The diffusion coefficient at the temperature of 1200 K is [tex]\mathbf{ 4.68 \times 10^{-13} \ m^2/s}[/tex]
The diffusion coefficient can be calculated by using the formula;
[tex]\mathbf{D_o = \dfrac{D}{exp(\dfrac{-Q}{R\times T})}}[/tex]
From the given information;
- the activation energy Q = 193,000 J/mol
- diffusion coefficient D at temp 1000 k = 1.0 × 10⁻¹⁴ m²/s
∴
[tex]\mathbf{D_o = \dfrac{1 \times 10 ^{-14} \ m^2/s}{exp(\dfrac{-193000 \ J/mol}{8.314 \ J/mo.K\times 1000 \ K})}}[/tex]
[tex]\mathbf{D_o = \dfrac{1 \times 10 ^{-14} \ m^2/s}{8.28 \times 10^{-11}}}[/tex]
[tex]\mathbf{D_o = 1.2 \times 10^{-4} \ m^2/s}[/tex]
Thus, the diffusion constant [tex]\mathbf{D_o = 1.2 \times 10^{-4} \ m^2/s}[/tex]
Now, since the diffusion constant is known, we can use the same above formula to determine the diffusion coefficient at 1200 K (927C).
∴
[tex]\mathbf{D_o = \dfrac{D}{exp(\dfrac{-Q}{R\times T})}}[/tex]
[tex]\mathbf{D = {D_o} \times {exp(\dfrac{-Q}{R\times T})}}[/tex]
[tex]\mathbf{D =1.2 \times 10^{-4} \ m^2/s \times {exp(\dfrac{-193000 \ J/mol}{8.314 \ J.mol/K \times 1200 \ K})}}[/tex]
[tex]\mathbf{D =1.2 \times 10^{-4} \ m^2/s \times 3.9 \times 10^{-9}}}[/tex]
[tex]\mathbf{D = 4.68 \times 10^{-13} \ m^2/s}[/tex]
Therefore, we can conclude that the diffusion coefficient at the temperature of 1200 K is [tex]\mathbf{ 4.68 \times 10^{-13} \ m^2/s}[/tex]
Learn more about activation energy here:
https://brainly.com/question/12558986?referrer=searchResults
