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: \(1 / 2 \mathrm{Cl}_{2}(\mathrm{~g}) \stackrel{1 / 2 \Delta \mathrm{H}_{\mathrm{Dias}}}{\longrightarrow} \mathrm{Cl}(\mathrm{g}) \stackrel{\Delta_{\mathrm{cg}} \mathrm{H}^{-}}{\longrightarrow}\) \(\mathrm{Cl}^{-}(\mathrm{g}) \quad \stackrel{\Delta_{\mathrm{ln} \mathrm{d}} \mathrm{H}}{\mathrm{H}} \longrightarrow \mathrm{Cl}^{-}(\mathrm{aq})\) The energy involved in the conversion of \(1 / 2 \mathrm{Cl}_{2}(\mathrm{~g})\) to \(\mathrm{Cl}^{-}(\mathrm{g})\) (Using the data, \(\Delta \mathrm{H}_{\mathrm{C}_{2}}=240 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\mathrm{cg}} \mathrm{H}-\mathrm{Cl}=\) \(\left.-349 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\mathrm{lnd}} \mathrm{H}-\mathrm{Cl}=-381 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)\) will be (a) \(+152 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-610 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-850 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(+120 \mathrm{~kJ} \mathrm{~mol}^{-\mathrm{t}}\)

Step-by-Step Solution

Answer

The correct answer is (b) \(-610 \mathrm{~kJ/mol}\).

Step 1: Write the Reaction Series

The process from \( \frac{1}{2} \mathrm{Cl}_2(\mathrm{g}) \) to \( \mathrm{Cl}^- (\mathrm{aq}) \) involves the following steps:1. \( \frac{1}{2} \mathrm{Cl}_2(\mathrm{g}) \rightarrow \mathrm{Cl}(\mathrm{g}) \)2. \( \mathrm{Cl}(\mathrm{g}) \rightarrow \mathrm{Cl}^-(\mathrm{g}) \)3. \( \mathrm{Cl}^-(\mathrm{g}) \rightarrow \mathrm{Cl}^-(\mathrm{aq}) \)

Step 2: Identify the Enthalpy Changes

For each step, we have the following enthalpy changes:- \( \frac{1}{2} \mathrm{Cl}_2(\mathrm{g}) \rightarrow \mathrm{Cl}(\mathrm{g}) \): \( \frac{1}{2} \Delta \mathrm{H}_{\mathrm{C}_2} = \frac{1}{2} \times 240 \mathrm{~kJ/mol} = 120 \mathrm{~kJ/mol} \)- \( \mathrm{Cl}(\mathrm{g}) \rightarrow \mathrm{Cl}^-(\mathrm{g}) \): \( \Delta_{\mathrm{cg}} \mathrm{H}^-= -349 \mathrm{~kJ/mol} \)- \( \mathrm{Cl}^-(\mathrm{g}) \rightarrow \mathrm{Cl}^-(\mathrm{aq}) \): \( \Delta_{\mathrm{lnd}} \mathrm{H}= -381 \mathrm{~kJ/mol} \)

Step 3: Calculate the Total Enthalpy Change

The total enthalpy change for the conversion from \( \frac{1}{2} \mathrm{Cl}_2(\mathrm{g}) \) to \( \mathrm{Cl}^- (\mathrm{aq}) \) is the sum of all the enthalpy changes:\[\Delta \mathrm{H}_{\text{total}} = \frac{1}{2} \Delta \mathrm{H}_\mathrm{C_2} + \Delta_{\mathrm{cg}} \mathrm{H}^- + \Delta_{\mathrm{lnd}} \mathrm{H} = 120 \mathrm{~kJ/mol} - 349 \mathrm{~kJ/mol} - 381 \mathrm{~kJ/mol}\]Calculating this gives:\[\Delta \mathrm{H}_{\text{total}} = 120 - 349 - 381 = -610 \mathrm{~kJ/mol}\]

Step 4: Determine the Correct Answer

The calculated total enthalpy change is \(-610 \mathrm{~kJ/mol}\). This corresponds with option (b).

Key Concepts

Enthalpy Changes in Chemical Reactions

Enthalpy changes are crucial to understanding energy exchanges in chemical reactions. In the context of chlorine conversion, we look at how energy is absorbed or released in each step of the transition from chlorine gas to aqueous chloride ions. Understanding these changes is pivotal for grasping why some reactions are exothermic (release energy) while others are endothermic (absorb energy).

- **Determination of Enthalpy**: Enthalpy changes (9H) can be calculated by assessing the differences in energy between reactants and products. In our case, chlorine undergoes several transformations, each with a specific enthalpy change.
- The dissociation of one mole of chlorine gas, 9H_C2, is given as 240 kJ/mol. Since we begin with half a mole, this value is halved to 120 kJ/mol for step one.
- **Sum of Enthalpy Changes**: The total enthalpy change for a series of reactions is the sum of individual enthalpies: dissociation of chlorine gas, electron gain by Cl, and solvation of chloride ion.
- **Exothermic Reaction Outcome**: A negative total enthalpy change (9H_total = -610 kJ/mol) suggests the overall process releases energy, typical of exothermic reactions, leading to stable products.

Chlorine Ionization

Chlorine ionization refers to the process where chlorine gas molecules lose electrons to form chlorine ions. This step in our exercise illustrates how chlorine, initially neutral as Cl_2, transforms into chloride ions (Cl^-). Understanding this process is vital, as it underpins the oxidizing power of chlorine.

- **Chlorine Dissociation**: Initially, chlorine molecules split into individual Cl atoms, an endothermic step requiring energy input (9H = 120 kJ/mol).
- **Electron Gain**: A Cl atom becomes a chloride ion by gaining an electron. Notably, this step (9H_cg) is highly exothermic, releasing 349 kJ/mol. This release indicates that energy is freed when the electron is added, showing the ion's stability.
- **Role in Oxidation**: As a potent oxidizing agent, chlorine readily adopts electrons in various reactions. This ability is what gives chlorine its unique oxidizing power, making it useful for disinfection and other applications.
- **Stable Ion Formation**: The negative enthalpy change during ionization signals that chloride ions formed are energetically favorable and stable in comparison to isolated chlorine atoms.

Chlorine Aqueous Solution Chemistry

Chlorine's behavior in aqueous solutions is fascinating due to its chemical transformations. Upon solvation, gaseous chloride ions become aqueous, an important step in water treatment and disinfection processes.

- **Solvation Process**: For chlorine ions transitioning from gas to aqueous (Cl^- (aq)), the enthalpy of solvation (9H_lnd) is -381 kJ/mol. This significant exothermic step releases energy as chloride ions interact with water molecules.
- **Interaction with Water**: The resulting solvent-stabilized ions prove critical for applications like bleaching and disinfection. The strong hydration of chloride ions enhances their solubility and reaction potential in water.
- **Chemical Stability**: The energy released during solvation contributes to the stability of chloride ions in solution, making them a less reactive, yet essential species in aqueous environments.
- **Practical Relevance**: Understanding chlorine's aqueous chemistry is key for industries relying on its oxidizing power, such as water purification, where its effectiveness is harnessed to destroy harmful microorganisms.