{"id":1082,"date":"2016-01-11T20:02:55","date_gmt":"2016-01-11T20:02:55","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/measuring-entropy-and-entropy-changes-2\/"},"modified":"2020-04-20T16:45:49","modified_gmt":"2020-04-20T16:45:49","slug":"measuring-entropy-and-entropy-changes","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/measuring-entropy-and-entropy-changes\/","title":{"raw":"Measuring Entropy and Entropy Changes","rendered":"Measuring Entropy and Entropy Changes"},"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 methods of measuring entropy and entropy change.<\/li>\r\n<\/ul>\r\n<\/div>\r\nAs the temperature of a sample decreases, its\u00a0kinetic energy decreases and, correspondingly, the number of microstates possible decreases. The third law of thermodynamics states:\u00a0<em>at absolute zero (0 K), the entropy of a pure, perfect crystal is zero<\/em>. In other words, at absolute zero, there is only one microstate and according to Boltzmann\u2019s equation:\r\n\r\n<em>S<\/em> = <em>k<\/em> ln <em>W<\/em> = <em>k<\/em> ln 1 =\u00a00\r\n\r\nUsing this as a reference point, the entropy of a substance can be obtained by measuring the heat required to raise the temperature a given amount, using a reversible process. Reversible heating requires very slow and very small increases in heat.\r\n\r\n\u0394<em>S<\/em> = $latex \\frac{q_{rev}}{T}$\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>Example 1<\/strong>\r\n\r\nDetermine the change in entropy (in J\/K) of water when 425 kJ of heat is applied to it at 50<sup>o<\/sup>C. Assume the change is reversible and the temperature remains constant.\r\n\r\nSolution\r\n\r\n\u0394<em>S<\/em> = $latex \\frac{q_{rev}}{T}$ = $latex \\frac{425\\ kJ}{323.15\\ K}$ = $latex \\frac{4.25\\ x\\ {10}^5\\ J}{323.15\\ K}$ = 1.32 x $${10}^5$$ J\/K\r\n\r\n<\/div>\r\n<h2>Standard Molar Entropy, <em>S<\/em><sup>o<\/sup><\/h2>\r\nThe <a class=\"glossterm\">standard molar entropy<\/a>, <em>S<\/em><sup>o<\/sup>, is the entropy of 1 mole of a substance in its standard state, at 1 atm of pressure. These values have been tabulated, and selected substances are listed in Table 18.1 \"Standard Molar Entropies of Selected Substances\u00a0at 298 K.\"\r\n\r\nTable 18.1. Standard Molar Entropies of Selected Substances at 298 K[footnote]Courtesy UC Davis ChemWiki by University of California\\CC-BY-SA-3.0[\/footnote]\r\n\r\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/06\/Standard_Molar_Entropy_Table_.png\"><img class=\"alignnone wp-image-1080\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/291\/2016\/01\/Standard_Molar_Entropy_Table_-1.png\" alt=\"Table #.#. Standard molar entropies of selected substances at 298 K.\" width=\"400\" height=\"159\" \/><\/a>\r\n\r\nSeveral trends emerge from standard molar entropy data:\r\n<ul>\r\n \t<li>Larger, more complex molecules have higher standard molar enthalpy values than smaller or simpler molecules. There are more possible arrangements of atoms in space for larger, more complex molecules, increasing the number of\u00a0possible microstates.<\/li>\r\n \t<li>Gases tend to have much larger standard molar enthalpies than liquids, and liquids tend to have larger values than solids, when comparing the same or similar substances.<\/li>\r\n \t<li>The standard molar entropy of any substance increases as the temperature increases. This can be seen in Figure 18.3 \"Entropy vs.\u00a0Temperature of a\u00a0Single\u00a0Substance.\" Large jumps in entropy occur at the phase changes: solid to liquid and liquid to gas. These large increases occur due to sudden increased molecular mobility and larger available volumes associated with the phase changes.<\/li>\r\n<\/ul>\r\n<span class=\"Apple-style-span\">Figure 18.3. Entropy vs.\u00a0Temperature of a Single Substance<\/span>\r\n\r\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/06\/generalized-plot-of-entropy-versus-temperature-of-a-single-substance.jpg\"><img class=\"alignnone wp-image-1081\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/291\/2019\/08\/generalized-plot-of-entropy-versus-temperature-of-a-single-substance-1.jpg\" alt=\"Figure #.#. Generalized plot of entropy versus temperature of a single substance.\" width=\"400\" height=\"429\" \/><\/a>\r\n\r\n<span class=\"Apple-style-span\">This is a generalized plot of entropy versus temperature for\u00a0a single substance.<\/span>[footnote]Courtesy UC Davis ChemWiki by University of California\\CC-BY-SA-3.0[\/footnote]\r\n<h2>Standard Entropy Change of a Reaction,\u00a0\u0394<em>S<\/em>\u2070<\/h2>\r\nThe entropy change of a reaction where the reactants and products are in their standard state can be determined using the following equation:\r\n\r\n\u0394<em>S<\/em>\u2070 = \u2211<em>nS<\/em>\u2070(products) - \u2211<em>mS<\/em>\u2070(reactants)\r\n\r\nwhere <em>n<\/em> and <em>m<\/em> are the coefficients found in the balanced chemical equation of the reaction.\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>Example 2<\/strong>\r\n\r\nDetermine the change in the standard entropy, \u0394<em>S<\/em>\u2070, for the synthesis of carbon dioxide from graphite and oxygen:\r\n\r\nC(s) + O<sub>2<\/sub>(g) \u2192 CO<sub>2<\/sub>(g)\r\n\r\nSolution\r\n\r\n\u0394<em>S<\/em>\u2070 = \u2211<em>nS<\/em>\u2070(products) - \u2211<em>mS<\/em>\u2070(reactants)\r\n\r\n\u0394<em>S<\/em>\u2070 = (213.8 J\/mol K) \u2013 (205.2 J\/mol K + 5.7 J\/mol K)\r\n\r\n\u0394<em>S<\/em>\u2070 = +2.9 J\/mol K\r\n\r\n<\/div>\r\n<h2>Entropy Changes in the Surroundings<\/h2>\r\nThe second law of thermodynamics states that a spontaneous reaction will result in an increase of entropy in the universe. The universe comprises both the system being examined and its surroundings.\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\nStandard entropy change can also be calculated by the following:\r\n\r\n\u0394<em>S<\/em>\u2070<sub>universe<\/sub> = \u0394<em>S<\/em>\u2070<sub>sys<\/sub> + \u0394<em>S<\/em>\u2070<sub>surr<\/sub>\r\n\r\nThe change in entropy of the surroundings is essentially just a measure of how much energy is being taken in or given off by the system. Under isothermal conditions, we can express the entropy change of the surroundings as:\r\n\r\n\u0394<em>S<\/em><sub>surr<\/sub> =\u00a0$latex \\frac{{-q}_{sys}}{T}$\u00a0 or \u0394<em>S<\/em><sub>surr<\/sub> =\u00a0 $latex \\frac{{-{\\rm \\ }\\Delta {\\rm H}}_{sys}}{T}$ (at constant pressure)\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>Example 3<\/strong>\r\n\r\nFor the previous example, the change in the standard entropy, \u0394<em>S<\/em>\u2070, for the synthesis of carbon dioxide from graphite and oxygen, use the previously calculated \u0394<em>S<\/em>\u2070<sub>sys<\/sub>\u00a0and standard enthalpy of formation values to determine <em>S<\/em>\u2070<sub>surr<\/sub>\u00a0and \u0394<em>S<\/em>\u2070<sub>universe<\/sub>.\r\n\r\nSolution\r\n\r\nFirst we should solve for the \u0394<em>H<\/em>\u2070<sub>sys<\/sub> using the standard enthalpies of formation values:\r\n\r\n\u0394<em>H<\/em>\u2070<sub>sys<\/sub> = \u0394<em>H<\/em>\u2070<sub>f<\/sub> [CO<sub>2<\/sub>(g)] - \u0394<em>H<\/em>\u2070<sub>f<\/sub> [C(s) + O<sub>2<\/sub>(g)]\r\n\r\n\u0394<em>H<\/em>\u2070<sub>sys<\/sub> = (-393.5 kJ\/mol) \u2013 (0 kJ\/mol + 0 kJ\/mol)\r\n\r\n\u0394<em>H<\/em>\u2070<sub>sys<\/sub> = -393.5 kJ\/mol\r\n\r\nNow we can convert this to the \u0394<em>S<\/em>\u2070<sub>surr<\/sub>:\r\n\r\n\u0394<em>S<\/em>\u2070<sub>surr<\/sub>\u00a0= \u00a0$latex \\frac{{-{\\rm \\ }\\Delta {\\rm H}}_{sys}}{T}$\u00a0= (-393.5 kJ\/mol)\/(298 K) = -1.32 kJ\/mol K\r\n\r\nFinally, solve for \u0394<em>S<\/em>\u2070<sub>universe<\/sub>:\r\n\r\n\u0394<em>S<\/em>\u2070<sub>universe<\/sub> = \u0394<em>S<\/em>\u2070<sub>sys<\/sub> + \u0394<em>S<\/em>\u2070<sub>surr<\/sub>\r\n\r\n\u0394<em>S<\/em>\u2070<sub>universe<\/sub> = (+2.9 J\/mol K) + (-1.32 x 10<sup>3<\/sup> J\/mol K)\r\n\r\n\u0394<em>S<\/em>\u2070<sub>universe<\/sub> = -1.3 x 10<sup>3<\/sup> J\/mol K\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>At absolute zero (0 K), the entropy of a pure, perfect crystal is zero.<\/li>\r\n \t<li>The entropy of a substance can be obtained by measuring the heat required to raise the temperature a given amount, using a reversible process.<\/li>\r\n \t<li>The standard molar entropy, <em>S<\/em><sup>o<\/sup>, is the entropy of 1 mole of a substance in its standard state, at 1 atm of pressure.<\/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 methods of measuring entropy and entropy change.<\/li>\n<\/ul>\n<\/div>\n<p>As the temperature of a sample decreases, its\u00a0kinetic energy decreases and, correspondingly, the number of microstates possible decreases. The third law of thermodynamics states:\u00a0<em>at absolute zero (0 K), the entropy of a pure, perfect crystal is zero<\/em>. In other words, at absolute zero, there is only one microstate and according to Boltzmann\u2019s equation:<\/p>\n<p><em>S<\/em> = <em>k<\/em> ln <em>W<\/em> = <em>k<\/em> ln 1 =\u00a00<\/p>\n<p>Using this as a reference point, the entropy of a substance can be obtained by measuring the heat required to raise the temperature a given amount, using a reversible process. Reversible heating requires very slow and very small increases in heat.<\/p>\n<p>\u0394<em>S<\/em> = $latex \\frac{q_{rev}}{T}$<\/p>\n<div class=\"textbox shaded\">\n<p><strong>Example 1<\/strong><\/p>\n<p>Determine the change in entropy (in J\/K) of water when 425 kJ of heat is applied to it at 50<sup>o<\/sup>C. Assume the change is reversible and the temperature remains constant.<\/p>\n<p>Solution<\/p>\n<p>\u0394<em>S<\/em> = $latex \\frac{q_{rev}}{T}$ = $latex \\frac{425\\ kJ}{323.15\\ K}$ = $latex \\frac{4.25\\ x\\ {10}^5\\ J}{323.15\\ K}$ = 1.32 x $${10}^5$$ J\/K<\/p>\n<\/div>\n<h2>Standard Molar Entropy, <em>S<\/em><sup>o<\/sup><\/h2>\n<p>The <a class=\"glossterm\">standard molar entropy<\/a>, <em>S<\/em><sup>o<\/sup>, is the entropy of 1 mole of a substance in its standard state, at 1 atm of pressure. These values have been tabulated, and selected substances are listed in Table 18.1 &#8220;Standard Molar Entropies of Selected Substances\u00a0at 298 K.&#8221;<\/p>\n<p>Table 18.1. Standard Molar Entropies of Selected Substances at 298 K<a class=\"footnote\" title=\"Courtesy UC Davis ChemWiki by University of California\\CC-BY-SA-3.0\" id=\"return-footnote-1082-1\" href=\"#footnote-1082-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a><\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/06\/Standard_Molar_Entropy_Table_.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1080\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/291\/2016\/01\/Standard_Molar_Entropy_Table_-1.png\" alt=\"Table #.#. Standard molar entropies of selected substances at 298 K.\" width=\"400\" height=\"159\" srcset=\"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2016\/01\/Standard_Molar_Entropy_Table_-1.png 1070w, https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2016\/01\/Standard_Molar_Entropy_Table_-1-300x119.png 300w, https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2016\/01\/Standard_Molar_Entropy_Table_-1-768x304.png 768w, https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2016\/01\/Standard_Molar_Entropy_Table_-1-1024x406.png 1024w, https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2016\/01\/Standard_Molar_Entropy_Table_-1-65x26.png 65w, https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2016\/01\/Standard_Molar_Entropy_Table_-1-225x89.png 225w, https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2016\/01\/Standard_Molar_Entropy_Table_-1-350x139.png 350w\" sizes=\"auto, (max-width: 400px) 100vw, 400px\" \/><\/a><\/p>\n<p>Several trends emerge from standard molar entropy data:<\/p>\n<ul>\n<li>Larger, more complex molecules have higher standard molar enthalpy values than smaller or simpler molecules. There are more possible arrangements of atoms in space for larger, more complex molecules, increasing the number of\u00a0possible microstates.<\/li>\n<li>Gases tend to have much larger standard molar enthalpies than liquids, and liquids tend to have larger values than solids, when comparing the same or similar substances.<\/li>\n<li>The standard molar entropy of any substance increases as the temperature increases. This can be seen in Figure 18.3 &#8220;Entropy vs.\u00a0Temperature of a\u00a0Single\u00a0Substance.&#8221; Large jumps in entropy occur at the phase changes: solid to liquid and liquid to gas. These large increases occur due to sudden increased molecular mobility and larger available volumes associated with the phase changes.<\/li>\n<\/ul>\n<p><span class=\"Apple-style-span\">Figure 18.3. Entropy vs.\u00a0Temperature of a Single Substance<\/span><\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/06\/generalized-plot-of-entropy-versus-temperature-of-a-single-substance.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1081\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/291\/2019\/08\/generalized-plot-of-entropy-versus-temperature-of-a-single-substance-1.jpg\" alt=\"Figure #.#. Generalized plot of entropy versus temperature of a single substance.\" width=\"400\" height=\"429\" srcset=\"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2019\/08\/generalized-plot-of-entropy-versus-temperature-of-a-single-substance-1.jpg 1161w, https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2019\/08\/generalized-plot-of-entropy-versus-temperature-of-a-single-substance-1-280x300.jpg 280w, https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2019\/08\/generalized-plot-of-entropy-versus-temperature-of-a-single-substance-1-768x824.jpg 768w, https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2019\/08\/generalized-plot-of-entropy-versus-temperature-of-a-single-substance-1-955x1024.jpg 955w, https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2019\/08\/generalized-plot-of-entropy-versus-temperature-of-a-single-substance-1-65x70.jpg 65w, https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2019\/08\/generalized-plot-of-entropy-versus-temperature-of-a-single-substance-1-225x241.jpg 225w, https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-content\/uploads\/sites\/291\/2019\/08\/generalized-plot-of-entropy-versus-temperature-of-a-single-substance-1-350x375.jpg 350w\" sizes=\"auto, (max-width: 400px) 100vw, 400px\" \/><\/a><\/p>\n<p><span class=\"Apple-style-span\">This is a generalized plot of entropy versus temperature for\u00a0a single substance.<\/span><a class=\"footnote\" title=\"Courtesy UC Davis ChemWiki by University of California\\CC-BY-SA-3.0\" id=\"return-footnote-1082-2\" href=\"#footnote-1082-2\" aria-label=\"Footnote 2\"><sup class=\"footnote\">[2]<\/sup><\/a><\/p>\n<h2>Standard Entropy Change of a Reaction,\u00a0\u0394<em>S<\/em>\u2070<\/h2>\n<p>The entropy change of a reaction where the reactants and products are in their standard state can be determined using the following equation:<\/p>\n<p>\u0394<em>S<\/em>\u2070 = \u2211<em>nS<\/em>\u2070(products) &#8211; \u2211<em>mS<\/em>\u2070(reactants)<\/p>\n<p>where <em>n<\/em> and <em>m<\/em> are the coefficients found in the balanced chemical equation of the reaction.<\/p>\n<div class=\"textbox shaded\">\n<p><strong>Example 2<\/strong><\/p>\n<p>Determine the change in the standard entropy, \u0394<em>S<\/em>\u2070, for the synthesis of carbon dioxide from graphite and oxygen:<\/p>\n<p>C(s) + O<sub>2<\/sub>(g) \u2192 CO<sub>2<\/sub>(g)<\/p>\n<p>Solution<\/p>\n<p>\u0394<em>S<\/em>\u2070 = \u2211<em>nS<\/em>\u2070(products) &#8211; \u2211<em>mS<\/em>\u2070(reactants)<\/p>\n<p>\u0394<em>S<\/em>\u2070 = (213.8 J\/mol K) \u2013 (205.2 J\/mol K + 5.7 J\/mol K)<\/p>\n<p>\u0394<em>S<\/em>\u2070 = +2.9 J\/mol K<\/p>\n<\/div>\n<h2>Entropy Changes in the Surroundings<\/h2>\n<p>The second law of thermodynamics states that a spontaneous reaction will result in an increase of entropy in the universe. The universe comprises both the system being examined and its surroundings.<\/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>Standard entropy change can also be calculated by the following:<\/p>\n<p>\u0394<em>S<\/em>\u2070<sub>universe<\/sub> = \u0394<em>S<\/em>\u2070<sub>sys<\/sub> + \u0394<em>S<\/em>\u2070<sub>surr<\/sub><\/p>\n<p>The change in entropy of the surroundings is essentially just a measure of how much energy is being taken in or given off by the system. Under isothermal conditions, we can express the entropy change of the surroundings as:<\/p>\n<p>\u0394<em>S<\/em><sub>surr<\/sub> =\u00a0$latex \\frac{{-q}_{sys}}{T}$\u00a0 or \u0394<em>S<\/em><sub>surr<\/sub> =\u00a0 $latex \\frac{{-{\\rm \\ }\\Delta {\\rm H}}_{sys}}{T}$ (at constant pressure)<\/p>\n<div class=\"textbox shaded\">\n<p><strong>Example 3<\/strong><\/p>\n<p>For the previous example, the change in the standard entropy, \u0394<em>S<\/em>\u2070, for the synthesis of carbon dioxide from graphite and oxygen, use the previously calculated \u0394<em>S<\/em>\u2070<sub>sys<\/sub>\u00a0and standard enthalpy of formation values to determine <em>S<\/em>\u2070<sub>surr<\/sub>\u00a0and \u0394<em>S<\/em>\u2070<sub>universe<\/sub>.<\/p>\n<p>Solution<\/p>\n<p>First we should solve for the \u0394<em>H<\/em>\u2070<sub>sys<\/sub> using the standard enthalpies of formation values:<\/p>\n<p>\u0394<em>H<\/em>\u2070<sub>sys<\/sub> = \u0394<em>H<\/em>\u2070<sub>f<\/sub> [CO<sub>2<\/sub>(g)] &#8211; \u0394<em>H<\/em>\u2070<sub>f<\/sub> [C(s) + O<sub>2<\/sub>(g)]<\/p>\n<p>\u0394<em>H<\/em>\u2070<sub>sys<\/sub> = (-393.5 kJ\/mol) \u2013 (0 kJ\/mol + 0 kJ\/mol)<\/p>\n<p>\u0394<em>H<\/em>\u2070<sub>sys<\/sub> = -393.5 kJ\/mol<\/p>\n<p>Now we can convert this to the \u0394<em>S<\/em>\u2070<sub>surr<\/sub>:<\/p>\n<p>\u0394<em>S<\/em>\u2070<sub>surr<\/sub>\u00a0= \u00a0$latex \\frac{{-{\\rm \\ }\\Delta {\\rm H}}_{sys}}{T}$\u00a0= (-393.5 kJ\/mol)\/(298 K) = -1.32 kJ\/mol K<\/p>\n<p>Finally, solve for \u0394<em>S<\/em>\u2070<sub>universe<\/sub>:<\/p>\n<p>\u0394<em>S<\/em>\u2070<sub>universe<\/sub> = \u0394<em>S<\/em>\u2070<sub>sys<\/sub> + \u0394<em>S<\/em>\u2070<sub>surr<\/sub><\/p>\n<p>\u0394<em>S<\/em>\u2070<sub>universe<\/sub> = (+2.9 J\/mol K) + (-1.32 x 10<sup>3<\/sup> J\/mol K)<\/p>\n<p>\u0394<em>S<\/em>\u2070<sub>universe<\/sub> = -1.3 x 10<sup>3<\/sup> J\/mol K<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul>\n<li>At absolute zero (0 K), the entropy of a pure, perfect crystal is zero.<\/li>\n<li>The entropy of a substance can be obtained by measuring the heat required to raise the temperature a given amount, using a reversible process.<\/li>\n<li>The standard molar entropy, <em>S<\/em><sup>o<\/sup>, is the entropy of 1 mole of a substance in its standard state, at 1 atm of pressure.<\/li>\n<\/ul>\n<\/div>\n<hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-1082-1\">Courtesy UC Davis ChemWiki by University of California\\CC-BY-SA-3.0 <a href=\"#return-footnote-1082-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><li id=\"footnote-1082-2\">Courtesy UC Davis ChemWiki by University of California\\CC-BY-SA-3.0 <a href=\"#return-footnote-1082-2\" class=\"return-footnote\" aria-label=\"Return to footnote 2\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":124,"menu_order":3,"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-1082","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\/1082","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\/1082\/revisions"}],"predecessor-version":[{"id":1458,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters\/1082\/revisions\/1458"}],"part":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/parts\/1071"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapters\/1082\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/media?parent=1082"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/pressbooks\/v2\/chapter-type?post=1082"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/contributor?post=1082"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistryclone\/wp-json\/wp\/v2\/license?post=1082"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}