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 required to reduce the nitrogen concentration in the cobalt plate, we can use Fick's Second Law of diffusion. This law relates the diffusion rate of a substance to its concentration gradient, diffusion coefficient, and diffusion distance.

The equation for Fick's Second Law is as follows:

∂C/∂t = D * (∂^2C/∂x^2)

Where:
- ∂C/∂t represents the rate of change of concentration with respect to time,
- D is the diffusion coefficient,
- ∂^2C/∂x^2 represents the second derivative of the concentration profile with respect to the distance x.

In this case, we can assume that the diffusion of nitrogen in cobalt follows a one-dimensional model, as the plate is only 1.0 cm thick and we are looking at a depth of 10μm (0.01 cm).

First, let's calculate the diffusion coefficient (D) using the given activation energy (Ea) and preexponential value (Do) using the Arrhenius equation:

D = Do * exp(-Ea/RT)

Where:
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (417 + 273 = 690 K)

Plugging in the values:
D = 0.01 cm^2/sec * exp(-100,000 J/mol / (8.314 J/(mol·K) * 690 K))

D = 0.01 cm^2/sec * 0.2505

D ≈ 0.0025 cm^2/sec

Now, let's calculate the concentration gradient at a depth of 10μm (∆C/∆x) using the initial and final concentrations:

∆C/∆x = (C_final - C_initial) / x

Where:
- C_final is the final concentration (1662 ppm)
- C_initial is the initial concentration (3091 ppm)
- x is the depth (10 μm = 0.01 cm)

∆C/∆x = (1662 ppm - 3091 ppm) / 0.01 cm

∆C/∆x = -1429 ppm/cm

Next, we'll substitute the values into Fick's Second Law:

∂C/∂t = D * (∂^2C/∂x^2)

∂C/∂t = 0.0025 cm^2/sec * (∂^2C/∂x^2)

Since the problem states the concentration is fixed at the surface to 111 ppm, we can assume the concentration gradient is zero at x = 0. This allows us to simplify the equation:

∂C/∂t = 0.0025 cm^2/sec * (∂^2C/∂x^2)

∂C/∂t = 0.0025 cm^2/sec * 0

∂C/∂t = 0

This implies that there is no change in concentration with respect to time (∂C/∂t = 0), meaning the concentration at the depth of 10μm (∂C/∂x) will remain constant at 1662 ppm.

Therefore, it will take an infinite amount of time to reduce the nitrogen concentration to 1662 ppm at a depth of 10μm.

In conclusion, the time required to reduce the nitrogen concentration to 1662 ppm at a depth of 10μm is infinite.