Question 5.2: Activation Energy For Interstitial Atoms Suppose that inters......
Activation Energy For Interstitial Atoms
Suppose that interstitial atoms are found to move from one site to another at the rates of 5 × 10^{8} jumps/s at 500 °C and 8 × 10^{10} jumps/s at 800 °C. Calculate the activation energy Q for the process.
Figure 5-4 represents the data on a ln(Rate) versus 1/T plot; the slope of this line, as calculated in the figure, gives Q/R = 14,032 K, or Q = 27,880 cal/mol. Alternately, we could write two simultaneous equations:
Rate=c_{0} \exp (\frac{-Q}{RT} )
5\times 10^8 \frac{jumps}{s}=c_{0} \exp \left[\frac{-Q}{(1.987\frac{cal}{mol \cdot K} )(500 \ K+273 \ K)} \right]
=c_{0} \exp (-0.000651 \ Q)
8\times 10^{10} \frac{jumps}{s}=c_{0} \exp \left[\frac{-Q}{(1.987\frac{cal}{mol \cdot K} )(800 \ K+273 \ K)} \right]
=c_{0} \exp (-0.000469 \ Q)
Note the temperatures were converted into K. Since
c_{0} =\frac{5\times 10^8}{\exp (-0.000651 \ Q)}(\frac{jumps}{s})then
8\times 10^{10} =\frac{(5\times 10^8) \exp (-0.000469 \ Q) }{\exp (-0.000651 \ Q)}
160=\exp [(0.000651-0.000469)Q]=\exp (0.000182 \ Q)
\ln (160)=5.075=0.000182 \ Q
Q=\frac{5.075}{0.000182}=27,880 \ cal/mol
