Question

Optical rotation of sucrose in $1$ $N$ $HCl$ at various times was found as shown below:
$\begin{array}{cccccc}Time\left(sec\right)& 0& 7.18& 18.0& 27.05& \infty \\ Rotation\left(deg\right)& +24.09& +21.7& +17.7& +15.0& -10.74\end{array}$
Show that the inversion of sucrose is a first order reaction.

Answer 1

For inversion of sucrose to be first order reaction above data should confirm to the equation.

$K=\frac{1}{t}ln\frac{r_o-r_{\infty }}{r_t-r_{\infty }}$

$r_o\to$rotation at time O, $r_t\to$ rotation at time t, $r_{\infty }\to$ rotation at completion of reaction.

$r_o-r_{\infty }=+24.09-(-10.74)=34.83$

Time	$r_t$	$r_t-r_{\infty }$	$\frac{1}{t}ln\frac{r_o-r_{\infty }}{r_t-r_{\infty }}=k$
7.18 min	+21.7	32.44	$\frac{1}{7.18}ln\frac{34.83}{32.44}=0.019{min}^{-1}$
18.0 min	+17.7	28.44	$\frac{1}{18}ln\frac{34.83}{28.44}=0.011{min}^{-1}$
27.05 min	+15.0	25.74	$\frac{1}{15}ln\frac{34.83}{25.74}=0.020{min}^{-1}$

Value of k is fairly constant, hence it is first older reaction.