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Chapter 8: Problem 75

Given the following bond-dissociation energies, calculate the average bond enthalpy for the Ti-Cl bond. \begin{tabular}{ll} \hline & \(\Delta H(\mathbf{k J} /\) mol \()\) \\ \hline \(\mathrm{TiCl}_{4}(g) \longrightarrow \mathrm{TiCl}_{3}(g)+\mathrm{Cl}(g)\) & 335 \\ \(\mathrm{TiCl}_{3}(g) \longrightarrow \mathrm{TiCl}_{2}(g)+\mathrm{Cl}(g)\) & 423 \\ \(\mathrm{TiCl}_{2}(g) \longrightarrow \mathrm{TiCl}(g)+\mathrm{Cl}(g)\) & 444 \\\ \(\mathrm{TiCl}(g) \longrightarrow \mathrm{Ti}(g)+\mathrm{Cl}(g)\) & 519 \\ \hline \end{tabular}

Short Answer

Expert verified
The average bond enthalpy for the Ti-Cl bond is approximately \(430.25 kJ/mol\).

Step by step solution

01

Identify the number of Ti-Cl bonds broken in each reaction

We will determine the number of Ti-Cl bonds being broken in each reaction as follows: 1. TiCl4(g) → TiCl3(g) + Cl(g) - In this reaction, 1 Ti-Cl bond is being broken (4 bonds in TiCl4 to 3 bonds in TiCl3) 2. TiCl3(g) → TiCl2(g) + Cl(g) - In this reaction, 1 Ti-Cl bond is being broken (3 bonds in TiCl3 to 2 bonds in TiCl2) 3. TiCl2(g)→ TiCl(g) + Cl(g) - In this reaction, 1 Ti-Cl bond is being broken (2 bonds in TiCl2 to 1 bond in TiCl) 4. TiCl(g) → Ti(g) + Cl(g) - In this reaction, 1 Ti-Cl bond is being broken (1 bond in TiCl to 0 bonds in Ti)
02

Calculate the cumulative bond dissociation energies for all Ti-Cl bonds

To calculate the cumulative bond dissociation energies, we need to multiply the bond dissociation energy for each reaction by the number of Ti-Cl bonds broken in the corresponding reaction. Then we need to sum the values for each reaction to find the total energy. Total Energy = (1 x 335 kJ/mol) + (1 x 423 kJ/mol) + (1 x 444 kJ/mol) + (1 x 519 kJ/mol) Total Energy = 335 kJ/mol + 423 kJ/mol + 444 kJ/mol + 519 kJ/mol = 1721 kJ/mol
03

Calculate the average bond enthalpy for the Ti-Cl bond

To calculate the average bond enthalpy for the Ti-Cl bond, divide the total cumulative bond dissociation energy by the total number of Ti-Cl bonds broken in all reactions. Average Bond Enthalpy = Total Energy / Total Number of Ti-Cl Bonds Broken There are a total of 1 + 1 + 1 + 1 = 4 Ti-Cl bonds broken over the four reactions. So, Average Bond Enthalpy = 1721 kJ/mol / 4 Average Bond Enthalpy = 430.25 kJ/mol The average bond enthalpy for the Ti-Cl bond is approximately 430.25 kJ/mol.

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Most popular questions from this chapter

(a) How does a polar molecule differ from a nonpolar one? (b) Atoms \(\mathrm{X}\) and \(\mathrm{Y}\) have different electronegativities. Will the diatomic molecule \(\mathrm{X}-\mathrm{Y}\) necessarily be polar? Explain. (c) What factors affect the size of the dipole moment of a diatomic molecule?

A common form of elemental phosphorus is the tetrahedral \(\mathrm{P}_{4}\) molecule, where all four phosphorus atoms are equivalent: At room temperature phosphorus is a solid. (a) Do you think there are any unshared pairs of electrons in the \(\mathrm{P}_{4}\) molecule? (b) How many \(\mathrm{P}-\mathrm{P}\) bonds are there in the molecule? (c) Can you draw a Lewis structure for a linear \(\mathrm{P}_{4}\) molecule that satisfies the octet rule? (d) Using formal charges, what can you say about the stability of the linear molecule versus that of the tetrahedral molecule?

The electron affinity of oxygen is \(-141 \mathrm{~kJ} / \mathrm{mol}\), corresponding to the reaction $$ \mathrm{O}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{O}^{-}(g) $$ The lattice energy of \(\mathrm{K}_{2} \mathrm{O}(s)\) is \(2238 \mathrm{~kJ} / \mathrm{mol}\). Use these data along with data in Appendix \(\mathrm{C}\) and Figure 7.9 to calculate the "second electron affinity" of oxygen, corresponding to the reaction $$ \mathrm{O}^{-}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{O}^{2-}(g) $$

(a) Explain the following trend in lattice energy: \(\mathrm{BeH}_{2}\), \(3205 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{MgH}_{2}, 2791 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{CaH}_{2}, 2410 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{SrH}_{2}\) \(2250 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{BaH}_{2}, 2121 \mathrm{~kJ} / \mathrm{mol}\) (b) The lattice energy of \(\mathrm{ZnH}_{2}\) is \(2870 \mathrm{~kJ} / \mathrm{mol}\). Based on the data given in part (a), the radius of the \(\mathrm{Zn}^{2+}\) ion is expected to be closest to that of which group \(2 \mathrm{~A}\) element?

Use Table 8.4 to estimate the enthalpy change for each of the following reactions: (a) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{O}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{O}-\mathrm{Cl}(g)\) (b) \(\mathrm{H}_{2} \mathrm{O}_{2}(g)+2 \mathrm{CO}(g) \longrightarrow \mathrm{H}_{2}(g)+2 \mathrm{CO}_{2}(g)\) (c) \(3 \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12}(g)\) (the six carbon atoms form a six-membered ring with two \(\mathrm{H}\) atoms on each \(\mathrm{C}\) atom \()\)
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