Question

Oil/water partitioning of drugs. In the process of partitioning of a drug from oil into water, \(\Delta s^{*}=\) \(-50 \mathrm{cal} \mathrm{deg}^{-1} \mathrm{~mol}^{-1}\) and \(\Delta h^{*}=0\) at \(T=300 \mathrm{~K}\). (a) What is \(\Delta \mu^{*}\) at \(T=300 \mathrm{~K}\) ? (b) What is the partition coefficient from oil to water, \(K_{\mathrm{oll}}^{\text {water }}\) at \(T=300 \mathrm{~K} ?\) (c) Assume that \(\Delta s^{*}\) and \(\Delta h^{*}\) are independent of temperature. Calculate \(K_{0}^{W}\) at \(T=320 \mathrm{~K}\).

Short answer

a) \Delta \mu^{*} = 15000 \mathrm{cal \cdot mol^{-1}} b) \ K_{\mathrm{oil}}^{\text {water }} \approx 1.20 \times 10^{-11} c) \ K_{0}^{W} \approx 1.25 \times 10^{-11}

Step-by-step solution

Understanding the Given Data

We have the following information: - Entropy change, \(\Delta s^{*} = -50 \mathrm{cal} \mathrm{deg}^{-1} \mathrm{~mol}^{-1}\) - Enthalpy change, \(\Delta h^{*} = 0\) - Temperature, \(\ T = 300 \mathrm{~K}\).We need to find \(\Delta \mu^{*}\), the partition coefficient \(\K_{\mathrm{oil}}^{\text {water }}\) at \(\T = 300 \mathrm{~K}\), and \(\K_{0}^{\W}\) at \(\T = 320 \mathrm{~K}\).

Calculate \Delta \mu^{*} at \T = 300 \mathrm{~K}

The change in chemical potential \(\Delta \mu^{*}\) is given by the equation: \[\Delta \mu^{*} = \Delta h^{*} - T \Delta s^{*} \] Given \Delta h^{*} = 0 \ \mathrm{cal \cdot mol^{-1}} and \ T = 300 \mathrm{~K}, we can substitute these values into the equation: \[\Delta \mu^{*} = 0 - 300 \mathrm{~K} \cdot (-50 \mathrm{cal \cdot deg^{-1} \cdot mol^{-1}}) = 0 + 15000 \mathrm{cal \cdot mol^{-1}} \] Thus, \Delta \mu^{*} = 15000 \mathrm{cal \cdot mol^{-1}}.

Calculate \K_{\mathrm{oil}}^{\text {water }} at \T = 300 \mathrm{~K}

The partition coefficient \ K_{\mathrm{oil}}^{\text {water }} is related to \Delta \mu^{*} by the equation: \[\ K_{\mathrm{oil}}^{\text {water }} = \exp\left(\frac{-\Delta \mu^{*}}{RT}\right) \] Given \(\Delta \mu^{*} = 15000 \mathrm{cal \cdot mol^{-1}}\), \(\ T = 300 \mathrm{~K}\), and the gas constant \(\ R = 1.987 \mathrm{cal \cdot mol^{-1} \cdot K^{-1}}\), we can substitute these values into the equation: \[\ K_{\mathrm{oil}}^{\text {water }} = \exp\left(\frac{-15000}{1.987 \times 300}\right) = \exp\left(\frac{-15000}{596.1}\right) = \exp\left(-25.15\right)\] Thus, \ K_{\mathrm{oil}}^{\text {water }} \approx 1.20 \times 10^{-11}.

Calculate \ K_{0}^{W} at \ T = 320 \mathrm{~K}

Given that \Delta s^{*} and \Delta h^{*} are independent of temperature, we use the same equations with \ T = 320 \mathrm{~K}. First, we calculate the new \Delta \mu^{*}: \[\Delta \mu^{*} = \Delta h^{*} - T \Delta s^{*} = 0 - 320 \ \mathrm{~K} \cdot (-50 \mathrm{cal \cdot deg^{-1} \cdot mol^{-1}}) = 0 + 16000 \mathrm{cal \cdot mol^{-1}} \] Thus, \Delta \mu^{*} = 16000 \mathrm{cal \cdot mol^{-1}}. Now, we find the partition coefficient: \[ K_{0}^{W} = \exp\left(\frac{-\Delta \mu^{*}}{RT}\right) = \exp\left(\frac{-16000}{1.987 \times 320}\right) = \exp\left(\frac{-16000}{635.84}\right) = \exp\left(-25.16\right)\] Thus, \ K_{0}^{W} \approx 1.25 \times 10^{-11}.

Key concepts

chemical potential

Chemical potential is a fundamental concept in thermodynamics and chemistry.
It represents the potential of a substance to undergo a chemical reaction or phase change.
For a drug partitioning between oil and water, the chemical potential change \( \Delta \mu^{*} \) is crucial. It's calculated as follows: \[ \Delta \mu^{*} = \Delta \h^{*} - T \Delta \s^{*} \] where:

• \Delta \h^{*} is the enthalpy change,
• \Delta \s^{*} is the entropy change,
• T is the absolute temperature in Kelvin.

In our problem, \Delta \h^{*} = 0 \mathrm{cal \cdot mol^{-1}} and \Delta \s^{*} = -50 \mathrm{cal \cdot deg^{-1} \cdot mol^{-1}} at T = 300 \mathrm{~K}. By substituting these values into the equation, we get: \[ \Delta \mu^{*} = 0 - 300 \mathrm{~K \cdot (-50 \mathrm{cal \cdot deg^{-1} \cdot mol^{-1}})} = 0 + 15000 \mathrm{cal \cdot mol^{-1}} \]
This indicates the energy change when a drug molecule moves from oil to water.

partition coefficient

The partition coefficient \( K_{\mathrm{oil}}^{\text {water }} \) measures how a drug distributes itself between oil and water.
It's given by the equation: \[ K_{\mathrm{oil}}^{\text {water }} = \exp\left(\frac{-\Delta \mu^{*}}{RT}\right) \] where:

• \Delta \mu^{*} is the chemical potential change,
• R is the gas constant (1.987 \mathrm{cal \cdot mol^{-1} \cdot K^{-1}}),
• T is the temperature in Kelvin.

At T = 300 \mathrm{~K}, using \Delta \mu^{*} = 15000 \mathrm{cal \cdot mol^{-1}}, we get: \[ K_{\mathrm{oil}}^{\text {water }} = \exp\left(\frac{-15000}{1.987 \times 300}\right) = \exp\left(\frac{-15000}{596.1}\right) = \exp(-25.15) \]
This gives approximately K_{\mathrm{oil}}^{\text {water }} ≈ 1.20 \times 10^{-11}, representing how the drug partitions between oil and water at 300 \mathrm{~K}.

thermodynamics

Thermodynamics helps us understand the energy transformations when a drug moves between oil and water.
This involves enthalpy (\Delta \h^{*}) and entropy (\Delta \s^{*}). Enthalpy (\Delta \h^{*}) reflects the heat absorbed or released, while entropy (\Delta \s^{*}) reflects disorder changes. For the given problem, \Delta \h^{*} = 0 and \Delta \s^{*} = -50 \mathrm{cal \cdot deg^{-1} \cdot mol^{-1}}.
When T = 320 \mathrm{~K}, the new chemical potential change \Delta \mu^{*} is: \[ \Delta \mu^{*} = \Delta \h^{*} - T \Delta \s^{*} = 0 - 320 \mathrm{~K \cdot (-50 \mathrm{cal \cdot deg^{-1} \cdot mol^{-1}})} = 0 + 16000 \mathrm{cal \cdot mol^{-1}} \]
Then, the new partition coefficient K_{0}^{W} at 320 \mathrm{~K} is: \[ K_{0}^{W} = \exp\left(\frac{-16000}{1.987 \times 320}\right) = \exp\left(\frac{-16000}{635.84}\right) = \exp(-25.16) \]
Thus, K_{0}^{W} ≈ 1.25 \times 10^{-11}, showing how slightly increased temperature influences drug distribution.