Inversion of a sugar follows first order rate equation which can be followed by noting the change in rotation of the plane of polarisation of light in a polarimeter. If $r_{\infty}, r_{t}$ and $r_{0}$ are the rotations at $t = \infty, t = t$ and $t = 0$ , then first order reaction can be written as:

k = \frac{1}{t} {log}_{e} \frac{r_{t} - r_{\infty}}{r_{0} - r_{\infty}}

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k = \frac{1}{t} {log}_{e} \frac{r_{0} - r_{\infty}}{r_{t} - r_{0}}

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k = \frac{1}{t} {log}_{e} \frac{r_{\infty} - r_{0}}{r_{\infty} - r_{t}}

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k = \frac{1}{t} {log}_{e} \frac{r_{\infty} - r_{t}}{r_{\infty} - r_{0}}

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Solution

The correct option is B $k = \frac{1}{t} {log}_{e} \frac{r_{0} - r_{\infty}}{r_{t} - r_{0}}$
$(r_{t} - r_{0}) = (r_{0} - r_{\infty}) e^{- k t} ln (\frac{r_{t} - r_{0}}{r_{0} - r_{\infty}}) = - k t k = \frac{1}{t} ln (\frac{r_{t} - r_{0}}{r_{0} - r_{\infty}})$

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