Question 4 of 8: Diffusion Activation Energy from Two Temperatures

Question 4: Diffusion Activation Energy 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. Boltzmann-type fraction $n/N=Me^{-E/kT}$; $T_1=500^\circ$C$=773$ K, $f_1=1\times10^{-10}$; $T_2=600^\circ$C$=873$ K, $f_2=1\times10^{-9}$; $k=1.38\times10^{-23}$ J/atom·K (the paper's $13.8\times10^{-24}$ is the identical value written with a different exponent split); $N_A=6.02\times10^{23}$ mol$^{-1}$; 1 cal $=4.18$ J.

Find. (a) Activation energy $E$ in eV/atom and cal/mol. (b) Fraction of atoms with enough energy at $T_3=700^\circ$C$=973$ K.

Approach

Taking the ratio of the Boltzmann-type expression at two temperatures eliminates the unknown pre-exponential constant $M$, leaving one equation in the one unknown $E$. The same relation, now with $E$ known, is then evaluated at the third temperature.

1. Eliminate $M$. Dividing $f_2=Me^{-E/kT_2}$ by $f_1=Me^{-E/kT_1}$, $$\frac{f_2}{f_1}=\exp\!\left[-\frac{E}{k}\left(\frac1{T_2}-\frac1{T_1}\right)\right] \ \Rightarrow\ E=\frac{k\,\ln(f_2/f_1)}{1/T_1-1/T_2}.$$
2. Evaluate $E$. With $f_2/f_1=10$, $1/T_1-1/T_2=1/773-1/873=1.4819\times10^{-4}\ \text{K}^{-1}$: $$E=\frac{(1.38\times10^{-23})(\ln10)}{1.4819\times10^{-4}}=2.144\times10^{-19}\ \text{J/atom}.$$ Converting: $E=2.144\times10^{-19}/1.6\times10^{-19}=\boxed{1.34\ \text{eV/atom}}$; per mole, $E\,N_A=1.291\times10^5\ \text{J/mol}=1.291\times10^5/4.18=\boxed{30{,}880\ \text{cal/mol}}$.
3. (b) Fraction at 700°C. Using $T_1$ as the reference, $$f_3=f_1\exp\!\left[-\frac{E}{k}\left(\frac1{T_3}-\frac1{T_1}\right)\right] =10^{-10}\exp\!\left[15{,}538\left(\frac1{773}-\frac1{973}\right)\right]=10^{-10}\times e^{4.132} =\boxed{6.23\times10^{-9}}.$$ (About 1 atom in every $1.6\times10^8$.)

The size of that swing is the physical point of the question. A 200 K rise, from 500 °C to 700 °C, multiplies the population of sufficiently energetic atoms by a factor of about 62, even though the absolute temperature rises by only 26 %. That is the signature of an exponential Boltzmann factor: what matters is the dimensionless ratio $E/kT$, and at $E=1.34$ eV the barrier is 16–20 times $kT$ across this temperature range, so a modest change in $T$ moves the exponent by several units. The same arithmetic underlies the strong temperature dependence of every diffusion-controlled process in this subject — carburizing, homogenizing, creep and precipitate coarsening all accelerate the same way, which is why diffusion data are always quoted as an Arrhenius pair $(D_0,\ Q_d)$ rather than as a single rate.