Problem 183 - Introductory Chemistry Atoms First

Question

Whether or not a solute dissolves in a solvent is decided not only by changes in energy but also by changes in

Step-by-Step Solution

Answer

The entropy change when 1 mole of a solute dissolves in a solvent is 0 J/mol Kelvin.

Step 1: Calculate the partition function (q)

Using the given formula, the partition function (q) can be calculated as follows: \(q = k \cdot V^{2/3}\) Given that k = 1 and V = 1 liter, the partition function (q) can be calculated as: \[q = 1 \cdot (1)^{2/3}\] \[q = 1\]

Step 2: Calculate the entropy (S)

Using the partition function (q) calculated in Step 1, the entropy (S) can be calculated using the given formula: \(S = R \cdot ln(q)\) Given R = 8.314 J/mol Kelvin and q = 1, the entropy (S) can be calculated as: \[S = 8.314 \cdot ln(1)\] \[S = 8.314 \cdot 0\] Since ln(1) = 0, the entropy (S) is 0 J/mol Kelvin. Therefore, the entropy change when 1 mole of a solute dissolves in a solvent is 0 J/mol Kelvin.

Key Concepts

Partition Function in Solubility

The concept of the partition function is critical in understanding the behavior of molecules in a solution. It represents the number of different states a solute can be arranged in when it is dissolved in a solvent. This function is vital in thermodynamics and statistical mechanics, as it helps determine how molecules distribute themselves in different energy levels.
The formula for the partition function is given by:

• \( q = k \cdot V^{2/3} \)

Where:

• \( k \) is a constant.
• \( V \) is the volume in liters.

By calculating the partition function, we can understand the potential energy configurations for the solute in the solution. In the example provided, with \( k = 1 \) and volume \( V = 1 \text{ liter} \), the calculation simplifies to:

• \( q = 1 \cdot (1)^{2/3} = 1 \)

Knowing the partition function helps predict how a solute behaves when it interacts with a solvent, giving insight into how readily a solute may dissolve.

Entropy Change in Solubility

Entropy is a measure of disorder or randomness in a system. In the context of solubility, it refers to the degree of disorder introduced when a solute dissolves in a solvent. Entropy change is an important aspect of the solubility process because it can influence whether a solute will dissolve or not.
The entropy \( S \) can be calculated using the partition function \( q \):

• \( S = R \cdot \ln(q) \)

Where:

• \( R \) is the universal gas constant, \( 8.314 \text{ J/mol K} \).
• \( \ln \) denotes the natural logarithm.

In our example, when the partition function \( q = 1 \), the entropy calculates to:

• \( S = 8.314 \cdot \ln(1) = 0 \text{ J/mol K} \)

This result indicates there is no change in entropy when the solute dissolves, suggesting a very organized or stable process that might involve energy being conserved rather than dispersed.

Solute-Solvent Interaction

When a solute dissolves in a solvent, it undergoes specific interactions that can affect its solubility. These interactions depend on the nature of the solute and the solvent. The key interactions include:

• **Ionic interactions**: Strong forces between charged particles, often seen in salts dissolving in water.
• **Dipole interactions**: Occur between molecules with permanent dipoles, like water and alcohols.
• **Hydrogen bonding**: A special type of dipole interaction, vital for water solubility of organic compounds.
• **Van der Waals forces**: Weak interactions that can also influence solubility, especially in nonpolar substances.

The strength and type of these interactions determine how well a solute will dissolve in a given solvent. Solutes that form strong interactions with a solvent, like hydrogen bonds in water, have higher solubility. Conversely, substances with weak solvent interactions might not dissolve as easily. Understanding solute-solvent interactions helps predict and manipulate the solubility of various substances.