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 solve this problem, we can use Fick's second law of diffusion, which relates the concentration of a diffusing species with time and position. The equation is as follows:
∂C/∂t = D * (∂²C/∂x²)
Where:
- ∂C/∂t is the rate of change of concentration with time,
- D is the diffusion coefficient, given by D = Do * exp(-Q/RT),
- ∂²C/∂x² is the second derivative of concentration with respect to position,
- Do is the preexponential value of the diffusion coefficient,
- Q is the activation energy,
- R is the universal gas constant, and
- T is the temperature in Kelvin.
First, let's calculate the diffusion coefficient at 417°C (690 K):
D = 0.01 cm²/sec * exp(-(100,000 J/mol) / (8.314 J/(mol·K) * 690 K)) = 0.01 cm²/sec * exp(-16.325) ≈ 0.0023 cm²/sec
Next, we'll find the rate of change of concentration (∂C/∂t) at a depth of 10 μm:
∂C/∂t = D * (∂²C/∂x²)
Since the concentration is changing with time and depth, we'll assume a 1-dimensional model of diffusion and equate (∂²C/∂x²) to (∂²C/∂z²), where z is the depth.
∂C/∂t = D * (∂²C/∂z²)
Now, we can use the given initial concentration of 3091 ppm and the desired final concentration of 1662 ppm to find the time it takes to achieve this at a depth of 10 μm.
Let's solve the diffusion equation using separable variables:
∂C/∂t = D * (∂²C/∂z²)
Separating variables:
∫ dC / C = D ∫ (∂²C/∂z²) dt
Integrating both sides:
ln(C) = D * (∂C/∂z) * t + ln(k),
where k is a constant of integration.
Now, we'll assume steady-state diffusion (∂C/∂t = 0) when the surface concentration is fixed at 111 ppm. At z = 0 (surface), C = 111 ppm, and at z = 10 μm, C = 1662 ppm.
Plugging in the given values:
ln(1662 ppm) = D * (∂C/∂z) * t + ln(k)
ln(111 ppm) = 0 + ln(k)
Subtracting the second equation from the first equation:
ln(1662 ppm / 111 ppm) = D * (∂C/∂z) * t
Simplifying the left side:
ln(14.972) = D * (∂C/∂z) * t
Taking the exponential of both sides:
14.972 = exp(D * (∂C/∂z) * t)
Now, we can calculate the time it takes to reach a concentration of 1662 ppm from the initial concentration of 3091 ppm at a depth of 10 μm:
t = ln(14.972) / (D * (∂C/∂z))
For a 1 cm thick plate, the depth is 10 μm = 0.001 cm.
Plugging in the known values:
t = ln(14.972) / (0.0023 cm²/sec * (∂C/∂z))
Now, we need to find (∂C/∂z), which represents the concentration gradient with respect to depth.
According to Fick's first law, the flux (J) can be defined as:
J = -D * (∂C/∂z)
By rearranging this equation, we find:
(∂C/∂z) = -J / D
The flux (J) can be calculated using the following equation:
J = (C2 - C1) / Δz,
where C1 and C2 are the concentrations at neighbouring depths and Δz is the distance between them.
Using the given values:
J = (1662 ppm - 3091 ppm) / (0.001 cm - 0 cm)
J = -1429 ppm / 0.001 cm
J = -1.429 x 10^6 ppm/cm
Now, we can calculate (∂C/∂z):
(∂C/∂z) = -J / D
(∂C/∂z) = -(-1.429 x 10^6 ppm/cm) / 0.0023 cm²/sec
(∂C/∂z) = 6.208 x 10^8 ppm/(cm²·sec)
Finally, we can substitute all the known values into the equation for time:
t = ln(14.972) / (0.0023 cm²/sec * 6.208 x 10^8 ppm/(cm²·sec))
Evaluating this expression gives us:
t ≈ ln(14.972) / (0.0143 cm²/sec)
t ≈ 0.2087 seconds
Therefore, it will take approximately 0.2087 seconds to reduce the nitrogen concentration to 1662 ppm at a depth of 10 μm in the cobalt plate.