Equation (3.3.39) indicates that the natural logarithm of the difference in extents of reactions at times t and t+\Delta should be linear in t. The difference between the extents of reaction at times t and t+\Delta is
\xi -\xi ^{'}=\frac{h-h^{'}}{h_{\infty }-h_{0}}\xi _{\infty } (A)
Combination of equations (A) and (3.3.39) gives
\ln (h-h^{'})=\nu _{A}kt+\ln \left ( \frac{h_{\infty }-h_{0}}{\xi _{\infty }} \right )+ constant
If one notes that for this reaction \nu _{A}=-1, and that the quan-tity \ln ((h_{\infty }-h_{0})/\xi _{\infty }) is itself a constant,
\ln (h-h^{'})=-kt+a new constant (B)
Note that k can be determined from the slope of a plot of the left side of equation (B) versus time and that this plot can be prepared without a knowledge of the dilatometer readings at times zero and infinity. The data are worked up below and plotted in Figure I3.4 using semi-logarithmic coordinates.
| Time, t (min) |
h −h′ |
| 0 |
5.34 |
| 10 |
4.56 |
| 20 |
4 |
| 30 |
3.51 |
| 40 |
3.06 |
| 50 |
2.69 |
| 60 |
2.38 |
| 70 |
2.07 |
From the slope of the plot,
k=\frac{\ln (5.26/2.06)}{70}=1.34\times 10^{-2}min^{-1}
This value compares favorably with the literature value of 1.32\times 10^{-2}min^{-1}, which is based on much more data. The half-life of this reaction is given by
t_{1/2}=\frac{0.693}{k}=\frac{0.693}{1.34\times 10^{-2}}=51.7 min
The time \Delta between readings (2 h) is thus greater than the reaction half-life, and the numerical averaging procedure leading to equation (3.3.24) could be used for an analysis of these data if an estimate of the reading at infinite time can be obtained. To manipulate this equation to a form that can make use of the available data, one can note that:
1. \frac{C_{Ai}}{C_{Ai}^{'}}=\frac{C_{A0}-\xi ^{\ast }}{C_{A0}-\xi ^{\ast '}}=\frac{1-(\xi ^{\ast }/C_{A0})}{1-(\xi ^{\ast '}/C_{A0})}
2. \frac{h-h_{0}}{h_{\infty }-h_{0}}=\frac{\xi }{\xi _{\infty }}=\frac{\xi ^{\ast }}{\xi _{\infty }^{\ast }}=\frac{\xi ^{\ast }}{C_{A0}}
3. 1-\frac{\xi ^{\ast }}{C_{A0}}=\frac{h_{\infty }-h}{h_{\infty }-h_{0}}
4. \frac{C_{Ai}}{C_{Ai}^{'}}=\frac{h_{\infty }-h}{h_{\infty }-h^{'}} (from 1 and 3)
Thus, the average value of the reaction rate constant can be calculated numerically from
\bar{k}=\frac{1}{(n+1)\Delta }\sum_{i=0}^{i=n}\ln \left ( \frac{h_{\infty }-h_{i}}{h_{\infty }-h_{i}^{'}} \right )
where n is the number of pairs of data points. Numerical calculations indicate that
\bar{k}=\frac{1}{(7+1)(120)}(12.768)=1.33\times 10^{-2}min^{-1}
This value is consistent with that obtained above.
