(a) Certain alloys such as LaNi5 can store hydrogen at room temperature. A plate of LaNi5 containing no hydrogen is placed in a chamber filled with pure hydrogen and maintained at a constant pressure. At what depth from the surface will the concentration of hydrogen be 1/2 the surface concentration after 1 hour? Assume the diffusivity of hydrogen in the alloy is 3.091×10−6 cm2/s. Use the approximation erf(x)≅x for 0<x<0.6 if appropriate.
Give your answer in unit centimeters with up to three decimal spaces
(b) The activation energy for H diffusion in LaNi5 is 0.25 eV. How far will the H diffuse (where will it reach 1/2 the surface concentration) if the experiment in part a is repeated at 35∘ C?
Give your answer in unit centimeters with up to three decimal spaces
To solve this problem, we will use Fick's second law of diffusion, which relates the concentration of a diffusing species to its diffusivity and the rate of diffusion.
(a) To find the depth from the surface where the concentration of hydrogen is 1/2 the surface concentration after 1 hour, we can use the following formula:
x = sqrt(2 * D * t * ln(c0 / c1))
where:
- x is the depth from the surface
- D is the diffusivity of hydrogen in the alloy (given as 3.091×10−6 cm2/s)
- t is the time in seconds (1 hour = 3600 seconds)
- c0 is the initial concentration at the surface (the surface concentration)
- c1 is the target concentration (1/2 the surface concentration)
Now let's plug in the values:
x = sqrt(2 * (3.091×10−6 cm2/s) * (3600 seconds) * ln(1/c1))
To find c1, we need to first determine the surface concentration. Since the plate of LaNi5 contains no hydrogen initially, the concentration at the surface is zero.
So, c0 = 0
Plugging in c0 = 0, we have:
x = sqrt(2 * (3.091×10−6 cm2/s) * (3600 seconds) * ln(1/2))
To find ln(1/2), we can use the approximation erf(x) ≅ x for 0 < x < 0.6. Since ln(x) = -ln(1/x), we can calculate ln(1/2) as -ln(2).
x = sqrt(2 * (3.091×10−6 cm2/s) * (3600 seconds) * (-ln(2)))
Evaluating the expression, we get:
x ≈ 0.006 cm
Hence, the concentration of hydrogen will be 1/2 the surface concentration at a depth of approximately 0.006 cm after 1 hour.
(b) To determine how far the hydrogen will diffuse at 35∘ C, we need to take into account the temperature dependency of the diffusivity. The Arrhenius equation can be used to relate the diffusivity at different temperatures:
D2 / D1 = exp(-Q / (R * T2) + Q / (R * T1))
where:
- D1 is the diffusivity at the initial temperature T1
- D2 is the diffusivity at the final temperature T2
- Q is the activation energy for H diffusion in LaNi5 (given as 0.25 eV)
- R is the ideal gas constant (8.314 J/(mol·K))
- T1 and T2 are the initial and final temperatures (in Kelvin)
We want to find the diffusivity at 35∘ C (308.15 K). Since we're given the diffusivity at room temperature, we can rewrite the equation as:
D2 = D1 * exp(-Q / (R * T2) + Q / (R * T1))
Plugging in the values:
D2 = (3.091×10−6 cm2/s) * exp(-0.25 eV / (8.314 J/(mol·K) * (308.15 K)) + 0.25 eV / (8.314 J/(mol·K) * (298.15 K))
Converting eV to J:
D2 ≈ (3.091×10−6 cm2/s) * exp(-0.25 * 1.602×10−19 J / (8.314 J/(mol·K) * (308.15 K)) + 0.25 * 1.602×10−19 J / (8.314 J/(mol·K) * (298.15 K))
Evaluating the expression using the given values and calculating D2, we can use the formula from part (a) to find the depth at 35∘ C where the concentration will be 1/2 the surface concentration.