Problem 112 - Chemistry and Chemical Reactivity

Question

Several standard enthalpies of formation (from Appendix L) are given below. Use these data to calculate (a) the standard enthalpy of vaporization of bromine. (b) the energy required for the reaction \(\mathrm{Br}_{2}(\mathrm{g}) \rightarrow\) \(2 \mathrm{Br}(g) .\) (This is the Br \(-\mathrm{Br}\) bond dissociation enthalpy.) $$\begin{aligned} &\text { Species } \quad \Delta_{f} H^{\circ}(\mathrm{kJ} / \mathrm{mol})\\\ &\begin{array}{lc} \hline B r(g) & 111.9 \\ B r_{2}(\ell) & 0 \\ B r_{2}(g) & 30.9 \end{array} \end{aligned}$$

Step-by-Step Solution

Answer

(a) 30.9 kJ/mol, (b) 192.9 kJ/mol.

Step 1: Understand the Given Data

We have the following standard enthalpies of formation: \( \Delta_{f} H^{\circ}(\text{Br}(g)) = 111.9 \, \text{kJ/mol} \), \( \Delta_{f} H^{\circ}(\text{Br}_2(\ell)) = 0 \, \text{kJ/mol} \), and \( \Delta_{f} H^{\circ}(\text{Br}_2(g)) = 30.9 \, \text{kJ/mol} \). Our task is to find two quantities: (a) the standard enthalpy of vaporization of bromine, and (b) the energy required for the dissociation of \( \text{Br}_2(g) \) into \( 2\text{Br}(g) \).

Step 2: Calculate the Enthalpy of Vaporization

The enthalpy of vaporization \( \Delta_{vap}H^{\circ} \) is the change in enthalpy when one mole of a substance transitions from liquid to gas. Thus, we calculate: \[ \Delta_{vap}H^{\circ} = \Delta_{f} H^{\circ}(\text{Br}_2(g)) - \Delta_{f} H^{\circ}(\text{Br}_2(\ell)) = 30.9 \, \text{kJ/mol} - 0 \, \text{kJ/mol} = 30.9 \, \text{kJ/mol}. \]

Step 3: Calculate the Bond Dissociation Enthalpy

The bond dissociation enthalpy is the energy for the reaction \( \text{Br}_2(g) \rightarrow 2\text{Br}(g) \). We calculate it as the difference in enthalpy between the products and the reactants:\[ \Delta_{diss}H^{\circ} = 2 \times \Delta_{f} H^{\circ}(\text{Br}(g)) - \Delta_{f} H^{\circ}(\text{Br}_2(g)). \] Substitute the values:\[ \Delta_{diss}H^{\circ} = 2 \times 111.9 \, \text{kJ/mol} - 30.9 \, \text{kJ/mol} = 223.8 \, \text{kJ/mol} - 30.9 \, \text{kJ/mol} = 192.9 \, \text{kJ/mol}. \]

Key Concepts

Enthalpy of Vaporization

Enthalpy of vaporization refers to the heat required to convert one mole of a substance from the liquid phase to the gas phase at constant pressure. It's a crucial concept in understanding phase changes. For bromine, we use the formula:

• \( \Delta_{vap}H^{\circ} = \Delta_{f} H^{\circ}(\text{Br}_2(g)) - \Delta_{f} H^{\circ}(\text{Br}_2(\ell)) \)

Given that \( \Delta_{f} H^{\circ}(\text{Br}_2(g)) = 30.9 \, \text{kJ/mol} \) and \( \Delta_{f} H^{\circ}(\text{Br}_2(\ell)) = 0 \, \text{kJ/mol} \), the enthalpy of vaporization for bromine calculates to 30.9 kJ/mol.
This measure tells us how much energy is necessary for bromine to transform from liquid to gas, emphasizing the energy input needed to break intermolecular forces.

Bond Dissociation Enthalpy

Bond dissociation enthalpy is the energy required to break a specific chemical bond. In this exercise, it refers to the energy needed to dissociate one mole of bromine gas \( \text{Br}_2(g) \) into two moles of bromine atoms \( 2\text{Br}(g) \).
We calculate it using the formula:

• \( \Delta_{diss}H^{\circ} = 2 \times \Delta_{f} H^{\circ}(\text{Br}(g)) - \Delta_{f} H^{\circ}(\text{Br}_2(g)) \)

Given \( \Delta_{f} H^{\circ}(\text{Br}(g)) = 111.9 \, \text{kJ/mol} \) and \( \Delta_{f} H^{\circ}(\text{Br}_2(g)) = 30.9 \, \text{kJ/mol} \), the bond dissociation enthalpy calculations show:

• \( 223.8 \, \text{kJ/mol} - 30.9 \, \text{kJ/mol} = 192.9 \, \text{kJ/mol} \)

This illustrates how much energy is required to split bromine molecules into separate atoms, making it pivotal for reactions involving bond breaking.

Chemical Thermodynamics

Chemical thermodynamics deals with the study of energy changes during chemical reactions and processes, such as phase changes or chemical bond formation/breaking.
For enthalpy changes:

• Enthalpy of formation: Energy change when one mole of a compound is formed from its elements in their standard states.
• Enthalpy of vaporization: Energy needed for a substance to go from liquid to gas.
• Bond dissociation energy: Required energy to break a chemical bond, producing separated atoms.

These concepts highlight how energy is absorbed or released, serving as a foundation to predict reaction feasibility and behavior.
By understanding these principles, we can comprehend how energy is conserved and transferred in various chemical contexts.