The activation energy for the diffusion of carbon in coper is

Question

The activation energy for the diffusion of carbon in copper is 200,000 J/mol. Calculate the diffusion coefficient in Å²/s at 300°C, given that the diffusion coefficient at 1000°C is 5.1x10^-11 m²/s.

1 Å = 10^-10 m

Your final answer should be in the format *.**

Answer 1

I believe this can be approached using the Arrhenius Equation:

ln (D2/D1) = -Ea/R (1/T2 - 1/T1)

D1 = 5.1x10^-11 m²/s

D2 = ?

T1 = 1000C + 273 = 1273K

T2 = 300C + 273 = 573K

Ea = 200,000 J/mol

R = 8.314 J/Kmol

Plugging in the values, and solving for D2 (in m²/s):

ln(D2/x5.1x10^-11) = -(200,000/8.314)(1/573 - 1/1273)

ln(D2/x5.1x10^-11) = -24056 (1.75x10^-3 - 7.86x10^-4)

ln(D2/x5.1x10^-11) = -24056 * 9.64x10^-4

ln(D2/x5.1x10^-11) = -23.19

D2 / 5.1x10^-11 = 8.49x10^-11

D2 = 4.3x10^-21 m²/s

Since 1Aº = 10^-10 m, then 1A² = 1x10^-20 m²

4.3x10^-21 m²/s x 1A² / 1x10^-20 m² = 0.43 A² / s

(be sure to check all of the math)