Question

Calculate \(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\) for chloride ion from the following data: \(1 / 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I} / 2 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{HCl}(\mathrm{g})\) \(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}=-92.4 \mathrm{~kJ}\) \(\mathrm{HCl}(\mathrm{g})+\mathrm{nH}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\) \(\Delta \mathrm{H}_{\mathrm{Hyd}}=-74.8 \mathrm{~kJ}\) \(\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{f}}\left[\mathrm{H}^{+}\right]=0.0 \mathrm{~kJ}\) (a) \(-189 \mathrm{~kJ}\) (b) \(-167 \mathrm{~kJ}\) (c) \(+167 \mathrm{~kJ}\) (d) \(-191 \mathrm{~kJ}\)

Short answer

The correct answer is (b) -167 kJ/mol.

Step-by-step solution

Understanding the Problem

We need to calculate the standard enthalpy of formation (\( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \)) for the chloride ion (\( \mathrm{Cl}^{-} \)) using given data for two reactions and enthalpies. We are given \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \) for the formation of \( \mathrm{HCl} \) and the hydration enthalpy \( \Delta \mathrm{H}_{\mathrm{Hyd}} \) for the ionization of \( \mathrm{HCl} \).

Writing the Target Reaction

The target reaction for which we are calculating \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \) is: \[ \frac{1}{2} \mathrm{H}_2(\mathrm{~g}) + \frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \rightarrow \mathrm{H}^{+}(\mathrm{aq}) + \mathrm{Cl}^{-}(\mathrm{aq}) \]

Breaking Down the Given Reactions

1. \( \frac{1}{2} \mathrm{H}_2(\mathrm{~g}) + \frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \rightarrow \mathrm{HCl}(\mathrm{g}) \) - \( \Delta \mathrm{H} = -92.4 \mathrm{~kJ} \)2. \( \mathrm{HCl}(\mathrm{g}) + \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}^{+}(\mathrm{aq}) + \mathrm{Cl}^{-}(\mathrm{aq}) \) - \( \Delta \mathrm{H} = -74.8 \mathrm{~kJ} \)Combine these reactions to write the full reaction: \[ \frac{1}{2} \mathrm{H}_2(\mathrm{~g}) + \frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \rightarrow \mathrm{H}^{+}(\mathrm{aq}) + \mathrm{Cl}^{-}(\mathrm{aq}) \]

Calculating \( \Delta \mathrm{H}_{\text{target}} \)

Use Hess's Law to calculate the enthalpy change for the target reaction:\[ \Delta \mathrm{H}_{\text{target}} = \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HCl}) + \Delta \mathrm{H}_{\mathrm{Hyd}} \]Substitute the known values:\[ \Delta \mathrm{H}_{\text{target}} = (-92.4 \text{ kJ}) + (-74.8 \text{ kJ}) \]Calculate:\[ \Delta \mathrm{H}_{\text{target}} = -167.2 \text{ kJ} \]Typically, we select the nearest integer value given in the options, which is \(-167 \text{ kJ/mol} \).

Verify and Select the Closest Answer

The calculated \( \Delta \mathrm{H}_{\text{target}} = -167.2 \mathrm{~kJ/mol} \) closely matches option (b) (-167 \mathrm{~kJ/mol}). Therefore, the correct answer is (b).

Key concepts

Hess's Law

Hess's Law is a powerful principle in thermodynamics which states that the total enthalpy change for a given chemical reaction is the same, regardless of the pathway taken, provided the initial and final conditions are the same. This means we can calculate the enthalpy change of a complex reaction by breaking it down into a series of simpler steps. Each step has an associated enthalpy change, which, when summed, provides the total enthalpy change for the entire process.
This concept is particularly useful when direct measurement of the enthalpy change is challenging or impossible.

• Hess's Law allows us to determine the enthalpy change by utilizing known data of comparable reactions.
• We can rearrange the equations of these reactions to fit our target reaction.

In our exercise, Hess's Law is applied to determine the enthalpy change for the formation of the chloride ion in aqueous solution from standard conditions. By summing the enthalpies of the formation and hydration reactions, we achieve the desired result for the caloric change of interest.

Hydration Enthalpy

Hydration enthalpy refers to the energy change associated with the process of dissolving ions in water, forming hydrated ions. This is an important concept in the field of chemistry as it helps in understanding how ions interact with solvent molecules, thus driving the solubility process.
When a gaseous ion dissolves in water, it attracts water molecules, a process which releases energy. This release of energy is what we call the hydration enthalpy.

• A negative hydration enthalpy indicates an exothermic process where energy is released.
• The magnitude of hydration enthalpy depends on the size and charge of the ion; smaller and highly charged ions typically release more energy.

In our exercise, we see that the hydration enthalpy for the ionization of HCl into aqueous ions involves an energy release of \(-74.8 \text{ kJ/mol}\). This value contributes to our calculation of the overall enthalpy change according to Hess's Law.

Enthalpy Change Calculation

Calculating enthalpy change is a crucial practice in chemistry, as enthalpy is a measure of the total energy of a thermodynamic system. To calculate the enthalpy change for a chemical reaction, we often employ data from tabulated standard enthalpies of formation and Hess's Law, similar to our exercise.
Here’s a simple breakdown of the process:

• Identify known values: For instance, the standard enthalpy of formation \(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\) for HCl and hydration enthalpy \(\Delta \mathrm{H}_{\mathrm{Hyd}}\).
• Calculate the enthalpy change for the target reaction: Use the relationship presented by Hess's Law to sum the known changes.
• Verify against known options: Compare the calculated value with given answer choices to find the best match.

In the original exercise, we used these principles to calculate the enthalpy change of \(-167.2 \text{ kJ/mol}\) and matched it to the closest provided option, \(-167 \text{ kJ/mol}\). This straightforward approach allows for a clear understanding of how energy dynamics work in ionic reactions.