Respuesta :
Answer : The Activation energy, Ea = 201.820 KJ/mol
The Pre-exponential constant, [tex]D_{0}[/tex] = 1.7658 × [tex]10^{-9}[/tex] [tex]m^{2}/s[/tex]
Solution : Given,
Diffusivity of Ni at [tex]1200^{0}C[/tex] , [tex]D_{1}[/tex] = 1.23 × [tex]10^{-16}[/tex] [tex]m^{2}/s[/tex]
Diffusivity of Ni at [tex]1800^{0}C[/tex] , [tex]D_{1}[/tex] = 1.45 × [tex]10^{-14}[/tex] [tex]m^{2}/s[/tex]
Temperature, [tex]T_{1}[/tex] = [tex]1200^{0}C[/tex] = 1200 + 273 = 1473 K
Temperature, [tex]T_{2}[/tex] = [tex]1800^{0}C[/tex] = 1800 + 273 = 2073 K
Value of R = 8.314 J/mol/K
Formula used :
[tex]D=D_{0}\times\text{exp}\left (\frac{-Ea}{RT} \right )[/tex] ..........(1)
This formula convert into logarithm term for the calculation of activation energy. So the formula is,
[tex]ln\frac{D_{1}}{D_{2}}=\frac{Ea}{R}\left [\frac{1}{T_{2}}-\frac{1}{T_{1}} \right ][/tex] ........(2)
Now put all the values in above formula (2), we get
[tex]ln\frac{1.23\times 10^{-16}}{1.45\times 10^{-14}}=\frac{Ea}{8.314}\left [\frac{1}{2073}-\frac{1}{1473} \right ][/tex]
Rearranging the terms, we get the value of Activation energy, (Ea) as
Ea = 201.820 KJ/mol
Now, we have to calculate the value of Pre-exponential constant,[tex]D_{0}[/tex] bye using formula (1)
[tex]D=D_{0}\times\text{exp}\left (\frac{-Ea}{RT} \right )[/tex]
now put all the values in formula,we get
[tex]1.23\times10^{-16}m^{2}/s =D_{0}\times\text{exp}\left (\frac{-201820J/mol} {8.314J/mol/K\times1473K} \right )[/tex]
Rearranging the terms, we get the value of Pre-exponential constant, [tex]D_{0}[/tex] as
[tex]D_{0}[/tex] = 1.7658 × [tex]10^{-9} m^{2}/s[/tex]
