{"id":1328,"date":"2020-06-23T16:44:18","date_gmt":"2020-06-23T20:44:18","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/chbe220\/?post_type=chapter&#038;p=1328"},"modified":"2020-08-20T14:00:47","modified_gmt":"2020-08-20T18:00:47","slug":"reactive-energy-balances","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/chbe220\/chapter\/reactive-energy-balances\/","title":{"raw":"Reactive Energy Balances","rendered":"Reactive Energy Balances"},"content":{"raw":"<div class=\"textbox textbox--learning-objectives\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Learning Objectives<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nBy the end of this section, you should be able to:\r\n<p id=\"Explain:\"><strong>Explain<\/strong>\u00a0heats of reaction as well as endothermic and exothermic reactions<\/p>\r\n<p id=\"Determine:\"><strong>Determine<\/strong>\u00a0the standard heat of reaction given other heats of reaction or heats of formation (Hess's Law)<\/p>\r\n<p id=\"Analyze:\"><strong>Analyze<\/strong>\u00a0energy balances involving reactive systems<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<h2 id=\"What-happens-in-reactions-in-terms-of-chemical-bonds?\">What happens in reactions in terms of chemical bonds?<\/h2>\r\nBonds can be formed and broken. Breaking bonds <strong>takes<\/strong> energy and forming bonds <strong>releases<\/strong> energy.\r\n<ul>\r\n \t<li>If more energy is released in forming bonds than absorbed in breaking bonds, then the reaction is <strong>exothermic<\/strong>.<\/li>\r\n \t<li>If more energy is absorbed in breaking bonds than released in forming bonds, then the reaction is <strong>endothermic<\/strong>.<\/li>\r\n<\/ul>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example: Chemical Bonds<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nConsider the simple reaction of the formation of water from hydrogen and oxygen:\r\n<p style=\"text-align: center\">[latex]2H_{2} + O_{2} \u2192 2H_{2}O[\/latex]<\/p>\r\n<img class=\" wp-image-1055 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/WaterRxn-300x180.png\" alt=\"\" width=\"420\" height=\"252\" \/>\r\n<p style=\"text-align: center\">Image from\u00a0 <a title=\"via Wikimedia Commons\" href=\"https:\/\/commons.wikimedia.org\/wiki\/File:H2%2BO2%3DH2O.svg\">\u3059\u3058\u306b\u304f\u30b7\u30c1\u30e5\u30fc<\/a> \/ CC0<\/p>\r\n<em>The formation of bonds takes energy while the breakage of bonds releases energy<\/em>\r\n<blockquote>In this reaction:<\/blockquote>\r\n<ul>\r\n \t<li>\r\n<blockquote>2 H-H bonds are broken<\/blockquote>\r\n<\/li>\r\n \t<li>\r\n<blockquote>\u00a01 O-O bonds are broken<\/blockquote>\r\n<\/li>\r\n \t<li>\r\n<blockquote>4 H-O bonds are formed<\/blockquote>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<h2 id=\"Heat-of-Reaction-$\\Delta-H_{r}(T,P)$\">Heat of Reaction [latex]\\Delta H_{r}(T,P)[\/latex]<\/h2>\r\n<strong>Heat of Reaction [latex]\\Delta H_{r}(T,P)[\/latex]<\/strong>: the stoichiometric enthalpy difference when reactants react completely to form products at a specified constant temperature and pressure.\r\n<p style=\"text-align: center\">[latex]\\Delta H_{r}(T,P) = H_{products} - H_{reactants}[\/latex]<\/p>\r\n\r\n<blockquote>In <strong>exothermic reactions<\/strong>, [latex]\\Delta H_{r}(T,P)&lt;0[\/latex]\r\n<ul>\r\n \t<li>[latex]H_{products} &lt; H_{reactants}[\/latex]<\/li>\r\n<\/ul>\r\nIn <strong>endothermic reactions<\/strong>, [latex]\\Delta H_{r}(T,P)&gt;0[\/latex]\r\n<ul>\r\n \t<li><span style=\"color: #333333\">[latex]H_{products} &gt; H_{reactants}[\/latex]<\/span><\/li>\r\n<\/ul>\r\n<\/blockquote>\r\nHeats of reaction are <em>directly proportional to the amount of reactants or products<\/em> in a reaction.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example: Formation of Nitrogen<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nConsider the formation of nitrogen dioxide\r\n<p style=\"text-align: center\">[latex]\\frac{1}{2} N_{2} (g) + O_{2}(g) \u2192 NO_{2} (g)[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta H_{r} = 33.2 kJ[\/latex]<\/p>\r\nThe formation of 1 mole of [latex]NO_{2}[\/latex] yields an enthalpy change of 33.2 kJ\r\n<p style=\"text-align: center\">[latex]N_{2} (g) + 2O_{2}(g) \u2192 2NO_{2} (g)[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta H_{r} = 66.4 kJ[\/latex]<\/p>\r\nThe formation of 2 moles of [latex]NO_{2}[\/latex] yields an enthalpy change of 66.2 kJ\r\n\r\nHeat of reactions for a reaction in the forward direction is *equal to the negative heat of reaction* for the backward reaction.\r\n<blockquote>Consider the formation of hydrogen chloride (or the corresponding decomposition):\r\n<p style=\"text-align: center\">[latex]H_{2} (g) + Cl_{2} (g) \u2192 2 HCl (g)[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta H_{r} = -184.6 kJ[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]2HCl(g) \u2192 H_{2} (g) + Cl_{2} (g) [\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta H_{r} = +184.6 kJ[\/latex]<\/p>\r\n<\/blockquote>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise: Heat of Reaction<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<div>\r\n\r\nConsider the oxidation of ammonia taking place in an isothermal and constant pressure reactor:\r\n<p style=\"text-align: center\">[latex] 4 NH_{3}(g) + 5 O_{2}(g) \u2192 4 NO(g) + 6 H_{2}O(v)[\/latex]<\/p>\r\nwhere the heat of reaction is [latex]\\Delta\\hat{H}^{\\circ}_{r} = -904.7 kJ[\/latex]\r\n\r\nThe ammonia is fed into the reactor at [latex]100 mol\/s[\/latex] and the oxygen enters at [latex]200 mol\/s[\/latex]. Assuming the limiting reactant is completely consumed, what is the enthalpy change for this reaction?\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Solution<\/h3>\r\n<strong>Step 1:<\/strong> Find the limiting reactant by finding the smallest reactant extent if all of a reactant is consumed.\r\n<p style=\"text-align: center\">[latex]\\text{if } N\\!H_{3} \\text{ is limiting: } \\xi=\\frac{100mol\/s}{4}=25\\frac{mol}{s}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\text{if } O_{2} \\text{ is limiting: } \\xi=\\frac{200mol\/s}{5}=40\\frac{mol}{s}[\/latex]<\/p>\r\n[latex]N\\!H_{3}[\/latex] being the limiting reagent yields the smaller reaction extent, therefore [latex]N\\!H_{3}[\/latex] is the limiting reagent. The reaction extent is [latex]25\\frac{mol}{s}[\/latex].\r\n\r\n<strong>Step 2:<\/strong> Multiply the reaction extent by the [latex]\\Delta\\hat{H}^{\\circ}_{r} = -904.7 \\frac{kJ}{molNH_{3}}[\/latex] to obtain [latex]\\Delta \\dot{H}_{r}[\/latex]\r\n<p style=\"text-align: center\">[latex]\\Delta \\dot{H}_{r}=\\xi*\\Delta\\hat{H}^{\\circ}_{r}[\/latex]\r\n[latex]\\Delta \\dot{H}_{r}=25\\frac{mol}{s}*-904.7\\frac{kJ}{mol}[\/latex]\r\n[latex]\\Delta \\dot{H}_{r}=-22620 \\frac{kJ}{s}[\/latex]<\/p>\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n&nbsp;\r\n<h3 id=\"Hess's-Law\">Hess's Law<\/h3>\r\nSome reactions may be difficult to reproduce in a laboratory setting. Instead, for analyzing a certain overall reaction, we may break the reaction down in multiple steps:\r\n\r\nFor example, say we want to find the standard heat of reaction for the following hypothetical reaction:\r\n<p style=\"text-align: center\">[latex] A + 0.5 B \u2192 C [\/latex]<\/p>\r\nIf this reaction is difficult to carry out in a laboratory, we might break it up into 2 reactions, that are easier to carry out and where we can determine specified standard heats of reaction:\r\n<blockquote>Reaction 1: [latex]A + B \u2192 D[\/latex] with [latex]\\Delta H^{\\circ}_{rxn1}[\/latex]\r\n\r\nReaction 2: [latex]C + 0.5 B \u2192 D[\/latex] with [latex]\\Delta H^{\\circ}_{rxn2}[\/latex]<\/blockquote>\r\nWe can combine reactions 1 and 2 to obtain our desired reaction by subtracting reaction 2 from reaction 1:\r\n<p style=\"text-align: center\">[latex]Reaction 1 - Reaction 2[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex] (A + B) - (C + 0.5 B) \u2192 D - D [\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex] A + 0.5 B - C \u2192 [\/latex]<\/p>\r\nWe obtain our desired reaction:\r\n<p style=\"text-align: center\">[latex] A + 0.5 B \u2192 C [\/latex]<\/p>\r\nTherefore, the heat of reaction of the desired reaction will be:\r\n<blockquote>\r\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ} = \\Delta H^{\\circ}_{rxn1} - \\Delta H^{\\circ}_{rxn2}[\/latex]<\/p>\r\n<\/blockquote>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise: Hess's Law<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n\r\nConsider the formation of chlorine trifluoride from chlorine monofluoride and fluorine:\r\n<p style=\"text-align: center\">[latex] ClF(g) + F_{2}(g) \u2192 ClF_{3}(g) [\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ} = ?[\/latex]<\/p>\r\nData on the following reactions are available:\r\n<table class=\"grid aligncenter\" style=\"height: 90px\">\r\n<thead>\r\n<tr style=\"height: 30px\">\r\n<th style=\"width: 69.45px;text-align: center;height: 30px\">Number<\/th>\r\n<th style=\"width: 539.85px;text-align: center;height: 30px\">Reaction<\/th>\r\n<th style=\"width: 179.05px;text-align: center;height: 30px\">[latex]\\Delta H^{\\circ}[\/latex] (kJ)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 69.45px;text-align: center;height: 15px\">1<\/td>\r\n<td style=\"width: 539.85px;text-align: center;height: 15px\">[latex]2OF_{2}(g) \u2192 O_{2}(g) + 2F_{2}(g)[\/latex]<\/td>\r\n<td style=\"width: 179.05px;text-align: center;height: 15px\">-49.4<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 69.45px;text-align: center;height: 15px\">2<\/td>\r\n<td style=\"width: 539.85px;text-align: center;height: 15px\">[latex]2ClF(g) + O_{2}(g) \u2192 Cl_{2}O(g) + OF_{2}(g)[\/latex]<\/td>\r\n<td style=\"width: 179.05px;text-align: center;height: 15px\">214.0<\/td>\r\n<\/tr>\r\n<tr style=\"height: 30px\">\r\n<td style=\"width: 69.45px;text-align: center;height: 30px\">3<\/td>\r\n<td style=\"width: 539.85px;text-align: center;height: 30px\">[latex]ClF_{3}(g) + O_{2}(g) \u2192 \\frac{1}{2} Cl_{2}O(g) + \\frac{3}{2}OF_{2}(g)[\/latex]<\/td>\r\n<td style=\"width: 179.05px;text-align: center;height: 30px\">236.2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhat is the [latex]\\Delta H^{\\circ}[\/latex] for the desired reaction?\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Solution<\/h3>\r\n<strong>Step 1:<\/strong> Determine what reaction will give us [latex]ClF[\/latex] in the reactants.\r\n\r\nSince [latex]ClF(g)[\/latex] is needed as a reactant, we can multiply reaction 2 by [latex]\\frac{1}{2}[\/latex] to obtain:\r\n<p style=\"text-align: center\">[latex]ClF(g) + \\frac{1}{2}O_{2}(g) \u2192 \\frac{1}{2}Cl_{2}O(g) + \\frac{1}{2}OF_{2}(g)[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ} = \\frac{1}{2}(214.0 kJ) = 107.0 kJ[\/latex]<\/p>\r\n<strong>Step 2:<\/strong> Determine what reaction will give us [latex]F_{2}[\/latex] in the reactants.\r\n\r\nSince [latex]F_{2}[\/latex] is needed as a reactant, we can multiply reaction 1 by [latex]-\\frac{1}{2}[\/latex] to obtain:\r\n<p style=\"text-align: center\">[latex]F_{2}(g) + \\frac{1}{2}O_{2}(g) \u2192 OF_{2}(g)[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ} = -\\frac{1}{2}(-49.4 kJ) = 24.7 kJ[\/latex]<\/p>\r\n<strong>Step 3:<\/strong> Determine what reaction will give us [latex]ClF_{3}[\/latex] as a product.\r\n\r\nSince [latex]ClF_{3}[\/latex] is needed as a product, we can multiply reaction 3 by -1 to obtain:\r\n<p style=\"text-align: center\">[latex]\\frac{1}{2} Cl_{2}O(g) + \\frac{3}{2}OF_{2}(g) \u2192 ClF_{3}(g) + O_{2}(g) [\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ} = -(236.2kJ) = -236.2 kJ[\/latex]<\/p>\r\n<strong>Step 4:<\/strong> Add the reactions up to ensure that the desired reaction is obtained:\r\n<p style=\"text-align: center\">[latex]ClF(g) + \\frac{1}{2}O_{2}(g) \u2192 \\frac{1}{2}Cl_{2}O(g) + \\frac{1}{2}OF_{2}(g)[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]F_{2}(g) + \\frac{1}{2}O_{2}(g) \u2192 OF_{2}(g)[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\frac{1}{2} Cl_{2}O(g) + \\frac{3}{2}OF_{2}(g) \u2192 ClF_{3}(g) + O_{2}(g)[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\u2193(+)[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex] ClF(g) + F_{2}(g) \u2192 ClF_{3}(g) [\/latex]<\/p>\r\n<strong>Step 5:<\/strong> Add up all the manipulated heat of reactions to obtain the desired [latex]\\Delta H^{\\circ}[\/latex]\r\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ} = (107.0 + 24.7 + -236.2)kJ[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ} = -104.5 kJ[\/latex]<\/p>\r\n&nbsp;\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n<h2 id=\"Heat-of-reaction-method-for-analyzing-energy-use-in-a-process\">Heat of Reaction Method for Analyzing Energy Use in a Process<\/h2>\r\nNow let's try applying the heat of reaction to determine energy use in a process, where reactants and products are coming in and exiting at a given temperature. The following process path is taken for the heat of reaction method, where the reference state is at [latex]25^{\\circ}C[\/latex]:\r\n\r\n<img class=\" wp-image-1056 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/HoR-300x156.png\" alt=\"\" width=\"423\" height=\"220\" \/>\r\n\r\n<span style=\"text-align: initial;font-size: 1em\">Generally, [latex]\\Delta H ^{\\circ}_{r}[\/latex] refers to standard state conditions at [latex]25^{\\circ}C[\/latex] and 1 atm, but always double-check whether a different standard state condition is used.<\/span>\r\n\r\nIn this process path, the enthalpy calculations are done in three steps:\r\n<ol>\r\n \t<li>Temperature change: calculate the [latex]\\Delta\\hat{H}_{reactants}[\/latex] using the heat capacities. The temperature change will be the difference between the inlet temperature of the reactants and the standard or reference temperature ([latex]25^{\\circ}C[\/latex] in this case).<\/li>\r\n \t<li>Reaction enthalpy: calculate the enthalpy of the reaction [latex]\\Delta H^{\\circ}_{r}[\/latex]. This can be done using Hess's Law or the Heat of Formation method (discussed below).<\/li>\r\n \t<li>Temperature change: calculate the [latex]\\Delta\\hat{H}_{products}[\/latex] using the heat capacities. The temperature change will be the difference between the outlet temperature of the products and the standard or reference temperature ([latex]25^{\\circ}C[\/latex] in this case).<\/li>\r\n<\/ol>\r\nThe calculated enthalpy changes for each step are then added to obtain [latex]\\Delta\\dot{H}[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"prompt input_prompt\">\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example: Heat of Reaction Method for Energy Use<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nConsider a propane combustion chamber, where [latex]100 mol\/s[\/latex] of propane are feed to the chamber at [latex]25^{\\circ}C[\/latex] and air is fed at [latex]300^{\\circ}C[\/latex] ([latex]600 mol\/s O_{2}(g)[\/latex] and [latex]2256 mol\/s N_{2}(g)[\/latex]). The products stream exits at [latex]1000^{\\circ}C[\/latex] and consists of [latex]100 mol\/s O_{2}(g)[\/latex], [latex]2256 mol\/s N_{2}(g)[\/latex], [latex]300 mol\/s CO_{2}(g)[\/latex], and [latex]400 mol\/s H_{2}O(v)[\/latex]. How much heatis released by this combustion chamber? Assumming atmoshpere pressure, so water boils at 100\u00b0C.\r\n<p style=\"text-align: center\">[latex]C_{3}H_{8} (g) + 5 O_{2} \u2192 3 CO_{2} (g) + 4 H_{2}O (l)[\/latex]\r\n[latex]\\Delta H^{\\circ}_{r} = -2220 kJ[\/latex]<\/p>\r\nThe following information is provided:\r\n<p style=\"text-align: left\">The values listed under each compound are specific enthalpies in kJ\/mol<\/p>\r\n\r\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 88.4861%;height: 75px\" border=\"0\">\r\n<tbody>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 20%;height: 15px;text-align: center\"><strong>Temperature ([latex]^{\\circ}C[\/latex])<\/strong><\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\"><strong>[latex]O_{2}[\/latex]<\/strong><\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\"><strong>[latex]N_{2}[\/latex]<\/strong><\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\"><strong>[latex]CO_{2}[\/latex]<\/strong><\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\"><strong>[latex]H_{2}O[\/latex]<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 20%;height: 15px;text-align: center\">25<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">0<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">0<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">0<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">0<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 20%;height: 15px;text-align: center\">100<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">2.24<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">2.19<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">2.90<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">2.54<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 20%;height: 15px;text-align: center\">300<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">8.47<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">8.12<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">11.58<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">9.57<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 20%;height: 15px;text-align: center\">1000<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">32.47<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">30.56<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">48.60<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">37.69<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<ul>\r\n \t<li>[latex]C_{P,l(H_{2}O)} = 75.4*10^{-3} \\frac{kJ}{molK}[\/latex]<\/li>\r\n \t<li>[latex]C_{P,v(H_{2}O)} = 33.46*10^{-3}+0.688*10^{-5}*T + 0.7604*10^{-8}*T^{2} - 3.593*10^{-12}*T^{3} (J\/molK)[\/latex]<\/li>\r\n \t<li>[latex]\\Delta\\hat{H}_{vap}(100^{\\circ}C) = 40.66 kJ\/mol[\/latex]<\/li>\r\n<\/ul>\r\n<h4>How can we solve the system's energy balance?<\/h4>\r\n1 - Solve material balances as much as possible\r\n<blockquote>This example already has the material balances solved<\/blockquote>\r\n2 - Choose reference states for energy calculations. Like we say in the previous figure we want to calculate the changes in energy associated with bringing our reactants to our reference state, calculating the energy of the reaction at the reference state, and then the energy associated with bringing the products to their final state. We will choose reference states based on the information we have to try to make these calculations as easy as possible.\r\n<blockquote>For reacting species:\r\nSince [latex]\\Delta H^{\\circ}_{r}[\/latex] is given, we will assume this is at [latex]25^{\\circ}C[\/latex] and 1 atm (as there is nothing indicating otherwise), and we will use this as our reference state.\r\n\r\nFor non-reacting species:\r\nWe can use any convenient temperature and pressure as a reference state (inlet temperature, outlet temperature, temperature in enthalpy table). With this example, the enthalpy table given uses a reference state of [latex]25^{\\circ}C[\/latex], and we will assume everything is at 1 atm (as no other values for pressure are given).<\/blockquote>\r\n3 - Calculate the extent of reaction for all reactions (in this case we just have the one reaction)\r\n<blockquote>\r\n<p style=\"text-align: center\">[latex] \\xi = \\frac{(\\dot{n}_{i})_{out}-(\\dot{n}_{i})_{in}}{\\dot{\\nu}_{i}}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\xi = \\frac{(\\dot{n}_{CO_{2}})_{out}-(\\dot{n}_{CO_{2}})_{in}}{\\dot{\\nu}_{CO_{2}}}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\xi = \\frac{300\\frac{mol}{s} - 0 \\frac{mol}{s}}{3} = 100 mol\/s[\/latex]<\/p>\r\n<\/blockquote>\r\n<p style=\"text-align: left\">4 - Prepare an inlet-out enthalpy table (this will show what enthalpies we need to calculate associated with energy changes in the reactants or products)<\/p>\r\n\r\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 95.2026%;height: 180px\" border=\"0\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 20%\"><strong>Substance<\/strong><\/td>\r\n<td style=\"width: 20%\"><strong>[latex]\\dot{n}_{in}[\/latex](mol\/s)<\/strong><\/td>\r\n<td style=\"width: 20%\"><strong>[latex]\\hat{H}_{in}[\/latex] (kJ\/mol)<\/strong><\/td>\r\n<td style=\"width: 20%\"><strong>[latex]\\dot{n}_{out}[\/latex] (mol\/s)<\/strong><\/td>\r\n<td style=\"width: 20%\"><strong>[latex]\\hat{H}_{out}[\/latex] (kJ\/mol)<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%\">[latex]C_{3}H_{8}[\/latex]<\/td>\r\n<td style=\"width: 20%\">100<\/td>\r\n<td style=\"width: 20%\">[latex]\\hat{H}_{1}[\/latex]<\/td>\r\n<td style=\"width: 20%\">-<\/td>\r\n<td style=\"width: 20%\">-<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%\">[latex]O_{2}[\/latex]<\/td>\r\n<td style=\"width: 20%\">600<\/td>\r\n<td style=\"width: 20%\">[latex]\\hat{H}_{2}[\/latex]<\/td>\r\n<td style=\"width: 20%\">100<\/td>\r\n<td style=\"width: 20%\">[latex]\\hat{H}_{4}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%\">[latex]N_{2}[\/latex]<\/td>\r\n<td style=\"width: 20%\">2256<\/td>\r\n<td style=\"width: 20%\">[latex]\\hat{H}_{3}[\/latex]<\/td>\r\n<td style=\"width: 20%\">2256<\/td>\r\n<td style=\"width: 20%\">[latex]\\hat{H}_{5}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%\">[latex]CO_{2}[\/latex]<\/td>\r\n<td style=\"width: 20%\">-<\/td>\r\n<td style=\"width: 20%\">-<\/td>\r\n<td style=\"width: 20%\">300<\/td>\r\n<td style=\"width: 20%\">[latex]\\hat{H}_{6}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%\">[latex]H_{2}O[\/latex]<\/td>\r\n<td style=\"width: 20%\">-<\/td>\r\n<td style=\"width: 20%\">-<\/td>\r\n<td style=\"width: 20%\">400<\/td>\r\n<td style=\"width: 20%\">[latex]\\hat{H}_{7}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n&nbsp;\r\n\r\n5 - Calculate all the enthalpies. For [latex]\\hat{H_{1}}[\/latex] to [latex]\\hat{H_{6}}[\/latex], the specific enthalpies at different temperatures are given as the difference in enthalpy from the reference state (25\u00b0C, 1atm, which is the same as what we choose as reference state):\r\n<blockquote>\r\n<p style=\"text-align: center\">[latex]\\hat{H}_{1}=\\Delta\\hat{H}_{C_{3}H_{8}} (25^{\\circ}C \u2192 25^{\\circ}C) = 0 kJ\/mol[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\hat{H}_{2}=\\Delta\\hat{H}_{O_{2}} (25^{\\circ}C \u2192 300^{\\circ}C) = (8.47 - 0) kJ\/mol[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\hat{H}_{3}=\\Delta\\hat{H}_{N_{2}} (25^{\\circ}C \u2192 300^{\\circ}C) = (8.12 - 0) kJ\/mol[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\hat{H}_{4}=\\Delta\\hat{H}_{O_{2}} (25^{\\circ}C \u2192 1000^{\\circ}C) = (32.47 - 0) kJ\/mol[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\hat{H}_{5}=\\Delta\\hat{H}_{N_{2}} (25^{\\circ}C \u2192 1000^{\\circ}C) = (30.56 - 0) kJ\/mol[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\hat{H}_{6}=\\Delta\\hat{H}_{CO_{2}} (25^{\\circ}C \u2192 1000^{\\circ}C) = (48.60 - 0) kJ\/mol[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\hat{H}_{7}=\\Delta\\hat{H}_{H_{2}O} (l,25^{\\circ}C \u2192 H_{2}O (v,1000^{\\circ}C))[\/latex]<\/p>\r\n<\/blockquote>\r\n6 - For [latex]\\hat{H_{7}}[\/latex], the specific enthalpy for water is calculated. From the reference state of 25\u00b0C to the stream temperature of 1000\u00b0C, water is first heated to 100\u00b0C, which is its natural boiling point at 1atm, then vaporized, then heated in vapor phase to 1000\u00b0C. The enthalpy change of each process is calculated separately and added together to get [latex]\\hat{H_{7}}[\/latex].\r\n<blockquote><strong>Note:<\/strong> The [latex]T[\/latex] in the formulas to calculate [latex]C_{P}[\/latex] is given in kelvin. This doesn't make a difference when [latex]C_{P}[\/latex] is given as a number, as the scales for one degree of Celcius and kelvin are the same, but the temperature must be converted to kelvin when [latex]T^2[\/latex] is higher power of [latex]T[\/latex] is used in calculation.\r\n<p style=\"text-align: center\">[latex]100\u00b0C=373K, 1000\u00b0C=1273K[\/latex]<\/p>\r\n\\begin{align*}\r\n\\hat{H}_{7}&amp; = \\int^{100^{\\circ}C}_{25^{\\circ}C} C_{P,l}dT + \\Delta \\hat{H}_{vap}(100^{\\circ}C) + \\int^{1000^{\\circ}C}_{100^{\\circ}C} C_{P,v}dT\\\\&amp; = \\int^{100^{\\circ}C}_{25^{\\circ}C} 75.4*10^{-3}dT + \\Delta \\hat{H}_{vap}(100^{\\circ}C) \\\\&amp; \\;\\;\\;\\;+\\int^{1273K}_{373K} (33.46*10^{-3}+0.688*10^{-5}*T + 0.7604*10^{-8}*T^{2} - 3.593*10^{-12}*T^{3})dT\\\\&amp; = 75.4*10^{-3}*(100-25)+ 40.66 \\\\&amp; \\;\\;\\;\\; +(33.46*10^{-3}*T+\\frac{1}{2}*0.688*10^{-5}*T^2 + \\frac{1}{3}*0.7604*10^{-8}*T^{3} - \\frac{1}{4}*3.593*10^{-12}*T^{4})\\bigg\\vert^{1273K}_{373K}\\\\&amp; = (5.65 + 40.66 + 35.1)kJ\/mol\\\\&amp; = 81.46 kJ\/mol\r\n\\end{align*}\r\n<p style=\"text-align: center\">[latex]\\hat{H}_{7}= (5.65 + 40.66 + 35.1)kJ\/mol = 81.46\\frac{kJ}{mol}[\/latex]<\/p>\r\nTherefore, we have calculated all the specific enthalpy for the reactants and products:<\/blockquote>\r\n<table class=\"grid\" style=\"border-collapse: collapse;width: 95.2026%;height: 120px\" border=\"0\">\r\n<tbody>\r\n<tr style=\"height: 30px\">\r\n<td style=\"width: 20%;height: 30px;text-align: center\"><strong>Substance<\/strong><\/td>\r\n<td style=\"width: 20%;height: 30px;text-align: center\"><strong>[latex]\\dot{n}_{in}[\/latex](mol\/s)<\/strong><\/td>\r\n<td style=\"width: 20%;height: 30px;text-align: center\"><strong>[latex]\\hat{H}_{in}[\/latex] (kJ\/mol)<\/strong><\/td>\r\n<td style=\"width: 20%;height: 30px;text-align: center\"><strong>[latex]\\dot{n}_{out}[\/latex] (mol\/s)<\/strong><\/td>\r\n<td style=\"width: 20%;height: 30px;text-align: center\"><strong>[latex]\\hat{H}_{out}[\/latex] (kJ\/mol)<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 30px\">\r\n<td style=\"width: 20%;height: 30px;text-align: center\">[latex]C_{3}H_{8}[\/latex]<\/td>\r\n<td style=\"width: 20%;height: 30px;text-align: center\">100<\/td>\r\n<td style=\"width: 20%;height: 30px;text-align: center\">0<\/td>\r\n<td style=\"width: 20%;height: 30px;text-align: center\">-<\/td>\r\n<td style=\"width: 20%;height: 30px;text-align: center\">-<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 20%;height: 15px;text-align: center\">[latex]O_{2}[\/latex]<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">600<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">8.47<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">100<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">32.47<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 20%;height: 15px;text-align: center\">[latex]N_{2}[\/latex]<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">2256<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">8.12<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">2256<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">30.56<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 20%;height: 15px;text-align: center\">[latex]CO_{2}[\/latex]<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">-<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">-<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">300<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">48.60<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 20%;height: 15px;text-align: center\">[latex]H_{2}O[\/latex]<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">-<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">-<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">400<\/td>\r\n<td style=\"width: 20%;height: 15px;text-align: center\">81.46<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n&nbsp;\r\n\r\n8 - Finally, solve the energy balance\r\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} + \\Delta\\dot{E}_{k} +\\Delta\\dot{E}_{p} = \\dot{Q} + \\dot{W}_{s}[\/latex]<\/p>\r\n[latex]\\Delta\\dot{E}_{k}[\/latex], [latex]\\Delta\\dot{E}_{p}[\/latex], and [latex]\\dot{W}_{s}[\/latex] are assumed negligible for this system (as no information is provided on these)\r\n<p style=\"text-align: center\">[latex]\\dot{Q} = \\Delta\\dot{H}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\dot{Q} = -126 MW[\/latex]<\/p>\r\nThe combustion chamber releases [latex]126 MW[\/latex] of heat to the environment\r\n\r\n<\/div>\r\n<\/div>\r\n<h2 id=\"Heat-of-Formation-$\\Delta-H^{\\circ}_{f}$\">Heat of Formation [latex]\\Delta H^{\\circ}_{f}[\/latex]<\/h2>\r\n<strong>Formation Reaction<\/strong>: a reaction in which the compound is formed from its elemental constituents as they would normally occur in nature (eg. [latex]O_{2}[\/latex] rather than [latex]O[\/latex]).\r\n\r\nFor elemental consistituents, the energy of formation is [latex]\\Delta H_{f}^{\\circ}[\/latex] = 0, since they would be forming themselves: [latex]O_{2}\u2192O_{2}[\/latex]\r\n<p style=\"text-align: left\"><strong>Standard specific heat of formation [latex]\\Delta H^{\\circ}_{f}[\/latex]<\/strong>: the enthalpy change associated with forming 1 mole of the compound of interest at standard temperature ([latex]25^{\\circ}C[\/latex]) and pressure (1 atm)<\/p>\r\n\r\n<table class=\"grid aligncenter\" style=\"height: 138px\">\r\n<thead>\r\n<tr style=\"height: 15px\">\r\n<th style=\"height: 15px;width: 75.05px;text-align: center\">Compound<\/th>\r\n<th style=\"height: 15px;width: 355.05px;text-align: center\">Reaction<\/th>\r\n<th style=\"height: 15px;width: 272.65px;text-align: center\">[latex]\\Delta\\hat{H}^{\\circ}_{f,i}[\/latex] (kJ\/mol)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px;width: 75.05px;text-align: center\">Water<\/td>\r\n<td style=\"height: 15px;width: 355.85px;text-align: center\">[latex]H_{2}(g) + \\frac{1}{2}O_{2}(g) \u2192 H_{2}O(l)[\/latex]<\/td>\r\n<td style=\"height: 15px;width: 273.45px;text-align: center\">-285.83<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px;width: 75.05px;text-align: center\">Methane<\/td>\r\n<td style=\"height: 15px;width: 355.85px;text-align: center\">[latex]C(s) + 2H_{2}(g) \u2192 CH_{4}(g)[\/latex]<\/td>\r\n<td style=\"height: 15px;width: 273.45px;text-align: center\">-74.8936<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px;width: 75.05px;text-align: center\">Ethane<\/td>\r\n<td style=\"height: 15px;width: 355.85px;text-align: center\">[latex]2C(s) + 3H_{2}(g) \u2192 C_{2}H_{6}(g)[\/latex]<\/td>\r\n<td style=\"height: 15px;width: 273.45px;text-align: center\">-83.82<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px;width: 75.05px;text-align: center\">Propane<\/td>\r\n<td style=\"height: 15px;width: 355.85px;text-align: center\">[latex]3C(s) + 4H_{2}(g) \u2192 C_{3}H_{8}(g)[\/latex]<\/td>\r\n<td style=\"height: 15px;width: 273.45px;text-align: center\">-104.68<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px;width: 75.05px;text-align: center\">Benzene<\/td>\r\n<td style=\"height: 15px;width: 355.85px;text-align: center\">[latex]6C(s) + 3H_{2}(g) \u2192 C_{6}H_{6}(l)[\/latex]<\/td>\r\n<td style=\"height: 15px;width: 273.45px;text-align: center\">82.88<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px;width: 75.05px;text-align: center\">Toluene<\/td>\r\n<td style=\"height: 15px;width: 355.85px;text-align: center\">[latex]7C(s) + 4H_{2}(g) \u2192 C_{7}H_{8}(l)[\/latex]<\/td>\r\n<td style=\"height: 15px;width: 273.45px;text-align: center\">50.17<\/td>\r\n<\/tr>\r\n<tr style=\"height: 18px\">\r\n<td style=\"height: 18px;width: 75.05px;text-align: center\">Oxygen<\/td>\r\n<td style=\"height: 18px;width: 355.85px;text-align: center\">[latex]O_{2}(g) \u2192 O_{2}(g)[\/latex]<\/td>\r\n<td style=\"height: 18px;width: 273.45px;text-align: center\">0<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px;width: 75.05px;text-align: center\">Nitrogen<\/td>\r\n<td style=\"height: 15px;width: 355.85px;text-align: center\">[latex]N_{2}(g) \u2192 N_{2}(g)[\/latex]<\/td>\r\n<td style=\"height: 15px;width: 273.45px;text-align: center\">0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: left\">We can use these heats of formation and Hess's law to find the heat of reaction for a given reaction:<\/p>\r\n\r\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 96.7839%;height: 93px\" border=\"0\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 100%;text-align: center\"><span style=\"font-size: 16px\">[latex]\\Delta H^{\\circ}_{r} = \\Sigma_{i}\\nu_{i}\\Delta\\hat{H}^{\\circ}_{f,i} = \\Sigma_{products}|\\nu_{i}|\\Delta\\hat{H}^{\\circ}_{f,i}-\\Sigma_{reactants}|\\nu_{i}|\\Delta\\hat{H}^{\\circ}_{f,i}[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example: Heat of Formation<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<div class=\"cell border-box-sizing text_cell rendered\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n\r\nConsider the combustion of ethane:\r\n<p style=\"text-align: center\">[latex]C_{2}H_{6} (g) + \\frac{7}{2} O_{2}(g) \u2192 2 CO_{2} (g) + 3 H_{2}O (l)[\/latex]<\/p>\r\nwhere [latex]\\Delta H^{\\circ}_{r} = -1560 kJ\/mol[\/latex]\r\n\r\nTo calculate the enthalpy using the heat of formation method, the following steps are taken:\r\n<p style=\"text-align: center\">[latex] \\Delta H^{\\circ}_{r} = (3*\\Delta\\hat{H}^{\\circ}_{f,H_{2}O} + 2*\\Delta\\hat{H}^{\\circ}_{f,CO_{2}}) - (1*\\Delta\\hat{H}^{\\circ}_{f,C_{2}H_{6}}+\\frac{7}{2}*\\Delta\\hat{H}^{\\circ}_{f,O_{2}})[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex] \\Delta H^{\\circ}_{r} = (3*-285.83 + 2*-393.51)kJ\/mol - (1*-83.82+\\frac{7}{2}*0)kJ\/mol[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ}_{r} = -1561 kJ\/mol[\/latex]<\/p>\r\n<em>The enthalpy of reaction using the heat of formation method is very close to the enthalpy of reaction (combustion).<\/em>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 id=\"Where-to-find-Heat-of-Formation-data\">Where to find Heat of Formation Data<\/h3>\r\nOne source of heat of formation data is the <a href=\"https:\/\/webbook.nist.gov\/chemistry\/name-ser\/\"> National Institute of Standards and Technology (NIST) Webbook NIST<\/a>\r\n\r\nTo find the heat of formation data at this site:\r\n\r\n1 - Look up the compound of interest\r\n\r\n2 - Heat of formation data will be under \"condensed phase thermochemistry data\" (for liquids or solids) or \"gas phase thermochemistry data\" (for gases).\r\n\r\n<em>Heats of formation can also be found in Appendix E of \"Introductory Chemical Engineering Thermodynamics\" by J. Richard Elliot and Carl T. Lira.<\/em>\r\n\r\n<\/div>\r\n<h2 id=\"Heat-of-formation-method-for-analyzing-energy-use-in-a-process\">Heat of Formation Method for Analyzing Energy Use in a Process<\/h2>\r\n<ol>\r\n \t<li>Solve the material balance as much as possible: this can include solving for mass or molar flows using stoichiometry or mass balances<\/li>\r\n \t<li>Choose reference states for energy calculations: reference states provide a basis for enthalpy calculations. Choose reference states that make your calculations convenient or reference states that match the available data. Most data is provided at [latex]25^{\\circ}C[\/latex] and 1 atm, therefore this is a common reference state.<\/li>\r\n \t<li>Prepare and inlet-outlet enthalpy table: this table will include all the compounds involved in the system, the inlet molar flow, the inlet enthalpy values, the outlet molar flow, and the outlet enthalpy values. Fill out the table with all the known values and number the enthalpy values accordingly.<\/li>\r\n \t<li>Calculate all the enthalpies: use process paths to calculate the enthalpies listed in the table in step 3. The enthalpy change of each compound is calculated by summing the heat of formation, enthalpy change due to temperature change in the same state (which can be calculated using the [latex]C_{P}[\/latex] values), and enthalpy change due to phase change when phase change is involved.<\/li>\r\n \t<li>Calculate the [latex]\\Delta\\dot{H}[\/latex] for the system: this is done by multiplying each molar flow by the corresponding enthalpy and using the following expression. Remember that the stoichiometric coefficients are negative for reactants and positive for products: [latex] \\Delta\\dot{H} = \\Sigma\\dot{n}_{out}*\\hat{H}_{out} - \\Sigma\\dot{n}_{in}*\\hat{H}_{in}[\/latex]<\/li>\r\n \t<li>Solve the energy balance: determine which energy terms are present in the system and solve accordingly.<\/li>\r\n<\/ol>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example: Heat of Formation Method for Energy Use<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nLet's consider the propane combustion chamber problem we analyzed before with the heat of reaction method, and let's see if we can get a similar answer using the heat of formation method.\r\n\r\n1 - Solve the material balance as much as possible\r\n<blockquote>This example already has the material balances solved<\/blockquote>\r\n2 - Choose reference states for energy calculations\r\n<blockquote>For reacting species: elemental species that make up reacting species at standard conditions; we will choose [latex]25^{\\circ}C[\/latex] at 1 atm with [latex]C(s)[\/latex], [latex]H_{2}(g)[\/latex], and [latex]O_{2}(g)[\/latex].\r\n\r\nFor non-reacting species (same as [latex]\\Delta H_{r}[\/latex]): Use any convenient temperature (inlet temperature, outlet temperature, temperature in enthalpy table) Here, [latex]25^{\\circ}C[\/latex] and 1 atm works because of our enthalpy table values<\/blockquote>\r\n3 - Prepare an inlet-outlet enthalpy table\r\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 95.2026%;height: 180px\" border=\"0\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 20%\"><strong>Substance<\/strong><\/td>\r\n<td style=\"width: 20%\"><strong>[latex]\\dot{n}_{in}[\/latex](mol\/s)<\/strong><\/td>\r\n<td style=\"width: 20%\"><strong>[latex]\\hat{H}_{in}[\/latex] (kJ\/mol)<\/strong><\/td>\r\n<td style=\"width: 20%\"><strong>[latex]\\dot{n}_{out}[\/latex] (mol\/s)<\/strong><\/td>\r\n<td style=\"width: 20%\"><strong>[latex]\\hat{H}_{out}[\/latex] (kJ\/mol)<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%\">[latex]C_{3}H_{8}[\/latex]<\/td>\r\n<td style=\"width: 20%\">100<\/td>\r\n<td style=\"width: 20%\">[latex]\\hat{H}_{1}[\/latex]<\/td>\r\n<td style=\"width: 20%\">-<\/td>\r\n<td style=\"width: 20%\">-<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%\">[latex]O_{2}[\/latex]<\/td>\r\n<td style=\"width: 20%\">600<\/td>\r\n<td style=\"width: 20%\">[latex]\\hat{H}_{2}[\/latex]<\/td>\r\n<td style=\"width: 20%\">100<\/td>\r\n<td style=\"width: 20%\">[latex]\\hat{H}_{4}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%\">[latex]N_{2}[\/latex]<\/td>\r\n<td style=\"width: 20%\">2256<\/td>\r\n<td style=\"width: 20%\">[latex]\\hat{H}_{3}[\/latex]<\/td>\r\n<td style=\"width: 20%\">2256<\/td>\r\n<td style=\"width: 20%\">[latex]\\hat{H}_{5}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%\">[latex]CO_{2}[\/latex]<\/td>\r\n<td style=\"width: 20%\">-<\/td>\r\n<td style=\"width: 20%\">-<\/td>\r\n<td style=\"width: 20%\">300<\/td>\r\n<td style=\"width: 20%\">[latex]\\hat{H}_{6}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%\">[latex]H_{2}O[\/latex]<\/td>\r\n<td style=\"width: 20%\">-<\/td>\r\n<td style=\"width: 20%\">-<\/td>\r\n<td style=\"width: 20%\">400<\/td>\r\n<td style=\"width: 20%\">[latex]\\hat{H}_{7}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n&nbsp;\r\n<p style=\"text-align: left\">4 - Calculate all the enthalpies<\/p>\r\n\r\n<table class=\"grid aligncenter\" style=\"height: 90px\">\r\n<thead>\r\n<tr style=\"height: 15px\">\r\n<th style=\"height: 15px;width: 128.65px;text-align: center\">Substance<\/th>\r\n<th style=\"height: 15px;width: 311.85px;text-align: center\">[latex]\\Delta H^{\\circ}_{f}[\/latex] (kJ\/mol)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px;width: 128.65px;text-align: center\">[latex]C_{3}H_{8}[\/latex]<\/td>\r\n<td style=\"height: 15px;width: 312.65px;text-align: center\">-103.8<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px;width: 128.65px;text-align: center\">[latex]O_{2}[\/latex]<\/td>\r\n<td style=\"height: 15px;width: 312.65px;text-align: center\">0<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px;width: 128.65px;text-align: center\">[latex]N_{2}[\/latex]<\/td>\r\n<td style=\"height: 15px;width: 312.65px;text-align: center\">0<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px;width: 128.65px;text-align: center\">[latex]CO_{2}[\/latex]<\/td>\r\n<td style=\"height: 15px;width: 312.65px;text-align: center\">-393.51<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px;width: 128.65px;text-align: center\">[latex]H_{2}O (v)[\/latex]<\/td>\r\n<td style=\"height: 15px;width: 312.65px;text-align: center\">-241.835<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<blockquote>\r\n<p style=\"text-align: center\">[latex]\\hat{H}_{1} = \\Delta\\hat{H}^{\\circ}_{f,C_{3}H_{8}(g)} = -103.8 kJ\/mol[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\hat{H}_{2}=\\Delta\\hat{H} (O_{2} (25^{\\circ}C \u2192 300^{\\circ}C) = (8.47 - 0) kJ\/mol[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\hat{H}_{3}=\\Delta\\hat{H} (N_{2} (25^{\\circ}C \u2192 300^{\\circ}C) = (8.12 - 0) kJ\/mol[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\hat{H}_{4}=\\Delta\\hat{H} (O_{2} (25^{\\circ}C \u2192 1000^{\\circ}C) = (32.47 - 0) kJ\/mol[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\hat{H}_{5}=\\Delta\\hat{H} (N_{2} (25^{\\circ}C \u2192 1000^{\\circ}C) = (30.56 - 0) kJ\/mol[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\hat{H}_{6}=\\Delta\\hat{H}^{\\circ}_{f,CO_{2}(g)} + \\int^{1000^{\\circ}C}_{25^{\\circ}C} C_{p,CO_{2}(g)}dT = -344.9 kJ\/mol[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\hat{H}_{7}=\\Delta\\hat{H}^{\\circ}_{f,H_{2}O(v)} + \\int^{1000^{\\circ}C}_{25^{\\circ}C} C_{p,H_{2}O(v)}dT = -204.1 kJ\/mol[\/latex]<\/p>\r\nHere, you assume that water forms as vapor directly in the reaction. Therefore, there is no need to account for heat of vaporization.<\/blockquote>\r\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 95.2026%;height: 180px\" border=\"0\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 20%;text-align: center\"><strong>Substance<\/strong><\/td>\r\n<td style=\"width: 20%;text-align: center\"><strong>[latex]\\dot{n}_{in}[\/latex](mol\/s)<\/strong><\/td>\r\n<td style=\"width: 20%;text-align: center\"><strong>[latex]\\hat{H}_{in}[\/latex] (kJ\/mol)<\/strong><\/td>\r\n<td style=\"width: 20%;text-align: center\"><strong>[latex]\\dot{n}_{out}[\/latex] (mol\/s)<\/strong><\/td>\r\n<td style=\"width: 20%;text-align: center\"><strong>[latex]\\hat{H}_{out}[\/latex] (kJ\/mol)<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%;text-align: center\">[latex]C_{3}H_{8}[\/latex]<\/td>\r\n<td style=\"width: 20%;text-align: center\">100<\/td>\r\n<td style=\"width: 20%;text-align: center\">-103.8<\/td>\r\n<td style=\"width: 20%;text-align: center\">-<\/td>\r\n<td style=\"width: 20%;text-align: center\">-<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%;text-align: center\">[latex]O_{2}[\/latex]<\/td>\r\n<td style=\"width: 20%;text-align: center\">600<\/td>\r\n<td style=\"width: 20%;text-align: center\">8.47<\/td>\r\n<td style=\"width: 20%;text-align: center\">100<\/td>\r\n<td style=\"width: 20%;text-align: center\">32.47<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%;text-align: center\">[latex]N_{2}[\/latex]<\/td>\r\n<td style=\"width: 20%;text-align: center\">2256<\/td>\r\n<td style=\"width: 20%;text-align: center\">8.12<\/td>\r\n<td style=\"width: 20%;text-align: center\">2256<\/td>\r\n<td style=\"width: 20%;text-align: center\">30.56<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%;text-align: center\">[latex]CO_{2}[\/latex]<\/td>\r\n<td style=\"width: 20%;text-align: center\">-<\/td>\r\n<td style=\"width: 20%;text-align: center\">-<\/td>\r\n<td style=\"width: 20%;text-align: center\">300<\/td>\r\n<td style=\"width: 20%;text-align: center\">-344.9<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 20%;text-align: center\">[latex]H_{2}O[\/latex]<\/td>\r\n<td style=\"width: 20%;text-align: center\">-<\/td>\r\n<td style=\"width: 20%;text-align: center\">-<\/td>\r\n<td style=\"width: 20%;text-align: center\">400<\/td>\r\n<td style=\"width: 20%;text-align: center\">-204.1<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n&nbsp;\r\n\r\n5 - Calculate $\\Delta\\dot{H}$ for the reactor\r\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = \\Sigma\\dot{n}_{out}*\\hat{H}_{out} - \\Sigma\\dot{n}_{in}*\\hat{H}_{in}=1.26x10^{5} kJ\/s[\/latex]<\/p>\r\n<p style=\"text-align: left\">6 - Finally, solve the energy balance<\/p>\r\n<p style=\"text-align: center\">[latex]\\dot{Q} = \\Delta\\dot{H}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\dot{Q} = -126 MW[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise: Heat of Formation<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<div class=\"inner_cell\">\r\n<div class=\"text_cell_render border-box-sizing rendered_html\">\r\n\r\nConsider a reactor where the following reaction reaction takes place :\r\n<p style=\"text-align: center\">[latex] 3 NO_{2}(g) + H_{2}O(l) \u2192 2HNO_{3}(aq) + NO(g)[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ} = ?[\/latex]<\/p>\r\nThe reactor uses an 0.8 kW impeller (mixer for liquids) that mixes the reactor contents. The reaction occurs at standard temperature and pressure. A feed consisting of 100 mol\/s and 300 mol\/s of water and [latex]NO_{2}[\/latex], respectively, enters the reactor. The reaction goes to completion and the products stream consists of 200 mol\/s of [latex]HNO_{3}[\/latex] and 100 mol\/s of [latex]NO[\/latex].\r\n<p style=\"text-align: left\">The following heats of formation are available:<\/p>\r\n\r\n<table class=\"grid aligncenter\" style=\"height: 174px;width: 667px\" width=\"864\">\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center;width: 56.1364px\">Number<\/th>\r\n<th style=\"text-align: center;width: 342.682px\">Reaction<\/th>\r\n<th style=\"text-align: center;width: 226.318px\">[latex] \\Delta H^{\\circ}_{f,i}[\/latex] (kJ\/mol)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td style=\"text-align: center;width: 56.1364px\">1<\/td>\r\n<td style=\"text-align: center;width: 342.682px\">[latex] \\frac{1}{2}N_{2}(g) + O_{2}(g) \u2192 NO_{2}(g)[\/latex]<\/td>\r\n<td style=\"text-align: center;width: 226.318px\">33.2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: center;width: 56.1364px\">2<\/td>\r\n<td style=\"text-align: center;width: 342.682px\">[latex] H_{2}(g) + \\frac{1}{2}O_{2}(g) \u2192 H_{2}O(l)[\/latex]<\/td>\r\n<td style=\"text-align: center;width: 226.318px\">-285.8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: center;width: 56.1364px\">3<\/td>\r\n<td style=\"text-align: center;width: 342.682px\">[latex] \\frac{1}{2}H_{2}(g) + \\frac{1}{2}N_{2}(g) + \\frac{3}{2}O_{2}(g) \u2192 HNO_{3}O(aq)[\/latex]<\/td>\r\n<td style=\"text-align: center;width: 226.318px\">-207.4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: center;width: 56.1364px\">4<\/td>\r\n<td style=\"text-align: center;width: 342.682px\">[latex] \\frac{1}{2}N_{2}(g) + \\frac{1}{2}O_{2}(g) \u2192 NO(g)[\/latex]<\/td>\r\n<td style=\"text-align: center;width: 226.318px\">90.2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nHow much heat must be removed from the reactor in one day for it to remain at standard temperature?\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"prompt input_prompt\">\r\n<div class=\"textbox\">\r\n<h3 class=\"prompt input_prompt\">Solution<\/h3>\r\n<div>\r\n\r\n<strong>Step 1:<\/strong> Determine what combination of the given reactions will give the desired reaction (Hess's Law)\r\n\r\nFor this desired reaction:\r\n\\begin{align*}\r\n&amp; -3*Reaction1\\\\\r\n&amp; -1*Reaction2\\\\\r\n&amp; +2*Reaction3\\\\\r\n&amp; +1*Reaction4\r\n\\end{align*}\r\n\r\nSumming all the reactions above yields:\r\n<p style=\"text-align: center\">[latex] 3 NO_{2}(g) + H_{2}O(l) \u2192 2HNO_{3}(aq) + NO(g) [\/latex]<\/p>\r\n&nbsp;\r\n\r\n<strong>Step 2<\/strong>: Add all the individual reaction enthalpies:\r\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ}_{r} = -3*\\Delta H^{\\circ}_{1} + -1*\\Delta H^{\\circ}_{2} + 2*\\Delta H^{\\circ}_{3} + \\Delta H^{\\circ}_{4}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ}_{r} = -3*33.2 kJ\/mol + -1*-285.8 kJ\/mol + 2*-207.4 kJ\/mol + 90.2 kJ\/mol[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ}_{r}= (-99.6+285.8+-414.8+90.2)kJ\/mol = -138.4 kJ\/mol[\/latex]<\/p>\r\n&nbsp;\r\n\r\n<strong>Step 3:<\/strong> Calculate the overall change in enthalpy using the extent of the reaction:\r\n\r\n\\begin{align*}\r\n\\xi &amp;= \\frac{|(\\dot{n}_{i})_{out}-(\\dot{n}_{i})_{in}|}{|(\\dot{\\nu}_{i})|}hat\\\\&amp; = \\frac{|(\\dot{n}_{H_{2}O})_{out}-(\\dot{n}_{H_{2}O})_{in}|}{|(\\dot{\\nu}_{H_{2}O})|} \\\\&amp; =\\frac{|0\\frac{mol}{s} - 100 \\frac{mol}{s}|}{|-1|}\\\\ &amp;= 100 mol\/s\r\n\\end{align*}\r\n\r\n&nbsp;\r\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = \\xi*\\Delta H^{\\circ}_{r}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = 100\\frac{mol}{s}*-138.4\\frac{kJ}{mol} = -138400 \\frac{kJ}{s}[\/latex]<\/p>\r\n&nbsp;\r\n\r\n<strong>Step 4:<\/strong> Calculate how much work is done on the reactor by the impeller in one day:\r\n<p style=\"text-align: center\">[latex]W_{s} = 0.8\\frac{kJ}{s}*3600\\frac{s}{hr}*24hr[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]W_{s} = 69120 kJ[\/latex]<\/p>\r\n&nbsp;\r\n\r\n<strong>Step 5<\/strong>: Perform an energy balance on the reactor\r\n<p style=\"text-align: center\">[latex]\\dot{Q}=\\Delta H^{\\circ}-W_{s}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\dot{Q} = (-138400 - 69120 ) kJ = -207520 kJ[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<h2 id=\"Summary-of-Heat-of-Reaction-and-Formation-Methods\" style=\"text-align: left\">Summary of Heat of Reaction and Formation Methods<\/h2>\r\n<h3 id=\"Heat-of-Reaction-Method\" style=\"text-align: left\">Heat of Reaction Method<\/h3>\r\n<p style=\"text-align: center\">[latex] \\Delta\\dot{H} = \\xi\\Delta\\dot{H}^{\\circ}_{r} + \\Sigma\\dot{n}_{out}*\\hat{H}_{out} - \\Sigma\\dot{n}_{in}*\\hat{H}_{in}[\/latex]<\/p>\r\n<p style=\"text-align: center\">where [latex]\\hat{H}_{i} = \\int^{T}_{T_{ref}}C_{P}dT[\/latex] assuming there are no phase changes<\/p>\r\n\r\n<blockquote>\r\n<p style=\"text-align: left\">If a phase change occurs, an additional heat term (i.e. heat of vapourization) will be added to the [latex]\\hat{H}_{i}[\/latex] term in both methods<\/p>\r\n<\/blockquote>\r\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 93.2663%;height: 91px\" border=\"0\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 100%;text-align: center\"><span style=\"font-size: 16px\">[latex] \\Delta\\dot{H} = \\xi\\Delta\\dot{H}_{r} + \\Sigma\\dot{n}_{out}*\\int^{T_{out}}_{T_{ref}}C_{P}dT - \\Sigma\\dot{n}_{in}*\\int^{T_{in}}_{T_{ref}}C_{P}dT[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h3 id=\"Heat-of-Formation-Method\">Heat of Formation Method<\/h3>\r\n<p style=\"text-align: center\">[latex] \\Delta\\dot{H} = \\Sigma\\dot{n}_{out}*\\hat{H}_{out} - \\Sigma\\dot{n}_{in}*\\hat{H}_{in}[\/latex]<\/p>\r\nwhere [latex]\\hat{H}_{i} = \\hat{H}^{\\circ}_{f,i} + \\int^{T}_{T_{ref}}C_{P}dT[\/latex] assuming there are no phase changes\r\n<p style=\"text-align: center\">[latex] \\Delta\\dot{H} = \\Sigma\\dot{n}_{out}*(\\hat{H}^{\\circ}_{f,i} + \\int^{T_{out}}_{T_{ref}}C_{P}dT) - \\Sigma\\dot{n}_{in}*(\\hat{H}^{\\circ}_{f,i} + \\int^{T_{in}}_{T_{ref}}C_{P}dT)[\/latex]<\/p>\r\n<p style=\"text-align: left\">For the same system, equating [latex] \\Delta\\dot{H}[\/latex] for both methods, we can see that the [latex]C_{P}[\/latex] terms cancel. We obtain the following:<\/p>\r\n\r\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 88.7437%;height: 43px\" border=\"0\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 100%;text-align: center\"><span style=\"font-size: 16px\">[latex]\\xi\\Delta\\dot{H}^{\\circ}_{r} = \\Sigma\\dot{n}_{out}*\\hat{H}^{\\circ}_{f,i} - \\Sigma\\dot{n}_{in}*\\hat{H}^{\\circ}_{f,i} [\/latex]<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nOverall, the final [latex] \\Delta\\dot{H}[\/latex] for both methods will be the <strong>same<\/strong>. This means we can choose either method based on which is easiest to use with the information we have available.\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise: Comparing Both Methods<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nEthane and oxygen are fed into a furnace at 100 kmol\/hr and 500 kmol\/hr, respectively. Ethane goes through complete combustion in the furnace. Oxygen, carbon dioxide, and water exit the furnace at 150 kmol\/hr, 200 kmol\/hr, and 300 kmol\/hr, respectively.\r\n<p style=\"text-align: center\">[latex]C_{2}H_{6}(g) + \\frac{7}{2}O_{2}(g) \u2192 2CO_{2}(g) + 2H_{2}O(l) [\/latex]<\/p>\r\nwhere the heat of reaction is: [latex]\\Delta H^{\\circ}_{r} = -1560.7 kJ\/mol[\/latex]\r\n<p style=\"text-align: left\">The following data on heats of formations is available:<\/p>\r\n\r\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 65.8848%;height: 75px\" border=\"0\">\r\n<tbody>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 50%;height: 15px;text-align: center\"><strong>Compound<\/strong><\/td>\r\n<td style=\"width: 50%;height: 15px;text-align: center\"><strong>[latex]\\Delta H^{\\circ}_{f,i}[\/latex] (kJ\/mol)<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 50%;height: 15px;text-align: center\">Ethane<\/td>\r\n<td style=\"width: 50%;height: 15px;text-align: center\">-83.82<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 50%;height: 15px;text-align: center\">Oxygen<\/td>\r\n<td style=\"width: 50%;height: 15px;text-align: center\">0<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 50%;height: 15px;text-align: center\">Carbon Dioxide<\/td>\r\n<td style=\"width: 50%;height: 15px;text-align: center\">-393.51<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 50%;height: 15px;text-align: center\">Water (liquid)<\/td>\r\n<td style=\"width: 50%;height: 15px;text-align: center\">-285.83<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAssume the reaction takes place in the furnace at standard temperature and pressure. What is the [latex]\\Delta H^{\\circ}[\/latex] for the desired reaction? Calculate the change in enthalpy using both the heat of reaction and heat of formation method\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Solution<\/h3>\r\n<strong>Heat of Reaction Method<\/strong>\r\n\r\n<strong>Step 1<\/strong>: Determine the extent of the reaction:\r\n\r\n\\begin{align*}\r\n\\xi &amp;= \\frac{|(\\dot{n}_{i})_{out}-(\\dot{n}_{i})_{in}|}{|(\\dot{\\nu}_{i})|}hat\\\\&amp; = \\frac{|(\\dot{n}_{C_{2}H_{6}})_{out}-(\\dot{n}_{C_{2}H_{6}})_{in}|}{|(\\dot{\\nu}_{C_{2}H_{6}})|} \\\\&amp; =\\frac{|0\\frac{kmol}{hr} - 100 \\frac{kmol}{hr}|}{|-1|}\\\\ &amp;= 100 kmol\/hr\r\n\\end{align*}\r\n\r\n&nbsp;\r\n\r\n<strong>Step 2<\/strong>: Multiply the extent of the reaction by the heat of reaction for combusion\r\n<p style=\"text-align: center\">[latex] \\Delta \\dot{H} = \\xi*\\Delta H^{\\circ}_{r}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta \\dot{H} = 100\\frac{kmol}{hr}*-1560.7 \\frac{kJ}{mol} * 1000 \\frac{mol}{kmol} [\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex] \\Delta \\dot{H} = -156 MJ\/hr[\/latex]<\/p>\r\n&nbsp;\r\n\r\n<strong>Heat of Formation Method<\/strong>\r\n\r\n<strong>Step 1<\/strong>: Set up the heat of formation formula\r\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = \\Sigma\\dot{n}_{out}*\\hat{H}_{out} - \\Sigma\\dot{n}_{in}*\\hat{H}_{in}[\/latex]<\/p>\r\n&nbsp;\r\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = (\\dot{n}_{out,ethane}*\\hat{H}_{ethane})+(\\dot{n}_{out,oxygen}*\\hat{H}_{oxygen}) + (\\dot{n}_{out,water}*\\hat{H}_{water}) + (\\dot{n}_{out,carbon dioxide}*\\hat{H}_{carbon dioxide}) - (\\dot{n}_{in,ethane}*\\hat{H}_{ethane})+(\\dot{n}_{in,oxygen}*\\hat{H}_{oxygen}) + (\\dot{n}_{in,water}*\\hat{H}_{water}) + (\\dot{n}_{in,carbon dioxide}*\\hat{H}_{carbon dioxide})[\/latex]<\/p>\r\n&nbsp;\r\n<p style=\"text-align: left\"><strong>Step 2<\/strong>: Plug in the enthalpies of formation and calculate the change in enthalpy<\/p>\r\n\r\n<table class=\"grid aligncenter\">\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center\">Compound<\/th>\r\n<th style=\"text-align: center\">[latex]n_{in}[\/latex] (kmol\/hr)<\/th>\r\n<th style=\"text-align: center\">[latex]n_{out}[\/latex] (kmol\/hr)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td style=\"text-align: center\">Ethane<\/td>\r\n<td style=\"text-align: center\">100<\/td>\r\n<td style=\"text-align: center\">0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: center\">Oxygen<\/td>\r\n<td style=\"text-align: center\">500<\/td>\r\n<td style=\"text-align: center\">150<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: center\">Carbon Dioxide<\/td>\r\n<td style=\"text-align: center\">0<\/td>\r\n<td style=\"text-align: center\">200<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: center\">Water<\/td>\r\n<td style=\"text-align: center\">0<\/td>\r\n<td style=\"text-align: center\">300<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n&nbsp;\r\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = ((0\\frac{kmol}{hr}*-83.82\\frac{kJ}{mol})+(150\\frac{kmol}{hr}*0\\frac{kJ}{mol}) + (200\\frac{kmol}{hr}*-393.51\\frac{kJ}{mol}) + (300\\frac{kmol}{hr}*-285.83\\frac{kJ}{mol}) - (100\\frac{kmol}{hr}*-83.82\\frac{kJ}{mol}) - (500\\frac{kmol}{hr}*0\\frac{kJ}{mol}) - (0\\frac{kmol}{hr}*-393.51\\frac{kJ}{mol}) - (0\\frac{kmol}{hr}*-285.83\\frac{kJ}{mol}))*1000\\frac{mol}{kmol} [\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = -150 MJ[\/latex]<\/p>\r\n<strong>Notice that the methods aren't exactly equal. This is due to slight differences in available enthalpy data and assumptions made regarding temperature<\/strong>\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>","rendered":"<div class=\"textbox textbox--learning-objectives\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Learning Objectives<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>By the end of this section, you should be able to:<\/p>\n<p id=\"Explain:\"><strong>Explain<\/strong>\u00a0heats of reaction as well as endothermic and exothermic reactions<\/p>\n<p id=\"Determine:\"><strong>Determine<\/strong>\u00a0the standard heat of reaction given other heats of reaction or heats of formation (Hess&#8217;s Law)<\/p>\n<p id=\"Analyze:\"><strong>Analyze<\/strong>\u00a0energy balances involving reactive systems<\/p>\n<\/div>\n<\/div>\n<h2 id=\"What-happens-in-reactions-in-terms-of-chemical-bonds?\">What happens in reactions in terms of chemical bonds?<\/h2>\n<p>Bonds can be formed and broken. Breaking bonds <strong>takes<\/strong> energy and forming bonds <strong>releases<\/strong> energy.<\/p>\n<ul>\n<li>If more energy is released in forming bonds than absorbed in breaking bonds, then the reaction is <strong>exothermic<\/strong>.<\/li>\n<li>If more energy is absorbed in breaking bonds than released in forming bonds, then the reaction is <strong>endothermic<\/strong>.<\/li>\n<\/ul>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example: Chemical Bonds<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Consider the simple reaction of the formation of water from hydrogen and oxygen:<\/p>\n<p style=\"text-align: center\">[latex]2H_{2} + O_{2} \u2192 2H_{2}O[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1055 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/WaterRxn-300x180.png\" alt=\"\" width=\"420\" height=\"252\" srcset=\"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/WaterRxn-300x180.png 300w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/WaterRxn-65x39.png 65w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/WaterRxn-225x135.png 225w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/WaterRxn-350x211.png 350w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/WaterRxn.png 482w\" sizes=\"auto, (max-width: 420px) 100vw, 420px\" \/><\/p>\n<p style=\"text-align: center\">Image from\u00a0 <a title=\"via Wikimedia Commons\" href=\"https:\/\/commons.wikimedia.org\/wiki\/File:H2%2BO2%3DH2O.svg\">\u3059\u3058\u306b\u304f\u30b7\u30c1\u30e5\u30fc<\/a> \/ CC0<\/p>\n<p><em>The formation of bonds takes energy while the breakage of bonds releases energy<\/em><\/p>\n<blockquote><p>In this reaction:<\/p><\/blockquote>\n<ul>\n<li>\n<blockquote><p>2 H-H bonds are broken<\/p><\/blockquote>\n<\/li>\n<li>\n<blockquote><p>\u00a01 O-O bonds are broken<\/p><\/blockquote>\n<\/li>\n<li>\n<blockquote><p>4 H-O bonds are formed<\/p><\/blockquote>\n<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<h2 id=\"Heat-of-Reaction-$\\Delta-H_{r}(T,P)$\">Heat of Reaction [latex]\\Delta H_{r}(T,P)[\/latex]<\/h2>\n<p><strong>Heat of Reaction [latex]\\Delta H_{r}(T,P)[\/latex]<\/strong>: the stoichiometric enthalpy difference when reactants react completely to form products at a specified constant temperature and pressure.<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H_{r}(T,P) = H_{products} - H_{reactants}[\/latex]<\/p>\n<blockquote><p>In <strong>exothermic reactions<\/strong>, [latex]\\Delta H_{r}(T,P)<0[\/latex]\n\n\n<ul>\n<li>[latex]H_{products} < H_{reactants}[\/latex]<\/li>\n<\/ul>\n<p>In <strong>endothermic reactions<\/strong>, [latex]\\Delta H_{r}(T,P)>0[\/latex]<\/p>\n<ul>\n<li><span style=\"color: #333333\">[latex]H_{products} > H_{reactants}[\/latex]<\/span><\/li>\n<\/ul>\n<\/blockquote>\n<p>Heats of reaction are <em>directly proportional to the amount of reactants or products<\/em> in a reaction.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example: Formation of Nitrogen<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Consider the formation of nitrogen dioxide<\/p>\n<p style=\"text-align: center\">[latex]\\frac{1}{2} N_{2} (g) + O_{2}(g) \u2192 NO_{2} (g)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H_{r} = 33.2 kJ[\/latex]<\/p>\n<p>The formation of 1 mole of [latex]NO_{2}[\/latex] yields an enthalpy change of 33.2 kJ<\/p>\n<p style=\"text-align: center\">[latex]N_{2} (g) + 2O_{2}(g) \u2192 2NO_{2} (g)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H_{r} = 66.4 kJ[\/latex]<\/p>\n<p>The formation of 2 moles of [latex]NO_{2}[\/latex] yields an enthalpy change of 66.2 kJ<\/p>\n<p>Heat of reactions for a reaction in the forward direction is *equal to the negative heat of reaction* for the backward reaction.<\/p>\n<blockquote><p>Consider the formation of hydrogen chloride (or the corresponding decomposition):<\/p>\n<p style=\"text-align: center\">[latex]H_{2} (g) + Cl_{2} (g) \u2192 2 HCl (g)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H_{r} = -184.6 kJ[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]2HCl(g) \u2192 H_{2} (g) + Cl_{2} (g)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H_{r} = +184.6 kJ[\/latex]<\/p>\n<\/blockquote>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise: Heat of Reaction<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<div>\n<p>Consider the oxidation of ammonia taking place in an isothermal and constant pressure reactor:<\/p>\n<p style=\"text-align: center\">[latex]4 NH_{3}(g) + 5 O_{2}(g) \u2192 4 NO(g) + 6 H_{2}O(v)[\/latex]<\/p>\n<p>where the heat of reaction is [latex]\\Delta\\hat{H}^{\\circ}_{r} = -904.7 kJ[\/latex]<\/p>\n<p>The ammonia is fed into the reactor at [latex]100 mol\/s[\/latex] and the oxygen enters at [latex]200 mol\/s[\/latex]. Assuming the limiting reactant is completely consumed, what is the enthalpy change for this reaction?<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Solution<\/h3>\n<p><strong>Step 1:<\/strong> Find the limiting reactant by finding the smallest reactant extent if all of a reactant is consumed.<\/p>\n<p style=\"text-align: center\">[latex]\\text{if } N\\!H_{3} \\text{ is limiting: } \\xi=\\frac{100mol\/s}{4}=25\\frac{mol}{s}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\text{if } O_{2} \\text{ is limiting: } \\xi=\\frac{200mol\/s}{5}=40\\frac{mol}{s}[\/latex]<\/p>\n<p>[latex]N\\!H_{3}[\/latex] being the limiting reagent yields the smaller reaction extent, therefore [latex]N\\!H_{3}[\/latex] is the limiting reagent. The reaction extent is [latex]25\\frac{mol}{s}[\/latex].<\/p>\n<p><strong>Step 2:<\/strong> Multiply the reaction extent by the [latex]\\Delta\\hat{H}^{\\circ}_{r} = -904.7 \\frac{kJ}{molNH_{3}}[\/latex] to obtain [latex]\\Delta \\dot{H}_{r}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta \\dot{H}_{r}=\\xi*\\Delta\\hat{H}^{\\circ}_{r}[\/latex]<br \/>\n[latex]\\Delta \\dot{H}_{r}=25\\frac{mol}{s}*-904.7\\frac{kJ}{mol}[\/latex]<br \/>\n[latex]\\Delta \\dot{H}_{r}=-22620 \\frac{kJ}{s}[\/latex]<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<h3 id=\"Hess's-Law\">Hess&#8217;s Law<\/h3>\n<p>Some reactions may be difficult to reproduce in a laboratory setting. Instead, for analyzing a certain overall reaction, we may break the reaction down in multiple steps:<\/p>\n<p>For example, say we want to find the standard heat of reaction for the following hypothetical reaction:<\/p>\n<p style=\"text-align: center\">[latex]A + 0.5 B \u2192 C[\/latex]<\/p>\n<p>If this reaction is difficult to carry out in a laboratory, we might break it up into 2 reactions, that are easier to carry out and where we can determine specified standard heats of reaction:<\/p>\n<blockquote><p>Reaction 1: [latex]A + B \u2192 D[\/latex] with [latex]\\Delta H^{\\circ}_{rxn1}[\/latex]<\/p>\n<p>Reaction 2: [latex]C + 0.5 B \u2192 D[\/latex] with [latex]\\Delta H^{\\circ}_{rxn2}[\/latex]<\/p><\/blockquote>\n<p>We can combine reactions 1 and 2 to obtain our desired reaction by subtracting reaction 2 from reaction 1:<\/p>\n<p style=\"text-align: center\">[latex]Reaction 1 - Reaction 2[\/latex]<\/p>\n<p style=\"text-align: center\">[latex](A + B) - (C + 0.5 B) \u2192 D - D[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]A + 0.5 B - C \u2192[\/latex]<\/p>\n<p>We obtain our desired reaction:<\/p>\n<p style=\"text-align: center\">[latex]A + 0.5 B \u2192 C[\/latex]<\/p>\n<p>Therefore, the heat of reaction of the desired reaction will be:<\/p>\n<blockquote>\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ} = \\Delta H^{\\circ}_{rxn1} - \\Delta H^{\\circ}_{rxn2}[\/latex]<\/p>\n<\/blockquote>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise: Hess&#8217;s Law<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<p>Consider the formation of chlorine trifluoride from chlorine monofluoride and fluorine:<\/p>\n<p style=\"text-align: center\">[latex]ClF(g) + F_{2}(g) \u2192 ClF_{3}(g)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ} = ?[\/latex]<\/p>\n<p>Data on the following reactions are available:<\/p>\n<table class=\"grid aligncenter\" style=\"height: 90px\">\n<thead>\n<tr style=\"height: 30px\">\n<th style=\"width: 69.45px;text-align: center;height: 30px\">Number<\/th>\n<th style=\"width: 539.85px;text-align: center;height: 30px\">Reaction<\/th>\n<th style=\"width: 179.05px;text-align: center;height: 30px\">[latex]\\Delta H^{\\circ}[\/latex] (kJ)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 15px\">\n<td style=\"width: 69.45px;text-align: center;height: 15px\">1<\/td>\n<td style=\"width: 539.85px;text-align: center;height: 15px\">[latex]2OF_{2}(g) \u2192 O_{2}(g) + 2F_{2}(g)[\/latex]<\/td>\n<td style=\"width: 179.05px;text-align: center;height: 15px\">-49.4<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 69.45px;text-align: center;height: 15px\">2<\/td>\n<td style=\"width: 539.85px;text-align: center;height: 15px\">[latex]2ClF(g) + O_{2}(g) \u2192 Cl_{2}O(g) + OF_{2}(g)[\/latex]<\/td>\n<td style=\"width: 179.05px;text-align: center;height: 15px\">214.0<\/td>\n<\/tr>\n<tr style=\"height: 30px\">\n<td style=\"width: 69.45px;text-align: center;height: 30px\">3<\/td>\n<td style=\"width: 539.85px;text-align: center;height: 30px\">[latex]ClF_{3}(g) + O_{2}(g) \u2192 \\frac{1}{2} Cl_{2}O(g) + \\frac{3}{2}OF_{2}(g)[\/latex]<\/td>\n<td style=\"width: 179.05px;text-align: center;height: 30px\">236.2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>What is the [latex]\\Delta H^{\\circ}[\/latex] for the desired reaction?<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Solution<\/h3>\n<p><strong>Step 1:<\/strong> Determine what reaction will give us [latex]ClF[\/latex] in the reactants.<\/p>\n<p>Since [latex]ClF(g)[\/latex] is needed as a reactant, we can multiply reaction 2 by [latex]\\frac{1}{2}[\/latex] to obtain:<\/p>\n<p style=\"text-align: center\">[latex]ClF(g) + \\frac{1}{2}O_{2}(g) \u2192 \\frac{1}{2}Cl_{2}O(g) + \\frac{1}{2}OF_{2}(g)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ} = \\frac{1}{2}(214.0 kJ) = 107.0 kJ[\/latex]<\/p>\n<p><strong>Step 2:<\/strong> Determine what reaction will give us [latex]F_{2}[\/latex] in the reactants.<\/p>\n<p>Since [latex]F_{2}[\/latex] is needed as a reactant, we can multiply reaction 1 by [latex]-\\frac{1}{2}[\/latex] to obtain:<\/p>\n<p style=\"text-align: center\">[latex]F_{2}(g) + \\frac{1}{2}O_{2}(g) \u2192 OF_{2}(g)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ} = -\\frac{1}{2}(-49.4 kJ) = 24.7 kJ[\/latex]<\/p>\n<p><strong>Step 3:<\/strong> Determine what reaction will give us [latex]ClF_{3}[\/latex] as a product.<\/p>\n<p>Since [latex]ClF_{3}[\/latex] is needed as a product, we can multiply reaction 3 by -1 to obtain:<\/p>\n<p style=\"text-align: center\">[latex]\\frac{1}{2} Cl_{2}O(g) + \\frac{3}{2}OF_{2}(g) \u2192 ClF_{3}(g) + O_{2}(g)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ} = -(236.2kJ) = -236.2 kJ[\/latex]<\/p>\n<p><strong>Step 4:<\/strong> Add the reactions up to ensure that the desired reaction is obtained:<\/p>\n<p style=\"text-align: center\">[latex]ClF(g) + \\frac{1}{2}O_{2}(g) \u2192 \\frac{1}{2}Cl_{2}O(g) + \\frac{1}{2}OF_{2}(g)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]F_{2}(g) + \\frac{1}{2}O_{2}(g) \u2192 OF_{2}(g)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\frac{1}{2} Cl_{2}O(g) + \\frac{3}{2}OF_{2}(g) \u2192 ClF_{3}(g) + O_{2}(g)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\u2193(+)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]ClF(g) + F_{2}(g) \u2192 ClF_{3}(g)[\/latex]<\/p>\n<p><strong>Step 5:<\/strong> Add up all the manipulated heat of reactions to obtain the desired [latex]\\Delta H^{\\circ}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ} = (107.0 + 24.7 + -236.2)kJ[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ} = -104.5 kJ[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<h2 id=\"Heat-of-reaction-method-for-analyzing-energy-use-in-a-process\">Heat of Reaction Method for Analyzing Energy Use in a Process<\/h2>\n<p>Now let&#8217;s try applying the heat of reaction to determine energy use in a process, where reactants and products are coming in and exiting at a given temperature. The following process path is taken for the heat of reaction method, where the reference state is at [latex]25^{\\circ}C[\/latex]:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1056 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/HoR-300x156.png\" alt=\"\" width=\"423\" height=\"220\" srcset=\"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/HoR-300x156.png 300w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/HoR-65x34.png 65w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/HoR-225x117.png 225w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/HoR-350x182.png 350w, https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-content\/uploads\/sites\/1010\/2020\/05\/HoR.png 736w\" sizes=\"auto, (max-width: 423px) 100vw, 423px\" \/><\/p>\n<p><span style=\"text-align: initial;font-size: 1em\">Generally, [latex]\\Delta H ^{\\circ}_{r}[\/latex] refers to standard state conditions at [latex]25^{\\circ}C[\/latex] and 1 atm, but always double-check whether a different standard state condition is used.<\/span><\/p>\n<p>In this process path, the enthalpy calculations are done in three steps:<\/p>\n<ol>\n<li>Temperature change: calculate the [latex]\\Delta\\hat{H}_{reactants}[\/latex] using the heat capacities. The temperature change will be the difference between the inlet temperature of the reactants and the standard or reference temperature ([latex]25^{\\circ}C[\/latex] in this case).<\/li>\n<li>Reaction enthalpy: calculate the enthalpy of the reaction [latex]\\Delta H^{\\circ}_{r}[\/latex]. This can be done using Hess&#8217;s Law or the Heat of Formation method (discussed below).<\/li>\n<li>Temperature change: calculate the [latex]\\Delta\\hat{H}_{products}[\/latex] using the heat capacities. The temperature change will be the difference between the outlet temperature of the products and the standard or reference temperature ([latex]25^{\\circ}C[\/latex] in this case).<\/li>\n<\/ol>\n<p>The calculated enthalpy changes for each step are then added to obtain [latex]\\Delta\\dot{H}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"prompt input_prompt\">\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example: Heat of Reaction Method for Energy Use<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Consider a propane combustion chamber, where [latex]100 mol\/s[\/latex] of propane are feed to the chamber at [latex]25^{\\circ}C[\/latex] and air is fed at [latex]300^{\\circ}C[\/latex] ([latex]600 mol\/s O_{2}(g)[\/latex] and [latex]2256 mol\/s N_{2}(g)[\/latex]). The products stream exits at [latex]1000^{\\circ}C[\/latex] and consists of [latex]100 mol\/s O_{2}(g)[\/latex], [latex]2256 mol\/s N_{2}(g)[\/latex], [latex]300 mol\/s CO_{2}(g)[\/latex], and [latex]400 mol\/s H_{2}O(v)[\/latex]. How much heatis released by this combustion chamber? Assumming atmoshpere pressure, so water boils at 100\u00b0C.<\/p>\n<p style=\"text-align: center\">[latex]C_{3}H_{8} (g) + 5 O_{2} \u2192 3 CO_{2} (g) + 4 H_{2}O (l)[\/latex]<br \/>\n[latex]\\Delta H^{\\circ}_{r} = -2220 kJ[\/latex]<\/p>\n<p>The following information is provided:<\/p>\n<p style=\"text-align: left\">The values listed under each compound are specific enthalpies in kJ\/mol<\/p>\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 88.4861%;height: 75px\">\n<tbody>\n<tr style=\"height: 15px\">\n<td style=\"width: 20%;height: 15px;text-align: center\"><strong>Temperature ([latex]^{\\circ}C[\/latex])<\/strong><\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\"><strong>[latex]O_{2}[\/latex]<\/strong><\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\"><strong>[latex]N_{2}[\/latex]<\/strong><\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\"><strong>[latex]CO_{2}[\/latex]<\/strong><\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\"><strong>[latex]H_{2}O[\/latex]<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 20%;height: 15px;text-align: center\">25<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">0<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">0<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">0<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">0<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 20%;height: 15px;text-align: center\">100<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">2.24<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">2.19<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">2.90<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">2.54<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 20%;height: 15px;text-align: center\">300<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">8.47<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">8.12<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">11.58<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">9.57<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 20%;height: 15px;text-align: center\">1000<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">32.47<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">30.56<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">48.60<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">37.69<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ul>\n<li>[latex]C_{P,l(H_{2}O)} = 75.4*10^{-3} \\frac{kJ}{molK}[\/latex]<\/li>\n<li>[latex]C_{P,v(H_{2}O)} = 33.46*10^{-3}+0.688*10^{-5}*T + 0.7604*10^{-8}*T^{2} - 3.593*10^{-12}*T^{3} (J\/molK)[\/latex]<\/li>\n<li>[latex]\\Delta\\hat{H}_{vap}(100^{\\circ}C) = 40.66 kJ\/mol[\/latex]<\/li>\n<\/ul>\n<h4>How can we solve the system&#8217;s energy balance?<\/h4>\n<p>1 &#8211; Solve material balances as much as possible<\/p>\n<blockquote><p>This example already has the material balances solved<\/p><\/blockquote>\n<p>2 &#8211; Choose reference states for energy calculations. Like we say in the previous figure we want to calculate the changes in energy associated with bringing our reactants to our reference state, calculating the energy of the reaction at the reference state, and then the energy associated with bringing the products to their final state. We will choose reference states based on the information we have to try to make these calculations as easy as possible.<\/p>\n<blockquote><p>For reacting species:<br \/>\nSince [latex]\\Delta H^{\\circ}_{r}[\/latex] is given, we will assume this is at [latex]25^{\\circ}C[\/latex] and 1 atm (as there is nothing indicating otherwise), and we will use this as our reference state.<\/p>\n<p>For non-reacting species:<br \/>\nWe can use any convenient temperature and pressure as a reference state (inlet temperature, outlet temperature, temperature in enthalpy table). With this example, the enthalpy table given uses a reference state of [latex]25^{\\circ}C[\/latex], and we will assume everything is at 1 atm (as no other values for pressure are given).<\/p><\/blockquote>\n<p>3 &#8211; Calculate the extent of reaction for all reactions (in this case we just have the one reaction)<\/p>\n<blockquote>\n<p style=\"text-align: center\">[latex]\\xi = \\frac{(\\dot{n}_{i})_{out}-(\\dot{n}_{i})_{in}}{\\dot{\\nu}_{i}}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\xi = \\frac{(\\dot{n}_{CO_{2}})_{out}-(\\dot{n}_{CO_{2}})_{in}}{\\dot{\\nu}_{CO_{2}}}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\xi = \\frac{300\\frac{mol}{s} - 0 \\frac{mol}{s}}{3} = 100 mol\/s[\/latex]<\/p>\n<\/blockquote>\n<p style=\"text-align: left\">4 &#8211; Prepare an inlet-out enthalpy table (this will show what enthalpies we need to calculate associated with energy changes in the reactants or products)<\/p>\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 95.2026%;height: 180px\">\n<tbody>\n<tr>\n<td style=\"width: 20%\"><strong>Substance<\/strong><\/td>\n<td style=\"width: 20%\"><strong>[latex]\\dot{n}_{in}[\/latex](mol\/s)<\/strong><\/td>\n<td style=\"width: 20%\"><strong>[latex]\\hat{H}_{in}[\/latex] (kJ\/mol)<\/strong><\/td>\n<td style=\"width: 20%\"><strong>[latex]\\dot{n}_{out}[\/latex] (mol\/s)<\/strong><\/td>\n<td style=\"width: 20%\"><strong>[latex]\\hat{H}_{out}[\/latex] (kJ\/mol)<\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%\">[latex]C_{3}H_{8}[\/latex]<\/td>\n<td style=\"width: 20%\">100<\/td>\n<td style=\"width: 20%\">[latex]\\hat{H}_{1}[\/latex]<\/td>\n<td style=\"width: 20%\">&#8211;<\/td>\n<td style=\"width: 20%\">&#8211;<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%\">[latex]O_{2}[\/latex]<\/td>\n<td style=\"width: 20%\">600<\/td>\n<td style=\"width: 20%\">[latex]\\hat{H}_{2}[\/latex]<\/td>\n<td style=\"width: 20%\">100<\/td>\n<td style=\"width: 20%\">[latex]\\hat{H}_{4}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%\">[latex]N_{2}[\/latex]<\/td>\n<td style=\"width: 20%\">2256<\/td>\n<td style=\"width: 20%\">[latex]\\hat{H}_{3}[\/latex]<\/td>\n<td style=\"width: 20%\">2256<\/td>\n<td style=\"width: 20%\">[latex]\\hat{H}_{5}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%\">[latex]CO_{2}[\/latex]<\/td>\n<td style=\"width: 20%\">&#8211;<\/td>\n<td style=\"width: 20%\">&#8211;<\/td>\n<td style=\"width: 20%\">300<\/td>\n<td style=\"width: 20%\">[latex]\\hat{H}_{6}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%\">[latex]H_{2}O[\/latex]<\/td>\n<td style=\"width: 20%\">&#8211;<\/td>\n<td style=\"width: 20%\">&#8211;<\/td>\n<td style=\"width: 20%\">400<\/td>\n<td style=\"width: 20%\">[latex]\\hat{H}_{7}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<p>5 &#8211; Calculate all the enthalpies. For [latex]\\hat{H_{1}}[\/latex] to [latex]\\hat{H_{6}}[\/latex], the specific enthalpies at different temperatures are given as the difference in enthalpy from the reference state (25\u00b0C, 1atm, which is the same as what we choose as reference state):<\/p>\n<blockquote>\n<p style=\"text-align: center\">[latex]\\hat{H}_{1}=\\Delta\\hat{H}_{C_{3}H_{8}} (25^{\\circ}C \u2192 25^{\\circ}C) = 0 kJ\/mol[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\hat{H}_{2}=\\Delta\\hat{H}_{O_{2}} (25^{\\circ}C \u2192 300^{\\circ}C) = (8.47 - 0) kJ\/mol[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\hat{H}_{3}=\\Delta\\hat{H}_{N_{2}} (25^{\\circ}C \u2192 300^{\\circ}C) = (8.12 - 0) kJ\/mol[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\hat{H}_{4}=\\Delta\\hat{H}_{O_{2}} (25^{\\circ}C \u2192 1000^{\\circ}C) = (32.47 - 0) kJ\/mol[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\hat{H}_{5}=\\Delta\\hat{H}_{N_{2}} (25^{\\circ}C \u2192 1000^{\\circ}C) = (30.56 - 0) kJ\/mol[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\hat{H}_{6}=\\Delta\\hat{H}_{CO_{2}} (25^{\\circ}C \u2192 1000^{\\circ}C) = (48.60 - 0) kJ\/mol[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\hat{H}_{7}=\\Delta\\hat{H}_{H_{2}O} (l,25^{\\circ}C \u2192 H_{2}O (v,1000^{\\circ}C))[\/latex]<\/p>\n<\/blockquote>\n<p>6 &#8211; For [latex]\\hat{H_{7}}[\/latex], the specific enthalpy for water is calculated. From the reference state of 25\u00b0C to the stream temperature of 1000\u00b0C, water is first heated to 100\u00b0C, which is its natural boiling point at 1atm, then vaporized, then heated in vapor phase to 1000\u00b0C. The enthalpy change of each process is calculated separately and added together to get [latex]\\hat{H_{7}}[\/latex].<\/p>\n<blockquote><p><strong>Note:<\/strong> The [latex]T[\/latex] in the formulas to calculate [latex]C_{P}[\/latex] is given in kelvin. This doesn&#8217;t make a difference when [latex]C_{P}[\/latex] is given as a number, as the scales for one degree of Celcius and kelvin are the same, but the temperature must be converted to kelvin when [latex]T^2[\/latex] is higher power of [latex]T[\/latex] is used in calculation.<\/p>\n<p style=\"text-align: center\">[latex]100\u00b0C=373K, 1000\u00b0C=1273K[\/latex]<\/p>\n<p>\\begin{align*}<br \/>\n\\hat{H}_{7}&amp; = \\int^{100^{\\circ}C}_{25^{\\circ}C} C_{P,l}dT + \\Delta \\hat{H}_{vap}(100^{\\circ}C) + \\int^{1000^{\\circ}C}_{100^{\\circ}C} C_{P,v}dT\\\\&amp; = \\int^{100^{\\circ}C}_{25^{\\circ}C} 75.4*10^{-3}dT + \\Delta \\hat{H}_{vap}(100^{\\circ}C) \\\\&amp; \\;\\;\\;\\;+\\int^{1273K}_{373K} (33.46*10^{-3}+0.688*10^{-5}*T + 0.7604*10^{-8}*T^{2} &#8211; 3.593*10^{-12}*T^{3})dT\\\\&amp; = 75.4*10^{-3}*(100-25)+ 40.66 \\\\&amp; \\;\\;\\;\\; +(33.46*10^{-3}*T+\\frac{1}{2}*0.688*10^{-5}*T^2 + \\frac{1}{3}*0.7604*10^{-8}*T^{3} &#8211; \\frac{1}{4}*3.593*10^{-12}*T^{4})\\bigg\\vert^{1273K}_{373K}\\\\&amp; = (5.65 + 40.66 + 35.1)kJ\/mol\\\\&amp; = 81.46 kJ\/mol<br \/>\n\\end{align*}<\/p>\n<p style=\"text-align: center\">[latex]\\hat{H}_{7}= (5.65 + 40.66 + 35.1)kJ\/mol = 81.46\\frac{kJ}{mol}[\/latex]<\/p>\n<p>Therefore, we have calculated all the specific enthalpy for the reactants and products:<\/p><\/blockquote>\n<table class=\"grid\" style=\"border-collapse: collapse;width: 95.2026%;height: 120px\">\n<tbody>\n<tr style=\"height: 30px\">\n<td style=\"width: 20%;height: 30px;text-align: center\"><strong>Substance<\/strong><\/td>\n<td style=\"width: 20%;height: 30px;text-align: center\"><strong>[latex]\\dot{n}_{in}[\/latex](mol\/s)<\/strong><\/td>\n<td style=\"width: 20%;height: 30px;text-align: center\"><strong>[latex]\\hat{H}_{in}[\/latex] (kJ\/mol)<\/strong><\/td>\n<td style=\"width: 20%;height: 30px;text-align: center\"><strong>[latex]\\dot{n}_{out}[\/latex] (mol\/s)<\/strong><\/td>\n<td style=\"width: 20%;height: 30px;text-align: center\"><strong>[latex]\\hat{H}_{out}[\/latex] (kJ\/mol)<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 30px\">\n<td style=\"width: 20%;height: 30px;text-align: center\">[latex]C_{3}H_{8}[\/latex]<\/td>\n<td style=\"width: 20%;height: 30px;text-align: center\">100<\/td>\n<td style=\"width: 20%;height: 30px;text-align: center\">0<\/td>\n<td style=\"width: 20%;height: 30px;text-align: center\">&#8211;<\/td>\n<td style=\"width: 20%;height: 30px;text-align: center\">&#8211;<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 20%;height: 15px;text-align: center\">[latex]O_{2}[\/latex]<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">600<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">8.47<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">100<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">32.47<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 20%;height: 15px;text-align: center\">[latex]N_{2}[\/latex]<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">2256<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">8.12<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">2256<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">30.56<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 20%;height: 15px;text-align: center\">[latex]CO_{2}[\/latex]<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">&#8211;<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">&#8211;<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">300<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">48.60<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 20%;height: 15px;text-align: center\">[latex]H_{2}O[\/latex]<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">&#8211;<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">&#8211;<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">400<\/td>\n<td style=\"width: 20%;height: 15px;text-align: center\">81.46<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<p>8 &#8211; Finally, solve the energy balance<\/p>\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} + \\Delta\\dot{E}_{k} +\\Delta\\dot{E}_{p} = \\dot{Q} + \\dot{W}_{s}[\/latex]<\/p>\n<p>[latex]\\Delta\\dot{E}_{k}[\/latex], [latex]\\Delta\\dot{E}_{p}[\/latex], and [latex]\\dot{W}_{s}[\/latex] are assumed negligible for this system (as no information is provided on these)<\/p>\n<p style=\"text-align: center\">[latex]\\dot{Q} = \\Delta\\dot{H}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\dot{Q} = -126 MW[\/latex]<\/p>\n<p>The combustion chamber releases [latex]126 MW[\/latex] of heat to the environment<\/p>\n<\/div>\n<\/div>\n<h2 id=\"Heat-of-Formation-$\\Delta-H^{\\circ}_{f}$\">Heat of Formation [latex]\\Delta H^{\\circ}_{f}[\/latex]<\/h2>\n<p><strong>Formation Reaction<\/strong>: a reaction in which the compound is formed from its elemental constituents as they would normally occur in nature (eg. [latex]O_{2}[\/latex] rather than [latex]O[\/latex]).<\/p>\n<p>For elemental consistituents, the energy of formation is [latex]\\Delta H_{f}^{\\circ}[\/latex] = 0, since they would be forming themselves: [latex]O_{2}\u2192O_{2}[\/latex]<\/p>\n<p style=\"text-align: left\"><strong>Standard specific heat of formation [latex]\\Delta H^{\\circ}_{f}[\/latex]<\/strong>: the enthalpy change associated with forming 1 mole of the compound of interest at standard temperature ([latex]25^{\\circ}C[\/latex]) and pressure (1 atm)<\/p>\n<table class=\"grid aligncenter\" style=\"height: 138px\">\n<thead>\n<tr style=\"height: 15px\">\n<th style=\"height: 15px;width: 75.05px;text-align: center\">Compound<\/th>\n<th style=\"height: 15px;width: 355.05px;text-align: center\">Reaction<\/th>\n<th style=\"height: 15px;width: 272.65px;text-align: center\">[latex]\\Delta\\hat{H}^{\\circ}_{f,i}[\/latex] (kJ\/mol)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px;width: 75.05px;text-align: center\">Water<\/td>\n<td style=\"height: 15px;width: 355.85px;text-align: center\">[latex]H_{2}(g) + \\frac{1}{2}O_{2}(g) \u2192 H_{2}O(l)[\/latex]<\/td>\n<td style=\"height: 15px;width: 273.45px;text-align: center\">-285.83<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px;width: 75.05px;text-align: center\">Methane<\/td>\n<td style=\"height: 15px;width: 355.85px;text-align: center\">[latex]C(s) + 2H_{2}(g) \u2192 CH_{4}(g)[\/latex]<\/td>\n<td style=\"height: 15px;width: 273.45px;text-align: center\">-74.8936<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px;width: 75.05px;text-align: center\">Ethane<\/td>\n<td style=\"height: 15px;width: 355.85px;text-align: center\">[latex]2C(s) + 3H_{2}(g) \u2192 C_{2}H_{6}(g)[\/latex]<\/td>\n<td style=\"height: 15px;width: 273.45px;text-align: center\">-83.82<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px;width: 75.05px;text-align: center\">Propane<\/td>\n<td style=\"height: 15px;width: 355.85px;text-align: center\">[latex]3C(s) + 4H_{2}(g) \u2192 C_{3}H_{8}(g)[\/latex]<\/td>\n<td style=\"height: 15px;width: 273.45px;text-align: center\">-104.68<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px;width: 75.05px;text-align: center\">Benzene<\/td>\n<td style=\"height: 15px;width: 355.85px;text-align: center\">[latex]6C(s) + 3H_{2}(g) \u2192 C_{6}H_{6}(l)[\/latex]<\/td>\n<td style=\"height: 15px;width: 273.45px;text-align: center\">82.88<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px;width: 75.05px;text-align: center\">Toluene<\/td>\n<td style=\"height: 15px;width: 355.85px;text-align: center\">[latex]7C(s) + 4H_{2}(g) \u2192 C_{7}H_{8}(l)[\/latex]<\/td>\n<td style=\"height: 15px;width: 273.45px;text-align: center\">50.17<\/td>\n<\/tr>\n<tr style=\"height: 18px\">\n<td style=\"height: 18px;width: 75.05px;text-align: center\">Oxygen<\/td>\n<td style=\"height: 18px;width: 355.85px;text-align: center\">[latex]O_{2}(g) \u2192 O_{2}(g)[\/latex]<\/td>\n<td style=\"height: 18px;width: 273.45px;text-align: center\">0<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px;width: 75.05px;text-align: center\">Nitrogen<\/td>\n<td style=\"height: 15px;width: 355.85px;text-align: center\">[latex]N_{2}(g) \u2192 N_{2}(g)[\/latex]<\/td>\n<td style=\"height: 15px;width: 273.45px;text-align: center\">0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: left\">We can use these heats of formation and Hess&#8217;s law to find the heat of reaction for a given reaction:<\/p>\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 96.7839%;height: 93px\">\n<tbody>\n<tr>\n<td style=\"width: 100%;text-align: center\"><span style=\"font-size: 16px\">[latex]\\Delta H^{\\circ}_{r} = \\Sigma_{i}\\nu_{i}\\Delta\\hat{H}^{\\circ}_{f,i} = \\Sigma_{products}|\\nu_{i}|\\Delta\\hat{H}^{\\circ}_{f,i}-\\Sigma_{reactants}|\\nu_{i}|\\Delta\\hat{H}^{\\circ}_{f,i}[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example: Heat of Formation<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<div class=\"cell border-box-sizing text_cell rendered\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<p>Consider the combustion of ethane:<\/p>\n<p style=\"text-align: center\">[latex]C_{2}H_{6} (g) + \\frac{7}{2} O_{2}(g) \u2192 2 CO_{2} (g) + 3 H_{2}O (l)[\/latex]<\/p>\n<p>where [latex]\\Delta H^{\\circ}_{r} = -1560 kJ\/mol[\/latex]<\/p>\n<p>To calculate the enthalpy using the heat of formation method, the following steps are taken:<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ}_{r} = (3*\\Delta\\hat{H}^{\\circ}_{f,H_{2}O} + 2*\\Delta\\hat{H}^{\\circ}_{f,CO_{2}}) - (1*\\Delta\\hat{H}^{\\circ}_{f,C_{2}H_{6}}+\\frac{7}{2}*\\Delta\\hat{H}^{\\circ}_{f,O_{2}})[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ}_{r} = (3*-285.83 + 2*-393.51)kJ\/mol - (1*-83.82+\\frac{7}{2}*0)kJ\/mol[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ}_{r} = -1561 kJ\/mol[\/latex]<\/p>\n<p><em>The enthalpy of reaction using the heat of formation method is very close to the enthalpy of reaction (combustion).<\/em><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 id=\"Where-to-find-Heat-of-Formation-data\">Where to find Heat of Formation Data<\/h3>\n<p>One source of heat of formation data is the <a href=\"https:\/\/webbook.nist.gov\/chemistry\/name-ser\/\"> National Institute of Standards and Technology (NIST) Webbook NIST<\/a><\/p>\n<p>To find the heat of formation data at this site:<\/p>\n<p>1 &#8211; Look up the compound of interest<\/p>\n<p>2 &#8211; Heat of formation data will be under &#8220;condensed phase thermochemistry data&#8221; (for liquids or solids) or &#8220;gas phase thermochemistry data&#8221; (for gases).<\/p>\n<p><em>Heats of formation can also be found in Appendix E of &#8220;Introductory Chemical Engineering Thermodynamics&#8221; by J. Richard Elliot and Carl T. Lira.<\/em><\/p>\n<\/div>\n<h2 id=\"Heat-of-formation-method-for-analyzing-energy-use-in-a-process\">Heat of Formation Method for Analyzing Energy Use in a Process<\/h2>\n<ol>\n<li>Solve the material balance as much as possible: this can include solving for mass or molar flows using stoichiometry or mass balances<\/li>\n<li>Choose reference states for energy calculations: reference states provide a basis for enthalpy calculations. Choose reference states that make your calculations convenient or reference states that match the available data. Most data is provided at [latex]25^{\\circ}C[\/latex] and 1 atm, therefore this is a common reference state.<\/li>\n<li>Prepare and inlet-outlet enthalpy table: this table will include all the compounds involved in the system, the inlet molar flow, the inlet enthalpy values, the outlet molar flow, and the outlet enthalpy values. Fill out the table with all the known values and number the enthalpy values accordingly.<\/li>\n<li>Calculate all the enthalpies: use process paths to calculate the enthalpies listed in the table in step 3. The enthalpy change of each compound is calculated by summing the heat of formation, enthalpy change due to temperature change in the same state (which can be calculated using the [latex]C_{P}[\/latex] values), and enthalpy change due to phase change when phase change is involved.<\/li>\n<li>Calculate the [latex]\\Delta\\dot{H}[\/latex] for the system: this is done by multiplying each molar flow by the corresponding enthalpy and using the following expression. Remember that the stoichiometric coefficients are negative for reactants and positive for products: [latex]\\Delta\\dot{H} = \\Sigma\\dot{n}_{out}*\\hat{H}_{out} - \\Sigma\\dot{n}_{in}*\\hat{H}_{in}[\/latex]<\/li>\n<li>Solve the energy balance: determine which energy terms are present in the system and solve accordingly.<\/li>\n<\/ol>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example: Heat of Formation Method for Energy Use<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Let&#8217;s consider the propane combustion chamber problem we analyzed before with the heat of reaction method, and let&#8217;s see if we can get a similar answer using the heat of formation method.<\/p>\n<p>1 &#8211; Solve the material balance as much as possible<\/p>\n<blockquote><p>This example already has the material balances solved<\/p><\/blockquote>\n<p>2 &#8211; Choose reference states for energy calculations<\/p>\n<blockquote><p>For reacting species: elemental species that make up reacting species at standard conditions; we will choose [latex]25^{\\circ}C[\/latex] at 1 atm with [latex]C(s)[\/latex], [latex]H_{2}(g)[\/latex], and [latex]O_{2}(g)[\/latex].<\/p>\n<p>For non-reacting species (same as [latex]\\Delta H_{r}[\/latex]): Use any convenient temperature (inlet temperature, outlet temperature, temperature in enthalpy table) Here, [latex]25^{\\circ}C[\/latex] and 1 atm works because of our enthalpy table values<\/p><\/blockquote>\n<p>3 &#8211; Prepare an inlet-outlet enthalpy table<\/p>\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 95.2026%;height: 180px\">\n<tbody>\n<tr>\n<td style=\"width: 20%\"><strong>Substance<\/strong><\/td>\n<td style=\"width: 20%\"><strong>[latex]\\dot{n}_{in}[\/latex](mol\/s)<\/strong><\/td>\n<td style=\"width: 20%\"><strong>[latex]\\hat{H}_{in}[\/latex] (kJ\/mol)<\/strong><\/td>\n<td style=\"width: 20%\"><strong>[latex]\\dot{n}_{out}[\/latex] (mol\/s)<\/strong><\/td>\n<td style=\"width: 20%\"><strong>[latex]\\hat{H}_{out}[\/latex] (kJ\/mol)<\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%\">[latex]C_{3}H_{8}[\/latex]<\/td>\n<td style=\"width: 20%\">100<\/td>\n<td style=\"width: 20%\">[latex]\\hat{H}_{1}[\/latex]<\/td>\n<td style=\"width: 20%\">&#8211;<\/td>\n<td style=\"width: 20%\">&#8211;<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%\">[latex]O_{2}[\/latex]<\/td>\n<td style=\"width: 20%\">600<\/td>\n<td style=\"width: 20%\">[latex]\\hat{H}_{2}[\/latex]<\/td>\n<td style=\"width: 20%\">100<\/td>\n<td style=\"width: 20%\">[latex]\\hat{H}_{4}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%\">[latex]N_{2}[\/latex]<\/td>\n<td style=\"width: 20%\">2256<\/td>\n<td style=\"width: 20%\">[latex]\\hat{H}_{3}[\/latex]<\/td>\n<td style=\"width: 20%\">2256<\/td>\n<td style=\"width: 20%\">[latex]\\hat{H}_{5}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%\">[latex]CO_{2}[\/latex]<\/td>\n<td style=\"width: 20%\">&#8211;<\/td>\n<td style=\"width: 20%\">&#8211;<\/td>\n<td style=\"width: 20%\">300<\/td>\n<td style=\"width: 20%\">[latex]\\hat{H}_{6}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%\">[latex]H_{2}O[\/latex]<\/td>\n<td style=\"width: 20%\">&#8211;<\/td>\n<td style=\"width: 20%\">&#8211;<\/td>\n<td style=\"width: 20%\">400<\/td>\n<td style=\"width: 20%\">[latex]\\hat{H}_{7}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<p style=\"text-align: left\">4 &#8211; Calculate all the enthalpies<\/p>\n<table class=\"grid aligncenter\" style=\"height: 90px\">\n<thead>\n<tr style=\"height: 15px\">\n<th style=\"height: 15px;width: 128.65px;text-align: center\">Substance<\/th>\n<th style=\"height: 15px;width: 311.85px;text-align: center\">[latex]\\Delta H^{\\circ}_{f}[\/latex] (kJ\/mol)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px;width: 128.65px;text-align: center\">[latex]C_{3}H_{8}[\/latex]<\/td>\n<td style=\"height: 15px;width: 312.65px;text-align: center\">-103.8<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px;width: 128.65px;text-align: center\">[latex]O_{2}[\/latex]<\/td>\n<td style=\"height: 15px;width: 312.65px;text-align: center\">0<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px;width: 128.65px;text-align: center\">[latex]N_{2}[\/latex]<\/td>\n<td style=\"height: 15px;width: 312.65px;text-align: center\">0<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px;width: 128.65px;text-align: center\">[latex]CO_{2}[\/latex]<\/td>\n<td style=\"height: 15px;width: 312.65px;text-align: center\">-393.51<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px;width: 128.65px;text-align: center\">[latex]H_{2}O (v)[\/latex]<\/td>\n<td style=\"height: 15px;width: 312.65px;text-align: center\">-241.835<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<blockquote>\n<p style=\"text-align: center\">[latex]\\hat{H}_{1} = \\Delta\\hat{H}^{\\circ}_{f,C_{3}H_{8}(g)} = -103.8 kJ\/mol[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\hat{H}_{2}=\\Delta\\hat{H} (O_{2} (25^{\\circ}C \u2192 300^{\\circ}C) = (8.47 - 0) kJ\/mol[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\hat{H}_{3}=\\Delta\\hat{H} (N_{2} (25^{\\circ}C \u2192 300^{\\circ}C) = (8.12 - 0) kJ\/mol[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\hat{H}_{4}=\\Delta\\hat{H} (O_{2} (25^{\\circ}C \u2192 1000^{\\circ}C) = (32.47 - 0) kJ\/mol[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\hat{H}_{5}=\\Delta\\hat{H} (N_{2} (25^{\\circ}C \u2192 1000^{\\circ}C) = (30.56 - 0) kJ\/mol[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\hat{H}_{6}=\\Delta\\hat{H}^{\\circ}_{f,CO_{2}(g)} + \\int^{1000^{\\circ}C}_{25^{\\circ}C} C_{p,CO_{2}(g)}dT = -344.9 kJ\/mol[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\hat{H}_{7}=\\Delta\\hat{H}^{\\circ}_{f,H_{2}O(v)} + \\int^{1000^{\\circ}C}_{25^{\\circ}C} C_{p,H_{2}O(v)}dT = -204.1 kJ\/mol[\/latex]<\/p>\n<p>Here, you assume that water forms as vapor directly in the reaction. Therefore, there is no need to account for heat of vaporization.<\/p><\/blockquote>\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 95.2026%;height: 180px\">\n<tbody>\n<tr>\n<td style=\"width: 20%;text-align: center\"><strong>Substance<\/strong><\/td>\n<td style=\"width: 20%;text-align: center\"><strong>[latex]\\dot{n}_{in}[\/latex](mol\/s)<\/strong><\/td>\n<td style=\"width: 20%;text-align: center\"><strong>[latex]\\hat{H}_{in}[\/latex] (kJ\/mol)<\/strong><\/td>\n<td style=\"width: 20%;text-align: center\"><strong>[latex]\\dot{n}_{out}[\/latex] (mol\/s)<\/strong><\/td>\n<td style=\"width: 20%;text-align: center\"><strong>[latex]\\hat{H}_{out}[\/latex] (kJ\/mol)<\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%;text-align: center\">[latex]C_{3}H_{8}[\/latex]<\/td>\n<td style=\"width: 20%;text-align: center\">100<\/td>\n<td style=\"width: 20%;text-align: center\">-103.8<\/td>\n<td style=\"width: 20%;text-align: center\">&#8211;<\/td>\n<td style=\"width: 20%;text-align: center\">&#8211;<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%;text-align: center\">[latex]O_{2}[\/latex]<\/td>\n<td style=\"width: 20%;text-align: center\">600<\/td>\n<td style=\"width: 20%;text-align: center\">8.47<\/td>\n<td style=\"width: 20%;text-align: center\">100<\/td>\n<td style=\"width: 20%;text-align: center\">32.47<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%;text-align: center\">[latex]N_{2}[\/latex]<\/td>\n<td style=\"width: 20%;text-align: center\">2256<\/td>\n<td style=\"width: 20%;text-align: center\">8.12<\/td>\n<td style=\"width: 20%;text-align: center\">2256<\/td>\n<td style=\"width: 20%;text-align: center\">30.56<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%;text-align: center\">[latex]CO_{2}[\/latex]<\/td>\n<td style=\"width: 20%;text-align: center\">&#8211;<\/td>\n<td style=\"width: 20%;text-align: center\">&#8211;<\/td>\n<td style=\"width: 20%;text-align: center\">300<\/td>\n<td style=\"width: 20%;text-align: center\">-344.9<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 20%;text-align: center\">[latex]H_{2}O[\/latex]<\/td>\n<td style=\"width: 20%;text-align: center\">&#8211;<\/td>\n<td style=\"width: 20%;text-align: center\">&#8211;<\/td>\n<td style=\"width: 20%;text-align: center\">400<\/td>\n<td style=\"width: 20%;text-align: center\">-204.1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<p>5 &#8211; Calculate $\\Delta\\dot{H}$ for the reactor<\/p>\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = \\Sigma\\dot{n}_{out}*\\hat{H}_{out} - \\Sigma\\dot{n}_{in}*\\hat{H}_{in}=1.26x10^{5} kJ\/s[\/latex]<\/p>\n<p style=\"text-align: left\">6 &#8211; Finally, solve the energy balance<\/p>\n<p style=\"text-align: center\">[latex]\\dot{Q} = \\Delta\\dot{H}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\dot{Q} = -126 MW[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise: Heat of Formation<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<div class=\"inner_cell\">\n<div class=\"text_cell_render border-box-sizing rendered_html\">\n<p>Consider a reactor where the following reaction reaction takes place :<\/p>\n<p style=\"text-align: center\">[latex]3 NO_{2}(g) + H_{2}O(l) \u2192 2HNO_{3}(aq) + NO(g)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ} = ?[\/latex]<\/p>\n<p>The reactor uses an 0.8 kW impeller (mixer for liquids) that mixes the reactor contents. The reaction occurs at standard temperature and pressure. A feed consisting of 100 mol\/s and 300 mol\/s of water and [latex]NO_{2}[\/latex], respectively, enters the reactor. The reaction goes to completion and the products stream consists of 200 mol\/s of [latex]HNO_{3}[\/latex] and 100 mol\/s of [latex]NO[\/latex].<\/p>\n<p style=\"text-align: left\">The following heats of formation are available:<\/p>\n<table class=\"grid aligncenter\" style=\"height: 174px;width: 667px; width: 864px;\">\n<thead>\n<tr>\n<th style=\"text-align: center;width: 56.1364px\">Number<\/th>\n<th style=\"text-align: center;width: 342.682px\">Reaction<\/th>\n<th style=\"text-align: center;width: 226.318px\">[latex]\\Delta H^{\\circ}_{f,i}[\/latex] (kJ\/mol)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"text-align: center;width: 56.1364px\">1<\/td>\n<td style=\"text-align: center;width: 342.682px\">[latex]\\frac{1}{2}N_{2}(g) + O_{2}(g) \u2192 NO_{2}(g)[\/latex]<\/td>\n<td style=\"text-align: center;width: 226.318px\">33.2<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;width: 56.1364px\">2<\/td>\n<td style=\"text-align: center;width: 342.682px\">[latex]H_{2}(g) + \\frac{1}{2}O_{2}(g) \u2192 H_{2}O(l)[\/latex]<\/td>\n<td style=\"text-align: center;width: 226.318px\">-285.8<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;width: 56.1364px\">3<\/td>\n<td style=\"text-align: center;width: 342.682px\">[latex]\\frac{1}{2}H_{2}(g) + \\frac{1}{2}N_{2}(g) + \\frac{3}{2}O_{2}(g) \u2192 HNO_{3}O(aq)[\/latex]<\/td>\n<td style=\"text-align: center;width: 226.318px\">-207.4<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;width: 56.1364px\">4<\/td>\n<td style=\"text-align: center;width: 342.682px\">[latex]\\frac{1}{2}N_{2}(g) + \\frac{1}{2}O_{2}(g) \u2192 NO(g)[\/latex]<\/td>\n<td style=\"text-align: center;width: 226.318px\">90.2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>How much heat must be removed from the reactor in one day for it to remain at standard temperature?<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"prompt input_prompt\">\n<div class=\"textbox\">\n<h3 class=\"prompt input_prompt\">Solution<\/h3>\n<div>\n<p><strong>Step 1:<\/strong> Determine what combination of the given reactions will give the desired reaction (Hess&#8217;s Law)<\/p>\n<p>For this desired reaction:<br \/>\n\\begin{align*}<br \/>\n&amp; -3*Reaction1\\\\<br \/>\n&amp; -1*Reaction2\\\\<br \/>\n&amp; +2*Reaction3\\\\<br \/>\n&amp; +1*Reaction4<br \/>\n\\end{align*}<\/p>\n<p>Summing all the reactions above yields:<\/p>\n<p style=\"text-align: center\">[latex]3 NO_{2}(g) + H_{2}O(l) \u2192 2HNO_{3}(aq) + NO(g)[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Step 2<\/strong>: Add all the individual reaction enthalpies:<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ}_{r} = -3*\\Delta H^{\\circ}_{1} + -1*\\Delta H^{\\circ}_{2} + 2*\\Delta H^{\\circ}_{3} + \\Delta H^{\\circ}_{4}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ}_{r} = -3*33.2 kJ\/mol + -1*-285.8 kJ\/mol + 2*-207.4 kJ\/mol + 90.2 kJ\/mol[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta H^{\\circ}_{r}= (-99.6+285.8+-414.8+90.2)kJ\/mol = -138.4 kJ\/mol[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Step 3:<\/strong> Calculate the overall change in enthalpy using the extent of the reaction:<\/p>\n<p>\\begin{align*}<br \/>\n\\xi &amp;= \\frac{|(\\dot{n}_{i})_{out}-(\\dot{n}_{i})_{in}|}{|(\\dot{\\nu}_{i})|}hat\\\\&amp; = \\frac{|(\\dot{n}_{H_{2}O})_{out}-(\\dot{n}_{H_{2}O})_{in}|}{|(\\dot{\\nu}_{H_{2}O})|} \\\\&amp; =\\frac{|0\\frac{mol}{s} &#8211; 100 \\frac{mol}{s}|}{|-1|}\\\\ &amp;= 100 mol\/s<br \/>\n\\end{align*}<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = \\xi*\\Delta H^{\\circ}_{r}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = 100\\frac{mol}{s}*-138.4\\frac{kJ}{mol} = -138400 \\frac{kJ}{s}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Step 4:<\/strong> Calculate how much work is done on the reactor by the impeller in one day:<\/p>\n<p style=\"text-align: center\">[latex]W_{s} = 0.8\\frac{kJ}{s}*3600\\frac{s}{hr}*24hr[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]W_{s} = 69120 kJ[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Step 5<\/strong>: Perform an energy balance on the reactor<\/p>\n<p style=\"text-align: center\">[latex]\\dot{Q}=\\Delta H^{\\circ}-W_{s}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\dot{Q} = (-138400 - 69120 ) kJ = -207520 kJ[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<h2 id=\"Summary-of-Heat-of-Reaction-and-Formation-Methods\" style=\"text-align: left\">Summary of Heat of Reaction and Formation Methods<\/h2>\n<h3 id=\"Heat-of-Reaction-Method\" style=\"text-align: left\">Heat of Reaction Method<\/h3>\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = \\xi\\Delta\\dot{H}^{\\circ}_{r} + \\Sigma\\dot{n}_{out}*\\hat{H}_{out} - \\Sigma\\dot{n}_{in}*\\hat{H}_{in}[\/latex]<\/p>\n<p style=\"text-align: center\">where [latex]\\hat{H}_{i} = \\int^{T}_{T_{ref}}C_{P}dT[\/latex] assuming there are no phase changes<\/p>\n<blockquote>\n<p style=\"text-align: left\">If a phase change occurs, an additional heat term (i.e. heat of vapourization) will be added to the [latex]\\hat{H}_{i}[\/latex] term in both methods<\/p>\n<\/blockquote>\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 93.2663%;height: 91px\">\n<tbody>\n<tr>\n<td style=\"width: 100%;text-align: center\"><span style=\"font-size: 16px\">[latex]\\Delta\\dot{H} = \\xi\\Delta\\dot{H}_{r} + \\Sigma\\dot{n}_{out}*\\int^{T_{out}}_{T_{ref}}C_{P}dT - \\Sigma\\dot{n}_{in}*\\int^{T_{in}}_{T_{ref}}C_{P}dT[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h3 id=\"Heat-of-Formation-Method\">Heat of Formation Method<\/h3>\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = \\Sigma\\dot{n}_{out}*\\hat{H}_{out} - \\Sigma\\dot{n}_{in}*\\hat{H}_{in}[\/latex]<\/p>\n<p>where [latex]\\hat{H}_{i} = \\hat{H}^{\\circ}_{f,i} + \\int^{T}_{T_{ref}}C_{P}dT[\/latex] assuming there are no phase changes<\/p>\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = \\Sigma\\dot{n}_{out}*(\\hat{H}^{\\circ}_{f,i} + \\int^{T_{out}}_{T_{ref}}C_{P}dT) - \\Sigma\\dot{n}_{in}*(\\hat{H}^{\\circ}_{f,i} + \\int^{T_{in}}_{T_{ref}}C_{P}dT)[\/latex]<\/p>\n<p style=\"text-align: left\">For the same system, equating [latex]\\Delta\\dot{H}[\/latex] for both methods, we can see that the [latex]C_{P}[\/latex] terms cancel. We obtain the following:<\/p>\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 88.7437%;height: 43px\">\n<tbody>\n<tr>\n<td style=\"width: 100%;text-align: center\"><span style=\"font-size: 16px\">[latex]\\xi\\Delta\\dot{H}^{\\circ}_{r} = \\Sigma\\dot{n}_{out}*\\hat{H}^{\\circ}_{f,i} - \\Sigma\\dot{n}_{in}*\\hat{H}^{\\circ}_{f,i}[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Overall, the final [latex]\\Delta\\dot{H}[\/latex] for both methods will be the <strong>same<\/strong>. This means we can choose either method based on which is easiest to use with the information we have available.<\/p>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise: Comparing Both Methods<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Ethane and oxygen are fed into a furnace at 100 kmol\/hr and 500 kmol\/hr, respectively. Ethane goes through complete combustion in the furnace. Oxygen, carbon dioxide, and water exit the furnace at 150 kmol\/hr, 200 kmol\/hr, and 300 kmol\/hr, respectively.<\/p>\n<p style=\"text-align: center\">[latex]C_{2}H_{6}(g) + \\frac{7}{2}O_{2}(g) \u2192 2CO_{2}(g) + 2H_{2}O(l)[\/latex]<\/p>\n<p>where the heat of reaction is: [latex]\\Delta H^{\\circ}_{r} = -1560.7 kJ\/mol[\/latex]<\/p>\n<p style=\"text-align: left\">The following data on heats of formations is available:<\/p>\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 65.8848%;height: 75px\">\n<tbody>\n<tr style=\"height: 15px\">\n<td style=\"width: 50%;height: 15px;text-align: center\"><strong>Compound<\/strong><\/td>\n<td style=\"width: 50%;height: 15px;text-align: center\"><strong>[latex]\\Delta H^{\\circ}_{f,i}[\/latex] (kJ\/mol)<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 50%;height: 15px;text-align: center\">Ethane<\/td>\n<td style=\"width: 50%;height: 15px;text-align: center\">-83.82<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 50%;height: 15px;text-align: center\">Oxygen<\/td>\n<td style=\"width: 50%;height: 15px;text-align: center\">0<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 50%;height: 15px;text-align: center\">Carbon Dioxide<\/td>\n<td style=\"width: 50%;height: 15px;text-align: center\">-393.51<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 50%;height: 15px;text-align: center\">Water (liquid)<\/td>\n<td style=\"width: 50%;height: 15px;text-align: center\">-285.83<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Assume the reaction takes place in the furnace at standard temperature and pressure. What is the [latex]\\Delta H^{\\circ}[\/latex] for the desired reaction? Calculate the change in enthalpy using both the heat of reaction and heat of formation method<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Solution<\/h3>\n<p><strong>Heat of Reaction Method<\/strong><\/p>\n<p><strong>Step 1<\/strong>: Determine the extent of the reaction:<\/p>\n<p>\\begin{align*}<br \/>\n\\xi &amp;= \\frac{|(\\dot{n}_{i})_{out}-(\\dot{n}_{i})_{in}|}{|(\\dot{\\nu}_{i})|}hat\\\\&amp; = \\frac{|(\\dot{n}_{C_{2}H_{6}})_{out}-(\\dot{n}_{C_{2}H_{6}})_{in}|}{|(\\dot{\\nu}_{C_{2}H_{6}})|} \\\\&amp; =\\frac{|0\\frac{kmol}{hr} &#8211; 100 \\frac{kmol}{hr}|}{|-1|}\\\\ &amp;= 100 kmol\/hr<br \/>\n\\end{align*}<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Step 2<\/strong>: Multiply the extent of the reaction by the heat of reaction for combusion<\/p>\n<p style=\"text-align: center\">[latex]\\Delta \\dot{H} = \\xi*\\Delta H^{\\circ}_{r}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta \\dot{H} = 100\\frac{kmol}{hr}*-1560.7 \\frac{kJ}{mol} * 1000 \\frac{mol}{kmol}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta \\dot{H} = -156 MJ\/hr[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Heat of Formation Method<\/strong><\/p>\n<p><strong>Step 1<\/strong>: Set up the heat of formation formula<\/p>\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = \\Sigma\\dot{n}_{out}*\\hat{H}_{out} - \\Sigma\\dot{n}_{in}*\\hat{H}_{in}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = (\\dot{n}_{out,ethane}*\\hat{H}_{ethane})+(\\dot{n}_{out,oxygen}*\\hat{H}_{oxygen}) + (\\dot{n}_{out,water}*\\hat{H}_{water}) + (\\dot{n}_{out,carbon dioxide}*\\hat{H}_{carbon dioxide}) - (\\dot{n}_{in,ethane}*\\hat{H}_{ethane})+(\\dot{n}_{in,oxygen}*\\hat{H}_{oxygen}) + (\\dot{n}_{in,water}*\\hat{H}_{water}) + (\\dot{n}_{in,carbon dioxide}*\\hat{H}_{carbon dioxide})[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: left\"><strong>Step 2<\/strong>: Plug in the enthalpies of formation and calculate the change in enthalpy<\/p>\n<table class=\"grid aligncenter\">\n<thead>\n<tr>\n<th style=\"text-align: center\">Compound<\/th>\n<th style=\"text-align: center\">[latex]n_{in}[\/latex] (kmol\/hr)<\/th>\n<th style=\"text-align: center\">[latex]n_{out}[\/latex] (kmol\/hr)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"text-align: center\">Ethane<\/td>\n<td style=\"text-align: center\">100<\/td>\n<td style=\"text-align: center\">0<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\">Oxygen<\/td>\n<td style=\"text-align: center\">500<\/td>\n<td style=\"text-align: center\">150<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\">Carbon Dioxide<\/td>\n<td style=\"text-align: center\">0<\/td>\n<td style=\"text-align: center\">200<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center\">Water<\/td>\n<td style=\"text-align: center\">0<\/td>\n<td style=\"text-align: center\">300<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = ((0\\frac{kmol}{hr}*-83.82\\frac{kJ}{mol})+(150\\frac{kmol}{hr}*0\\frac{kJ}{mol}) + (200\\frac{kmol}{hr}*-393.51\\frac{kJ}{mol}) + (300\\frac{kmol}{hr}*-285.83\\frac{kJ}{mol}) - (100\\frac{kmol}{hr}*-83.82\\frac{kJ}{mol}) - (500\\frac{kmol}{hr}*0\\frac{kJ}{mol}) - (0\\frac{kmol}{hr}*-393.51\\frac{kJ}{mol}) - (0\\frac{kmol}{hr}*-285.83\\frac{kJ}{mol}))*1000\\frac{mol}{kmol}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Delta\\dot{H} = -150 MJ[\/latex]<\/p>\n<p><strong>Notice that the methods aren&#8217;t exactly equal. This is due to slight differences in available enthalpy data and assumptions made regarding temperature<\/strong><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n","protected":false},"author":949,"menu_order":16,"comment_status":"closed","ping_status":"closed","template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1328","chapter","type-chapter","status-publish","hentry"],"part":1313,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapters\/1328","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/users\/949"}],"replies":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/comments?post=1328"}],"version-history":[{"count":10,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapters\/1328\/revisions"}],"predecessor-version":[{"id":2810,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapters\/1328\/revisions\/2810"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/parts\/1313"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapters\/1328\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/media?parent=1328"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/pressbooks\/v2\/chapter-type?post=1328"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/contributor?post=1328"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chbe220\/wp-json\/wp\/v2\/license?post=1328"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}