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.

Question

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.

Answer 1

To solve this problem, we can use Fick's second law of diffusion. This law states that the rate at which a substance diffuses through a material is proportional to the concentration gradient and the diffusion coefficient. The equation for Fick's second law is:

dC/dt = (D * d^2C/dx^2)

Where:
- dC/dt is the rate of change of concentration with respect to time,
- D is the diffusion coefficient,
- d^2C/dx^2 is the second derivative of concentration with respect to distance.

We can rearrange this equation to solve for the time it takes to achieve a certain concentration at a given depth:

dt = (dx^2 / D) * dC / d^2C

Given:
- D (diffusion coefficient) = 0.01 cm^2/sec
- C0 (initial concentration) = 3091 ppm
- Cs (surface concentration) = 111 ppm
- Cf (final concentration) = 1662 ppm
- d (thickness of the plate) = 1.0 cm
- dx (depth at which we want to measure the concentration) = 10 μm = 10^-3 cm

First, let's calculate the concentration gradient:

ΔC = Cf - Cs
ΔC = 1662 ppm - 111 ppm
ΔC = 1551 ppm

Next, let's calculate the second derivative of concentration:

d^2C = (Cf - C0) / d^2
d^2C = (1662 ppm - 3091 ppm) / (10^-3 cm^2)
d^2C = - 1429 ppm/cm^2

Now, we can plug these values into the equation to find the time:

dt = (dx^2 / D) * dC / d^2C
dt = ((10^-3 cm)^2 / 0.01 cm^2/sec) * (1551 ppm) / (- 1429 ppm/cm^2)
dt = (10^-6 cm^2) / 0.01 cm^2/sec * (1551/1429) sec
dt = 1.551 seconds

Therefore, it will take approximately 1.551 seconds to reduce the nitrogen concentration to 1662 ppm at a depth of 10 μm in the cobalt plate.