Diffusion Rate Calculator (Fick's First Law)
This calculator determines the rate of diffusion (flux) based on Fick's First Law. It calculates how fast a substance moves from an area of higher concentration to an area of lower concentration across a specific distance.
Understanding Diffusion and Fick's First Law
Diffusion is the net movement of anything (for example, atoms, ions, molecules, energy) from a region of higher concentration to a region of lower concentration. Diffusion is driven by a gradient in concentration.
The fundamental equation describing this process in one dimension is Fick's First Law. It states that the flux (rate of flow per unit area) of a substance is proportional to the concentration gradient.
The formula is generally written as:
$J = -D \frac{dC}{dx}$
Where:
- $J$ is the diffusion flux: This is the amount of substance that flows through a unit area per unit time. Common units are mol/(m²·s) or g/(cm²·s).
- $D$ is the diffusion coefficient (or diffusivity): A proportionality constant that depends on the diffusing substance, the medium it's moving through, and the temperature. Its units are area per time, such as m²/s or cm²/s.
- $C$ is the concentration: The amount of substance per unit volume, e.g., mol/m³ or g/cm³.
- $x$ is the position (distance): The length unit along the direction of diffusion, e.g., m or cm.
- $dC/dx$ is the concentration gradient: The change in concentration over a distance. The negative sign indicates that diffusion occurs in the direction of decreasing concentration (down the gradient).
For practical calculation between two distinct points, we can simplify the gradient term to $(C_{final} – C_{initial}) / \Delta x$. To find the magnitude of the flow from high to low concentration, we use:
$J = D \frac{(C_{high} – C_{low})}{\Delta x}$
Example Calculation
Let's calculate the diffusion of glucose through a membrane of thickness 0.01 cm.
- The diffusion coefficient ($D$) of glucose in the membrane is $5 \times 10^{-7}$ cm²/s.
- The concentration on the high side ($C_1$) is $0.2$ mol/cm³.
- The concentration on the low side ($C_2$) is $0.05$ mol/cm³.
- The distance ($dx$) is $0.01$ cm.
Using the formula:
$J = (5 \times 10^{-7} \text{ cm²/s}) \times \frac{(0.2 – 0.05) \text{ mol/cm³}}{0.01 \text{ cm}}$
$J = (5 \times 10^{-7}) \times \frac{0.15}{0.01} \text{ mol} \cdot \text{cm}^{-2} \cdot \text{s}^{-1}$
$J = (5 \times 10^{-7}) \times 15 \text{ mol} \cdot \text{cm}^{-2} \cdot \text{s}^{-1}$
$J = 7.5 \times 10^{-6} \text{ mol}/(\text{cm}^2 \cdot \text{s})$
The rate of glucose diffusion across the membrane is $7.5 \times 10^{-6}$ moles per square centimeter per second.