Question

1- The preexponential and activation energy for the diffusion of iron in cobalt are 1.1×10-5 m 2 /s and 253,300 J/mol, respectively. At what temperature will the diffusion coefficient have a value of 2.1×10-5 m 2 /s ?

Answer 1

The temperature at which the diffusion coefficient have a value of 2.1×10-5 m 2 /s is  -47078 K.

Using the relation;

logD = logDo - Ea/2.303RT

D = diffusion coefficient

Do =  preexponential

Ea = activation energy

R = gas constant

T = temperature

Substituting values;

log(2.1×10-5)= log (1.1×10-5 ) - 253,300/2.303 × 8.314 × T

log(2.1×10-5) -  log (1.1×10-5 ) =  - 253,300/2.303 × 8.314 × T

log[2.1×10-5/1.1×10-5] = - 253,300/2.303 × 8.314 × T

0.281 × (2.303 × 8.314 × T) = - 253,300

T =  - 253,300/2.303 × 0.281 × 8.314

T = -47078 K

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