Question

Observe the following reactions: 
\( AB(g) + 25 H_2O(l) \rightarrow AB(H_2S{O_4}) \quad \Delta H = x \, {kJ/mol}^{-1} \) 
\( AB(g) + 50 H_2O(l) \rightarrow AB(H_2SO_4) \quad \Delta H = y \, {kJ/mol}^{-1} \) 
The enthalpy of dilution, \( \Delta H_{dil} \) in kJ/mol\(^{-1}\), is:

• A. \( (y - x) \)
• B. \( (y + x) \)
• C. \( \frac{y}{x} \)
• D. \( \frac{x}{y} \)

Hint:

In thermodynamics, the enthalpy of dilution refers to the change in enthalpy when the volume of the solution increases. It is typically calculated as the difference in enthalpy between the two reactions involved.

Answer 1

The Correct Option is A

We are given two reactions, and the enthalpy change for each reaction is denoted as \( x \) and \( y \).
The enthalpy change for dilution can be obtained from the difference in enthalpy between the second and the first reactions, as dilution refers to the increase in volume, which leads to a change in the enthalpy.
Thus, the enthalpy of dilution \( \Delta H_{dil} \) is: \[ \Delta H_{dil} = y - x \] Thus, the correct option is \( (y - x) \).

Answer 2

To solve the problem of finding the enthalpy of dilution \( \Delta H_{dil} \), we must understand the reactions and changes in enthalpy values. Here's the breakdown:

Given Reactions:

1. \( AB(g) + 25 H_2O(l) \rightarrow AB(H_2SO_4), \quad \Delta H = x \, \text{kJ/mol}^{-1} \)

2. \( AB(g) + 50 H_2O(l) \rightarrow AB(H_2SO_4), \quad \Delta H = y \, \text{kJ/mol}^{-1} \)

These reactions illustrate the dissolution of a gaseous compound \( AB \) into sulfuric acid solutions with differing amounts of water. The enthalpy change, \( \Delta H \), indicates the energy change during the process.

Concept of Enthalpy of Dilution:

The enthalpy of dilution \( \Delta H_{dil} \) is defined as the change in enthalpy when a solution at a certain concentration is diluted. Here, it essentially represents the enthalpy change by adding more water to the solution, transitioning from 25 moles of water to 50 moles. To calculate, we assess the enthalpy difference between these two states.

Calculation:

The reaction with 50 moles of water compared to 25 moles implies the addition of 25 more moles of water. The change in enthalpy due to this dilution is:

\( \Delta H_{dil} = y - x \)

Here, \( y \) represents the enthalpy change incorporating 50 moles of water, and \( x \) represents the same with 25 moles. The difference gives the enthalpy of dilution for those additional 25 moles of water.

Conclusion:

Thus, the enthalpy of dilution \( \Delta H_{dil} \) in kJ/mol\(^{-1}\) is:

\( (y - x) \)