Problem 48 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The dissolution of \(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) in a large volume of water is endothermic to the extent of \(3.5 \mathrm{kcal} / \mathrm{mol}\). For the reaction, \(\mathrm{CaCl}_{2}(\mathrm{~s})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) \(\Delta H\) is \(-23.2\) kcal. The heat of solution of anhydrous \(\mathrm{CaCl}_{2}\) in large quantity of water will be (a) \(-26.7 \mathrm{kcal} \mathrm{mol}^{-1}\) (b) \(-19.7\) kcal \(\mathrm{mol}^{-1}\) (c) \(19.7 \mathrm{kcal} \mathrm{mol}^{-}\) (d) \(26.7 \mathrm{kcal} \mathrm{mol}^{-1}\)

Step-by-Step Solution

Answer

\(\Delta H_{\text{solution}} = -19.7 \text{ kcal/mol}\), so the correct answer is (b) \-19.7\text{ kcal mol}^{-1}\.

Step 1: Identify the Known Values

We know the enthalpy change for the dissolution of \(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) is 3.5 kcal/mol (endothermic, hence it absorbs heat). We also have the enthalpy change \(\Delta H\) for the reaction \(\mathrm{CaCl}_{2}(\mathrm{~s})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) as -23.2 kcal.

Step 2: Write Down the Enthalpy Changes for Processes

Let's denote the enthalpy change for the dissolution of anhydrous \(\mathrm{CaCl}_{2}\) as \(\Delta H_{\text{anhyd}}\). We know: \[\Delta H_{\text{hydrate}} = 3.5 \text{ kcal/mol}\] and \[\Delta H_{\text{reaction}} = -23.2 \text{ kcal/mol}\]. We need to find \(\Delta H_{\text{anhyd}}\).

Step 3: Use Hess's Law

Hess's Law states that the total enthalpy change for a reaction is the same regardless of the number of steps taken. So, we can add the enthalpy of the dissolution step of the hydrate to the enthalpy of the reaction to find the enthalpy of solution of anhydrous \(\mathrm{CaCl}_{2}\). \[\Delta H_{\text{solution}} = \Delta H_{\text{reaction}} + \Delta H_{\text{hydrate}}\]

Step 4: Calculate the Heat of Solution

Substituting the known values for \(\Delta H_{\text{reaction}}\) and \(\Delta H_{\text{hydrate}}\): \[\Delta H_{\text{solution}} = -23.2 \text{ kcal/mol} + 3.5 \text{ kcal/mol} = -19.7 \text{ kcal/mol}\]. Since the reaction is endothermic, the heat of solution should be expressed as a negative value.

Key Concepts

Enthalpy Change

Understanding enthalpy change is fundamental in the study of energy transformations in chemical reactions. It represents the heat content change in a system when a reaction occurs under constant pressure. Enthalpy change, usually denoted as \( \Delta H \), can be either positive or negative.

When \( \Delta H \) is positive, the process is endothermic—the system absorbs heat from its surroundings. Conversely, a negative \( \Delta H \) signifies an exothermic reaction, where the system releases heat. In our textbook example, the dissolution of \( \mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O} \) is endothermic with a \( \Delta H \) of 3.5 kcal/mol, indicating that the solution absorbs heat.

Hess's Law

Next, let's turn our attention to Hess's Law, a principle which is a real game changer when it comes to calculating enthalpies. This law asserts that the total enthalpy change for a chemical reaction is the same, no matter how many steps it takes to get from reactants to products.

It allows us to add together the enthalpy changes of individual steps to get the overall enthalpy change for the reaction. This is incredibly useful when direct measurement is impractical. In the example of \( \mathrm{CaCl}_{2} \) dissolution, we cleverly apply Hess's Law to determine the heat of solution for anhydrous \( \mathrm{CaCl}_{2} \) by adding the enthalpy change of the hydrate's dissolution to the enthalpy change of forming the hydrate from the anhydrous salt.

Endothermic Reaction

Focusing on Endothermic Reactions, these occur when the energy absorbed from the surrounding environment is greater than the energy released during the formation of new bonds. An endothermic reaction feels cold to the touch, since it's absorbing heat.

This contrasts with exothermic reactions that release heat, warming up the surroundings. In the process of dissolving \( \mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O} \) in water, the system's absorption of 3.5 kcal/mol of energy is what makes it endothermic.

Thermochemistry

Finally, the broader field in which all these concepts are interrelated is Thermochemistry. It's the study of the heat evolved or absorbed in chemical reactions. Thermochemistry involves the application of thermodynamic principles to chemical reactions, giving us insights into energy changes during these reactions.

It provides a basis for making predictions about energy transfers, understanding reaction spontaneity, and optimizing conditions for industrial processes. Our study of the heat of solution for \( \mathrm{CaCl}_{2} \) utilizes thermochemical principles to calculate the energy involved in the dissolution process, resulting in our solution of \( -19.7 \text{ kcal/mol} \) as the heat of solution for anhydrous \( \mathrm{CaCl}_{2} \) in water.