Question

Inversion of a sugar folllows first order rate equation which can be followed by nothing the change in rotation of the plane of polarization of light in the polarimeter. If $r_{\infty ,}{r}_{t}and{r}_0$ are the rotations at
$t=\infty ,t=tandt=0,$ then, first order reaction can be written as:

• A. $k=\frac{1}{t}log\frac{r_t-r_{\infty }}{r_0-r_{\infty }}$
• B. $k=\frac{1}{t}ln\frac{r_0-r_{\infty }}{r_t-r_0}$
• C. $k=\frac{1}{t}ln\frac{r_{\infty }-r_0}{r_{\infty }-r_t}$
• D. None of these

Answer 1

The correct option is C $k=\frac{1}{t}ln\frac{r_{\infty }-r_0}{r_{\infty }-r_t}$

$\underset{d-Sucrose}{C_{12}{H}_{22}{O}_{11}}+H_{2}O\stackrel{H+}{\to }\underset{d-Glucose}{C_6{H}_{12}{O}_6}+\underset{l-Fructose}{C_6{H}_{12}{O}_6}$

Initially a Excess 0 0

After time t a-x Constant x x

At infinity 0 Constant a a

If $r_0,r_t$ and $r_{\infty }$ be the observed angle of rotations of the sample at zero times, times $t$ and infinity respectively, and $k_1,k_2$ and $k_3$ proportionate in terms of sucrose,glucose and fructose, respectively.

Then,

$r_0=k_{1}a$

$r_t=k_1(a-x)+k_{2}x+k_{3}x$

$r_{\infty }=k_{2}a+k_{3}a$

From these equations it can be shown that

$\frac{a}{a-x}=\frac{r_0-r_{\infty }}{r_t-r_{\infty }}$

So, the expression for the rate constant rate of this reaction in terms of the optical rotational data may be

put as $k=\frac{2.303}{t}\log \frac{r_0-r_{\infty }}{r_t-r_{\infty }}$