Respuesta :
Answer:
(a). The value of the activation energy is 171565.80 J/mol.
(b). The value of D₀ is [tex]1.652\times10^{10^{-8}}\ m^2/s[/tex]
(c). The magnitude of D at 1160°C is [tex]5.067\times10^{-8}\ m^2/s[/tex]
Explanation:
Given that,
Temperatures,
[tex]T_{1}=1030^{\circ}C[/tex]
[tex]T_{2}=1250^{\circ}C[/tex]
Diffusion coefficients are
[tex]D_{1}=6.11\times10^{-17}[/tex]
[tex]D_{2}=6.02\times10^{-16}[/tex]
(a). We need to calculate the value of the activation energy
Using relation between diffusion coefficient and temper
[tex]ln(D_{1})=ln(D_{0})-(\dfrac{Q_{d}}{RT_{1}})[/tex]....(I)
[tex]ln(D_{2})=ln(D_{0})-(\dfrac{Q_{d}}{RT_{2}})[/tex].....(II)
Divided equation (II) by (I)
[tex]ln(\dfrac{D_{2}}{D_{1}})=\dfrac{Q_{d}}{R}\times(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}})[/tex]
Put the value into the formula
[tex]ln(\dfrac{6.02\times10^{-16}}{6.11\times10^{-17}})=\dfrac{Q_{d}}{8.314}\times(\dfrac{1}{1030+273}-\dfrac{1}{1250+273})[/tex]
[tex]Q_{d}=\dfrac{2.2877\times8.314\times1984469}{220}[/tex]
[tex]Q_{d}=171565.80\ J/mol[/tex]
(b). We need to calculate the value of D₀
Using relation between diffusion coefficient and temperature
[tex]ln(D_{0})=ln(D)+(\dfrac{Q_{d}}{RT_{1}})[/tex]
Put the value of in to the formula
[tex]ln(D_{0})=ln(6.11\times10^{-17})+(\dfrac{171565.80}{8.314\times1030+273})[/tex]
[tex]ln(D_{0})=-17.9182[/tex]
[tex]D_{0}=e^{-17.9182}[/tex]
[tex]D_{0}=1.652\times10^{10^{-8}}\ m^2/s[/tex]
Now again put the value in the equation (II)
[tex]ln(D)=ln(1.652\times10^{10^{-8}})-(\dfrac{171565.80}{8.314\times1160+273})[/tex]
[tex]ln(D_{0})=-16.79776[/tex]
[tex]D_{0}=e^{-16.79776}[/tex]
[tex]D_{0}=5.067\times10^{-8}\ m^2/s[/tex]
Hence, (a). The value of the activation energy is 171565.80 J/mol.
(b). The value of D₀ is [tex]1.652\times10^{10^{-8}}\ m^2/s[/tex]
(c). The magnitude of D at 1160°C is [tex]5.067\times10^{-8}\ m^2/s[/tex]
