Question

The optical rotations of sucrose in $0.5MHCl$ at ${35}^o\text{C}$ at various time intervals are given below. The reaction of sucrose with HCl follows first order reaction. Find the rate constant of the given reaction.

Time (minutes)	0	10	20	30	40	$\infty$
Rotation (degrees)	+32.4	+28.8	+25.5	+22.4	+19.6	-11.1

Take, $ln1.09=0.086$

• A. $16.2\times {10}^{-4}$
• B. $8.6\times {10}^{-3}$
• C. $8.6\times {10}^{-2}$
• D. $4.3\times {10}^{-2}$

Answer 1

The correct option is B $8.6\times {10}^{-3}$

The inversion of sucrose is a first-order reaction, hence,

$k_1=\frac{\text{l}}{t}\text{ln}\frac{r_0-r_{\infty }}{r_t-r_{\infty }}$

where $r_0,r_t\text{and}{r}_{\infty }$ represent optical rotations at the commencement of the reaction, after time $t$ and at the completion of the reaction, respectively.

In this case, $a_0=r_0-r_{\infty }=+32.4-\left(11.1\right)=+43.5$
The value of $k_1$ at different times are calculated as follows :

Time	$r_t$	$r_t-r_{\infty }$	$\frac{1}{t}\text{ln}\frac{r_0-r_{\infty }}{r_t-r_{\infty }}=k_1$
10 min	+28.8	39.9	$\frac{1}{10}\text{ln}\frac{43.5}{39.9}=0.008625{\text{min}}^{-1}$

Thus, the value of rate constant is $8.6\times {10}^{-3}$