{"id":1914,"date":"2015-04-22T20:09:44","date_gmt":"2015-04-22T20:09:44","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=1914"},"modified":"2016-10-27T17:17:13","modified_gmt":"2016-10-27T17:17:13","slug":"strengths-of-ionic-and-covalent-bonds","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/chapter\/strengths-of-ionic-and-covalent-bonds\/","title":{"raw":"Strengths of Ionic and Covalent Bonds","rendered":"Strengths of Ionic and Covalent Bonds"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Describe the energetics of covalent and ionic bond formation and breakage<\/li>\r\n \t<li>Use the Born-Haber cycle to compute lattice energies for ionic compounds<\/li>\r\n \t<li>Use average covalent bond energies to estimate enthalpies of reaction<\/li>\r\n<\/ul>\r\n<\/div>\r\nA bond\u2019s strength describes how strongly each atom is joined to another atom, and therefore how much energy is required to break the bond between the two atoms. In this section, you will learn about the bond strength of covalent bonds, and then compare that to the strength of ionic bonds, which is related to the lattice energy of a compound.\r\n<h2 data-type=\"title\">Bond Strength: Covalent Bonds<\/h2>\r\nStable molecules exist because covalent bonds hold the atoms together. We measure the strength of a covalent bond by the energy required to break it, that is, the energy necessary to separate the bonded atoms. Separating any pair of bonded atoms requires energy (see Figure 1). The stronger a bond, the greater the energy required to break it.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"600\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211345\/CNX_Chem_07_02_Morse1.jpg\" alt=\"A graph is shown with the x-axis labeled, \u201cInternuclear distance ( p m )\u201d while the y-axis is labeled, \u201cEnergy ( J ).\u201d One value, \u201c0,\u201d is labeled midway up the y-axis and two values: \u201c0\u201d at the far left and \u201c0.74\u201d to the left, are labeled on the x-axis. The point \u201c0.74\u201d is labeled, \u201cH bond H distance.\u201d A line is graphed that begins near the top of the y-axis and to the far left on the x-axis and drops steeply to a point labeled, \u201cnegative 7.24 times 10 superscript negative 19 J\u201d on the y-axis and 0.74 on the x-axis. This low point on the graph corresponds to a drawing of two spheres that overlap considerably. The line then rises to zero on the y-axis and levels out. The point where it almost reaches zero corresponds to two spheres that overlap slightly. The line at zero on the y-axis corresponds to two spheres that are far from one another.\" width=\"600\" height=\"589\" data-media-type=\"image\/jpeg\" \/> Figure 1. The potential energy of two separate hydrogen atoms (right) decreases as they approach each other, and the single electrons on each atom are shared to form a covalent bond. The bond length is the internuclear distance at which the lowest potential energy is achieved.[\/caption]\r\n\r\nThe energy required to break a specific covalent bond in one mole of gaseous molecules is called the bond energy or the bond dissociation energy. The bond energy for a diatomic molecule, D<sub>X\u2013Y<\/sub>, is defined as the standard enthalpy change for the endothermic reaction:\r\n<p style=\"text-align: center;\">[latex]\\text{XY}\\left(g\\right)\\rightarrow\\text{X}\\left(g\\right)+\\text{Y}\\left(g\\right){\\text{D}}_{\\text{X-Y}}=\\Delta H^{\\circ}[\/latex]<\/p>\r\nFor example, the bond energy of the pure covalent H\u2013H bond, D<sub>H\u2013H<\/sub>, is 436 kJ per mole of H\u2013H bonds broken:\r\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{2}\\left(g\\right)\\rightarrow 2\\text{H}\\left(g\\right){\\text{D}}_{\\text{H-H}}=\\Delta {H}^{\\circ}=436\\text{ kJ}[\/latex]<\/p>\r\nMolecules with three or more atoms have two or more bonds. The sum of all bond energies in such a molecule is equal to the standard enthalpy change for the endothermic reaction that breaks all the bonds in the molecule. For example, the sum of the four C\u2013H bond energies in CH<sub>4<\/sub>, 1660 kJ, is equal to the standard enthalpy change of the reaction:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211644\/CNX_Chem_07_05_CH4bond_img1.jpg\" alt=\"A reaction is shown with Lewis structures. The first structure shows a carbon atom single bonded to four hydrogen atoms with the symbol, \u201c( g )\u201d written next to it. A right-facing arrow points to the letter \u201cC\u201d and the symbol \u201c( g ),\u201d which is followed by a plus sign. Next is the number 4, the letter \u201cH\u201d and the symbol, \u201c( g ).\u201d To the right of this equation is another equation: capital delta H superscript degree symbol equals 1660 k J.\" data-media-type=\"image\/jpeg\" \/>\r\n\r\nThe average C\u2013H bond energy, D<sub>C\u2013H<\/sub>, is 1660\/4 = 415 kJ\/mol because there are four moles of C\u2013H bonds broken per mole of the reaction. Although the four C\u2013H bonds are equivalent in the original molecule, they do not each require the same energy to break; once the first bond is broken (which requires 439 kJ\/mol), the remaining bonds are easier to break. The 415 kJ\/mol value is the average, not the exact value required to break any one bond.\r\n\r\nThe strength of a bond between two atoms increases as the number of electron pairs in the bond increases. Generally, as the bond strength increases, the bond length decreases. Thus, we find that triple bonds are stronger and shorter than double bonds between the same two atoms; likewise, double bonds are stronger and shorter than single bonds between the same two atoms. Average bond energies for some common bonds appear in Table 1, and a comparison of bond lengths and bond strengths for some common bonds appears in Table 2. When one atom bonds to various atoms in a group, the bond strength typically decreases as we move down the group. For example, C\u2013F is 439 kJ\/mol, C\u2013Cl is 330 kJ\/mol, and C\u2013Br is 275 kJ\/mol.\r\n<table summary=\"This table has six columns and twenty-four rows. The first row is a header row that labels the columns: \u201cBond,\u201d \u201cBond Energy,\u201d \u201cBond,\u201d \u201cBond Energy,\u201d \u201cBond,\u201d and, \u201cBond Energy.\u201d Under the first \u201cBond\u201d column are the values: H bond to H with a single bond; H bonds to C with a single bond; H bonds to N with a single bond; H bonds to O with a single bond; H bonds to F with a single bond; H bonds to S i with a single bond; H bonds to P with a single bond; H bonds to S with a single bond; H bonds to C l with a single bond; H bonds to B r with a single bond; H bonds to I with a single bond; C bonds to C with a single bond; C bonds to C with a double bond; C bonds to C with a triple bond; C bonds to N with a single bond; C bonds to N with a double bond; C bonds to N with a triple bond; C bonds to O with a single bond; C bonds to O with a double bond; C bonds to O with a triple bond; C bonds to F with a single bond; C bonds to S i with a single bond; and C bonds to P with a single bond. Under the first \u201cBond Energy\u201d column are the values: 436; 415; 390; 464; 569; 395; 320; 340; 432; 370; 295; 345; 611; 837; 290; 615; 891; 350; 741; 1080; 439; 360; and 265. Under the second \u201cBond\u201d column are the values: C bonds to S with a single bond; C bonds to C l with a single bond; C bonds to B r with a single bond; C bonds to I with a single bond; N bonds to N with a single bond; N bonds to N with a double bond; N bonds to N with a triple bond; N bonds to O with a single bond; N bonds to F with a single bond; N bonds to P with a single bond; N bonds to C l with a single bond; N bonds to B r with a single bond; O bonds to O with a single bond; O bonds to O with a double bond; O bonds to F with a single bond; O bonds to S i with a single bond; O bonds to P with a single bond; O bonds to C l with a single bond; O bonds to I with a single bond; F bonds to F with a single bond; F bonds to S i with a single bond; F bonds to P with a single bond; and F bonds to S with a single bond. Under the second \u201cBond Energy\u201d column are the values: 260; 330; 275; 240; 160; 418; 946; 200; 270; 210; 200; 245; 140; 498; 160; 370; 350; 205; 200; 160; 540; 489; and 285. Under the third \u201cBond\u201d column are the values: F bonds to C l with a single bond; F bonds to B r with a single bond; S i bonds to S i with a single bond; S i bonds to P with a single bond; S i bonds to S with a single bond; S i bonds to C l with a single bond; S i bonds to B r with a single bond; S i bonds to I with a single bond; P bonds to P with a single bond; P bonds to S with a single bond; P bonds to C l with a single bond; P bonds to B r with a single bond; P bonds to I with a single bond; S bonds to S with a single bond; S bonds to C l with a single bond; S bonds to B r with a single bond; C l bonds to C l with a single bond; C l bonds to B r with a single bond; C l bonds to I with a single bond; B r bonds to B r with a single bond; B r bonds to I with a single bond; I bonds to I with a single bond; and the last cell in the column is empty. Under the third \u201cBond Energy\u201d column are the values: 255; 235; 230; 215; 225; 359; 290; 215; 215; 230; 330; 270; 215; 215; 250; 215; 243; 220; 210; 190; 180; 150; and the last cell in the column is empty.\">\r\n<thead>\r\n<tr>\r\n<th colspan=\"6\">Table 1. Bond Energies (kJ\/mol)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Bond<\/th>\r\n<th>Bond Energy<\/th>\r\n<th>Bond<\/th>\r\n<th>Bond Energy<\/th>\r\n<th>Bond<\/th>\r\n<th>Bond Energy<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>H\u2013H<\/td>\r\n<td>436<\/td>\r\n<td>C\u2013S<\/td>\r\n<td>260<\/td>\r\n<td>F\u2013Cl<\/td>\r\n<td>255<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H\u2013C<\/td>\r\n<td>415<\/td>\r\n<td>C\u2013Cl<\/td>\r\n<td>330<\/td>\r\n<td>F\u2013Br<\/td>\r\n<td>235<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H\u2013N<\/td>\r\n<td>390<\/td>\r\n<td>C\u2013Br<\/td>\r\n<td>275<\/td>\r\n<td>Si\u2013Si<\/td>\r\n<td>230<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H\u2013O<\/td>\r\n<td>464<\/td>\r\n<td>C\u2013I<\/td>\r\n<td>240<\/td>\r\n<td>Si\u2013P<\/td>\r\n<td>215<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H\u2013F<\/td>\r\n<td>569<\/td>\r\n<td>N\u2013N<\/td>\r\n<td>160<\/td>\r\n<td>Si\u2013S<\/td>\r\n<td>225<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H\u2013Si<\/td>\r\n<td>395<\/td>\r\n<td>[latex]\\text{N}=\\text{N}[\/latex]<\/td>\r\n<td>418<\/td>\r\n<td>Si\u2013Cl<\/td>\r\n<td>359<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H\u2013P<\/td>\r\n<td>320<\/td>\r\n<td>[latex]\\text{N}\\equiv \\text{N}[\/latex]<\/td>\r\n<td>946<\/td>\r\n<td>Si\u2013Br<\/td>\r\n<td>290<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H\u2013S<\/td>\r\n<td>340<\/td>\r\n<td>N\u2013O<\/td>\r\n<td>200<\/td>\r\n<td>Si\u2013I<\/td>\r\n<td>215<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H\u2013Cl<\/td>\r\n<td>432<\/td>\r\n<td>N\u2013F<\/td>\r\n<td>270<\/td>\r\n<td>P\u2013P<\/td>\r\n<td>215<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H\u2013Br<\/td>\r\n<td>370<\/td>\r\n<td>N\u2013P<\/td>\r\n<td>210<\/td>\r\n<td>P\u2013S<\/td>\r\n<td>230<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H\u2013I<\/td>\r\n<td>295<\/td>\r\n<td>N\u2013Cl<\/td>\r\n<td>200<\/td>\r\n<td>P\u2013Cl<\/td>\r\n<td>330<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>C\u2013C<\/td>\r\n<td>345<\/td>\r\n<td>N\u2013Br<\/td>\r\n<td>245<\/td>\r\n<td>P\u2013Br<\/td>\r\n<td>270<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\text{C}=\\text{C}[\/latex]<\/td>\r\n<td>611<\/td>\r\n<td>O\u2013O<\/td>\r\n<td>140<\/td>\r\n<td>P\u2013I<\/td>\r\n<td>215<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\text{C}\\equiv \\text{C}[\/latex]<\/td>\r\n<td>837<\/td>\r\n<td>[latex]\\text{O}=\\text{O}[\/latex]<\/td>\r\n<td>498<\/td>\r\n<td>S\u2013S<\/td>\r\n<td>215<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>C\u2013N<\/td>\r\n<td>290<\/td>\r\n<td>O\u2013F<\/td>\r\n<td>160<\/td>\r\n<td>S\u2013Cl<\/td>\r\n<td>250<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\text{C}=\\text{N}[\/latex]<\/td>\r\n<td>615<\/td>\r\n<td>O\u2013Si<\/td>\r\n<td>370<\/td>\r\n<td>S\u2013Br<\/td>\r\n<td>215<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\text{C}\\equiv \\text{N}[\/latex]<\/td>\r\n<td>891<\/td>\r\n<td>O\u2013P<\/td>\r\n<td>350<\/td>\r\n<td>Cl\u2013Cl<\/td>\r\n<td>243<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>C\u2013O<\/td>\r\n<td>350<\/td>\r\n<td>O\u2013Cl<\/td>\r\n<td>205<\/td>\r\n<td>Cl\u2013Br<\/td>\r\n<td>220<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\text{C}=\\text{O}[\/latex]<\/td>\r\n<td>741<\/td>\r\n<td>O\u2013I<\/td>\r\n<td>200<\/td>\r\n<td>Cl\u2013I<\/td>\r\n<td>210<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\text{C}\\equiv \\text{O}[\/latex]<\/td>\r\n<td>1080<\/td>\r\n<td>F\u2013F<\/td>\r\n<td>160<\/td>\r\n<td>Br\u2013Br<\/td>\r\n<td>190<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>C\u2013F<\/td>\r\n<td>439<\/td>\r\n<td>F\u2013Si<\/td>\r\n<td>540<\/td>\r\n<td>Br\u2013I<\/td>\r\n<td>180<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>C\u2013Si<\/td>\r\n<td>360<\/td>\r\n<td>F\u2013P<\/td>\r\n<td>489<\/td>\r\n<td>I\u2013I<\/td>\r\n<td>150<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>C\u2013P<\/td>\r\n<td>265<\/td>\r\n<td>F\u2013S<\/td>\r\n<td>285<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<table summary=\"This table has three columns and ten rows. The first row is a header row that labels the columns: \u201cBond,\u201d \u201cBond Length in angstroms,\u201d and, \u201cBond Energy in k J \/ mol.\u201d Under the column \u201cBond\u201d are the values: C bonds to C with a single bond; C bonds to C with a double bond; C bonds to C with a triple bond; C bonds to N with a single bond; C bonds to N with a double bond; C bonds to N with a triple bond; C bonds to O with a single bond; C bonds to O with a double bond; and C bonds to O with a triple bond. Under the column \u201cBond Length in angstroms\u201d are the values: 1.54; 1.34; 1.20; 1.43; 1.38; 1.16; 1.43; 1.23; and 1.13. Under the column \u201cBond Energy in k J \/ mol\u201d are the values: 345; 611; 837; 290; 615; 891; 350; 741; and 1080.\">\r\n<thead>\r\n<tr>\r\n<th colspan=\"3\">Table 2. Average Bond Lengths and Bond Energies for Some Common Bonds<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Bond<\/th>\r\n<th>Bond Length (\u00c5)<\/th>\r\n<th>Bond Energy (kJ\/mol)<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>C\u2013C<\/td>\r\n<td>1.54<\/td>\r\n<td>345<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\text{C}=\\text{C}[\/latex]<\/td>\r\n<td>1.34<\/td>\r\n<td>611<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\text{C}\\equiv \\text{C}[\/latex]<\/td>\r\n<td>1.20<\/td>\r\n<td>837<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>C\u2013N<\/td>\r\n<td>1.43<\/td>\r\n<td>290<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\text{C}=\\text{N}[\/latex]<\/td>\r\n<td>1.38<\/td>\r\n<td>615<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\text{C}\\equiv \\text{N}[\/latex]<\/td>\r\n<td>1.16<\/td>\r\n<td>891<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>C\u2013O<\/td>\r\n<td>1.43<\/td>\r\n<td>350<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\text{C}=\\text{O}[\/latex]<\/td>\r\n<td>1.23<\/td>\r\n<td>741<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\text{C}\\equiv \\text{O}[\/latex]<\/td>\r\n<td>1.13<\/td>\r\n<td>1080<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe can use bond energies to calculate approximate enthalpy changes for reactions where enthalpies of formation are not available. Calculations of this type will also tell us whether a reaction is exothermic or endothermic. An exothermic reaction (\u0394<em>H<\/em> negative, heat produced) results when the bonds in the products are stronger than the bonds in the reactants. An endothermic reaction (\u0394<em>H<\/em> positive, heat absorbed) results when the bonds in the products are weaker than those in the reactants.\r\n\r\nThe enthalpy change, \u0394<em>H<\/em>, for a chemical reaction is approximately equal to the sum of the energy required to break all bonds in the reactants (energy \u201cin,\u201d positive sign) plus the energy released when all bonds are formed in the products (energy \u201cout,\u201d negative sign). This can be expressed mathematically in the following way:\r\n<p style=\"text-align: center;\">[latex]\\Delta H={\\Sigma{D}}_{\\text{bonds broken}}-{\\Sigma{D}}_{\\text{bonds formed}}[\/latex]<\/p>\r\nIn this expression, the symbol \u01a9 means \u201cthe sum of\u201d and D represents the bond energy in kilojoules per mole, which is always a positive number. The bond energy is obtained from a table (like Table 2) and will depend on whether the particular bond is a single, double, or triple bond. Thus, in calculating enthalpies in this manner, it is important that we consider the bonding in all reactants and products. Because D values are typically averages for one type of bond in many different molecules, this calculation provides a rough estimate, not an exact value, for the enthalpy of reaction.\r\n\r\nConsider the following reaction:\r\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{2}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\rightarrow 2\\text{HCl}\\left(g\\right)[\/latex]\r\nor\r\n[latex]\\text{H-H}\\left(g\\right)+\\text{Cl-Cl}\\left(g\\right)\\rightarrow 2\\text{H-Cl}\\left(g\\right)[\/latex]<\/p>\r\nTo form two moles of HCl, one mole of H\u2013H bonds and one mole of Cl\u2013Cl bonds must be broken. The energy required to break these bonds is the sum of the bond energy of the H\u2013H bond (436 kJ\/mol) and the Cl\u2013Cl bond (243 kJ\/mol). During the reaction, two moles of H\u2013Cl bonds are formed (bond energy = 432 kJ\/mol), releasing 2 \u00d7 432 kJ; or 864 kJ. Because the bonds in the products are stronger than those in the reactants, the reaction releases more energy than it consumes:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}\\hfill \\Delta H&amp; =&amp; {\\Sigma{D}}_{\\text{bonds broken}}-{\\Sigma{D}}_{\\text{bonds formed}}\\hfill \\\\ \\hfill \\Delta H&amp; =&amp; \\left[{\\text{D}}_{\\text{H-H}}+{\\text{D}}_{\\text{Cl-Cl}}\\right]-2{\\text{D}}_{\\text{H-Cl}}\\hfill \\\\ &amp; =&amp; \\left[436+243\\right]-2\\left(432\\right)=-185\\text{ kJ}\\hfill \\end{array}[\/latex]<\/p>\r\nThis excess energy is released as heat, so the reaction is exothermic. <a title=\"Standard Thermodynamic Properties for Selected Substances\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\">Standard Thermodynamic Properties for Selected Substances<\/a> gives a value for the standard molar enthalpy of formation of HCl(g), [latex]\\Delta{H}_{\\text{f}}^{\\circ},[\/latex] of \u201392.307 kJ\/mol. Twice that value is \u2013184.6 kJ, which agrees well with the answer obtained earlier for the formation of two moles of HCl.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0Using Bond Energies to Calculate Approximate Enthalpy Changes<\/h3>\r\nMethanol, CH<sub>3<\/sub>OH, may be an excellent alternative fuel. The high-temperature reaction of steam and carbon produces a mixture of the gases carbon monoxide, CO, and hydrogen, H<sub>2<\/sub>, from which methanol can be produced. Using the bond energies in Table 2, calculate the approximate enthalpy change, \u0394<em>H<\/em>, for the reaction here:\r\n<p style=\"text-align: center;\">[latex]\\text{CO}\\left(g\\right)+2{\\text{H}}_{2}\\left(g\\right)\\rightarrow{\\text{CH}}_{3}\\text{OH}\\left(g\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"231938\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"231938\"]\r\n\r\nFirst, we need to write the Lewis structures of the reactants and the products:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211645\/CNX_Chem_07_05_CH3OHLew_img1.jpg\" alt=\"A set of Lewis diagrams show a chemical reaction. The first structure shows a carbon atom with a lone pair of electrons triple bonded to an oxygen with a lone pair of electrons. To the right of this structure is a plus sign, then the number 2 followed by a hydrogen atom single bonded to a hydrogen atom. To the right of this structure is a right-facing arrow followed by a hydrogen atom single bonded to a carbon atom that is single bonded to two hydrogen atoms and an oxygen atom with two lone pairs of electrons. The oxygen atom is also single bonded to a hydrogen atom.\" data-media-type=\"image\/jpeg\" \/>\r\n\r\nFrom this, we see that \u0394<em>H<\/em> for this reaction involves the energy required to break a C\u2013O triple bond and two H\u2013H single bonds, as well as the energy produced by the formation of three C\u2013H single bonds, a C\u2013O single bond, and an O\u2013H single bond. We can express this as follows:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}{}\\Delta{H}&amp;=&amp;\\Sigma_{\\text{bonds broken}}-\\Sigma{D}_{\\text{bonds formed}}\\\\\\Delta{H}&amp;=&amp;\\left[\\text{D}_{\\text{C}\\equiv\\text{O}}+2\\left(\\text{D}_{\\text{H}-\\text{H}}\\right)\\right]-\\left[3\\left(\\text{D}_{\\text{C}-\\text{H}}\\right)+\\text{D}_{\\text{C-O}}+{\\text{D}}_{\\text{O-H}}\\right]\\end{array}[\/latex]<\/p>\r\nUsing the bond energy values in Table 2, we obtain:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\hfill \\Delta H&amp; =\\left[1080+2\\left(436\\right)\\right]-\\left[3\\left(415\\right)+350+464\\right]\\\\ &amp; =-107\\text{kJ}\\end{array}[\/latex]<\/p>\r\nWe can compare this value to the value calculated based on [latex]\\Delta{H}_{\\text{f}}^{\\circ}[\/latex] data from\u00a0<a title=\"Standard Thermodynamic Properties for Selected Substances\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\">Standard Thermodynamic Properties for Selected Substances<\/a>:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\hfill \\Delta H&amp; =\\left[\\Delta{H}_{\\text{f}}^{\\circ}{\\text{CH}}_{3}\\text{OH}\\left(g\\right)\\right]-\\left[\\Delta{H}_{\\text{f}}^{\\circ}\\text{CO}\\left(g\\right)+2\\times \\Delta{H}_{\\text{f}}^{\\circ}{\\text{H}}_{2}\\right]\\\\ &amp; =\\left[-201.0\\right]-\\left[-110.52+2\\times 0\\right]\\\\ &amp; =-90.5\\text{kJ}\\end{array}[\/latex]<\/p>\r\nNote that there is a fairly significant gap between the values calculated using the two different methods. This occurs because D values are the <em>average<\/em> of different bond strengths; therefore, they often give only rough agreement with other data.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nEthyl alcohol, CH<sub>3<\/sub>CH<sub>2<\/sub>OH, was one of the first organic chemicals deliberately synthesized by humans. It has many uses in industry, and it is the alcohol contained in alcoholic beverages. It can be obtained by the fermentation of sugar or synthesized by the hydration of ethylene in the following reaction:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211646\/CNX_Chem_07_05_Ethanol_img1.jpg\" alt=\"A set of Lewis structures show a chemical reaction. The first structure shows two carbon atoms that are double bonded together and are each single bonded to two hydrogen atoms. This structure is followed by a plus sign, then an oxygen atom with two lone pairs of electrons single bonded to two hydrogen atoms. A right-facing arrow leads to a carbon atom single bonded to three hydrogen atoms and a second carbon atom. The second carbon atom is single bonded to two hydrogen atoms and an oxygen atom with two lone pairs of electrons. The oxygen atom is single bonded to a hydrogen atom as well.\" data-media-type=\"image\/jpeg\" \/>\r\n\r\nUsing the bond energies in Table 2, calculate an approximate enthalpy change, \u0394<em>H<\/em>, for this reaction.\r\n\r\n[reveal-answer q=\"893975\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"893975\"]\u201335 kJ[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Ionic Bond Strength and Lattice Energy<\/h2>\r\nAn ionic compound is stable because of the electrostatic attraction between its positive and negative ions. The lattice energy of a compound is a measure of the strength of this attraction. The lattice energy (\u0394<em>H<\/em><sub>lattice<\/sub>) of an ionic compound is defined as the energy required to separate one mole of the solid into its component gaseous ions. For the ionic solid MX, the lattice energy is the enthalpy change of the process:\r\n<p style=\"text-align: center;\">[latex]\\text{MX}\\left(s\\right)\\rightarrow{\\text{M}}^{n\\text{+}}\\left(g\\right)+{\\text{X}}^{n-}\\left(g\\right)\\Delta{H}_{\\text{lattice}}[\/latex]<\/p>\r\nNote that we are using the convention where the ionic solid is separated into ions, so our lattice energies will be <em>endothermic<\/em> (positive values). Some texts use the equivalent but opposite convention, defining lattice energy as the energy released when separate ions combine to form a lattice and giving negative (exothermic) values. Thus, if you are looking up lattice energies in another reference, be certain to check which definition is being used. In both cases, a larger magnitude for lattice energy indicates a more stable ionic compound. For sodium chloride, \u0394<em>H<\/em><sub>lattice<\/sub> = 769 kJ. Thus, it requires 769 kJ to separate one mole of solid NaCl into gaseous Na<sup>+<\/sup> and Cl<sup>\u2013<\/sup> ions. When one mole each of gaseous Na<sup>+<\/sup> and Cl<sup>\u2013<\/sup> ions form solid NaCl, 769 kJ of heat is released.\r\n\r\nThe lattice energy \u0394<em>H<\/em><sub>lattice<\/sub> of an ionic crystal can be expressed by the following equation (derived from Coulomb\u2019s law, governing the forces between electric charges):\r\n<p style=\"text-align: center;\">[latex]\\Delta{H}_{\\text{lattice}}=\\frac{\\text{C}\\left({\\text{Z}}^{\\text{+}}\\right)\\left({\\text{Z}}^{-}\\right)}{{\\text{R}}_{\\text{o}}}[\/latex]<\/p>\r\nin which C is a constant that depends on the type of crystal structure; Z<sup>+<\/sup> and Z<sup>\u2013<\/sup> are the charges on the ions; and R<sub>o<\/sub> is the interionic distance (the sum of the radii of the positive and negative ions). Thus, the lattice energy of an ionic crystal increases rapidly as the charges of the ions increase and the sizes of the ions decrease. When all other parameters are kept constant, doubling the charge of both the cation and anion quadruples the lattice energy. For example, the lattice energy of LiF (Z<sup>+<\/sup> and Z<sup>\u2013<\/sup> = 1) is 1023 kJ\/mol, whereas that of MgO (Z<sup>+<\/sup> and Z<sup>\u2013<\/sup> = 2) is 3900 kJ\/mol (R<sub>o<\/sub> is nearly the same\u2014about 200 pm for both compounds).\r\n\r\nDifferent interatomic distances produce different lattice energies. For example, we can compare the lattice energy of MgF<sub>2<\/sub> (2957 kJ\/mol) to that of MgI<sub>2<\/sub> (2327 kJ\/mol) to observe the effect on lattice energy of the smaller ionic size of F<sup>\u2013<\/sup> as compared to I<sup>\u2013<\/sup>.\r\n<div class=\"textbox examples\">\r\n<h3>Example 2:\u00a0Lattice Energy Comparisons<\/h3>\r\nThe precious gem ruby is aluminum oxide, Al<sub>2<\/sub>O<sub>3<\/sub>, containing traces of Cr<sup>3+<\/sup>. The compound Al<sub>2<\/sub>Se<sub>3<\/sub> is used in the fabrication of some semiconductor devices. Which has the larger lattice energy, Al<sub>2<\/sub>O<sub>3<\/sub> or Al<sub>2<\/sub>Se<sub>3<\/sub>?\r\n\r\n[reveal-answer q=\"664779\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"664779\"]In these two ionic compounds, the charges Z<sup>+<\/sup> and Z<sup>\u2013<\/sup> are the same, so the difference in lattice energy will depend upon R<sub>o<\/sub>. The O<sup>2\u2013<\/sup> ion is smaller than the Se<sup>2\u2013<\/sup> ion. Thus, Al<sub>2<\/sub>O<sub>3<\/sub> would have a shorter interionic distance than Al<sub>2<\/sub>Se<sub>3<\/sub>, and Al<sub>2<\/sub>O<sub>3<\/sub> would have the larger lattice energy.[\/hidden-answer]\r\n<h4>Check Your Understanding<\/h4>\r\nZinc oxide, ZnO, is a very effective sunscreen. How would the lattice energy of ZnO compare to that of NaCl?\r\n\r\n[reveal-answer q=\"508439\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"508439\"]ZnO would have the larger lattice energy because the Z values of both the cation and the anion in ZnO are greater, and the interionic distance of ZnO is smaller than that of NaCl.[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">The Born-Haber Cycle<\/h2>\r\nIt is not possible to measure lattice energies directly. However, the lattice energy can be calculated using the equation given in the previous section or by using a thermochemical cycle. The <strong>Born-Haber cycle<\/strong> is an application of Hess\u2019s law that breaks down the formation of an ionic solid into a series of individual steps:\r\n<ul data-bullet-=\"\">\r\n \t<li>[latex]\\Delta{H}_{\\text{f}}^{\\circ},[\/latex] the standard enthalpy of formation of the compound<\/li>\r\n \t<li><em>IE<\/em>, the ionization energy of the metal<\/li>\r\n \t<li><em>EA<\/em>, the electron affinity of the nonmetal<\/li>\r\n \t<li>[latex]\\Delta{H}_{s}^{\\circ},[\/latex] the enthalpy of sublimation of the metal<\/li>\r\n \t<li><em>D<\/em>, the bond dissociation energy of the nonmetal<\/li>\r\n \t<li>\u0394<em>H<\/em><sub>lattice<\/sub>, the lattice energy of the compound<\/li>\r\n<\/ul>\r\nFigure 2 diagrams the Born-Haber cycle for the formation of solid cesium fluoride.\r\n\r\n<figure>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211648\/CNX_Chem_07_05_BornHaber1.jpg\" alt=\"A diagram is shown. An upward facing arrow is drawn to the far left of the chart and is labeled \u201cH increasing.\u201d A horizontal line is drawn at the bottom of the chart. A downward-facing, vertical arrow to the left side of this line is labeled, \u201cOverall change.\u201d Beside this arrow is another label, \u201ccapital delta H subscript f, equals negative 553.5 k J per mol, ( Enthalpy of formation ).\u201d Three horizontal lines, one above the other, and all above the bottom line, are labeled, from bottom to top, as: \u201cC s ( s ), plus sign, one half F subscript 2, ( g ),\u201d \u201cC s ( g ), plus sign, one half F subscript 2, ( g ),\u201d and \u201cC s, superscript positive sign, ( g ), plus sign, one half F subscript 2, ( g ).\u201d Each of these lines is connected by an upward-facing vertical arrow. Each arrow is labeled, \u201ccapital delta H subscript 1, equals 76.5 k J per mol, ( Enthalpy of sublimation ),\u201d \u201ccapital delta H subscript 2, equals 375.7 k J per mol, ( ionization energy ),\u201d and \u201ccapital delta H subscript 3 equals 79.4 k J \/ mol ( one half dissociation energy ).\u201d Another horizontal line is drawn in the center top portion of the diagram and is labeled \u201cC s, superscript positive sign, ( g ), plus sign, F, ( g ).\u201d There is one more horizontal line drawn to the right of the overall diagram and located halfway down the image. An arrow connects the top line to this line and is labeled, \u201ccapital delta H equals negative 328.2 k J \/ mol ( electron affinity ).\u201d The line is labeled, \u201cC s superscript positive sign ( g ) plus F superscript negative sign ( g ).\u201d The arrow connecting this line to the bottom line is labeled, \u201cnegative capital delta H subscript lattice equals negative 756.9 k J \/ mol.\u201d The arrow points to a label on the bottom line which reads, \u201cC s F ( s ).\u201d\" width=\"975\" height=\"528\" data-media-type=\"image\/jpeg\" \/> Figure 2. The Born-Haber cycle shows the relative energies of each step involved in the formation of an ionic solid from the necessary elements in their reference states.[\/caption]\r\n\r\n<\/figure>We begin with the elements in their most common states, Cs(<em>s<\/em>) and F<sub>2<\/sub>(<em>g<\/em>). The [latex]\\Delta{H}_{s}^{\\circ}[\/latex] represents the conversion of solid cesium into a gas, and then the ionization energy converts the gaseous cesium atoms into cations. In the next step, we account for the energy required to break the F-F bond to produce fluorine atoms. Converting one mole of fluorine atoms into fluoride ions is an exothermic process, so this step gives off energy (the electron affinity) and is shown as decreasing along the <em>y<\/em>-axis. We now have one mole of Cs cations and one mole of F anions. These ions combine to produce solid cesium fluoride. The enthalpy change in this step is the negative of the lattice energy, so it is also an exothermic quantity. The total energy involved in this conversion is equal to the experimentally determined enthalpy of formation, [latex]\\Delta{H}_{\\text{f}}^{\\circ},[\/latex] of the compound from its elements. In this case, the overall change is exothermic.\r\n\r\nHess\u2019s law can also be used to show the relationship between the enthalpies of the individual steps and the enthalpy of formation. Table 3 shows this for cesium chloride, CsCl<sub>2<\/sub>.\r\n<table summary=\"This table has two columns and six rows. The first row is labeled, \u201cEnthalpy of sublimation of C s ( s )\u201d and the enthalpy reaction is C s ( s ) yields C s ( g ). Beside this equation is capital delta H which equals capital delta H subscript s superscript degree symbol which also equals 76.5 k J. The second row is labeled, \u201cOne-half of the bond energy of C l subscript 2.\u201d The equation for this is one half C l subscript 2 ( g ) yields C l ( g ). Beside this equation is capital delta H equals one half D which also equals 122 k J. The third row is labeled, \u201cIonization Energy of N a ( g ).\u201d The equation for the ionization energy of N a ( g ) is N a ( g ) yields N a superscript positive sign ( g ) plus lower case e superscript negative sign. Beside this equation is capital delta H equals I E which also equals 496 k J. The fourth row is labeled, \u201cNegative of the electron affinity of C l.\u201d The equation for this is C l ( g ) plus lowercase e superscript negative sign yields C l superscript negative sign ( g ). Beside this equation is capital delta H equals negative E A which also equals negative 368 k J. The fifth row is labeled \u201cNegative of the lattice energy of N a C l ( s ).\u201d The equation for this is N a superscript positive sign ( g ) plus C l superscript negative sign ( g ) yields N a C l ( s ). Beside this equation is capital delta H equals negative capital delta H subscript lattice which also equals unknown. The sixth and final row is labeled, \u201cEnthalpy of formation of N a C l ( s ), add steps 1 - 5.\u201d The equation for this is capital delta H equals capital delta H subscript f superscript degree symbol which also equals capital delta H subscript s superscript degree symbol plus one-half D plus I E plus negative E A plus negative capital delta H subscript lattice. Underneath that equation, is another which is N a ( s ) plus one-half C l subscript 2 ( g ) yields N a C l ( s ) which equals negative 411 k J.\" data-label=\"\">\r\n<thead>\r\n<tr>\r\n<th colspan=\"2\">Table 3<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Enthalpy of sublimation of Cs(<em>s<\/em>)<\/td>\r\n<td>[latex]\\text{Cs}\\left(s\\right)\\rightarrow\\text{Cs}\\left(g\\right)\\Delta H=\\Delta{H}_{s}^{\\circ}=76.5\\text{kJ}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>One-half of the bond energy of Cl<sub>2<\/sub><\/td>\r\n<td>[latex]\\frac{1}{2}{\\text{Cl}}_{2}\\text{(}g\\text{)}\\rightarrow\\text{Cl}\\left(g\\right)\\Delta H=\\frac{1}{2}D=122\\text{kJ}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Ionization energy of Na(<em>g<\/em>)<\/td>\r\n<td>[latex]\\text{Na}\\left(g\\right)\\rightarrow{\\text{Na}}^{\\text{+}}\\text{(}g\\text{)}+{\\text{e}}^{-}\\Delta H=IE=496\\text{kJ}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Negative of the electron affinity of Cl<\/td>\r\n<td>[latex]\\text{Cl}\\left(g\\right)+{\\text{e}}^{-}\\rightarrow{\\text{Cl}}^{-}\\text{(}g\\text{)}\\Delta H=-EA=-368\\text{kJ}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Negative of the lattice energy of NaCl(<em>s<\/em>)<\/td>\r\n<td>[latex]{\\text{Na}}^{\\text{+}}\\text{(}g\\text{)}+{\\text{Cl}}^{-}\\text{(}g\\text{)}\\rightarrow\\text{NaCl}\\left(s\\right)\\Delta H=-{\\Delta{H}}_{\\text{lattice}}=?[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Enthalpy of formation of NaCl(<em>s<\/em>), add steps 1\u20135<\/td>\r\n<td>[latex]\\begin{array}{l}\\Delta H=\\Delta{H}_{f}^{\\circ}=\\Delta{H}_{s}^{\\circ}+\\frac{1}{2}D+IE+\\left(-EA\\right)+\\left(-{\\Delta{H}}_{\\text{lattice}}\\right)\\\\ \\text{Na}\\left(s\\right)+\\frac{1}{2}{\\text{Cl}}_{2}\\left(g\\right)\\rightarrow\\text{NaCl}\\left(s\\right)=-411\\text{kJ}\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThus, the lattice energy can be calculated from other values. For cesium chloride, using this data, the lattice energy is:\r\n<p style=\"text-align: center;\">[latex]\\Delta{H}_{\\text{lattice}}=\\left(411+109+122+496+368\\right)\\text{kJ}=770\\text{kJ}[\/latex]<\/p>\r\nThe Born-Haber cycle may also be used to calculate any one of the other quantities in the equation for lattice energy, provided that the remainder is known. For example, if the relevant enthalpy of sublimation [latex]\\Delta{H}_{s}^{\\circ},[\/latex] ionization energy (IE), bond dissociation enthalpy (D), lattice energy \u0394<em>H<\/em><sub>lattice,<\/sub> and standard enthalpy of formation [latex]\\Delta{H}_{\\text{f}}^{\\circ}[\/latex] are known, the Born-Haber cycle can be used to determine the electron affinity of an atom.\r\n\r\nLattice energies calculated for ionic compounds are typically much higher than bond dissociation energies measured for covalent bonds. Whereas lattice energies typically fall in the range of 600\u20134000 kJ\/mol (some even higher), covalent bond dissociation energies are typically between 150\u2013400 kJ\/mol for single bonds. Keep in mind, however, that these are not directly comparable values. For ionic compounds, lattice energies are associated with many interactions, as cations and anions pack together in an extended lattice. For covalent bonds, the bond dissociation energy is associated with the interaction of just two atoms.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nThe strength of a covalent bond is measured by its bond dissociation energy, that is, the amount of energy required to break that particular bond in a mole of molecules. Multiple bonds are stronger than single bonds between the same atoms. The enthalpy of a reaction can be estimated based on the energy input required to break bonds and the energy released when new bonds are formed. For ionic bonds, the lattice energy is the energy required to separate one mole of a compound into its gas phase ions. Lattice energy increases for ions with higher charges and shorter distances between ions. Lattice energies are often calculated using the Born-Haber cycle, a thermochemical cycle including all of the energetic steps involved in converting elements into an ionic compound.\r\n<h4>Key Equations<\/h4>\r\n<ul>\r\n \t<li>Bond energy for a diatomic molecule: [latex]\\text{XY}\\left(g\\right)\\rightarrow\\text{X}\\left(g\\right)+\\text{Y}\\left(g\\right){\\text{D}}_{\\text{X-Y}}=\\Delta H^{\\circ}[\/latex]<\/li>\r\n \t<li>Enthalpy change: \u0394<em>H<\/em> = \u01a9D<sub>bonds broken<\/sub> \u2013 \u01a9D<sub>bonds formed<\/sub><\/li>\r\n \t<li>Lattice energy for a solid MX: [latex]\\text{MX}\\left(s\\right)\\rightarrow{\\text{M}}^{n\\text{+}}\\left(g\\right)+{\\text{X}}^{n-}\\left(g\\right)\\Delta{H}_{\\text{lattice}}[\/latex]<\/li>\r\n \t<li>Lattice energy for an ionic crystal: [latex]\\Delta{H}_{\\text{lattice}}=\\frac{\\text{C}\\left({\\text{Z}}^{\\text{+}}\\right)\\left({\\text{Z}}^{-}\\right)}{{\\text{R}}_{\\text{o}}}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<ol>\r\n \t<li>Which bond in each of the following pairs of bonds is the strongest?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>C\u2013C or [latex]\\text{C}=\\text{C}[\/latex]<\/li>\r\n \t<li>C\u2013N or [latex]\\text{C}\\equiv \\text{N}[\/latex]<\/li>\r\n \t<li>[latex]\\text{C}\\equiv \\text{O}[\/latex] or [latex]\\text{C}=\\text{O}[\/latex]<\/li>\r\n \t<li>H\u2013F or H\u2013Cl<\/li>\r\n \t<li>C\u2013H or O\u2013H<\/li>\r\n \t<li>C\u2013N or C\u2013O<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Using the bond energies in Table 1, determine the approximate enthalpy change for each of the following reactions:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{\\text{H}}_{2}\\left(g\\right)+{\\text{Br}}_{2}\\left(g\\right)\\rightarrow 2\\text{HBr}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{CH}}_{4}\\left(g\\right)+{\\text{I}}_{2}\\left(g\\right)\\rightarrow{\\text{CH}}_{3}\\text{I}\\left(g\\right)+\\text{HI}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{C}}_{2}{\\text{H}}_{4}\\left(g\\right)+3{\\text{O}}_{2}\\left(g\\right)\\rightarrow 2{\\text{CO}}_{2}\\left(g\\right)+2{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Using the bond energies in Table 1, determine the approximate enthalpy change for each of the following reactions:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{\\text{Cl}}_{2}\\left(g\\right)+3{\\text{F}}_{2}\\left(g\\right)\\rightarrow 2{\\text{ClF}}_{3}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{H}}_{2}\\text{C}={\\text{CH}}_{2}\\left(g\\right)+{\\text{H}}_{2}\\left(g\\right)\\rightarrow{\\text{H}}_{3}{\\text{CCH}}_{3}\\left(g\\right)[\/latex]<\/li>\r\n \t<li>[latex]2{\\text{C}}_{2}{\\text{H}}_{6}\\left(g\\right)+7{\\text{O}}_{2}\\left(g\\right)\\rightarrow 4{\\text{CO}}_{2}\\left(g\\right)+6{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>When a molecule can form two different structures, the structure with the stronger bonds is usually the more stable form. Use bond energies to predict the correct structure of the hydroxylamine molecule: <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211650\/CNX_Chem_07_05_Hydroxya_img1.jpg\" alt=\"Two Lewis structures are shows with the word \u201cor\u201d written in between them. The left structure shows a nitrogen atom with one lone pair of electrons single bonded to two hydrogen atoms. It is also bonded to an oxygen atom with two lone pairs of electrons. The oxygen atom is bonded to a hydrogen atom. The right structure shows a nitrogen atom single bonded to three hydrogen atoms and an oxygen atom with three lone pairs of electrons.\" data-media-type=\"image\/jpeg\" \/><\/li>\r\n \t<li>How does the bond energy of HCl(<em>g<\/em>) differ from the standard enthalpy of formation of HCl(<em>g<\/em>)?<\/li>\r\n \t<li>Using the standard enthalpy of formation data in\u00a0<a title=\"Standard Thermodynamic Properties for Selected Substances\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\">Standard Thermodynamic Properties for Selected Substances<\/a>, show how can the standard enthalpy of formation of HCl(g) can be used to determine the bond energy.<\/li>\r\n \t<li>Using the standard enthalpy of formation data in\u00a0<a title=\"Standard Thermodynamic Properties for Selected Substances\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\">Standard Thermodynamic Properties for Selected Substances<\/a>, calculate the bond energy of the carbon-sulfur double bond in CS<sub>2<\/sub>.<\/li>\r\n \t<li>Using the standard enthalpy of formation data in\u00a0<a title=\"Standard Thermodynamic Properties for Selected Substances\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\">Standard Thermodynamic Properties for Selected Substances<\/a>, determine which bond is stronger: the S\u2013F bond in SF<sub>4<\/sub>(<em>g<\/em>) or in SF<sub>6<\/sub>(<em>g<\/em>)?<\/li>\r\n \t<li>Using the standard enthalpy of formation data in\u00a0<a title=\"Standard Thermodynamic Properties for Selected Substances\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\">Standard Thermodynamic Properties for Selected Substances<\/a>, determine which bond is stronger: the P\u2013Cl bond in PCl<sub>3<\/sub>(<em>g<\/em>) or in PCl<sub>5<\/sub>(<em>g<\/em>)?<\/li>\r\n \t<li>Complete the following Lewis structure by adding bonds (not atoms), and then indicate the longest bond:<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211651\/CNX_Chem_07_05_C6H8Lew_img1.jpg\" alt=\"A Lewis structure is shown that is missing its bonds. It shows a horizontal row of six carbon atoms, equally spaced. Three hydrogen atoms are drawn around the first carbon, two around the second, one above the fifth, and two by the sixth.\" data-media-type=\"image\/jpeg\" \/><\/li>\r\n \t<li>Use the bond energy to calculate an approximate value of \u0394<em>H<\/em> for the following reaction. Which is the more stable form of FNO<sub>2<\/sub>?\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211653\/CNX_Chem_07_05_FNO2_img1.jpg\" alt=\"Two Lewis structures are shown with a right-facing arrow in between. The left structure shows a nitrogen atom double bonded to an oxygen atom with two lone pairs of electrons. It is also bonded to a fluorine atom and another oxygen atom, each with three lone pairs of electrons. The right structure shows an oxygen atom with two lone pairs of electrons double bonded to a nitrogen atom with one lone pair of electrons. This nitrogen atom is single bonded to an oxygen with two lone pairs of electrons. The oxygen atom is single bonded to a fluorine atom with three lone pairs of electrons.\" data-media-type=\"image\/jpeg\" \/><\/li>\r\n \t<li>Use principles of atomic structure to answer each of the following:<sup>1\u00a0<\/sup>\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>The radius of the Ca atom is 197 pm; the radius of the Ca<sup>2+<\/sup> ion is 99 pm. Account for the difference.<\/li>\r\n \t<li>The lattice energy of CaO(s) is \u20133460 kJ\/mol; the lattice energy of K<sub>2<\/sub>O is \u20132240 kJ\/mol. Account for the difference.<\/li>\r\n \t<li>Given these ionization values, explain the difference between Ca and K with regard to their first and second ionization energies.\r\n<table summary=\"This table has three columns and three rows. The first row is a header row that labels the columns: \u201cElement,\u201d \u201cFirst Ionization Energy in k J \/ mol,\u201d and \u201cSecond Ionization Energy in k J \/ mol.\u201d Under the column \u201cElement\u201d are the letters: K and C a. Under the column \u201cFirst Ionization Energy in k J \/ mol\u201d are the values: 419 and 590. Under the column \u201cSecond Ionization Energy in k J \/ mol\u201d are the values: 3050 and 1140.\" data-label=\"\">\r\n<thead>\r\n<tr>\r\n<th>Element<\/th>\r\n<th>First Ionization Energy (kJ\/mol)<\/th>\r\n<th>Second Ionization Energy (kJ\/mol)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>K<\/td>\r\n<td>419<\/td>\r\n<td>3050<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Ca<\/td>\r\n<td>590<\/td>\r\n<td>1140<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t<li>The first ionization energy of Mg is 738 kJ\/mol and that of Al is 578 kJ\/mol. Account for this difference.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The lattice energy of LiF is 1023 kJ\/mol, and the Li\u2013F distance is 200.8 pm. NaF crystallizes in the same structure as LiF but with a Na\u2013F distance of 231 pm. Which of the following values most closely approximates the lattice energy of NaF: 510, 890, 1023, 1175, or 4090 kJ\/mol? Explain your choice.<\/li>\r\n \t<li>For which of the following substances is the least energy required to convert one mole of the solid into separate ions?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>MgO<\/li>\r\n \t<li>SrO<\/li>\r\n \t<li>KF<\/li>\r\n \t<li>CsF<\/li>\r\n \t<li>MgF<sub>2<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The reaction of a metal, M, with a halogen, X<sub>2<\/sub>, proceeds by an exothermic reaction as indicated by this equation: [latex]\\text{M}\\left(s\\right)+{\\text{X}}_{2}\\left(g\\right)\\rightarrow{\\text{MX}}_{2}\\left(s\\right).[\/latex] For each of the following, indicate which option will make the reaction more exothermic. Explain your answers.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>a large radius vs. a small radius for M<sup>+2\u00a0<\/sup><\/li>\r\n \t<li>a high ionization energy vs. a low ionization energy for M<\/li>\r\n \t<li>an increasing bond energy for the halogen<\/li>\r\n \t<li>a decreasing electron affinity for the halogen<\/li>\r\n \t<li>an increasing size of the anion formed by the halogen<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The lattice energy of LiF is 1023 kJ\/mol, and the Li\u2013F distance is 201 pm. MgO crystallizes in the same structure as LiF but with a Mg\u2013O distance of 205 pm. Which of the following values most closely approximates the lattice energy of MgO: 256 kJ\/mol, 512 kJ\/mol, 1023 kJ\/mol, 2046 kJ\/mol, or 4090 kJ\/mol? Explain your choice.<\/li>\r\n \t<li>Which compound in each of the following pairs has the larger lattice energy? Note: Mg<sup>2+<\/sup> and Li<sup>+<\/sup> have similar radii; O<sup>2\u2013<\/sup> and F<sup>\u2013<\/sup> have similar radii. Explain your choices.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>MgO or MgSe<\/li>\r\n \t<li>LiF or MgO<\/li>\r\n \t<li>Li<sub>2<\/sub>O or LiCl<\/li>\r\n \t<li>Li<sub>2<\/sub>Se or MgO<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Which compound in each of the following pairs has the larger lattice energy? Note: Ba<sup>2+<\/sup> and K<sup>+<\/sup> have similar radii; S<sup>2\u2013<\/sup> and Cl<sup>\u2013<\/sup> have similar radii. Explain your choices.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>K<sub>2<\/sub>O or Na<sub>2<\/sub>O<\/li>\r\n \t<li>K<sub>2<\/sub>S or BaS<\/li>\r\n \t<li>KCl or BaS<\/li>\r\n \t<li>BaS or BaCl<sub>2<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>MgO<\/li>\r\n \t<li>SrO<\/li>\r\n \t<li>KF<\/li>\r\n \t<li>CsF<\/li>\r\n \t<li>MgF<sub>2<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>K<sub>2<\/sub>S<\/li>\r\n \t<li>K<sub>2<\/sub>O<\/li>\r\n \t<li>CaS<\/li>\r\n \t<li>Cs<sub>2<\/sub>S<\/li>\r\n \t<li>CaO<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The lattice energy of KF is 794 kJ\/mol, and the interionic distance is 269 pm. The Na\u2013F distance in NaF, which has the same structure as KF, is 231 pm. Which of the following values is the closest approximation of the lattice energy of NaF: 682 kJ\/mol, 794 kJ\/mol, 924 kJ\/mol, 1588 kJ\/mol, or 3175 kJ\/mol? Explain your answer.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"888536\"]Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"888536\"]\r\n\r\n1. In general, a multiple bond between the same two elements is stronger than a single bond. The greater the electronegativity difference between two similar elements, the greater the bond energy\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\text{C}=\\text{C}[\/latex]<\/li>\r\n \t<li>[latex]\\text{C}\\equiv \\text{N}[\/latex]<\/li>\r\n \t<li>[latex]\\text{C}\\equiv \\text{O}[\/latex]<\/li>\r\n \t<li>H\u2013F<\/li>\r\n \t<li>O\u2013H<\/li>\r\n \t<li>C\u2013O<\/li>\r\n<\/ol>\r\n3.\u00a0The approximate enthalpy changes are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\begin{array}{ll}\\hfill DH^{\\circ}&amp; ={\\Sigma{D}}_{\\text{bonds broken}}-{\\Sigma{D}}_{\\text{bonds formed}}\\\\ &amp; =2{D}_{\\text{Cl-Cl}}+3{D}_{\\text{F-F}}-6{D}_{\\text{Cl-F}}\\\\ &amp; =-564\\text{kJ}\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{ll}\\hfill DH^{\\circ}&amp; ={\\Sigma{D}}_{\\text{bonds broken}}-{\\Sigma{D}}_{\\text{bonds formed}}\\\\ &amp; ={D}_{\\text{C-C}}+4{D}_{\\text{C-H}}+{D}_{\\text{H-H}}-{D}_{\\text{C-C}}-6{D}_{\\text{C-H}}\\\\ &amp; =611+4\\left(415\\right)+436-345-6\\left(415\\right)\\\\ &amp; =-128\\text{kJ}\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{ll}\\hfill DH^{\\circ}&amp; ={\\Sigma{D}}_{\\text{bonds broken}}-{\\Sigma{D}}_{\\text{bonds formed}}\\\\ &amp; =2{D}_{\\text{C-C}}+12{D}_{\\text{C-H}}+7{D}_{\\text{O-O}}-8{D}_{\\text{C-O}}-12{D}_{\\text{O-H}}\\\\ &amp; =2\\left(345\\right)+12\\left(415\\right)+7\\left(496\\right)-8\\left(741\\right)-12\\left(464\\right)\\\\ &amp; =-2354\\text{kJ}\\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n5.\u00a0The bond energy involves breaking HCl into H and Cl atoms. The enthalpy of formation involves making HCl from H<sub>2<\/sub> and Cl<sub>2<\/sub> molecules.\r\n\r\n7. This problem can be solved as follows:\r\n<p style=\"text-align: center;\">[latex]{\\text{CS}}_{2}\\left(g\\right)\\rightarrow\\text{C}\\left(\\text{graphite}\\right)+2\\text{S}\\left(s\\right){\\Delta{H}}_{1}^{\\circ}=\\Delta{H}_{\\text{f}\\left[{\\text{CS}}_{2}\\left(g\\right)\\right]}^{\\circ}[\/latex]\r\n[latex]\\text{C}\\left(\\text{graphite}\\right)\\rightarrow\\text{C}\\left(g\\right)\\Delta{H}_{2}^{\\circ }=\\Delta{H}_{\\text{fC}\\left(g\\right)}^{\\circ}[\/latex]\r\n[latex]2\\text{S}\\left(s\\right)\\rightarrow 2\\text{S}\\left(g\\right)2\\Delta{H}_{3}^{\\circ }=2\\Delta{H}_{\\text{fS}\\left(g\\right)}^{\\circ}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\\\ \\hfill {D}_{{\\text{CS}}_{2}}&amp; =\\Delta H^{\\circ}=-{\\Delta{H}}_{\\text{f}\\left[\\left({\\text{CS}}_{2}\\left(g\\right)\\right)\\right]}^{\\circ}+\\Delta{H}_{\\text{fC}\\left(g\\right)}^{\\circ}+2\\Delta{H}_{\\text{fS}\\left(g\\right)}^{\\circ}\\\\ \\hfill &amp; =-116.9+716.681+2\\left(278.81\\right)\\\\ \\hfill &amp; =1157.4\\text{ kJ}{\\text{mol}}^{-1}\\\\ \\hfill {D}_{\\text{C }=\\text{S}}&amp; =\\frac{1157.4}{2}=578.7\\text{kJ}{\\text{mol}}^{-1}\\text{of}\\text{C=S}\\text{ bonds}\\end{array}[\/latex]<\/p>\r\n9. This problem can be solved as follows:\r\n<p style=\"text-align: center;\">[latex]{\\text{PCl}}_{3}\\left(g\\right)\\rightarrow\\frac{1}{4}{\\text{P}}_{4}\\left(s\\right)+\\frac{3}{2}{\\text{Cl}}_{2}\\left(g\\right)\\Delta{H}_{1}^{\\circ}=-{\\Delta{H}}_{\\text{f}\\left[\\left({\\text{PCl}}_{3}\\left(g\\right)\\right)\\right]}^{\\circ}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{4}{\\text{P}}_{4}\\left(s\\right)\\rightarrow\\text{P}\\left(g\\right)\\Delta{H}_{2}^{\\circ}=\\Delta{H}_{\\text{fP}\\left(g\\right)}^{\\circ}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{3}{2}{\\text{Cl}}_{2}\\left(g\\right)\\rightarrow 3\\text{Cl}\\left(g\\right)3\\Delta{H}_{3}^{\\circ}=3\\Delta{H}_{\\text{fCl}\\left(g\\right)}^{\\circ}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\\\{D}_{{\\text{PCl}}_{3}}&amp; =\\Delta H^{\\circ}=-{\\Delta{H}}_{\\text{f}\\left[\\left({\\text{PCl}}_{3}\\left(g\\right)\\right)\\right]}^{\\circ}+\\Delta{H}_{\\text{fP}\\left(g\\right)}^{\\circ}+3\\Delta{H}_{\\text{fCl}\\left(g\\right)}^{\\circ}\\\\ \\hfill &amp; =287.0+314.64+3\\left(121.3\\right)=965.54{\\text{kJ mol}}^{-\\text{1}}\\\\ \\hfill {D}_{{\\text{PCl}}_{3}}&amp; =\\frac{965.54\\text{kJ}}{3}=321.8{\\text{kJ per mol}}^{{-1}}\\text{of bonds}\\end{array}[\/latex]<\/p>\r\nProceeding in the same manners, [latex]-{\\Delta{H}}_{{\\text{f[PCl}}_{\\text{5}}\\text{(g)]}}=\\text{374.9 kJ}{\\text{mol}}^{-1}[\/latex]. The P(<em>g<\/em>) and the 5Cl(<em>g<\/em>) contribute 921.14 kJ; then [latex]{\\text{DF}}_{{\\text{PCl}}_{\\text{5}}}=\\text{1296.04 kJ}[\/latex] and [latex]{D}_{\\text{P-Cl}}=\\frac{\\text{1296.04 kJ\/mol}}{5}=\\text{259.2 kJ\/mol}[\/latex] of bonds. The P\u2013Cl bond in PCl<sub>3<\/sub> is stronger.\r\n\r\n10.The C\u2013C single bonds are longest.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211652\/CNX_Chem_07_05_C6H8ans_img1.jpg\" alt=\"A Lewis structure is shown. A carbon atom that is single bonded to three hydrogen atoms is bonded to a second carbon atom. The second carbon atom is single bonded to two hydrogen atoms. The second carbon atom is single bonded to a third carbon atom that is triple bonded to a fourth carbon atom and single bonded to a fifth carbon atom. The fifth carbon atom is single bonded to a hydrogen atom and double bonded to a sixth carbon atom that is single bonded to two hydrogen atoms.\" data-media-type=\"image\/jpeg\" \/>\r\n\r\n11. the left hand arrangement: [latex]\\text{O}=\\text{N}[\/latex] not listed, N-F 270, N-O 200; the right hand arrangement: [latex]\\text{O}=\\text{N}[\/latex] not listed, N-O 200, O-F 185; the bond energy of [latex]\\text{O}=\\text{N}[\/latex] does not matter because it must be the same in both cases, the form on the right has a bond energy of <em>X<\/em> +470; that on the right, <em>X<\/em> +385; the form on the left is more stable.\r\n\r\n12. The answers are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>When two electrons are removed from the valence shell, the Ca radius loses the outermost energy level and reverts to the lower <em>n<\/em> = 3 level, which is much smaller in radius.<\/li>\r\n \t<li>The +2 charge on calcium pulls the oxygen much closer compared with K, thereby increasing the lattice energy relative to a less charged ion.<\/li>\r\n \t<li>Removal of the 4<em>s<\/em> electron in Ca requires more energy than removal of the 4<em>s<\/em> electron in K, because of the stronger attraction of the nucleus and the extra energy required to break the pairing of the electrons. The second ionization energy for K requires that an electron be removed from a lower energy level, where the attraction is much stronger from the nucleus for the electron. In addition, energy is required to unpair two electrons in a full orbital. For Ca, the second ionization potential requires removing only a lone electron in the exposed outer energy level.<\/li>\r\n \t<li>In Al, the removed electron is relatively unprotected and unpaired in a <em>p<\/em> orbital. The higher energy for Mg mainly reflects the unpairing of the 2<em>s<\/em> electron.<\/li>\r\n<\/ol>\r\n13. The lattice energy is given by [latex]U=C\\left(\\frac{{Z}^{+}{Z}^{-}}{{R}_{o}}\\right),[\/latex] where <em>R<\/em><sub>o<\/sub> is the interatomic distance. The charges are the same in both LiF and NaF. The major difference is expected to be the interatomic distance 2.008 \u00c5 versus 2.31 \u00c5. From the data for LiF, with <em>Z<\/em><sup>+<\/sup><em>Z<\/em><sup>\u2013<\/sup> = \u20131, [latex]C=\\frac{U{R}_{\\text{o}}}{{Z}^{+}{Z}^{-}}=\\frac{1023\\times 2.008}{-1}=-2054\\text{ kJ}\\text{A}{\\text{mol}}^{-1}[\/latex].\r\n\r\nThen, [latex]{U}_{\\text{NaF}}=\\frac{-2054\\text{kJ}\\text{A}{\\text{mol}}^{-1}\\left(-1\\right)}{2.31\\text{A}}=889\\text{kJ}{\\text{mol}}^{-1}\\text{or }890{\\text{ kJ mol}}^{-1}[\/latex].\r\n\r\n14. The lattice energy, <em>U<\/em>, is the energy required to convert the solid into separate ions. <em>U<\/em> may be calculated from the Born-Haber cycle.\r\n\r\nThe values in kJ\/mol are approximately (a) 3791; (b) 3223; (c) 821; (d) 740; and (e) 2957.\r\n\r\nThe answer is (d), which requires about 740 kJ\/mol.\r\n\r\n15.\u00a0In each case, think about how it would affect the Born-Haber cycle. Recall that the more negative the overall value, the more exothermic the reaction is.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>The smaller the radius of the cation, the shorter the interionic distance and the greater the lattice energy would be. Since the lattice energy is negative in the Born-Haber cycle, this would lead to a more exothermic reaction.<\/li>\r\n \t<li>A lower ionization energy is a lower positive energy in the Born-Haber cycle. This would make the reaction more exothermic, as a smaller positive value is \u201cmore exothermic.\u201d<\/li>\r\n \t<li>As in part (b), the bond energy is a positive energy. The lower it is, the more exothermic the reaction will be.<\/li>\r\n \t<li>A higher electron affinity is more negative. In the Born-Haber cycle, the more negative the electron affinity, the more exothermic the overall reaction.<\/li>\r\n \t<li>The smaller the radius of the anion, the shorter the interionic distance and the greater the lattice energy would be. Since the lattice energy is negative in the Born-Haber cycle, this would lead to a more exothermic reaction.<\/li>\r\n<\/ol>\r\n16.\u00a04008 kJ\/mol; both ions in MgO have twice the charge of the ions in LiF; the bond length is very similar and both have the same structure; a quadrupling of the energy is expected based on the equation for lattice energy\r\n\r\n17. The compounds with the larger lattice energy are\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>MgO; selenium has larger radius than oxygen and, therefore, a larger interionic distance and thus, a larger smaller lattice energy than MgO<\/li>\r\n \t<li>MgO; the higher charges on Mg and O, given the similar radii of the ions, leads to a larger lattice energy<\/li>\r\n \t<li>Li<sub>2<\/sub>O; the higher charge on O<sup>2\u2013<\/sup> leads to a larger energy; additionally, Cl<sup>\u2013<\/sup> is larger than O<sup>2\u2013<\/sup>; this leads to a larger interionic distance in LiCl and a lower lattice energy<\/li>\r\n \t<li>MgO; the higher charge on Mg leads to a larger lattice energy<\/li>\r\n<\/ol>\r\n18.\u00a0The compounds with the larger lattice energy are\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Na<sub>2<\/sub>O; Na<sup>+<\/sup> has a smaller radius than K<sup>+<\/sup><\/li>\r\n \t<li>BaS; Ba has a larger charge than K<\/li>\r\n \t<li>BaS; Ba and S have larger charges<\/li>\r\n \t<li>BaS; S has a larger charge<\/li>\r\n<\/ol>\r\n19.\u00a0MgO\r\n\r\n20.\u00a0CaO\r\n\r\n21.\u00a0924 kJ\/mol\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>bond energy:\u00a0<\/strong>(also, bond dissociation energy) energy required to break a covalent bond in a gaseous substance\r\n\r\n<strong>Born-Haber cycle:\u00a0<\/strong>thermochemical cycle relating the various energetic steps involved in the formation of an ionic solid from the relevant elements\r\n\r\n<strong>lattice energy (\u0394<em>H<\/em><sub>lattice<\/sub>):\u00a0<\/strong>energy required to separate one mole of an ionic solid into its component gaseous ions","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Describe the energetics of covalent and ionic bond formation and breakage<\/li>\n<li>Use the Born-Haber cycle to compute lattice energies for ionic compounds<\/li>\n<li>Use average covalent bond energies to estimate enthalpies of reaction<\/li>\n<\/ul>\n<\/div>\n<p>A bond\u2019s strength describes how strongly each atom is joined to another atom, and therefore how much energy is required to break the bond between the two atoms. In this section, you will learn about the bond strength of covalent bonds, and then compare that to the strength of ionic bonds, which is related to the lattice energy of a compound.<\/p>\n<h2 data-type=\"title\">Bond Strength: Covalent Bonds<\/h2>\n<p>Stable molecules exist because covalent bonds hold the atoms together. We measure the strength of a covalent bond by the energy required to break it, that is, the energy necessary to separate the bonded atoms. Separating any pair of bonded atoms requires energy (see Figure 1). The stronger a bond, the greater the energy required to break it.<\/p>\n<div style=\"width: 610px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211345\/CNX_Chem_07_02_Morse1.jpg\" alt=\"A graph is shown with the x-axis labeled, \u201cInternuclear distance ( p m )\u201d while the y-axis is labeled, \u201cEnergy ( J ).\u201d One value, \u201c0,\u201d is labeled midway up the y-axis and two values: \u201c0\u201d at the far left and \u201c0.74\u201d to the left, are labeled on the x-axis. The point \u201c0.74\u201d is labeled, \u201cH bond H distance.\u201d A line is graphed that begins near the top of the y-axis and to the far left on the x-axis and drops steeply to a point labeled, \u201cnegative 7.24 times 10 superscript negative 19 J\u201d on the y-axis and 0.74 on the x-axis. This low point on the graph corresponds to a drawing of two spheres that overlap considerably. The line then rises to zero on the y-axis and levels out. The point where it almost reaches zero corresponds to two spheres that overlap slightly. The line at zero on the y-axis corresponds to two spheres that are far from one another.\" width=\"600\" height=\"589\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The potential energy of two separate hydrogen atoms (right) decreases as they approach each other, and the single electrons on each atom are shared to form a covalent bond. The bond length is the internuclear distance at which the lowest potential energy is achieved.<\/p>\n<\/div>\n<p>The energy required to break a specific covalent bond in one mole of gaseous molecules is called the bond energy or the bond dissociation energy. The bond energy for a diatomic molecule, D<sub>X\u2013Y<\/sub>, is defined as the standard enthalpy change for the endothermic reaction:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{XY}\\left(g\\right)\\rightarrow\\text{X}\\left(g\\right)+\\text{Y}\\left(g\\right){\\text{D}}_{\\text{X-Y}}=\\Delta H^{\\circ}[\/latex]<\/p>\n<p>For example, the bond energy of the pure covalent H\u2013H bond, D<sub>H\u2013H<\/sub>, is 436 kJ per mole of H\u2013H bonds broken:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{2}\\left(g\\right)\\rightarrow 2\\text{H}\\left(g\\right){\\text{D}}_{\\text{H-H}}=\\Delta {H}^{\\circ}=436\\text{ kJ}[\/latex]<\/p>\n<p>Molecules with three or more atoms have two or more bonds. The sum of all bond energies in such a molecule is equal to the standard enthalpy change for the endothermic reaction that breaks all the bonds in the molecule. For example, the sum of the four C\u2013H bond energies in CH<sub>4<\/sub>, 1660 kJ, is equal to the standard enthalpy change of the reaction:<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211644\/CNX_Chem_07_05_CH4bond_img1.jpg\" alt=\"A reaction is shown with Lewis structures. The first structure shows a carbon atom single bonded to four hydrogen atoms with the symbol, \u201c( g )\u201d written next to it. A right-facing arrow points to the letter \u201cC\u201d and the symbol \u201c( g ),\u201d which is followed by a plus sign. Next is the number 4, the letter \u201cH\u201d and the symbol, \u201c( g ).\u201d To the right of this equation is another equation: capital delta H superscript degree symbol equals 1660 k J.\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p>The average C\u2013H bond energy, D<sub>C\u2013H<\/sub>, is 1660\/4 = 415 kJ\/mol because there are four moles of C\u2013H bonds broken per mole of the reaction. Although the four C\u2013H bonds are equivalent in the original molecule, they do not each require the same energy to break; once the first bond is broken (which requires 439 kJ\/mol), the remaining bonds are easier to break. The 415 kJ\/mol value is the average, not the exact value required to break any one bond.<\/p>\n<p>The strength of a bond between two atoms increases as the number of electron pairs in the bond increases. Generally, as the bond strength increases, the bond length decreases. Thus, we find that triple bonds are stronger and shorter than double bonds between the same two atoms; likewise, double bonds are stronger and shorter than single bonds between the same two atoms. Average bond energies for some common bonds appear in Table 1, and a comparison of bond lengths and bond strengths for some common bonds appears in Table 2. When one atom bonds to various atoms in a group, the bond strength typically decreases as we move down the group. For example, C\u2013F is 439 kJ\/mol, C\u2013Cl is 330 kJ\/mol, and C\u2013Br is 275 kJ\/mol.<\/p>\n<table summary=\"This table has six columns and twenty-four rows. The first row is a header row that labels the columns: \u201cBond,\u201d \u201cBond Energy,\u201d \u201cBond,\u201d \u201cBond Energy,\u201d \u201cBond,\u201d and, \u201cBond Energy.\u201d Under the first \u201cBond\u201d column are the values: H bond to H with a single bond; H bonds to C with a single bond; H bonds to N with a single bond; H bonds to O with a single bond; H bonds to F with a single bond; H bonds to S i with a single bond; H bonds to P with a single bond; H bonds to S with a single bond; H bonds to C l with a single bond; H bonds to B r with a single bond; H bonds to I with a single bond; C bonds to C with a single bond; C bonds to C with a double bond; C bonds to C with a triple bond; C bonds to N with a single bond; C bonds to N with a double bond; C bonds to N with a triple bond; C bonds to O with a single bond; C bonds to O with a double bond; C bonds to O with a triple bond; C bonds to F with a single bond; C bonds to S i with a single bond; and C bonds to P with a single bond. Under the first \u201cBond Energy\u201d column are the values: 436; 415; 390; 464; 569; 395; 320; 340; 432; 370; 295; 345; 611; 837; 290; 615; 891; 350; 741; 1080; 439; 360; and 265. Under the second \u201cBond\u201d column are the values: C bonds to S with a single bond; C bonds to C l with a single bond; C bonds to B r with a single bond; C bonds to I with a single bond; N bonds to N with a single bond; N bonds to N with a double bond; N bonds to N with a triple bond; N bonds to O with a single bond; N bonds to F with a single bond; N bonds to P with a single bond; N bonds to C l with a single bond; N bonds to B r with a single bond; O bonds to O with a single bond; O bonds to O with a double bond; O bonds to F with a single bond; O bonds to S i with a single bond; O bonds to P with a single bond; O bonds to C l with a single bond; O bonds to I with a single bond; F bonds to F with a single bond; F bonds to S i with a single bond; F bonds to P with a single bond; and F bonds to S with a single bond. Under the second \u201cBond Energy\u201d column are the values: 260; 330; 275; 240; 160; 418; 946; 200; 270; 210; 200; 245; 140; 498; 160; 370; 350; 205; 200; 160; 540; 489; and 285. Under the third \u201cBond\u201d column are the values: F bonds to C l with a single bond; F bonds to B r with a single bond; S i bonds to S i with a single bond; S i bonds to P with a single bond; S i bonds to S with a single bond; S i bonds to C l with a single bond; S i bonds to B r with a single bond; S i bonds to I with a single bond; P bonds to P with a single bond; P bonds to S with a single bond; P bonds to C l with a single bond; P bonds to B r with a single bond; P bonds to I with a single bond; S bonds to S with a single bond; S bonds to C l with a single bond; S bonds to B r with a single bond; C l bonds to C l with a single bond; C l bonds to B r with a single bond; C l bonds to I with a single bond; B r bonds to B r with a single bond; B r bonds to I with a single bond; I bonds to I with a single bond; and the last cell in the column is empty. Under the third \u201cBond Energy\u201d column are the values: 255; 235; 230; 215; 225; 359; 290; 215; 215; 230; 330; 270; 215; 215; 250; 215; 243; 220; 210; 190; 180; 150; and the last cell in the column is empty.\">\n<thead>\n<tr>\n<th colspan=\"6\">Table 1. Bond Energies (kJ\/mol)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Bond<\/th>\n<th>Bond Energy<\/th>\n<th>Bond<\/th>\n<th>Bond Energy<\/th>\n<th>Bond<\/th>\n<th>Bond Energy<\/th>\n<\/tr>\n<tr>\n<td>H\u2013H<\/td>\n<td>436<\/td>\n<td>C\u2013S<\/td>\n<td>260<\/td>\n<td>F\u2013Cl<\/td>\n<td>255<\/td>\n<\/tr>\n<tr>\n<td>H\u2013C<\/td>\n<td>415<\/td>\n<td>C\u2013Cl<\/td>\n<td>330<\/td>\n<td>F\u2013Br<\/td>\n<td>235<\/td>\n<\/tr>\n<tr>\n<td>H\u2013N<\/td>\n<td>390<\/td>\n<td>C\u2013Br<\/td>\n<td>275<\/td>\n<td>Si\u2013Si<\/td>\n<td>230<\/td>\n<\/tr>\n<tr>\n<td>H\u2013O<\/td>\n<td>464<\/td>\n<td>C\u2013I<\/td>\n<td>240<\/td>\n<td>Si\u2013P<\/td>\n<td>215<\/td>\n<\/tr>\n<tr>\n<td>H\u2013F<\/td>\n<td>569<\/td>\n<td>N\u2013N<\/td>\n<td>160<\/td>\n<td>Si\u2013S<\/td>\n<td>225<\/td>\n<\/tr>\n<tr>\n<td>H\u2013Si<\/td>\n<td>395<\/td>\n<td>[latex]\\text{N}=\\text{N}[\/latex]<\/td>\n<td>418<\/td>\n<td>Si\u2013Cl<\/td>\n<td>359<\/td>\n<\/tr>\n<tr>\n<td>H\u2013P<\/td>\n<td>320<\/td>\n<td>[latex]\\text{N}\\equiv \\text{N}[\/latex]<\/td>\n<td>946<\/td>\n<td>Si\u2013Br<\/td>\n<td>290<\/td>\n<\/tr>\n<tr>\n<td>H\u2013S<\/td>\n<td>340<\/td>\n<td>N\u2013O<\/td>\n<td>200<\/td>\n<td>Si\u2013I<\/td>\n<td>215<\/td>\n<\/tr>\n<tr>\n<td>H\u2013Cl<\/td>\n<td>432<\/td>\n<td>N\u2013F<\/td>\n<td>270<\/td>\n<td>P\u2013P<\/td>\n<td>215<\/td>\n<\/tr>\n<tr>\n<td>H\u2013Br<\/td>\n<td>370<\/td>\n<td>N\u2013P<\/td>\n<td>210<\/td>\n<td>P\u2013S<\/td>\n<td>230<\/td>\n<\/tr>\n<tr>\n<td>H\u2013I<\/td>\n<td>295<\/td>\n<td>N\u2013Cl<\/td>\n<td>200<\/td>\n<td>P\u2013Cl<\/td>\n<td>330<\/td>\n<\/tr>\n<tr>\n<td>C\u2013C<\/td>\n<td>345<\/td>\n<td>N\u2013Br<\/td>\n<td>245<\/td>\n<td>P\u2013Br<\/td>\n<td>270<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\text{C}=\\text{C}[\/latex]<\/td>\n<td>611<\/td>\n<td>O\u2013O<\/td>\n<td>140<\/td>\n<td>P\u2013I<\/td>\n<td>215<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\text{C}\\equiv \\text{C}[\/latex]<\/td>\n<td>837<\/td>\n<td>[latex]\\text{O}=\\text{O}[\/latex]<\/td>\n<td>498<\/td>\n<td>S\u2013S<\/td>\n<td>215<\/td>\n<\/tr>\n<tr>\n<td>C\u2013N<\/td>\n<td>290<\/td>\n<td>O\u2013F<\/td>\n<td>160<\/td>\n<td>S\u2013Cl<\/td>\n<td>250<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\text{C}=\\text{N}[\/latex]<\/td>\n<td>615<\/td>\n<td>O\u2013Si<\/td>\n<td>370<\/td>\n<td>S\u2013Br<\/td>\n<td>215<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\text{C}\\equiv \\text{N}[\/latex]<\/td>\n<td>891<\/td>\n<td>O\u2013P<\/td>\n<td>350<\/td>\n<td>Cl\u2013Cl<\/td>\n<td>243<\/td>\n<\/tr>\n<tr>\n<td>C\u2013O<\/td>\n<td>350<\/td>\n<td>O\u2013Cl<\/td>\n<td>205<\/td>\n<td>Cl\u2013Br<\/td>\n<td>220<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\text{C}=\\text{O}[\/latex]<\/td>\n<td>741<\/td>\n<td>O\u2013I<\/td>\n<td>200<\/td>\n<td>Cl\u2013I<\/td>\n<td>210<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\text{C}\\equiv \\text{O}[\/latex]<\/td>\n<td>1080<\/td>\n<td>F\u2013F<\/td>\n<td>160<\/td>\n<td>Br\u2013Br<\/td>\n<td>190<\/td>\n<\/tr>\n<tr>\n<td>C\u2013F<\/td>\n<td>439<\/td>\n<td>F\u2013Si<\/td>\n<td>540<\/td>\n<td>Br\u2013I<\/td>\n<td>180<\/td>\n<\/tr>\n<tr>\n<td>C\u2013Si<\/td>\n<td>360<\/td>\n<td>F\u2013P<\/td>\n<td>489<\/td>\n<td>I\u2013I<\/td>\n<td>150<\/td>\n<\/tr>\n<tr>\n<td>C\u2013P<\/td>\n<td>265<\/td>\n<td>F\u2013S<\/td>\n<td>285<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table summary=\"This table has three columns and ten rows. The first row is a header row that labels the columns: \u201cBond,\u201d \u201cBond Length in angstroms,\u201d and, \u201cBond Energy in k J \/ mol.\u201d Under the column \u201cBond\u201d are the values: C bonds to C with a single bond; C bonds to C with a double bond; C bonds to C with a triple bond; C bonds to N with a single bond; C bonds to N with a double bond; C bonds to N with a triple bond; C bonds to O with a single bond; C bonds to O with a double bond; and C bonds to O with a triple bond. Under the column \u201cBond Length in angstroms\u201d are the values: 1.54; 1.34; 1.20; 1.43; 1.38; 1.16; 1.43; 1.23; and 1.13. Under the column \u201cBond Energy in k J \/ mol\u201d are the values: 345; 611; 837; 290; 615; 891; 350; 741; and 1080.\">\n<thead>\n<tr>\n<th colspan=\"3\">Table 2. Average Bond Lengths and Bond Energies for Some Common Bonds<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Bond<\/th>\n<th>Bond Length (\u00c5)<\/th>\n<th>Bond Energy (kJ\/mol)<\/th>\n<\/tr>\n<tr>\n<td>C\u2013C<\/td>\n<td>1.54<\/td>\n<td>345<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\text{C}=\\text{C}[\/latex]<\/td>\n<td>1.34<\/td>\n<td>611<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\text{C}\\equiv \\text{C}[\/latex]<\/td>\n<td>1.20<\/td>\n<td>837<\/td>\n<\/tr>\n<tr>\n<td>C\u2013N<\/td>\n<td>1.43<\/td>\n<td>290<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\text{C}=\\text{N}[\/latex]<\/td>\n<td>1.38<\/td>\n<td>615<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\text{C}\\equiv \\text{N}[\/latex]<\/td>\n<td>1.16<\/td>\n<td>891<\/td>\n<\/tr>\n<tr>\n<td>C\u2013O<\/td>\n<td>1.43<\/td>\n<td>350<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\text{C}=\\text{O}[\/latex]<\/td>\n<td>1.23<\/td>\n<td>741<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\text{C}\\equiv \\text{O}[\/latex]<\/td>\n<td>1.13<\/td>\n<td>1080<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We can use bond energies to calculate approximate enthalpy changes for reactions where enthalpies of formation are not available. Calculations of this type will also tell us whether a reaction is exothermic or endothermic. An exothermic reaction (\u0394<em>H<\/em> negative, heat produced) results when the bonds in the products are stronger than the bonds in the reactants. An endothermic reaction (\u0394<em>H<\/em> positive, heat absorbed) results when the bonds in the products are weaker than those in the reactants.<\/p>\n<p>The enthalpy change, \u0394<em>H<\/em>, for a chemical reaction is approximately equal to the sum of the energy required to break all bonds in the reactants (energy \u201cin,\u201d positive sign) plus the energy released when all bonds are formed in the products (energy \u201cout,\u201d negative sign). This can be expressed mathematically in the following way:<\/p>\n<p style=\"text-align: center;\">[latex]\\Delta H={\\Sigma{D}}_{\\text{bonds broken}}-{\\Sigma{D}}_{\\text{bonds formed}}[\/latex]<\/p>\n<p>In this expression, the symbol \u01a9 means \u201cthe sum of\u201d and D represents the bond energy in kilojoules per mole, which is always a positive number. The bond energy is obtained from a table (like Table 2) and will depend on whether the particular bond is a single, double, or triple bond. Thus, in calculating enthalpies in this manner, it is important that we consider the bonding in all reactants and products. Because D values are typically averages for one type of bond in many different molecules, this calculation provides a rough estimate, not an exact value, for the enthalpy of reaction.<\/p>\n<p>Consider the following reaction:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{2}\\left(g\\right)+{\\text{Cl}}_{2}\\left(g\\right)\\rightarrow 2\\text{HCl}\\left(g\\right)[\/latex]<br \/>\nor<br \/>\n[latex]\\text{H-H}\\left(g\\right)+\\text{Cl-Cl}\\left(g\\right)\\rightarrow 2\\text{H-Cl}\\left(g\\right)[\/latex]<\/p>\n<p>To form two moles of HCl, one mole of H\u2013H bonds and one mole of Cl\u2013Cl bonds must be broken. The energy required to break these bonds is the sum of the bond energy of the H\u2013H bond (436 kJ\/mol) and the Cl\u2013Cl bond (243 kJ\/mol). During the reaction, two moles of H\u2013Cl bonds are formed (bond energy = 432 kJ\/mol), releasing 2 \u00d7 432 kJ; or 864 kJ. Because the bonds in the products are stronger than those in the reactants, the reaction releases more energy than it consumes:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}\\hfill \\Delta H& =& {\\Sigma{D}}_{\\text{bonds broken}}-{\\Sigma{D}}_{\\text{bonds formed}}\\hfill \\\\ \\hfill \\Delta H& =& \\left[{\\text{D}}_{\\text{H-H}}+{\\text{D}}_{\\text{Cl-Cl}}\\right]-2{\\text{D}}_{\\text{H-Cl}}\\hfill \\\\ & =& \\left[436+243\\right]-2\\left(432\\right)=-185\\text{ kJ}\\hfill \\end{array}[\/latex]<\/p>\n<p>This excess energy is released as heat, so the reaction is exothermic. <a title=\"Standard Thermodynamic Properties for Selected Substances\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\">Standard Thermodynamic Properties for Selected Substances<\/a> gives a value for the standard molar enthalpy of formation of HCl(g), [latex]\\Delta{H}_{\\text{f}}^{\\circ},[\/latex] of \u201392.307 kJ\/mol. Twice that value is \u2013184.6 kJ, which agrees well with the answer obtained earlier for the formation of two moles of HCl.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0Using Bond Energies to Calculate Approximate Enthalpy Changes<\/h3>\n<p>Methanol, CH<sub>3<\/sub>OH, may be an excellent alternative fuel. The high-temperature reaction of steam and carbon produces a mixture of the gases carbon monoxide, CO, and hydrogen, H<sub>2<\/sub>, from which methanol can be produced. Using the bond energies in Table 2, calculate the approximate enthalpy change, \u0394<em>H<\/em>, for the reaction here:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{CO}\\left(g\\right)+2{\\text{H}}_{2}\\left(g\\right)\\rightarrow{\\text{CH}}_{3}\\text{OH}\\left(g\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q231938\">Show Answer<\/span><\/p>\n<div id=\"q231938\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we need to write the Lewis structures of the reactants and the products:<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211645\/CNX_Chem_07_05_CH3OHLew_img1.jpg\" alt=\"A set of Lewis diagrams show a chemical reaction. The first structure shows a carbon atom with a lone pair of electrons triple bonded to an oxygen with a lone pair of electrons. To the right of this structure is a plus sign, then the number 2 followed by a hydrogen atom single bonded to a hydrogen atom. To the right of this structure is a right-facing arrow followed by a hydrogen atom single bonded to a carbon atom that is single bonded to two hydrogen atoms and an oxygen atom with two lone pairs of electrons. The oxygen atom is also single bonded to a hydrogen atom.\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p>From this, we see that \u0394<em>H<\/em> for this reaction involves the energy required to break a C\u2013O triple bond and two H\u2013H single bonds, as well as the energy produced by the formation of three C\u2013H single bonds, a C\u2013O single bond, and an O\u2013H single bond. We can express this as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}{}\\Delta{H}&=&\\Sigma_{\\text{bonds broken}}-\\Sigma{D}_{\\text{bonds formed}}\\\\\\Delta{H}&=&\\left[\\text{D}_{\\text{C}\\equiv\\text{O}}+2\\left(\\text{D}_{\\text{H}-\\text{H}}\\right)\\right]-\\left[3\\left(\\text{D}_{\\text{C}-\\text{H}}\\right)+\\text{D}_{\\text{C-O}}+{\\text{D}}_{\\text{O-H}}\\right]\\end{array}[\/latex]<\/p>\n<p>Using the bond energy values in Table 2, we obtain:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\hfill \\Delta H& =\\left[1080+2\\left(436\\right)\\right]-\\left[3\\left(415\\right)+350+464\\right]\\\\ & =-107\\text{kJ}\\end{array}[\/latex]<\/p>\n<p>We can compare this value to the value calculated based on [latex]\\Delta{H}_{\\text{f}}^{\\circ}[\/latex] data from\u00a0<a title=\"Standard Thermodynamic Properties for Selected Substances\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\">Standard Thermodynamic Properties for Selected Substances<\/a>:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\hfill \\Delta H& =\\left[\\Delta{H}_{\\text{f}}^{\\circ}{\\text{CH}}_{3}\\text{OH}\\left(g\\right)\\right]-\\left[\\Delta{H}_{\\text{f}}^{\\circ}\\text{CO}\\left(g\\right)+2\\times \\Delta{H}_{\\text{f}}^{\\circ}{\\text{H}}_{2}\\right]\\\\ & =\\left[-201.0\\right]-\\left[-110.52+2\\times 0\\right]\\\\ & =-90.5\\text{kJ}\\end{array}[\/latex]<\/p>\n<p>Note that there is a fairly significant gap between the values calculated using the two different methods. This occurs because D values are the <em>average<\/em> of different bond strengths; therefore, they often give only rough agreement with other data.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Ethyl alcohol, CH<sub>3<\/sub>CH<sub>2<\/sub>OH, was one of the first organic chemicals deliberately synthesized by humans. It has many uses in industry, and it is the alcohol contained in alcoholic beverages. It can be obtained by the fermentation of sugar or synthesized by the hydration of ethylene in the following reaction:<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211646\/CNX_Chem_07_05_Ethanol_img1.jpg\" alt=\"A set of Lewis structures show a chemical reaction. The first structure shows two carbon atoms that are double bonded together and are each single bonded to two hydrogen atoms. This structure is followed by a plus sign, then an oxygen atom with two lone pairs of electrons single bonded to two hydrogen atoms. A right-facing arrow leads to a carbon atom single bonded to three hydrogen atoms and a second carbon atom. The second carbon atom is single bonded to two hydrogen atoms and an oxygen atom with two lone pairs of electrons. The oxygen atom is single bonded to a hydrogen atom as well.\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p>Using the bond energies in Table 2, calculate an approximate enthalpy change, \u0394<em>H<\/em>, for this reaction.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q893975\">Show Answer<\/span><\/p>\n<div id=\"q893975\" class=\"hidden-answer\" style=\"display: none\">\u201335 kJ<\/div>\n<\/div>\n<\/div>\n<h2 data-type=\"title\">Ionic Bond Strength and Lattice Energy<\/h2>\n<p>An ionic compound is stable because of the electrostatic attraction between its positive and negative ions. The lattice energy of a compound is a measure of the strength of this attraction. The lattice energy (\u0394<em>H<\/em><sub>lattice<\/sub>) of an ionic compound is defined as the energy required to separate one mole of the solid into its component gaseous ions. For the ionic solid MX, the lattice energy is the enthalpy change of the process:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{MX}\\left(s\\right)\\rightarrow{\\text{M}}^{n\\text{+}}\\left(g\\right)+{\\text{X}}^{n-}\\left(g\\right)\\Delta{H}_{\\text{lattice}}[\/latex]<\/p>\n<p>Note that we are using the convention where the ionic solid is separated into ions, so our lattice energies will be <em>endothermic<\/em> (positive values). Some texts use the equivalent but opposite convention, defining lattice energy as the energy released when separate ions combine to form a lattice and giving negative (exothermic) values. Thus, if you are looking up lattice energies in another reference, be certain to check which definition is being used. In both cases, a larger magnitude for lattice energy indicates a more stable ionic compound. For sodium chloride, \u0394<em>H<\/em><sub>lattice<\/sub> = 769 kJ. Thus, it requires 769 kJ to separate one mole of solid NaCl into gaseous Na<sup>+<\/sup> and Cl<sup>\u2013<\/sup> ions. When one mole each of gaseous Na<sup>+<\/sup> and Cl<sup>\u2013<\/sup> ions form solid NaCl, 769 kJ of heat is released.<\/p>\n<p>The lattice energy \u0394<em>H<\/em><sub>lattice<\/sub> of an ionic crystal can be expressed by the following equation (derived from Coulomb\u2019s law, governing the forces between electric charges):<\/p>\n<p style=\"text-align: center;\">[latex]\\Delta{H}_{\\text{lattice}}=\\frac{\\text{C}\\left({\\text{Z}}^{\\text{+}}\\right)\\left({\\text{Z}}^{-}\\right)}{{\\text{R}}_{\\text{o}}}[\/latex]<\/p>\n<p>in which C is a constant that depends on the type of crystal structure; Z<sup>+<\/sup> and Z<sup>\u2013<\/sup> are the charges on the ions; and R<sub>o<\/sub> is the interionic distance (the sum of the radii of the positive and negative ions). Thus, the lattice energy of an ionic crystal increases rapidly as the charges of the ions increase and the sizes of the ions decrease. When all other parameters are kept constant, doubling the charge of both the cation and anion quadruples the lattice energy. For example, the lattice energy of LiF (Z<sup>+<\/sup> and Z<sup>\u2013<\/sup> = 1) is 1023 kJ\/mol, whereas that of MgO (Z<sup>+<\/sup> and Z<sup>\u2013<\/sup> = 2) is 3900 kJ\/mol (R<sub>o<\/sub> is nearly the same\u2014about 200 pm for both compounds).<\/p>\n<p>Different interatomic distances produce different lattice energies. For example, we can compare the lattice energy of MgF<sub>2<\/sub> (2957 kJ\/mol) to that of MgI<sub>2<\/sub> (2327 kJ\/mol) to observe the effect on lattice energy of the smaller ionic size of F<sup>\u2013<\/sup> as compared to I<sup>\u2013<\/sup>.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2:\u00a0Lattice Energy Comparisons<\/h3>\n<p>The precious gem ruby is aluminum oxide, Al<sub>2<\/sub>O<sub>3<\/sub>, containing traces of Cr<sup>3+<\/sup>. The compound Al<sub>2<\/sub>Se<sub>3<\/sub> is used in the fabrication of some semiconductor devices. Which has the larger lattice energy, Al<sub>2<\/sub>O<sub>3<\/sub> or Al<sub>2<\/sub>Se<sub>3<\/sub>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q664779\">Show Answer<\/span><\/p>\n<div id=\"q664779\" class=\"hidden-answer\" style=\"display: none\">In these two ionic compounds, the charges Z<sup>+<\/sup> and Z<sup>\u2013<\/sup> are the same, so the difference in lattice energy will depend upon R<sub>o<\/sub>. The O<sup>2\u2013<\/sup> ion is smaller than the Se<sup>2\u2013<\/sup> ion. Thus, Al<sub>2<\/sub>O<sub>3<\/sub> would have a shorter interionic distance than Al<sub>2<\/sub>Se<sub>3<\/sub>, and Al<sub>2<\/sub>O<sub>3<\/sub> would have the larger lattice energy.<\/div>\n<\/div>\n<h4>Check Your Understanding<\/h4>\n<p>Zinc oxide, ZnO, is a very effective sunscreen. How would the lattice energy of ZnO compare to that of NaCl?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q508439\">Show Answer<\/span><\/p>\n<div id=\"q508439\" class=\"hidden-answer\" style=\"display: none\">ZnO would have the larger lattice energy because the Z values of both the cation and the anion in ZnO are greater, and the interionic distance of ZnO is smaller than that of NaCl.<\/div>\n<\/div>\n<\/div>\n<h2 data-type=\"title\">The Born-Haber Cycle<\/h2>\n<p>It is not possible to measure lattice energies directly. However, the lattice energy can be calculated using the equation given in the previous section or by using a thermochemical cycle. The <strong>Born-Haber cycle<\/strong> is an application of Hess\u2019s law that breaks down the formation of an ionic solid into a series of individual steps:<\/p>\n<ul data-bullet-=\"\">\n<li>[latex]\\Delta{H}_{\\text{f}}^{\\circ},[\/latex] the standard enthalpy of formation of the compound<\/li>\n<li><em>IE<\/em>, the ionization energy of the metal<\/li>\n<li><em>EA<\/em>, the electron affinity of the nonmetal<\/li>\n<li>[latex]\\Delta{H}_{s}^{\\circ},[\/latex] the enthalpy of sublimation of the metal<\/li>\n<li><em>D<\/em>, the bond dissociation energy of the nonmetal<\/li>\n<li>\u0394<em>H<\/em><sub>lattice<\/sub>, the lattice energy of the compound<\/li>\n<\/ul>\n<p>Figure 2 diagrams the Born-Haber cycle for the formation of solid cesium fluoride.<\/p>\n<figure>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211648\/CNX_Chem_07_05_BornHaber1.jpg\" alt=\"A diagram is shown. An upward facing arrow is drawn to the far left of the chart and is labeled \u201cH increasing.\u201d A horizontal line is drawn at the bottom of the chart. A downward-facing, vertical arrow to the left side of this line is labeled, \u201cOverall change.\u201d Beside this arrow is another label, \u201ccapital delta H subscript f, equals negative 553.5 k J per mol, ( Enthalpy of formation ).\u201d Three horizontal lines, one above the other, and all above the bottom line, are labeled, from bottom to top, as: \u201cC s ( s ), plus sign, one half F subscript 2, ( g ),\u201d \u201cC s ( g ), plus sign, one half F subscript 2, ( g ),\u201d and \u201cC s, superscript positive sign, ( g ), plus sign, one half F subscript 2, ( g ).\u201d Each of these lines is connected by an upward-facing vertical arrow. Each arrow is labeled, \u201ccapital delta H subscript 1, equals 76.5 k J per mol, ( Enthalpy of sublimation ),\u201d \u201ccapital delta H subscript 2, equals 375.7 k J per mol, ( ionization energy ),\u201d and \u201ccapital delta H subscript 3 equals 79.4 k J \/ mol ( one half dissociation energy ).\u201d Another horizontal line is drawn in the center top portion of the diagram and is labeled \u201cC s, superscript positive sign, ( g ), plus sign, F, ( g ).\u201d There is one more horizontal line drawn to the right of the overall diagram and located halfway down the image. An arrow connects the top line to this line and is labeled, \u201ccapital delta H equals negative 328.2 k J \/ mol ( electron affinity ).\u201d The line is labeled, \u201cC s superscript positive sign ( g ) plus F superscript negative sign ( g ).\u201d The arrow connecting this line to the bottom line is labeled, \u201cnegative capital delta H subscript lattice equals negative 756.9 k J \/ mol.\u201d The arrow points to a label on the bottom line which reads, \u201cC s F ( s ).\u201d\" width=\"975\" height=\"528\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The Born-Haber cycle shows the relative energies of each step involved in the formation of an ionic solid from the necessary elements in their reference states.<\/p>\n<\/div>\n<\/figure>\n<p>We begin with the elements in their most common states, Cs(<em>s<\/em>) and F<sub>2<\/sub>(<em>g<\/em>). The [latex]\\Delta{H}_{s}^{\\circ}[\/latex] represents the conversion of solid cesium into a gas, and then the ionization energy converts the gaseous cesium atoms into cations. In the next step, we account for the energy required to break the F-F bond to produce fluorine atoms. Converting one mole of fluorine atoms into fluoride ions is an exothermic process, so this step gives off energy (the electron affinity) and is shown as decreasing along the <em>y<\/em>-axis. We now have one mole of Cs cations and one mole of F anions. These ions combine to produce solid cesium fluoride. The enthalpy change in this step is the negative of the lattice energy, so it is also an exothermic quantity. The total energy involved in this conversion is equal to the experimentally determined enthalpy of formation, [latex]\\Delta{H}_{\\text{f}}^{\\circ},[\/latex] of the compound from its elements. In this case, the overall change is exothermic.<\/p>\n<p>Hess\u2019s law can also be used to show the relationship between the enthalpies of the individual steps and the enthalpy of formation. Table 3 shows this for cesium chloride, CsCl<sub>2<\/sub>.<\/p>\n<table summary=\"This table has two columns and six rows. The first row is labeled, \u201cEnthalpy of sublimation of C s ( s )\u201d and the enthalpy reaction is C s ( s ) yields C s ( g ). Beside this equation is capital delta H which equals capital delta H subscript s superscript degree symbol which also equals 76.5 k J. The second row is labeled, \u201cOne-half of the bond energy of C l subscript 2.\u201d The equation for this is one half C l subscript 2 ( g ) yields C l ( g ). Beside this equation is capital delta H equals one half D which also equals 122 k J. The third row is labeled, \u201cIonization Energy of N a ( g ).\u201d The equation for the ionization energy of N a ( g ) is N a ( g ) yields N a superscript positive sign ( g ) plus lower case e superscript negative sign. Beside this equation is capital delta H equals I E which also equals 496 k J. The fourth row is labeled, \u201cNegative of the electron affinity of C l.\u201d The equation for this is C l ( g ) plus lowercase e superscript negative sign yields C l superscript negative sign ( g ). Beside this equation is capital delta H equals negative E A which also equals negative 368 k J. The fifth row is labeled \u201cNegative of the lattice energy of N a C l ( s ).\u201d The equation for this is N a superscript positive sign ( g ) plus C l superscript negative sign ( g ) yields N a C l ( s ). Beside this equation is capital delta H equals negative capital delta H subscript lattice which also equals unknown. The sixth and final row is labeled, \u201cEnthalpy of formation of N a C l ( s ), add steps 1 - 5.\u201d The equation for this is capital delta H equals capital delta H subscript f superscript degree symbol which also equals capital delta H subscript s superscript degree symbol plus one-half D plus I E plus negative E A plus negative capital delta H subscript lattice. Underneath that equation, is another which is N a ( s ) plus one-half C l subscript 2 ( g ) yields N a C l ( s ) which equals negative 411 k J.\" data-label=\"\">\n<thead>\n<tr>\n<th colspan=\"2\">Table 3<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Enthalpy of sublimation of Cs(<em>s<\/em>)<\/td>\n<td>[latex]\\text{Cs}\\left(s\\right)\\rightarrow\\text{Cs}\\left(g\\right)\\Delta H=\\Delta{H}_{s}^{\\circ}=76.5\\text{kJ}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>One-half of the bond energy of Cl<sub>2<\/sub><\/td>\n<td>[latex]\\frac{1}{2}{\\text{Cl}}_{2}\\text{(}g\\text{)}\\rightarrow\\text{Cl}\\left(g\\right)\\Delta H=\\frac{1}{2}D=122\\text{kJ}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Ionization energy of Na(<em>g<\/em>)<\/td>\n<td>[latex]\\text{Na}\\left(g\\right)\\rightarrow{\\text{Na}}^{\\text{+}}\\text{(}g\\text{)}+{\\text{e}}^{-}\\Delta H=IE=496\\text{kJ}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Negative of the electron affinity of Cl<\/td>\n<td>[latex]\\text{Cl}\\left(g\\right)+{\\text{e}}^{-}\\rightarrow{\\text{Cl}}^{-}\\text{(}g\\text{)}\\Delta H=-EA=-368\\text{kJ}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Negative of the lattice energy of NaCl(<em>s<\/em>)<\/td>\n<td>[latex]{\\text{Na}}^{\\text{+}}\\text{(}g\\text{)}+{\\text{Cl}}^{-}\\text{(}g\\text{)}\\rightarrow\\text{NaCl}\\left(s\\right)\\Delta H=-{\\Delta{H}}_{\\text{lattice}}=?[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Enthalpy of formation of NaCl(<em>s<\/em>), add steps 1\u20135<\/td>\n<td>[latex]\\begin{array}{l}\\Delta H=\\Delta{H}_{f}^{\\circ}=\\Delta{H}_{s}^{\\circ}+\\frac{1}{2}D+IE+\\left(-EA\\right)+\\left(-{\\Delta{H}}_{\\text{lattice}}\\right)\\\\ \\text{Na}\\left(s\\right)+\\frac{1}{2}{\\text{Cl}}_{2}\\left(g\\right)\\rightarrow\\text{NaCl}\\left(s\\right)=-411\\text{kJ}\\end{array}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Thus, the lattice energy can be calculated from other values. For cesium chloride, using this data, the lattice energy is:<\/p>\n<p style=\"text-align: center;\">[latex]\\Delta{H}_{\\text{lattice}}=\\left(411+109+122+496+368\\right)\\text{kJ}=770\\text{kJ}[\/latex]<\/p>\n<p>The Born-Haber cycle may also be used to calculate any one of the other quantities in the equation for lattice energy, provided that the remainder is known. For example, if the relevant enthalpy of sublimation [latex]\\Delta{H}_{s}^{\\circ},[\/latex] ionization energy (IE), bond dissociation enthalpy (D), lattice energy \u0394<em>H<\/em><sub>lattice,<\/sub> and standard enthalpy of formation [latex]\\Delta{H}_{\\text{f}}^{\\circ}[\/latex] are known, the Born-Haber cycle can be used to determine the electron affinity of an atom.<\/p>\n<p>Lattice energies calculated for ionic compounds are typically much higher than bond dissociation energies measured for covalent bonds. Whereas lattice energies typically fall in the range of 600\u20134000 kJ\/mol (some even higher), covalent bond dissociation energies are typically between 150\u2013400 kJ\/mol for single bonds. Keep in mind, however, that these are not directly comparable values. For ionic compounds, lattice energies are associated with many interactions, as cations and anions pack together in an extended lattice. For covalent bonds, the bond dissociation energy is associated with the interaction of just two atoms.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>The strength of a covalent bond is measured by its bond dissociation energy, that is, the amount of energy required to break that particular bond in a mole of molecules. Multiple bonds are stronger than single bonds between the same atoms. The enthalpy of a reaction can be estimated based on the energy input required to break bonds and the energy released when new bonds are formed. For ionic bonds, the lattice energy is the energy required to separate one mole of a compound into its gas phase ions. Lattice energy increases for ions with higher charges and shorter distances between ions. Lattice energies are often calculated using the Born-Haber cycle, a thermochemical cycle including all of the energetic steps involved in converting elements into an ionic compound.<\/p>\n<h4>Key Equations<\/h4>\n<ul>\n<li>Bond energy for a diatomic molecule: [latex]\\text{XY}\\left(g\\right)\\rightarrow\\text{X}\\left(g\\right)+\\text{Y}\\left(g\\right){\\text{D}}_{\\text{X-Y}}=\\Delta H^{\\circ}[\/latex]<\/li>\n<li>Enthalpy change: \u0394<em>H<\/em> = \u01a9D<sub>bonds broken<\/sub> \u2013 \u01a9D<sub>bonds formed<\/sub><\/li>\n<li>Lattice energy for a solid MX: [latex]\\text{MX}\\left(s\\right)\\rightarrow{\\text{M}}^{n\\text{+}}\\left(g\\right)+{\\text{X}}^{n-}\\left(g\\right)\\Delta{H}_{\\text{lattice}}[\/latex]<\/li>\n<li>Lattice energy for an ionic crystal: [latex]\\Delta{H}_{\\text{lattice}}=\\frac{\\text{C}\\left({\\text{Z}}^{\\text{+}}\\right)\\left({\\text{Z}}^{-}\\right)}{{\\text{R}}_{\\text{o}}}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol>\n<li>Which bond in each of the following pairs of bonds is the strongest?\n<ol style=\"list-style-type: lower-alpha;\">\n<li>C\u2013C or [latex]\\text{C}=\\text{C}[\/latex]<\/li>\n<li>C\u2013N or [latex]\\text{C}\\equiv \\text{N}[\/latex]<\/li>\n<li>[latex]\\text{C}\\equiv \\text{O}[\/latex] or [latex]\\text{C}=\\text{O}[\/latex]<\/li>\n<li>H\u2013F or H\u2013Cl<\/li>\n<li>C\u2013H or O\u2013H<\/li>\n<li>C\u2013N or C\u2013O<\/li>\n<\/ol>\n<\/li>\n<li>Using the bond energies in Table 1, determine the approximate enthalpy change for each of the following reactions:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{\\text{H}}_{2}\\left(g\\right)+{\\text{Br}}_{2}\\left(g\\right)\\rightarrow 2\\text{HBr}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]{\\text{CH}}_{4}\\left(g\\right)+{\\text{I}}_{2}\\left(g\\right)\\rightarrow{\\text{CH}}_{3}\\text{I}\\left(g\\right)+\\text{HI}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]{\\text{C}}_{2}{\\text{H}}_{4}\\left(g\\right)+3{\\text{O}}_{2}\\left(g\\right)\\rightarrow 2{\\text{CO}}_{2}\\left(g\\right)+2{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>Using the bond energies in Table 1, determine the approximate enthalpy change for each of the following reactions:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{\\text{Cl}}_{2}\\left(g\\right)+3{\\text{F}}_{2}\\left(g\\right)\\rightarrow 2{\\text{ClF}}_{3}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]{\\text{H}}_{2}\\text{C}={\\text{CH}}_{2}\\left(g\\right)+{\\text{H}}_{2}\\left(g\\right)\\rightarrow{\\text{H}}_{3}{\\text{CCH}}_{3}\\left(g\\right)[\/latex]<\/li>\n<li>[latex]2{\\text{C}}_{2}{\\text{H}}_{6}\\left(g\\right)+7{\\text{O}}_{2}\\left(g\\right)\\rightarrow 4{\\text{CO}}_{2}\\left(g\\right)+6{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>When a molecule can form two different structures, the structure with the stronger bonds is usually the more stable form. Use bond energies to predict the correct structure of the hydroxylamine molecule: <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211650\/CNX_Chem_07_05_Hydroxya_img1.jpg\" alt=\"Two Lewis structures are shows with the word \u201cor\u201d written in between them. The left structure shows a nitrogen atom with one lone pair of electrons single bonded to two hydrogen atoms. It is also bonded to an oxygen atom with two lone pairs of electrons. The oxygen atom is bonded to a hydrogen atom. The right structure shows a nitrogen atom single bonded to three hydrogen atoms and an oxygen atom with three lone pairs of electrons.\" data-media-type=\"image\/jpeg\" \/><\/li>\n<li>How does the bond energy of HCl(<em>g<\/em>) differ from the standard enthalpy of formation of HCl(<em>g<\/em>)?<\/li>\n<li>Using the standard enthalpy of formation data in\u00a0<a title=\"Standard Thermodynamic Properties for Selected Substances\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\">Standard Thermodynamic Properties for Selected Substances<\/a>, show how can the standard enthalpy of formation of HCl(g) can be used to determine the bond energy.<\/li>\n<li>Using the standard enthalpy of formation data in\u00a0<a title=\"Standard Thermodynamic Properties for Selected Substances\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\">Standard Thermodynamic Properties for Selected Substances<\/a>, calculate the bond energy of the carbon-sulfur double bond in CS<sub>2<\/sub>.<\/li>\n<li>Using the standard enthalpy of formation data in\u00a0<a title=\"Standard Thermodynamic Properties for Selected Substances\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\">Standard Thermodynamic Properties for Selected Substances<\/a>, determine which bond is stronger: the S\u2013F bond in SF<sub>4<\/sub>(<em>g<\/em>) or in SF<sub>6<\/sub>(<em>g<\/em>)?<\/li>\n<li>Using the standard enthalpy of formation data in\u00a0<a title=\"Standard Thermodynamic Properties for Selected Substances\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\">Standard Thermodynamic Properties for Selected Substances<\/a>, determine which bond is stronger: the P\u2013Cl bond in PCl<sub>3<\/sub>(<em>g<\/em>) or in PCl<sub>5<\/sub>(<em>g<\/em>)?<\/li>\n<li>Complete the following Lewis structure by adding bonds (not atoms), and then indicate the longest bond:<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211651\/CNX_Chem_07_05_C6H8Lew_img1.jpg\" alt=\"A Lewis structure is shown that is missing its bonds. It shows a horizontal row of six carbon atoms, equally spaced. Three hydrogen atoms are drawn around the first carbon, two around the second, one above the fifth, and two by the sixth.\" data-media-type=\"image\/jpeg\" \/><\/li>\n<li>Use the bond energy to calculate an approximate value of \u0394<em>H<\/em> for the following reaction. Which is the more stable form of FNO<sub>2<\/sub>?<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211653\/CNX_Chem_07_05_FNO2_img1.jpg\" alt=\"Two Lewis structures are shown with a right-facing arrow in between. The left structure shows a nitrogen atom double bonded to an oxygen atom with two lone pairs of electrons. It is also bonded to a fluorine atom and another oxygen atom, each with three lone pairs of electrons. The right structure shows an oxygen atom with two lone pairs of electrons double bonded to a nitrogen atom with one lone pair of electrons. This nitrogen atom is single bonded to an oxygen with two lone pairs of electrons. The oxygen atom is single bonded to a fluorine atom with three lone pairs of electrons.\" data-media-type=\"image\/jpeg\" \/><\/li>\n<li>Use principles of atomic structure to answer each of the following:<sup>1\u00a0<\/sup>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>The radius of the Ca atom is 197 pm; the radius of the Ca<sup>2+<\/sup> ion is 99 pm. Account for the difference.<\/li>\n<li>The lattice energy of CaO(s) is \u20133460 kJ\/mol; the lattice energy of K<sub>2<\/sub>O is \u20132240 kJ\/mol. Account for the difference.<\/li>\n<li>Given these ionization values, explain the difference between Ca and K with regard to their first and second ionization energies.<br \/>\n<table summary=\"This table has three columns and three rows. The first row is a header row that labels the columns: \u201cElement,\u201d \u201cFirst Ionization Energy in k J \/ mol,\u201d and \u201cSecond Ionization Energy in k J \/ mol.\u201d Under the column \u201cElement\u201d are the letters: K and C a. Under the column \u201cFirst Ionization Energy in k J \/ mol\u201d are the values: 419 and 590. Under the column \u201cSecond Ionization Energy in k J \/ mol\u201d are the values: 3050 and 1140.\" data-label=\"\">\n<thead>\n<tr>\n<th>Element<\/th>\n<th>First Ionization Energy (kJ\/mol)<\/th>\n<th>Second Ionization Energy (kJ\/mol)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>K<\/td>\n<td>419<\/td>\n<td>3050<\/td>\n<\/tr>\n<tr>\n<td>Ca<\/td>\n<td>590<\/td>\n<td>1140<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>The first ionization energy of Mg is 738 kJ\/mol and that of Al is 578 kJ\/mol. Account for this difference.<\/li>\n<\/ol>\n<\/li>\n<li>The lattice energy of LiF is 1023 kJ\/mol, and the Li\u2013F distance is 200.8 pm. NaF crystallizes in the same structure as LiF but with a Na\u2013F distance of 231 pm. Which of the following values most closely approximates the lattice energy of NaF: 510, 890, 1023, 1175, or 4090 kJ\/mol? Explain your choice.<\/li>\n<li>For which of the following substances is the least energy required to convert one mole of the solid into separate ions?\n<ol style=\"list-style-type: lower-alpha;\">\n<li>MgO<\/li>\n<li>SrO<\/li>\n<li>KF<\/li>\n<li>CsF<\/li>\n<li>MgF<sub>2<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>The reaction of a metal, M, with a halogen, X<sub>2<\/sub>, proceeds by an exothermic reaction as indicated by this equation: [latex]\\text{M}\\left(s\\right)+{\\text{X}}_{2}\\left(g\\right)\\rightarrow{\\text{MX}}_{2}\\left(s\\right).[\/latex] For each of the following, indicate which option will make the reaction more exothermic. Explain your answers.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>a large radius vs. a small radius for M<sup>+2\u00a0<\/sup><\/li>\n<li>a high ionization energy vs. a low ionization energy for M<\/li>\n<li>an increasing bond energy for the halogen<\/li>\n<li>a decreasing electron affinity for the halogen<\/li>\n<li>an increasing size of the anion formed by the halogen<\/li>\n<\/ol>\n<\/li>\n<li>The lattice energy of LiF is 1023 kJ\/mol, and the Li\u2013F distance is 201 pm. MgO crystallizes in the same structure as LiF but with a Mg\u2013O distance of 205 pm. Which of the following values most closely approximates the lattice energy of MgO: 256 kJ\/mol, 512 kJ\/mol, 1023 kJ\/mol, 2046 kJ\/mol, or 4090 kJ\/mol? Explain your choice.<\/li>\n<li>Which compound in each of the following pairs has the larger lattice energy? Note: Mg<sup>2+<\/sup> and Li<sup>+<\/sup> have similar radii; O<sup>2\u2013<\/sup> and F<sup>\u2013<\/sup> have similar radii. Explain your choices.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>MgO or MgSe<\/li>\n<li>LiF or MgO<\/li>\n<li>Li<sub>2<\/sub>O or LiCl<\/li>\n<li>Li<sub>2<\/sub>Se or MgO<\/li>\n<\/ol>\n<\/li>\n<li>Which compound in each of the following pairs has the larger lattice energy? Note: Ba<sup>2+<\/sup> and K<sup>+<\/sup> have similar radii; S<sup>2\u2013<\/sup> and Cl<sup>\u2013<\/sup> have similar radii. Explain your choices.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>K<sub>2<\/sub>O or Na<sub>2<\/sub>O<\/li>\n<li>K<sub>2<\/sub>S or BaS<\/li>\n<li>KCl or BaS<\/li>\n<li>BaS or BaCl<sub>2<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?\n<ol style=\"list-style-type: lower-alpha;\">\n<li>MgO<\/li>\n<li>SrO<\/li>\n<li>KF<\/li>\n<li>CsF<\/li>\n<li>MgF<sub>2<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?\n<ol style=\"list-style-type: lower-alpha;\">\n<li>K<sub>2<\/sub>S<\/li>\n<li>K<sub>2<\/sub>O<\/li>\n<li>CaS<\/li>\n<li>Cs<sub>2<\/sub>S<\/li>\n<li>CaO<\/li>\n<\/ol>\n<\/li>\n<li>The lattice energy of KF is 794 kJ\/mol, and the interionic distance is 269 pm. The Na\u2013F distance in NaF, which has the same structure as KF, is 231 pm. Which of the following values is the closest approximation of the lattice energy of NaF: 682 kJ\/mol, 794 kJ\/mol, 924 kJ\/mol, 1588 kJ\/mol, or 3175 kJ\/mol? Explain your answer.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q888536\">Selected Answers<\/span><\/p>\n<div id=\"q888536\" class=\"hidden-answer\" style=\"display: none\">\n<p>1. In general, a multiple bond between the same two elements is stronger than a single bond. The greater the electronegativity difference between two similar elements, the greater the bond energy<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\text{C}=\\text{C}[\/latex]<\/li>\n<li>[latex]\\text{C}\\equiv \\text{N}[\/latex]<\/li>\n<li>[latex]\\text{C}\\equiv \\text{O}[\/latex]<\/li>\n<li>H\u2013F<\/li>\n<li>O\u2013H<\/li>\n<li>C\u2013O<\/li>\n<\/ol>\n<p>3.\u00a0The approximate enthalpy changes are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\begin{array}{ll}\\hfill DH^{\\circ}& ={\\Sigma{D}}_{\\text{bonds broken}}-{\\Sigma{D}}_{\\text{bonds formed}}\\\\ & =2{D}_{\\text{Cl-Cl}}+3{D}_{\\text{F-F}}-6{D}_{\\text{Cl-F}}\\\\ & =-564\\text{kJ}\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{ll}\\hfill DH^{\\circ}& ={\\Sigma{D}}_{\\text{bonds broken}}-{\\Sigma{D}}_{\\text{bonds formed}}\\\\ & ={D}_{\\text{C-C}}+4{D}_{\\text{C-H}}+{D}_{\\text{H-H}}-{D}_{\\text{C-C}}-6{D}_{\\text{C-H}}\\\\ & =611+4\\left(415\\right)+436-345-6\\left(415\\right)\\\\ & =-128\\text{kJ}\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{ll}\\hfill DH^{\\circ}& ={\\Sigma{D}}_{\\text{bonds broken}}-{\\Sigma{D}}_{\\text{bonds formed}}\\\\ & =2{D}_{\\text{C-C}}+12{D}_{\\text{C-H}}+7{D}_{\\text{O-O}}-8{D}_{\\text{C-O}}-12{D}_{\\text{O-H}}\\\\ & =2\\left(345\\right)+12\\left(415\\right)+7\\left(496\\right)-8\\left(741\\right)-12\\left(464\\right)\\\\ & =-2354\\text{kJ}\\end{array}[\/latex]<\/li>\n<\/ol>\n<p>5.\u00a0The bond energy involves breaking HCl into H and Cl atoms. The enthalpy of formation involves making HCl from H<sub>2<\/sub> and Cl<sub>2<\/sub> molecules.<\/p>\n<p>7. This problem can be solved as follows:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{CS}}_{2}\\left(g\\right)\\rightarrow\\text{C}\\left(\\text{graphite}\\right)+2\\text{S}\\left(s\\right){\\Delta{H}}_{1}^{\\circ}=\\Delta{H}_{\\text{f}\\left[{\\text{CS}}_{2}\\left(g\\right)\\right]}^{\\circ}[\/latex]<br \/>\n[latex]\\text{C}\\left(\\text{graphite}\\right)\\rightarrow\\text{C}\\left(g\\right)\\Delta{H}_{2}^{\\circ }=\\Delta{H}_{\\text{fC}\\left(g\\right)}^{\\circ}[\/latex]<br \/>\n[latex]2\\text{S}\\left(s\\right)\\rightarrow 2\\text{S}\\left(g\\right)2\\Delta{H}_{3}^{\\circ }=2\\Delta{H}_{\\text{fS}\\left(g\\right)}^{\\circ}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\\\ \\hfill {D}_{{\\text{CS}}_{2}}& =\\Delta H^{\\circ}=-{\\Delta{H}}_{\\text{f}\\left[\\left({\\text{CS}}_{2}\\left(g\\right)\\right)\\right]}^{\\circ}+\\Delta{H}_{\\text{fC}\\left(g\\right)}^{\\circ}+2\\Delta{H}_{\\text{fS}\\left(g\\right)}^{\\circ}\\\\ \\hfill & =-116.9+716.681+2\\left(278.81\\right)\\\\ \\hfill & =1157.4\\text{ kJ}{\\text{mol}}^{-1}\\\\ \\hfill {D}_{\\text{C }=\\text{S}}& =\\frac{1157.4}{2}=578.7\\text{kJ}{\\text{mol}}^{-1}\\text{of}\\text{C=S}\\text{ bonds}\\end{array}[\/latex]<\/p>\n<p>9. This problem can be solved as follows:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{PCl}}_{3}\\left(g\\right)\\rightarrow\\frac{1}{4}{\\text{P}}_{4}\\left(s\\right)+\\frac{3}{2}{\\text{Cl}}_{2}\\left(g\\right)\\Delta{H}_{1}^{\\circ}=-{\\Delta{H}}_{\\text{f}\\left[\\left({\\text{PCl}}_{3}\\left(g\\right)\\right)\\right]}^{\\circ}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{4}{\\text{P}}_{4}\\left(s\\right)\\rightarrow\\text{P}\\left(g\\right)\\Delta{H}_{2}^{\\circ}=\\Delta{H}_{\\text{fP}\\left(g\\right)}^{\\circ}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{3}{2}{\\text{Cl}}_{2}\\left(g\\right)\\rightarrow 3\\text{Cl}\\left(g\\right)3\\Delta{H}_{3}^{\\circ}=3\\Delta{H}_{\\text{fCl}\\left(g\\right)}^{\\circ}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\\\{D}_{{\\text{PCl}}_{3}}& =\\Delta H^{\\circ}=-{\\Delta{H}}_{\\text{f}\\left[\\left({\\text{PCl}}_{3}\\left(g\\right)\\right)\\right]}^{\\circ}+\\Delta{H}_{\\text{fP}\\left(g\\right)}^{\\circ}+3\\Delta{H}_{\\text{fCl}\\left(g\\right)}^{\\circ}\\\\ \\hfill & =287.0+314.64+3\\left(121.3\\right)=965.54{\\text{kJ mol}}^{-\\text{1}}\\\\ \\hfill {D}_{{\\text{PCl}}_{3}}& =\\frac{965.54\\text{kJ}}{3}=321.8{\\text{kJ per mol}}^{{-1}}\\text{of bonds}\\end{array}[\/latex]<\/p>\n<p>Proceeding in the same manners, [latex]-{\\Delta{H}}_{{\\text{f[PCl}}_{\\text{5}}\\text{(g)]}}=\\text{374.9 kJ}{\\text{mol}}^{-1}[\/latex]. The P(<em>g<\/em>) and the 5Cl(<em>g<\/em>) contribute 921.14 kJ; then [latex]{\\text{DF}}_{{\\text{PCl}}_{\\text{5}}}=\\text{1296.04 kJ}[\/latex] and [latex]{D}_{\\text{P-Cl}}=\\frac{\\text{1296.04 kJ\/mol}}{5}=\\text{259.2 kJ\/mol}[\/latex] of bonds. The P\u2013Cl bond in PCl<sub>3<\/sub> is stronger.<\/p>\n<p>10.The C\u2013C single bonds are longest.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211652\/CNX_Chem_07_05_C6H8ans_img1.jpg\" alt=\"A Lewis structure is shown. A carbon atom that is single bonded to three hydrogen atoms is bonded to a second carbon atom. The second carbon atom is single bonded to two hydrogen atoms. The second carbon atom is single bonded to a third carbon atom that is triple bonded to a fourth carbon atom and single bonded to a fifth carbon atom. The fifth carbon atom is single bonded to a hydrogen atom and double bonded to a sixth carbon atom that is single bonded to two hydrogen atoms.\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p>11. the left hand arrangement: [latex]\\text{O}=\\text{N}[\/latex] not listed, N-F 270, N-O 200; the right hand arrangement: [latex]\\text{O}=\\text{N}[\/latex] not listed, N-O 200, O-F 185; the bond energy of [latex]\\text{O}=\\text{N}[\/latex] does not matter because it must be the same in both cases, the form on the right has a bond energy of <em>X<\/em> +470; that on the right, <em>X<\/em> +385; the form on the left is more stable.<\/p>\n<p>12. The answers are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>When two electrons are removed from the valence shell, the Ca radius loses the outermost energy level and reverts to the lower <em>n<\/em> = 3 level, which is much smaller in radius.<\/li>\n<li>The +2 charge on calcium pulls the oxygen much closer compared with K, thereby increasing the lattice energy relative to a less charged ion.<\/li>\n<li>Removal of the 4<em>s<\/em> electron in Ca requires more energy than removal of the 4<em>s<\/em> electron in K, because of the stronger attraction of the nucleus and the extra energy required to break the pairing of the electrons. The second ionization energy for K requires that an electron be removed from a lower energy level, where the attraction is much stronger from the nucleus for the electron. In addition, energy is required to unpair two electrons in a full orbital. For Ca, the second ionization potential requires removing only a lone electron in the exposed outer energy level.<\/li>\n<li>In Al, the removed electron is relatively unprotected and unpaired in a <em>p<\/em> orbital. The higher energy for Mg mainly reflects the unpairing of the 2<em>s<\/em> electron.<\/li>\n<\/ol>\n<p>13. The lattice energy is given by [latex]U=C\\left(\\frac{{Z}^{+}{Z}^{-}}{{R}_{o}}\\right),[\/latex] where <em>R<\/em><sub>o<\/sub> is the interatomic distance. The charges are the same in both LiF and NaF. The major difference is expected to be the interatomic distance 2.008 \u00c5 versus 2.31 \u00c5. From the data for LiF, with <em>Z<\/em><sup>+<\/sup><em>Z<\/em><sup>\u2013<\/sup> = \u20131, [latex]C=\\frac{U{R}_{\\text{o}}}{{Z}^{+}{Z}^{-}}=\\frac{1023\\times 2.008}{-1}=-2054\\text{ kJ}\\text{A}{\\text{mol}}^{-1}[\/latex].<\/p>\n<p>Then, [latex]{U}_{\\text{NaF}}=\\frac{-2054\\text{kJ}\\text{A}{\\text{mol}}^{-1}\\left(-1\\right)}{2.31\\text{A}}=889\\text{kJ}{\\text{mol}}^{-1}\\text{or }890{\\text{ kJ mol}}^{-1}[\/latex].<\/p>\n<p>14. The lattice energy, <em>U<\/em>, is the energy required to convert the solid into separate ions. <em>U<\/em> may be calculated from the Born-Haber cycle.<\/p>\n<p>The values in kJ\/mol are approximately (a) 3791; (b) 3223; (c) 821; (d) 740; and (e) 2957.<\/p>\n<p>The answer is (d), which requires about 740 kJ\/mol.<\/p>\n<p>15.\u00a0In each case, think about how it would affect the Born-Haber cycle. Recall that the more negative the overall value, the more exothermic the reaction is.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>The smaller the radius of the cation, the shorter the interionic distance and the greater the lattice energy would be. Since the lattice energy is negative in the Born-Haber cycle, this would lead to a more exothermic reaction.<\/li>\n<li>A lower ionization energy is a lower positive energy in the Born-Haber cycle. This would make the reaction more exothermic, as a smaller positive value is \u201cmore exothermic.\u201d<\/li>\n<li>As in part (b), the bond energy is a positive energy. The lower it is, the more exothermic the reaction will be.<\/li>\n<li>A higher electron affinity is more negative. In the Born-Haber cycle, the more negative the electron affinity, the more exothermic the overall reaction.<\/li>\n<li>The smaller the radius of the anion, the shorter the interionic distance and the greater the lattice energy would be. Since the lattice energy is negative in the Born-Haber cycle, this would lead to a more exothermic reaction.<\/li>\n<\/ol>\n<p>16.\u00a04008 kJ\/mol; both ions in MgO have twice the charge of the ions in LiF; the bond length is very similar and both have the same structure; a quadrupling of the energy is expected based on the equation for lattice energy<\/p>\n<p>17. The compounds with the larger lattice energy are<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>MgO; selenium has larger radius than oxygen and, therefore, a larger interionic distance and thus, a larger smaller lattice energy than MgO<\/li>\n<li>MgO; the higher charges on Mg and O, given the similar radii of the ions, leads to a larger lattice energy<\/li>\n<li>Li<sub>2<\/sub>O; the higher charge on O<sup>2\u2013<\/sup> leads to a larger energy; additionally, Cl<sup>\u2013<\/sup> is larger than O<sup>2\u2013<\/sup>; this leads to a larger interionic distance in LiCl and a lower lattice energy<\/li>\n<li>MgO; the higher charge on Mg leads to a larger lattice energy<\/li>\n<\/ol>\n<p>18.\u00a0The compounds with the larger lattice energy are<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Na<sub>2<\/sub>O; Na<sup>+<\/sup> has a smaller radius than K<sup>+<\/sup><\/li>\n<li>BaS; Ba has a larger charge than K<\/li>\n<li>BaS; Ba and S have larger charges<\/li>\n<li>BaS; S has a larger charge<\/li>\n<\/ol>\n<p>19.\u00a0MgO<\/p>\n<p>20.\u00a0CaO<\/p>\n<p>21.\u00a0924 kJ\/mol<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>bond energy:\u00a0<\/strong>(also, bond dissociation energy) energy required to break a covalent bond in a gaseous substance<\/p>\n<p><strong>Born-Haber cycle:\u00a0<\/strong>thermochemical cycle relating the various energetic steps involved in the formation of an ionic solid from the relevant elements<\/p>\n<p><strong>lattice energy (\u0394<em>H<\/em><sub>lattice<\/sub>):\u00a0<\/strong>energy required to separate one mole of an ionic solid into its component gaseous ions<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1914\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry. <strong>Provided by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/openstaxcollege.org\">http:\/\/openstaxcollege.org<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17,"menu_order":12,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry\",\"author\":\"\",\"organization\":\"OpenStax College\",\"url\":\"http:\/\/openstaxcollege.org\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1914","chapter","type-chapter","status-publish","hentry"],"part":3023,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/1914","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":43,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/1914\/revisions"}],"predecessor-version":[{"id":5998,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/1914\/revisions\/5998"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/parts\/3023"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/1914\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/media?parent=1914"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapter-type?post=1914"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/contributor?post=1914"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/license?post=1914"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}