{"id":3622,"date":"2015-05-06T03:51:00","date_gmt":"2015-05-06T03:51:00","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=3622"},"modified":"2016-10-20T20:21:22","modified_gmt":"2016-10-20T20:21:22","slug":"the-second-and-third-laws-of-thermodynamics","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/chapter\/the-second-and-third-laws-of-thermodynamics\/","title":{"raw":"The Second and Third Laws of Thermodynamics","rendered":"The Second and Third Laws of Thermodynamics"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>State and explain the second and third laws of thermodynamics<\/li>\r\n \t<li>Calculate entropy changes for phase transitions and chemical reactions under standard conditions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>The Second Law of Thermodynamics<\/h2>\r\nIn the quest to identify a property that may reliably predict the spontaneity of a process, we have identified a very promising candidate: entropy. Processes that involve an increase in entropy <em>of the system<\/em> (\u0394<em>S<\/em> &gt; 0) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include <em>the surroundings<\/em>, we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true:\r\n<p style=\"text-align: center;\">[latex]\\Delta {S}_{\\text{univ}}=\\Delta {S}_{\\text{sys}}+\\Delta {S}_{\\text{surr}}[\/latex]<\/p>\r\nTo illustrate this relation, consider again the process of heat flow between two objects, one identified as the system and the other as the surroundings. There are three possibilities for such a process:\r\n<ol>\r\n \t<li>The objects are at different temperatures, and heat flows from the hotter to the cooler object. <em>This is always observed to occur spontaneously.<\/em> Designating the hotter object as the system and invoking the definition of entropy yields the following:\r\n[latex]\\Delta {S}_{\\text{sys}}=\\frac{-{q}_{\\text{rev}}}{{T}_{\\text{sys}}}\\text{and}\\Delta {S}_{\\text{surr}}=\\frac{{q}_{\\text{rev}}}{{T}_{\\text{surr}}}[\/latex]\r\nThe arithmetic signs of <em>q<\/em><sub>rev<\/sub> denote the loss of heat by the system and the gain of heat by the surroundings. Since <em>T<\/em><sub>sys<\/sub> &gt; <em>T<\/em><sub>surr<\/sub> in this scenario, the magnitude of the entropy change for the surroundings will be greater than that for the system, and so the sum of \u0394<em>S<\/em><sub>sys<\/sub> and \u0394<em>S<\/em><sub>surr<\/sub> will yield a positive value for \u0394<em>S<\/em><sub>univ<\/sub>. <em>This process involves an increase in the entropy of the universe.<\/em><\/li>\r\n \t<li>The objects are at different temperatures, and heat flows from the cooler to the hotter object. <em>This is never observed to occur spontaneously.<\/em> Again designating the hotter object as the system and invoking the definition of entropy yields the following:\r\n[latex]\\Delta {S}_{\\text{sys}}=\\frac{{q}_{\\text{rev}}}{{T}_{\\text{sys}}}\\text{and}\\Delta {S}_{\\text{surr}}=\\frac{-{q}_{\\text{rev}}}{{T}_{\\text{surr}}}[\/latex]\r\nThe arithmetic signs of <em>q<\/em><sub>rev<\/sub> denote the gain of heat by the system and the loss of heat by the surroundings. The magnitude of the entropy change for the surroundings will again be greater than that for the system, but in this case, the signs of the heat changes will yield a negative value for \u0394<em>S<\/em><sub>univ<\/sub>. <em>This process involves a decrease in the entropy of the universe.<\/em><\/li>\r\n \t<li>The temperature difference between the objects is infinitesimally small, <em>T<\/em><sub>sys<\/sub> \u2248 <em>T<\/em><sub>surr<\/sub>, and so the heat flow is thermodynamically reversible. See the previous section\u2019s discussion). In this case, the system and surroundings experience entropy changes that are equal in magnitude and therefore sum to yield a value of zero for \u0394<em>S<\/em><sub>univ<\/sub>. <em>This process involves no change in the entropy of the universe.<\/em><\/li>\r\n<\/ol>\r\nThese results lead to a profound statement regarding the relation between entropy and spontaneity known as the <strong>second law of thermodynamics<\/strong>: <em>all spontaneous changes cause an increase in the entropy of the universe.<\/em> A summary of these three relations is provided in Table\u00a01.\r\n<table id=\"fs-idp41455824\" class=\"span-all\" summary=\"This table contains two 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: \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=\"2\">Table\u00a01. The Second Law of Thermodynamics<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>\u0394<em>S<\/em><sub>univ<\/sub> &gt; 0<\/td>\r\n<td>spontaneous<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>\u0394<em>S<\/em><sub>univ<\/sub> &lt; 0<\/td>\r\n<td>nonspontaneous (spontaneous in opposite direction)<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>\u0394<em>S<\/em><sub>univ<\/sub> = 0<\/td>\r\n<td>reversible (system is at equilibrium)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFor many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. As a result, <em>q<\/em><sub>surr<\/sub> is a good approximation of <em>q<\/em><sub>rev<\/sub>, and the second law may be stated as the following:\r\n<p style=\"text-align: center;\">[latex]\\Delta {S}_{\\text{univ}}=\\Delta {S}_{\\text{sys}}+\\Delta {S}_{\\text{surr}}=\\Delta {S}_{\\text{sys}}+\\frac{{q}_{\\text{surr}}}{T}[\/latex]<\/p>\r\nWe may use this equation to predict the spontaneity of a process as illustrated in Example 1.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0<strong>Will Ice Spontaneously Melt?<\/strong><\/h3>\r\nThe entropy change for the process is\u00a0[latex]{\\text{H}}_{2}\\text{O}\\left(s\\right)\\longrightarrow {\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]\r\n\r\nis 22.1 J\/K and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at \u221210.00 \u00b0C? Is it spontaneous at +10.00 \u00b0C?\r\n\r\n[reveal-answer q=\"692410\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"692410\"]\r\n\r\nWe can assess the spontaneity of the process by calculating the entropy change of the universe. If \u0394<em>S<\/em><sub>univ<\/sub> is positive, then the process is spontaneous. At both temperatures, \u0394<em>S<\/em><sub>sys<\/sub> = 22.1 J\/K and <em>q<\/em><sub>surr<\/sub> = \u22126.00 kJ.\r\n\r\nAt \u221210.00 \u00b0C (263.15 K), the following is true:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\Delta {S}_{\\text{univ}}&amp; =\\Delta {S}_{\\text{sys}}+\\Delta {S}_{\\text{surr}}=\\Delta {S}_{\\text{sys}}+\\frac{{q}_{\\text{surr}}}{T}\\hfill \\\\ &amp; =\\text{22.1 J\/K}+\\frac{-6.00\\times {10}^{3}\\text{J}}{\\text{263.15 K}}=-0.7\\text{J\/K}\\hfill \\end{array}[\/latex]<\/p>\r\n<em>S<\/em><sub>univ<\/sub> &lt; 0, so melting is nonspontaneous (<em>not<\/em> spontaneous) at \u221210.0 \u00b0C.\r\n\r\nAt 10.00 \u00b0C (283.15 K), the following is true:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ \\Delta {S}_{\\text{univ}}&amp;=&amp;\\Delta {S}_{\\text{sys}}+\\frac{{q}_{\\text{surr}}}{T}\\\\{}&amp;=&amp;22.1\\text{J\/K}+\\frac{-6.00\\times {10}^{3}\\text{J}}{\\text{283.15 K}}=\\text{+0.9 J\/K}\\end{array}[\/latex]<\/p>\r\n<em>S<\/em><sub>univ<\/sub> &gt; 0, so melting <em>is<\/em> spontaneous at 10.00 \u00b0C.\r\n\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idp54808912\"><strong>Check Your Learning<\/strong><\/h4>\r\nUsing this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of <em>S<\/em><sub>univ<\/sub>?\r\n\r\n[reveal-answer q=\"686682\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"686682\"]Entropy is a state function, and freezing is the opposite of melting. At \u221210.00 \u00b0C spontaneous, +0.7 J\/K; at +10.00 \u00b0C nonspontaneous, \u22120.9 J\/K.[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>The Third Law of Thermodynamics<\/h2>\r\nThe previous section described the various contributions of matter and energy dispersal that contribute to the entropy of a system. With these contributions in mind, consider the entropy of a pure, perfectly crystalline solid possessing no kinetic energy (that is, at a temperature of absolute zero, 0 K). This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion means there is but one possible location for each identical atom or molecule comprising the crystal (<em>W<\/em> = 1). According to the Boltzmann equation, the entropy of this system is zero.\r\n<p style=\"text-align: center;\">[latex]S=k\\text{ln}W=k\\text{ln}\\left(1\\right)=0[\/latex]<\/p>\r\nThis limiting condition for a system\u2019s entropy represents the <strong>third law of thermodynamics<\/strong>: <em>the entropy of a pure, perfect crystalline substance at 0 K is zero.<\/em>\r\n\r\nWe can make careful calorimetric measurements to determine the temperature dependence of a substance\u2019s entropy and to derive absolute entropy values under specific conditions. <strong>Standard entropies<\/strong> are given the label [latex]{S}_{298}^{\\circ }[\/latex] for values determined for one mole of substance at a pressure of 1 bar and a temperature of 298 K. The <strong>standard entropy change (\u0394<em>S<\/em>\u00b0)<\/strong> for any process may be computed from the standard entropies of its reactant and product species like the following:\r\n<p style=\"text-align: center;\">[latex]\\Delta S\\text{\\circ }=\\sum \\nu {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu {S}_{298}^{\\circ }\\text{(reactants)}[\/latex]<\/p>\r\nHere, \u03bd represents stoichiometric coefficients in the balanced equation representing the process. For example, \u0394<em>S<\/em>\u00b0 for the following reaction at room temperature\u00a0[latex]m\\text{A}+n\\text{B}\\longrightarrow x\\text{C}+y\\text{D}[\/latex]\u00a0is computed as the following:\r\n<p style=\"text-align: center;\">[latex]=\\left[x{S}_{298}^{\\circ }\\left(\\text{C}\\right)+y{S}_{298}^{\\circ }\\left(\\text{D}\\right)\\right]-\\left[m{S}_{298}^{\\circ }\\left(\\text{A}\\right)+n{S}_{298}^{\\circ }\\left(\\text{B}\\right)\\right][\/latex]<\/p>\r\nTable\u00a02\u00a0lists some standard entropies at 298.15 K. You can find additional standard entropies in <a class=\"target-chapter\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\">Standard Thermodynamic Properties for Selected Substances<\/a>.\r\n<table id=\"fs-idm78597984\" class=\"span-all\" summary=\"The table has two columns and twenty rows. The first row is a header row and it labels the columns, \u201cSubstance,\u201d and \u201cS subscript 298 superscript degree symbol ( J mol superscript negative 1 K superscript negative 1 ).\u201d The second row spans both columns and contains the word, \u201cCarbon.\u201d Under the \u201cSubstance\u201d column for carbon are the following: C ( s, graphite ), C ( s, diamond ), C O ( g ), C O subscript 2 ( g ), C H subscript 4 ( g ), C subscript 2 H subscript 4 ( g ), C subscript 2 H subscript 6 ( g ), C H subscript 3 O H ( l ), and C subscript 2 H subscript 5 O H ( l ). Under the \u201cS subscript 298 superscript degree symbol ( J mol superscript negative 1 K superscript negative 1 )\u201d column for carbon are the following: 5.740, 2.38, 197.7, 213.8, 186.3, 219.5, 229.5, 126.8, and 160.7. The twelfth row spans both columns and contains the word, \u201cHydrogen.\u201d Under the \u201cSubstance\u201d column for hydrogen are the following: H subscript 2 ( g ), H ( g ), H subscript 2 O ( g ), H subscript 2 O ( l ), H C I ( g ), and H subscript 2 S ( g ). Under the \u201cS subscript 298 superscript degree symbol ( J mol superscript negative 1 K superscript negative 1 )\u201d column for hydrogen are the following: 130.57, 114.6, 188.71, 69.91, 186.8, and 205.7. The nineteenth row spans both columns and contains the word, \u201cOxygen.\u201d Under the \u201cSubstance\u201d column for oxygen is O subscript 2 ( g ). Under the \u201cS subscript 298 superscript degree symbol ( J mol superscript negative 1 K superscript negative 1 )\u201d column for oxygen is 205.03.\">\r\n<thead>\r\n<tr>\r\n<th colspan=\"2\">Table\u00a02. Standard Entropies (at 298.15 K, 1 atm)<\/th>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<th>Substance<\/th>\r\n<th>[latex]{S}_{298}^{\\circ }[\/latex] (J mol<sup>\u22121<\/sup> K<sup>\u22121<\/sup>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td colspan=\"2\">carbon<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>C(<em>s<\/em>, graphite)<\/td>\r\n<td>5.740<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>C(<em>s<\/em>, diamond)<\/td>\r\n<td>2.38<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>CO(<em>g<\/em>)<\/td>\r\n<td>197.7<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>CO<sub>2<\/sub>(<em>g<\/em>)<\/td>\r\n<td>213.8<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>CH<sub>4<\/sub>(<em>g<\/em>)<\/td>\r\n<td>186.3<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>C<sub>2<\/sub>H<sub>4<\/sub>(<em>g<\/em>)<\/td>\r\n<td>219.5<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>C<sub>2<\/sub>H<sub>6<\/sub>(<em>g<\/em>)<\/td>\r\n<td>229.5<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>CH<sub>3<\/sub>OH(<em>l<\/em>)<\/td>\r\n<td>126.8<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>C<sub>2<\/sub>H<sub>5<\/sub>OH(<em>l<\/em>)<\/td>\r\n<td>160.7<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td colspan=\"2\">hydrogen<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>H<sub>2<\/sub>(<em>g<\/em>)<\/td>\r\n<td>130.57<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>H(<em>g<\/em>)<\/td>\r\n<td>114.6<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>H<sub>2<\/sub>O(<em>g<\/em>)<\/td>\r\n<td>188.71<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>H<sub>2<\/sub>O(<em>l<\/em>)<\/td>\r\n<td>69.91<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>HCI(<em>g<\/em>)<\/td>\r\n<td>186.8<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>H<sub>2<\/sub>S(<em>g<\/em>)<\/td>\r\n<td>205.7<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td colspan=\"2\">oxygen<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>O<sub>2<\/sub>(<em>g<\/em>)<\/td>\r\n<td>205.03<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"textbox examples\">\r\n<h3>Example 2:\u00a0<strong>Determination of \u0394<em>S<\/em>\u00b0<\/strong><\/h3>\r\nCalculate the standard entropy change for the following process:\r\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{2}\\text{O}\\left(g\\right)\\longrightarrow {\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"63375\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"63375\"]\r\n\r\nThe value of the standard entropy change at room temperature, [latex]\\Delta {S}_{298}^{\\circ }[\/latex], is the difference between the standard entropy of the product, H<sub>2<\/sub>O(<em>l<\/em>), and the standard entropy of the reactant, H<sub>2<\/sub>O(<em>g<\/em>).\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ \\hfill \\Delta {S}_{298}^{\\circ }&amp; ={S}_{298}^{\\circ }\\left({\\text{H}}_{2}\\text{O}\\left(l\\right)\\right)-{S}_{298}^{\\circ }\\left({\\text{H}}_{2}\\text{O}\\left(g\\right)\\right)\\hfill \\\\ &amp; =\\left(\\text{70.0 J}{\\text{mol}}^{-1}{\\text{K}}^{-1}\\right)-\\left(\\text{188.8 J}{\\text{mol}}^{-1}{\\text{K}}^{-1}\\right)=-118.8\\text{J}{\\text{mol}}^{-1}{\\text{K}}^{-1}\\hfill \\end{array}[\/latex]<\/p>\r\nThe value for [latex]\\Delta {S}_{298}^{\\circ }[\/latex] is negative, as expected for this phase transition (condensation), which the previous section discussed.\r\n\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idp319200\"><strong>Check Your Learning<\/strong><\/h4>\r\nCalculate the standard entropy change for the following process:\r\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{2}\\left(g\\right)+{\\text{C}}_{2}{\\text{H}}_{4}\\left(g\\right)\\longrightarrow {\\text{C}}_{2}{\\text{H}}_{6}\\left(g\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"697101\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"697101\"]\u2212120.6 J mol<sup>\u22121<\/sup> K<sup>\u22121<\/sup>[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 3:\u00a0<strong>Determination of \u0394<em>S<\/em>\u00b0<\/strong><\/h3>\r\nCalculate the standard entropy change for the combustion of methanol, CH<sub>3<\/sub>OH:\r\n<p style=\"text-align: center;\">[latex]2{\\text{CH}}_{3}\\text{OH}\\left(l\\right)+3{\\text{O}}_{2}\\left(g\\right)\\longrightarrow 2{\\text{CO}}_{2}\\left(g\\right)+4{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"856373\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"856373\"]\r\n\r\nThe value of the standard entropy change is equal to the difference between the standard entropies of the products and the entropies of the reactants scaled by their stoichiometric coefficients.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}{}\\Delta S^{\\circ }=\\Delta {S}_{298}^{\\circ }&amp;=&amp;\\sum \\nu {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu {S}_{298}^{\\circ }\\text{(reactants)}\\\\{}&amp;=&amp;\\left[2{S}_{298}^{\\circ }\\left({\\text{CO}}_{2}\\left(g\\right)\\right)+4{S}_{298}^{\\circ }\\left({\\text{H}}_{2}\\text{O}\\left(l\\right)\\right)\\right]-\\left[2{S}_{298}^{\\circ }\\left({\\text{CH}}_{3}\\text{OH}\\left(l\\right)\\right)+3{S}_{298}^{\\circ }\\left({\\text{O}}_{2}\\left(g\\right)\\right)\\right]\\\\{}&amp;=&amp;\\left\\{\\left[213.8+4\\times 70.0\\right]-\\left[2\\left(126.8\\right)+3\\left(205.03\\right)\\right]\\right\\}=-371.6\\text{J\/mol}\\cdot\\text{ K}\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idm27377968\"><strong>Check Your Learning<\/strong><\/h4>\r\nCalculate the standard entropy change for the following reaction:\r\n<p style=\"text-align: center;\">[latex]\\text{Ca}{\\left(\\text{OH}\\right)}_{2}\\left(\\text{s}\\right)\\longrightarrow \\text{CaO}\\left(s\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"677489\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"677489\"]24.7 J\/mol\u00b7K[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nThe second law of thermodynamics states that a spontaneous process increases the entropy of the universe, <em>S<\/em><sub>univ<\/sub> &gt; 0. If \u0394<em>S<\/em><sub>univ<\/sub> &lt; 0, the process is nonspontaneous, and if \u0394<em>S<\/em><sub>univ<\/sub> = 0, the system is at equilibrium. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K. With only one possible microstate, the entropy is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process.\r\n<h4>Key Equations<\/h4>\r\n<ul>\r\n \t<li>[latex]\\Delta S\\text{\\circ }=\\Delta {S}_{298}^{\\circ }=\\sum \\nu {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu {S}_{298}^{\\circ }\\text{(reactants)}[\/latex]<\/li>\r\n \t<li>[latex]\\Delta S=\\frac{{q}_{\\text{rev}}}{T}[\/latex]<\/li>\r\n \t<li>\u0394<em>S<\/em><sub>univ<\/sub> = \u0394<em>S<\/em><sub>sys<\/sub> + \u0394<em>S<\/em><sub>surr<\/sub><\/li>\r\n \t<li>[latex]\\Delta {S}_{\\text{univ}}=\\Delta {S}_{\\text{sys}}+\\Delta {S}_{\\text{surr}}=\\Delta {S}_{\\text{sys}}+\\frac{{q}_{\\text{surr}}}{T}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<ol>\r\n \t<li id=\"fs-idp50899456\">What is the difference between \u0394<em>S<\/em>, \u0394<em>S<\/em>\u00b0, and [latex]\\Delta {S}_{298}^{\\circ }[\/latex] for a chemical change?<\/li>\r\n \t<li>Calculate [latex]\\Delta {S}_{298}^{\\circ }[\/latex] for the following changes.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{\\text{SnCl}}_{4}\\left(l\\right)\\longrightarrow {\\text{SnCl}}_{4}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{CS}}_{2}\\left(g\\right)\\longrightarrow {\\text{CS}}_{2}\\left(l\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\text{Cu}\\left(s\\right)\\longrightarrow \\text{Cu}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]2{\\text{H}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\longrightarrow 2{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/li>\r\n \t<li>[latex]2\\text{HCl}\\left(g\\right)+\\text{Pb}\\left(s\\right)\\longrightarrow {\\text{PbCl}}_{2}\\left(s\\right)+{\\text{H}}_{2}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\text{Zn}\\left(s\\right)+{\\text{CuSO}}_{4}\\left(s\\right)\\longrightarrow \\text{Cu}\\left(s\\right)+{\\text{ZnSO}}_{4}\\left(s\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Determine the entropy change for the combustion of liquid ethanol, C<sub>2<\/sub>H<sub>5<\/sub>OH, under standard state conditions to give gaseous carbon dioxide and liquid water.<\/li>\r\n \t<li>Determine the entropy change for the combustion of gaseous propane, C<sub>3<\/sub>H<sub>8<\/sub>, under standard state conditions to give gaseous carbon dioxide and water.<\/li>\r\n \t<li>\u201cThermite\u201d reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is [latex]{\\text{Fe}}_{2}{\\text{O}}_{3}\\left(s\\right)+2\\text{Al}\\left(s\\right)\\longrightarrow {\\text{Al}}_{2}{\\text{O}}_{3}\\left(s\\right)+2\\text{Fe}\\left(s\\right)[\/latex].<\/li>\r\n \t<li>Using the relevant [latex]{S}_{298}^{\\circ }[\/latex] values listed in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\">Standard Thermodynamic Properties for Selected Substances<\/a>, calculate [latex]{S}_{298}^{\\circ }[\/latex]\u00a0for the following changes:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\longrightarrow 2{\\text{NH}}_{3}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{N}}_{2}\\left(g\\right)+\\frac{5}{2}{\\text{O}}_{2}\\left(g\\right)\\longrightarrow {\\text{N}}_{2}{\\text{O}}_{5}\\left(g\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>From the following information, determine [latex]\\Delta {S}_{298}^{\\circ }[\/latex] for the following:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\text{N}\\left(g\\right)+\\text{O}\\left(g\\right)\\longrightarrow \\text{NO}\\left(g\\right)\\,\\,\\,\\,{;}\\,\\,\\,\\,\\Delta {S}_{298}^{\\circ }=?[\/latex]<\/li>\r\n \t<li>[latex]{\\text{N}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\longrightarrow 2\\text{NO}\\left(g\\right)\\,\\,\\,\\,{;}\\,\\,\\,\\,\\Delta {S}_{298}^{\\circ }=\\text{24.8 J\/K}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{N}}_{2}\\left(g\\right)\\longrightarrow 2\\text{N}\\left(g\\right)\\,\\,\\,\\,{;}\\,\\,\\,\\,\\Delta {S}_{298}^{\\circ }=\\text{115.0 J\/K}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{O}}_{2}\\left(g\\right)\\longrightarrow 2\\text{O}\\left(g\\right)\\,\\,\\,\\,{;}\\,\\,\\,\\,\\Delta {S}_{298}^{\\circ }=\\text{117.0 J\/K}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>By calculating \u0394<em>S<\/em><sub>univ<\/sub> at each temperature, determine if the melting of 1 mole of NaCl(<em>s<\/em>) is spontaneous at 500 \u00b0C and at 700 \u00b0C.\r\n[latex]{S}_{\\text{NaCl}\\left(s\\right)}^{\\circ }=72.11\\frac{\\text{J}}{\\text{mol}\\cdot\\text{ K}}{S}_{\\text{NaCl}\\left(l\\right)}^{\\circ }=95.06\\frac{\\text{J}}{\\text{mol}\\cdot\\text{ K}}\\Delta {H}_{\\text{fusion}}^{\\circ }=\\text{27.95 kJ\/mol}[\/latex]\r\nWhat assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"377854\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"377854\"]\r\n\r\n2. [latex]\\Delta {S}_{298}^{\\circ }[\/latex] for each change is as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\begin{array}{rll}\\\\ \\\\ \\Delta {S}_{298}^{\\circ }&amp;=&amp;\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(reactants)}\\\\{}&amp;=&amp;1\\Delta {S}_{298}^{\\circ }{\\text{SnCl}}_{4}\\left(g\\right)-1\\Delta {S}_{298}^{}{\\text{SnCl}}_{4}\\left(l\\right)\\\\{}&amp;=&amp;\\left[\\text{1 mol}\\left(366\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(259\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=\\text{107 J\/K}\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{rll}\\\\ \\\\ \\Delta {S}_{298}^{\\circ }&amp;=&amp;\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(reactants)}\\\\{}&amp;=&amp;1\\Delta {S}_{298}^{\\circ }{\\text{Cs}}_{2}\\left(l\\right)-1\\Delta {S}_{298}^{\\circ }{\\text{Cs}}_{2}\\left(g\\right)\\\\{}&amp;=&amp;\\left[\\text{1 mol}\\left(151.3\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(237.7\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=-\\text{86.4 J\/K}\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{rll}\\\\ \\\\ \\Delta{S}_{298}^{\\circ }&amp;=&amp;\\sum\\nu\\Delta{S}_{298}^{\\circ}\\left(\\text{products}\\right)-\\sum\\nu\\Delta{S}_{298}^{\\circ}\\left(\\text{reactants}\\right)\\\\{}&amp;=&amp;1\\Delta{S}_{298}^{\\circ}\\text{Cu}\\left(g\\right)-1\\Delta {S}_{298}^{\\circ}\\text{Cu}\\left(s\\right)\\\\{}&amp;=&amp;\\left[\\text{1 mol}\\left(166.3\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(33.15\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=133.2\\text{ J\/K}\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{rll}\\\\ \\\\ \\Delta {S}_{298}^{\\circ}&amp;=&amp;\\sum\\nu\\Delta{S}_{298}^{\\circ}\\left(\\text{products}\\right)-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\left(\\text{reactants}\\right)\\\\{}&amp;=&amp;1\\Delta {S}_{298}^{\\circ}\\text{H}_{2}\\text{O}\\left(g\\right)-1\\Delta{S}_{298}^{\\circ}\\text{H}_{2}\\text{O}\\left(l\\right)\\\\{}&amp;=&amp;\\left[\\text{1 mol}\\left(188.8\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(70.0\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=\\text{118.8 J\/K}\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{rll}\\\\ \\\\ \\Delta {S}_{298}^{\\circ }&amp;=&amp;\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(reactants)}\\\\{}&amp;=&amp;2\\Delta {S}_{298}^{\\circ }{\\text{H}}_{2}\\text{O}\\left(l\\right)-\\left[1\\Delta {S}_{298}^{\\circ }{\\text{O}}_{2}\\left(g\\right)+2\\Delta {S}_{298}^{\\circ }{\\text{H}}_{2}\\left(g\\right)\\right]\\\\{}&amp;=&amp;\\left[\\text{2 mol}\\left(70.0\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(205.2\\frac{\\text{J}}{\\text{mol K}}\\right)+\\text{2 mol}\\left(130.7\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=-\\text{326.6 J\/K}\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{rll}\\Delta {S}_{298}^{\\circ }&amp;=&amp;\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(reactants)}\\\\{}&amp;=&amp;\\left[1\\Delta {S}_{298}^{\\circ }{\\text{PbCl}}_{2}\\left(s\\right)+1\\Delta {S}_{298}^{\\circ }{\\text{H}}_{2}\\left(g\\right)\\right]-\\left[1\\Delta {S}_{298}^{\\circ }\\text{Pb}\\left(s\\right)+2\\Delta {S}_{298}^{\\circ }\\text{HCl}\\left(g\\right)\\right]\\\\{}&amp;=&amp;\\left[\\text{1 mol}\\left(136.0\\frac{\\text{J}}{\\text{mol K}}\\right)+\\text{1 mol}\\left(130.7\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(64.81\\frac{\\text{J}}{\\text{mol K}}\\right)+\\text{2 mol}\\left(186.9\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=-\\text{171.9 J\/K}\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{rll}\\\\ \\\\ \\Delta {S}_{298}^{\\circ }&amp;=&amp;\\sum \\nu \\Delta {S}_{298}^{\\circ }\\left(\\text{products}\\right)-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\left(\\text{reactants}\\right)\\\\{}&amp;=&amp;\\left[1\\Delta {S}_{298}^{\\circ }\\text{Cu}\\left(s\\right)+1\\Delta {S}_{298}^{\\circ }{\\text{ZnSO}}_{4}\\left(s\\right)\\right]-\\left[1\\Delta {S}_{298}^{\\circ }\\text{Zn}\\left(s\\right)+1\\Delta {S}_{298}^{\\circ }{\\text{CuSO}}_{4}\\left(s\\right)\\right]\\\\{}&amp;=&amp;\\left[\\text{1 mol}\\left(33.15\\frac{\\text{J}}{\\text{mol K}}\\right)+\\text{1 mol}\\left(110.5\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(41.6\\frac{\\text{J}}{\\text{mol K}}\\right)+\\text{1 mol}\\left(109.2\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=-\\text{7.2 J\/K}\\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n4. The reaction is [latex]{\\text{C}}_{3}{\\text{H}}_{8}\\left(g\\right)+5{\\text{O}}_{2}\\left(g\\right)\\longrightarrow 3{\\text{CO}}_{2}\\left(g\\right)+4{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}\\\\ \\Delta {S}_{298}^{\\circ }&amp;=&amp;\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(reactants)}\\\\ &amp;=&amp;\\left[3\\Delta {S}_{298}^{\\circ }{\\text{CO}}_{2}\\left(g\\right)+4\\Delta {S}_{298}^{\\circ }{\\text{H}}_{2}\\text{O}\\left(g\\right)\\right]-\\left[1\\Delta {S}_{298}^{\\circ }{\\text{C}}_{3}{\\text{H}}_{8}\\left(g\\right)+5\\Delta {S}_{298}^{\\circ }{\\text{O}}_{2}\\left(g\\right)\\right]\\\\ &amp;=&amp;\\left[\\text{3 mol}\\left(213.6\\frac{\\text{J}}{\\text{mol K}}\\right)+\\text{4 mol}\\left(69.91\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(269.9\\frac{\\text{J}}{\\text{mol K}}\\right)+\\text{5 mol}\\left(205.03\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=\\text{100.6 J\/K}\\end{array}[\/latex]<\/p>\r\n6. [latex]\\Delta {S}_{298}^{\\circ }[\/latex] for each change is as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\begin{array}{rll}\\\\ \\\\ \\Delta {S}_{\\text{sys}}^{\\circ }&amp;=&amp;\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(reactants)}\\\\\\Delta {S}_{\\text{sys}}^{\\circ }&amp;=&amp;\\left(2\\Delta {S}_{298}^{\\circ }{\\text{NH}}_{3}\\right)-\\left(1\\Delta {S}_{298}^{\\circ }{\\text{N}}_{2}+3\\Delta {S}_{298}^{\\circ }{\\text{H}}_{2}\\right)\\\\\\Delta {S}_{\\text{sys}}^{\\circ }&amp;=&amp;\\left[\\text{2 mol}\\left(192.8\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(191.6\\frac{\\text{J}}{\\text{mol K}}\\right)+\\text{3 mol}\\left(130.7\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=-\\text{198.1 J\/K}\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{rll}\\\\ \\\\ \\Delta {S}_{\\text{sys}}^{\\circ }&amp;=&amp;\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(reactants)}\\\\ \\Delta {S}_{\\text{sys}}^{\\circ }&amp;=&amp;\\left(2\\Delta {S}_{298}^{\\circ }{\\text{N}}_{2}{\\text{O}}_{5}\\right)-\\left(1\\Delta {S}_{298}^{\\circ }{\\text{N}}_{2}+\\frac{5}{2}\\times \\Delta {S}_{298}^{\\circ }{\\text{O}}_{2}\\right)\\\\\\Delta {S}_{\\text{sys}}^{\\circ }&amp;=&amp;\\left[\\text{1 mol}\\left(355.7\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(191.6\\frac{\\text{J}}{\\text{mol K}}\\right)+\\frac{5}{2}\\text{mol}\\left(205.2\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=-\\text{348.9 J\/K}\\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n8. The process is [latex]\\text{NaCl}\\left(s\\right)\\longrightarrow \\text{NaCl}\\left(l\\right)[\/latex].\u00a0At 500 \u00b0C, the following is true:\r\n<p style=\"text-align: center;\">[latex]\\Delta {S}_{\\text{univ}}=\\Delta {S}_{\\text{sys}}+\\frac{{q}_{\\text{surr}}}{T}=\\left(95.06 - 72.11\\right)\\frac{\\text{J}}{\\text{mol}\\cdot\\text{ K}}+\\frac{-27.95\\times {10}^{3}\\frac{\\text{J}}{\\text{mol}}}{500+273.15}=-13.2\\frac{\\text{J}}{\\text{mol}\\cdot\\text{ K}}[\/latex]<\/p>\r\nAt 700 \u00b0C, the following is true:\r\n<p style=\"text-align: center;\">[latex]\\Delta {S}_{\\text{univ}}=\\Delta {S}_{\\text{sys}}+\\frac{{q}_{\\text{surr}}}{T}=\\left(95.06 - 72.11\\right)\\frac{\\text{J}}{\\text{mol}\\cdot\\text{ K}}+\\frac{-27.95\\times {10}^{3}\\frac{\\text{J}}{\\text{mol}}}{700+273.15}=-5.8\\frac{\\text{J}}{\\text{mol}\\cdot\\text{ K}}[\/latex]<\/p>\r\nAs \u0394<em>S<\/em><sub>univ<\/sub> &lt; 0 at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K. It is assumed that these do not change significantly at the higher temperatures used in the problem.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>second law of thermodynamics: <\/strong>entropy of the universe increases for a spontaneous process\r\n\r\n<strong>standard entropy (<em>S<\/em>\u00b0): <\/strong>entropy for a substance at 1 bar pressure; tabulated values are usually determined at 298.15 K and denoted [latex]{S}_{298}^{\\circ }[\/latex]\r\n\r\n<strong>standard entropy change (\u0394<em>S<\/em>\u00b0): <\/strong>change in entropy for a reaction calculated using the standard entropies, usually at room temperature and denoted [latex]\\Delta {S}_{298}^{\\circ }[\/latex]\r\n\r\n<strong>third law of thermodynamics: <\/strong>entropy of a perfect crystal at absolute zero (0 K) is zero","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>State and explain the second and third laws of thermodynamics<\/li>\n<li>Calculate entropy changes for phase transitions and chemical reactions under standard conditions<\/li>\n<\/ul>\n<\/div>\n<h2>The Second Law of Thermodynamics<\/h2>\n<p>In the quest to identify a property that may reliably predict the spontaneity of a process, we have identified a very promising candidate: entropy. Processes that involve an increase in entropy <em>of the system<\/em> (\u0394<em>S<\/em> &gt; 0) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include <em>the surroundings<\/em>, we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true:<\/p>\n<p style=\"text-align: center;\">[latex]\\Delta {S}_{\\text{univ}}=\\Delta {S}_{\\text{sys}}+\\Delta {S}_{\\text{surr}}[\/latex]<\/p>\n<p>To illustrate this relation, consider again the process of heat flow between two objects, one identified as the system and the other as the surroundings. There are three possibilities for such a process:<\/p>\n<ol>\n<li>The objects are at different temperatures, and heat flows from the hotter to the cooler object. <em>This is always observed to occur spontaneously.<\/em> Designating the hotter object as the system and invoking the definition of entropy yields the following:<br \/>\n[latex]\\Delta {S}_{\\text{sys}}=\\frac{-{q}_{\\text{rev}}}{{T}_{\\text{sys}}}\\text{and}\\Delta {S}_{\\text{surr}}=\\frac{{q}_{\\text{rev}}}{{T}_{\\text{surr}}}[\/latex]<br \/>\nThe arithmetic signs of <em>q<\/em><sub>rev<\/sub> denote the loss of heat by the system and the gain of heat by the surroundings. Since <em>T<\/em><sub>sys<\/sub> &gt; <em>T<\/em><sub>surr<\/sub> in this scenario, the magnitude of the entropy change for the surroundings will be greater than that for the system, and so the sum of \u0394<em>S<\/em><sub>sys<\/sub> and \u0394<em>S<\/em><sub>surr<\/sub> will yield a positive value for \u0394<em>S<\/em><sub>univ<\/sub>. <em>This process involves an increase in the entropy of the universe.<\/em><\/li>\n<li>The objects are at different temperatures, and heat flows from the cooler to the hotter object. <em>This is never observed to occur spontaneously.<\/em> Again designating the hotter object as the system and invoking the definition of entropy yields the following:<br \/>\n[latex]\\Delta {S}_{\\text{sys}}=\\frac{{q}_{\\text{rev}}}{{T}_{\\text{sys}}}\\text{and}\\Delta {S}_{\\text{surr}}=\\frac{-{q}_{\\text{rev}}}{{T}_{\\text{surr}}}[\/latex]<br \/>\nThe arithmetic signs of <em>q<\/em><sub>rev<\/sub> denote the gain of heat by the system and the loss of heat by the surroundings. The magnitude of the entropy change for the surroundings will again be greater than that for the system, but in this case, the signs of the heat changes will yield a negative value for \u0394<em>S<\/em><sub>univ<\/sub>. <em>This process involves a decrease in the entropy of the universe.<\/em><\/li>\n<li>The temperature difference between the objects is infinitesimally small, <em>T<\/em><sub>sys<\/sub> \u2248 <em>T<\/em><sub>surr<\/sub>, and so the heat flow is thermodynamically reversible. See the previous section\u2019s discussion). In this case, the system and surroundings experience entropy changes that are equal in magnitude and therefore sum to yield a value of zero for \u0394<em>S<\/em><sub>univ<\/sub>. <em>This process involves no change in the entropy of the universe.<\/em><\/li>\n<\/ol>\n<p>These results lead to a profound statement regarding the relation between entropy and spontaneity known as the <strong>second law of thermodynamics<\/strong>: <em>all spontaneous changes cause an increase in the entropy of the universe.<\/em> A summary of these three relations is provided in Table\u00a01.<\/p>\n<table id=\"fs-idp41455824\" class=\"span-all\" summary=\"This table contains two 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: \u201cSpontaneous,\u201d \u201cnonspontaneous ( spontaneous in opposite direction ),\u201d and, \u201creversible ( system is at equilibrium ).\u201d\">\n<thead>\n<tr>\n<th colspan=\"2\">Table\u00a01. The Second Law of Thermodynamics<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>\u0394<em>S<\/em><sub>univ<\/sub> &gt; 0<\/td>\n<td>spontaneous<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>\u0394<em>S<\/em><sub>univ<\/sub> &lt; 0<\/td>\n<td>nonspontaneous (spontaneous in opposite direction)<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>\u0394<em>S<\/em><sub>univ<\/sub> = 0<\/td>\n<td>reversible (system is at equilibrium)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. As a result, <em>q<\/em><sub>surr<\/sub> is a good approximation of <em>q<\/em><sub>rev<\/sub>, and the second law may be stated as the following:<\/p>\n<p style=\"text-align: center;\">[latex]\\Delta {S}_{\\text{univ}}=\\Delta {S}_{\\text{sys}}+\\Delta {S}_{\\text{surr}}=\\Delta {S}_{\\text{sys}}+\\frac{{q}_{\\text{surr}}}{T}[\/latex]<\/p>\n<p>We may use this equation to predict the spontaneity of a process as illustrated in Example 1.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0<strong>Will Ice Spontaneously Melt?<\/strong><\/h3>\n<p>The entropy change for the process is\u00a0[latex]{\\text{H}}_{2}\\text{O}\\left(s\\right)\\longrightarrow {\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\n<p>is 22.1 J\/K and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at \u221210.00 \u00b0C? Is it spontaneous at +10.00 \u00b0C?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q692410\">Show Answer<\/span><\/p>\n<div id=\"q692410\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can assess the spontaneity of the process by calculating the entropy change of the universe. If \u0394<em>S<\/em><sub>univ<\/sub> is positive, then the process is spontaneous. At both temperatures, \u0394<em>S<\/em><sub>sys<\/sub> = 22.1 J\/K and <em>q<\/em><sub>surr<\/sub> = \u22126.00 kJ.<\/p>\n<p>At \u221210.00 \u00b0C (263.15 K), the following is true:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\Delta {S}_{\\text{univ}}& =\\Delta {S}_{\\text{sys}}+\\Delta {S}_{\\text{surr}}=\\Delta {S}_{\\text{sys}}+\\frac{{q}_{\\text{surr}}}{T}\\hfill \\\\ & =\\text{22.1 J\/K}+\\frac{-6.00\\times {10}^{3}\\text{J}}{\\text{263.15 K}}=-0.7\\text{J\/K}\\hfill \\end{array}[\/latex]<\/p>\n<p><em>S<\/em><sub>univ<\/sub> &lt; 0, so melting is nonspontaneous (<em>not<\/em> spontaneous) at \u221210.0 \u00b0C.<\/p>\n<p>At 10.00 \u00b0C (283.15 K), the following is true:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ \\Delta {S}_{\\text{univ}}&=&\\Delta {S}_{\\text{sys}}+\\frac{{q}_{\\text{surr}}}{T}\\\\{}&=&22.1\\text{J\/K}+\\frac{-6.00\\times {10}^{3}\\text{J}}{\\text{283.15 K}}=\\text{+0.9 J\/K}\\end{array}[\/latex]<\/p>\n<p><em>S<\/em><sub>univ<\/sub> &gt; 0, so melting <em>is<\/em> spontaneous at 10.00 \u00b0C.<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-idp54808912\"><strong>Check Your Learning<\/strong><\/h4>\n<p>Using this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of <em>S<\/em><sub>univ<\/sub>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q686682\">Show Answer<\/span><\/p>\n<div id=\"q686682\" class=\"hidden-answer\" style=\"display: none\">Entropy is a state function, and freezing is the opposite of melting. At \u221210.00 \u00b0C spontaneous, +0.7 J\/K; at +10.00 \u00b0C nonspontaneous, \u22120.9 J\/K.<\/div>\n<\/div>\n<\/div>\n<h2>The Third Law of Thermodynamics<\/h2>\n<p>The previous section described the various contributions of matter and energy dispersal that contribute to the entropy of a system. With these contributions in mind, consider the entropy of a pure, perfectly crystalline solid possessing no kinetic energy (that is, at a temperature of absolute zero, 0 K). This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion means there is but one possible location for each identical atom or molecule comprising the crystal (<em>W<\/em> = 1). According to the Boltzmann equation, the entropy of this system is zero.<\/p>\n<p style=\"text-align: center;\">[latex]S=k\\text{ln}W=k\\text{ln}\\left(1\\right)=0[\/latex]<\/p>\n<p>This limiting condition for a system\u2019s entropy represents the <strong>third law of thermodynamics<\/strong>: <em>the entropy of a pure, perfect crystalline substance at 0 K is zero.<\/em><\/p>\n<p>We can make careful calorimetric measurements to determine the temperature dependence of a substance\u2019s entropy and to derive absolute entropy values under specific conditions. <strong>Standard entropies<\/strong> are given the label [latex]{S}_{298}^{\\circ }[\/latex] for values determined for one mole of substance at a pressure of 1 bar and a temperature of 298 K. The <strong>standard entropy change (\u0394<em>S<\/em>\u00b0)<\/strong> for any process may be computed from the standard entropies of its reactant and product species like the following:<\/p>\n<p style=\"text-align: center;\">[latex]\\Delta S\\text{\\circ }=\\sum \\nu {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu {S}_{298}^{\\circ }\\text{(reactants)}[\/latex]<\/p>\n<p>Here, \u03bd represents stoichiometric coefficients in the balanced equation representing the process. For example, \u0394<em>S<\/em>\u00b0 for the following reaction at room temperature\u00a0[latex]m\\text{A}+n\\text{B}\\longrightarrow x\\text{C}+y\\text{D}[\/latex]\u00a0is computed as the following:<\/p>\n<p style=\"text-align: center;\">[latex]=\\left[x{S}_{298}^{\\circ }\\left(\\text{C}\\right)+y{S}_{298}^{\\circ }\\left(\\text{D}\\right)\\right]-\\left[m{S}_{298}^{\\circ }\\left(\\text{A}\\right)+n{S}_{298}^{\\circ }\\left(\\text{B}\\right)\\right][\/latex]<\/p>\n<p>Table\u00a02\u00a0lists some standard entropies at 298.15 K. You can find additional standard entropies in <a class=\"target-chapter\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\">Standard Thermodynamic Properties for Selected Substances<\/a>.<\/p>\n<table id=\"fs-idm78597984\" class=\"span-all\" summary=\"The table has two columns and twenty rows. The first row is a header row and it labels the columns, \u201cSubstance,\u201d and \u201cS subscript 298 superscript degree symbol ( J mol superscript negative 1 K superscript negative 1 ).\u201d The second row spans both columns and contains the word, \u201cCarbon.\u201d Under the \u201cSubstance\u201d column for carbon are the following: C ( s, graphite ), C ( s, diamond ), C O ( g ), C O subscript 2 ( g ), C H subscript 4 ( g ), C subscript 2 H subscript 4 ( g ), C subscript 2 H subscript 6 ( g ), C H subscript 3 O H ( l ), and C subscript 2 H subscript 5 O H ( l ). Under the \u201cS subscript 298 superscript degree symbol ( J mol superscript negative 1 K superscript negative 1 )\u201d column for carbon are the following: 5.740, 2.38, 197.7, 213.8, 186.3, 219.5, 229.5, 126.8, and 160.7. The twelfth row spans both columns and contains the word, \u201cHydrogen.\u201d Under the \u201cSubstance\u201d column for hydrogen are the following: H subscript 2 ( g ), H ( g ), H subscript 2 O ( g ), H subscript 2 O ( l ), H C I ( g ), and H subscript 2 S ( g ). Under the \u201cS subscript 298 superscript degree symbol ( J mol superscript negative 1 K superscript negative 1 )\u201d column for hydrogen are the following: 130.57, 114.6, 188.71, 69.91, 186.8, and 205.7. The nineteenth row spans both columns and contains the word, \u201cOxygen.\u201d Under the \u201cSubstance\u201d column for oxygen is O subscript 2 ( g ). Under the \u201cS subscript 298 superscript degree symbol ( J mol superscript negative 1 K superscript negative 1 )\u201d column for oxygen is 205.03.\">\n<thead>\n<tr>\n<th colspan=\"2\">Table\u00a02. Standard Entropies (at 298.15 K, 1 atm)<\/th>\n<\/tr>\n<tr valign=\"top\">\n<th>Substance<\/th>\n<th>[latex]{S}_{298}^{\\circ }[\/latex] (J mol<sup>\u22121<\/sup> K<sup>\u22121<\/sup>)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td colspan=\"2\">carbon<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>C(<em>s<\/em>, graphite)<\/td>\n<td>5.740<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>C(<em>s<\/em>, diamond)<\/td>\n<td>2.38<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>CO(<em>g<\/em>)<\/td>\n<td>197.7<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>CO<sub>2<\/sub>(<em>g<\/em>)<\/td>\n<td>213.8<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>CH<sub>4<\/sub>(<em>g<\/em>)<\/td>\n<td>186.3<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>C<sub>2<\/sub>H<sub>4<\/sub>(<em>g<\/em>)<\/td>\n<td>219.5<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>C<sub>2<\/sub>H<sub>6<\/sub>(<em>g<\/em>)<\/td>\n<td>229.5<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>CH<sub>3<\/sub>OH(<em>l<\/em>)<\/td>\n<td>126.8<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>C<sub>2<\/sub>H<sub>5<\/sub>OH(<em>l<\/em>)<\/td>\n<td>160.7<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td colspan=\"2\">hydrogen<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>H<sub>2<\/sub>(<em>g<\/em>)<\/td>\n<td>130.57<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>H(<em>g<\/em>)<\/td>\n<td>114.6<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>H<sub>2<\/sub>O(<em>g<\/em>)<\/td>\n<td>188.71<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>H<sub>2<\/sub>O(<em>l<\/em>)<\/td>\n<td>69.91<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>HCI(<em>g<\/em>)<\/td>\n<td>186.8<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>H<sub>2<\/sub>S(<em>g<\/em>)<\/td>\n<td>205.7<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td colspan=\"2\">oxygen<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>O<sub>2<\/sub>(<em>g<\/em>)<\/td>\n<td>205.03<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"textbox examples\">\n<h3>Example 2:\u00a0<strong>Determination of \u0394<em>S<\/em>\u00b0<\/strong><\/h3>\n<p>Calculate the standard entropy change for the following process:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{2}\\text{O}\\left(g\\right)\\longrightarrow {\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q63375\">Show Answer<\/span><\/p>\n<div id=\"q63375\" class=\"hidden-answer\" style=\"display: none\">\n<p>The value of the standard entropy change at room temperature, [latex]\\Delta {S}_{298}^{\\circ }[\/latex], is the difference between the standard entropy of the product, H<sub>2<\/sub>O(<em>l<\/em>), and the standard entropy of the reactant, H<sub>2<\/sub>O(<em>g<\/em>).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ \\hfill \\Delta {S}_{298}^{\\circ }& ={S}_{298}^{\\circ }\\left({\\text{H}}_{2}\\text{O}\\left(l\\right)\\right)-{S}_{298}^{\\circ }\\left({\\text{H}}_{2}\\text{O}\\left(g\\right)\\right)\\hfill \\\\ & =\\left(\\text{70.0 J}{\\text{mol}}^{-1}{\\text{K}}^{-1}\\right)-\\left(\\text{188.8 J}{\\text{mol}}^{-1}{\\text{K}}^{-1}\\right)=-118.8\\text{J}{\\text{mol}}^{-1}{\\text{K}}^{-1}\\hfill \\end{array}[\/latex]<\/p>\n<p>The value for [latex]\\Delta {S}_{298}^{\\circ }[\/latex] is negative, as expected for this phase transition (condensation), which the previous section discussed.<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-idp319200\"><strong>Check Your Learning<\/strong><\/h4>\n<p>Calculate the standard entropy change for the following process:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{2}\\left(g\\right)+{\\text{C}}_{2}{\\text{H}}_{4}\\left(g\\right)\\longrightarrow {\\text{C}}_{2}{\\text{H}}_{6}\\left(g\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q697101\">Show Answer<\/span><\/p>\n<div id=\"q697101\" class=\"hidden-answer\" style=\"display: none\">\u2212120.6 J mol<sup>\u22121<\/sup> K<sup>\u22121<\/sup><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 3:\u00a0<strong>Determination of \u0394<em>S<\/em>\u00b0<\/strong><\/h3>\n<p>Calculate the standard entropy change for the combustion of methanol, CH<sub>3<\/sub>OH:<\/p>\n<p style=\"text-align: center;\">[latex]2{\\text{CH}}_{3}\\text{OH}\\left(l\\right)+3{\\text{O}}_{2}\\left(g\\right)\\longrightarrow 2{\\text{CO}}_{2}\\left(g\\right)+4{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q856373\">Show Answer<\/span><\/p>\n<div id=\"q856373\" class=\"hidden-answer\" style=\"display: none\">\n<p>The value of the standard entropy change is equal to the difference between the standard entropies of the products and the entropies of the reactants scaled by their stoichiometric coefficients.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}{}\\Delta S^{\\circ }=\\Delta {S}_{298}^{\\circ }&=&\\sum \\nu {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu {S}_{298}^{\\circ }\\text{(reactants)}\\\\{}&=&\\left[2{S}_{298}^{\\circ }\\left({\\text{CO}}_{2}\\left(g\\right)\\right)+4{S}_{298}^{\\circ }\\left({\\text{H}}_{2}\\text{O}\\left(l\\right)\\right)\\right]-\\left[2{S}_{298}^{\\circ }\\left({\\text{CH}}_{3}\\text{OH}\\left(l\\right)\\right)+3{S}_{298}^{\\circ }\\left({\\text{O}}_{2}\\left(g\\right)\\right)\\right]\\\\{}&=&\\left\\{\\left[213.8+4\\times 70.0\\right]-\\left[2\\left(126.8\\right)+3\\left(205.03\\right)\\right]\\right\\}=-371.6\\text{J\/mol}\\cdot\\text{ K}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-idm27377968\"><strong>Check Your Learning<\/strong><\/h4>\n<p>Calculate the standard entropy change for the following reaction:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Ca}{\\left(\\text{OH}\\right)}_{2}\\left(\\text{s}\\right)\\longrightarrow \\text{CaO}\\left(s\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q677489\">Show Answer<\/span><\/p>\n<div id=\"q677489\" class=\"hidden-answer\" style=\"display: none\">24.7 J\/mol\u00b7K<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, <em>S<\/em><sub>univ<\/sub> &gt; 0. If \u0394<em>S<\/em><sub>univ<\/sub> &lt; 0, the process is nonspontaneous, and if \u0394<em>S<\/em><sub>univ<\/sub> = 0, the system is at equilibrium. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K. With only one possible microstate, the entropy is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process.<\/p>\n<h4>Key Equations<\/h4>\n<ul>\n<li>[latex]\\Delta S\\text{\\circ }=\\Delta {S}_{298}^{\\circ }=\\sum \\nu {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu {S}_{298}^{\\circ }\\text{(reactants)}[\/latex]<\/li>\n<li>[latex]\\Delta S=\\frac{{q}_{\\text{rev}}}{T}[\/latex]<\/li>\n<li>\u0394<em>S<\/em><sub>univ<\/sub> = \u0394<em>S<\/em><sub>sys<\/sub> + \u0394<em>S<\/em><sub>surr<\/sub><\/li>\n<li>[latex]\\Delta {S}_{\\text{univ}}=\\Delta {S}_{\\text{sys}}+\\Delta {S}_{\\text{surr}}=\\Delta {S}_{\\text{sys}}+\\frac{{q}_{\\text{surr}}}{T}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol>\n<li id=\"fs-idp50899456\">What is the difference between \u0394<em>S<\/em>, \u0394<em>S<\/em>\u00b0, and [latex]\\Delta {S}_{298}^{\\circ }[\/latex] for a chemical change?<\/li>\n<li>Calculate [latex]\\Delta {S}_{298}^{\\circ }[\/latex] for the following changes.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{\\text{SnCl}}_{4}\\left(l\\right)\\longrightarrow {\\text{SnCl}}_{4}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]{\\text{CS}}_{2}\\left(g\\right)\\longrightarrow {\\text{CS}}_{2}\\left(l\\right)[\/latex]<\/li>\n<li>[latex]\\text{Cu}\\left(s\\right)\\longrightarrow \\text{Cu}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]2{\\text{H}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\longrightarrow 2{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/li>\n<li>[latex]2\\text{HCl}\\left(g\\right)+\\text{Pb}\\left(s\\right)\\longrightarrow {\\text{PbCl}}_{2}\\left(s\\right)+{\\text{H}}_{2}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]\\text{Zn}\\left(s\\right)+{\\text{CuSO}}_{4}\\left(s\\right)\\longrightarrow \\text{Cu}\\left(s\\right)+{\\text{ZnSO}}_{4}\\left(s\\right)[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>Determine the entropy change for the combustion of liquid ethanol, C<sub>2<\/sub>H<sub>5<\/sub>OH, under standard state conditions to give gaseous carbon dioxide and liquid water.<\/li>\n<li>Determine the entropy change for the combustion of gaseous propane, C<sub>3<\/sub>H<sub>8<\/sub>, under standard state conditions to give gaseous carbon dioxide and water.<\/li>\n<li>\u201cThermite\u201d reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is [latex]{\\text{Fe}}_{2}{\\text{O}}_{3}\\left(s\\right)+2\\text{Al}\\left(s\\right)\\longrightarrow {\\text{Al}}_{2}{\\text{O}}_{3}\\left(s\\right)+2\\text{Fe}\\left(s\\right)[\/latex].<\/li>\n<li>Using the relevant [latex]{S}_{298}^{\\circ }[\/latex] values listed in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\">Standard Thermodynamic Properties for Selected Substances<\/a>, calculate [latex]{S}_{298}^{\\circ }[\/latex]\u00a0for the following changes:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{\\text{N}}_{2}\\left(g\\right)+3{\\text{H}}_{2}\\left(g\\right)\\longrightarrow 2{\\text{NH}}_{3}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]{\\text{N}}_{2}\\left(g\\right)+\\frac{5}{2}{\\text{O}}_{2}\\left(g\\right)\\longrightarrow {\\text{N}}_{2}{\\text{O}}_{5}\\left(g\\right)[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>From the following information, determine [latex]\\Delta {S}_{298}^{\\circ }[\/latex] for the following:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\text{N}\\left(g\\right)+\\text{O}\\left(g\\right)\\longrightarrow \\text{NO}\\left(g\\right)\\,\\,\\,\\,{;}\\,\\,\\,\\,\\Delta {S}_{298}^{\\circ }=?[\/latex]<\/li>\n<li>[latex]{\\text{N}}_{2}\\left(g\\right)+{\\text{O}}_{2}\\left(g\\right)\\longrightarrow 2\\text{NO}\\left(g\\right)\\,\\,\\,\\,{;}\\,\\,\\,\\,\\Delta {S}_{298}^{\\circ }=\\text{24.8 J\/K}[\/latex]<\/li>\n<li>[latex]{\\text{N}}_{2}\\left(g\\right)\\longrightarrow 2\\text{N}\\left(g\\right)\\,\\,\\,\\,{;}\\,\\,\\,\\,\\Delta {S}_{298}^{\\circ }=\\text{115.0 J\/K}[\/latex]<\/li>\n<li>[latex]{\\text{O}}_{2}\\left(g\\right)\\longrightarrow 2\\text{O}\\left(g\\right)\\,\\,\\,\\,{;}\\,\\,\\,\\,\\Delta {S}_{298}^{\\circ }=\\text{117.0 J\/K}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>By calculating \u0394<em>S<\/em><sub>univ<\/sub> at each temperature, determine if the melting of 1 mole of NaCl(<em>s<\/em>) is spontaneous at 500 \u00b0C and at 700 \u00b0C.<br \/>\n[latex]{S}_{\\text{NaCl}\\left(s\\right)}^{\\circ }=72.11\\frac{\\text{J}}{\\text{mol}\\cdot\\text{ K}}{S}_{\\text{NaCl}\\left(l\\right)}^{\\circ }=95.06\\frac{\\text{J}}{\\text{mol}\\cdot\\text{ K}}\\Delta {H}_{\\text{fusion}}^{\\circ }=\\text{27.95 kJ\/mol}[\/latex]<br \/>\nWhat assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q377854\">Show Selected Answers<\/span><\/p>\n<div id=\"q377854\" class=\"hidden-answer\" style=\"display: none\">\n<p>2. [latex]\\Delta {S}_{298}^{\\circ }[\/latex] for each change is as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\begin{array}{rll}\\\\ \\\\ \\Delta {S}_{298}^{\\circ }&=&\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(reactants)}\\\\{}&=&1\\Delta {S}_{298}^{\\circ }{\\text{SnCl}}_{4}\\left(g\\right)-1\\Delta {S}_{298}^{}{\\text{SnCl}}_{4}\\left(l\\right)\\\\{}&=&\\left[\\text{1 mol}\\left(366\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(259\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=\\text{107 J\/K}\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{rll}\\\\ \\\\ \\Delta {S}_{298}^{\\circ }&=&\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(reactants)}\\\\{}&=&1\\Delta {S}_{298}^{\\circ }{\\text{Cs}}_{2}\\left(l\\right)-1\\Delta {S}_{298}^{\\circ }{\\text{Cs}}_{2}\\left(g\\right)\\\\{}&=&\\left[\\text{1 mol}\\left(151.3\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(237.7\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=-\\text{86.4 J\/K}\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{rll}\\\\ \\\\ \\Delta{S}_{298}^{\\circ }&=&\\sum\\nu\\Delta{S}_{298}^{\\circ}\\left(\\text{products}\\right)-\\sum\\nu\\Delta{S}_{298}^{\\circ}\\left(\\text{reactants}\\right)\\\\{}&=&1\\Delta{S}_{298}^{\\circ}\\text{Cu}\\left(g\\right)-1\\Delta {S}_{298}^{\\circ}\\text{Cu}\\left(s\\right)\\\\{}&=&\\left[\\text{1 mol}\\left(166.3\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(33.15\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=133.2\\text{ J\/K}\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{rll}\\\\ \\\\ \\Delta {S}_{298}^{\\circ}&=&\\sum\\nu\\Delta{S}_{298}^{\\circ}\\left(\\text{products}\\right)-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\left(\\text{reactants}\\right)\\\\{}&=&1\\Delta {S}_{298}^{\\circ}\\text{H}_{2}\\text{O}\\left(g\\right)-1\\Delta{S}_{298}^{\\circ}\\text{H}_{2}\\text{O}\\left(l\\right)\\\\{}&=&\\left[\\text{1 mol}\\left(188.8\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(70.0\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=\\text{118.8 J\/K}\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{rll}\\\\ \\\\ \\Delta {S}_{298}^{\\circ }&=&\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(reactants)}\\\\{}&=&2\\Delta {S}_{298}^{\\circ }{\\text{H}}_{2}\\text{O}\\left(l\\right)-\\left[1\\Delta {S}_{298}^{\\circ }{\\text{O}}_{2}\\left(g\\right)+2\\Delta {S}_{298}^{\\circ }{\\text{H}}_{2}\\left(g\\right)\\right]\\\\{}&=&\\left[\\text{2 mol}\\left(70.0\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(205.2\\frac{\\text{J}}{\\text{mol K}}\\right)+\\text{2 mol}\\left(130.7\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=-\\text{326.6 J\/K}\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{rll}\\Delta {S}_{298}^{\\circ }&=&\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(reactants)}\\\\{}&=&\\left[1\\Delta {S}_{298}^{\\circ }{\\text{PbCl}}_{2}\\left(s\\right)+1\\Delta {S}_{298}^{\\circ }{\\text{H}}_{2}\\left(g\\right)\\right]-\\left[1\\Delta {S}_{298}^{\\circ }\\text{Pb}\\left(s\\right)+2\\Delta {S}_{298}^{\\circ }\\text{HCl}\\left(g\\right)\\right]\\\\{}&=&\\left[\\text{1 mol}\\left(136.0\\frac{\\text{J}}{\\text{mol K}}\\right)+\\text{1 mol}\\left(130.7\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(64.81\\frac{\\text{J}}{\\text{mol K}}\\right)+\\text{2 mol}\\left(186.9\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=-\\text{171.9 J\/K}\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{rll}\\\\ \\\\ \\Delta {S}_{298}^{\\circ }&=&\\sum \\nu \\Delta {S}_{298}^{\\circ }\\left(\\text{products}\\right)-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\left(\\text{reactants}\\right)\\\\{}&=&\\left[1\\Delta {S}_{298}^{\\circ }\\text{Cu}\\left(s\\right)+1\\Delta {S}_{298}^{\\circ }{\\text{ZnSO}}_{4}\\left(s\\right)\\right]-\\left[1\\Delta {S}_{298}^{\\circ }\\text{Zn}\\left(s\\right)+1\\Delta {S}_{298}^{\\circ }{\\text{CuSO}}_{4}\\left(s\\right)\\right]\\\\{}&=&\\left[\\text{1 mol}\\left(33.15\\frac{\\text{J}}{\\text{mol K}}\\right)+\\text{1 mol}\\left(110.5\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(41.6\\frac{\\text{J}}{\\text{mol K}}\\right)+\\text{1 mol}\\left(109.2\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=-\\text{7.2 J\/K}\\end{array}[\/latex]<\/li>\n<\/ol>\n<p>4. The reaction is [latex]{\\text{C}}_{3}{\\text{H}}_{8}\\left(g\\right)+5{\\text{O}}_{2}\\left(g\\right)\\longrightarrow 3{\\text{CO}}_{2}\\left(g\\right)+4{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}\\\\ \\Delta {S}_{298}^{\\circ }&=&\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(reactants)}\\\\ &=&\\left[3\\Delta {S}_{298}^{\\circ }{\\text{CO}}_{2}\\left(g\\right)+4\\Delta {S}_{298}^{\\circ }{\\text{H}}_{2}\\text{O}\\left(g\\right)\\right]-\\left[1\\Delta {S}_{298}^{\\circ }{\\text{C}}_{3}{\\text{H}}_{8}\\left(g\\right)+5\\Delta {S}_{298}^{\\circ }{\\text{O}}_{2}\\left(g\\right)\\right]\\\\ &=&\\left[\\text{3 mol}\\left(213.6\\frac{\\text{J}}{\\text{mol K}}\\right)+\\text{4 mol}\\left(69.91\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(269.9\\frac{\\text{J}}{\\text{mol K}}\\right)+\\text{5 mol}\\left(205.03\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=\\text{100.6 J\/K}\\end{array}[\/latex]<\/p>\n<p>6. [latex]\\Delta {S}_{298}^{\\circ }[\/latex] for each change is as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\begin{array}{rll}\\\\ \\\\ \\Delta {S}_{\\text{sys}}^{\\circ }&=&\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(reactants)}\\\\\\Delta {S}_{\\text{sys}}^{\\circ }&=&\\left(2\\Delta {S}_{298}^{\\circ }{\\text{NH}}_{3}\\right)-\\left(1\\Delta {S}_{298}^{\\circ }{\\text{N}}_{2}+3\\Delta {S}_{298}^{\\circ }{\\text{H}}_{2}\\right)\\\\\\Delta {S}_{\\text{sys}}^{\\circ }&=&\\left[\\text{2 mol}\\left(192.8\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(191.6\\frac{\\text{J}}{\\text{mol K}}\\right)+\\text{3 mol}\\left(130.7\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=-\\text{198.1 J\/K}\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{rll}\\\\ \\\\ \\Delta {S}_{\\text{sys}}^{\\circ }&=&\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(products)}-\\sum \\nu \\Delta {S}_{298}^{\\circ }\\text{(reactants)}\\\\ \\Delta {S}_{\\text{sys}}^{\\circ }&=&\\left(2\\Delta {S}_{298}^{\\circ }{\\text{N}}_{2}{\\text{O}}_{5}\\right)-\\left(1\\Delta {S}_{298}^{\\circ }{\\text{N}}_{2}+\\frac{5}{2}\\times \\Delta {S}_{298}^{\\circ }{\\text{O}}_{2}\\right)\\\\\\Delta {S}_{\\text{sys}}^{\\circ }&=&\\left[\\text{1 mol}\\left(355.7\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]-\\left[\\text{1 mol}\\left(191.6\\frac{\\text{J}}{\\text{mol K}}\\right)+\\frac{5}{2}\\text{mol}\\left(205.2\\frac{\\text{J}}{\\text{mol K}}\\right)\\right]=-\\text{348.9 J\/K}\\end{array}[\/latex]<\/li>\n<\/ol>\n<p>8. The process is [latex]\\text{NaCl}\\left(s\\right)\\longrightarrow \\text{NaCl}\\left(l\\right)[\/latex].\u00a0At 500 \u00b0C, the following is true:<\/p>\n<p style=\"text-align: center;\">[latex]\\Delta {S}_{\\text{univ}}=\\Delta {S}_{\\text{sys}}+\\frac{{q}_{\\text{surr}}}{T}=\\left(95.06 - 72.11\\right)\\frac{\\text{J}}{\\text{mol}\\cdot\\text{ K}}+\\frac{-27.95\\times {10}^{3}\\frac{\\text{J}}{\\text{mol}}}{500+273.15}=-13.2\\frac{\\text{J}}{\\text{mol}\\cdot\\text{ K}}[\/latex]<\/p>\n<p>At 700 \u00b0C, the following is true:<\/p>\n<p style=\"text-align: center;\">[latex]\\Delta {S}_{\\text{univ}}=\\Delta {S}_{\\text{sys}}+\\frac{{q}_{\\text{surr}}}{T}=\\left(95.06 - 72.11\\right)\\frac{\\text{J}}{\\text{mol}\\cdot\\text{ K}}+\\frac{-27.95\\times {10}^{3}\\frac{\\text{J}}{\\text{mol}}}{700+273.15}=-5.8\\frac{\\text{J}}{\\text{mol}\\cdot\\text{ K}}[\/latex]<\/p>\n<p>As \u0394<em>S<\/em><sub>univ<\/sub> &lt; 0 at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K. It is assumed that these do not change significantly at the higher temperatures used in the problem.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>second law of thermodynamics: <\/strong>entropy of the universe increases for a spontaneous process<\/p>\n<p><strong>standard entropy (<em>S<\/em>\u00b0): <\/strong>entropy for a substance at 1 bar pressure; tabulated values are usually determined at 298.15 K and denoted [latex]{S}_{298}^{\\circ }[\/latex]<\/p>\n<p><strong>standard entropy change (\u0394<em>S<\/em>\u00b0): <\/strong>change in entropy for a reaction calculated using the standard entropies, usually at room temperature and denoted [latex]\\Delta {S}_{298}^{\\circ }[\/latex]<\/p>\n<p><strong>third law of thermodynamics: <\/strong>entropy of a perfect crystal at absolute zero (0 K) is zero<\/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-3622\">\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>Chemistry. <strong>Provided by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/openstaxcollege.org\">http:\/\/openstaxcollege.org<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get<\/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":17,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry\",\"author\":\"\",\"organization\":\"OpenStax College\",\"url\":\"http:\/\/openstaxcollege.org\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3622","chapter","type-chapter","status-publish","hentry"],"part":2977,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/3622","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":12,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/3622\/revisions"}],"predecessor-version":[{"id":5877,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/3622\/revisions\/5877"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/parts\/2977"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/3622\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/media?parent=3622"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapter-type?post=3622"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/contributor?post=3622"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/license?post=3622"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}