{"id":956,"date":"2017-12-14T21:52:03","date_gmt":"2017-12-14T21:52:03","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/spontaneity-free-energy-and-temperature\/"},"modified":"2017-12-14T21:52:03","modified_gmt":"2017-12-14T21:52:03","slug":"spontaneity-free-energy-and-temperature","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/chapter\/spontaneity-free-energy-and-temperature\/","title":{"raw":"Spontaneity: Free Energy and Temperature","rendered":"Spontaneity: Free Energy and Temperature"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ul><li>To gain an understanding of the relationship between spontaneity, free energy, and temperature.<\/li>\n\t<li>To be able to calculate the temperature at which a process is at equilibrium under standard conditions.<\/li>\n<\/ul><\/div>\nIn the Gibbs free energy change equation, the only part we as scientists can control\u00a0is the temperature. We have seen how we can calculate the standard change in Gibbs free energy, \u0394<em>G<\/em>\u2070, but not all reactions we are interested in occur at exactly 298 K. The temperature plays an important role in determining the Gibbs free energy and spontaneity of a reaction.\n\n\u0394<em>G<\/em> = \u0394<em>H<\/em> - <em>T<\/em>\u0394<em>S<\/em>\n\nIf we examine the Gibbs free energy change equation, we can cluster the components to create two general terms, an enthalpy term, \u0394<em>H<\/em>, and an entropy term, \u2013<em>T<\/em>\u0394<em>S<\/em>. Depending on the sign and magnitude of each, the sum of these terms determines the sign of \u0394<em>G<\/em> and therefore the spontaneity (Table 18.2 \"Spontaneity and the\u00a0Signs of\u00a0Enthalpy and\u00a0Entropy\u00a0Terms\").\n\nTable 18.2. Spontaneity and the Signs of Enthalpy and Entropy Terms\n\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/06\/signs_of_enthalpy_and_entropy_terms_and_spontaneity.png\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14215202\/signs_of_enthalpy_and_entropy_terms_and_spontaneity-1.png\" alt=\"Table #.#. Spontaneity and the signs of enthalpy and entropy terms.\" class=\"alignnone wp-image-2656\" height=\"117\" width=\"400\"\/><\/a>\n\nSince all temperature values are positive in the Kelvin scale, the temperature affects the magnitude of the entropy term. As shown in Table 18.2 \"Spontaneity and the\u00a0Signs of\u00a0Enthalpy and\u00a0Entropy\u00a0Terms,\" the temperature can be the deciding factor in spontaneity when the enthalpy and entropy terms have opposite signs. If \u0394<em>H<\/em> is negative, and \u2013<em>T<\/em>\u0394<em>S<\/em> positive, the reaction will be spontaneous at low temperatures (decreasing the magnitude of the entropy term). If\u00a0\u0394<em>H<\/em> is positive, and \u2013<em>T<\/em>\u0394<em>S<\/em> negative, the reaction will be spontaneous at high temperatures (increasing the magnitude of the entropy term).\n\nSometimes it can be helpful to determine the temperature when \u0394<em>G<\/em>\u2070 = 0 and the process is at equilibrium. Knowing this value, we can adjust the temperature to drive the process to spontaneity or alternatively to prevent the process from occurring spontaneously. Remember that, at equilibrium:\n\n\u0394<em>G<\/em>\u2070 = 0 = \u0394<em>H<\/em>\u2070- <em>T<\/em>\u0394<em>S<\/em>\u2070\n\nWe can rearrange and solve for the temperature <em>T<\/em>:\n\n<em>T<\/em>\u0394<em>S<\/em>\u2070 =\u0394<em>H<\/em>\u2070\n\n<em>T<\/em> =$latex \\frac{{{\\rm \\ }\\Delta {\\rm H}}^{o}}{{\\rm \\ }\\Delta {\\rm S}^{o}}$\n<div class=\"textbox shaded\">\n\n<strong>Example 6<\/strong>\n\nUsing the appendix table of standard thermodynamic quantities, determine the temperature at which the following process is at equilibrium:\n\nCHCl<sub>3<\/sub>(<span class=\"Apple-style-span\" style=\"color: #1f1f1d\">\u2113<\/span>) \u21cc CHCl<sub>3<\/sub>(g)\n\nHow does the value you calculated compare to the boiling point of chloroform given in the literature?\n\nSolution\n\nAt equilibrium: \u0394<em>G<\/em>\u2070 = 0 = \u0394<em>H<\/em>\u2070- <em>T<\/em>\u0394<em>S<\/em>\u2070\n\nWe must estimate \u0394<em>H<\/em>\u2070 and <em>S<\/em>\u2070 from their enthalpies of formation and standard molar entropies, respectively.\n\n\u0394<em>H<\/em>\u2070 = \u2211<em>n\u0394H<\/em>\u2070<em><sub>f<\/sub><\/em>(products) - \u2211<em>m\u0394H<\/em>\u2070<em><sub>f<\/sub><\/em>(reactants)\n\n\u0394<em>H<\/em>\u2070=<em> <span class=\"Apple-style-span\">\u2013<\/span><\/em>102.7 kJ\/mol \u2013 (<span class=\"Apple-style-span\">\u2013<\/span>134.1 kJ\/mol)\n\n\u0394<em>H<\/em>\u2070 = + 31.4 kJ\/mol\n\n\u0394<em>S<\/em>\u2070 = \u2211<em>n\u0394S<\/em>\u2070 (products) - \u2211<em>m\u0394S<\/em>\u2070 (reactants)\n\n\u0394<em>S<\/em>\u2070 = 295.7 J\/mol K \u2013 (201.7 J\/mol K)\n\n\u0394<em>S<\/em>\u2070 = 94.0 J\/mol K (or 94.0 x 10<sup>-3<\/sup> kJ\/mol K)\n\nNow we can use these values to solve for the temperature:\n\n<em>T<\/em> =$latex \\frac{{{\\rm \\ }\\Delta {\\rm H}}^{o}}{{\\rm \\ }\\Delta {\\rm S}^{o}}$\n\n<em>T<\/em> = <em>\u00a0$latex \\textit{ } \\frac{31.4\\ kJ\/mol}{94.0\\ x\\ {10}^{-3}\\ kJ\/mol\\ K} \\textit{ }$<\/em>\n<em>T\u00a0<\/em>= 334 K = 60.9\u2070C\n\nThe literature boiling point of chloroform is 61.2\u2070C. The value we have calculated is very close but slightly lower due to the assumption that \u0394<em>H<\/em>\u2070 and <em>S<\/em>\u2070 do not change with temperature when we estimate the \u0394<em>H<\/em>\u2070 and <em>S<\/em>\u2070 from their enthalpies of formation and standard molar entropies.\n\n<\/div>\n\u00a0\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul><li>The temperature can be the deciding factor in spontaneity when the enthalpy and entropy terms have opposite signs:<\/li>\n<\/ul><p style=\"padding-left: 60px\">- If\u00a0\u0394<em>H<\/em>\u00a0is negative, and <em>\u2013T<\/em>\u0394<em>S\u00a0<\/em>positive, the reaction will be spontaneous at low temperatures (decreasing the magnitude of the entropy term).\n- If\u00a0\u0394<em>H<\/em>\u00a0is positive, and <em>\u2013T<\/em>\u0394<em>S\u00a0<\/em>negative, the reaction will be spontaneous at high temperatures (increasing the magnitude of the entropy term).<\/p>\n\n<\/div>\n\u00a0","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>To gain an understanding of the relationship between spontaneity, free energy, and temperature.<\/li>\n<li>To be able to calculate the temperature at which a process is at equilibrium under standard conditions.<\/li>\n<\/ul>\n<\/div>\n<p>In the Gibbs free energy change equation, the only part we as scientists can control\u00a0is the temperature. We have seen how we can calculate the standard change in Gibbs free energy, \u0394<em>G<\/em>\u2070, but not all reactions we are interested in occur at exactly 298 K. The temperature plays an important role in determining the Gibbs free energy and spontaneity of a reaction.<\/p>\n<p>\u0394<em>G<\/em> = \u0394<em>H<\/em> &#8211; <em>T<\/em>\u0394<em>S<\/em><\/p>\n<p>If we examine the Gibbs free energy change equation, we can cluster the components to create two general terms, an enthalpy term, \u0394<em>H<\/em>, and an entropy term, \u2013<em>T<\/em>\u0394<em>S<\/em>. Depending on the sign and magnitude of each, the sum of these terms determines the sign of \u0394<em>G<\/em> and therefore the spontaneity (Table 18.2 &#8220;Spontaneity and the\u00a0Signs of\u00a0Enthalpy and\u00a0Entropy\u00a0Terms&#8221;).<\/p>\n<p>Table 18.2. Spontaneity and the Signs of Enthalpy and Entropy Terms<\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/06\/signs_of_enthalpy_and_entropy_terms_and_spontaneity.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14215202\/signs_of_enthalpy_and_entropy_terms_and_spontaneity-1.png\" alt=\"Table #.#. Spontaneity and the signs of enthalpy and entropy terms.\" class=\"alignnone wp-image-2656\" height=\"117\" width=\"400\" \/><\/a><\/p>\n<p>Since all temperature values are positive in the Kelvin scale, the temperature affects the magnitude of the entropy term. As shown in Table 18.2 &#8220;Spontaneity and the\u00a0Signs of\u00a0Enthalpy and\u00a0Entropy\u00a0Terms,&#8221; the temperature can be the deciding factor in spontaneity when the enthalpy and entropy terms have opposite signs. If \u0394<em>H<\/em> is negative, and \u2013<em>T<\/em>\u0394<em>S<\/em> positive, the reaction will be spontaneous at low temperatures (decreasing the magnitude of the entropy term). If\u00a0\u0394<em>H<\/em> is positive, and \u2013<em>T<\/em>\u0394<em>S<\/em> negative, the reaction will be spontaneous at high temperatures (increasing the magnitude of the entropy term).<\/p>\n<p>Sometimes it can be helpful to determine the temperature when \u0394<em>G<\/em>\u2070 = 0 and the process is at equilibrium. Knowing this value, we can adjust the temperature to drive the process to spontaneity or alternatively to prevent the process from occurring spontaneously. Remember that, at equilibrium:<\/p>\n<p>\u0394<em>G<\/em>\u2070 = 0 = \u0394<em>H<\/em>\u2070- <em>T<\/em>\u0394<em>S<\/em>\u2070<\/p>\n<p>We can rearrange and solve for the temperature <em>T<\/em>:<\/p>\n<p><em>T<\/em>\u0394<em>S<\/em>\u2070 =\u0394<em>H<\/em>\u2070<\/p>\n<p><em>T<\/em> =[latex]\\frac{{{\\rm \\ }\\Delta {\\rm H}}^{o}}{{\\rm \\ }\\Delta {\\rm S}^{o}}[\/latex]<\/p>\n<div class=\"textbox shaded\">\n<p><strong>Example 6<\/strong><\/p>\n<p>Using the appendix table of standard thermodynamic quantities, determine the temperature at which the following process is at equilibrium:<\/p>\n<p>CHCl<sub>3<\/sub>(<span class=\"Apple-style-span\" style=\"color: #1f1f1d\">\u2113<\/span>) \u21cc CHCl<sub>3<\/sub>(g)<\/p>\n<p>How does the value you calculated compare to the boiling point of chloroform given in the literature?<\/p>\n<p>Solution<\/p>\n<p>At equilibrium: \u0394<em>G<\/em>\u2070 = 0 = \u0394<em>H<\/em>\u2070- <em>T<\/em>\u0394<em>S<\/em>\u2070<\/p>\n<p>We must estimate \u0394<em>H<\/em>\u2070 and <em>S<\/em>\u2070 from their enthalpies of formation and standard molar entropies, respectively.<\/p>\n<p>\u0394<em>H<\/em>\u2070 = \u2211<em>n\u0394H<\/em>\u2070<em><sub>f<\/sub><\/em>(products) &#8211; \u2211<em>m\u0394H<\/em>\u2070<em><sub>f<\/sub><\/em>(reactants)<\/p>\n<p>\u0394<em>H<\/em>\u2070=<em> <span class=\"Apple-style-span\">\u2013<\/span><\/em>102.7 kJ\/mol \u2013 (<span class=\"Apple-style-span\">\u2013<\/span>134.1 kJ\/mol)<\/p>\n<p>\u0394<em>H<\/em>\u2070 = + 31.4 kJ\/mol<\/p>\n<p>\u0394<em>S<\/em>\u2070 = \u2211<em>n\u0394S<\/em>\u2070 (products) &#8211; \u2211<em>m\u0394S<\/em>\u2070 (reactants)<\/p>\n<p>\u0394<em>S<\/em>\u2070 = 295.7 J\/mol K \u2013 (201.7 J\/mol K)<\/p>\n<p>\u0394<em>S<\/em>\u2070 = 94.0 J\/mol K (or 94.0 x 10<sup>-3<\/sup> kJ\/mol K)<\/p>\n<p>Now we can use these values to solve for the temperature:<\/p>\n<p><em>T<\/em> =[latex]\\frac{{{\\rm \\ }\\Delta {\\rm H}}^{o}}{{\\rm \\ }\\Delta {\\rm S}^{o}}[\/latex]<\/p>\n<p><em>T<\/em> = <em>\u00a0[latex]\\textit{ } \\frac{31.4\\ kJ\/mol}{94.0\\ x\\ {10}^{-3}\\ kJ\/mol\\ K} \\textit{ }[\/latex]<\/em><br \/>\n<em>T\u00a0<\/em>= 334 K = 60.9\u2070C<\/p>\n<p>The literature boiling point of chloroform is 61.2\u2070C. The value we have calculated is very close but slightly lower due to the assumption that \u0394<em>H<\/em>\u2070 and <em>S<\/em>\u2070 do not change with temperature when we estimate the \u0394<em>H<\/em>\u2070 and <em>S<\/em>\u2070 from their enthalpies of formation and standard molar entropies.<\/p>\n<\/div>\n<p>\u00a0<\/p>\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul>\n<li>The temperature can be the deciding factor in spontaneity when the enthalpy and entropy terms have opposite signs:<\/li>\n<\/ul>\n<p style=\"padding-left: 60px\">&#8211; If\u00a0\u0394<em>H<\/em>\u00a0is negative, and <em>\u2013T<\/em>\u0394<em>S\u00a0<\/em>positive, the reaction will be spontaneous at low temperatures (decreasing the magnitude of the entropy term).<br \/>\n&#8211; If\u00a0\u0394<em>H<\/em>\u00a0is positive, and <em>\u2013T<\/em>\u0394<em>S\u00a0<\/em>negative, the reaction will be spontaneous at high temperatures (increasing the magnitude of the entropy term).<\/p>\n<\/div>\n<p>\u00a0<\/p>\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-956\">\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>","protected":false},"author":23485,"menu_order":6,"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-956","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\/956","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\/956\/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\/956\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/wp\/v2\/media?parent=956"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=956"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/wp\/v2\/contributor?post=956"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/wp-json\/wp\/v2\/license?post=956"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}