At room temperature, $298~\mathrm{K}$, the diffusivity of carbon in iron is $9.06\cdot 10^{-26}\frac{m^2}{s}$.
At $573~\mathrm{K}$ the diffusivity is $1.66\cdot 10^{-15}\frac{m^2}{s}$. What is the activation energy, $Q$?

I am wondering if my solution is correct:

$D=D_0\exp\left\{\frac{-Q}{RT}\right\}$ and $D_0$ is constant (is this correct), thus

$\displaystyle \frac{D_{298}}{\exp\left\{-Q/(RT_{298})\right\}}=\frac{D_{573}}{\exp\left\{-Q/(RT_{573})\right\}} \implies Q\approx 122~\mathrm{\frac{kJ}{mol}}$

I would be very greatful if someone could check my math here as well, if the logic is correct, since this was a test problem.


1 Answer 1


Maybe I'm wrong but my calculation has given another result: \begin{align} D_{298} &= D_0\cdot\exp\left\{\frac{-Q}{298~\mathrm{K}\cdot\mathcal{R}}\right\}\\ D_{573} &= D_0\cdot\exp\left\{\frac{-Q}{573~\mathrm{K}\cdot\mathcal{R}}\right\}\\ \end{align}

Thus, going through the $\ln$ form, you get \begin{align} \ln\{D_{298}\} &= \ln\{D_0\} - \frac{Q}{298~\mathrm{K}\cdot\mathcal{R}}\\ \ln\{D_{573}\} &= \ln\{D_0\} - \frac{Q}{573~\mathrm{K}\cdot\mathcal{R}}\\ \ln\{D_{298}\} - \ln\{D_{573}\} &= -\frac{Q}{\mathcal{R}}\cdot\left(\frac{1}{298}-\frac{1}{573}\right) \end{align}

And then I obtained $Q\approx 38~\mathrm{kJ/mol}$.


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