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Determine the activation energy for diffusing copper into gold if you know that the diffusion coefficient is 3.98 x 10-13 m2/s at 980 degreeC and the diffusion coefficient is 3.55 x 10-16 m2/s at 650degreeC.

Respuesta :

Answer:

[tex]Q_d = 0.0166 J/mol[/tex]

Explanation:

given data:

diffusion coefficient [tex]= 3.98 \times 10^{-13} m^2/s[/tex]

[tex]T_1 = 980 Degree\ celcius = 1253 K[/tex]

[tex]T_2 = 650 Degree\ celcius = 923 K[/tex]

[tex]D_2 = 3.55\times 10^{-16} m^2/s[/tex]

Activation energy is given as[tex] = -2.3R \frac{ \Delta log D}{\Delta\frac{1}{T}}[/tex]

[tex] Q_d = -2.3 R [\frac{logD_1 - logD_2}{ \frac{1}{T_1} - \frac{1}{T_2}}][/tex]

[tex]Q_d = -2.3 \times 8.31  \frac{log(3.98*10^{-13} - log(3.55*10^{-16}}{ \frac{1}{1253} - \frac{1}{923}}[/tex]

[tex]Q_d = 0.0166 J/mol[/tex]

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