You wish to remove nitrogen from the near-surface region of a plate of cobalt which is 1.0 cm thick. The plate is placed in a furnace at 417∘ C where an atmosphere of H2 and NH3 reacts with the nitrogen in the cobalt and fixes the surface concentration to 111 ppm (parts per million by mass). If the initial concentration is 3091ppm, how long will it take to reduce the nitrogen concentration to 1662 ppm at a depth of 10μm? The diffusion of nitrogen in cobalt has an activation energy of 100 kJ/mol and a preexponential value (Do) of 0.01 cm2/sec.
Give your answer in units of seconds.
To calculate the time it will take to reduce the nitrogen concentration in the plate of cobalt, we can use Fick's second law of diffusion. This law relates the rate of diffusion of a substance to its concentration gradient and diffusion coefficients.
The equation for Fick's second law of diffusion is:
∂C/∂t = D * (∂²C/∂x²)
Where ∂C/∂t is the rate of change of the concentration with respect to time, D is the diffusion coefficient, and (∂²C/∂x²) is the second derivative of the concentration with respect to distance.
We are given that the diffusion coefficient for nitrogen in cobalt is 0.01 cm2/sec.
To simplify the equation, we can assume that the diffusion is one-dimensional since we are only interested in the depth of 10μm from the surface. This means that the concentration gradient ∂²C/∂x² can be represented as (∂C/∂x)², where (∂C/∂x) represents the concentration gradient.
We can rewrite Fick's second law as:
∂C/∂t = D * (∂C/∂x)²
To solve this differential equation, we need to assume an initial condition. In this case, the initial condition is that the concentration at the surface is fixed at 111 ppm.
We can rewrite the equation as:
∂C/∂t = D * (∂C/∂x)²
∂C/∂t = D * (111 ppm - C)/(0.01 cm)
Where C represents the concentration at a depth of 10μm.
To solve for time, we can rearrange the equation as:
∂t = (∂C/∂C) * (0.01 cm)/(111 ppm - C)
Integrating both sides:
∫ ∂t = ∫ (0.01 cm)/(111 ppm - C) dC
The integration limits are from the initial concentration of 3091 ppm to the final concentration of 1662 ppm.
∫ ∂t = ∫ (0.01 cm)/(111 ppm - C) dC
∫ ∂t = ∫ (0.01 cm)/(111 - C) dC
∫ ∂t = -ln|111 - C|
Now, we can integrate again to solve for time:
t = -∫ ln|111 - C| dC
Evaluating the integral from the initial concentration to the final concentration:
t = -∫ ln|111 - C| dC (from C=3091 ppm to C=1662 ppm)
To evaluate the integral, we can use a symbolic calculator or software. The result will be the time it takes to reduce the nitrogen concentration to 1662 ppm at a depth of 10μm in units of seconds.