Question 2 of 7: Activation Energy for Interstitial Diffusion from Two Temperatures

Question 2: Activation Energy for Interstitial Diffusion from Two Temperatures (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $T_1=500^\circ$C$=773$ K, $f_1=10^{-10}$; $T_2=600^\circ$C$=873$ K, $f_2=10^{-9}$; $T_3=700^\circ$C$=973$ K; gas constant $R=N_Ak=8.31$ J/(mol·K) (from the page-1 Boltzmann constant); $1\,\text{eV}=1.6\times10^{-19}$ J, $1\,\text{cal}=4.18$ J (page-1).

Find. (a) Activation energy $Q$, in eV/atom and cal/mol. (b) Fraction of atoms with enough energy at $700^\circ$C.

Approach

The fraction of atoms with enough thermal energy to make the jump follows an Arrhenius-type Boltzmann relation, $f=A\exp(-Q/RT)$. Taking the ratio of $f$ at the two known temperatures eliminates the unknown pre-exponential constant $A$ and leaves one equation in the one unknown, $Q$; once $Q$ is known, the same relation (referenced back to either known point) predicts $f$ at any other temperature.

1. Eliminate $A$ by taking the ratio at the two known temperatures. $$\frac{f_2}{f_1}=\exp\!\left[-\frac{Q}{R}\left(\frac1{T_2}-\frac1{T_1}\right)\right] \ \Rightarrow\ \ln\frac{f_2}{f_1}=\frac{Q}{R}\left(\frac1{T_1}-\frac1{T_2}\right)$$
2. Solve for $Q$. $$\frac1{T_1}-\frac1{T_2}=\frac1{773}-\frac1{873}=1.482\times10^{-4}\ \text{K}^{-1},\qquad \ln\frac{f_2}{f_1}=\ln(10)=2.3026$$ $$Q=\frac{R\ln(f_2/f_1)}{1/T_1-1/T_2}=\frac{(8.31)(2.3026)}{1.482\times10^{-4}}$$ $$\boxed{Q\approx1.29\times10^{5}\ \text{J/mol}=129.1\ \text{kJ/mol}}$$
3. Convert to eV/atom and cal/mol. $$Q_{eV}=\frac{Q}{N_A\,(1.6\times10^{-19}\,\text{J/eV})} =\frac{129{,}100}{(6.02\times10^{23})(1.6\times10^{-19})}$$ $$\boxed{Q\approx1.34\ \text{eV/atom}}$$ $$Q_{cal}=\frac{Q}{4.18}=\frac{129{,}100}{4.18}$$ $$\boxed{Q\approx3.09\times10^4\ \text{cal/mol}\ (30.9\ \text{kcal/mol})}$$
4. (b) Fraction with enough energy at $700^\circ$C. Reference back to the $T_1=773$ K point using the same Arrhenius form: $$f_3=f_1\exp\!\left[\frac{Q}{R}\left(\frac1{T_1}-\frac1{T_3}\right)\right],\qquad \frac1{773}-\frac1{973}=2.659\times10^{-4}\ \text{K}^{-1}$$ $$f_3=10^{-10}\exp\!\left[\frac{129{,}100}{8.31}(2.659\times10^{-4})\right] =10^{-10}\exp(4.131)=10^{-10}(62.2)$$ $$\boxed{f_3\approx6.2\times10^{-9}\ \ (\text{about 1 atom in }1.6\times10^8)}$$ As a sanity check, $f_3>f_2>f_1$, exactly as expected since $700^\circ$C is above both reference temperatures and the fraction rises steeply (exponentially) with $T$.