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 calculate the time required to reduce the nitrogen concentration in the cobalt plate, we can use Fick's second law of diffusion, which describes the diffusion of species in a medium:

∂C/∂t = D * (∂^2C/∂x^2)

where:
C = concentration of nitrogen in cobalt (ppm)
t = time (seconds)
D = diffusion coefficient (cm^2/sec)
x = distance from the surface of the cobalt plate (cm)

In this case, we want to find the time it takes for the nitrogen concentration to decrease from 3091 ppm to 1662 ppm at a depth of 10μm (0.01 cm).

First, we need to find the diffusion coefficient D using the diffusion equation:

D = Do * e^(-Ea/RT)

where:
Do = preexponential value for diffusion (cm^2/sec)
Ea = activation energy for diffusion (kJ/mol)
R = gas constant (8.314 J/(mol·K))
T = temperature (Kelvin)

Let's convert the given values to the appropriate units:

Do = 0.01 cm^2/sec
Ea = 100 kJ/mol = 100,000 J/mol
R = 8.314 J/(mol·K)
T = 417 °C + 273.15 = 690.15 K

Now, substitute these values into the equation to find D:

D = 0.01 * e^(-100,000/(8.314 * 690.15))

Next, we can use this diffusion coefficient in Fick's second law of diffusion to solve for the time required for the concentration to decrease from 3091 ppm to 1662 ppm at a depth of 10 μm:

∂C/∂t = D * (∂^2C/∂x^2)

By rearranging the equation and integrating, we can solve for t:

∫[3091-1662]/∫t = ∫[0.01 * (∂^2C/∂x^2)]/∫D

Integrating gives:

[(1662-3091)/t] = [(0.01 * ∂C/∂x)]/D

Simplifying further:

[(1662-3091)/t] = (0.01/D) * ∂C/∂x

Finally, since we are interested in the time required to reduce the concentration at a depth of 10 μm, we can substitute ∂C/∂x = (1662-3091)/(0.01 cm) into the equation:

[(1662-3091)/t] = (0.01/D) * (1662-3091)/(0.01 cm)

t = [(0.01 cm) / (0.01/D)] * [(1662-3091) / (1662-3091)]
t = 1/D

Now, substitute the value of D into the equation to determine the time required:

t = 1 / (0.01 * e^(-100,000/(8.314 * 690.15)))

Calculate this expression to find the time required in seconds.