For the inversion of cane sugar, the measured polarimeter readings closely recorded at , , and are , — Chemical Kinetics Chemistry Question
Question
For the inversion of cane sugar, the measured polarimeter readings closely recorded at $t=0$, $t=30 \text{ min}$, and $t=\infty$ are $30^\circ$, $20^\circ$, and $-15^\circ$ respectively. Considering it rigidly follows pseudo-first-order kinetics, what is the mathematically calculated half-life of this particular reaction? (Given: $\log 2=0.3, \log 3=0.48, \log 7=0.84$)
💡 Solution & Explanation
Using the polarimetric pseudo-first-order formula: $k = \frac{2.303}{t} \log \frac{r_0 - r_\infty}{r_t - r_\infty}$. Initial angle difference: $30 - (-15) = 45$. Time $t$ angle difference: $20 - (-15) = 35$. $k = \frac{2.303}{30} \log\left(\frac{45}{35}\right) = \frac{2.303}{30} \log\left(\frac{9}{7}\right)$. Using logs: $\log 9 - \log 7 = 0.96 - 0.84 = 0.12$. So $k = \frac{2.303 \times 0.12}{30}$. Half-life $t_{1/2} = \frac{0.693}{k} = \frac{0.3 \times 2.303 \times 30}{2.303 \times 0.12} = \frac{9}{0.12} = 75 \text{ min}$.