{"id":242,"date":"2021-07-23T09:19:23","date_gmt":"2021-07-23T13:19:23","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/enthalpy\/"},"modified":"2022-06-23T08:57:45","modified_gmt":"2022-06-23T12:57:45","slug":"enthalpy","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/enthalpy\/","title":{"raw":"5.3 Enthalpy","rendered":"5.3 Enthalpy"},"content":{"raw":"<strong><span style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em;background-color: #cbd4b6;color: #000000\">Learning Objectives<\/span><\/strong>\r\n<div class=\"textbox textbox--learning-objectives\">\r\n\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>State the first law of thermodynamics<\/li>\r\n \t<li>Define enthalpy and explain its classification as a state function<\/li>\r\n \t<li>Write and balance thermochemical equations<\/li>\r\n \t<li>Calculate enthalpy changes for various chemical reactions<\/li>\r\n \t<li>Explain Hess\u2019s law and use it to compute reaction enthalpies<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idp160276224\">Thermochemistry is a branch of <strong>chemical thermodynamics<\/strong>, the science that deals with the relationships between heat, work, and other forms of energy in the context of chemical and physical processes. As we concentrate on thermochemistry in this chapter, we need to consider some widely used concepts of thermodynamics.<\/p>\r\n<p id=\"fs-idp5509104\">Substances act as reservoirs of energy, meaning that energy can be added to them or removed from them. Energy is stored in a substance when the kinetic or potential energy of its atoms or molecules is raised.\u00a0 The total of all possible kinds of energy present in a substance is called the <strong>internal energy (<em data-effect=\"italics\">U<\/em>)<\/strong>, sometimes symbolized as <em data-effect=\"italics\">E<\/em>.<\/p>\r\n<p id=\"fs-idm1690784\">As a system undergoes a change, its internal energy can change, and energy can be transferred from the system to the surroundings, or from the surroundings to the system. Energy is transferred into a system when it absorbs heat (<em data-effect=\"italics\">q<\/em>) from the surroundings or when the surroundings do work (<em data-effect=\"italics\">w<\/em>) on the system. For example, energy is transferred into room-temperature metal wire if it is immersed in hot water (the wire absorbs heat from the water), or if you rapidly bend the wire back and forth (the wire becomes warmer because of the work done on it). Both processes increase the internal energy of the wire. Conversely, energy is transferred out of a system when heat is lost from the system, or when the system does work on the surroundings.<\/p>\r\n<p id=\"fs-idp16545808\">The relationship between internal energy, heat, and work can be represented by the equation:<\/p>\r\n\r\n<div id=\"fs-idm150109936\" style=\"text-align: center\" data-type=\"equation\">\u0394<em>U<\/em> = <em>q<\/em> + <em>w<\/em><\/div>\r\n<p id=\"fs-idp146392048\">as shown in <a class=\"autogenerated-content\" href=\"#CNX_Chem_05_03_Systemqw\">(Figure)<\/a>. This is one version of the <span data-type=\"term\">first law of thermodynamics<\/span>, and it shows that the internal energy of a system changes through heat flow into or out of the system (positive <em data-effect=\"italics\">q<\/em> is heat flow in; negative <em data-effect=\"italics\">q<\/em> is heat flow out) or work done on or by the system. The work, <em data-effect=\"italics\">w<\/em>, is positive if it is done on the system and negative if it is done by the system.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_05_03_Systemqw\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">The internal energy, <em data-effect=\"italics\">U<\/em>, of a system can be changed by heat flow and work. If heat flows into the system, <em data-effect=\"italics\">q<\/em><sub>in<\/sub>, or work is done on the system, <em data-effect=\"italics\">w<\/em><sub>on<\/sub>, its internal energy increases, \u0394<em data-effect=\"italics\">U<\/em> &gt; 0. If heat flows out of the system, <em data-effect=\"italics\">q<\/em><sub>out<\/sub>, or work is done by the system, <em data-effect=\"italics\">w<\/em><sub>by<\/sub>, its internal energy decreases, \u0394<em data-effect=\"italics\">U<\/em> &lt; 0.<span id=\"fs-idp1427264\" data-type=\"media\">\u00a0<img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_05_03_Systemqw-1.jpg\" data-media-type=\"image\/jpeg\" \/>\u00a0<\/span><\/div>\r\n<\/div>\r\n<p id=\"fs-idp6896176\">A type of work called <span data-type=\"term\">expansion work<\/span> (or pressure-volume work) occurs when a system pushes back the surroundings against a restraining pressure, or when the surroundings compress the system. An example of this occurs during the operation of an internal combustion engine. The reaction of gasoline and oxygen is exothermic. Some of this energy is given off as heat, and some does work pushing the piston in the cylinder. The substances involved in the reaction are the system, and the engine and the rest of the universe are the surroundings. The system loses energy by both heating and doing work on the surroundings, and its internal energy decreases. (The engine is able to keep the car moving because this process is repeated many times per second while the engine is running.) We will consider how to determine the amount of work involved in a chemical or physical change in the chapter on thermodynamics.<\/p>\r\n\r\n<div id=\"fs-idp889984\" class=\"chemistry link-to-learning\" data-type=\"note\">\r\n<p id=\"fs-idp13166816\">This view of <a href=\"http:\/\/openstaxcollege.org\/l\/16combustion\">an internal combustion engine<\/a> illustrates the conversion of energy produced by the exothermic combustion reaction of a fuel such as gasoline into energy of motion.<\/p>\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-idp120420048\">As discussed, the relationship between internal energy, heat, and work can be represented as \u0394<em data-effect=\"italics\">U<\/em> = <em data-effect=\"italics\">q<\/em> + <em data-effect=\"italics\">w<\/em>. Internal energy is an example of a<strong> state function <\/strong>(or state variable), whereas heat and work are not state functions. The value of a state function depends only on the state that a system is in, and not on how that state is reached. If a quantity is not a state function, then its value <em data-effect=\"italics\">does<\/em> depend on how the state is reached. An example of a state function is altitude or elevation. If you stand on the summit of Mt. Kilimanjaro, you are at an altitude of 5895 m, and it does not matter whether you hiked there or parachuted there. The distance you traveled to the top of Kilimanjaro, however, is not a state function. You could climb to the summit by a direct route or by a more roundabout, circuitous path (<a class=\"autogenerated-content\" href=\"#CNX_Chem_05_03_Summit\">(Figure)<\/a>). The distances traveled would differ (distance is not a state function) but the elevation reached would be the same (altitude is a state function).<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_05_03_Summit\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">Paths X and Y represent two different routes to the summit of Mt. Kilimanjaro. Both have the same change in elevation (altitude or elevation on a mountain is a state function; it does not depend on path), but they have very different distances traveled (distance walked is not a state function; it depends on the path). (credit: modification of work by Paul Shaffner)<\/div>\r\n<span id=\"fs-idp164506448\" data-type=\"media\" data-alt=\"An aerial photo depicts a view of Mount Kilimanjaro. A straight, green arrow labeled X is drawn from the term \u201cbase,\u201d written at the bottom of the mountain, to the term \u201cSummit,\u201d written at the top of the mountain. Another arrow labeled Y is draw from the base to the summit alongside the green arrow, but this arrow is pink and has three large S-shaped curves along its length.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_05_03_Summit-1.jpg\" alt=\"An aerial photo depicts a view of Mount Kilimanjaro. A straight, green arrow labeled X is drawn from the term \u201cbase,\u201d written at the bottom of the mountain, to the term \u201cSummit,\u201d written at the top of the mountain. Another arrow labeled Y is draw from the base to the summit alongside the green arrow, but this arrow is pink and has three large S-shaped curves along its length.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idp107461968\">Chemists ordinarily use a property known as <span data-type=\"term\">enthalpy (<em data-effect=\"italics\">H<\/em>)<\/span> to describe the thermodynamics of chemical and physical processes. Enthalpy is defined as the sum of a system\u2019s internal energy (<em data-effect=\"italics\">U<\/em>) and the mathematical product of its pressure (<em data-effect=\"italics\">P<\/em>) and volume (<em data-effect=\"italics\">V<\/em>):<\/p>\r\n<p style=\"text-align: center\"><em>H<\/em> = <em>U<\/em> + <em>PV<\/em><\/p>\r\n<p id=\"fs-idm20812336\">Enthalpy is also a state function. Enthalpy values for specific substances cannot be measured directly; only enthalpy <em data-effect=\"italics\">changes<\/em> for chemical or physical processes can be determined. For processes that take place at constant pressure (a common condition for many chemical and physical changes), the <span data-type=\"term\">enthalpy change (\u0394<em data-effect=\"italics\">H<\/em>)<\/span> is:<\/p>\r\n\r\n<div id=\"fs-idm182172016\" style=\"text-align: center\" data-type=\"equation\">\u0394<em>H <\/em>= \u0394<em>U <\/em>+ <em>P<\/em>\u0394<em>V\u00a0 \u00a0 \u00a0<\/em>(at constant <em>P<\/em>)<\/div>\r\n<p id=\"fs-idm15694224\">The mathematical product <em data-effect=\"italics\">P<\/em>\u0394<em data-effect=\"italics\">V<\/em> represents work (<em data-effect=\"italics\">w<\/em>), namely, expansion or pressure-volume work as noted. By their definitions, the arithmetic signs of \u0394<em data-effect=\"italics\">V<\/em> and <em data-effect=\"italics\">w<\/em> will always be opposite:<\/p>\r\n\r\n<div id=\"fs-idm223126448\" style=\"text-align: center\" data-type=\"equation\"><em>P<\/em>\u0394<em>V<\/em> = \u2212<em>w<\/em><\/div>\r\n<p id=\"fs-idp7052304\">Substituting this equation and the definition of internal energy into the enthalpy-change equation yields:<\/p>\r\n\r\n<div id=\"fs-idm223599728\" style=\"text-align: center\" data-type=\"equation\">\u0394<em>H<\/em> = \u0394<em>U<\/em> + <em>P<\/em>\u0394<em>V<\/em> = <em>q<\/em><sub>p<\/sub> + <em>w<\/em> - <em>w <\/em>= <em>q<\/em><sub>p<\/sub><\/div>\r\n<p id=\"fs-idp58958304\">where <em data-effect=\"italics\">q<sub>p<\/sub><\/em> is the heat of reaction under conditions of constant pressure.<\/p>\r\n<p id=\"fs-idm50104768\">And so, if a chemical or physical process is carried out at constant pressure with the only work done caused by expansion or contraction, then the heat flow (<em data-effect=\"italics\">q<sub>p<\/sub><\/em>) and enthalpy change (\u0394<em data-effect=\"italics\">H<\/em>) for the process are equal.<\/p>\r\n<p id=\"fs-idm22504976\">The heat given off when you operate a Bunsen burner is equal to the enthalpy change of the methane combustion reaction that takes place, since it occurs at the essentially constant pressure of the atmosphere. On the other hand, the heat produced by a reaction measured in a bomb calorimeter is not equal to \u0394<em data-effect=\"italics\">H<\/em> because the closed, constant-volume metal container prevents the pressure from remaining constant (it may increase or decrease if the reaction yields increased or decreased amounts of gaseous species). Chemists usually perform experiments under normal atmospheric conditions, at constant external pressure with <em data-effect=\"italics\">q<\/em> = \u0394<em data-effect=\"italics\">H<\/em>, which makes enthalpy the most convenient choice for determining heat changes for chemical reactions.<\/p>\r\n<p id=\"fs-idp25166224\">The following conventions apply when using \u0394<em data-effect=\"italics\">H<\/em>:<\/p>\r\n\r\n<ul id=\"fs-idm49861360\" data-bullet-style=\"bullet\">\r\n \t<li>\r\n<p id=\"fs-idm11576944\">A negative value of an enthalpy change, \u0394<em data-effect=\"italics\">H<\/em> &lt; 0, indicates an exothermic reaction; a positive value, \u0394<em data-effect=\"italics\">H<\/em> &gt; 0, indicates an endothermic reaction. If the direction of a chemical equation is reversed, the arithmetic sign of its \u0394<em data-effect=\"italics\">H<\/em> is changed (a process that is endothermic in one direction is exothermic in the opposite direction).<\/p>\r\n<\/li>\r\n \t<li>\r\n<p id=\"fs-idp165042832\">Chemists use a <strong>thermochemical equation<\/strong> to represent the changes in both matter and energy. In a thermochemical equation, the enthalpy change of a reaction is shown as a \u0394<em data-effect=\"italics\">H<\/em> value following the equation for the reaction. This \u0394<em data-effect=\"italics\">H<\/em> value indicates the amount of heat associated with the reaction involving the number of moles of reactants and products <em data-effect=\"italics\">as shown in the chemical equation<\/em>. For example, consider this equation:<\/p>\r\n\r\n<div id=\"fs-idp13211824\" data-type=\"equation\"><img class=\"alignnone size-medium wp-image-1232 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3a-300x42.png\" alt=\"\" width=\"300\" height=\"42\" \/><\/div>\r\n<p id=\"fs-idm35882976\">This equation indicates that when 1 mole of hydrogen gas and \u00bd mole of oxygen gas at some temperature and pressure change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released to the surroundings. If the coefficients of the chemical equation are multiplied by some factor, the enthalpy change must be multiplied by that same factor (\u0394<em data-effect=\"italics\">H<\/em> is an extensive property):<\/p>\r\n\r\n<div id=\"fs-idm23265328\" style=\"text-align: center\" data-type=\"equation\"><img class=\"alignnone wp-image-1233\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3b-300x59.png\" alt=\"\" width=\"462\" height=\"91\" \/><\/div><\/li>\r\n \t<li>\r\n<p id=\"fs-idm72030880\">The enthalpy change of a reaction depends on the physical states of the reactants and products, so these must be shown. For example, when 1 mole of hydrogen gas and \u00bd mole of oxygen gas change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released. If gaseous water forms, only 242 kJ of heat are released.<\/p>\r\n\r\n<div id=\"fs-idp4587936\" data-type=\"equation\"><img class=\"alignnone wp-image-1234 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3c-300x37.png\" alt=\"\" width=\"349\" height=\"43\" \/><\/div><\/li>\r\n<\/ul>\r\n<div id=\"fs-idm9240992\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp52228448\"><strong>Writing Thermochemical Equations:<\/strong><\/p>\r\nWhen 0.0500 mol of HCl(<em data-effect=\"italics\">aq<\/em>) reacts with 0.0500 mol of NaOH(<em data-effect=\"italics\">aq<\/em>) to form 0.0500 mol of NaCl(<em data-effect=\"italics\">aq<\/em>), 2.9 kJ of heat are produced. Write a balanced thermochemical equation for the reaction of one mole of HCl.?\r\n<div id=\"fs-idp149553440\" data-type=\"equation\"><img class=\"alignnone wp-image-1235 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3d-300x35.png\" alt=\"\" width=\"326\" height=\"38\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm10084896\"><strong>Solution:<\/strong><\/p>\r\nFor the reaction of 0.0500 mol acid (HCl), <em data-effect=\"italics\">q<\/em> = \u22122.9 kJ. The reactants are provided in stoichiometric amounts (same molar ratio as in the balanced equation), and so the amount of acid may be used to calculate a molar enthalpy change. Since \u0394<em data-effect=\"italics\">H<\/em> is an extensive property, it is proportional to the amount of acid neutralized:\r\n<div id=\"fs-idp125057232\" data-type=\"equation\"><img class=\"alignnone wp-image-1236 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3e-300x42.png\" alt=\"\" width=\"314\" height=\"44\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp13720784\">The thermochemical equation is then<\/p>\r\n\r\n<div id=\"fs-idm13850464\" data-type=\"equation\"><img class=\"wp-image-1237 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3f-300x25.png\" alt=\"\" width=\"468\" height=\"39\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp6958160\"><strong>Check Your Learning:<\/strong><\/p>\r\nWhen 1.34 g Zn(<em data-effect=\"italics\">s<\/em>) reacts with 60.0 mL of 0.750 M HCl(<em data-effect=\"italics\">aq<\/em>), 3.14 kJ of heat are produced. Determine the enthalpy change per mole of zinc reacting for the reaction:\r\n<div id=\"fs-idp6791056\" data-type=\"equation\"><img class=\"alignnone size-medium wp-image-1238 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3g-300x39.png\" alt=\"\" width=\"300\" height=\"39\" \/><\/div>\r\n<div id=\"fs-idp58943984\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp35719536\">\u0394<em data-effect=\"italics\">H<\/em> = \u2212153 kJ<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idm14305808\">Be sure to take both stoichiometry and limiting reactants into account when determining the \u0394<em data-effect=\"italics\">H<\/em> for a chemical reaction.<\/p>\r\n\r\n<div id=\"fs-idp107963312\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm6526496\"><strong>Writing Thermochemical Equations:<\/strong><\/p>\r\nA gummy bear contains 2.67 g sucrose, C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub>. When it reacts with 7.19 g potassium chlorate, KClO<sub>3<\/sub>, 43.7 kJ of heat are produced. Write a thermochemical equation for the reaction of one mole of sucrose:\r\n<div id=\"fs-idp12841552\" data-type=\"equation\"><img class=\"alignnone wp-image-1240 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3h-300x23.png\" alt=\"\" width=\"443\" height=\"34\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm42247408\"><strong>Solution:<\/strong><\/p>\r\nUnlike the previous example exercise, this one does not involve the reaction of stoichiometric amounts of reactants, and so the <em data-effect=\"italics\">limiting reactant<\/em> must be identified (it limits the yield of the reaction and the amount of thermal energy produced or consumed).\r\n<p id=\"fs-idm378999344\">The provided amounts of the two reactants are<\/p>\r\n\r\n<div id=\"fs-idm372152464\" data-type=\"equation\"><img class=\"alignnone wp-image-1242 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3i-300x41.png\" alt=\"\" width=\"359\" height=\"49\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm329792512\">The provided molar ratio of perchlorate-to-sucrose is then<\/p>\r\n\r\n<div id=\"fs-idm378911456\" data-type=\"equation\"><img class=\"alignnone wp-image-1241 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3j-300x22.png\" alt=\"\" width=\"355\" height=\"26\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm379230752\">The balanced equation indicates 8 mol KClO<sub>3<\/sub> are required for reaction with 1 mol C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub>. Since the provided amount of KClO<sub>3<\/sub> is less than the stoichiometric amount, it is the limiting reactant and may be used to compute the enthalpy change:<\/p>\r\n\r\n<div id=\"fs-idm352441488\" data-type=\"equation\"><img class=\"wp-image-1244 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3k-300x19.png\" alt=\"\" width=\"442\" height=\"28\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm386464528\">Because the equation, as written, represents the reaction of 8 mol KClO<sub>3<\/sub>, the enthalpy change is<\/p>\r\n\r\n<div id=\"fs-idm389610336\" data-type=\"equation\"><img class=\" wp-image-1247 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3l-300x29.png\" alt=\"\" width=\"341\" height=\"33\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm379229776\">The enthalpy change for this reaction is \u22125960 kJ, and the thermochemical equation is:<\/p>\r\n\r\n<div id=\"fs-idp90097408\" data-type=\"equation\"><img class=\" wp-image-1248 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3m-300x18.png\" alt=\"\" width=\"499\" height=\"30\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp82549456\"><strong>Check Your Learning:<\/strong><\/p>\r\nWhen 1.42 g of iron reacts with 1.80 g of chlorine, 3.22 g of FeCl<sub>2<\/sub>(<em data-effect=\"italics\">s<\/em>) and 8.60 kJ of heat is produced. What is the enthalpy change for the reaction when 1 mole of FeCl<sub>2<\/sub>(<em data-effect=\"italics\">s<\/em>) is produced?\r\n\r\n&nbsp;\r\n<div id=\"fs-idp120447376\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm17649024\">\u0394<em data-effect=\"italics\">H<\/em> = \u2212338 kJ<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idm18196832\">Enthalpy changes are typically tabulated for reactions in which both the reactants and products are at the same conditions. A <strong>standard state<\/strong> is a commonly accepted set of conditions used as a reference point for the determination of properties under other different conditions. For chemists, the IUPAC standard state refers to materials under a pressure of 1 bar and solutions at 1 M, and does not specify a temperature. Many thermochemical tables list values with a standard state of 1 atm. Because the \u0394<em data-effect=\"italics\">H<\/em> of a reaction changes very little with such small changes in pressure (1 bar = 0.987 atm), \u0394<em data-effect=\"italics\">H<\/em> values (except for the most precisely measured values) are essentially the same under both sets of standard conditions. We will include a superscripted \u201co\u201d in the enthalpy change symbol to designate standard state. Since the usual (but not technically standard) temperature is 298.15 K, this temperature will be assumed unless some other temperature is specified. Thus, the symbol \u0394<em>H<\/em>\u00b0 is used to indicate an enthalpy change for a process occurring under these conditions. (The symbol \u0394<em data-effect=\"italics\">H<\/em> is used to indicate an enthalpy change for a reaction occurring under nonstandard conditions.)<\/p>\r\n<p id=\"fs-idp148022240\">The enthalpy changes for many types of chemical and physical processes are available in the reference literature, including those for combustion reactions, phase transitions, and formation reactions. As we discuss these quantities, it is important to pay attention to the <em data-effect=\"italics\">extensive<\/em> nature of enthalpy and enthalpy changes. Since the enthalpy change for a given reaction is proportional to the amounts of substances involved, it may be reported on that basis (i.e., as the \u0394<em data-effect=\"italics\">H<\/em> for specific amounts of reactants). However, we often find it more useful to divide one extensive property (\u0394<em data-effect=\"italics\">H<\/em>) by another (amount of substance), and report a per-amount <em data-effect=\"italics\">intensive<\/em> value of \u0394<em data-effect=\"italics\">H<\/em>, often \u201cnormalized\u201d to a per-mole basis. (Note that this is similar to determining the intensive property specific heat from the extensive property heat capacity, as seen previously.)<\/p>\r\n\r\n<div id=\"fs-idp4580336\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Standard Enthalpy of Combustion<\/strong><\/h3>\r\n<p id=\"fs-idp8281184\"><span data-type=\"term\">Standard enthalpy of combustion <\/span>(\u0394<em>H<\/em><sub>c<\/sub>\u00b0) is the enthalpy change when 1 mole of a substance burns (combines vigorously with oxygen) under standard state conditions; it is sometimes called \u201cheat of combustion.\u201d For example, the enthalpy of combustion of ethanol, \u22121366.8 kJ\/mol, is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 \u00b0C and 1 atmosphere pressure, yielding products also at 25 \u00b0C and 1 atm.<\/p>\r\n\r\n<div id=\"fs-idm27865968\" data-type=\"equation\"><img class=\"alignnone wp-image-1250 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3n-300x16.png\" alt=\"\" width=\"451\" height=\"24\" \/><\/div>\r\n<p id=\"fs-idm75327968\">Enthalpies of combustion for many substances have been measured; a few of these are listed in <a class=\"autogenerated-content\" href=\"#fs-idp98710048\">(Figure)<\/a>. Many readily available substances with large enthalpies of combustion are used as fuels, including hydrogen, carbon (as coal or charcoal), and <span data-type=\"term\">hydrocarbons<\/span> (compounds containing only hydrogen and carbon), such as methane, propane, and the major components of gasoline.<\/p>\r\n\r\n<table id=\"fs-idp98710048\" class=\"top-titled\" style=\"height: 311px\" summary=\"A data table is shown that has three columns and eleven rows. The header row reads \u201cSubstance,\u201d \u201cCombustion reaction,\u201d and \u201cEnthalpy of Combustion, \u0394 H subscript C superscript degree symbol (k J\/ mol at 25 \u00b0 C).\u201d The first column contains entries reading \u201ccarbon,\u201d \u201chydrogen,\u201d \u201cmagnesium,\u201d \u201csulfur,\u201d \u201ccarbon monoxide,\u201d \u201cmethane,\u201d \u201cacetylene,\u201d \u201cethanol,\u201d \u201cmethanol,\u201d and \u201cisooctane.\u201d The second column contains the equations \u201cC (s) + O (g) right-facing arrow C O subscript 2 (g),\u201d \u201cH subscript 2 (g) + one half O subscript 2 (g) right-facing arrow H subscript 2 O (l),\u201d \u201cM g (s) + one half O subscript 2 (g) right-facing arrow M g O (s),\u201d \u201cS (s) + O subscript 2 (g) right-facing arrow S O subscript 2 (g),\u201d \u201cC O (g) + one half O subscript 2 (g) right-facing arrow C O subscript 2 (g),\u201d \u201cC H subscript 4 (g) + 2 O subscript 2 (g) right-facing arrow C O subscript 2 (g) + 2 H subscript 2 O (g),\u201d \u201cC subscript 2 H subscript 2 (g) + five halves O subscript 2 (g) right-facing arrow 2 C O subscript 2 (g) + H subscript 2 O (l),\u201d \u201cC subscript 2 H subscript 5 O H (l) + 2 O subscript 2 (g) right-facing arrow C O subscript 2 (g) + 3 H subscript 2 O (l),\u201d \u201cC H subscript 3 O H (l) + three halves O subscript 2 (g) right-facing arrow C O subscript 2 (g) + 2 H subscript 2 O (l),\u201d and \u201cC subscript 8 H subscript 18 (l) + twenty five halves O subscript 2 (g) right-facing arrow 8 C O subscript 2 (g) + 9 H subscript 2 O (l).\u201d The final column contains the values \u201c\u2013393.5,\u201d \u201c\u2013285.8,\u201d \u201c\u2013601.6,\u201d \u201c\u2013296.8,\u201d \u201c\u2013283.0,\u201d \u201c\u2013890.8,\u201d \u201c\u20131301.1,\u201d \u201c\u20131366.8,\u201d \u201c\u2013726.1,\u201d and \u201c\u20135461.\u201d\">\r\n<thead>\r\n<tr style=\"height: 15px\">\r\n<th style=\"height: 15px;width: 541.85px\" colspan=\"3\" data-align=\"center\">Standard Molar Enthalpies of Combustion<\/th>\r\n<\/tr>\r\n<tr style=\"height: 93px\" valign=\"middle\">\r\n<th style=\"height: 93px;width: 76.5px\" data-align=\"left\">Substance<\/th>\r\n<th style=\"height: 93px;width: 272.7px\" data-align=\"left\">Combustion Reaction<\/th>\r\n<th style=\"height: 93px;width: 164.75px\" data-align=\"left\">Enthalpy of Combustion, <img class=\"alignnone wp-image-1251\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3o.png\" alt=\"\" width=\"126\" height=\"26\" \/><\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 30px\" valign=\"middle\">\r\n<td style=\"height: 30px;width: 77px\" data-align=\"left\">carbon<\/td>\r\n<td style=\"height: 30px;width: 273.7px\" data-align=\"left\">C(<em>s<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u27f6 CO<sub>2<\/sub>(<em>g<\/em>)<\/td>\r\n<td style=\"height: 30px;width: 165.25px\" data-align=\"left\">\u2212393.5<\/td>\r\n<\/tr>\r\n<tr style=\"height: 16px\" valign=\"middle\">\r\n<td style=\"height: 16px;width: 77px\" data-align=\"left\">hydrogen<\/td>\r\n<td style=\"height: 16px;width: 273.7px\" data-align=\"left\">H<sub>2<\/sub>(<em>g<\/em>) + (1\/2)O<sub>2<\/sub>(<em>g<\/em>) \u27f6 H<sub>2<\/sub>O(<em>l<\/em>)<\/td>\r\n<td style=\"height: 16px;width: 165.25px\" data-align=\"left\">\u2212285.8<\/td>\r\n<\/tr>\r\n<tr style=\"height: 16px\" valign=\"middle\">\r\n<td style=\"height: 16px;width: 77px\" data-align=\"left\">magnesium<\/td>\r\n<td style=\"height: 16px;width: 273.7px\" data-align=\"left\">Mg(<em>s<\/em>) + (1\/2)O<sub>2<\/sub>(<em>g<\/em>) \u27f6 MgO(<em>s<\/em>)<\/td>\r\n<td style=\"height: 16px;width: 165.25px\" data-align=\"left\">\u2212601.6<\/td>\r\n<\/tr>\r\n<tr style=\"height: 16px\" valign=\"middle\">\r\n<td style=\"height: 16px;width: 77px\" data-align=\"left\">sulfur<\/td>\r\n<td style=\"height: 16px;width: 273.7px\" data-align=\"left\">S(<em>s<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u27f6 SO<sub>2<\/sub>(<em>g<\/em>)<\/td>\r\n<td style=\"height: 16px;width: 165.25px\" data-align=\"left\">\u2212296.8<\/td>\r\n<\/tr>\r\n<tr style=\"height: 31px\" valign=\"middle\">\r\n<td style=\"height: 31px;width: 77px\" data-align=\"left\">carbon monoxide<\/td>\r\n<td style=\"height: 31px;width: 273.7px\" data-align=\"left\">CO(<em>g<\/em>) + (1\/2)O<sub>2<\/sub>(<em>g<\/em>) \u27f6 CO<sub>2<\/sub>(<em>g<\/em>)<\/td>\r\n<td style=\"height: 31px;width: 165.25px\" data-align=\"left\">\u2212283.0<\/td>\r\n<\/tr>\r\n<tr style=\"height: 16px\" valign=\"middle\">\r\n<td style=\"height: 16px;width: 77px\" data-align=\"left\">methane<\/td>\r\n<td style=\"height: 16px;width: 273.7px\" data-align=\"left\">CH<sub>4<\/sub>(<em>g<\/em>) + 2O<sub>2<\/sub>(<em>g<\/em>) \u27f6 CO<sub>2<\/sub>(<em>g<\/em>) + 2H<sub>2<\/sub>O(<em>l<\/em>)<\/td>\r\n<td style=\"height: 16px;width: 165.25px\" data-align=\"left\">\u2212890.8<\/td>\r\n<\/tr>\r\n<tr style=\"height: 16px\" valign=\"middle\">\r\n<td style=\"height: 16px;width: 77px\" data-align=\"left\">acetylene<\/td>\r\n<td style=\"height: 16px;width: 273.7px\" data-align=\"left\">C<sub>2<\/sub>H<sub>2<\/sub>(<em>g<\/em>) + (5\/2)O<sub>2<\/sub>(<em>g<\/em>) \u27f6 2CO<sub>2<\/sub>(<em>g<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)<\/td>\r\n<td style=\"height: 16px;width: 165.25px\" data-align=\"left\">\u22121301.1<\/td>\r\n<\/tr>\r\n<tr style=\"height: 16px\" valign=\"middle\">\r\n<td style=\"height: 16px;width: 77px\" data-align=\"left\">ethanol<\/td>\r\n<td style=\"height: 16px;width: 273.7px\" data-align=\"left\">C<sub>2<\/sub>H<sub>5<\/sub>OH(<em>l<\/em>) + 3O<sub>2<\/sub>(<em>g<\/em>)\u27f6 2CO<sub>2<\/sub>(<em>g<\/em>) + 3H<sub>2<\/sub>O(<em>l<\/em>)<\/td>\r\n<td style=\"height: 16px;width: 165.25px\" data-align=\"left\">\u22121366.8<\/td>\r\n<\/tr>\r\n<tr style=\"height: 21px\" valign=\"middle\">\r\n<td style=\"height: 21px;width: 77px\" data-align=\"left\">methanol<\/td>\r\n<td style=\"height: 21px;width: 273.7px\" data-align=\"left\">CH<sub>3<\/sub>OH(<em>l<\/em>) + (3\/2)O<sub>2<\/sub>(<em>g<\/em>)\u27f6 CO<sub>2<\/sub>(<em>g<\/em>) + 2H<sub>2<\/sub>O(<em>l<\/em>)<\/td>\r\n<td style=\"height: 21px;width: 165.25px\" data-align=\"left\">\u2212726.1<\/td>\r\n<\/tr>\r\n<tr style=\"height: 25px\" valign=\"middle\">\r\n<td style=\"height: 25px;width: 77px\" data-align=\"left\">isooctane<\/td>\r\n<td style=\"height: 25px;width: 273.7px\" data-align=\"left\">C<sub>8<\/sub>H<sub>18<\/sub>(<em>l<\/em>) + (25\/2)O<sub>2<\/sub>(<em>g<\/em>)\u27f6 8CO<sub>2<\/sub>(<em>g<\/em>) + 9H<sub>2<\/sub>O(<em>l<\/em>)<\/td>\r\n<td style=\"height: 25px;width: 165.25px\" data-align=\"left\">\u22125460<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-idp90124816\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp186633648\"><strong>Using Enthalpy of Combustion:<\/strong><\/p>\r\nAs <a class=\"autogenerated-content\" href=\"#CNX_Chem_05_03_GasBurning\">(Figure)<\/a> suggests, the combustion of gasoline is a highly exothermic process. Let us determine the approximate amount of heat produced by burning 1.00 L of gasoline, assuming the enthalpy of combustion of gasoline is the same as that of isooctane, a common component of gasoline. The density of isooctane is 0.692 g\/mL.\r\n\r\n&nbsp;\r\n<div id=\"CNX_Chem_05_03_GasBurning\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">The combustion of gasoline is very exothermic. (credit: modification of work by \u201cAlexEagle\u201d\/Flickr)<\/div>\r\n<span id=\"fs-idm3605664\" data-type=\"media\" data-alt=\"A picture shows a large ball of fire burning on a road. A fire truck and fireman are shown in the foreground.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_05_03_GasBurning-1.jpg\" alt=\"A picture shows a large ball of fire burning on a road. A fire truck and fireman are shown in the foreground.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idp177773984\"><strong>Solution:<\/strong><\/p>\r\nStarting with a known amount (1.00 L of isooctane), we can perform conversions between units until we arrive at the desired amount of heat or energy. The enthalpy of combustion of isooctane provides one of the necessary conversions. <a class=\"autogenerated-content\" href=\"#fs-idp98710048\">(Figure)<\/a> gives this value as \u22125460 kJ per 1 mole of isooctane (C<sub>8<\/sub>H<sub>18<\/sub>).\r\n<p id=\"fs-idp6149360\">Using these data,<\/p>\r\n\r\n<div id=\"fs-idp6149744\" data-type=\"equation\"><img class=\"alignnone wp-image-1253 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3p-300x24.png\" alt=\"\" width=\"525\" height=\"42\" \/><\/div>\r\n<p id=\"fs-idp145151008\">The combustion of 1.00 L of isooctane produces 33,100 kJ of heat. (This amount of energy is enough to melt 99.2 kg, or about 218 lb, of ice.)<\/p>\r\n<p id=\"fs-idp94999600\">Note: If you do this calculation one step at a time, you would find:<\/p>\r\n\r\n<div id=\"fs-idp94999984\" data-type=\"equation\"><img class=\"alignnone size-medium wp-image-1254 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3q-300x106.png\" alt=\"\" width=\"300\" height=\"106\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp49733232\"><strong>Check Your Learning:<\/strong><\/p>\r\nHow much heat is produced by the combustion of 125 g of acetylene?\r\n\r\n&nbsp;\r\n<div id=\"fs-idp49733872\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp49734384\">6.25 \u00d7 10<sup>3<\/sup> kJ<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp90130208\" class=\"chemistry everyday-life\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div data-type=\"title\"><strong>Emerging Algae-Based Energy Technologies (Biofuels)<\/strong><\/div>\r\n<p id=\"fs-idp90130976\">As reserves of fossil fuels diminish and become more costly to extract, the search is ongoing for replacement fuel sources for the future. Among the most promising biofuels are those derived from algae (<a class=\"autogenerated-content\" href=\"#CNX_Chem_05_03_AlgalFuel1\">(Figure)<\/a>). The species of algae used are nontoxic, biodegradable, and among the world\u2019s fastest growing organisms. About 50% of algal weight is oil, which can be readily converted into fuel such as biodiesel. Algae can yield 26,000 gallons of <span class=\"no-emphasis\" data-type=\"term\">biofuel<\/span> per hectare\u2014much more energy per acre than other crops. Some strains of algae can flourish in brackish water that is not usable for growing other crops. Algae can produce biodiesel, biogasoline, ethanol, butanol, methane, and even jet fuel.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_05_03_AlgalFuel1\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">(a) Tiny algal organisms can be (b) grown in large quantities and eventually (c) turned into a useful fuel such as biodiesel. (credit a: modification of work by Micah Sittig; credit b: modification of work by Robert Kerton; credit c: modification of work by John F. Williams)<\/div>\r\n<span id=\"fs-idp157749040\" data-type=\"media\" data-alt=\"Three pictures are shown and labeled a, b, and c. Picture a shows a microscopic view of algal organisms. They are brown, multipart strands and net-like structures on a background of light violet. Picture b shows five large tubs full of a brown liquid containing these algal organisms. Picture c depicts a cylinder full of green liquid in the foreground and a poster in the background that has the title \u201cFrom Field to Fleet.\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_05_03_AlgalFuel1-1.jpg\" alt=\"Three pictures are shown and labeled a, b, and c. Picture a shows a microscopic view of algal organisms. They are brown, multipart strands and net-like structures on a background of light violet. Picture b shows five large tubs full of a brown liquid containing these algal organisms. Picture c depicts a cylinder full of green liquid in the foreground and a poster in the background that has the title \u201cFrom Field to Fleet.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idp157750544\">According to the US Department of Energy, only 39,000 square kilometers (about 0.4% of the land mass of the US or less than 1\/7 of the area used to grow corn) can produce enough algal fuel to replace all the petroleum-based fuel used in the US. The cost of algal fuels is becoming more competitive\u2014for instance, the US Air Force is producing jet fuel from algae at a total cost of under $5 per gallon.<sup data-type=\"footnote-number\"><a href=\"#footnote1\" data-type=\"footnote-link\">1<\/a><\/sup> The process used to produce algal fuel is as follows: grow the algae (which use sunlight as their energy source and CO<sub>2<\/sub> as a raw material); harvest the algae; extract the fuel compounds (or precursor compounds); process as necessary (e.g., perform a transesterification reaction to make biodiesel); purify; and distribute (<a class=\"autogenerated-content\" href=\"#CNX_Chem_05_03_AlgalFuel2\">(Figure)<\/a>).<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_05_03_AlgalFuel2\" class=\"bc-figure figure\">\r\n<div class=\"bc-figcaption figcaption\">Algae convert sunlight and carbon dioxide into oil that is harvested, extracted, purified, and transformed into a variety of renewable fuels.<\/div>\r\n<span id=\"fs-idp167165312\" data-type=\"media\" data-alt=\"A flowchart is shown that contains pictures and words. Reading from left to right, the terms \u201cGrow,\u201d \u201cHarvest,\u201d \u201cExtract,\u201d \u201cProcess and purify,\u201d and \u201cJet fuel gasoline diesel\u201d are shown with right-facing arrows in between each. Above each term, respectively, are diagrams of three containers, three cylinders lying side-by-side, a pyramid-like container with liquid inside, a factory, and a fuel pump. In the space above all of the diagrams and to the left of the images is a diagram of the sun.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_05_03_AlgalFuel2-1.jpg\" alt=\"A flowchart is shown that contains pictures and words. Reading from left to right, the terms \u201cGrow,\u201d \u201cHarvest,\u201d \u201cExtract,\u201d \u201cProcess and purify,\u201d and \u201cJet fuel gasoline diesel\u201d are shown with right-facing arrows in between each. Above each term, respectively, are diagrams of three containers, three cylinders lying side-by-side, a pyramid-like container with liquid inside, a factory, and a fuel pump. In the space above all of the diagrams and to the left of the images is a diagram of the sun.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp93919792\" class=\"chemistry link-to-learning\" data-type=\"note\">\r\n<p id=\"fs-idp93921200\">Click <a href=\"http:\/\/openstaxcollege.org\/l\/16biofuel\">here<\/a> to learn more about the process of creating algae biofuel.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp93922352\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Standard Enthalpy of Formation<\/strong><\/h3>\r\n<p id=\"fs-idp93922992\">A <span data-type=\"term\"><strong>standard enthalpy of formation<\/strong>, \u0394<em>H<\/em><sub>f<\/sub>\u00b0,<\/span>\u00a0is an enthalpy change for a reaction in which exactly 1 mole of a pure substance is formed from free elements in their most stable states under standard state conditions. These values are especially useful for computing or predicting enthalpy changes for chemical reactions that are impractical or dangerous to carry out, or for processes for which it is difficult to make measurements. If we have values for the appropriate standard enthalpies of formation, we can determine the enthalpy change for any reaction, which we will practice in the next section on Hess\u2019s law.<\/p>\r\n<p id=\"fs-idp185900096\">The standard enthalpy of formation of CO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) is \u2212393.5 kJ\/mol. This is the enthalpy change for the exothermic reaction:<\/p>\r\n\r\n<div id=\"fs-idp185901552\" data-type=\"equation\"><img class=\" wp-image-1255 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3r-300x27.png\" alt=\"\" width=\"422\" height=\"38\" \/><\/div>\r\n<p id=\"fs-idp165684992\">starting with the reactants at a pressure of 1 atm and 25 \u00b0C (with the carbon present as graphite, the most stable form of carbon under these conditions) and ending with one mole of CO<sub>2<\/sub>, also at 1 atm and 25 \u00b0C. For nitrogen dioxide, NO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>), <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span> is 33.2 kJ\/mol. This is the enthalpy change for the reaction:<\/p>\r\n\r\n<div id=\"fs-idp185944512\" data-type=\"equation\"><img class=\"wp-image-1256 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3s-300x32.png\" alt=\"\" width=\"441\" height=\"47\" \/><\/div>\r\n<p id=\"fs-idp173428336\">A reaction equation with \u00bd mole of N<sub>2<\/sub> and 1 mole of O<sub>2<\/sub> is correct in this case because the standard enthalpy of formation always refers to 1 mole of product, NO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>).<\/p>\r\n<p id=\"fs-idp173432432\">You will find a table of standard enthalpies of formation of many common substances in <a class=\"target-chapter\" href=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/back-matter\/standard-thermodynamic-properties-for-selected-substances\/\">Appendix G<\/a>. These values indicate that formation reactions range from highly exothermic (such as \u22122984 kJ\/mol for the formation of P<sub>4<\/sub>O<sub>10<\/sub>) to strongly endothermic (such as +226.7 kJ\/mol for the formation of acetylene, C<sub>2<\/sub>H<sub>2<\/sub>). By definition, the standard enthalpy of formation of an element in its most stable form is equal to zero under standard conditions, which is 1 atm for gases and 1 M for solutions.<\/p>\r\n\r\n<div id=\"fs-idp167759616\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp167759872\"><strong>Evaluating an Enthalpy of Formation:<\/strong><\/p>\r\nOzone, O<sub>3<\/sub>(<em data-effect=\"italics\">g<\/em>), forms from oxygen, O<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>), by an endothermic process. Ultraviolet radiation is the source of the energy that drives this reaction in the upper atmosphere. Assuming that both the reactants and products of the reaction are in their standard states, determine the standard enthalpy of formation, <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0,<\/span>\u00a0of ozone from the following information:\r\n\r\n<img class=\"wp-image-1273 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3t-300x23.png\" alt=\"\" width=\"352\" height=\"27\" \/>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp166467632\"><strong>Solution:<\/strong><\/p>\r\n<span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span> is the enthalpy change for the formation of one mole of a substance in its standard state from the elements in their standard states. Thus, <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0 <\/span>for O<sub>3<\/sub>(<em data-effect=\"italics\">g<\/em>) is the enthalpy change for the reaction:\r\n<div id=\"fs-idp166474448\" data-type=\"equation\"><img class=\"wp-image-1274 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3u.png\" alt=\"\" width=\"111\" height=\"34\" \/><\/div>\r\n<p id=\"fs-idp95008768\">For the formation of 2 mol of O<sub>3<\/sub>(<em data-effect=\"italics\">g<\/em>), \u0394<em>H<\/em>\u00b0 = +286 kJ. This ratio, <img class=\"alignnone wp-image-1275\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3v.png\" alt=\"\" width=\"55\" height=\"25\" \/>, can be used as a conversion factor to find the heat produced when 1 mole of O<sub>3<\/sub>(<em data-effect=\"italics\">g<\/em>) is formed, which is the enthalpy of formation for O<sub>3<\/sub>(<em data-effect=\"italics\">g<\/em>):<\/p>\r\n\r\n<div id=\"fs-idp90148272\" data-type=\"equation\"><img class=\"alignnone wp-image-1276 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3x-300x34.png\" alt=\"\" width=\"318\" height=\"36\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp186495680\">Therefore, <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span>[O<sub>3<\/sub>(<em>g<\/em>)]= +143 kJ\/mol.<\/p>\r\n<p id=\"fs-idp167644640\"><strong>Check Your Learning:<\/strong><\/p>\r\nHydrogen gas, H<sub>2<\/sub>, reacts explosively with gaseous chlorine, Cl<sub>2<\/sub>, to form hydrogen chloride, HCl(<em data-effect=\"italics\">g<\/em>). What is the enthalpy change for the reaction of 1 mole of H<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) with 1 mole of Cl<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) if both the reactants and products are at standard state conditions? The standard enthalpy of formation of HCl(<em data-effect=\"italics\">g<\/em>) is \u221292.3 kJ\/mol.\r\n\r\n&nbsp;\r\n<div id=\"fs-idp167648992\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp167649504\">For the reaction <img class=\"alignnone size-medium wp-image-1277\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3y-300x19.png\" alt=\"\" width=\"300\" height=\"19\" \/><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp186597312\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp186597568\"><strong>Writing Reaction Equations for \u0394<em>H<\/em><sub>f<\/sub>\u00b0:<\/strong><\/p>\r\n\u00a0Write the heat of formation reaction equations for:\r\n<p id=\"fs-idp186600640\">(a) C<sub>2<\/sub>H<sub>5<\/sub>OH(<em data-effect=\"italics\">l<\/em>)<\/p>\r\n<p id=\"fs-idp186602304\">(b) Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>(<em data-effect=\"italics\">s<\/em>)<\/p>\r\n<p id=\"fs-idp186604352\"><strong>Solution:<\/strong><\/p>\r\nRemembering that <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span> reaction equations are for forming 1 mole of the compound from its constituent elements under standard conditions, we have:\r\n<p id=\"fs-idp166486208\">(a) 2C(<em>s,graphite<\/em>) + 3H<sub>2<\/sub>(<em>g<\/em>) + (1\/2)O<sub>2<\/sub>(<em>g<\/em>) \u27f6 C<sub>2<\/sub>H<sub>5<\/sub>OH(<em data-effect=\"italics\">l<\/em>)<\/p>\r\n<p id=\"fs-idp145132528\">(b)\u00a0 3Ca(<em>s<\/em>) + (1\/2)P<sub>4<\/sub>(<em>s<\/em>) + 4O<sub>2<\/sub>(<em>g<\/em>) \u27f6 Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>(<em data-effect=\"italics\">s<\/em>)<\/p>\r\n<p id=\"fs-idp166524800\">Note: The standard state of carbon is graphite, and phosphorus exists as P<sub>4<\/sub>.<\/p>\r\n<p id=\"fs-idp166525568\"><strong>Check Your Learning:<\/strong><\/p>\r\nWrite the heat of formation reaction equations for:\r\n<p id=\"fs-idp166526208\">(a) C<sub>2<\/sub>H<sub>5<\/sub>OC<sub>2<\/sub>H<sub>5<\/sub>(<em data-effect=\"italics\">l<\/em>)<\/p>\r\n<p id=\"fs-idp185932768\">(b) Na<sub>2<\/sub>CO<sub>3<\/sub>(<em data-effect=\"italics\">s<\/em>)<\/p>\r\n\r\n<div id=\"fs-idp185934432\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp185934944\">(a) 4C(<em>s,graphite<\/em>) + 5H<sub>2<\/sub>(<em>g<\/em>) + (1\/2)O<sub>2<\/sub>(<em>g<\/em>) \u27f6 C<sub>2<\/sub>H<sub>5<\/sub>OC<sub>2<\/sub>H<sub>5<\/sub>(<em data-effect=\"italics\">l<\/em>)<\/p>\r\n(b) 2Na(<em>s<\/em>) + C(<em>s,graphite<\/em>)+ (3\/2)O<sub>2<\/sub>(<em>g<\/em>) \u27f6 Na<sub>2<\/sub>CO<sub>3<\/sub>(<em data-effect=\"italics\">s<\/em>)\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp218911984\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Hess\u2019s Law<\/strong><\/h3>\r\n<p id=\"fs-idp218912624\">There are two ways to determine the amount of heat involved in a chemical change: measure it experimentally, or calculate it from other experimentally determined enthalpy changes. Some reactions are difficult, if not impossible, to investigate and make accurate measurements for experimentally. And even when a reaction is not hard to perform or measure, it is convenient to be able to determine the heat involved in a reaction without having to perform an experiment.<\/p>\r\n<p id=\"fs-idp177508528\">This type of calculation usually involves the use of <span data-type=\"term\">Hess\u2019s law<\/span>, which states: <em data-effect=\"italics\">If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps<\/em>. Hess\u2019s law is valid because enthalpy is a state function: Enthalpy changes depend only on where a chemical process starts and ends, but not on the path it takes from start to finish. For example, we can think of the reaction of carbon with oxygen to form carbon dioxide as occurring either directly or by a two-step process. The direct process is written:<\/p>\r\n\r\n<div id=\"fs-idp177510624\" data-type=\"equation\"><img class=\"alignnone size-medium wp-image-1280 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3aa-300x34.png\" alt=\"\" width=\"300\" height=\"34\" \/><\/div>\r\n<p id=\"fs-idp51924752\">In the two-step process, first carbon monoxide is formed:<\/p>\r\n\r\n<div id=\"fs-idp51925136\" data-type=\"equation\"><img class=\"size-medium wp-image-1281 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ab-300x35.png\" alt=\"\" width=\"300\" height=\"35\" \/><\/div>\r\n<p id=\"fs-idp161438416\">Then, carbon monoxide reacts further to form carbon dioxide:<\/p>\r\n\r\n<div id=\"fs-idp161438800\" data-type=\"equation\"><img class=\"size-medium wp-image-1283 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ac-300x36.png\" alt=\"\" width=\"300\" height=\"36\" \/><\/div>\r\n<p id=\"fs-idp186572000\">The equation describing the overall reaction is the sum of these two chemical changes:<\/p>\r\n\r\n<div id=\"fs-idp186572384\" data-type=\"equation\"><img class=\"size-medium wp-image-1282 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ad-300x61.png\" alt=\"\" width=\"300\" height=\"61\" \/><\/div>\r\n<p id=\"fs-idp40118928\">Because the CO produced in Step 1 is consumed in Step 2, the net change is:<\/p>\r\n\r\n<div id=\"fs-idp40119312\" data-type=\"equation\"><img class=\" wp-image-1284 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ae-300x61.png\" alt=\"\" width=\"172\" height=\"35\" \/><\/div>\r\n<p id=\"fs-idp128370320\">According to Hess\u2019s law, the enthalpy change of the reaction will equal the sum of the enthalpy changes of the steps.<\/p>\r\n\r\n<div id=\"fs-idp128371600\" data-type=\"equation\"><img class=\"size-medium wp-image-1285 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3af-300x69.png\" alt=\"\" width=\"300\" height=\"69\" \/><\/div>\r\n<p id=\"fs-idp10365136\">The result is shown in <a class=\"autogenerated-content\" href=\"#CNX_Chem_05_03_HessCO2\">(Figure)<\/a>. We see that \u0394<em data-effect=\"italics\">H<\/em> of the overall reaction is the same whether it occurs in one step or two. This finding (overall \u0394<em data-effect=\"italics\">H<\/em> for the reaction = sum of \u0394<em data-effect=\"italics\">H<\/em> values for reaction \u201csteps\u201d in the overall reaction) is true in general for chemical and physical processes.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_05_03_HessCO2\" class=\"bc-figure figure\">\r\n<div class=\"bc-figcaption figcaption\">The formation of CO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) from its elements can be thought of as occurring in two steps, which sum to the overall reaction, as described by Hess\u2019s law. The horizontal blue lines represent enthalpies. For an exothermic process, the products are at lower enthalpy than are the reactants.<\/div>\r\n<span id=\"fs-idp167194048\" data-type=\"media\" data-alt=\"A diagram is shown. A long arrow faces upward on the left with the phrase \u201cH increasing.\u201d A horizontal line at the bottom of the diagram is shown with the formula \u201cC O subscript 2 (g)\u201d below it. A horizontal line at the top of the diagram has the formulas \u201cC (s) + O subscript 2 (g)\u201d above it. The top and bottom lines are connected by a downward facing arrow with the value \u201c\u0394 H = \u2013394 k J\u201d written beside it. Below and to the right of the top horizontal line is a second horizontal line with the equations \u201cC O (g) + one half O subscript 2 (g)\u201d above it. This line and the bottom line are connected by a downward facing arrow with the value \u201c\u0394 H = \u2013283 k J\u201d written beside it. The same line and the top line are connected by a downward facing arrow with the value \u201c\u0394 H = \u2013111 k J\u201d written beside it. There are three brackets to the right of the diagram. The first bracket runs from the top horizontal line to the second horizontal line. It is labeled, \u201cEnthalpy of reactants.\u201d The second bracket runs from the second horizontal line to the bottom horizontal line. It is labeled, \u201cEnthalpy of products.\u201d Both of these brackets are included in the third bracket which runs from the top to the bottom of the diagram. It is labeled, \u201cEnthalpy change of exothermic reaction in 1 or 2 steps.\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_05_03_HessCO2-1.jpg\" alt=\"A diagram is shown. A long arrow faces upward on the left with the phrase \u201cH increasing.\u201d A horizontal line at the bottom of the diagram is shown with the formula \u201cC O subscript 2 (g)\u201d below it. A horizontal line at the top of the diagram has the formulas \u201cC (s) + O subscript 2 (g)\u201d above it. The top and bottom lines are connected by a downward facing arrow with the value \u201c\u0394 H = \u2013394 k J\u201d written beside it. Below and to the right of the top horizontal line is a second horizontal line with the equations \u201cC O (g) + one half O subscript 2 (g)\u201d above it. This line and the bottom line are connected by a downward facing arrow with the value \u201c\u0394 H = \u2013283 k J\u201d written beside it. The same line and the top line are connected by a downward facing arrow with the value \u201c\u0394 H = \u2013111 k J\u201d written beside it. There are three brackets to the right of the diagram. The first bracket runs from the top horizontal line to the second horizontal line. It is labeled, \u201cEnthalpy of reactants.\u201d The second bracket runs from the second horizontal line to the bottom horizontal line. It is labeled, \u201cEnthalpy of products.\u201d Both of these brackets are included in the third bracket which runs from the top to the bottom of the diagram. It is labeled, \u201cEnthalpy change of exothermic reaction in 1 or 2 steps.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idp167197264\">Before we further practice using Hess\u2019s law, let us recall two important features of \u0394<em data-effect=\"italics\">H<\/em>.<\/p>\r\n\r\n<ol id=\"fs-idp167198464\" type=\"1\">\r\n \t<li>\r\n<p id=\"fs-idp167199328\">\u0394<em data-effect=\"italics\">H<\/em> is directly proportional to the quantities of reactants or products. For example, the enthalpy change for the reaction forming 1 mole of NO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) is +33.2 kJ:<\/p>\r\n\r\n<div id=\"fs-idp167201504\" data-type=\"equation\"><img class=\"alignnone wp-image-1286 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ba-300x41.png\" alt=\"\" width=\"315\" height=\"43\" \/><\/div>\r\n<p id=\"fs-idp160715680\">When 2 moles of NO<sub>2<\/sub> (twice as much) are formed, the \u0394<em data-effect=\"italics\">H<\/em> will be twice as large:<\/p>\r\n\r\n<div id=\"fs-idp160717072\" data-type=\"equation\"><img class=\"alignnone wp-image-1287 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bb-300x27.png\" alt=\"\" width=\"322\" height=\"29\" \/><\/div>\r\n<p id=\"fs-idp146331072\">In general, if we multiply or divide an equation by a number, then the enthalpy change should also be multiplied or divided by the same number.<\/p>\r\n<\/li>\r\n \t<li>\r\n<p id=\"fs-idp146331744\">\u0394<em data-effect=\"italics\">H<\/em> for a reaction in one direction is equal in magnitude and opposite in sign to \u0394<em data-effect=\"italics\">H<\/em> for the reaction in the reverse direction. For example, given that:<\/p>\r\n\r\n<div id=\"fs-idp146333552\" data-type=\"equation\"><img class=\"alignnone wp-image-1288 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bc-300x24.png\" alt=\"\" width=\"313\" height=\"25\" \/><\/div>\r\n<p id=\"fs-idp146344672\">Then, for the \u201creverse\u201d reaction, the enthalpy change is also \u201creversed\u201d:<\/p>\r\n\r\n<div id=\"fs-idp146345216\" data-type=\"equation\"><img class=\"alignnone wp-image-1289 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bd-300x24.png\" alt=\"\" width=\"325\" height=\"26\" \/><\/div><\/li>\r\n<\/ol>\r\n<div id=\"fs-idp185548816\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp185549072\"><strong>Stepwise Calculation of \u0394<em>H<\/em><sub>f<\/sub>\u00b0 Using Hess\u2019s Law:<\/strong><\/p>\r\nDetermine the enthalpy of formation, <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span>, of FeCl<sub>3<\/sub>(<em data-effect=\"italics\">s<\/em>) from the enthalpy changes of the following two-step process that occurs under standard state conditions:\r\n<div id=\"fs-idp103964560\" style=\"text-align: center\" data-type=\"equation\"><img class=\"alignnone wp-image-1290\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ca-300x63.png\" alt=\"\" width=\"348\" height=\"73\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp190748768\"><strong>Solution:<\/strong><\/p>\r\nWe are trying to find the standard enthalpy of formation of FeCl<sub>3<\/sub>(<em data-effect=\"italics\">s<\/em>), which is equal to \u0394<em data-effect=\"italics\">H<\/em>\u00b0 for the reaction:\r\n<div id=\"fs-idp190751040\" data-type=\"equation\"><img class=\"alignnone size-medium wp-image-1291 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cb-300x43.png\" alt=\"\" width=\"300\" height=\"43\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp190763840\">Looking at the reactions, we see that the reaction for which we want to find \u0394<em data-effect=\"italics\">H<\/em>\u00b0 is the sum of the two reactions with known \u0394<em data-effect=\"italics\">H<\/em> values, so we must sum their \u0394<em data-effect=\"italics\">H<\/em>s:<\/p>\r\n\r\n<div id=\"fs-idp133790064\" data-type=\"equation\"><img class=\"alignnone wp-image-1292 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cc-300x60.png\" alt=\"\" width=\"325\" height=\"65\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp149506592\">The enthalpy of formation, <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span>, of FeCl<sub>3<\/sub>(<em data-effect=\"italics\">s<\/em>) is \u2212399.5 kJ\/mol.<\/p>\r\n&nbsp;\r\n<p id=\"fs-idp149511040\"><strong>Check Your Learning:<\/strong><\/p>\r\nCalculate \u0394<em data-effect=\"italics\">H<\/em> for the process:\r\n<div id=\"fs-idp149512304\" data-type=\"equation\"><img class=\"wp-image-1293 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cd-300x44.png\" alt=\"\" width=\"218\" height=\"32\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp164462848\">from the following information:<\/p>\r\n\r\n<div id=\"fs-idp164463232\" data-type=\"equation\"><img class=\"alignnone wp-image-1294 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ce-300x64.png\" alt=\"\" width=\"356\" height=\"76\" \/><\/div>\r\n<div id=\"fs-idp146898608\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp146899120\">66.4 kJ<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idp146899760\">Here is a less straightforward example that illustrates the thought process involved in solving many Hess\u2019s law problems. It shows how we can find many standard enthalpies of formation (and other values of \u0394<em data-effect=\"italics\">H<\/em>) if they are difficult to determine experimentally.<\/p>\r\n\r\n<div id=\"fs-idp146901200\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp146901456\"><strong>A More Challenging Problem Using Hess\u2019s Law:<\/strong><\/p>\r\nChlorine monofluoride can react with fluorine to form chlorine trifluoride:\r\n<p id=\"fs-idp146902208\"><img class=\"alignnone size-medium wp-image-1296\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3da-300x32.png\" alt=\"\" width=\"300\" height=\"32\" \/><\/p>\r\n<p id=\"fs-idp146912304\">Use the reactions here to determine the \u0394<em data-effect=\"italics\">H<\/em>\u00b0 for reaction <em data-effect=\"italics\">(i)<\/em>:<\/p>\r\n<p id=\"fs-idp146913936\"><img class=\"alignnone wp-image-1297\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3db-300x53.png\" alt=\"\" width=\"413\" height=\"73\" \/><\/p>\r\n<p id=\"fs-idp167247104\"><strong>Solution:<\/strong><\/p>\r\nOur goal is to manipulate and combine reactions <em data-effect=\"italics\">(ii)<\/em>, <em data-effect=\"italics\">(iii)<\/em>, and <em data-effect=\"italics\">(iv)<\/em> such that they add up to reaction <em data-effect=\"italics\">(i)<\/em>. Going from left to right in <em data-effect=\"italics\">(i)<\/em>, we first see that ClF(<em data-effect=\"italics\">g<\/em>) is needed as a reactant. This can be obtained by multiplying reaction <em data-effect=\"italics\">(iii)<\/em> by \u00bd, which means that the \u0394<em data-effect=\"italics\">H<\/em>\u00b0 change is also multiplied by \u00bd:\r\n<div id=\"fs-idp167256800\" data-type=\"equation\"><img class=\"wp-image-1298 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dc-300x21.png\" alt=\"\" width=\"443\" height=\"31\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp160398800\">Next, we see that F<sub>2<\/sub> is also needed as a reactant. To get this, reverse and halve reaction <em data-effect=\"italics\">(ii)<\/em>, which means that the \u0394<em data-effect=\"italics\">H<\/em>\u00b0 changes sign and is halved:<\/p>\r\n\r\n<div id=\"fs-idp160400816\" data-type=\"equation\"><img class=\"alignnone wp-image-1299 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dd-300x37.png\" alt=\"\" width=\"324\" height=\"40\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp134606352\">To get ClF<sub>3<\/sub> as a product, reverse <em data-effect=\"italics\">(iv)<\/em>, changing the sign of \u0394<em data-effect=\"italics\">H<\/em>\u00b0:<\/p>\r\n\r\n<div id=\"fs-idp134608368\" data-type=\"equation\"><img class=\"alignnone wp-image-1300 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3de-300x28.png\" alt=\"\" width=\"343\" height=\"32\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp134626368\">Now check to make sure that these reactions add up to the reaction we want:<\/p>\r\n\r\n<div id=\"fs-idp134626752\" data-type=\"equation\"><img class=\"alignnone wp-image-1301 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3df-300x62.png\" alt=\"\" width=\"334\" height=\"69\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp154356656\">Reactants \u00bdO<sub>2<\/sub> and \u00bdO<sub>2<\/sub> cancel out product O<sub>2<\/sub>; product \u00bdCl<sub>2<\/sub>O cancels reactant \u00bdCl<sub>2<\/sub>O; and reactant (3\/2)OF<sub>2<\/sub> is cancelled by products \u00bdOF<sub>2<\/sub> and OF<sub>2<\/sub>. This leaves only reactants ClF(<em data-effect=\"italics\">g<\/em>) and F<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) and product ClF<sub>3<\/sub>(<em data-effect=\"italics\">g<\/em>), which are what we want. Since summing these three modified reactions yields the reaction of interest, summing the three modified \u0394<em data-effect=\"italics\">H<\/em>\u00b0 values will give the desired \u0394<em data-effect=\"italics\">H<\/em>\u00b0:<\/p>\r\n\r\n<div id=\"fs-idm259960768\" data-type=\"equation\"><img class=\"alignnone wp-image-1302 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dg-300x27.png\" alt=\"\" width=\"333\" height=\"30\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp154380128\"><strong>Check Your Learning:<\/strong><\/p>\r\nAluminum chloride can be formed from its elements:\r\n<p id=\"fs-idp154380768\"><img class=\"alignnone size-medium wp-image-1303\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ea-300x26.png\" alt=\"\" width=\"300\" height=\"26\" \/><\/p>\r\n<p id=\"fs-idp154392528\">Use the reactions here to determine the \u0394<em data-effect=\"italics\">H<\/em>\u00b0 for reaction <em data-effect=\"italics\">(i)<\/em>:<\/p>\r\n<p id=\"fs-idp154394160\"><img class=\"alignnone wp-image-1304\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3eb-300x69.png\" alt=\"\" width=\"417\" height=\"96\" \/><\/p>\r\n\r\n<div id=\"fs-idp120712352\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp120712864\">\u22121407 kJ<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idp120713504\">We also can use <span class=\"no-emphasis\" data-type=\"term\">Hess\u2019s law<\/span> to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products are available. The stepwise reactions we consider are: (i) decompositions of the reactants into their component elements (for which the enthalpy changes are proportional to the negative of the enthalpies of formation of the reactants), followed by (ii) re-combinations of the elements to give the products (with the enthalpy changes proportional to the enthalpies of formation of the products). The standard enthalpy change of the overall reaction is therefore equal to: (ii) the sum of the standard enthalpies of formation of all the products plus (i) the sum of the negatives of the standard enthalpies of formation of the reactants. This is usually rearranged slightly to be written as follows, with \u2211 representing \u201cthe sum of\u201d and <em data-effect=\"italics\">n<\/em> standing for the stoichiometric coefficients:<\/p>\r\n\r\n<div id=\"fs-idp120716832\" data-type=\"equation\"><img class=\"alignnone wp-image-1305 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fa-300x33.png\" alt=\"\" width=\"355\" height=\"39\" \/><\/div>\r\n<p id=\"fs-idp120729664\">The following example shows in detail why this equation is valid, and how to use it to calculate the enthalpy change for a reaction of interest.<\/p>\r\n\r\n<div id=\"fs-idp120730208\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp134549264\"><strong>Using Hess\u2019s Law:<\/strong><\/p>\r\nWhat is the standard enthalpy change for the reaction:\r\n<div id=\"fs-idp134549904\" data-type=\"equation\"><img class=\"alignnone wp-image-1306 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fb-300x24.png\" alt=\"\" width=\"338\" height=\"27\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp146127120\"><strong>Solution:<\/strong><\/p>\r\nUse the special form of Hess\u2019s law given previously, and values from <a class=\"target-chapter\" href=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/back-matter\/standard-thermodynamic-properties-for-selected-substances\/\">Appendix G<\/a>:\r\n<div id=\"fs-idp146127936\" data-type=\"equation\"><img class=\"wp-image-1307 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fc-300x176.png\" alt=\"\" width=\"394\" height=\"231\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp134564208\"><span data-type=\"title\"><strong>Solution:<\/strong> <\/span><strong>Supporting Why the General Equation Is Valid<\/strong><\/p>\r\nAlternatively, we can write this reaction as the sum of the decompositions of 3NO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) and 1H<sub>2<\/sub>O(<em data-effect=\"italics\">l<\/em>) into their constituent elements, and the formation of 2HNO<sub>3<\/sub>(<em data-effect=\"italics\">aq<\/em>) and 1NO(<em data-effect=\"italics\">g<\/em>) from their constituent elements. Writing out these reactions, and noting their relationships to the <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span> values for these compounds (from <a class=\"target-chapter\" href=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/back-matter\/standard-thermodynamic-properties-for-selected-substances\/\">Appendix G<\/a> ), we have:\r\n<div id=\"fs-idp134571776\" data-type=\"equation\"><img class=\"alignnone wp-image-1309 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ga-300x97.png\" alt=\"\" width=\"405\" height=\"131\" \/><\/div>\r\n<div id=\"fs-idp194526576\" data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp155650064\">Summing these reaction equations gives the reaction we are interested in:<\/p>\r\n\r\n<div id=\"fs-idp155650448\" data-type=\"equation\"><img class=\"alignnone size-medium wp-image-1310 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3gb-300x33.png\" alt=\"\" width=\"300\" height=\"33\" \/><\/div>\r\n<p id=\"fs-idp155662480\">Summing their enthalpy changes gives the value we want to determine:<\/p>\r\n\r\n<div id=\"fs-idp155662864\" data-type=\"equation\"><img class=\"alignnone wp-image-1311 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3gc-300x23.png\" alt=\"\" width=\"534\" height=\"41\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp146119648\">So the standard enthalpy change for this reaction is \u0394<em data-effect=\"italics\">H<\/em>\u00b0 = \u2212138.4 kJ.<\/p>\r\n<p id=\"fs-idp146120784\">Note that this result was obtained by (1) multiplying the <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span> of each product by its stoichiometric coefficient and summing those values, (2) multiplying the <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span> of each reactant by its stoichiometric coefficient and summing those values, and then (3) subtracting the result found in (2) from the result found in (1). This is also the procedure in using the general equation, as shown.<\/p>\r\n&nbsp;\r\n<p id=\"fs-idp165556368\"><strong>Check Your Learning:<\/strong><\/p>\r\nCalculate the heat of combustion of 1 mole of ethanol, C<sub>2<\/sub>H<sub>5<\/sub>OH(<em data-effect=\"italics\">l<\/em>), when H<sub>2<\/sub>O(<em data-effect=\"italics\">l<\/em>) and CO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) are formed. Use the following enthalpies of formation: C<sub>2<\/sub>H<sub>5<\/sub>OH(<em data-effect=\"italics\">l<\/em>), \u2212278 kJ\/mol; H<sub>2<\/sub>O(<em data-effect=\"italics\">l<\/em>), \u2212286 kJ\/mol; and CO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>), \u2212394 kJ\/mol.\r\n\r\n&nbsp;\r\n<div id=\"fs-idp165563824\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp165564336\">\u22121368 kJ\/mol<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp165565104\" class=\"summary\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\r\n<p id=\"fs-idp165565984\">If a chemical change is carried out at constant pressure and the only work done is caused by expansion or contraction, <em data-effect=\"italics\">q<\/em> for the change is called the enthalpy change with the symbol \u0394<em data-effect=\"italics\">H<\/em>, or <span data-type=\"term\">\u0394<em>H<\/em>\u00b0<\/span> for reactions occurring under standard state conditions at 298 K. The value of \u0394<em data-effect=\"italics\">H<\/em> for a reaction in one direction is equal in magnitude, but opposite in sign, to \u0394<em data-effect=\"italics\">H<\/em> for the reaction in the opposite direction, and \u0394<em data-effect=\"italics\">H<\/em> is directly proportional to the quantity of reactants and products. The standard enthalpy of formation, <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0,<\/span> is the enthalpy change accompanying the formation of 1 mole of a substance from the elements in their most stable states at 1 bar. If the enthalpies of formation are available for the reactants and products of a reaction, the enthalpy change can be calculated using Hess\u2019s law: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idp165577648\" class=\"key-equations\" data-depth=\"1\"><\/div>\r\n<div id=\"fs-idp167491136\" class=\"exercises\" data-depth=\"1\">\r\n<div id=\"fs-idp165246448\" data-type=\"exercise\">\r\n<div id=\"fs-idp165246704\" data-type=\"problem\">\r\n<p id=\"fs-idp165259776\"><strong><span style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em\">Footnotes<\/span><\/strong><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"footnote-refs\">\r\n<ul data-list-type=\"bulleted\" data-bullet-style=\"none\">\r\n \t<li data-type=\"footnote-ref\"><a href=\"#footnote-ref1\" data-type=\"footnote-ref-link\">1<\/a><span data-type=\"footnote-ref-content\">For more on algal fuel, see http:\/\/www.theguardian.com\/environment\/2010\/feb\/13\/algae-solve-pentagon-fuel-problem.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox shaded\" data-type=\"glossary\">\r\n<h3 data-type=\"glossary-title\"><strong>Glossary<\/strong><\/h3>\r\n<dl id=\"fs-idp102379584\">\r\n \t<dt>chemical thermodynamics<\/dt>\r\n \t<dd id=\"fs-idp102379968\">area of science that deals with the relationships between heat, work, and all forms of energy associated with chemical and physical processes<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp102380512\">\r\n \t<dt>enthalpy (<em data-effect=\"italics\">H<\/em>)<\/dt>\r\n \t<dd id=\"fs-idp102381520\">sum of a system\u2019s internal energy and the mathematical product of its pressure and volume<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp102382096\">\r\n \t<dt>enthalpy change (\u0394<em data-effect=\"italics\">H<\/em>)<\/dt>\r\n \t<dd id=\"fs-idp102383104\">heat released or absorbed by a system under constant pressure during a chemical or physical process<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp102383600\">\r\n \t<dt>expansion work (pressure-volume work)<\/dt>\r\n \t<dd id=\"fs-idp102383984\">work done as a system expands or contracts against external pressure<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp102384368\">\r\n \t<dt>first law of thermodynamics<\/dt>\r\n \t<dd id=\"fs-idp102384752\">internal energy of a system changes due to heat flow in or out of the system or work done on or by the system<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp102385264\">\r\n \t<dt>Hess\u2019s law<\/dt>\r\n \t<dd id=\"fs-idp102385648\">if a process can be represented as the sum of several steps, the enthalpy change of the process equals the sum of the enthalpy changes of the steps<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp102386960\">\r\n \t<dt>internal energy (<em data-effect=\"italics\">U<\/em>)<\/dt>\r\n \t<dd id=\"fs-idp102387968\">total of all possible kinds of energy present in a substance or substances<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp102388352\">\r\n \t<dt>standard enthalpy of combustion (\u0394<em>H<\/em><sub>c<\/sub>\u00b0)<\/dt>\r\n \t<dd id=\"fs-idp102391408\">heat released when one mole of a compound undergoes complete combustion under standard conditions<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp102391904\">\r\n \t<dt>standard enthalpy of formation <span style=\"font-size: 1em\">(\u0394<\/span><em>H<\/em><em style=\"font-size: 1em\"><sub>f<\/sub><\/em><span style=\"font-size: 1em\">\u00b0)<\/span><\/dt>\r\n \t<dd id=\"fs-idp102394960\">enthalpy change of a chemical reaction in which 1 mole of a pure substance is formed from its elements in their most stable states under standard state conditions<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp102395520\">\r\n \t<dt>standard state<\/dt>\r\n \t<dd id=\"fs-idp102395904\">set of physical conditions as accepted as common reference conditions for reporting thermodynamic properties; 1 bar of pressure, and solutions at 1 molar concentrations<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp102396512\">\r\n \t<dt>state function<\/dt>\r\n \t<dd id=\"fs-idp102396896\">property depending only on the state of a system, and not the path taken to reach that state<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<p><strong><span style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em;background-color: #cbd4b6;color: #000000\">Learning Objectives<\/span><\/strong><\/p>\n<div class=\"textbox textbox--learning-objectives\">\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>State the first law of thermodynamics<\/li>\n<li>Define enthalpy and explain its classification as a state function<\/li>\n<li>Write and balance thermochemical equations<\/li>\n<li>Calculate enthalpy changes for various chemical reactions<\/li>\n<li>Explain Hess\u2019s law and use it to compute reaction enthalpies<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idp160276224\">Thermochemistry is a branch of <strong>chemical thermodynamics<\/strong>, the science that deals with the relationships between heat, work, and other forms of energy in the context of chemical and physical processes. As we concentrate on thermochemistry in this chapter, we need to consider some widely used concepts of thermodynamics.<\/p>\n<p id=\"fs-idp5509104\">Substances act as reservoirs of energy, meaning that energy can be added to them or removed from them. Energy is stored in a substance when the kinetic or potential energy of its atoms or molecules is raised.\u00a0 The total of all possible kinds of energy present in a substance is called the <strong>internal energy (<em data-effect=\"italics\">U<\/em>)<\/strong>, sometimes symbolized as <em data-effect=\"italics\">E<\/em>.<\/p>\n<p id=\"fs-idm1690784\">As a system undergoes a change, its internal energy can change, and energy can be transferred from the system to the surroundings, or from the surroundings to the system. Energy is transferred into a system when it absorbs heat (<em data-effect=\"italics\">q<\/em>) from the surroundings or when the surroundings do work (<em data-effect=\"italics\">w<\/em>) on the system. For example, energy is transferred into room-temperature metal wire if it is immersed in hot water (the wire absorbs heat from the water), or if you rapidly bend the wire back and forth (the wire becomes warmer because of the work done on it). Both processes increase the internal energy of the wire. Conversely, energy is transferred out of a system when heat is lost from the system, or when the system does work on the surroundings.<\/p>\n<p id=\"fs-idp16545808\">The relationship between internal energy, heat, and work can be represented by the equation:<\/p>\n<div id=\"fs-idm150109936\" style=\"text-align: center\" data-type=\"equation\">\u0394<em>U<\/em> = <em>q<\/em> + <em>w<\/em><\/div>\n<p id=\"fs-idp146392048\">as shown in <a class=\"autogenerated-content\" href=\"#CNX_Chem_05_03_Systemqw\">(Figure)<\/a>. This is one version of the <span data-type=\"term\">first law of thermodynamics<\/span>, and it shows that the internal energy of a system changes through heat flow into or out of the system (positive <em data-effect=\"italics\">q<\/em> is heat flow in; negative <em data-effect=\"italics\">q<\/em> is heat flow out) or work done on or by the system. The work, <em data-effect=\"italics\">w<\/em>, is positive if it is done on the system and negative if it is done by the system.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_05_03_Systemqw\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">The internal energy, <em data-effect=\"italics\">U<\/em>, of a system can be changed by heat flow and work. If heat flows into the system, <em data-effect=\"italics\">q<\/em><sub>in<\/sub>, or work is done on the system, <em data-effect=\"italics\">w<\/em><sub>on<\/sub>, its internal energy increases, \u0394<em data-effect=\"italics\">U<\/em> &gt; 0. If heat flows out of the system, <em data-effect=\"italics\">q<\/em><sub>out<\/sub>, or work is done by the system, <em data-effect=\"italics\">w<\/em><sub>by<\/sub>, its internal energy decreases, \u0394<em data-effect=\"italics\">U<\/em> &lt; 0.<span id=\"fs-idp1427264\" data-type=\"media\">\u00a0<img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_05_03_Systemqw-1.jpg\" data-media-type=\"image\/jpeg\" alt=\"image\" \/>\u00a0<\/span><\/div>\n<\/div>\n<p id=\"fs-idp6896176\">A type of work called <span data-type=\"term\">expansion work<\/span> (or pressure-volume work) occurs when a system pushes back the surroundings against a restraining pressure, or when the surroundings compress the system. An example of this occurs during the operation of an internal combustion engine. The reaction of gasoline and oxygen is exothermic. Some of this energy is given off as heat, and some does work pushing the piston in the cylinder. The substances involved in the reaction are the system, and the engine and the rest of the universe are the surroundings. The system loses energy by both heating and doing work on the surroundings, and its internal energy decreases. (The engine is able to keep the car moving because this process is repeated many times per second while the engine is running.) We will consider how to determine the amount of work involved in a chemical or physical change in the chapter on thermodynamics.<\/p>\n<div id=\"fs-idp889984\" class=\"chemistry link-to-learning\" data-type=\"note\">\n<p id=\"fs-idp13166816\">This view of <a href=\"http:\/\/openstaxcollege.org\/l\/16combustion\">an internal combustion engine<\/a> illustrates the conversion of energy produced by the exothermic combustion reaction of a fuel such as gasoline into energy of motion.<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-idp120420048\">As discussed, the relationship between internal energy, heat, and work can be represented as \u0394<em data-effect=\"italics\">U<\/em> = <em data-effect=\"italics\">q<\/em> + <em data-effect=\"italics\">w<\/em>. Internal energy is an example of a<strong> state function <\/strong>(or state variable), whereas heat and work are not state functions. The value of a state function depends only on the state that a system is in, and not on how that state is reached. If a quantity is not a state function, then its value <em data-effect=\"italics\">does<\/em> depend on how the state is reached. An example of a state function is altitude or elevation. If you stand on the summit of Mt. Kilimanjaro, you are at an altitude of 5895 m, and it does not matter whether you hiked there or parachuted there. The distance you traveled to the top of Kilimanjaro, however, is not a state function. You could climb to the summit by a direct route or by a more roundabout, circuitous path (<a class=\"autogenerated-content\" href=\"#CNX_Chem_05_03_Summit\">(Figure)<\/a>). The distances traveled would differ (distance is not a state function) but the elevation reached would be the same (altitude is a state function).<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_05_03_Summit\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">Paths X and Y represent two different routes to the summit of Mt. Kilimanjaro. Both have the same change in elevation (altitude or elevation on a mountain is a state function; it does not depend on path), but they have very different distances traveled (distance walked is not a state function; it depends on the path). (credit: modification of work by Paul Shaffner)<\/div>\n<p><span id=\"fs-idp164506448\" data-type=\"media\" data-alt=\"An aerial photo depicts a view of Mount Kilimanjaro. A straight, green arrow labeled X is drawn from the term \u201cbase,\u201d written at the bottom of the mountain, to the term \u201cSummit,\u201d written at the top of the mountain. Another arrow labeled Y is draw from the base to the summit alongside the green arrow, but this arrow is pink and has three large S-shaped curves along its length.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_05_03_Summit-1.jpg\" alt=\"An aerial photo depicts a view of Mount Kilimanjaro. A straight, green arrow labeled X is drawn from the term \u201cbase,\u201d written at the bottom of the mountain, to the term \u201cSummit,\u201d written at the top of the mountain. Another arrow labeled Y is draw from the base to the summit alongside the green arrow, but this arrow is pink and has three large S-shaped curves along its length.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idp107461968\">Chemists ordinarily use a property known as <span data-type=\"term\">enthalpy (<em data-effect=\"italics\">H<\/em>)<\/span> to describe the thermodynamics of chemical and physical processes. Enthalpy is defined as the sum of a system\u2019s internal energy (<em data-effect=\"italics\">U<\/em>) and the mathematical product of its pressure (<em data-effect=\"italics\">P<\/em>) and volume (<em data-effect=\"italics\">V<\/em>):<\/p>\n<p style=\"text-align: center\"><em>H<\/em> = <em>U<\/em> + <em>PV<\/em><\/p>\n<p id=\"fs-idm20812336\">Enthalpy is also a state function. Enthalpy values for specific substances cannot be measured directly; only enthalpy <em data-effect=\"italics\">changes<\/em> for chemical or physical processes can be determined. For processes that take place at constant pressure (a common condition for many chemical and physical changes), the <span data-type=\"term\">enthalpy change (\u0394<em data-effect=\"italics\">H<\/em>)<\/span> is:<\/p>\n<div id=\"fs-idm182172016\" style=\"text-align: center\" data-type=\"equation\">\u0394<em>H <\/em>= \u0394<em>U <\/em>+ <em>P<\/em>\u0394<em>V\u00a0 \u00a0 \u00a0<\/em>(at constant <em>P<\/em>)<\/div>\n<p id=\"fs-idm15694224\">The mathematical product <em data-effect=\"italics\">P<\/em>\u0394<em data-effect=\"italics\">V<\/em> represents work (<em data-effect=\"italics\">w<\/em>), namely, expansion or pressure-volume work as noted. By their definitions, the arithmetic signs of \u0394<em data-effect=\"italics\">V<\/em> and <em data-effect=\"italics\">w<\/em> will always be opposite:<\/p>\n<div id=\"fs-idm223126448\" style=\"text-align: center\" data-type=\"equation\"><em>P<\/em>\u0394<em>V<\/em> = \u2212<em>w<\/em><\/div>\n<p id=\"fs-idp7052304\">Substituting this equation and the definition of internal energy into the enthalpy-change equation yields:<\/p>\n<div id=\"fs-idm223599728\" style=\"text-align: center\" data-type=\"equation\">\u0394<em>H<\/em> = \u0394<em>U<\/em> + <em>P<\/em>\u0394<em>V<\/em> = <em>q<\/em><sub>p<\/sub> + <em>w<\/em> &#8211; <em>w <\/em>= <em>q<\/em><sub>p<\/sub><\/div>\n<p id=\"fs-idp58958304\">where <em data-effect=\"italics\">q<sub>p<\/sub><\/em> is the heat of reaction under conditions of constant pressure.<\/p>\n<p id=\"fs-idm50104768\">And so, if a chemical or physical process is carried out at constant pressure with the only work done caused by expansion or contraction, then the heat flow (<em data-effect=\"italics\">q<sub>p<\/sub><\/em>) and enthalpy change (\u0394<em data-effect=\"italics\">H<\/em>) for the process are equal.<\/p>\n<p id=\"fs-idm22504976\">The heat given off when you operate a Bunsen burner is equal to the enthalpy change of the methane combustion reaction that takes place, since it occurs at the essentially constant pressure of the atmosphere. On the other hand, the heat produced by a reaction measured in a bomb calorimeter is not equal to \u0394<em data-effect=\"italics\">H<\/em> because the closed, constant-volume metal container prevents the pressure from remaining constant (it may increase or decrease if the reaction yields increased or decreased amounts of gaseous species). Chemists usually perform experiments under normal atmospheric conditions, at constant external pressure with <em data-effect=\"italics\">q<\/em> = \u0394<em data-effect=\"italics\">H<\/em>, which makes enthalpy the most convenient choice for determining heat changes for chemical reactions.<\/p>\n<p id=\"fs-idp25166224\">The following conventions apply when using \u0394<em data-effect=\"italics\">H<\/em>:<\/p>\n<ul id=\"fs-idm49861360\" data-bullet-style=\"bullet\">\n<li>\n<p id=\"fs-idm11576944\">A negative value of an enthalpy change, \u0394<em data-effect=\"italics\">H<\/em> &lt; 0, indicates an exothermic reaction; a positive value, \u0394<em data-effect=\"italics\">H<\/em> &gt; 0, indicates an endothermic reaction. If the direction of a chemical equation is reversed, the arithmetic sign of its \u0394<em data-effect=\"italics\">H<\/em> is changed (a process that is endothermic in one direction is exothermic in the opposite direction).<\/p>\n<\/li>\n<li>\n<p id=\"fs-idp165042832\">Chemists use a <strong>thermochemical equation<\/strong> to represent the changes in both matter and energy. In a thermochemical equation, the enthalpy change of a reaction is shown as a \u0394<em data-effect=\"italics\">H<\/em> value following the equation for the reaction. This \u0394<em data-effect=\"italics\">H<\/em> value indicates the amount of heat associated with the reaction involving the number of moles of reactants and products <em data-effect=\"italics\">as shown in the chemical equation<\/em>. For example, consider this equation:<\/p>\n<div id=\"fs-idp13211824\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1232 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3a-300x42.png\" alt=\"\" width=\"300\" height=\"42\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3a-300x42.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3a-65x9.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3a-225x31.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3a-350x49.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3a.png 746w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<p id=\"fs-idm35882976\">This equation indicates that when 1 mole of hydrogen gas and \u00bd mole of oxygen gas at some temperature and pressure change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released to the surroundings. If the coefficients of the chemical equation are multiplied by some factor, the enthalpy change must be multiplied by that same factor (\u0394<em data-effect=\"italics\">H<\/em> is an extensive property):<\/p>\n<div id=\"fs-idm23265328\" style=\"text-align: center\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1233\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3b-300x59.png\" alt=\"\" width=\"462\" height=\"91\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3b-300x59.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3b-1024x201.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3b-768x150.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3b-65x13.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3b-225x44.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3b-350x69.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3b.png 1103w\" sizes=\"auto, (max-width: 462px) 100vw, 462px\" \/><\/div>\n<\/li>\n<li>\n<p id=\"fs-idm72030880\">The enthalpy change of a reaction depends on the physical states of the reactants and products, so these must be shown. For example, when 1 mole of hydrogen gas and \u00bd mole of oxygen gas change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released. If gaseous water forms, only 242 kJ of heat are released.<\/p>\n<div id=\"fs-idp4587936\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1234 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3c-300x37.png\" alt=\"\" width=\"349\" height=\"43\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3c-300x37.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3c-768x95.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3c-65x8.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3c-225x28.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3c-350x43.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3c.png 798w\" sizes=\"auto, (max-width: 349px) 100vw, 349px\" \/><\/div>\n<\/li>\n<\/ul>\n<div id=\"fs-idm9240992\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp52228448\"><strong>Writing Thermochemical Equations:<\/strong><\/p>\n<p>When 0.0500 mol of HCl(<em data-effect=\"italics\">aq<\/em>) reacts with 0.0500 mol of NaOH(<em data-effect=\"italics\">aq<\/em>) to form 0.0500 mol of NaCl(<em data-effect=\"italics\">aq<\/em>), 2.9 kJ of heat are produced. Write a balanced thermochemical equation for the reaction of one mole of HCl.?<\/p>\n<div id=\"fs-idp149553440\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1235 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3d-300x35.png\" alt=\"\" width=\"326\" height=\"38\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3d-300x35.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3d-65x8.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3d-225x26.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3d-350x41.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3d.png 717w\" sizes=\"auto, (max-width: 326px) 100vw, 326px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm10084896\"><strong>Solution:<\/strong><\/p>\n<p>For the reaction of 0.0500 mol acid (HCl), <em data-effect=\"italics\">q<\/em> = \u22122.9 kJ. The reactants are provided in stoichiometric amounts (same molar ratio as in the balanced equation), and so the amount of acid may be used to calculate a molar enthalpy change. Since \u0394<em data-effect=\"italics\">H<\/em> is an extensive property, it is proportional to the amount of acid neutralized:<\/p>\n<div id=\"fs-idp125057232\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1236 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3e-300x42.png\" alt=\"\" width=\"314\" height=\"44\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3e-300x42.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3e-65x9.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3e-225x32.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3e-350x49.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3e.png 764w\" sizes=\"auto, (max-width: 314px) 100vw, 314px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp13720784\">The thermochemical equation is then<\/p>\n<div id=\"fs-idm13850464\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1237 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3f-300x25.png\" alt=\"\" width=\"468\" height=\"39\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3f-300x25.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3f-1024x85.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3f-768x64.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3f-65x5.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3f-225x19.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3f-350x29.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3f.png 1044w\" sizes=\"auto, (max-width: 468px) 100vw, 468px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp6958160\"><strong>Check Your Learning:<\/strong><\/p>\n<p>When 1.34 g Zn(<em data-effect=\"italics\">s<\/em>) reacts with 60.0 mL of 0.750 M HCl(<em data-effect=\"italics\">aq<\/em>), 3.14 kJ of heat are produced. Determine the enthalpy change per mole of zinc reacting for the reaction:<\/p>\n<div id=\"fs-idp6791056\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1238 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3g-300x39.png\" alt=\"\" width=\"300\" height=\"39\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3g-300x39.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3g-65x8.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3g-225x29.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3g-350x45.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3g.png 672w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<div id=\"fs-idp58943984\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp35719536\">\u0394<em data-effect=\"italics\">H<\/em> = \u2212153 kJ<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idm14305808\">Be sure to take both stoichiometry and limiting reactants into account when determining the \u0394<em data-effect=\"italics\">H<\/em> for a chemical reaction.<\/p>\n<div id=\"fs-idp107963312\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm6526496\"><strong>Writing Thermochemical Equations:<\/strong><\/p>\n<p>A gummy bear contains 2.67 g sucrose, C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub>. When it reacts with 7.19 g potassium chlorate, KClO<sub>3<\/sub>, 43.7 kJ of heat are produced. Write a thermochemical equation for the reaction of one mole of sucrose:<\/p>\n<div id=\"fs-idp12841552\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1240 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3h-300x23.png\" alt=\"\" width=\"443\" height=\"34\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3h-300x23.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3h-1024x78.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3h-768x59.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3h-65x5.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3h-225x17.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3h-350x27.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3h.png 1074w\" sizes=\"auto, (max-width: 443px) 100vw, 443px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm42247408\"><strong>Solution:<\/strong><\/p>\n<p>Unlike the previous example exercise, this one does not involve the reaction of stoichiometric amounts of reactants, and so the <em data-effect=\"italics\">limiting reactant<\/em> must be identified (it limits the yield of the reaction and the amount of thermal energy produced or consumed).<\/p>\n<p id=\"fs-idm378999344\">The provided amounts of the two reactants are<\/p>\n<div id=\"fs-idm372152464\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1242 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3i-300x41.png\" alt=\"\" width=\"359\" height=\"49\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3i-300x41.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3i-768x106.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3i-65x9.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3i-225x31.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3i-350x48.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3i.png 790w\" sizes=\"auto, (max-width: 359px) 100vw, 359px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm329792512\">The provided molar ratio of perchlorate-to-sucrose is then<\/p>\n<div id=\"fs-idm378911456\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1241 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3j-300x22.png\" alt=\"\" width=\"355\" height=\"26\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3j-300x22.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3j-768x56.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3j-65x5.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3j-225x16.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3j-350x25.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3j.png 771w\" sizes=\"auto, (max-width: 355px) 100vw, 355px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm379230752\">The balanced equation indicates 8 mol KClO<sub>3<\/sub> are required for reaction with 1 mol C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub>. Since the provided amount of KClO<sub>3<\/sub> is less than the stoichiometric amount, it is the limiting reactant and may be used to compute the enthalpy change:<\/p>\n<div id=\"fs-idm352441488\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1244 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3k-300x19.png\" alt=\"\" width=\"442\" height=\"28\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3k-300x19.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3k-768x49.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3k-65x4.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3k-225x14.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3k-350x23.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3k.png 886w\" sizes=\"auto, (max-width: 442px) 100vw, 442px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm386464528\">Because the equation, as written, represents the reaction of 8 mol KClO<sub>3<\/sub>, the enthalpy change is<\/p>\n<div id=\"fs-idm389610336\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1247 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3l-300x29.png\" alt=\"\" width=\"341\" height=\"33\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3l-300x29.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3l-65x6.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3l-225x22.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3l-350x34.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3l.png 706w\" sizes=\"auto, (max-width: 341px) 100vw, 341px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm379229776\">The enthalpy change for this reaction is \u22125960 kJ, and the thermochemical equation is:<\/p>\n<div id=\"fs-idp90097408\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1248 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3m-300x18.png\" alt=\"\" width=\"499\" height=\"30\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3m-300x18.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3m-1024x62.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3m-768x47.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3m-65x4.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3m-225x14.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3m-350x21.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3m.png 1149w\" sizes=\"auto, (max-width: 499px) 100vw, 499px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp82549456\"><strong>Check Your Learning:<\/strong><\/p>\n<p>When 1.42 g of iron reacts with 1.80 g of chlorine, 3.22 g of FeCl<sub>2<\/sub>(<em data-effect=\"italics\">s<\/em>) and 8.60 kJ of heat is produced. What is the enthalpy change for the reaction when 1 mole of FeCl<sub>2<\/sub>(<em data-effect=\"italics\">s<\/em>) is produced?<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp120447376\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm17649024\">\u0394<em data-effect=\"italics\">H<\/em> = \u2212338 kJ<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idm18196832\">Enthalpy changes are typically tabulated for reactions in which both the reactants and products are at the same conditions. A <strong>standard state<\/strong> is a commonly accepted set of conditions used as a reference point for the determination of properties under other different conditions. For chemists, the IUPAC standard state refers to materials under a pressure of 1 bar and solutions at 1 M, and does not specify a temperature. Many thermochemical tables list values with a standard state of 1 atm. Because the \u0394<em data-effect=\"italics\">H<\/em> of a reaction changes very little with such small changes in pressure (1 bar = 0.987 atm), \u0394<em data-effect=\"italics\">H<\/em> values (except for the most precisely measured values) are essentially the same under both sets of standard conditions. We will include a superscripted \u201co\u201d in the enthalpy change symbol to designate standard state. Since the usual (but not technically standard) temperature is 298.15 K, this temperature will be assumed unless some other temperature is specified. Thus, the symbol \u0394<em>H<\/em>\u00b0 is used to indicate an enthalpy change for a process occurring under these conditions. (The symbol \u0394<em data-effect=\"italics\">H<\/em> is used to indicate an enthalpy change for a reaction occurring under nonstandard conditions.)<\/p>\n<p id=\"fs-idp148022240\">The enthalpy changes for many types of chemical and physical processes are available in the reference literature, including those for combustion reactions, phase transitions, and formation reactions. As we discuss these quantities, it is important to pay attention to the <em data-effect=\"italics\">extensive<\/em> nature of enthalpy and enthalpy changes. Since the enthalpy change for a given reaction is proportional to the amounts of substances involved, it may be reported on that basis (i.e., as the \u0394<em data-effect=\"italics\">H<\/em> for specific amounts of reactants). However, we often find it more useful to divide one extensive property (\u0394<em data-effect=\"italics\">H<\/em>) by another (amount of substance), and report a per-amount <em data-effect=\"italics\">intensive<\/em> value of \u0394<em data-effect=\"italics\">H<\/em>, often \u201cnormalized\u201d to a per-mole basis. (Note that this is similar to determining the intensive property specific heat from the extensive property heat capacity, as seen previously.)<\/p>\n<div id=\"fs-idp4580336\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Standard Enthalpy of Combustion<\/strong><\/h3>\n<p id=\"fs-idp8281184\"><span data-type=\"term\">Standard enthalpy of combustion <\/span>(\u0394<em>H<\/em><sub>c<\/sub>\u00b0) is the enthalpy change when 1 mole of a substance burns (combines vigorously with oxygen) under standard state conditions; it is sometimes called \u201cheat of combustion.\u201d For example, the enthalpy of combustion of ethanol, \u22121366.8 kJ\/mol, is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 \u00b0C and 1 atmosphere pressure, yielding products also at 25 \u00b0C and 1 atm.<\/p>\n<div id=\"fs-idm27865968\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1250 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3n-300x16.png\" alt=\"\" width=\"451\" height=\"24\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3n-300x16.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3n-1024x53.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3n-768x40.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3n-65x3.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3n-225x12.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3n-350x18.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3n.png 1062w\" sizes=\"auto, (max-width: 451px) 100vw, 451px\" \/><\/div>\n<p id=\"fs-idm75327968\">Enthalpies of combustion for many substances have been measured; a few of these are listed in <a class=\"autogenerated-content\" href=\"#fs-idp98710048\">(Figure)<\/a>. Many readily available substances with large enthalpies of combustion are used as fuels, including hydrogen, carbon (as coal or charcoal), and <span data-type=\"term\">hydrocarbons<\/span> (compounds containing only hydrogen and carbon), such as methane, propane, and the major components of gasoline.<\/p>\n<table id=\"fs-idp98710048\" class=\"top-titled\" style=\"height: 311px\" summary=\"A data table is shown that has three columns and eleven rows. The header row reads \u201cSubstance,\u201d \u201cCombustion reaction,\u201d and \u201cEnthalpy of Combustion, \u0394 H subscript C superscript degree symbol (k J\/ mol at 25 \u00b0 C).\u201d The first column contains entries reading \u201ccarbon,\u201d \u201chydrogen,\u201d \u201cmagnesium,\u201d \u201csulfur,\u201d \u201ccarbon monoxide,\u201d \u201cmethane,\u201d \u201cacetylene,\u201d \u201cethanol,\u201d \u201cmethanol,\u201d and \u201cisooctane.\u201d The second column contains the equations \u201cC (s) + O (g) right-facing arrow C O subscript 2 (g),\u201d \u201cH subscript 2 (g) + one half O subscript 2 (g) right-facing arrow H subscript 2 O (l),\u201d \u201cM g (s) + one half O subscript 2 (g) right-facing arrow M g O (s),\u201d \u201cS (s) + O subscript 2 (g) right-facing arrow S O subscript 2 (g),\u201d \u201cC O (g) + one half O subscript 2 (g) right-facing arrow C O subscript 2 (g),\u201d \u201cC H subscript 4 (g) + 2 O subscript 2 (g) right-facing arrow C O subscript 2 (g) + 2 H subscript 2 O (g),\u201d \u201cC subscript 2 H subscript 2 (g) + five halves O subscript 2 (g) right-facing arrow 2 C O subscript 2 (g) + H subscript 2 O (l),\u201d \u201cC subscript 2 H subscript 5 O H (l) + 2 O subscript 2 (g) right-facing arrow C O subscript 2 (g) + 3 H subscript 2 O (l),\u201d \u201cC H subscript 3 O H (l) + three halves O subscript 2 (g) right-facing arrow C O subscript 2 (g) + 2 H subscript 2 O (l),\u201d and \u201cC subscript 8 H subscript 18 (l) + twenty five halves O subscript 2 (g) right-facing arrow 8 C O subscript 2 (g) + 9 H subscript 2 O (l).\u201d The final column contains the values \u201c\u2013393.5,\u201d \u201c\u2013285.8,\u201d \u201c\u2013601.6,\u201d \u201c\u2013296.8,\u201d \u201c\u2013283.0,\u201d \u201c\u2013890.8,\u201d \u201c\u20131301.1,\u201d \u201c\u20131366.8,\u201d \u201c\u2013726.1,\u201d and \u201c\u20135461.\u201d\">\n<thead>\n<tr style=\"height: 15px\">\n<th style=\"height: 15px;width: 541.85px\" colspan=\"3\" data-align=\"center\">Standard Molar Enthalpies of Combustion<\/th>\n<\/tr>\n<tr style=\"height: 93px\" valign=\"middle\">\n<th style=\"height: 93px;width: 76.5px\" data-align=\"left\">Substance<\/th>\n<th style=\"height: 93px;width: 272.7px\" data-align=\"left\">Combustion Reaction<\/th>\n<th style=\"height: 93px;width: 164.75px\" data-align=\"left\">Enthalpy of Combustion, <img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1251\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3o.png\" alt=\"\" width=\"126\" height=\"26\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3o.png 291w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3o-65x13.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3o-225x46.png 225w\" sizes=\"auto, (max-width: 126px) 100vw, 126px\" \/><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 30px\" valign=\"middle\">\n<td style=\"height: 30px;width: 77px\" data-align=\"left\">carbon<\/td>\n<td style=\"height: 30px;width: 273.7px\" data-align=\"left\">C(<em>s<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u27f6 CO<sub>2<\/sub>(<em>g<\/em>)<\/td>\n<td style=\"height: 30px;width: 165.25px\" data-align=\"left\">\u2212393.5<\/td>\n<\/tr>\n<tr style=\"height: 16px\" valign=\"middle\">\n<td style=\"height: 16px;width: 77px\" data-align=\"left\">hydrogen<\/td>\n<td style=\"height: 16px;width: 273.7px\" data-align=\"left\">H<sub>2<\/sub>(<em>g<\/em>) + (1\/2)O<sub>2<\/sub>(<em>g<\/em>) \u27f6 H<sub>2<\/sub>O(<em>l<\/em>)<\/td>\n<td style=\"height: 16px;width: 165.25px\" data-align=\"left\">\u2212285.8<\/td>\n<\/tr>\n<tr style=\"height: 16px\" valign=\"middle\">\n<td style=\"height: 16px;width: 77px\" data-align=\"left\">magnesium<\/td>\n<td style=\"height: 16px;width: 273.7px\" data-align=\"left\">Mg(<em>s<\/em>) + (1\/2)O<sub>2<\/sub>(<em>g<\/em>) \u27f6 MgO(<em>s<\/em>)<\/td>\n<td style=\"height: 16px;width: 165.25px\" data-align=\"left\">\u2212601.6<\/td>\n<\/tr>\n<tr style=\"height: 16px\" valign=\"middle\">\n<td style=\"height: 16px;width: 77px\" data-align=\"left\">sulfur<\/td>\n<td style=\"height: 16px;width: 273.7px\" data-align=\"left\">S(<em>s<\/em>) + O<sub>2<\/sub>(<em>g<\/em>) \u27f6 SO<sub>2<\/sub>(<em>g<\/em>)<\/td>\n<td style=\"height: 16px;width: 165.25px\" data-align=\"left\">\u2212296.8<\/td>\n<\/tr>\n<tr style=\"height: 31px\" valign=\"middle\">\n<td style=\"height: 31px;width: 77px\" data-align=\"left\">carbon monoxide<\/td>\n<td style=\"height: 31px;width: 273.7px\" data-align=\"left\">CO(<em>g<\/em>) + (1\/2)O<sub>2<\/sub>(<em>g<\/em>) \u27f6 CO<sub>2<\/sub>(<em>g<\/em>)<\/td>\n<td style=\"height: 31px;width: 165.25px\" data-align=\"left\">\u2212283.0<\/td>\n<\/tr>\n<tr style=\"height: 16px\" valign=\"middle\">\n<td style=\"height: 16px;width: 77px\" data-align=\"left\">methane<\/td>\n<td style=\"height: 16px;width: 273.7px\" data-align=\"left\">CH<sub>4<\/sub>(<em>g<\/em>) + 2O<sub>2<\/sub>(<em>g<\/em>) \u27f6 CO<sub>2<\/sub>(<em>g<\/em>) + 2H<sub>2<\/sub>O(<em>l<\/em>)<\/td>\n<td style=\"height: 16px;width: 165.25px\" data-align=\"left\">\u2212890.8<\/td>\n<\/tr>\n<tr style=\"height: 16px\" valign=\"middle\">\n<td style=\"height: 16px;width: 77px\" data-align=\"left\">acetylene<\/td>\n<td style=\"height: 16px;width: 273.7px\" data-align=\"left\">C<sub>2<\/sub>H<sub>2<\/sub>(<em>g<\/em>) + (5\/2)O<sub>2<\/sub>(<em>g<\/em>) \u27f6 2CO<sub>2<\/sub>(<em>g<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>)<\/td>\n<td style=\"height: 16px;width: 165.25px\" data-align=\"left\">\u22121301.1<\/td>\n<\/tr>\n<tr style=\"height: 16px\" valign=\"middle\">\n<td style=\"height: 16px;width: 77px\" data-align=\"left\">ethanol<\/td>\n<td style=\"height: 16px;width: 273.7px\" data-align=\"left\">C<sub>2<\/sub>H<sub>5<\/sub>OH(<em>l<\/em>) + 3O<sub>2<\/sub>(<em>g<\/em>)\u27f6 2CO<sub>2<\/sub>(<em>g<\/em>) + 3H<sub>2<\/sub>O(<em>l<\/em>)<\/td>\n<td style=\"height: 16px;width: 165.25px\" data-align=\"left\">\u22121366.8<\/td>\n<\/tr>\n<tr style=\"height: 21px\" valign=\"middle\">\n<td style=\"height: 21px;width: 77px\" data-align=\"left\">methanol<\/td>\n<td style=\"height: 21px;width: 273.7px\" data-align=\"left\">CH<sub>3<\/sub>OH(<em>l<\/em>) + (3\/2)O<sub>2<\/sub>(<em>g<\/em>)\u27f6 CO<sub>2<\/sub>(<em>g<\/em>) + 2H<sub>2<\/sub>O(<em>l<\/em>)<\/td>\n<td style=\"height: 21px;width: 165.25px\" data-align=\"left\">\u2212726.1<\/td>\n<\/tr>\n<tr style=\"height: 25px\" valign=\"middle\">\n<td style=\"height: 25px;width: 77px\" data-align=\"left\">isooctane<\/td>\n<td style=\"height: 25px;width: 273.7px\" data-align=\"left\">C<sub>8<\/sub>H<sub>18<\/sub>(<em>l<\/em>) + (25\/2)O<sub>2<\/sub>(<em>g<\/em>)\u27f6 8CO<sub>2<\/sub>(<em>g<\/em>) + 9H<sub>2<\/sub>O(<em>l<\/em>)<\/td>\n<td style=\"height: 25px;width: 165.25px\" data-align=\"left\">\u22125460<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-idp90124816\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp186633648\"><strong>Using Enthalpy of Combustion:<\/strong><\/p>\n<p>As <a class=\"autogenerated-content\" href=\"#CNX_Chem_05_03_GasBurning\">(Figure)<\/a> suggests, the combustion of gasoline is a highly exothermic process. Let us determine the approximate amount of heat produced by burning 1.00 L of gasoline, assuming the enthalpy of combustion of gasoline is the same as that of isooctane, a common component of gasoline. The density of isooctane is 0.692 g\/mL.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_05_03_GasBurning\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">The combustion of gasoline is very exothermic. (credit: modification of work by \u201cAlexEagle\u201d\/Flickr)<\/div>\n<p><span id=\"fs-idm3605664\" data-type=\"media\" data-alt=\"A picture shows a large ball of fire burning on a road. A fire truck and fireman are shown in the foreground.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_05_03_GasBurning-1.jpg\" alt=\"A picture shows a large ball of fire burning on a road. A fire truck and fireman are shown in the foreground.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idp177773984\"><strong>Solution:<\/strong><\/p>\n<p>Starting with a known amount (1.00 L of isooctane), we can perform conversions between units until we arrive at the desired amount of heat or energy. The enthalpy of combustion of isooctane provides one of the necessary conversions. <a class=\"autogenerated-content\" href=\"#fs-idp98710048\">(Figure)<\/a> gives this value as \u22125460 kJ per 1 mole of isooctane (C<sub>8<\/sub>H<sub>18<\/sub>).<\/p>\n<p id=\"fs-idp6149360\">Using these data,<\/p>\n<div id=\"fs-idp6149744\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1253 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3p-300x24.png\" alt=\"\" width=\"525\" height=\"42\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3p-300x24.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3p-1024x83.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3p-768x62.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3p-1536x124.png 1536w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3p-65x5.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3p-225x18.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3p-350x28.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3p.png 1642w\" sizes=\"auto, (max-width: 525px) 100vw, 525px\" \/><\/div>\n<p id=\"fs-idp145151008\">The combustion of 1.00 L of isooctane produces 33,100 kJ of heat. (This amount of energy is enough to melt 99.2 kg, or about 218 lb, of ice.)<\/p>\n<p id=\"fs-idp94999600\">Note: If you do this calculation one step at a time, you would find:<\/p>\n<div id=\"fs-idp94999984\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1254 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3q-300x106.png\" alt=\"\" width=\"300\" height=\"106\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3q-300x106.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3q-65x23.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3q-225x79.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3q-350x123.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3q.png 634w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp49733232\"><strong>Check Your Learning:<\/strong><\/p>\n<p>How much heat is produced by the combustion of 125 g of acetylene?<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp49733872\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp49734384\">6.25 \u00d7 10<sup>3<\/sup> kJ<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp90130208\" class=\"chemistry everyday-life\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div data-type=\"title\"><strong>Emerging Algae-Based Energy Technologies (Biofuels)<\/strong><\/div>\n<p id=\"fs-idp90130976\">As reserves of fossil fuels diminish and become more costly to extract, the search is ongoing for replacement fuel sources for the future. Among the most promising biofuels are those derived from algae (<a class=\"autogenerated-content\" href=\"#CNX_Chem_05_03_AlgalFuel1\">(Figure)<\/a>). The species of algae used are nontoxic, biodegradable, and among the world\u2019s fastest growing organisms. About 50% of algal weight is oil, which can be readily converted into fuel such as biodiesel. Algae can yield 26,000 gallons of <span class=\"no-emphasis\" data-type=\"term\">biofuel<\/span> per hectare\u2014much more energy per acre than other crops. Some strains of algae can flourish in brackish water that is not usable for growing other crops. Algae can produce biodiesel, biogasoline, ethanol, butanol, methane, and even jet fuel.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_05_03_AlgalFuel1\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">(a) Tiny algal organisms can be (b) grown in large quantities and eventually (c) turned into a useful fuel such as biodiesel. (credit a: modification of work by Micah Sittig; credit b: modification of work by Robert Kerton; credit c: modification of work by John F. Williams)<\/div>\n<p><span id=\"fs-idp157749040\" data-type=\"media\" data-alt=\"Three pictures are shown and labeled a, b, and c. Picture a shows a microscopic view of algal organisms. They are brown, multipart strands and net-like structures on a background of light violet. Picture b shows five large tubs full of a brown liquid containing these algal organisms. Picture c depicts a cylinder full of green liquid in the foreground and a poster in the background that has the title \u201cFrom Field to Fleet.\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_05_03_AlgalFuel1-1.jpg\" alt=\"Three pictures are shown and labeled a, b, and c. Picture a shows a microscopic view of algal organisms. They are brown, multipart strands and net-like structures on a background of light violet. Picture b shows five large tubs full of a brown liquid containing these algal organisms. Picture c depicts a cylinder full of green liquid in the foreground and a poster in the background that has the title \u201cFrom Field to Fleet.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idp157750544\">According to the US Department of Energy, only 39,000 square kilometers (about 0.4% of the land mass of the US or less than 1\/7 of the area used to grow corn) can produce enough algal fuel to replace all the petroleum-based fuel used in the US. The cost of algal fuels is becoming more competitive\u2014for instance, the US Air Force is producing jet fuel from algae at a total cost of under $5 per gallon.<sup data-type=\"footnote-number\"><a href=\"#footnote1\" data-type=\"footnote-link\">1<\/a><\/sup> The process used to produce algal fuel is as follows: grow the algae (which use sunlight as their energy source and CO<sub>2<\/sub> as a raw material); harvest the algae; extract the fuel compounds (or precursor compounds); process as necessary (e.g., perform a transesterification reaction to make biodiesel); purify; and distribute (<a class=\"autogenerated-content\" href=\"#CNX_Chem_05_03_AlgalFuel2\">(Figure)<\/a>).<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_05_03_AlgalFuel2\" class=\"bc-figure figure\">\n<div class=\"bc-figcaption figcaption\">Algae convert sunlight and carbon dioxide into oil that is harvested, extracted, purified, and transformed into a variety of renewable fuels.<\/div>\n<p><span id=\"fs-idp167165312\" data-type=\"media\" data-alt=\"A flowchart is shown that contains pictures and words. Reading from left to right, the terms \u201cGrow,\u201d \u201cHarvest,\u201d \u201cExtract,\u201d \u201cProcess and purify,\u201d and \u201cJet fuel gasoline diesel\u201d are shown with right-facing arrows in between each. Above each term, respectively, are diagrams of three containers, three cylinders lying side-by-side, a pyramid-like container with liquid inside, a factory, and a fuel pump. In the space above all of the diagrams and to the left of the images is a diagram of the sun.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_05_03_AlgalFuel2-1.jpg\" alt=\"A flowchart is shown that contains pictures and words. Reading from left to right, the terms \u201cGrow,\u201d \u201cHarvest,\u201d \u201cExtract,\u201d \u201cProcess and purify,\u201d and \u201cJet fuel gasoline diesel\u201d are shown with right-facing arrows in between each. Above each term, respectively, are diagrams of three containers, three cylinders lying side-by-side, a pyramid-like container with liquid inside, a factory, and a fuel pump. In the space above all of the diagrams and to the left of the images is a diagram of the sun.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp93919792\" class=\"chemistry link-to-learning\" data-type=\"note\">\n<p id=\"fs-idp93921200\">Click <a href=\"http:\/\/openstaxcollege.org\/l\/16biofuel\">here<\/a> to learn more about the process of creating algae biofuel.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp93922352\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Standard Enthalpy of Formation<\/strong><\/h3>\n<p id=\"fs-idp93922992\">A <span data-type=\"term\"><strong>standard enthalpy of formation<\/strong>, \u0394<em>H<\/em><sub>f<\/sub>\u00b0,<\/span>\u00a0is an enthalpy change for a reaction in which exactly 1 mole of a pure substance is formed from free elements in their most stable states under standard state conditions. These values are especially useful for computing or predicting enthalpy changes for chemical reactions that are impractical or dangerous to carry out, or for processes for which it is difficult to make measurements. If we have values for the appropriate standard enthalpies of formation, we can determine the enthalpy change for any reaction, which we will practice in the next section on Hess\u2019s law.<\/p>\n<p id=\"fs-idp185900096\">The standard enthalpy of formation of CO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) is \u2212393.5 kJ\/mol. This is the enthalpy change for the exothermic reaction:<\/p>\n<div id=\"fs-idp185901552\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1255 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3r-300x27.png\" alt=\"\" width=\"422\" height=\"38\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3r-300x27.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3r-768x69.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3r-65x6.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3r-225x20.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3r-350x31.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3r.png 937w\" sizes=\"auto, (max-width: 422px) 100vw, 422px\" \/><\/div>\n<p id=\"fs-idp165684992\">starting with the reactants at a pressure of 1 atm and 25 \u00b0C (with the carbon present as graphite, the most stable form of carbon under these conditions) and ending with one mole of CO<sub>2<\/sub>, also at 1 atm and 25 \u00b0C. For nitrogen dioxide, NO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>), <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span> is 33.2 kJ\/mol. This is the enthalpy change for the reaction:<\/p>\n<div id=\"fs-idp185944512\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1256 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3s-300x32.png\" alt=\"\" width=\"441\" height=\"47\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3s-300x32.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3s-768x82.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3s-65x7.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3s-225x24.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3s-350x37.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3s.png 951w\" sizes=\"auto, (max-width: 441px) 100vw, 441px\" \/><\/div>\n<p id=\"fs-idp173428336\">A reaction equation with \u00bd mole of N<sub>2<\/sub> and 1 mole of O<sub>2<\/sub> is correct in this case because the standard enthalpy of formation always refers to 1 mole of product, NO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>).<\/p>\n<p id=\"fs-idp173432432\">You will find a table of standard enthalpies of formation of many common substances in <a class=\"target-chapter\" href=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/back-matter\/standard-thermodynamic-properties-for-selected-substances\/\">Appendix G<\/a>. These values indicate that formation reactions range from highly exothermic (such as \u22122984 kJ\/mol for the formation of P<sub>4<\/sub>O<sub>10<\/sub>) to strongly endothermic (such as +226.7 kJ\/mol for the formation of acetylene, C<sub>2<\/sub>H<sub>2<\/sub>). By definition, the standard enthalpy of formation of an element in its most stable form is equal to zero under standard conditions, which is 1 atm for gases and 1 M for solutions.<\/p>\n<div id=\"fs-idp167759616\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp167759872\"><strong>Evaluating an Enthalpy of Formation:<\/strong><\/p>\n<p>Ozone, O<sub>3<\/sub>(<em data-effect=\"italics\">g<\/em>), forms from oxygen, O<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>), by an endothermic process. Ultraviolet radiation is the source of the energy that drives this reaction in the upper atmosphere. Assuming that both the reactants and products of the reaction are in their standard states, determine the standard enthalpy of formation, <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0,<\/span>\u00a0of ozone from the following information:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1273 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3t-300x23.png\" alt=\"\" width=\"352\" height=\"27\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3t-300x23.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3t-65x5.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3t-225x17.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3t-350x26.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3t.png 676w\" sizes=\"auto, (max-width: 352px) 100vw, 352px\" \/><\/p>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp166467632\"><strong>Solution:<\/strong><\/p>\n<p><span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span> is the enthalpy change for the formation of one mole of a substance in its standard state from the elements in their standard states. Thus, <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0 <\/span>for O<sub>3<\/sub>(<em data-effect=\"italics\">g<\/em>) is the enthalpy change for the reaction:<\/p>\n<div id=\"fs-idp166474448\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1274 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3u.png\" alt=\"\" width=\"111\" height=\"34\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3u.png 301w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3u-65x20.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3u-225x69.png 225w\" sizes=\"auto, (max-width: 111px) 100vw, 111px\" \/><\/div>\n<p id=\"fs-idp95008768\">For the formation of 2 mol of O<sub>3<\/sub>(<em data-effect=\"italics\">g<\/em>), \u0394<em>H<\/em>\u00b0 = +286 kJ. This ratio, <img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1275\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3v.png\" alt=\"\" width=\"55\" height=\"25\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3v.png 147w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3v-65x30.png 65w\" sizes=\"auto, (max-width: 55px) 100vw, 55px\" \/>, can be used as a conversion factor to find the heat produced when 1 mole of O<sub>3<\/sub>(<em data-effect=\"italics\">g<\/em>) is formed, which is the enthalpy of formation for O<sub>3<\/sub>(<em data-effect=\"italics\">g<\/em>):<\/p>\n<div id=\"fs-idp90148272\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1276 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3x-300x34.png\" alt=\"\" width=\"318\" height=\"36\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3x-300x34.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3x-768x88.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3x-65x7.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3x-225x26.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3x-350x40.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3x.png 931w\" sizes=\"auto, (max-width: 318px) 100vw, 318px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp186495680\">Therefore, <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span>[O<sub>3<\/sub>(<em>g<\/em>)]= +143 kJ\/mol.<\/p>\n<p id=\"fs-idp167644640\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Hydrogen gas, H<sub>2<\/sub>, reacts explosively with gaseous chlorine, Cl<sub>2<\/sub>, to form hydrogen chloride, HCl(<em data-effect=\"italics\">g<\/em>). What is the enthalpy change for the reaction of 1 mole of H<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) with 1 mole of Cl<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) if both the reactants and products are at standard state conditions? The standard enthalpy of formation of HCl(<em data-effect=\"italics\">g<\/em>) is \u221292.3 kJ\/mol.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp167648992\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp167649504\">For the reaction <img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1277\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3y-300x19.png\" alt=\"\" width=\"300\" height=\"19\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3y-300x19.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3y-768x50.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3y-65x4.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3y-225x15.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3y-350x23.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3y.png 817w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp186597312\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp186597568\"><strong>Writing Reaction Equations for \u0394<em>H<\/em><sub>f<\/sub>\u00b0:<\/strong><\/p>\n<p>\u00a0Write the heat of formation reaction equations for:<\/p>\n<p id=\"fs-idp186600640\">(a) C<sub>2<\/sub>H<sub>5<\/sub>OH(<em data-effect=\"italics\">l<\/em>)<\/p>\n<p id=\"fs-idp186602304\">(b) Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>(<em data-effect=\"italics\">s<\/em>)<\/p>\n<p id=\"fs-idp186604352\"><strong>Solution:<\/strong><\/p>\n<p>Remembering that <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span> reaction equations are for forming 1 mole of the compound from its constituent elements under standard conditions, we have:<\/p>\n<p id=\"fs-idp166486208\">(a) 2C(<em>s,graphite<\/em>) + 3H<sub>2<\/sub>(<em>g<\/em>) + (1\/2)O<sub>2<\/sub>(<em>g<\/em>) \u27f6 C<sub>2<\/sub>H<sub>5<\/sub>OH(<em data-effect=\"italics\">l<\/em>)<\/p>\n<p id=\"fs-idp145132528\">(b)\u00a0 3Ca(<em>s<\/em>) + (1\/2)P<sub>4<\/sub>(<em>s<\/em>) + 4O<sub>2<\/sub>(<em>g<\/em>) \u27f6 Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>(<em data-effect=\"italics\">s<\/em>)<\/p>\n<p id=\"fs-idp166524800\">Note: The standard state of carbon is graphite, and phosphorus exists as P<sub>4<\/sub>.<\/p>\n<p id=\"fs-idp166525568\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Write the heat of formation reaction equations for:<\/p>\n<p id=\"fs-idp166526208\">(a) C<sub>2<\/sub>H<sub>5<\/sub>OC<sub>2<\/sub>H<sub>5<\/sub>(<em data-effect=\"italics\">l<\/em>)<\/p>\n<p id=\"fs-idp185932768\">(b) Na<sub>2<\/sub>CO<sub>3<\/sub>(<em data-effect=\"italics\">s<\/em>)<\/p>\n<div id=\"fs-idp185934432\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp185934944\">(a) 4C(<em>s,graphite<\/em>) + 5H<sub>2<\/sub>(<em>g<\/em>) + (1\/2)O<sub>2<\/sub>(<em>g<\/em>) \u27f6 C<sub>2<\/sub>H<sub>5<\/sub>OC<sub>2<\/sub>H<sub>5<\/sub>(<em data-effect=\"italics\">l<\/em>)<\/p>\n<p>(b) 2Na(<em>s<\/em>) + C(<em>s,graphite<\/em>)+ (3\/2)O<sub>2<\/sub>(<em>g<\/em>) \u27f6 Na<sub>2<\/sub>CO<sub>3<\/sub>(<em data-effect=\"italics\">s<\/em>)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idp218911984\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Hess\u2019s Law<\/strong><\/h3>\n<p id=\"fs-idp218912624\">There are two ways to determine the amount of heat involved in a chemical change: measure it experimentally, or calculate it from other experimentally determined enthalpy changes. Some reactions are difficult, if not impossible, to investigate and make accurate measurements for experimentally. And even when a reaction is not hard to perform or measure, it is convenient to be able to determine the heat involved in a reaction without having to perform an experiment.<\/p>\n<p id=\"fs-idp177508528\">This type of calculation usually involves the use of <span data-type=\"term\">Hess\u2019s law<\/span>, which states: <em data-effect=\"italics\">If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps<\/em>. Hess\u2019s law is valid because enthalpy is a state function: Enthalpy changes depend only on where a chemical process starts and ends, but not on the path it takes from start to finish. For example, we can think of the reaction of carbon with oxygen to form carbon dioxide as occurring either directly or by a two-step process. The direct process is written:<\/p>\n<div id=\"fs-idp177510624\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1280 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3aa-300x34.png\" alt=\"\" width=\"300\" height=\"34\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3aa-300x34.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3aa-768x87.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3aa-65x7.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3aa-225x26.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3aa-350x40.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3aa.png 775w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<p id=\"fs-idp51924752\">In the two-step process, first carbon monoxide is formed:<\/p>\n<div id=\"fs-idp51925136\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-1281 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ab-300x35.png\" alt=\"\" width=\"300\" height=\"35\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ab-300x35.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ab-768x89.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ab-65x8.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ab-225x26.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ab-350x41.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ab.png 776w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<p id=\"fs-idp161438416\">Then, carbon monoxide reacts further to form carbon dioxide:<\/p>\n<div id=\"fs-idp161438800\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-1283 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ac-300x36.png\" alt=\"\" width=\"300\" height=\"36\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ac-300x36.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ac-768x93.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ac-65x8.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ac-225x27.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ac-350x43.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ac.png 814w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<p id=\"fs-idp186572000\">The equation describing the overall reaction is the sum of these two chemical changes:<\/p>\n<div id=\"fs-idp186572384\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-1282 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ad-300x61.png\" alt=\"\" width=\"300\" height=\"61\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ad-300x61.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ad-768x157.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ad-65x13.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ad-225x46.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ad-350x72.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ad.png 953w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<p id=\"fs-idp40118928\">Because the CO produced in Step 1 is consumed in Step 2, the net change is:<\/p>\n<div id=\"fs-idp40119312\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1284 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ae-300x61.png\" alt=\"\" width=\"172\" height=\"35\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ae-300x61.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ae-65x13.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ae-225x46.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ae-350x71.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ae.png 404w\" sizes=\"auto, (max-width: 172px) 100vw, 172px\" \/><\/div>\n<p id=\"fs-idp128370320\">According to Hess\u2019s law, the enthalpy change of the reaction will equal the sum of the enthalpy changes of the steps.<\/p>\n<div id=\"fs-idp128371600\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-1285 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3af-300x69.png\" alt=\"\" width=\"300\" height=\"69\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3af-300x69.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3af-65x15.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3af-225x52.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3af-350x81.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3af.png 705w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<p id=\"fs-idp10365136\">The result is shown in <a class=\"autogenerated-content\" href=\"#CNX_Chem_05_03_HessCO2\">(Figure)<\/a>. We see that \u0394<em data-effect=\"italics\">H<\/em> of the overall reaction is the same whether it occurs in one step or two. This finding (overall \u0394<em data-effect=\"italics\">H<\/em> for the reaction = sum of \u0394<em data-effect=\"italics\">H<\/em> values for reaction \u201csteps\u201d in the overall reaction) is true in general for chemical and physical processes.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_05_03_HessCO2\" class=\"bc-figure figure\">\n<div class=\"bc-figcaption figcaption\">The formation of CO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) from its elements can be thought of as occurring in two steps, which sum to the overall reaction, as described by Hess\u2019s law. The horizontal blue lines represent enthalpies. For an exothermic process, the products are at lower enthalpy than are the reactants.<\/div>\n<p><span id=\"fs-idp167194048\" data-type=\"media\" data-alt=\"A diagram is shown. A long arrow faces upward on the left with the phrase \u201cH increasing.\u201d A horizontal line at the bottom of the diagram is shown with the formula \u201cC O subscript 2 (g)\u201d below it. A horizontal line at the top of the diagram has the formulas \u201cC (s) + O subscript 2 (g)\u201d above it. The top and bottom lines are connected by a downward facing arrow with the value \u201c\u0394 H = \u2013394 k J\u201d written beside it. Below and to the right of the top horizontal line is a second horizontal line with the equations \u201cC O (g) + one half O subscript 2 (g)\u201d above it. This line and the bottom line are connected by a downward facing arrow with the value \u201c\u0394 H = \u2013283 k J\u201d written beside it. The same line and the top line are connected by a downward facing arrow with the value \u201c\u0394 H = \u2013111 k J\u201d written beside it. There are three brackets to the right of the diagram. The first bracket runs from the top horizontal line to the second horizontal line. It is labeled, \u201cEnthalpy of reactants.\u201d The second bracket runs from the second horizontal line to the bottom horizontal line. It is labeled, \u201cEnthalpy of products.\u201d Both of these brackets are included in the third bracket which runs from the top to the bottom of the diagram. It is labeled, \u201cEnthalpy change of exothermic reaction in 1 or 2 steps.\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_05_03_HessCO2-1.jpg\" alt=\"A diagram is shown. A long arrow faces upward on the left with the phrase \u201cH increasing.\u201d A horizontal line at the bottom of the diagram is shown with the formula \u201cC O subscript 2 (g)\u201d below it. A horizontal line at the top of the diagram has the formulas \u201cC (s) + O subscript 2 (g)\u201d above it. The top and bottom lines are connected by a downward facing arrow with the value \u201c\u0394 H = \u2013394 k J\u201d written beside it. Below and to the right of the top horizontal line is a second horizontal line with the equations \u201cC O (g) + one half O subscript 2 (g)\u201d above it. This line and the bottom line are connected by a downward facing arrow with the value \u201c\u0394 H = \u2013283 k J\u201d written beside it. The same line and the top line are connected by a downward facing arrow with the value \u201c\u0394 H = \u2013111 k J\u201d written beside it. There are three brackets to the right of the diagram. The first bracket runs from the top horizontal line to the second horizontal line. It is labeled, \u201cEnthalpy of reactants.\u201d The second bracket runs from the second horizontal line to the bottom horizontal line. It is labeled, \u201cEnthalpy of products.\u201d Both of these brackets are included in the third bracket which runs from the top to the bottom of the diagram. It is labeled, \u201cEnthalpy change of exothermic reaction in 1 or 2 steps.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idp167197264\">Before we further practice using Hess\u2019s law, let us recall two important features of \u0394<em data-effect=\"italics\">H<\/em>.<\/p>\n<ol id=\"fs-idp167198464\" type=\"1\">\n<li>\n<p id=\"fs-idp167199328\">\u0394<em data-effect=\"italics\">H<\/em> is directly proportional to the quantities of reactants or products. For example, the enthalpy change for the reaction forming 1 mole of NO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) is +33.2 kJ:<\/p>\n<div id=\"fs-idp167201504\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1286 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ba-300x41.png\" alt=\"\" width=\"315\" height=\"43\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ba-300x41.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ba-768x106.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ba-65x9.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ba-225x31.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ba-350x48.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ba.png 805w\" sizes=\"auto, (max-width: 315px) 100vw, 315px\" \/><\/div>\n<p id=\"fs-idp160715680\">When 2 moles of NO<sub>2<\/sub> (twice as much) are formed, the \u0394<em data-effect=\"italics\">H<\/em> will be twice as large:<\/p>\n<div id=\"fs-idp160717072\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1287 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bb-300x27.png\" alt=\"\" width=\"322\" height=\"29\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bb-300x27.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bb-768x68.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bb-65x6.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bb-225x20.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bb-350x31.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bb.png 826w\" sizes=\"auto, (max-width: 322px) 100vw, 322px\" \/><\/div>\n<p id=\"fs-idp146331072\">In general, if we multiply or divide an equation by a number, then the enthalpy change should also be multiplied or divided by the same number.<\/p>\n<\/li>\n<li>\n<p id=\"fs-idp146331744\">\u0394<em data-effect=\"italics\">H<\/em> for a reaction in one direction is equal in magnitude and opposite in sign to \u0394<em data-effect=\"italics\">H<\/em> for the reaction in the reverse direction. For example, given that:<\/p>\n<div id=\"fs-idp146333552\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1288 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bc-300x24.png\" alt=\"\" width=\"313\" height=\"25\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bc-300x24.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bc-768x61.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bc-65x5.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bc-225x18.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bc-350x28.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bc.png 817w\" sizes=\"auto, (max-width: 313px) 100vw, 313px\" \/><\/div>\n<p id=\"fs-idp146344672\">Then, for the \u201creverse\u201d reaction, the enthalpy change is also \u201creversed\u201d:<\/p>\n<div id=\"fs-idp146345216\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1289 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bd-300x24.png\" alt=\"\" width=\"325\" height=\"26\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bd-300x24.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bd-768x62.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bd-65x5.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bd-225x18.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bd-350x28.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3bd.png 806w\" sizes=\"auto, (max-width: 325px) 100vw, 325px\" \/><\/div>\n<\/li>\n<\/ol>\n<div id=\"fs-idp185548816\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp185549072\"><strong>Stepwise Calculation of \u0394<em>H<\/em><sub>f<\/sub>\u00b0 Using Hess\u2019s Law:<\/strong><\/p>\n<p>Determine the enthalpy of formation, <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span>, of FeCl<sub>3<\/sub>(<em data-effect=\"italics\">s<\/em>) from the enthalpy changes of the following two-step process that occurs under standard state conditions:<\/p>\n<div id=\"fs-idp103964560\" style=\"text-align: center\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1290\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ca-300x63.png\" alt=\"\" width=\"348\" height=\"73\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ca-300x63.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ca-768x160.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ca-65x14.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ca-225x47.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ca-350x73.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ca.png 900w\" sizes=\"auto, (max-width: 348px) 100vw, 348px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp190748768\"><strong>Solution:<\/strong><\/p>\n<p>We are trying to find the standard enthalpy of formation of FeCl<sub>3<\/sub>(<em data-effect=\"italics\">s<\/em>), which is equal to \u0394<em data-effect=\"italics\">H<\/em>\u00b0 for the reaction:<\/p>\n<div id=\"fs-idp190751040\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1291 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cb-300x43.png\" alt=\"\" width=\"300\" height=\"43\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cb-300x43.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cb-65x9.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cb-225x33.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cb-350x51.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cb.png 732w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp190763840\">Looking at the reactions, we see that the reaction for which we want to find \u0394<em data-effect=\"italics\">H<\/em>\u00b0 is the sum of the two reactions with known \u0394<em data-effect=\"italics\">H<\/em> values, so we must sum their \u0394<em data-effect=\"italics\">H<\/em>s:<\/p>\n<div id=\"fs-idp133790064\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1292 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cc-300x60.png\" alt=\"\" width=\"325\" height=\"65\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cc-300x60.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cc-768x153.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cc-65x13.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cc-225x45.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cc-350x70.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cc.png 784w\" sizes=\"auto, (max-width: 325px) 100vw, 325px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp149506592\">The enthalpy of formation, <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span>, of FeCl<sub>3<\/sub>(<em data-effect=\"italics\">s<\/em>) is \u2212399.5 kJ\/mol.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp149511040\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Calculate \u0394<em data-effect=\"italics\">H<\/em> for the process:<\/p>\n<div id=\"fs-idp149512304\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1293 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cd-300x44.png\" alt=\"\" width=\"218\" height=\"32\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cd-300x44.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cd-65x10.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cd-225x33.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cd-350x51.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3cd.png 476w\" sizes=\"auto, (max-width: 218px) 100vw, 218px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp164462848\">from the following information:<\/p>\n<div id=\"fs-idp164463232\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1294 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ce-300x64.png\" alt=\"\" width=\"356\" height=\"76\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ce-300x64.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ce-768x164.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ce-65x14.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ce-225x48.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ce-350x75.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ce.png 863w\" sizes=\"auto, (max-width: 356px) 100vw, 356px\" \/><\/div>\n<div id=\"fs-idp146898608\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp146899120\">66.4 kJ<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idp146899760\">Here is a less straightforward example that illustrates the thought process involved in solving many Hess\u2019s law problems. It shows how we can find many standard enthalpies of formation (and other values of \u0394<em data-effect=\"italics\">H<\/em>) if they are difficult to determine experimentally.<\/p>\n<div id=\"fs-idp146901200\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp146901456\"><strong>A More Challenging Problem Using Hess\u2019s Law:<\/strong><\/p>\n<p>Chlorine monofluoride can react with fluorine to form chlorine trifluoride:<\/p>\n<p id=\"fs-idp146902208\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1296\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3da-300x32.png\" alt=\"\" width=\"300\" height=\"32\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3da-300x32.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3da-65x7.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3da-225x24.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3da-350x37.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3da.png 698w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p id=\"fs-idp146912304\">Use the reactions here to determine the \u0394<em data-effect=\"italics\">H<\/em>\u00b0 for reaction <em data-effect=\"italics\">(i)<\/em>:<\/p>\n<p id=\"fs-idp146913936\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1297\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3db-300x53.png\" alt=\"\" width=\"413\" height=\"73\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3db-300x53.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3db-1024x182.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3db-768x137.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3db-65x12.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3db-225x40.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3db-350x62.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3db.png 1125w\" sizes=\"auto, (max-width: 413px) 100vw, 413px\" \/><\/p>\n<p id=\"fs-idp167247104\"><strong>Solution:<\/strong><\/p>\n<p>Our goal is to manipulate and combine reactions <em data-effect=\"italics\">(ii)<\/em>, <em data-effect=\"italics\">(iii)<\/em>, and <em data-effect=\"italics\">(iv)<\/em> such that they add up to reaction <em data-effect=\"italics\">(i)<\/em>. Going from left to right in <em data-effect=\"italics\">(i)<\/em>, we first see that ClF(<em data-effect=\"italics\">g<\/em>) is needed as a reactant. This can be obtained by multiplying reaction <em data-effect=\"italics\">(iii)<\/em> by \u00bd, which means that the \u0394<em data-effect=\"italics\">H<\/em>\u00b0 change is also multiplied by \u00bd:<\/p>\n<div id=\"fs-idp167256800\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1298 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dc-300x21.png\" alt=\"\" width=\"443\" height=\"31\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dc-300x21.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dc-1024x73.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dc-768x55.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dc-65x5.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dc-225x16.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dc-350x25.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dc.png 1279w\" sizes=\"auto, (max-width: 443px) 100vw, 443px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp160398800\">Next, we see that F<sub>2<\/sub> is also needed as a reactant. To get this, reverse and halve reaction <em data-effect=\"italics\">(ii)<\/em>, which means that the \u0394<em data-effect=\"italics\">H<\/em>\u00b0 changes sign and is halved:<\/p>\n<div id=\"fs-idp160400816\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1299 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dd-300x37.png\" alt=\"\" width=\"324\" height=\"40\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dd-300x37.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dd-768x94.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dd-65x8.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dd-225x27.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dd-350x43.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dd.png 819w\" sizes=\"auto, (max-width: 324px) 100vw, 324px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp134606352\">To get ClF<sub>3<\/sub> as a product, reverse <em data-effect=\"italics\">(iv)<\/em>, changing the sign of \u0394<em data-effect=\"italics\">H<\/em>\u00b0:<\/p>\n<div id=\"fs-idp134608368\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1300 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3de-300x28.png\" alt=\"\" width=\"343\" height=\"32\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3de-300x28.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3de-1024x95.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3de-768x71.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3de-65x6.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3de-225x21.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3de-350x32.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3de.png 1079w\" sizes=\"auto, (max-width: 343px) 100vw, 343px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp134626368\">Now check to make sure that these reactions add up to the reaction we want:<\/p>\n<div id=\"fs-idp134626752\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1301 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3df-300x62.png\" alt=\"\" width=\"334\" height=\"69\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3df-300x62.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3df-1024x210.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3df-768x158.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3df-65x13.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3df-225x46.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3df-350x72.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3df.png 1043w\" sizes=\"auto, (max-width: 334px) 100vw, 334px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp154356656\">Reactants \u00bdO<sub>2<\/sub> and \u00bdO<sub>2<\/sub> cancel out product O<sub>2<\/sub>; product \u00bdCl<sub>2<\/sub>O cancels reactant \u00bdCl<sub>2<\/sub>O; and reactant (3\/2)OF<sub>2<\/sub> is cancelled by products \u00bdOF<sub>2<\/sub> and OF<sub>2<\/sub>. This leaves only reactants ClF(<em data-effect=\"italics\">g<\/em>) and F<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) and product ClF<sub>3<\/sub>(<em data-effect=\"italics\">g<\/em>), which are what we want. Since summing these three modified reactions yields the reaction of interest, summing the three modified \u0394<em data-effect=\"italics\">H<\/em>\u00b0 values will give the desired \u0394<em data-effect=\"italics\">H<\/em>\u00b0:<\/p>\n<div id=\"fs-idm259960768\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1302 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dg-300x27.png\" alt=\"\" width=\"333\" height=\"30\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dg-300x27.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dg-768x68.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dg-65x6.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dg-225x20.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dg-350x31.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3dg.png 903w\" sizes=\"auto, (max-width: 333px) 100vw, 333px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp154380128\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Aluminum chloride can be formed from its elements:<\/p>\n<p id=\"fs-idp154380768\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1303\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ea-300x26.png\" alt=\"\" width=\"300\" height=\"26\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ea-300x26.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ea-768x67.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ea-65x6.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ea-225x20.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ea-350x30.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ea.png 783w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p id=\"fs-idp154392528\">Use the reactions here to determine the \u0394<em data-effect=\"italics\">H<\/em>\u00b0 for reaction <em data-effect=\"italics\">(i)<\/em>:<\/p>\n<p id=\"fs-idp154394160\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1304\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3eb-300x69.png\" alt=\"\" width=\"417\" height=\"96\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3eb-300x69.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3eb-1024x235.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3eb-768x176.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3eb-65x15.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3eb-225x52.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3eb-350x80.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3eb.png 1121w\" sizes=\"auto, (max-width: 417px) 100vw, 417px\" \/><\/p>\n<div id=\"fs-idp120712352\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp120712864\">\u22121407 kJ<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idp120713504\">We also can use <span class=\"no-emphasis\" data-type=\"term\">Hess\u2019s law<\/span> to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products are available. The stepwise reactions we consider are: (i) decompositions of the reactants into their component elements (for which the enthalpy changes are proportional to the negative of the enthalpies of formation of the reactants), followed by (ii) re-combinations of the elements to give the products (with the enthalpy changes proportional to the enthalpies of formation of the products). The standard enthalpy change of the overall reaction is therefore equal to: (ii) the sum of the standard enthalpies of formation of all the products plus (i) the sum of the negatives of the standard enthalpies of formation of the reactants. This is usually rearranged slightly to be written as follows, with \u2211 representing \u201cthe sum of\u201d and <em data-effect=\"italics\">n<\/em> standing for the stoichiometric coefficients:<\/p>\n<div id=\"fs-idp120716832\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1305 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fa-300x33.png\" alt=\"\" width=\"355\" height=\"39\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fa-300x33.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fa-768x84.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fa-65x7.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fa-225x24.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fa-350x38.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fa.png 993w\" sizes=\"auto, (max-width: 355px) 100vw, 355px\" \/><\/div>\n<p id=\"fs-idp120729664\">The following example shows in detail why this equation is valid, and how to use it to calculate the enthalpy change for a reaction of interest.<\/p>\n<div id=\"fs-idp120730208\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp134549264\"><strong>Using Hess\u2019s Law:<\/strong><\/p>\n<p>What is the standard enthalpy change for the reaction:<\/p>\n<div id=\"fs-idp134549904\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1306 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fb-300x24.png\" alt=\"\" width=\"338\" height=\"27\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fb-300x24.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fb-768x61.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fb-65x5.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fb-225x18.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fb-350x28.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fb.png 953w\" sizes=\"auto, (max-width: 338px) 100vw, 338px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp146127120\"><strong>Solution:<\/strong><\/p>\n<p>Use the special form of Hess\u2019s law given previously, and values from <a class=\"target-chapter\" href=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/back-matter\/standard-thermodynamic-properties-for-selected-substances\/\">Appendix G<\/a>:<\/p>\n<div id=\"fs-idp146127936\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1307 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fc-300x176.png\" alt=\"\" width=\"394\" height=\"231\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fc-300x176.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fc-1024x600.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fc-768x450.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fc-65x38.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fc-225x132.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fc-350x205.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3fc.png 1033w\" sizes=\"auto, (max-width: 394px) 100vw, 394px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp134564208\"><span data-type=\"title\"><strong>Solution:<\/strong> <\/span><strong>Supporting Why the General Equation Is Valid<\/strong><\/p>\n<p>Alternatively, we can write this reaction as the sum of the decompositions of 3NO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) and 1H<sub>2<\/sub>O(<em data-effect=\"italics\">l<\/em>) into their constituent elements, and the formation of 2HNO<sub>3<\/sub>(<em data-effect=\"italics\">aq<\/em>) and 1NO(<em data-effect=\"italics\">g<\/em>) from their constituent elements. Writing out these reactions, and noting their relationships to the <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span> values for these compounds (from <a class=\"target-chapter\" href=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/back-matter\/standard-thermodynamic-properties-for-selected-substances\/\">Appendix G<\/a> ), we have:<\/p>\n<div id=\"fs-idp134571776\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1309 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ga-300x97.png\" alt=\"\" width=\"405\" height=\"131\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ga-300x97.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ga-1024x331.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ga-768x248.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ga-65x21.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ga-225x73.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ga-350x113.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3ga.png 1307w\" sizes=\"auto, (max-width: 405px) 100vw, 405px\" \/><\/div>\n<div id=\"fs-idp194526576\" data-type=\"equation\"><\/div>\n<p id=\"fs-idp155650064\">Summing these reaction equations gives the reaction we are interested in:<\/p>\n<div id=\"fs-idp155650448\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1310 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3gb-300x33.png\" alt=\"\" width=\"300\" height=\"33\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3gb-300x33.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3gb-65x7.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3gb-225x25.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3gb-350x39.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3gb.png 695w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<p id=\"fs-idp155662480\">Summing their enthalpy changes gives the value we want to determine:<\/p>\n<div id=\"fs-idp155662864\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1311 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3gc-300x23.png\" alt=\"\" width=\"534\" height=\"41\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3gc-300x23.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3gc-1024x78.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3gc-768x58.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3gc-65x5.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3gc-225x17.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3gc-350x27.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/5.3gc.png 1445w\" sizes=\"auto, (max-width: 534px) 100vw, 534px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp146119648\">So the standard enthalpy change for this reaction is \u0394<em data-effect=\"italics\">H<\/em>\u00b0 = \u2212138.4 kJ.<\/p>\n<p id=\"fs-idp146120784\">Note that this result was obtained by (1) multiplying the <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span> of each product by its stoichiometric coefficient and summing those values, (2) multiplying the <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0<\/span> of each reactant by its stoichiometric coefficient and summing those values, and then (3) subtracting the result found in (2) from the result found in (1). This is also the procedure in using the general equation, as shown.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp165556368\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Calculate the heat of combustion of 1 mole of ethanol, C<sub>2<\/sub>H<sub>5<\/sub>OH(<em data-effect=\"italics\">l<\/em>), when H<sub>2<\/sub>O(<em data-effect=\"italics\">l<\/em>) and CO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) are formed. Use the following enthalpies of formation: C<sub>2<\/sub>H<sub>5<\/sub>OH(<em data-effect=\"italics\">l<\/em>), \u2212278 kJ\/mol; H<sub>2<\/sub>O(<em data-effect=\"italics\">l<\/em>), \u2212286 kJ\/mol; and CO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>), \u2212394 kJ\/mol.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idp165563824\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp165564336\">\u22121368 kJ\/mol<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idp165565104\" class=\"summary\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\n<p id=\"fs-idp165565984\">If a chemical change is carried out at constant pressure and the only work done is caused by expansion or contraction, <em data-effect=\"italics\">q<\/em> for the change is called the enthalpy change with the symbol \u0394<em data-effect=\"italics\">H<\/em>, or <span data-type=\"term\">\u0394<em>H<\/em>\u00b0<\/span> for reactions occurring under standard state conditions at 298 K. The value of \u0394<em data-effect=\"italics\">H<\/em> for a reaction in one direction is equal in magnitude, but opposite in sign, to \u0394<em data-effect=\"italics\">H<\/em> for the reaction in the opposite direction, and \u0394<em data-effect=\"italics\">H<\/em> is directly proportional to the quantity of reactants and products. The standard enthalpy of formation, <span data-type=\"term\">\u0394<em>H<\/em><sub>f<\/sub>\u00b0,<\/span> is the enthalpy change accompanying the formation of 1 mole of a substance from the elements in their most stable states at 1 bar. If the enthalpies of formation are available for the reactants and products of a reaction, the enthalpy change can be calculated using Hess\u2019s law: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps.<\/p>\n<\/div>\n<div id=\"fs-idp165577648\" class=\"key-equations\" data-depth=\"1\"><\/div>\n<div id=\"fs-idp167491136\" class=\"exercises\" data-depth=\"1\">\n<div id=\"fs-idp165246448\" data-type=\"exercise\">\n<div id=\"fs-idp165246704\" data-type=\"problem\">\n<p id=\"fs-idp165259776\"><strong><span style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em\">Footnotes<\/span><\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"footnote-refs\">\n<ul data-list-type=\"bulleted\" data-bullet-style=\"none\">\n<li data-type=\"footnote-ref\"><a href=\"#footnote-ref1\" data-type=\"footnote-ref-link\">1<\/a><span data-type=\"footnote-ref-content\">For more on algal fuel, see http:\/\/www.theguardian.com\/environment\/2010\/feb\/13\/algae-solve-pentagon-fuel-problem.<\/span><\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox shaded\" data-type=\"glossary\">\n<h3 data-type=\"glossary-title\"><strong>Glossary<\/strong><\/h3>\n<dl id=\"fs-idp102379584\">\n<dt>chemical thermodynamics<\/dt>\n<dd id=\"fs-idp102379968\">area of science that deals with the relationships between heat, work, and all forms of energy associated with chemical and physical processes<\/dd>\n<\/dl>\n<dl id=\"fs-idp102380512\">\n<dt>enthalpy (<em data-effect=\"italics\">H<\/em>)<\/dt>\n<dd id=\"fs-idp102381520\">sum of a system\u2019s internal energy and the mathematical product of its pressure and volume<\/dd>\n<\/dl>\n<dl id=\"fs-idp102382096\">\n<dt>enthalpy change (\u0394<em data-effect=\"italics\">H<\/em>)<\/dt>\n<dd id=\"fs-idp102383104\">heat released or absorbed by a system under constant pressure during a chemical or physical process<\/dd>\n<\/dl>\n<dl id=\"fs-idp102383600\">\n<dt>expansion work (pressure-volume work)<\/dt>\n<dd id=\"fs-idp102383984\">work done as a system expands or contracts against external pressure<\/dd>\n<\/dl>\n<dl id=\"fs-idp102384368\">\n<dt>first law of thermodynamics<\/dt>\n<dd id=\"fs-idp102384752\">internal energy of a system changes due to heat flow in or out of the system or work done on or by the system<\/dd>\n<\/dl>\n<dl id=\"fs-idp102385264\">\n<dt>Hess\u2019s law<\/dt>\n<dd id=\"fs-idp102385648\">if a process can be represented as the sum of several steps, the enthalpy change of the process equals the sum of the enthalpy changes of the steps<\/dd>\n<\/dl>\n<dl id=\"fs-idp102386960\">\n<dt>internal energy (<em data-effect=\"italics\">U<\/em>)<\/dt>\n<dd id=\"fs-idp102387968\">total of all possible kinds of energy present in a substance or substances<\/dd>\n<\/dl>\n<dl id=\"fs-idp102388352\">\n<dt>standard enthalpy of combustion (\u0394<em>H<\/em><sub>c<\/sub>\u00b0)<\/dt>\n<dd id=\"fs-idp102391408\">heat released when one mole of a compound undergoes complete combustion under standard conditions<\/dd>\n<\/dl>\n<dl id=\"fs-idp102391904\">\n<dt>standard enthalpy of formation <span style=\"font-size: 1em\">(\u0394<\/span><em>H<\/em><em style=\"font-size: 1em\"><sub>f<\/sub><\/em><span style=\"font-size: 1em\">\u00b0)<\/span><\/dt>\n<dd id=\"fs-idp102394960\">enthalpy change of a chemical reaction in which 1 mole of a pure substance is formed from its elements in their most stable states under standard state conditions<\/dd>\n<\/dl>\n<dl id=\"fs-idp102395520\">\n<dt>standard state<\/dt>\n<dd id=\"fs-idp102395904\">set of physical conditions as accepted as common reference conditions for reporting thermodynamic properties; 1 bar of pressure, and solutions at 1 molar concentrations<\/dd>\n<\/dl>\n<dl id=\"fs-idp102396512\">\n<dt>state function<\/dt>\n<dd id=\"fs-idp102396896\">property depending only on the state of a system, and not the path taken to reach that state<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":1392,"menu_order":4,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[48],"contributor":[],"license":[],"class_list":["post-242","chapter","type-chapter","status-publish","hentry","chapter-type-numberless"],"part":214,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/242","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/users\/1392"}],"version-history":[{"count":14,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/242\/revisions"}],"predecessor-version":[{"id":2124,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/242\/revisions\/2124"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/parts\/214"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/242\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/media?parent=242"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapter-type?post=242"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/contributor?post=242"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/license?post=242"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}