibbs<\/span>\n\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/06\/Gibbs_Josiah_Willard.jpg\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14215200\/Gibbs_Josiah_Willard-1.jpg\" alt=\"Figure #.#. A portrait of J. Willard Gibbs.\" class=\"alignnone wp-image-2643 size-full\" height=\"543\" width=\"362\"\/><\/a>\n\n<span class=\"Apple-style-span\">A portrait of J. Willard Gibbs<\/span>[footnote]Gibbs Josiah Willard\/Public Domain[\/footnote]\n\nJ. Willard Gibbs (1839-1903) proposed a single state function to determine spontaneity:\n\n<em>G = H - TS<\/em>\n\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>.\n\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:\n\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\n\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:\n\n\u0394<em>S<\/em><sub>universe<\/sub> = \u0394<em>S<\/em><sub>sys<\/sub> + \u0394<em>S<\/em><sub>surr<\/sub>\n\n$latex \\Delta\\textit{S}{}_{universe}$ = $latex \\Delta \\textit{S}{}_{sys}$ + ($latex \\frac{{-{\\rm \\ }\\Delta {\\rm H}}_{sys}}{T}$ )\n\nMultiply both sides of the equation by \u2013<em>T<\/em>:\n\n-<em>T<\/em>\u0394<em>S<\/em><sub>universe<\/sub>= $latex \\Delta\\textit{S}{}_{sys}$ + ($latex \\frac{{-{\\rm \\ }\\Delta {\\rm H}}_{sys}}{T}$ )\n\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>\n\nTherefore \u0394<em>G = <\/em>-<em>T<\/em>\u0394<em>S<\/em><sub>universe<\/sub>\n\nAs a result of this relationship, the sign of Gibbs free energy provides information on the spontaneity of a given reaction:\n\nIf \u0394<em>G <\/em>&gt; 0, the reaction is nonspontaneous in the direction written.\n\nIf \u0394<em>G <\/em>= 0, the reaction is in a state of equilibrium.\n\nIf \u0394<em>G <\/em>&lt; 0, the reaction is spontaneous in the direction written.\n\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>.\n\nIf <em>Q <\/em>&gt; <em>K<\/em>, the reaction is nonspontaneous in the direction written.\n\nIf <em>Q <\/em>=<em> K<\/em>, the reaction is in a state of equilibrium.\n\nIf <em>Q <\/em>&lt; <em>K<\/em>, the reaction is spontaneous in the direction written.\n<div class=\"textbox shaded\">\n\n<strong>Example 4<\/strong>\n\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?\n\nSolution\n\n\u0394<em>G<\/em>\u2070<em> = <\/em>\u0394<em>H<\/em>\u2070\u00a0<em>- T<\/em>\u0394<em>S<\/em>\u2070\n\n\u0394<em>G<\/em>\u2070<em> = <\/em>36.2 kJ<em> \u2013<\/em> (298 K x 123 J\/K)\n\n\u0394<em>G<\/em>\u2070<em> = <\/em>-0.4 kJ\n\nTherefore the reaction is spontaneous because \u0394<em>G<\/em>\u2070 is negative.\n\n<\/div>\n\u00a0\n<h2>Determining \u0394<em>G<\/em>\u2070 from Standard Free Energy of Formation<\/h2>\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>.\n\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)\n\nwhere <em>n<\/em> and <em>m<\/em> are the coefficients in the balanced chemical equation of the reaction.\n\nStandard free energies of formation values are listed in the appendix, \"Standard Thermodynamic Quantities for Chemical Substances at 25\u00b0C.\"\n<div class=\"textbox shaded\">\n\n<strong>Example 5<\/strong>\n\nCalculate the standard free energy change for the following reaction, using standard free energies of formation:\n\n5 C(s) + 2 SO<sub>2<\/sub>(g) \u2192 CS<sub>2<\/sub>(g) + 4 CO(g)\n\nIs this a spontaneous reaction?\n\nSolution\n\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)\n\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)]\n\n\u0394<em>G<\/em>\u2070= (-481.7 kJ\/mol) \u2013 (-600.2 kJ\/mol)\n\n\u0394<em>G<\/em>\u2070= 118.5 kJ\/mol\n\n\u0394<em>G<\/em>\u2070\u00a0has a positive value so this is not a spontaneous process.\n\n<\/div>\n\u00a0\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul><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\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>\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>\n<\/ul><\/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\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14215200\/Gibbs_Josiah_Willard-1.jpg\" alt=\"Figure #.#. A portrait of J. Willard Gibbs.\" class=\"alignnone wp-image-2643 size-full\" height=\"543\" width=\"362\" \/><\/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-954-1\" href=\"#footnote-954-1\" 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] = [latex]\\Delta \\textit{S}{}_{sys}[\/latex] + ([latex]\\frac{{-{\\rm \\ }\\Delta {\\rm H}}_{sys}}{T}[\/latex] )<\/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] + ([latex]\\frac{{-{\\rm \\ }\\Delta {\\rm H}}_{sys}}{T}[\/latex] )<\/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<p>\u00a0<\/p>\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<p>\u00a0<\/p>\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\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-954\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Introductory Chemistry- 1st Canadian Edition . <strong>Authored by<\/strong>: Jessie A. Key and David W. Ball. <strong>Provided by<\/strong>: BCCampus. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/opentextbc.ca\/introductorychemistry\/\">https:\/\/opentextbc.ca\/introductorychemistry\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Download this book for free at http:\/\/open.bccampus.ca<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-954-1\">Gibbs Josiah Willard\/Public Domain <a href=\"#return-footnote-954-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":23485,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Introductory Chemistry- 1st Canadian Edition \",\"author\":\"Jessie A. Key and David W. Ball\",\"organization\":\"BCCampus\",\"url\":\"https:\/\/opentextbc.ca\/introductorychemistry\/\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Download this book for free at http:\/\/open.bccampus.ca\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":["jessie-a-key"],"pb_section_license":"cc-by"},"chapter-type":[],"contributor":[59],"license":[50],"class_list":["post-954","chapter","type-chapter","status-publish","hentry","contributor-jessie-a-key","license-cc-by"],"part":944,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/pressbooks\/v2\/chapters\/954","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/wp\/v2\/users\/23485"}],"version-history":[{"count":0,"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/pressbooks\/v2\/chapters\/954\/revisions"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/pressbooks\/v2\/parts\/944"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/pressbooks\/v2\/chapters\/954\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/wp\/v2\/media?parent=954"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=954"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/wp\/v2\/contributor?post=954"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/wp\/v2\/license?post=954"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}