Question

The pre-exponential constant and activation energy for the diffusion of iron in cobalt are 1.7 x 10-5 m2/s and 273,300 J/mol, respectively. At what temperature will the diffusion coefficient have a value of 2.7 x 10-14 m2/s

Answer 1

Answer: The temperature of the system will be 1622 K

Explanation:

The equation relating the pre-exponential factor and activation energy follows:

$\log D=\log D_o-\frac{E_a}{2.303RT}$

where,

D = diffusion coefficient = $2.7\times 10^{-14}m^2/s$

$D_o$ = pre-exponential constant = $1.7\times 10^{-5}m^2/s$

$E_a$ = activation energy of iron in cobalt = 273,300 J/mol

R = Gas constant = 8.314 J/mol.K

T = temperature = ?

Putting values in above equation, we get:

$\log (2.7\times 10^{-14})=\log (1.7\times 10^{-5})-\frac{273,300}{2.303\times 8.314\times T}\\\\T=1622K$

Hence, the temperature of the system will be 1622 K