Question

The activation energy for the diffusion of carbon in chromium is 111,000 J/mol. Calculate the diffusion coefficient at 1100 K (827C), given that D at 1400 K (1127C) is 6.25 1011 m2/s

Answer 1

Explanation:

As the formula is as follows.

          D = $D_{o} exp (\frac{-Q_{d}}{RT})$

where,   D = diffusion coefficient

            $D_{o}$ = constant

            T = temperature

            R = gas constant

         $Q_{d}$ = activation energy

For T = 1400 K,

    $D_{1} = D_{o} exp (\frac{-Q_{d}}{R \times 1400})$

For T = 1100 K,

     $D_{1} = D_{o} exp (\frac{-Q_{d}}{R \times 1100})$

Now,

  $\frac{D_{1}}{D_{2}} = exp[\frac{-Q_{d}}{1400 R} + \frac{Q}{1100 R}]$

  $\frac{6.25 \times 10^{-11}}{D_{2}} = exp [\frac{-Q_{d}}{R}(\frac{-300}{1400 \times 1100})]$

     $\frac{6.25 \times 10^{-11}}{D_{2}}$ = exp (2.6)

                                 = 13.46

      $D_{2} = \frac{6.25 \times 10^{-11}}{13.46}$

                   = $4.64 \times 10^{-12} m^{2}/s$

Thus, we can conclude that the diffusion coefficient at 1100 K is $4.64 \times 10^{-12} m^{2}/s$.