Question

Cane sugar is gradually converted into dextrose and laevulose by dilute acid. The rate of inversion is observed by measuring the polarisation angle, at various times, when the following results are obtained:

Time (min)	$0$	$10$	$20$	$30$	$40$	$100$	$\infty$
Angle	$32.4$	$28.8$	$25.5$	$22.4$	$19.6$	$-6.1$	$-14.1$

Show that the reaction is of first order. Calculate the value of $t$, when the solution is optically inactive.

Answer 1

In case, the inversion of cane sugar is a first order change, then
$k=\frac{2.303}{t}{\log }_{10}\frac{{\left[A\right]}_0}{\left[A\right]}=\frac{2.303}{t}{\log }_{10}\frac{r_0-r_{\infty }}{r_t-r_{\infty }}$
When, $t=10$, $r_0=32.4$, $r_t=28.8$, $r_{\infty }=-14.1$
$k=\frac{2.303}{10}{\log }_{10}\frac{32.4-(-14.1)}{28.8-(-14.1)}$
$=\frac{2.303}{10}{\log }_{10}\frac{46.5}{42.9}=0.008{min}^{-1}$
When, $t=20$, $r_0=32.4$, $r_t=25.5$, $r_{\infty }=-14.1$
$k=\frac{2.303}{20}{\log }_{10}\frac{32.4-(-14.1)}{25.5-(-14.1)}=\frac{2.303}{20}\log \frac{46.5}{39.6}$
$=0.008{min}^{-1}$
When, $t=30$, $r_0=32.4$, $r_t=22.4$, $r_{\infty }=-14.1$
$k=\frac{2.303}{30}{\log }_{10}\frac{32.4-(-14.1)}{22.4-(-14.1)}=\frac{2.303}{30}{\log }_{10}\frac{46.5}{36.5}$
$=0.008{min}^{-1}$
Thus, the reaction is of first order as the value of $k$ is constant.
The solution will be optically inactive when half of the cane sugar is inverted.
$t_{1/2}=\frac{0.693}{k}=\frac{0.693}{0.008}=86.6$ min