Problem 71 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

The average bond enthalpy for the \(\mathrm{Ti}-\mathrm{Cl}\) bond is approximately 430.25 kJ/mol.

Step 1: Identify the number of bonds broken in each reaction

In each reaction, we can see that the number of \(\mathrm{Ti}-\mathrm{Cl}\) bonds broken is 1.

Step 2: Calculate the energy for each \(\mathrm{Ti}-\mathrm{Cl}\) bond.

Since there is only one \(\mathrm{Ti}-\mathrm{Cl}\) bond broken in each reaction, the energies provided are also the energy values for each \(\mathrm{Ti}-\mathrm{Cl}\) bond dissociation: - For \(\mathrm{TiCl}_4(g) \longrightarrow \mathrm{TiCl}_3(g) + \mathrm{Cl}(g)\), the bond energy is 335 kJ/mol - For \(\mathrm{TiCl}_3(g) \longrightarrow \mathrm{TiCl}_2(g) + \mathrm{Cl}(g)\), the bond energy is 423 kJ/mol - For \(\mathrm{TiCl}_2(g) \longrightarrow \mathrm{TiCl}(g) + \mathrm{Cl}(g)\), the bond energy is 444 kJ/mol - For \(\mathrm{TiCl}(g) \longrightarrow \mathrm{Ti}(g) + \mathrm{Cl}(g)\), the bond energy is 519 kJ/mol

Step 3: Calculate the sum of the bond energies.

Now, we will add all the bond energies found in Step 2: $$ \text{Sum of bond energies} = 335 + 423 + 444 + 519 = 1721\,\text{kJ/mol} $$

Step 4: Calculate the average bond enthalpy for the \(\mathrm{Ti}-\mathrm{Cl}\) bond.

There are a total of 4 \(\mathrm{Ti}-\mathrm{Cl}\) bonds across the given reactions, so we can calculate their average enthalpy by dividing the sum of bond energies by the number of bonds: $$ \text{Average bond enthalpy} = \frac{\text{Sum of bond energies}}{\text{Number of bonds}} = \frac{1721}{4} = 430.25\,\text{kJ/mol} $$ Thus, the average bond enthalpy for the \(\mathrm{Ti}-\mathrm{Cl}\) bond is approximately 430.25 kJ/mol.

Key Concepts

Bond-Dissociation Energy

Bond-dissociation energy, also known as bond-breaking energy, is the amount of energy required to break a specific chemical bond in one mole of gaseous molecules. It is a key concept for understanding the strength and stability of chemical bonds.

Consider a simple chemical reaction where a molecule undergoes a process resulting in the breaking of a bond and the formation of two separate atoms or fragments. The energy change associated with this process is the bond-dissociation energy. For diatomic molecules, this value is relatively straightforward to measure, but for larger molecules with multiple bonds, the bond-dissociation energy can differ for each bond.

To quantify this energy, chemists use the unit kilojoules per mole (kJ/mol). It is important to note that bond-dissociation energies are average values, meaning they are derived from the energy required to break similar bonds in a series of related compounds, since individual bond energies can vary depending on the molecule's environment.

Enthalpy Calculation

Enthalpy calculation is essential in thermochemistry to determine the heat change in chemical reactions. It provides insight into whether a reaction is endothermic (absorbing heat) or exothermic (releasing heat). The symbol \( \Delta H \) represents enthalpy change.

When calculating the enthalpy change for breaking bonds, we add up the bond-dissociation energies of all bonds broken and formed within the reaction. The overall enthalpy change can then be determined by subtracting the energy needed to break bonds from the energy released upon forming new bonds.

Mathematically, this is shown as:\[\Delta H = \text{Energy to break bonds} - \text{Energy to form bonds}\]In cases where there are multiple instances of the same bond breaking, as shown in our textbook problem, we can calculate the average bond enthalpy, which is the average amount of energy needed to break one mole of bonds in a gaseous molecule.

Chemical Bonds

Chemical bonds are the attractive forces that hold atoms together in molecules and compounds. They are fundamental to the structure and stability of all substances. There are several types of bonds, including ionic, covalent, metallic, and hydrogen bonds.

Ionic bonds occur between metals and nonmetals, where one atom donates electrons to another, forming oppositely charged ions that attract each other. Covalent bonds involve the sharing of electron pairs between nonmetals. Metallic bonds are present in metals, where positively charged ion cores are surrounded by a 'sea' of delocalized electrons. Hydrogen bonds are weak bonds that can form between a hydrogen atom in one molecule and an electronegative atom in another.

Each bond type has characteristic properties, and the strength of the bond is a crucial factor determining the substance's physical properties. Bond enthalpy is directly related to the strength of a bond; the higher the bond enthalpy, the stronger the bond and the more stable the molecule.