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Chapter 8: Problem 58

Use the following data to estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for magnesium fluoride. Lattice energy $$ \begin{aligned} \mathrm{Mg}(s)+\mathrm{F}_{2}(g) \longrightarrow & \mathrm{MgF}_{2}(s) \\ &-2913 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ First ionization energy of \(\mathrm{Mg} \quad 735 \mathrm{~kJ} / \mathrm{mol}\) Second ionization energy of \(\mathrm{Mg} \quad 1445 \mathrm{~kJ} / \mathrm{mol}\) \(\begin{array}{ll}\text { Electron affinity of } \mathrm{F} & -328 \mathrm{~kJ} / \mathrm{mol}\end{array}\) Bond energy of \(\mathrm{F}_{2}\) \(154 \mathrm{~kJ} / \mathrm{mol}\) Enthalpy of sublimation for \(\mathrm{Mg}\) 150. \(\mathrm{kJ} / \mathrm{mol}\)

Short Answer

Expert verified
The estimated standard enthalpy of formation for magnesium fluoride (MgF2) is \(3069 \, \mathrm{kJ/mol}\).

Step by step solution

01

Write down the Born-Haber cycle

The Born-Haber cycle relates lattice energy, ionization energies, electron affinities, bond energy and enthalpy of sublimation to the enthalpy of formation. For magnesium fluoride, the cycle can be written as: Sublimation of Mg(s) + First ionization energy of Mg + Second ionization energy of Mg + Bond energy of F2 + Electron affinity of F = Enthalpy of formation of MgF2 (ΔHf°) + Lattice energy
02

Replace the values in the equation

Use the given data to substitute the values in the Born-Haber cycle equation: 150 kJ/mol + 735 kJ/mol + 1445 kJ/mol + 154 kJ/mol - 328 kJ/mol = ΔHf° -2913 kJ/mol
03

Calculate ΔHf°

Now we solve the equation for ΔHf°: ΔHf° = 150 kJ/mol + 735 kJ/mol + 1445 kJ/mol + 154 kJ/mol - 328 kJ/mol + 2913 kJ/mol ΔHf° = 3069 kJ/mol The estimated standard enthalpy of formation for magnesium fluoride (MgF2) is 3069 kJ/mol.

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Most popular questions from this chapter

Which of the following statements is(are) true? Correct the false statements. a. The molecules \(\mathrm{SeS}_{3}, \mathrm{SeS}_{2}, \mathrm{PCl}_{5}, \mathrm{TeCl}_{4}, \mathrm{ICl}_{3}\), and \(\mathrm{XeCl}_{2}\) all exhibit at least one bond angle, which is approximately \(120^{\circ}\). b. The bond angle in \(\mathrm{SO}_{2}\) should be similar to the bond angle in \(\mathrm{CS}_{2}\) or \(\mathrm{SCl}_{2}\). c. Of the compounds \(\mathrm{CF}_{4}, \mathrm{KrF}_{4}\), and \(\mathrm{SeF}_{4}\), only \(\mathrm{SeF}_{4}\) exhibits an overall dipole moment (is polar). d. Central atoms in a molecule adopt a geometry of the bonded atoms and lone pairs about the central atom in order to maximize electron repulsions.

Write Lewis structures and predict whether each of the following is polar or nonpolar. a. HOCN (exists as \(\mathrm{HO}-\mathrm{CN}\) ) b. \(\operatorname{COS}\) c. \(\mathrm{XeF}_{2}\) d. \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) e. \(\mathrm{SeF}_{6}\) f. \(\mathrm{H}_{2} \mathrm{CO}(\mathrm{C}\) is the central atom \()\)

The compound hexaazaisowurtzitane is one of the highestenergy explosives known ( \(C\) \& E News, Jan. 17, 1994, p. 26). The compound, also known as CL-20, was first synthesized in 1987 . The method of synthesis and detailed performance data are still classified because of CL-20's potential military application in rocket boosters and in warheads of "smart" weapons. The structure of CL-20 is In such shorthand structures, each point where lines meet represents a carbon atom. In addition, the hydrogens attached to the carbon atoms are omitted; each of the six carbon atoms has one hydrogen atom attached. Finally, assume that the two \(\mathrm{O}\) atoms in the \(\mathrm{NO}_{2}\) groups are attached to \(\mathrm{N}\) with one single bond and one double bond. Three possible reactions for the explosive decomposition of \(\mathrm{CL}-20\) are i. \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{~N}_{12} \mathrm{O}_{12}(s) \rightarrow 6 \mathrm{CO}(g)+6 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g)+\frac{3}{2} \mathrm{O}_{2}(g)\) ii. \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{~N}_{12} \mathrm{O}_{12}(s) \rightarrow 3 \mathrm{CO}(g)+3 \mathrm{CO}_{2}(g)+6 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g)\) iii. \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{~N}_{12} \mathrm{O}_{12}(s) \rightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2}(g)\) a. Use bond energies to estimate \(\Delta H\) for these three reactions. b. Which of the above reactions releases the largest amount of energy per kilogram of CL-20?

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) has three possible Lewis structures: Given the following bond lengths, $$ \begin{array}{llll} \mathrm{N}-\mathrm{N} & 167 \mathrm{pm} & \mathrm{N}=\mathrm{O} & 115 \mathrm{pm} \\ \mathrm{N}=\mathrm{N} & 120 \mathrm{pm} & \mathrm{N}-\mathrm{O} & 147 \mathrm{pm} \\ \mathrm{N} \equiv \mathrm{N} & 110 \mathrm{pm} & & \end{array} $$ rationalize the observations that the \(\mathrm{N}-\mathrm{N}\) bond length in \(\mathrm{N}_{2} \mathrm{O}\) is \(112 \mathrm{pm}\) and that the \(\mathrm{N}-\mathrm{O}\) bond length is \(119 \mathrm{pm}\). Assign formal charges to the resonance structures for \(\mathrm{N}_{2} \mathrm{O}\). Can you eliminate any of the resonance structures on the basis of formal charges? Is this consistent with observation?

In general, the higher the charge on the ions in an ionic compound, the more favorable the lattice energy. Why do some stable ionic compounds have \(+1\) charged ions even though \(+4,+5\), and \(+6\) charged ions would have a more favorable lattice energy?
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