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6.1.1: Lattice Energies (Problems)

  • Page ID
    235785
  • 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. \(\ce{C2H4}(g)+\ce{3O2}(g)⟶\ce{2CO2}(g)+\ce{2H2O}(g)\)

    Answer a

    −114 kJ

    Answer b

    30 kJ

    Answer c

    −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. \(\ce{2C2H6}(g)+\ce{7O2}(g)⟶\ce{4CO2}(g)+\ce{6H2O}(g)\)

    Answer a

    -128 kJ

    Answer b

    -5175 kJ

    Click here to see a video of the solution

    PROBLEM \(\PageIndex{3}\)

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

    Answer

    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.

    Answer

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

     

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

    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)?

    Answer

    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:

    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.

    Answer

    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.

     

    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.

    Answer a

    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.

    Answer b

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

    Answer c

    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.

    Answer d

    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.

     

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