Problem 109 - A Complete Resource Book in Chemistry for JEE Main

Question

The enthalpies of solution of \(\mathrm{BaCl}_{2}\) (s) and \(\mathrm{BaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) (s) are \(-20.6\) and \(8.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. The enthalpy change for the hydration of \(\mathrm{BaCl}_{2}(\mathrm{~s})\) is (a) \(29.8 \mathrm{~kJ}\) (b) \(-11.8 \mathrm{~kJ}\) (c) \(-20.6 \mathrm{~kJ}\) (d) \(-29.4 \mathrm{~kJ}\).

Step-by-Step Solution

Answer

The enthalpy change for hydration is \(-29.4\, \mathrm{kJ}\, \mathrm{mol}^{-1}\). Answer is (d).

Step 1: Understanding the Problem

We need to find the enthalpy change for the hydration of \( \mathrm{BaCl}_{2}(\mathrm{~s}) \) into \( \mathrm{BaCl}_{2}\cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{~s}) \). The provided data includes the enthalpies of solution for \( \mathrm{BaCl}_{2}(\mathrm{~s}) \) and \( \mathrm{BaCl}_{2}\cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{~s}) \).

Step 2: Expressing the Given Enthalpies

We are given:1. The enthalpy of solution for \( \mathrm{BaCl}_{2}(\mathrm{~s}) \) is \(-20.6\, \mathrm{kJ}\, \mathrm{mol}^{-1}\).2. The enthalpy of solution for \( \mathrm{BaCl}_{2}\cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{~s}) \) is \(8.8\, \mathrm{kJ}\, \mathrm{mol}^{-1}\).

Step 3: Calculating the Hydration Enthalpy

To find the hydration enthalpy of \( \mathrm{BaCl}_{2}(\mathrm{~s}) \) to \( \mathrm{BaCl}_{2}\cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{~s}) \), subtract the enthalpy of solution of the hydrated compound from the anhydrous compound:\[\Delta H_{\text{hydration}} = -20.6\, \mathrm{kJ}\, \mathrm{mol}^{-1} - 8.8\, \mathrm{kJ}\, \mathrm{mol}^{-1} = -29.4\, \mathrm{kJ}\, \mathrm{mol}^{-1}\]

Step 4: Selecting the Correct Option

The calculated enthalpy change for the hydration is \(-29.4\, \mathrm{kJ}\, \mathrm{mol}^{-1}\). Comparing with the given options, the correct answer is (d) \(-29.4\, \mathrm{kJ}\).

Key Concepts

Hydration Enthalpy

Hydration enthalpy refers to the heat energy released when one mole of ions undergo hydration, that is, the ions dissolve in water. The ions interact with water molecules, forming a hydrate.
This process is generally exothermic, meaning it releases heat.In our example with \( \mathrm{BaCl}_{2} \), we begin with an anhydrous salt, which means it has no water molecules attached. When it hydrates to form \( \mathrm{BaCl}_{2}\cdot 2 \mathrm{H}_{2} \mathrm{O} \), it bonds with water, rendering the enthalpy change for the hydration as calculated in the exercise equal to \( -29.4 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \). The negative sign indicates an exothermic reaction.
Understanding hydration enthalpy is essential when discussing solubility, as the energy released during hydration affects a substance’s solubility in water, impacting chemical reactions in solutions.

Thermodynamics in Chemistry

Thermodynamics in chemistry focuses on the energy changes that occur during chemical reactions and processes. It is essential in understanding enthalpy, including hydration enthalpy, which illustrates how energy is exchanged when substances interact with their surroundings, particularly water.Some core concepts in thermodynamics include:

• System and Surroundings: The system is the part of the universe we are studying, while the surroundings are everything else. In hydration, the system includes the ions and water molecules.
• First Law of Thermodynamics: This law states that energy cannot be created or destroyed, only transferred or converted. This principle underlies enthalpy changes.
• Exothermic and Endothermic Processes: Exothermic reactions release heat into the surroundings, while endothermic ones absorb heat. Hydration of \( \mathrm{BaCl}_{2} \) is exothermic, as indicated by the negative \( \Delta H_{\text{hydration}} \).

Thermodynamics provides the framework to predict whether reactions like dissolving or hydration will occur spontaneously based on these energy changes.

Ionic Compounds Energy Calculations

Energy calculations involving ionic compounds are crucial to predicting the energetic outcome of various reactions. Ionic compounds consist of positively and negatively charged ions held together through ionic bonds.
They are often soluble in water, and understanding their energy properties is vital for determining their behavior in solution.When calculating enthalpies, such as the hydration enthalpy of \( \mathrm{BaCl}_{2} \), it involves:

• Determining the lattice energy, which is the energy required to separate one mole of a solid ionic compound into its gaseous ions. Though not directly involved in this exercise, it’s a concept closely related to hydration.
• Calculating the enthalpy of solution by combining the ionic interactions with solvent interactions, which are key in reactions and solution properties.
• Using Hess's Law to understand and simplify the complexities of these reactions — this law allows the calculation of enthalpy changes indirectly, by summing known enthalpy changes.

Energy calculations for ionic compounds are fundamental for chemists when analyzing reactions and processes in solutions, allowing the discovery and confirmation of reaction mechanisms through numeric energy values.