The diffusion coefficients for nickel in iron are given at two temperatures: $$\begin{array}{cc} \boldsymbol{T}(\boldsymbol{K}) & \boldsymbol{D}\left(\boldsymbol{m}^{2} / \boldsymbol{s}\right) \\ \hline 1473 & 2.2 \times 10^{-15} \\ 1673 & 4.8 \times 10^{-14} \end{array}$$ (a) Determine the values of \(D_{0}\) and the activation energy \(Q_{d}\) (b) What is the magnitude of \(D\) at \(1300^{\circ} \mathrm{C}\) \((1573 \mathrm{K}) ?\)

The Arrhenius equation for diffusivity is given by: $$D = D_{0} \cdot e^{-\frac{Q_d}{RT}}$$ where \(D\) is the diffusivity, \(D_0\) is the pre-exponential factor, \(Q_d\) is the activation energy for diffusion, \(R\) is the gas constant (\(8.314\,\text{J}\cdot \text{mol}^{-1}\cdot \text{K}^{-1}\)), and \(T\) is the temperature in Kelvin.

We have the diffusion coefficients for two different temperatures from the table: $$ \begin{cases} D_1 = D_{0} \cdot e^{-\frac{Q_d}{R T_1}} \\ D_2 = D_{0} \cdot e^{-\frac{Q_d}{R T_2}} \end{cases} $$ Plugging in the given values from the table, we get: $$ \begin{cases} 2.2 \times 10^{-15} = D_{0} \cdot e^{-\frac{Q_d}{8.314 \cdot 1473}} \\ 4.8 \times 10^{-14} = D_{0} \cdot e^{-\frac{Q_d}{8.314 \cdot 1673}} \end{cases} $$

Divide the second equation by the first equation to eliminate the \(D_0\): $$\frac{4.8 \times 10^{-14}}{2.2 \times 10^{-15}} = \frac{e^{-\frac{Q_d}{8.314 \cdot 1673}}}{e^{-\frac{Q_d}{8.314 \cdot 1473}}}$$ This simplifies to: $$21.818 = e^{\frac{Q_d}{8.314}(\frac{1}{1473} - \frac{1}{1673})}$$ Now take the natural logarithm: $$\ln(21.818) = \frac{Q_d}{8.314}(\frac{1}{1473} - \frac{1}{1673})$$ Which leads us to the value of \(Q_d\): $$Q_d = 8.314 \cdot \ln(21.818) \cdot \frac{1473 \cdot 1673}{1673 - 1473} \approx 183627\, \text{J} \cdot \text{mol}^{-1}$$ Now plug the value of \(Q_d\) back into one of our initial equations to find \(D_0\): $$2.2 \times 10^{-15} = D_{0}\cdot e^{-\frac{183627}{8.314 \cdot 1473}}$$ Solving for \(D_0\) gives: $$D_0 \approx 4.04 \times 10^{-8}\, \text{m}^{2}\cdot \text{s}^{-1}$$

Now that we have the values of \(D_0\) and \(Q_d\), we can calculate \(D\) at \(1300^{\circ}\,\text{C}\) (\(1573\,\text{K}\)) using the Arrhenius equation: $$D = 4.04 \times 10^{-8} \cdot e^{-\frac{183627}{8.314 \cdot 1573}} \approx 6.63 \times 10^{-15}\, \text{m}^{2}\cdot \text{s}^{-1}$$ Therefore, the diffusion coefficient \(D\) at \(1300^{\circ}\,\text{C}\) \((1573\,\text{K})\) is approximately \(6.63 \times 10^{-15}\, \text{m}^{2}\cdot \text{s}^{-1}\).