# Optical rotation in inversion of sucrose [closed]

In reading about the inversion of sucrose I came across the equation for the reaction rate constant below:

$$k = \left(\frac{2.303}{t}\right) \log\left[\frac{\alpha(0) - \alpha(\infty)}{\alpha(t) - \alpha(0)}\right] \tag{5}$$

What is the logic for difference of the rotations' sign for the logarithmic term? i.e conceptuatlly why is $\alpha(0)$ the minuend in the numerator and the subtrahend in the denominator?

I have seen other quantities like volume or pressure represented in this form but I don't understand the reason why reacting sugar would be presented this way.

## closed as unclear what you're asking by Mithoron, A.K., DrMoishe Pippik, Todd Minehardt, Jannis AndreskaAug 7 '18 at 14:26

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• In a first order reaction can we use anything directly proportional to concentration rather than concentration itself because these come as a ratio (and hence units cancel) as in the expression as above. Your particular case is described in the answer by MaxW below. – porphyrin Aug 5 '18 at 10:10
• The above formula is wrong. The denominator should be $\alpha(t) - \alpha(\infty)$ instead of $\alpha(t) - \alpha(0)$. – aventurin Sep 15 '18 at 22:36

• @harambe - Let's say that sucrose rotates $50^\circ$ to the right and the completely reacted product of fructose and glucose rotates $50^\circ$ to the left. There is thus a $100^\circ$ difference, so each 1% of the reaction will cause a $1^\circ$ change in rotation. – MaxW Aug 4 '18 at 16:00