{"id":2485,"date":"2018-06-19T20:32:56","date_gmt":"2018-06-19T20:32:56","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/homolytic-c-h-bond-dissociation-energies-of-organic-molecules\/"},"modified":"2018-08-06T11:59:03","modified_gmt":"2018-08-06T11:59:03","slug":"5-2-thermodynamics-of-reactions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/5-2-thermodynamics-of-reactions\/","title":{"raw":"5.2. Thermodynamics of reactions","rendered":"5.2. Thermodynamics of reactions"},"content":{"raw":"<h1>Fundamentals<\/h1>\r\n<section>Consider a simple equilibrium<\/section><section><img class=\"alignnone size-full wp-image-4404\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/31054118\/ABCDequilibrium.png\" alt=\"\" width=\"231\" height=\"35\" \/><\/section><section>The equilibrium constant will be given by<\/section><section><img class=\"alignnone size-full wp-image-4406\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/31060544\/ABCDequilibriumK.png\" alt=\"\" width=\"128\" height=\"67\" \/><\/section><section>If Keq &gt;1, the reaction favors the products.\u00a0 If K<sub>eq<\/sub> = 1, the reaction will be equally balanced.<\/section><section>Recall that the standard free energy change \u0394G<sup>o<\/sup> = -RTlnK<sub>eq<\/sub> where \u0394G<sup>o<\/sup> = G<sup>o<\/sup>(reactants) - G<sup>o<\/sup>(products)<\/section><section>\u0394G<sup>o<\/sup> represents the driving force for the reaction:<\/section>\r\n<ul>\r\n \t<li>\u0394G<sup>o<\/sup> = 0, K<sub>eq<\/sub> = 1, the reaction is equally balanced.<\/li>\r\n \t<li>\u0394G<sup>o<\/sup> is negative, K<sub>eq<\/sub> &gt; 1, the reaction favors the products<\/li>\r\n \t<li>\u0394G<sup>o<\/sup> is positive, K<sub>eq<\/sub> &lt; 1, the reaction favors the reactants<\/li>\r\n<\/ul>\r\nFree energy relates to the heat change through the equation (\u0394H)\r\n\r\n<img class=\"alignnone size-full wp-image-4409\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/31061727\/DeltaG.png\" alt=\"\" width=\"196\" height=\"33\" \/>\r\n\r\nThis tells us that a more negative \u0394H (exothermic reaction) will help favor the products.\u00a0 Also, a positive entropy change (increase in disorder) will also help form the products, especially at higher temperatures.\r\n\r\n<section><\/section><section><\/section>\r\n<h1>Bond dissociation energy<\/h1>\r\n<section class=\"mt-content-container\">The homolytic bond dissociation energy is the amount of energy needed to break apart one mole of covalently bonded gases into a pair of radicals. The <a title=\"SI Units\" href=\"https:\/\/chem.libretexts.org\/Core\/Analytical_Chemistry\/Quantifying_Nature\/Units_of_Measure\/SI_Units\" rel=\"internal\">SI units<\/a> used to describe bond energy are kiloJoules\u00a0per mole of bonds (kJ\/Mol). It indicates how strongly the atoms are bonded to each other.\r\n<div class=\"mt-section\">\r\n<h3 class=\"editable\">Introduction<\/h3>\r\nBreaking a <a title=\"Covalent Bonds\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Chemical_Bonding\/Fundamentals_of_Chemical_Bonding\/Covalent_Bonds\" rel=\"internal\">covalent bond <\/a>between two partners, A-B, can occur either heterolytically, where the shared pair of electron goes with one partner or another\r\n\r\n\\[A-B \\rightarrow A^+\u00a0\u00a0+ \u00a0B:^-\\]\r\n\r\nor\r\n\r\n\\[A-B \\rightarrow A:^- \u00a0+ \u00a0B^+\u00a0\u200b\\]\r\n\r\nor homolytically, where one electron stays with each partner.\r\n\r\n\\[A-B \\rightarrow A^\u2022\u00a0 +\u00a0 B^\u2022 \\]\r\n\r\nThe products of homolytic cleavage are <a title=\"Chapter 17: Radical reactions\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/17%3A_Radical_reactions\" rel=\"internal\">radicals<\/a> and the energy that is required to break the bond homolytically is called the <em>Bond Dissociation Energy<\/em> (BDE) and is a measure of the strength of the bond.\r\n\r\n<\/div>\r\n<div class=\"mt-section\">\r\n<h3 class=\"editable\">Calculation of the BDE<\/h3>\r\nThe BDE\u00a0for a molecule A-B is calculated as the difference in the <a title=\"Standard Enthalpy of Formation\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Thermodynamics\/Energies_and_Potentials\/Enthalpy\/Standard_Enthalpy_Of_Formation\" rel=\"internal\">enthalpies of formation<\/a> of the products and reactants for homolysis.\r\n\r\n<em>Officially, the <a title=\"Nomenclature\" href=\"https:\/\/chem.libretexts.org\/Core\/Organic_Chemistry\/Fundamentals\/Nomenclature\" rel=\"internal\">IUPAC<\/a> definition of bond dissociation energy refers to the energy change that occurs at 0 K, and the symbol is $$D_o$$. However, it is commonly referred to as BDE, the bond dissociation energy, and it is generally used, albeit imprecisely, interchangeably with the bond dissociation\u00a0enthalpy, which generally refers to the enthalpy change at room temperature (298K). Although there are technically differences between BDEs at 0 K and 298 K, those difference are not large and generally do not affect interpretations of chemical processes.<\/em>\r\n\r\n<\/div>\r\n<div class=\"mt-section\">\r\n<h3 class=\"editable\">Bond breakage\/formation<\/h3>\r\nBond dissociation energy (or enthalpy) is a <a title=\"State Functions\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Thermodynamics\/Energies_and_Potentials\" rel=\"internal\">state function<\/a> and consequently does not depend on the path by which it occurs.\u00a0 Therefore, the specific mechanism in how a bond breaks or is formed does not affect the BDE. Bond dissociation energies are useful in assessing the energetics of chemical processes. For chemical reactions, combining bond dissociation energies for bonds formed and bonds broken in a chemical reaction using <a title=\"Hess's Law\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Thermodynamics\/Thermodynamic_Cycles\/Hess's_Law\" rel=\"internal\">Hess's Law<\/a> can be used to estimate reaction enthalpies.\r\n<div class=\"textbox examples\">\r\n<h3>Example: Chlorination of Methane<\/h3>\r\nConsider the chlorination of methane\r\n\r\n<big>CH<sub>4<\/sub>\u00a0 +\u00a0 Cl<sub>2<\/sub>\u00a0 \u2192\u00a0 CH<sub>3<\/sub>Cl\u00a0 +\u00a0 HCl<\/big>\r\n\r\nThe overall reaction thermochemistry can be calculated exactly by combining the BDEs for the bonds broken and bonds formed, i.e., \u0394H = BDE(bonds broken) - BDE(bonds made)\r\n<em>The \"bonds made\" part of the equation is negative because it represents the opposite of bonds broken, the BDE.<\/em>\r\n\r\nCH<sub>4<\/sub> \u2192 CH<sub>3<\/sub>\u2022 + H\u2022\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 BDE(CH<sub>3<\/sub>-H)\r\n\r\nCl<sub>2<\/sub>\u00a0 \u2192 2Cl\u2022\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 BDE(Cl-Cl)\r\n\r\nH\u2022\u00a0 +\u00a0 Cl\u2022\u00a0 \u2192\u00a0 HCl\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -BDE(H-Cl)\r\n\r\nCH<sub>3<\/sub>\u2022\u00a0 +\u00a0 Cl\u2022\u00a0 \u2192\u00a0 CH<sub>3<\/sub>Cl\u00a0 -BDE(CH<sub>3<\/sub>-Cl)\r\n\r\n---------------------------------------------------\r\n\r\nCH<sub>4\u00a0 <\/sub>+\u00a0 Cl<sub>2\u00a0 <\/sub>\u2192\u00a0 CH<sub>3<\/sub>Cl\u00a0 +\u00a0 HCl\r\n\r\n\u0394H = BDE(CH<sub>3<\/sub>\u2212H) + BDE(Cl-Cl) \u2212 BDE(H-Cl) \u2212 BDE(CH<sub>3<\/sub>\u2212Cl)\r\n\r\nBecause reaction enthalpy is a state function, it does not matter what reactions are combined to make up the overall process using Hess's Law. However, BDEs are convenient to use because they are readily available.\r\n\r\n<\/div>\r\nAlternatively, BDEs can be used to assess individual steps of a mechanism. For example, an important step in<a class=\"mt-disabled\" title=\"Free Radical Reactions of Alkanes\" rel=\"broken\"> free radical chlorination of alkanes<\/a> is the abstraction of hydrogen from the alkane to form a free radical.\r\n\r\nRH\u00a0 +\u00a0 Cl\u2022\u00a0 \u2192 R\u2022\u00a0 +\u00a0 HCl\r\n\r\nThe energy change for this step is equal to the difference in the BDEs in RH and HCl\r\n\r\n\\[\\Delta H = BDE(R-H) - BDE(HCl)\\]\r\n\r\nThis relationship shows that the hydrogen abstraction step is more favorable when BDE(R-H) is smaller.\u00a0 The difference in energies accounts for the selectivity in the halogenation of hydrocarbons with different types of C-H bonds.\r\n\r\n<\/div>\r\n<div class=\"mt-section\">\r\n<h3 class=\"editable\">Representative C-H BDEs in Organic Molecules<\/h3>\r\n<table style=\"border-spacing: 1px\" border=\"1\" cellpadding=\"1\">\r\n<tbody>\r\n<tr>\r\n<td><strong>R-H<\/strong><\/td>\r\n<td><em>D<sub>o<\/sub>, <\/em>kJ\/mo<em>l<\/em><\/td>\r\n<td><em>D<sub>298<\/sub><\/em>\u200b, kJ\/mol<\/td>\r\n<td><strong>R-H<\/strong><\/td>\r\n<td><em>D<sub>o<\/sub><\/em>, kJ\/mol<\/td>\r\n<td><em>D<sub>298<\/sub><\/em>, kJ\/mol<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>CH<sub>3<\/sub>-H<\/td>\r\n<td>432.7\u00b10.1<\/td>\r\n<td>439.3\u00b10.4<\/td>\r\n<td>H<sub>2<\/sub>C=CH-H<\/td>\r\n<td>456.7\u00b12.7<\/td>\r\n<td>463.2\u00b12.9<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>CH<sub>3<\/sub>CH<sub>2<\/sub>-H<\/td>\r\n<td><\/td>\r\n<td>423.0\u00b11.7<\/td>\r\n<td>C<sub>6<\/sub>H<sub>5<\/sub>-H<\/td>\r\n<td>465.8\u00b11.9<\/td>\r\n<td>472.4\u00b12.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>(CH<sub>3<\/sub>)<sub>2<\/sub>CH-H<\/td>\r\n<td><\/td>\r\n<td>412.5\u00b11.7<\/td>\r\n<td>HCCH<\/td>\r\n<td>551.2\u00b10.1<\/td>\r\n<td>557.8\u00b10.3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>(CH<sub>3<\/sub>)<sub>3<\/sub>C-H<\/td>\r\n<td><\/td>\r\n<td>403.8\u00b11.7<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td>H<sub>2<\/sub>C=CHCH<sub>2<\/sub>-H<\/td>\r\n<td><\/td>\r\n<td>371.5\u00b11.7<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HC(O)-H<\/td>\r\n<td><\/td>\r\n<td>368.6\u00b10.8<\/td>\r\n<td>C<sub>6<\/sub>H<sub>5<\/sub>CH<sub>2<\/sub>-H<\/td>\r\n<td><\/td>\r\n<td>375.3\u00b12.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>CH<sub>3<\/sub>C(O)-H<\/td>\r\n<td><\/td>\r\n<td>374.0\u00b11.2<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"mt-section\"><\/div>\r\n<div class=\"mt-section\">\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"mt-section\">\r\n<div class=\"textbox examples\">\r\n<h3>Further Reading<\/h3>\r\n<em>MasterOrganicChemistry<\/em>\r\n\r\n<a class=\"external\" href=\"http:\/\/www.masterorganicchemistry.com\/2013\/08\/14\/bond-strengths-radical-stability\/\" target=\"_blank\" rel=\"external nofollow noopener\">Bond strengths and radical stability<\/a>\r\n\r\n<img class=\"size-thumbnail wp-image-4605 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/01142903\/static_qr_code_without_logo2-150x150.png\" alt=\"\" width=\"150\" height=\"150\" \/>\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/section>","rendered":"<h1>Fundamentals<\/h1>\n<section>Consider a simple equilibrium<\/section>\n<section><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4404\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/31054118\/ABCDequilibrium.png\" alt=\"\" width=\"231\" height=\"35\" \/><\/section>\n<section>The equilibrium constant will be given by<\/section>\n<section><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4406\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/31060544\/ABCDequilibriumK.png\" alt=\"\" width=\"128\" height=\"67\" \/><\/section>\n<section>If Keq &gt;1, the reaction favors the products.\u00a0 If K<sub>eq<\/sub> = 1, the reaction will be equally balanced.<\/section>\n<section>Recall that the standard free energy change \u0394G<sup>o<\/sup> = -RTlnK<sub>eq<\/sub> where \u0394G<sup>o<\/sup> = G<sup>o<\/sup>(reactants) &#8211; G<sup>o<\/sup>(products)<\/section>\n<section>\u0394G<sup>o<\/sup> represents the driving force for the reaction:<\/section>\n<ul>\n<li>\u0394G<sup>o<\/sup> = 0, K<sub>eq<\/sub> = 1, the reaction is equally balanced.<\/li>\n<li>\u0394G<sup>o<\/sup> is negative, K<sub>eq<\/sub> &gt; 1, the reaction favors the products<\/li>\n<li>\u0394G<sup>o<\/sup> is positive, K<sub>eq<\/sub> &lt; 1, the reaction favors the reactants<\/li>\n<\/ul>\n<p>Free energy relates to the heat change through the equation (\u0394H)<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4409\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/31061727\/DeltaG.png\" alt=\"\" width=\"196\" height=\"33\" \/><\/p>\n<p>This tells us that a more negative \u0394H (exothermic reaction) will help favor the products.\u00a0 Also, a positive entropy change (increase in disorder) will also help form the products, especially at higher temperatures.<\/p>\n<section><\/section>\n<section><\/section>\n<h1>Bond dissociation energy<\/h1>\n<section class=\"mt-content-container\">The homolytic bond dissociation energy is the amount of energy needed to break apart one mole of covalently bonded gases into a pair of radicals. The <a title=\"SI Units\" href=\"https:\/\/chem.libretexts.org\/Core\/Analytical_Chemistry\/Quantifying_Nature\/Units_of_Measure\/SI_Units\" rel=\"internal\">SI units<\/a> used to describe bond energy are kiloJoules\u00a0per mole of bonds (kJ\/Mol). It indicates how strongly the atoms are bonded to each other.<\/p>\n<div class=\"mt-section\">\n<h3 class=\"editable\">Introduction<\/h3>\n<p>Breaking a <a title=\"Covalent Bonds\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Chemical_Bonding\/Fundamentals_of_Chemical_Bonding\/Covalent_Bonds\" rel=\"internal\">covalent bond <\/a>between two partners, A-B, can occur either heterolytically, where the shared pair of electron goes with one partner or another<\/p>\n<p>\\[A-B \\rightarrow A^+\u00a0\u00a0+ \u00a0B:^-\\]<\/p>\n<p>or<\/p>\n<p>\\[A-B \\rightarrow A:^- \u00a0+ \u00a0B^+\u00a0\u200b\\]<\/p>\n<p>or homolytically, where one electron stays with each partner.<\/p>\n<p>\\[A-B \\rightarrow A^\u2022\u00a0 +\u00a0 B^\u2022 \\]<\/p>\n<p>The products of homolytic cleavage are <a title=\"Chapter 17: Radical reactions\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/17%3A_Radical_reactions\" rel=\"internal\">radicals<\/a> and the energy that is required to break the bond homolytically is called the <em>Bond Dissociation Energy<\/em> (BDE) and is a measure of the strength of the bond.<\/p>\n<\/div>\n<div class=\"mt-section\">\n<h3 class=\"editable\">Calculation of the BDE<\/h3>\n<p>The BDE\u00a0for a molecule A-B is calculated as the difference in the <a title=\"Standard Enthalpy of Formation\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Thermodynamics\/Energies_and_Potentials\/Enthalpy\/Standard_Enthalpy_Of_Formation\" rel=\"internal\">enthalpies of formation<\/a> of the products and reactants for homolysis.<\/p>\n<p><em>Officially, the <a title=\"Nomenclature\" href=\"https:\/\/chem.libretexts.org\/Core\/Organic_Chemistry\/Fundamentals\/Nomenclature\" rel=\"internal\">IUPAC<\/a> definition of bond dissociation energy refers to the energy change that occurs at 0 K, and the symbol is $$D_o$$. However, it is commonly referred to as BDE, the bond dissociation energy, and it is generally used, albeit imprecisely, interchangeably with the bond dissociation\u00a0enthalpy, which generally refers to the enthalpy change at room temperature (298K). Although there are technically differences between BDEs at 0 K and 298 K, those difference are not large and generally do not affect interpretations of chemical processes.<\/em><\/p>\n<\/div>\n<div class=\"mt-section\">\n<h3 class=\"editable\">Bond breakage\/formation<\/h3>\n<p>Bond dissociation energy (or enthalpy) is a <a title=\"State Functions\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Thermodynamics\/Energies_and_Potentials\" rel=\"internal\">state function<\/a> and consequently does not depend on the path by which it occurs.\u00a0 Therefore, the specific mechanism in how a bond breaks or is formed does not affect the BDE. Bond dissociation energies are useful in assessing the energetics of chemical processes. For chemical reactions, combining bond dissociation energies for bonds formed and bonds broken in a chemical reaction using <a title=\"Hess's Law\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Thermodynamics\/Thermodynamic_Cycles\/Hess's_Law\" rel=\"internal\">Hess&#8217;s Law<\/a> can be used to estimate reaction enthalpies.<\/p>\n<div class=\"textbox examples\">\n<h3>Example: Chlorination of Methane<\/h3>\n<p>Consider the chlorination of methane<\/p>\n<p><span style=\"font-size: larger;\">CH<sub>4<\/sub>\u00a0 +\u00a0 Cl<sub>2<\/sub>\u00a0 \u2192\u00a0 CH<sub>3<\/sub>Cl\u00a0 +\u00a0 HCl<\/span><\/p>\n<p>The overall reaction thermochemistry can be calculated exactly by combining the BDEs for the bonds broken and bonds formed, i.e., \u0394H = BDE(bonds broken) &#8211; BDE(bonds made)<br \/>\n<em>The &#8220;bonds made&#8221; part of the equation is negative because it represents the opposite of bonds broken, the BDE.<\/em><\/p>\n<p>CH<sub>4<\/sub> \u2192 CH<sub>3<\/sub>\u2022 + H\u2022\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 BDE(CH<sub>3<\/sub>-H)<\/p>\n<p>Cl<sub>2<\/sub>\u00a0 \u2192 2Cl\u2022\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 BDE(Cl-Cl)<\/p>\n<p>H\u2022\u00a0 +\u00a0 Cl\u2022\u00a0 \u2192\u00a0 HCl\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -BDE(H-Cl)<\/p>\n<p>CH<sub>3<\/sub>\u2022\u00a0 +\u00a0 Cl\u2022\u00a0 \u2192\u00a0 CH<sub>3<\/sub>Cl\u00a0 -BDE(CH<sub>3<\/sub>-Cl)<\/p>\n<p>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;<\/p>\n<p>CH<sub>4\u00a0 <\/sub>+\u00a0 Cl<sub>2\u00a0 <\/sub>\u2192\u00a0 CH<sub>3<\/sub>Cl\u00a0 +\u00a0 HCl<\/p>\n<p>\u0394H = BDE(CH<sub>3<\/sub>\u2212H) + BDE(Cl-Cl) \u2212 BDE(H-Cl) \u2212 BDE(CH<sub>3<\/sub>\u2212Cl)<\/p>\n<p>Because reaction enthalpy is a state function, it does not matter what reactions are combined to make up the overall process using Hess&#8217;s Law. However, BDEs are convenient to use because they are readily available.<\/p>\n<\/div>\n<p>Alternatively, BDEs can be used to assess individual steps of a mechanism. For example, an important step in<a class=\"mt-disabled\" title=\"Free Radical Reactions of Alkanes\" rel=\"broken\"> free radical chlorination of alkanes<\/a> is the abstraction of hydrogen from the alkane to form a free radical.<\/p>\n<p>RH\u00a0 +\u00a0 Cl\u2022\u00a0 \u2192 R\u2022\u00a0 +\u00a0 HCl<\/p>\n<p>The energy change for this step is equal to the difference in the BDEs in RH and HCl<\/p>\n<p>\\[\\Delta H = BDE(R-H) &#8211; BDE(HCl)\\]<\/p>\n<p>This relationship shows that the hydrogen abstraction step is more favorable when BDE(R-H) is smaller.\u00a0 The difference in energies accounts for the selectivity in the halogenation of hydrocarbons with different types of C-H bonds.<\/p>\n<\/div>\n<div class=\"mt-section\">\n<h3 class=\"editable\">Representative C-H BDEs in Organic Molecules<\/h3>\n<table style=\"border-spacing: 1px\" cellpadding=\"1\">\n<tbody>\n<tr>\n<td><strong>R-H<\/strong><\/td>\n<td><em>D<sub>o<\/sub>, <\/em>kJ\/mo<em>l<\/em><\/td>\n<td><em>D<sub>298<\/sub><\/em>\u200b, kJ\/mol<\/td>\n<td><strong>R-H<\/strong><\/td>\n<td><em>D<sub>o<\/sub><\/em>, kJ\/mol<\/td>\n<td><em>D<sub>298<\/sub><\/em>, kJ\/mol<\/td>\n<\/tr>\n<tr>\n<td>CH<sub>3<\/sub>-H<\/td>\n<td>432.7\u00b10.1<\/td>\n<td>439.3\u00b10.4<\/td>\n<td>H<sub>2<\/sub>C=CH-H<\/td>\n<td>456.7\u00b12.7<\/td>\n<td>463.2\u00b12.9<\/td>\n<\/tr>\n<tr>\n<td>CH<sub>3<\/sub>CH<sub>2<\/sub>-H<\/td>\n<td><\/td>\n<td>423.0\u00b11.7<\/td>\n<td>C<sub>6<\/sub>H<sub>5<\/sub>-H<\/td>\n<td>465.8\u00b11.9<\/td>\n<td>472.4\u00b12.5<\/td>\n<\/tr>\n<tr>\n<td>(CH<sub>3<\/sub>)<sub>2<\/sub>CH-H<\/td>\n<td><\/td>\n<td>412.5\u00b11.7<\/td>\n<td>HCCH<\/td>\n<td>551.2\u00b10.1<\/td>\n<td>557.8\u00b10.3<\/td>\n<\/tr>\n<tr>\n<td>(CH<sub>3<\/sub>)<sub>3<\/sub>C-H<\/td>\n<td><\/td>\n<td>403.8\u00b11.7<\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<td>H<sub>2<\/sub>C=CHCH<sub>2<\/sub>-H<\/td>\n<td><\/td>\n<td>371.5\u00b11.7<\/td>\n<\/tr>\n<tr>\n<td>HC(O)-H<\/td>\n<td><\/td>\n<td>368.6\u00b10.8<\/td>\n<td>C<sub>6<\/sub>H<sub>5<\/sub>CH<sub>2<\/sub>-H<\/td>\n<td><\/td>\n<td>375.3\u00b12.5<\/td>\n<\/tr>\n<tr>\n<td>CH<sub>3<\/sub>C(O)-H<\/td>\n<td><\/td>\n<td>374.0\u00b11.2<\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"mt-section\"><\/div>\n<div class=\"mt-section\">\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"mt-section\">\n<div class=\"textbox examples\">\n<h3>Further Reading<\/h3>\n<p><em>MasterOrganicChemistry<\/em><\/p>\n<p><a class=\"external\" href=\"http:\/\/www.masterorganicchemistry.com\/2013\/08\/14\/bond-strengths-radical-stability\/\" target=\"_blank\" rel=\"external nofollow noopener\">Bond strengths and radical stability<\/a><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-thumbnail wp-image-4605 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/01142903\/static_qr_code_without_logo2-150x150.png\" alt=\"\" width=\"150\" height=\"150\" \/><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/section>\n","protected":false},"author":311,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2485","chapter","type-chapter","status-publish","hentry"],"part":22,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/2485","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":28,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/2485\/revisions"}],"predecessor-version":[{"id":4783,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/2485\/revisions\/4783"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/parts\/22"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/2485\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/media?parent=2485"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=2485"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/contributor?post=2485"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/license?post=2485"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}