{"id":391,"date":"2017-08-21T06:47:35","date_gmt":"2017-08-21T06:47:35","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/chemistry\/?page_id=391"},"modified":"2017-09-20T18:12:48","modified_gmt":"2017-09-20T18:12:48","slug":"hesss-law-and-gibbs-free-energy","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/chemistry\/hesss-law-and-gibbs-free-energy\/","title":{"rendered":"Hess&#8217;s Law and Gibbs Free Energy"},"content":{"rendered":"<div class=\"twelve columns\" style=\"margin-top: 10%;\">\n<div class=\"advance\">\n<p><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/energy-work-and-heat-flow\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/reactions-and-reactivity\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/le-chateliers-principle-and-equilibrium-constants\">Next Lesson\u00a0\u27a1<\/a><\/p>\n<\/div>\n<p><!-- UPDATE NEXT\/PREVIOUS ABOVE --><\/p>\n<p><!-- CONTENT STARTS HERE --><\/p>\n<h1 id=\"title\">Hess&#8217;s Law and Gibbs Free Energy<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson we will review Hess\u2019s law and how it is used in calculations. We will also solve for\u00a0Gibbs Free Energy.<\/p>\n<h4>Previously we covered&#8230;<\/h4>\n<ul>\n<li>Enthalpy is the internal energy of a system plus the pressure of the system multiplied by its\u00a0volume.<\/li>\n<li>Entropy is the measure of disorder of a system.<\/li>\n<li>The free energy of a system is the minimum amount of energy available when a system is in\u00a0equilibrium with its surroundings.<\/li>\n<li>Performing calculations related to internal energy and work<\/li>\n<li>Performing calorimetry calculations using\u00a0<em>q <\/em>= <em>mC<sub>p<\/sub><\/em>\u0394<em>T<\/em><\/li>\n<li>Calculating energy of phase changes<\/li>\n<li>Using the Clausius-Clapeyron equation to determine vapor pressure and temperature<\/li>\n<\/ul>\n<section>\n<h3>Hess\u2019s Law<\/h3>\n<p>Hess\u2019s law states that in a reaction the\u00a0change in enthalpy is the same\u00a0whether the reaction takes place in one step or a series of steps.<\/p>\n<p>To determine the change in enthalpy, we may rearrange\u00a0the given equations to\u00a0obtain the desired equation, and then add\u00a0the corresponding changes in enthalpy to\u00a0determine\u00a0the enthalpy for the overall reaction.<\/p>\n<h3>Sample Problem<\/h3>\n<p>Given the following equation: B \u2192 E + 2C,\u00a0calculate \u2206H.<\/p>\n<p>The intermediate steps and their respective enthalpies are given below<\/p>\n<table class=\"gas_law_table\">\n<tbody>\n<tr>\n<td valign=\"top\" width=\"168\"><\/td>\n<td valign=\"top\" width=\"168\"><strong>\u2206<em>H<\/em> (kJ\/mol) <\/strong><\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" width=\"168\">(1\/2)A \u2192 B<\/td>\n<td valign=\"top\" width=\"168\">150<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" width=\"168\">3B \u2192 2C + D<\/td>\n<td valign=\"top\" width=\"168\">-125<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" width=\"168\">E + A \u2192 D<\/td>\n<td valign=\"top\" width=\"168\">350<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>To obtain the third given equation, select the substance that only appears in one of the intermediate equations, in\u00a0this case E. The\u00a0desired equation has E on the reactants side, but we want it to be on the\u00a0products side, so reversing the third given equation we get:<\/p>\n<p class=\"center\">D \u2192 E +\u00a0A, \u2206<em>H<\/em> = -350 kJ\/mol<\/p>\n<p>Then select another substance that only appears once in the given equations and that is in the desired equation, in<br \/>\nthis case C. 2C is desired to be on the products side in the final equation and it is in the second given equation,<br \/>\nso the second step stays the same.<\/p>\n<p class=\"center\">3B \u2192 2C + D, \u2206<em>H<\/em>=-125.<\/p>\n<p>Next, there is only B remaining in the desired equation. 3B is already a reactant, but only B is desired, so 2B<br \/>\nmust be on the products side to cancel out 2B from the reactant side. The first equation with B as a product must be<br \/>\nmultiplied by two to cancel out 2B from the reactants side. Remember also to multiply the change in enthalpy by 2<br \/>\nbecause the amount of products and reactants is doubled<\/p>\n<p class=\"center\">A \u2192 2B, \u2206<em>H<\/em> = 300<\/p>\n<p>Next, add the three reactions and the corresponding changes in enthalpy, then cancel as\u00a0necessary.<\/p>\n<table class=\"gas_law_table\">\n<thead>\n<tr>\n<th valign=\"top\" width=\"272\"><\/th>\n<th valign=\"top\" width=\"135\"><strong>\u2206<em>H<\/em> (kJ\/mol) <\/strong><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td valign=\"top\" width=\"272\">D \u2192 E + A<\/td>\n<td valign=\"top\" width=\"135\">-350<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" width=\"272\">3B \u2192 2C + D<\/td>\n<td valign=\"top\" width=\"135\">-125<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" width=\"272\">A \u2192 2B<\/td>\n<td valign=\"top\" width=\"135\">300<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" width=\"272\">D + 3B + A \u2192 E + A + 2C + D + 2B<\/td>\n<td valign=\"top\" width=\"135\">-175 kJ\/mol<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now, cancel the terms in the same amount that appear on both sides of the equation. One D is on both sides and\u00a0cancels. Two B\u2019s are on either side of the arrow and cancel as does the one A.<\/p>\n<p class=\"center\">\u2206<em>H<\/em> for B \u2192 E + 2C is -175 kJ\/mol<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Use Hess&#8217;s law to find the change in enthalpy at 25\u00b0C for the following equation:<\/p>\n<p class=\"center\">CaC<sub>2<\/sub>(<em>s<\/em>) + 2H<sub>2<\/sub>O(<em>l<\/em>) \u2192 Ca(OH)<sub>2<\/sub>(<em>aq<\/em>) + C<sub>2<\/sub>H<sub>2<\/sub>(<em>g<\/em>)<\/p>\n<p>Given:<\/p>\n<table class=\"gas_law_table\">\n<tbody>\n<tr>\n<td valign=\"top\" width=\"354\"><\/td>\n<td valign=\"top\" width=\"78\"><strong>\u2206<em>H<\/em> (kJ) <\/strong><\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">Ca(<em>s<\/em>) + 2C(graphite) \u2192 CaC<sub>2<\/sub>(<em>s<\/em>)<\/td>\n<td valign=\"top\">-62.8<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">Ca(<em>s<\/em>) + (1\/2)O<sub>2<\/sub>(<em>g<\/em>) \u2192 CaO(<em>s<\/em>)<\/td>\n<td valign=\"top\">-635.5<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">CaO(<em>s<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>) \u2192 Ca(OH)<sub>2<\/sub>(<em>aq<\/em>)<\/td>\n<td valign=\"top\">-653.1<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">C<sub>2<\/sub>H<sub>2<\/sub>(<em>g<\/em>) + (5\/2)O<sub>2<\/sub>(<em>g<\/em>) \u2192 2CO<sub>2<\/sub>(<em>g<\/em>)+<br \/>\nH<sub>2<\/sub>O(<em>l<\/em>)<\/td>\n<td valign=\"top\">-1300<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">C(graphite) + O<sub>2<\/sub>(<em>g<\/em>) \u2192 CO<sub>2<\/sub>(<em>g<\/em>)<\/td>\n<td valign=\"top\">-393.51<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol>\n<li>-3.0410<sup>3<\/sup>kJ<\/li>\n<li>-7.13\u00d710<sup>2<\/sup>kJ<\/li>\n<li>-3.19\u00d710<sup>2 <\/sup>kJ<\/li>\n<li>-8.38\u00d710<sup>2 <\/sup>kJ<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is B.\u00a0Solve the problem as follows:\u00a0CaC<sub>2<\/sub> is desired on the reactants side. The only equation with CaC<sub>2<\/sub> is\u00a0the first one, but to get calcium carbide as a reactant the equation\u00a0must be reversed. This also causes the enthalpy to become positive.<\/p>\n<p class=\"q-reveal center\">CaC<sub>2<\/sub>(<em>s<\/em>) \u2192 2C(graphite) + Ca(s), \u0394H = +62.8kJ<\/p>\n<p class=\"q-reveal\">Skip water since it appears in multiple places in the given equations.\u00a0Calcium hydroxide is a desired product and the only reaction with the\u00a0substance is the third equation, where it is a product. This reaction\u00a0will stay the same.<\/p>\n<p class=\"center q-reveal\">CaO(<em>s<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>) \u2192 Ca(OH)<sub>2<\/sub>(<em>aq<\/em>), \u0394H = \u2013653.1 kJ<\/p>\n<p class=\"q-reveal\">Acetylene is a desired product. It appears as a reactant in the fourth\u00a0equation, so that equation will need to be reversed as well as the sign\u00a0of the enthalpy.<\/p>\n<p class=\"center q-reveal\">2CO<sub>2<\/sub> + H<sub>2<\/sub>O(<em>l<\/em>) \u2192 C<sub>2<\/sub>H<sub>2<\/sub>(<em>g<\/em>)\u00a0+ (5\/2)O<sub>2<\/sub>(<em>g<\/em>), \u0394H = +1300 kJ<\/p>\n<p class=\"q-reveal\">Next,\u00a0two carbons in the form of graphite are currently products due to\u00a0reversing the first equation. Since carbon is not a desired product,\u00a0two carbons must be reactants to cancel the two that appear as\u00a0products. Hence, the fifth equation needs to be multiplied by two, as\u00a0will its enthalpy.<\/p>\n<p class=\"center q-reveal\">2C(graphite) + 2O<sub>2<\/sub>(<em>g<\/em>) \u2192 2CO<sub>2<\/sub>(<em>g<\/em>), \u0394H = \u2013787.02 kJ<\/p>\n<p class=\"q-reveal\">Taking a look at the second equation, if left as is the equation will\u00a0cancel the calcium and calcium oxide from the other equations that are\u00a0not a desired product. This reaction stays as is.<\/p>\n<p class=\"center q-reveal\">Ca(s) + (1\/2)O<sub>2<\/sub>(g) \u2192 CaO(<em>s<\/em>),\u00a0\u0394H = \u2013635.5 kJ<\/p>\n<p class=\"q-reveal\">Add the five equations, their respective enthalpies and cancel as necessary.<\/p>\n<p class=\"q-reveal\">After canceling, we get the following equation:<\/p>\n<p class=\"center q-reveal\">CaC<sub>2<\/sub>(<em>s<\/em>) + 2H<sub>2<\/sub>O(<em>l<\/em>) \u2192 Ca(OH)<sub>2<\/sub>(<em>aq<\/em>) +<br \/>\nC<sub>2<\/sub>H<sub>2<\/sub>(<em>g<\/em>) , \u0394<em>H<\/em> = -7.13 x 10<sup>2<\/sup> kJ<\/p>\n<\/section>\n<h3>Gibbs Free Energy<\/h3>\n<p>Free energy is a thermodynamic function that is equal in value to the enthalpy minus the Kelvin temperature multiplied by the entropy.<\/p>\n<p>The change in Gibbs Free Energy can be determined using the equation<\/p>\n<p class=\"center\">\u0394<em>G<\/em> =\u0394<em>H<\/em> \u2013 <em>T<\/em>\u0394<em>S<\/em><\/p>\n<p><em>G<\/em> is free energy, <em>H<\/em> is\u00a0enthalpy, <em>T<\/em> is temperature in\u00a0Kelvin,\u00a0and <em>S<\/em> is entropy. The values for \u0394<em>G<\/em> and \u0394<em>H<\/em> are measured in kJ, \u0394<em>S<\/em> is measured in joules, so the value must be divided by 1000 to get\u00a0kilojoules.<\/p>\n<p>To solve, we\u00a0must find the total enthalpy of the\u00a0products minus the enthalpy of the reactants. We can use Hess\u2019s Law to solve for changes in entropy and\u00a0enthalpy. Thus:<\/p>\n<p class=\"center\">\u0394<em>G<\/em> =\u03a3<em>G<\/em><sub>products<\/sub> \u2212 \u03a3<em>G<\/em><sub>reactants<\/sub><\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s9_006.gif\" width=\"365\" height=\"44\" \/><\/center>The sign of the values for the change in free energy (\u0394<em>G<\/em>), enthalpy\u00a0(\u0394<em>H<\/em>) and entropy (\u0394<em>S<\/em>) can be used to determine whether the equation is <abbr title=\"The reaction occurs without any outside influence. \">spontaneous<\/abbr>.<\/p>\n<table class=\"gas_law_table\">\n<tbody>\n<tr>\n<th valign=\"top\" width=\"12%\"><strong> \u0394<em>G<\/em><\/strong><\/th>\n<th valign=\"top\" width=\"12%\"><strong> \u0394<em>H<\/em><\/strong><\/th>\n<th valign=\"top\" width=\"12%\"><strong> \u0394<em>S<\/em><\/strong><\/th>\n<th valign=\"top\" width=\"66%\"><strong> Reaction <\/strong><\/th>\n<\/tr>\n<tr>\n<td valign=\"top\">(-)<\/td>\n<td valign=\"top\" width=\"12%\">(-)<\/td>\n<td valign=\"top\" width=\"12%\">(+)<\/td>\n<td valign=\"top\" width=\"66%\">The reaction is always spontaneous<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">(+ or -)<\/td>\n<td valign=\"top\" width=\"12%\">(+)<\/td>\n<td valign=\"top\" width=\"12%\">(+)<\/td>\n<td valign=\"top\" width=\"66%\">The reaction is spontaneous at high temperatures<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">(+ or -)<\/td>\n<td valign=\"top\" width=\"12%\">(-)<\/td>\n<td valign=\"top\" width=\"12%\">(-)<\/td>\n<td valign=\"top\" width=\"66%\">The reaction is spontaneous at low temperatures<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">(+)<\/td>\n<td valign=\"top\" width=\"12%\">(+)<\/td>\n<td valign=\"top\" width=\"12%\">(-)<\/td>\n<td valign=\"top\" width=\"66%\">The reaction is never spontaneous<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>When \u0394<em>H<\/em> = <em>T<\/em>\u0394<em>S<\/em>, the value for \u0394<em>G<\/em> = 0 and the reaction is at equilibrium.<\/p>\n<h4>Sample Problem<\/h4>\n<p>Find Gibbs free energy for the following reaction at 298 K.<\/p>\n<p class=\"center\">Fe<sub>3<\/sub>O<sub>4<\/sub>(<em>s<\/em>, magnetite) + Al(<em>s<\/em>) \u2192 Al<sub>2<\/sub>O<sub>3<\/sub>(<em>s<\/em>) + Fe(<em>s<\/em>)<\/p>\n<p>To find the values for the substances, look at a chart of thermodynamic data.<\/p>\n<table class=\"gas_law_table\">\n<tbody>\n<tr>\n<th valign=\"top\" width=\"148\">Substance<\/th>\n<th valign=\"top\" width=\"104\">\u0394<em>H<\/em>(kJ\/mol)<\/th>\n<th valign=\"top\" width=\"112\"><em>S<\/em>(J\/mol\u00b7K)<\/th>\n<\/tr>\n<tr>\n<td valign=\"top\">Al<sub>2<\/sub>O<sub>3<\/sub>(<em>s<\/em>)<\/td>\n<td valign=\"top\">-1676<\/td>\n<td valign=\"top\">51<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">Fe(<em>s<\/em>)<\/td>\n<td valign=\"top\">0<\/td>\n<td valign=\"top\">27<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">Fe<sub>3<\/sub>O<sub>4<\/sub>(<em>s<\/em>, magnetite)<\/td>\n<td valign=\"top\">-1117<\/td>\n<td valign=\"top\">146<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">Al(<em>s<\/em>)<\/td>\n<td valign=\"top\">0<\/td>\n<td valign=\"top\">28<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>First, remember\u00a0to balance the\u00a0equation:<\/p>\n<p class=\"center\">3Fe<sub>3<\/sub>O<sub>4<\/sub>(<em>s<\/em>, magnetite) + 8Al(<em>s<\/em>) \u2192 4Al<sub>2<\/sub>O<sub>3<\/sub>(<em>s<\/em>) + 9Fe(<em>s<\/em>)<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s9_006.gif\" width=\"365\" height=\"44\" \/><\/center><\/p>\n<p class=\"center\">\u0394<em>H<\/em> = (\u03a3<em>H<\/em><sub>products<\/sub> \u2212 \u03a3<em>H<\/em><sub>reactants<\/sub>)<\/p>\n<p class=\"center\">\u0394<em>H<\/em> = [4 mol(-1676 kJ\/mol) \u2013 9 mol(0 kJ\/mol)] \u2013 [8\u00a0mol(0 kJ\/mol) + 3 mol(-1117 kJ\/mol)]<\/p>\n<p style=\"text-align: center;\">\u0394<em>H<\/em> = -3353 kJ<\/p>\n<p class=\"center\">\u0394<em>S<\/em> = <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s9_008.gif\" width=\"150\" height=\"44\" \/><\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s9_009.gif\" width=\"638\" height=\"41\" \/><\/center><\/p>\n<p class=\"center\">\u0394<em>S<\/em> = -0.21 kJ\/K<br \/>\n\u0394<em>G<\/em> =\u0394<em>H<\/em> \u2013 <em>T<\/em>\u0394<em>S<\/em><br \/>\n\u0394<em>G<\/em> =-3353 kJ \u2013 (298 K)(-0.21 kJ\/K)<br \/>\n\u0394<em>G <\/em>= -3300 kJ<\/p>\n<p>Even though the value of the entropy change is negative the temperature is low enough that the negative enthalpy term still allows the free energy to be negative. Thus the reaction is spontaneous.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Determine the free energy and spontaneity for the following reaction that occurs at 333 K.<\/p>\n<p class=\"center\">KClO<sub>3<\/sub>(<em>s<\/em>)\u00a0 \u2192 KCl(<em>s<\/em>)\u00a0+\u00a0O<sub>2<\/sub>(<em>g<\/em>)<\/p>\n<table class=\"q_table\">\n<tbody>\n<tr>\n<th valign=\"top\" width=\"33%\">Substance<\/th>\n<th valign=\"top\" width=\"33%\">\u0394<em>H<\/em>(kJ\/mol)<\/th>\n<th valign=\"top\" width=\"117\"><em>S<\/em>(J\/mol\u00b7K)<\/th>\n<\/tr>\n<tr>\n<td valign=\"top\" width=\"33%\">KClO<sub>3<\/sub>(<em>s<\/em>)<\/td>\n<td valign=\"top\" width=\"33%\">-391<\/td>\n<td valign=\"top\">143<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" width=\"33%\">KCl(<em>s<\/em>)<\/td>\n<td valign=\"top\" width=\"33%\">\n<div align=\"center\">-436<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">83<\/div>\n<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" width=\"33%\">\n<div align=\"center\">KCl(<em>aq<\/em>)<\/div>\n<\/td>\n<td valign=\"top\" width=\"33%\">\n<div align=\"center\">-450<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">205<\/div>\n<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" width=\"33%\">\n<div align=\"center\">O<sub>2<\/sub>(<em>g<\/em>)<\/div>\n<\/td>\n<td valign=\"top\" width=\"33%\">\n<div align=\"center\">0<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">28<\/div>\n<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" width=\"33%\">\n<div align=\"center\">O<sub>2<\/sub>(<em>aq<\/em>)<\/div>\n<\/td>\n<td valign=\"top\" width=\"33%\">\n<div align=\"center\">-12<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">111<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol>\n<li>-67 kJ<\/li>\n<li>-78 kJ<\/li>\n<li>6.7 \u00d7 10<sup>2<\/sup> kJ<\/li>\n<li>7.8 \u00d7 10<sup>3<\/sup> kJ<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is B.<\/p>\n<p class=\"center q-reveal\" style=\"text-align: center;\">2KClO<sub>3<\/sub>(<em>s<\/em>) \u2192 2KCl(<em>s<\/em>)+ 3O<sub>2<\/sub>(<em>g<\/em>)<\/p>\n<p class=\"center q-reveal\">\u0394<em>H<\/em> = (\u03a3<em>H<\/em><sub>products<\/sub> \u2212 \u03a3<em>H<\/em><sub>reactants<\/sub>)<br \/>\n\u0394<em>H<\/em> = [3 mol(0 kJ\/mol) + 2 mol(-436 kJ\/mol)] \u2013 [2 mol(-391 kJ\/mol)]<br \/>\n\u0394H = -90 kJ<br \/>\n\u0394S = <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s9_008.gif\" width=\"150\" height=\"44\" \/><br \/>\n\u0394S = [3 mol(28 kJ\/mol) + 2 mol(83 kJ\/mol)] \u2013 [2 mol(143 kJ\/mol)]<br \/>\n\u0394S = -36 kJ<br \/>\n\u0394G = -90 kJ \u2013 (333 K)(-0.036 kJ)<br \/>\n\u0394G = -78 kJ<\/p>\n<p class=\"q-reveal\">The reaction is spontaneous.<\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Determine\u00a0Gibbs free energy\u00a0and the spontaneity for the\u00a0following reaction at 34.5\u00b0C.<\/p>\n<p class=\"center\">NH<sub>3<\/sub>(<em>g<\/em>)\u00a0+ HCl(<em>aq<\/em>) \u2192 NH<sub>4<\/sub>Cl(<em>aq<\/em>)<\/p>\n<table class=\"q_table\">\n<tbody>\n<tr>\n<td valign=\"top\" width=\"92\">\n<div align=\"center\"><strong> Substance <\/strong><\/div>\n<\/td>\n<td valign=\"top\" width=\"110\">\n<div align=\"center\"><strong> \u0394<em>H<\/em>(kJ\/mol) <\/strong><\/div>\n<\/td>\n<td valign=\"top\" width=\"112\">\n<div align=\"center\"><strong> \u0394<em>S<\/em>(J\/mol\u00b7K) <\/strong><\/div>\n<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">\n<div align=\"center\">NH<sub>3<\/sub>(<em>g<\/em>)<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">\u201346<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">193<\/div>\n<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">\n<div align=\"center\">NH<sub>3<\/sub>(<em>aq<\/em>)<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">\u201380<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">111<\/div>\n<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">\n<div align=\"center\">HCl(<em>aq<\/em>)<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">\u2013167<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">57<\/div>\n<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">\n<div align=\"center\">HCl(<em>g<\/em>)<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">\u201392<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">187<\/div>\n<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">\n<div align=\"center\">NH<sub>4<\/sub>Cl(<em>aq<\/em>)<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">\u2013300<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">170<\/div>\n<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">\n<div align=\"center\">NH<sub>4<\/sub>Cl(<em>s<\/em>)<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">\u2013314<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">96<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol>\n<li>-94 kJ<\/li>\n<li>-62 kJ<\/li>\n<li>2.6\u00d710<sup>3<\/sup> kJ<\/li>\n<li>2.4\u00d710<sup>4<\/sup> kJ<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is B.<\/p>\n<p class=\"center q-reveal\" style=\"text-align: center;\">NH<sub>3<\/sub>(<em>g<\/em>) + HCl(<em>aq<\/em>) \u2192 NH<sub>4<\/sub>Cl(<em>aq<\/em>)<\/p>\n<p class=\"center q-reveal\" style=\"text-align: center;\">\u0394<em>H<\/em> = (\u03a3<em>H<\/em><sub>products<\/sub> \u2212 \u03a3<em>H<\/em><sub>reactants<\/sub>)<\/p>\n<p class=\"center q-reveal\" style=\"text-align: center;\">\u0394<em>H<\/em> = [1 mol(-300 kJ\/mol)] &#8211; [1 mol (-46 kJ\/mol) + 1 mol (-167kJ\/mol)]<\/p>\n<p class=\"center q-reveal\" style=\"text-align: center;\">\u0394<em>H<\/em> = -87 kJ<\/p>\n<p class=\"center q-reveal\" style=\"text-align: center;\">\u0394S = <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s9_008.gif\" width=\"150\" height=\"44\" \/><\/p>\n<p class=\"center q-reveal\" style=\"text-align: center;\">\u0394S = <img loading=\"lazy\" decoding=\"async\" class=\"non_block_image\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s9_023.gif\" width=\"477\" height=\"41\" align=\"absmiddle\" \/><\/p>\n<p class=\"center q-reveal\" style=\"text-align: center;\">\u0394S = -0.080 kJ\/K<\/p>\n<p class=\"center q-reveal\" style=\"text-align: center;\">\u0394G = -87 kJ &#8211; (307.65 K)(-0.080 kJ\/K)<\/p>\n<p class=\"center q-reveal\" style=\"text-align: center;\">\u0394G = -62 kJ<\/p>\n<p class=\"q-reveal\">The reaction is spontaneous at low temperatures.<\/p>\n<\/section>\n<p><!-- CONTENT ENDS HERE --><\/p>\n<p><!-- UPDATE NEXT\/PREVIOUS BELOW --><\/p>\n<div class=\"advance\"><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/energy-work-and-heat-flow\">\u2b05 Previous\u00a0Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/reactions-and-reactivity\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/le-chateliers-principle-and-equilibrium-constants\">Next Lesson\u00a0\u27a1<\/a><\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/section>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>\u2b05 Previous Lesson\u00a0Workshop Index\u00a0Next Lesson\u00a0\u27a1 Hess&#8217;s Law and Gibbs Free Energy Objective In this lesson we will review Hess\u2019s law and how it is used in calculations. We will also solve for\u00a0Gibbs Free Energy. Previously we covered&#8230; Enthalpy is the internal energy of a system plus the pressure of the system multiplied by its\u00a0volume. Entropy [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-391","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/391","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/comments?post=391"}],"version-history":[{"count":16,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/391\/revisions"}],"predecessor-version":[{"id":934,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/391\/revisions\/934"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/media?parent=391"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}