Question

If S is the solubility of small particles of radius r, is the normal solubility (i.e., of a solid consisting of fairly large particles), is the interfacial energy, M is the molecular weight of the solid, ρ is the density of the bulk solid, R is the gas constant and T is the thermodynamic temperature, then which of the following equation indicates the changes in interfacial free energy that accompany the dissolution of particles of varying sizes causing the solubility of substance to increase with decreasing particle size?

• A. Log(S₀/S)=2γMr/2.303RTρ
• B. Log(S/S₀)=2γM/2.303RTρr
• C. Log(S/S₀)=2γMr/2.303RTρ
• D. Log(S₀/S)=2γMr/2.303RTρr

Answer 1

The Correct Option is B

The problem at hand concerns understanding how the solubility of particles is influenced by their size due to changes in interfacial energy. We need to identify the correct equation that describes this relationship.

The context provided is related to the principle of the Kelvin effect or Gibbs-Thomson effect. The solubility of small particles is often greater than that of larger ones due to the curved surface of small particles. This is mathematically expressed considering the interfacial energy, the molecular weight of the solid, density, gas constant, and temperature. 

Given:

• \(S\): Solubility of small particles.
• \(S_0\): Normal solubility.
• \(\gamma\): Interfacial energy.
• \(M\): Molecular weight of the solid.
• \(\rho\): Density of the bulk solid.
• \(R\): Gas constant.
• \(T\): Thermodynamic temperature.

We have to find the correct equation that links all these variables together.

The formula to describe this relationship is:

\(\text{Log}\left(\frac{S}{S_0}\right) = \frac{2\gamma M}{2.303RT\rho r}\)

This equation suggests that as the particle size \(r\) decreases, the solubility \(S\) increases, assuming all other parameters remain constant. The factor of \(2.303\) comes in when switching from natural logarithms to common logarithms.

Let's evaluate the options:

• Option 1: \(\text{Log}\left(\frac{S_0}{S}\right) = \frac{2\gamma Mr}{2.303RT\rho}\) - Incorrect due to incorrect placement of \(r\).
• Option 2: \(\text{Log}\left(\frac{S}{S_0}\right) = \frac{2\gamma M}{2.303RT\rho r}\) - Correct equation representing the correct relationship.
• Option 3: \(\text{Log}\left(\frac{S}{S_0}\right) = \frac{2\gamma Mr}{2.303RT\rho}\) - Incorrect due to incorrect placement of \(r\).
• Option 4: \(\text{Log}\left(\frac{S_0}{S}\right) = \frac{2\gamma Mr}{2.303RT\rho r}\) - Incorrect due to incorrect placement of \(r\) and \(S_0/S\).

Thus, the correct answer is: \(\text{Log}\left(\frac{S}{S_0}\right) = \frac{2\gamma M}{2.303RT\rho r}\)