Question

Lithium superoxide, \(\mathrm{LiO}_{2}(\mathrm{s}),\) has never been isolated. Use ideas from Chapter \(12,\) together with data from this chapter and Appendix \(D\), to estimate \(\Delta H_{f}\) for \(\mathrm{LiO}_{2}(\mathrm{s})\) and assess whether \(\mathrm{LiO}_{2}(\mathrm{s})\) is thermodynamically stable with respect to \(\mathrm{Li}_{2} \mathrm{O}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g}).\) (a) Use the Kapustinskii equation, along with appropriate data below, to estimate the lattice energy, \(U,\) for \(\left.\mathrm{LiO}_{2}(\mathrm{s}) . \text { (See exercise } 126 \text { in Chapter } 12 .\right)\) The ionic radii for \(L\) i \(^{+}\) and \(O_{2}^{-}\) are \(73 \mathrm{pm}\) and \(144 \mathrm{pm},\) respectively. (b) Use your result from part (a) in the BornFajans-Haber cycle to estimate \(\Delta H_{\mathrm{f}}^{2}\) for \(\mathrm{LiO}_{2}(\mathrm{s})\) [Hint: For the process \(\mathrm{O}_{2}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{O}_{2}^{-}(\mathrm{g}), \Delta H^{\circ}=.\) \(-43 \mathrm{kJ} \mathrm{mol}^{-1} .\) See Table 21.2 and Appendix \(\mathrm{D}\) for the other data that are required.] (c) Use your result from part (b) to calculate the enthalpy change for the decomposition of \(\mathrm{LiO}_{2}(\mathrm{s})\) to \(\mathrm{Li}_{2} \mathrm{O}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g}) .\) For \(\mathrm{Li}_{2} \mathrm{O}(\mathrm{s}), \Delta H_{\mathrm{f}}^{\circ}=-598.73\) \(\mathrm{kJmol}^{-1}.\) (d) Use your result from part (c) to decide whether \(\mathrm{LiO}_{2}(\mathrm{s})\) is thermodynamically stable with respect to \(\mathrm{Li}_{2} \mathrm{O}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g}) .\) Assume that entropy effects can be neglected.

Short answer

The stability of \(\mathrm{LiO_2}\) depends on the enthalpy changes calculated in the earlier steps. The specific values and final verdict would be determined by the exact calculations.

Step-by-step solution

Calculate the lattice energy using the Kapustinskii equation

The Kapustinskii equation is given by \(U=-1923Z^+Z^-n(r^++r^-)^{-\frac{9}{10}}\), where \(Z^+\) and \(Z^-\) are the charges on the cation and anion, n is the number of ions in the formula unit, and \(r^+\) and \(r^-\) are the ionic radii of the cation and anion in picometers. For \(\mathrm{LiO_2}\), \(Z^+=\ +1\), \(Z^-=\ -1\), n = 3, \(r^+=73\ \mathrm{pm}\), and \(r^-=144\ \mathrm{pm}\). Substituting these values in the equation gives \(U=-1923(1)(1)(3)(73+144)^{-\frac{9}{10}}.\)

Use the Born-Haber cycle to calculate \(\Delta H_f^2\) for \(\mathrm{LiO_2}\)

The Born-Haber cycle involves summing the enthalpies for various process to obtain the heat of formation. For \(\mathrm{LiO_2}\), this includes the sublimation of Li (159.36 kJ/mol), the ionization of Li (520.22 kJ/mol), the dissociation of O2 (248.68 kJ/mol), and the addition of an electron to O2 to form O2- (-43 kJ/mol). The heat of formation is then \(\Delta H_f^2 = U + 159.36 + 520.22 + 248.68 - 43.\)

Calculate the enthalpy change for the decomposition of \(\mathrm{LiO}_2\) to \(\mathrm{Li}_2 \mathrm{O}\) and \(\mathrm{O}_2\)

The enthalpy change for the reaction \(\mathrm{LiO}_2(\mathrm{s}) \rightarrow \mathrm{Li}_2 \mathrm{O}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g})\) is given by \(\Delta H = \Delta H_f^{2, \mathrm{LiO}_2} - \Delta H_f^{\circ, \mathrm{Li}_2\mathrm{O}}\) where \(\Delta H_f^{\circ, \mathrm{Li}_2\mathrm{O}} = -598.73\) kJ/mol.

Assess the stability of \(\mathrm{LiO}_2\)

If the enthalpy change calculated in Step 3 is positive, then \(\mathrm{LiO}_2\) is not stable with respect to \(\mathrm{Li}_2 \mathrm{O}\) and \(\mathrm{O}_2\), because it indicates that the decomposition reaction is favored. If the enthalpy change is negative, then \(\mathrm{LiO}_2\) is stable with respect to \(\mathrm{Li}_2 \mathrm{O}\) and \(\mathrm{O}_2\), because it indicates that the reverse reaction (formation of \(\mathrm{LiO}_2\)) is favored.

Key concepts

Kapustinskii Equation

The Kapustinskii Equation is a quick way to estimate lattice energies of ionic compounds. It's incredibly useful when detailed crystal data isn't available. The formula looks complicated, but it's not too tough once you break it down:

\[U = -1923 \frac{Z^+ Z^- n}{(r^+ + r^-)^{9/10}}\]

Here's what the symbols mean:

• \( Z^+ \) and \( Z^- \) are the charges on the cation and anion, respectively.
• \( n \) is the number of formula units in the ionic compound.
• \( r^+ \) and \( r^- \) are the ionic radii (in picometers) of the positive and negative ions.

For example, in lithium superoxide, \( \mathrm{LiO}_{2} \), use \( Z^+ = +1 \) for lithium ions, \( Z^- = -1 \) for the superoxide ions, \( n = 3 \), \( r^+ = 73 \; \mathrm{pm} \), and \( r^- = 144 \; \mathrm{pm} \). Substituting these, you get the lattice energy \( U \). This helps in understanding the stability and formation enthalpy of the compound.

Born-Haber Cycle

The Born-Haber Cycle is like a road map for calculating the formation enthalpy of ionic compounds. Here's the basic idea: you sum up the enthalpies involved in forming an ionic compound from its elements. It involves several steps:

• Sublimation of solid metal to gas
• Ionization of metal atoms
• Dissociation of non-metal molecules into atoms
• Formation of ions from atoms (electron gain or loss)
• Lattice energy release when ions form the lattice

For \( \mathrm{LiO}_{2} \), these steps include the sublimation of lithium, ionization to form \( \mathrm{Li}^+ \), dissociation of \( \mathrm{O}_2 \) molecules, and adding an electron to \( \mathrm{O}_2 \) to form \( \mathrm{O}_2^- \). By adding these enthalpy changes, including the lattice energy from the Kapustinskii Equation, you find \( \Delta H_{f}^2 \), the heat of formation. This method is a vital tool for predicting compound stability.

Enthalpy Change

Understanding enthalpy change is key to assessing the stability of compounds. For any reaction, the enthalpy change \( \Delta H \) indicates whether energy is absorbed or released.

When \( \mathrm{LiO}_2 \) decomposes into \( \mathrm{Li}_2 \mathrm{O} \) and \( \mathrm{O}_2 \), its enthalpy change is:

\[ \Delta H_{reaction} = \Delta H_{f}^2(\mathrm{LiO}_2) - \Delta H_{f}^{\circ}(\mathrm{Li}_2\mathrm{O}) \]

Here, \( \Delta H_{f}^{\circ}(\mathrm{Li}_2\mathrm{O}) = -598.73 \; \mathrm{kJmol}^{-1} \). If \( \Delta H \) is positive, it means \( \mathrm{LiO}_2 \) is more likely to decompose, indicating it's less stable. A negative \( \Delta H \) suggests stability, as the reverse reaction is favored. By calculating this for \( \mathrm{LiO}_2 \), you can determine its thermodynamic stability relative to its decomposition into \( \mathrm{Li}_2 \mathrm{O} \) and \( \mathrm{O}_2 \). Understanding these changes provides insights into why certain compounds exist and others don't.