Kinetics of inversion of cane sugar Question 1
The formula for concentration of cane sugar as a function of time is given by
$$k = \left(\frac{2.303}{t}\right) \log\left[\frac{\alpha(0) - \alpha(\infty)}{\alpha(t) - \alpha(\infty)}\right]$$
where $\alpha(t)$ is the optical rotation of the solution at any time t. This was the formula given in my textbook, but I feel it is wrong.
The optical rotation in the formula is the rotation of the solution, which includes rotation due to external factors which is taken care of by the $\alpha(\infty)$ term but it also includes rotation due to glucose and fructose , whereas the formula would only be valid if $\alpha(t)$ denoted only the rotation of cane sugar + external factors.
The reasoning given in the book is that the rotation due to glucose and fructose included in the external factors, but that is just wrong. The fundamental assumption is that rotation due to external factors remains constant and hence gets cancelled out, but if you include glucose and fructose in external factors, then their concentration and hence rotation is changing with time. Also they do not form a racemic mixture, so their rotations do not get cancelled out.
Question 2
Another question given in my textbook was to find the optical rotation if the reaction was 50% complete. Given rotations of cane sugar = 34, glucose = 10 , fructose = -20
The solution was $$\left(\frac{17+5-10}{3/2}\right)$$
Notice the 3/2 which represent the total moles in the denominator. If this is correct then the formula becomes even more incorrect as optical rotation also depends on total number of moles.
What is the correct formula? I am also not able to understand the calculation for total optical rotation as given in second question. Why did they divide by 3/2 and how does the optical rotation depend on the number of moles?
It seems as though your issue with the formula arises due to the fact that there is a feeling of not knowing what is considered to be factors that are internal or if they are external factors. To prove the formula, the method would be as follows.
The formula for the given reaction would be:
$$\ce{C12H22O11 + H2O -&gt;  C6H12O6 (glucose) +C6H12O6 (fructose)}$$
Drawing a modified version of a RICE table with optical rotation, and the case at t=$\infty$, we get:
\begin{array}{|l|c|} \hline
\text{R} &amp;\ce{C12H22O11   &amp;C6H12O6(glucose)  &amp;C6H12O6 (fructose)} &amp;\text{Optical Rotation} \\ \hline
\text{I} &amp; n &amp; 0&amp; 0 &amp; \alpha_{s}n+\mathrm E \\
\text{C} &amp; -x &amp;  x&amp;  x &amp; - \\
\text{E} &amp; n-x &amp;  x &amp; x &amp; \alpha_{s}(n-x)+ \alpha_{g}(x) + \alpha_{f}(x)+ \mathrm E \\
\infty &amp; 0 &amp; n &amp; n &amp; (\alpha_{g}+\alpha_{f})(n) +\mathrm E \\ \hline
\end{array}
Now, for a general first order reaction, the formula would be:
$$k=\dfrac{2.303}{t}\log \frac{n}{n-x}$$
Therefore, we need to find the value of $\dfrac{n}{n-x}$. Using the given data, we get:
$$\frac{\alpha(\mathrm I)-\alpha(\infty)}{\alpha(\mathrm t)-\alpha(\infty)}=\frac{n}{n-x}$$
Therefore the formula given is true, even considering an external factor. You consider glucose and fructose to be internal.
$$k=\dfrac{2.303}{t}\log \frac{\alpha(\mathrm I)-\alpha(\infty)}{\alpha(\mathrm t)-\alpha(\infty)}$$