Multiple bonds are stronger than single bonds between the same atoms. Check Your Learning The lattice energy of a compound is a measure of the strength of this attraction. (a) [latex]\begin{array}{lll}\text{2 N-H bonds}\hfill & =\hfill & \hfill 2\left(390\right)\\ \text{1 N-O bond}\hfill & =\hfill & \hfill 200\\ \text{1 O-H bond}\hfill & =\hfill & \hfill \underline{464}\\ \hfill & \hfill & \hfill \text{1444 kJ}\end{array};[/latex], (b) [latex]\begin{array}{lll}\text{3 N-H bonds}\hfill & =\hfill & \hfill 3\left(390\right)\\ \text{1 N-O bond}\hfill & =\hfill & \hfill \underline{200}\\ \hfill & \hfill & \hfill \text{1370 kJ}\end{array};[/latex] (c) As in part (b), the bond energy is a positive energy. Using the bond energies in Table 7.2, determine the approximate enthalpy change for each of the following reactions: (a) [latex]{\text{H}}_{2}\left(g\right)+{\text{Br}}_{2}\left(g\right)\rightarrow 2\text{HBr}\left(g\right)[/latex], (b) [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], (c) [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], (a) [latex]{\text{Cl}}_{2}\left(g\right)+3{\text{F}}_{2}\left(g\right)\rightarrow 2{\text{ClF}}_{3}\left(g\right)[/latex], (b) [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], (c) [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]. In the next step, we account for the energy required to break the F-F bond to produce fluorine atoms. The total energy involved in this conversion is equal to the experimentally determined enthalpy of formation, [latex]\Delta{H}_{\text{f}}^{\textdegree },[/latex] of the compound from its elements. U may be calculated from the Born-Haber cycle. The bond length is the internuclear distance at which the lowest potential energy is achieved. For example, if the relevant enthalpy of sublimation [latex]\Delta{H}_{s}^{\textdegree },[/latex] ionization energy (IE), bond dissociation enthalpy (D), lattice energy ΔHlattice, and standard enthalpy of formation [latex]\Delta{H}_{\text{f}}^{\textdegree }[/latex] are known, the Born-Haber cycle can be used to determine the electron affinity of an atom. We can use bond energies to calculate approximate enthalpy changes for reactions where enthalpies of formation are not available. [/latex], [latex]\begin{array}{l}\Delta H=\Delta{H}_{f}^{\textdegree }=\Delta{H}_{s}^{\textdegree }+\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], Describe the energetics of covalent and ionic bond formation and breakage, Use the Born-Haber cycle to compute lattice energies for ionic compounds, Use average covalent bond energies to estimate enthalpies of reaction, [latex]\Delta{H}_{\text{f}}^{\textdegree },[/latex] the standard enthalpy of formation of the compound, [latex]\Delta{H}_{s}^{\textdegree },[/latex] the enthalpy of sublimation of the metal, 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\text{\textdegree }[/latex], Lattice energy for a solid MX: [latex]\text{MX}\left(s\right)\rightarrow{\text{M}}^{n\text{+}}\left(g\right)+{\text{X}}^{n\text{-}}\left(g\right)\Delta{H}_{\text{lattice}}[/latex], Lattice energy for an ionic crystal: [latex]\Delta{H}_{\text{lattice}}=\frac{\text{C}\left({\text{Z}}^{\text{+}}\right)\left({\text{Z}}^{\text{-}}\right)}{{\text{R}}_{\text{o}}}[/latex]. These ions combine to produce solid cesium fluoride. 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. Predict whether CaO is ionic or covalent, based on the location of their constituent atoms in the periodic table. The 415 kJ/mol value is the average, not the exact value required to break any one bond. 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.

Whereas lattice energies typically fall in the range of 600–4000 kJ/mol (some even higher), covalent bond dissociation energies are typically between 150–400 kJ/mol for single bonds. We now have one mole of Cs cations and one mole of F anions. For the ionic solid MX, the lattice energy is the enthalpy change of the process: Note that we are using the convention where the ionic solid is separated into ions, so our lattice energies will be endothermic (positive values).

Thus, Al2O3 would have a shorter interionic distance than Al2Se3, and Al2O3 would have the larger lattice energy. Lattice energies calculated for ionic compounds are typically much higher than bond dissociation energies measured for covalent bonds. This would make the reaction more exothermic, as a smaller positive value is “more exothermic.”. the greater bond energy is for (a), and it is more stable. 15. 5.HBr - polar covalent. 6. In general, a multiple bond between the same two elements is stronger than a single bond. You can also practice Chemistry - OpenStax 2015th Edition practice problems. 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? Stable molecules exist because covalent bonds hold the atoms together. 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, it requires 769 kJ to separate one mole of solid NaCl into gaseous Na+ and Cl– ions. Zinc oxide, ZnO, is a very effective sunscreen. In each case, think about how it would affect the Born-Haber cycle. [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}^{\textdegree }=\text{-Delta }{H}_{\text{f}\left[\left({\text{PCl}}_{3}\left(g\right)\right)\right]}^{\textdegree }[/latex]. Different interatomic distances produce different lattice energies.

Keep in mind, however, that these are not directly comparable values. For ionic bonds, the lattice energy is the energy required to separate one mole of a compound into its gas phase ions. 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.

7.CaO - ionic. The Na–F distance in NaF, which has the same structure as KF, is 231 pm. The lattice energy of KF is 794 kJ/mol, and the interionic distance is 269 pm. We can determine if CaO is ionic or covalent by analyzing Ca and O 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. An endothermic reaction (ΔH positive, heat absorbed) results when the bonds in the products are weaker than those in the reactants.


4008 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. The compound Al2Se3 is used in the fabrication of some semiconductor devices. 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. Although the four C–H 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.

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