{"id":6387,"date":"2020-05-21T17:02:20","date_gmt":"2020-05-21T17:02:20","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/?post_type=chapter&#038;p=6387"},"modified":"2020-07-03T18:27:31","modified_gmt":"2020-07-03T18:27:31","slug":"lattice-energy-and-enthalpy-of-solution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/chapter\/lattice-energy-and-enthalpy-of-solution\/","title":{"raw":"Lattice Energy and Enthalpy of Solution","rendered":"Lattice Energy and Enthalpy of Solution"},"content":{"raw":"<h2>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>\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<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\" \/> 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]<\/figure>\r\nWe 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.\">\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 Takeaways<\/h3>\r\nLattice 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<ul>\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{k}\\left({\\text{Q}}^{\\text{+}}\\right)\\left({\\text{Q}}^{-}\\right)}{{\\text{d}}}[\/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>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. 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\n2. 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\n3. In 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\n4.\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\n5. 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\n6.\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\n7.\u00a0MgO\r\n\r\n8.\u00a0CaO\r\n\r\n9.\u00a0924 kJ\/mol\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\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\r\n\r\n&nbsp;","rendered":"<h2>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>\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\" \/><\/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.\">\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 Takeaways<\/h3>\n<p>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<ul>\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{k}\\left({\\text{Q}}^{\\text{+}}\\right)\\left({\\text{Q}}^{-}\\right)}{{\\text{d}}}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol>\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. 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>2. 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>3. In 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>4.\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>5. 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>6.\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>7.\u00a0MgO<\/p>\n<p>8.\u00a0CaO<\/p>\n<p>9.\u00a0924 kJ\/mol<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\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<p>&nbsp;<\/p>\n","protected":false},"author":264249,"menu_order":4,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-6387","chapter","type-chapter","status-publish","hentry"],"part":2995,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/pressbooks\/v2\/chapters\/6387","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/wp\/v2\/users\/264249"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/pressbooks\/v2\/chapters\/6387\/revisions"}],"predecessor-version":[{"id":6391,"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/pressbooks\/v2\/chapters\/6387\/revisions\/6391"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/pressbooks\/v2\/parts\/2995"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/pressbooks\/v2\/chapters\/6387\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/wp\/v2\/media?parent=6387"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=6387"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/wp\/v2\/contributor?post=6387"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/wp\/v2\/license?post=6387"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}