Chapter 4, Problem 4.20P

To determine

The activation enthalpy and pre-exponential constant for the diffusion of oxygen $\left({\text{O}}^{\text{18}}\right)$ into $\text{ZnO}$.

Answer to Problem 4.20P

The activation enthalpy is $3.83\times {10}^5\text{J}/\text{mole}\text{of}\text{O}\text{atoms}$ and the pre exponential constant is $0.40\times {10}^{-3}{\text{m}}^{\text{2}}/\text{s}$.

Explanation of Solution

Formula used:

The diffusion coefficient of point $1$ is given by,

$D_{0{\text{ZnO}}_{\text{1}}}=D_{0{\text{ZnO}}_{\text{0}}}\exp \left(-\frac{\Delta H_{D_{0\text{ZnO}}}}{k{T}_1}\right)$   ....... (I)

Here, $D_{{\text{0Zn0}}_{\text{1}}}$ is the diffusion coefficient of point $1$, $D_{\text{0ZnO}}$ is the pre-exponential constant for oxygen diffusion into $\text{ZnO}$, $k$ is the Boltzmann’s constant, $T_1$ is the temperature at point $1$ and $\Delta H_{{\text{D}}_{\text{0ZnO}}}$ is the activation enthalpy for oxygen into $\text{ZnO}$.

The diffusion coefficient of point $2$ is given by,

$D_{0{\text{ZnO}}_{\text{2}}}=D_{0{\text{ZnO}}_{\text{0}}}\exp \left(-\frac{\Delta H_{D_{0\text{ZnO}}}}{k{T}_2}\right)$   ....... (II)

Here, $D_{{\text{0Zn0}}_2}$ is the diffusion coefficient of point $1$ and $T_2$ is the temperature at point $2.$

The ratio of diffusion coefficient of points $1$ and $2$ is given by,

$\begin{array}{c}\frac{D_{0{\text{ZnO}}_{\text{1}}}}{D_{0{\text{ZnO}}_{\text{2}}}}=\frac{D_{0{\text{ZnO}}_{\text{0}}}\exp \left(-\frac{\Delta H_{D_{0\text{ZnO}}}}{k{T}_1}\right)}{D_{0{\text{ZnO}}_{\text{0}}}\exp \left(-\frac{\Delta H_{D_{0\text{ZnO}}}}{k{T}_2}\right)}\\ =\frac{\exp \left(-\frac{\Delta H_{D_{0\text{ZnO}}}}{k{T}_1}\right)}{\exp \left(-\frac{\Delta H_{D_{0\text{ZnO}}}}{k{T}_2}\right)}\\ =\exp \left[-\left(\frac{\Delta H_{D_{0\text{ZnO}}}}{k}\right)\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\right]\end{array}$   ....... (III)

Calculation:

The activation enthalpy is calculated as,

Substitute $1\times {10}^{-12}{\text{cm}}^{\text{2}}/\text{s}$ for $D_{{\text{0ZnO}}_{\text{1}}}$, $1\times {10}^{-18}{\text{cm}}^{\text{2}}/\text{s}$ for $D_{{\text{0ZnO}}_{\text{2}}}$, $8.62\times {10}^{-5}\text{eV}/\text{atom}\cdot \text{K}$ for $k$, $1587\text{K}$ for $T_1$ and $1075\text{K}$ for $T_2$ in equation (III).

$\begin{array}{c}\frac{1\times {10}^{-12}{\text{cm}}^{\text{2}}/\text{s}}{1\times {10}^{-18}{\text{cm}}^{\text{2}}/\text{s}}=\exp \left[-\left(\frac{\Delta H_{D_{0\text{ZnO}}}}{8.62\times {10}^{-5}\text{eV}/\text{atom}\cdot \text{K}}\right)\left(\frac{1}{1587\text{K}}-\frac{1}{1075\text{K}}\right)\right]\\ {10}^6=\exp \left[-\Delta H_{D_{0\text{ZnO}}}\left(-3.48\text{atom}/\text{eV}\right)\right]\\ \ln \left({10}^6\right)=\Delta H_{D_{0\text{ZnO}}}\left(3.48\text{atom}/\text{eV}\right)\\ \Delta H_{D_{0\text{ZnO}}}=\frac{\ln \left({10}^6\right)}{3.48\text{atom}/\text{eV}}\end{array}$

Solve further,

$\begin{array}{c}\Delta H_{D_{0\text{ZnO}}}=\frac{13.82}{3.48\text{atom}/\text{eV}}\\ =\left(3.97\text{eV}/\text{atom}\times \frac{96.521\times {10}^3\text{J}/\text{mole}\text{of}\text{O}\text{atoms}}{1\text{eV}/\text{atom}}\right)\\ =3.83\times {10}^5\text{J}/\text{mole}\text{of}\text{O}\text{atoms}\end{array}$

The diffusion coefficient is calculated as,

Substitute $1\times {10}^{-12}{\text{cm}}^{\text{2}}/\text{s}$ for $D_{{\text{0ZnO}}_{\text{1}}}$, $8.62\times {10}^{-5}\text{eV}/\text{atom}\cdot \text{K}$ for $k$, $1587\text{K}$ for $T_1$ and $3.97\text{eV}/\text{atom}$ for $\Delta H_{\text{D0ZnO}}$ in equation (I).

$\begin{array}{c}1\times {10}^{-12}{\text{cm}}^{\text{2}}/\text{s}=D_{0{\text{ZnO}}_{\text{0}}}\exp \left(-\frac{3.97\text{eV}/\text{atom}}{\left(8.62\times {10}^{-5}\text{eV}/\text{atom}\cdot \text{K}\right)\left(1587\text{K}\right)}\right)\\ D_{0{\text{ZnO}}_{\text{0}}}=\frac{\left(1\times {10}^{-12}{\text{cm}}^{\text{2}}/\text{s}\times \frac{{10}^{-4}{\text{m}}^{\text{2}}}{1{\text{cm}}^{\text{2}}}\right)}{\exp \left(-29.02\right)}\\ =\frac{1\times {10}^{-16}{\text{m}}^{\text{2}}/\text{s}}{2.32\times {10}^{-13}}\\ =0.40\times {10}^{-3}{\text{m}}^{\text{2}}/\text{s}\end{array}$

Conclusion:

Therefore, the activation enthalpy is $3.83\times {10}^5\text{J}/\text{mole}\text{of}\text{O}\text{atoms}$ and the pre exponential constant is $0.40\times {10}^{-3}{\text{m}}^{\text{2}}/\text{s}$.