Problem 168 - A Complete Resource Book in Chemistry for JEE Main

Question

Oxidizing power of chlorine in aqueous solution can be determined by the parameters indicated below: \(1 / 2 \mathrm{Cl}_{2}(\mathrm{~g}) \stackrel{1 / 2 \Delta \mathrm{H}_{\mathrm{Diss}}}{\longrightarrow} \mathrm{Cl}(\mathrm{g}) \stackrel{\Delta_{\mathrm{eg}} \mathrm{H}^{-}}{\longrightarrow}\) \(\mathbf{1 7 4}\). \(\mathrm{Cl}^{-}(\mathrm{g}) \quad \stackrel{\Delta_{\mathrm{hyd}} \mathrm{H}}{\longrightarrow} \mathrm{Cl}^{-}(\mathrm{aq})\) The energy involved in the conversion of \(1 / 2 \mathrm{Cl}_{2}(\mathrm{~g})\) to \(\mathrm{Cl}^{-}(\mathrm{g})\) (Using the data, \(\Delta \mathrm{H}_{\mathrm{Cl}_{2}}=240 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\mathrm{cg}} \mathrm{H}^{-\mathrm{Cl}}=\) \(-349 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\mathrm{liyd}} \mathrm{H} \mathrm{Cl}=-381 \mathrm{~kJ} \mathrm{~mol}^{-\mathrm{i}}\) ) will be [2008] (a) \(+152 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-610 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-850 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(+120 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The energy involved is \(-610 \text{ kJ/mol}\). This matches option (b).

Step 1: Understanding the Process

First, we need to decompose the process of converting half a mole of chlorine gas into chloride ions in a gaseous and then aqueous state. The reactions are split into three stages: dissociation of chlorine into chlorine atoms, electron gain by chlorine atoms to form chloride ions, and hydration of chloride ions into an aqueous solution.

Step 2: Calculating the Dissociation Energy Contribution

The dissociation half-reaction of chlorine is given by: \( \frac{1}{2} \text{Cl}_2 \rightarrow \text{Cl} \). The enthalpy change for this process, \( \Delta H_{\text{Diss}} \), can be calculated as half of the dissociation energy of \( \text{Cl}_2 \), which is given as 240 kJ/mol. Therefore, \( \frac{1}{2} \times 240 = 120 \text{ kJ/mol}. \)

Step 3: Adding Electron Gain Enthalpy

Next, we add the electron gain enthalpy for the formation of \( \text{Cl}^- \) ions from chlorine atoms: \( \text{Cl} + e^- \rightarrow \text{Cl}^- \). The enthalpy change \( \Delta_{eg} H^- \) is given as -349 kJ/mol. Since it's an exothermic reaction, we add this value directly.

Step 4: Calculating Energy of Conversion to Gaseous Ion

The total energy for forming gaseous chloride ions from chlorine gas is calculated by summing the energies from Step 2 and Step 3. \( 120 \text{ kJ/mol} - 349 \text{ kJ/mol} = -229 \text{ kJ/mol}. \)

Step 5: Incorporating Hydration Enthalpy

The hydration process converts \( \text{Cl}^- \) from gaseous to aqueous state: \( \text{Cl}^- (g) \rightarrow \text{Cl}^- (aq) \). The enthalpy change \( \Delta_{hyd} H \) for this is given as -381 kJ/mol. This value is added to the previous result because it too is exothermic.

Step 6: Final Calculation of Total Energy

To find the total energy involved in converting \( \frac{1}{2} \text{Cl}_2 \) to \( \text{Cl}^- (aq) \), sum the results from Steps 4 and 5. \( -229 \text{ kJ/mol} - 381 \text{ kJ/mol} = -610 \text{ kJ/mol}. \)

Key Concepts

Dissociation Energy

In the context of chemical reactions, dissociation energy refers to the amount of energy required to break a particular bond in a molecule. The dissociation energy for chlorine is key in understanding why chlorine is a potent oxidizer. Let's break it down simply:

• Dissociation energy is quantified as the energy needed to split a molecule into its individual atoms.
• Specifically, for chlorine gas (\( \text{Cl}_2 \)), this means dividing the energy needed to separate the chlorine-chlorine single bond.
• In our example with chlorine, we took half a mole of chlorine gas. Since the total dissociation energy for \( \text{Cl}_2 \) is given as 240 kJ/mol, half of this is 120 kJ/mol.

Dissociation is a crucial first step, as breaking chlorine molecules into atoms prepares them to gain electrons and form ions. This released energy indicates why the dissociation of chlorine is an endothermic step, meaning it absorbs energy, which is later balanced out by subsequent exothermic steps.

Electron Gain Enthalpy

Electron gain enthalpy is the energy change that occurs when an atom gains an electron. For chlorine, this process is crucial in converting chlorine atoms into chloride ions. Here's how it works:

• When an atom like chlorine gains an electron, it forms a negatively charged ion, known as chloride (\( \text{Cl}^- \)).
• The electron gain enthalpy represents this energy change. It's noted as \( \Delta_{eg} H^- = -349 \text{ kJ/mol} \) for chlorine.
• This negative value tells us that the process is exothermic, where energy is released when the electron is gained.

Why is this important? This exothermic nature of electron gain propels the overall oxidizing ability of chlorine, as energy release facilitates further interactions with other substances. The high electronegativity of chlorine ensures it efficiently captures electrons, making the electron gain enthalpy a driving force of its oxidizing power.

Hydration Enthalpy

Hydration enthalpy is the energy change that occurs when gaseous ions become aqueous, effectively incorporating them into a solution. This process completes the sequence of conversions from chlorine gas to chloride ions in solution.

• The hydration process for chloride involves moving from a gaseous state to an ion surrounded by water molecules in a solution.
• Hydration enthalpy is given as \( \Delta_{hyd} H = -381 \text{ kJ/mol} \) for chlorine ions. This negative value signifies an exothermic reaction here as well.

Why does this matter? The release of energy during hydration aids in stabilizing the ions in solution, making the reaction more favorable. The significant negative hydration enthalpy ensures the dissolved ions remain stable and active, enhancing their reactivity and thus, their utility in oxidation reactions. The entire step reinforces the energy profile of exothermic reactions that empower chlorine's oxidizing capabilities.