Question

A drug that has diffusion constant \(D=10^{-6} \mathrm{~cm}^{2} \mathrm{~s}^{-1}\) reacts at a rate \(k_{\mathrm{rx}}=10^{2} \mathrm{~s}^{-1}\) at the aqueous surface of a tablet. At what 'decay length' is the concentration of drug \(1 / e\) of its value inside the tablet?

Short answer

Decay length is \(10^{-4} \mathrm{~cm}\).

Step-by-step solution

Understand the problem

Determine the 'decay length' at which the concentration of the drug is reduced to \(\frac{1}{e}\) of its value inside the tablet.

Identify the given values

\(D = 10^{-6} \mathrm{~cm}^{2} \mathrm{~s}^{-1}\) (diffusion constant), \( k_{\mathrm{rx}} = 10^{2} \mathrm{~s}^{-1} \) (reaction rate)

Write the formula for decay length

The formula correlating diffusion constant and reaction rate to decay length is given by:\[ \lambda = \sqrt{\frac{D}{k_{\mathrm{rx}}}} \]

Substitute the given values

Plug in the values of \(D\) and \(k_{\mathrm{rx}}\) into the formula:\[ \lambda = \sqrt{\frac{10^{-6} \mathrm{~cm}^{2} \mathrm{~s}^{-1}}{10^{2} \mathrm{~s}^{-1}}} \]

Simplify the expression

Simplify under the square root:\[ \lambda = \sqrt{\frac{10^{-6}}{10^{2}}} \mathrm{~cm} \lambda = \sqrt{10^{-8}} \mathrm{~cm} \lambda = 10^{-4} \mathrm{~cm} \]

Key concepts

Diffusion Constant

The diffusion constant, often represented as the letter \(D\), indicates how quickly a substance spreads out or diffuses through another medium. In the context of drug diffusion, it specifies how fast a drug moves through a solvent.

For example, in our problem, the drug has a diffusion constant of \(D = 10^{-6} \text{ cm}^2 \text{ s}^{-1}\). This means the drug diffuses at a rate of \(10^{-6} \text{ cm}^2 \text{ per second}\).

To understand it better:

• High diffusion constant: The substance spreads out quickly.
• Low diffusion constant: The substance spreads out slowly.

The formula for the diffusion rate can be represented as: \[ J = -D \frac{\text{d}C}{\text{d}x} \] where \(J\) is the flux, \(C\) is the concentration, and \(x\) is the position. This equation helps to relate the concentration gradient with the speed of diffusion.

Reaction Rate

The reaction rate, symbolized as \(k_{\text{rx}}\), defines how quickly a chemical reaction occurs. In the case of drug diffusion, it tells how fast the drug is reacting or degrading over time.

In our example, the given reaction rate is \(k_{\text{rx}} = 10^{2} \text{ s}^{-1}\). This tells us that the drug reacts very quickly at the surface.

Key points to remember:

• High reaction rate: The reaction occurs rapidly.
• Low reaction rate: The reaction takes place slowly.

To quantify how fast the concentration of a substance changes due to the reaction, we can use: \( \frac{\text{d}C}{\text{d}t} = -k_{\text{rx}} C \) where \(C\) is the concentration. This indicates that the reaction rate is proportional to the concentration of the drug.

Decay Length Formula

The decay length, represented as \( \lambda \), is the distance over which the concentration of the drug drops to \( \frac{1}{e} \) of its initial value inside the tablet. It's a crucial concept in determining how far the drug will effectively diffuse before significantly degrading.

The decay length is calculated using the formula: \[ \lambda = \sqrt{\frac{D}{k_{\text{rx}}}} \]

Let's break it down with our example:

• \(D\) is the diffusion constant.
• \(k_{\text{rx}}\) is the reaction rate.

By substituting the given values: \( D = 10^{-6} \text{ cm}^2 \text{ s}^{-1} \) and \( k_{\text{rx}} = 10^2 \text{ s}^{-1} \):

\[ \lambda = \sqrt{\frac{10^{-6} \text{ cm}^2 \text{ s}^{-1}}{10^2 \text{ s}^{-1}}} \]
Simplifying it:

\[ \lambda = \sqrt{\frac{10^{-6}}{10^{2}}} \text{ cm} = \sqrt{10^{-8}} \text{ cm} = 10^{-4} \text{ cm} \]
So, the decay length \( \lambda \) is \( 10^{-4} \text{ cm} \), meaning beyond this distance, the drug concentration falls significantly.