Question

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.

Answer 1

`Q_d = 0.0166 J/mol`

Step-by-step explanation:

given data:

diffusion coefficient
`= 3.98 * 10^(-13) m^2/s`

`T_1 = 980 Degree\ celcius = 1253 K`

`T_2 = 650 Degree\ celcius = 923 K`

`D_2 = 3.55* 10^(-16) m^2/s`

Activation energy is given as
`= -2.3R ( \Delta log D)/(\Delta(1)/(T))`

`Q_d = -2.3 R [(logD_1 - logD_2)/( (1)/(T_1) - (1)/(T_2))]`

`Q_d = -2.3 * 8.31  (log(3.98*10^(-13) - log(3.55*10^(-16))/( (1)/(1253) - (1)/(923))`

`Q_d = 0.0166 J/mol`