# 3.6.2.0: Lattice Energies (Problems)

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PROBLEM $$\PageIndex{1}$$

Using the bond energies in Table 7.3.1, determine the approximate enthalpy change for each of the following reactions:

a. $$\ce{H2}(g)+\ce{Br2}(g)⟶\ce{2HBr}(g)$$
b. $$\ce{CH4}(g)+\ce{I2}(g)⟶\ce{CH3I}(g)+\ce{HI}(g)$$
c. $$\ce{C2H4}(g)+\ce{3O2}(g)⟶\ce{2CO2}(g)+\ce{2H2O}(g)$$

−114 kJ

30 kJ

−1055 kJ

PROBLEM $$\PageIndex{2}$$

Using the bond energies in Table 7.3.1, determine the approximate enthalpy change for each of the following reactions:

a. $$\mathrm{H_2C=CH_2}(g)+\ce{H2}(g)⟶\ce{H3CCH3}(g)$$
b. $$\ce{2C2H6}(g)+\ce{7O2}(g)⟶\ce{4CO2}(g)+\ce{6H2O}(g)$$

-128 kJ

-5175 kJ

PROBLEM $$\PageIndex{3}$$

How does the bond energy of HCl differ from the standard enthalpy of formation of HCl(g)?

The enthalpy of formation is -431.6 kJ, while the bond energy of H-Cl is -432 kJ. They are practically the same.

PROBLEM $$\PageIndex{4}$$

Using the standard enthalpy of formation data in Appendix G, show how the standard enthalpy of formation of HCl(g) can be used to determine the bond energy.

$$\ce{HCl}(g)⟶\dfrac{1}{2}\ce{H2}(g)+\dfrac{1}{2}\ce{Cl2}(g)\hspace{20px}ΔH^\circ_1=−ΔH^\circ_{\ce f[\ce{HCl}(g)]}\\ \dfrac{1}{2}\ce{H2}(g)⟶\ce{H}(g)\hspace{105px}ΔH^\circ_2=ΔH^\circ_{\ce f[\ce H(g)]}\\ \underline{\dfrac{1}{2}\ce{Cl2}(g)⟶\ce{Cl}(g)\hspace{99px}ΔH^\circ_3=ΔH^\circ_{\ce f[\ce{Cl}(g)]}}\\ \ce{HCl}(g)⟶\ce{H}(g)+\ce{Cl}(g)\hspace{58px}ΔH^\circ_{298}=ΔH^\circ_1+ΔH^\circ_2+ΔH^\circ_3$$

\begin{align} D_\ce{HCl}=ΔH^\circ_{298}&=ΔH^\circ_{\ce f[\ce{HCl}(g)]}+ΔH^\circ_{\ce f[\ce H(g)]}+ΔH^\circ_{\ce f[\ce{Cl}(g)]}\\ &=\mathrm{−(−92.307\:kJ)+217.97\:kJ+121.3\:kJ}\\ &=\mathrm{431.6\:kJ} \end{align}

PROBLEM $$\PageIndex{5}$$

Using the standard enthalpy of formation data in Appendix G, determine which bond is stronger: the S–F bond in SF4(g) or in SF6(g)?

The S–F bond in SF4 is stronger.

PROBLEM $$\PageIndex{6}$$

Complete the following Lewis structure by adding bonds (not atoms), and then indicate the longest bond:  The C–C single bonds are longest.

PROBLEM $$\PageIndex{7}$$

Use principles of atomic structure to answer each of the following:1

a. The radius of the Ca atom is 197 pm; the radius of the Ca2+ ion is 99 pm. Account for the difference.
b. The lattice energy of CaO(s) is –3460 kJ/mol; the lattice energy of K2O is –2240 kJ/mol. Account for the difference.
c. Given these ionization values, explain the difference between Ca and K with regard to their first and second ionization energies.

Element First Ionization Energy (kJ/mol) Second Ionization Energy (kJ/mol)
K 419 3050
Ca 590 1140

d. The first ionization energy of Mg is 738 kJ/mol and that of Al is 578 kJ/mol. Account for this difference.

When two electrons are removed from the valence shell, the Ca radius loses the outermost energy level and reverts to the lower n = 3 level, which is much smaller in radius.

The +2 charge on calcium pulls the oxygen much closer compared with K, thereby increasing the lattice energy relative to a less charged ion.

Removal of the 4s electron in Ca requires more energy than removal of the 4s 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.

In Al, the removed electron is relatively unprotected and unpaired in a p orbital. The higher energy for Mg mainly reflects the unpairing of the 2s electron.