Question

The optical rotation of cane sugar in 0.5 N lactic acid at ${25}^{0}C$ at various time intervals are given below

Time (min)	0	1435	11360	$\infty$
Rotation ${(}^{0}C)$	${34.50}^0$	${31.10}^0$	${13.98}^0$	$-{10.77}^0$

Show that the reaction is of first order.

Answer 1

The inversion of cane sugar takes place according to the equation:-

$C_{12}{H}_{22}{O}_{11}+H_{2}O\underrightarrow{H^{+}}{C}_6{H}_{12}{O}_6+C_6{H}_{12}{O}_6$

Sucrose (cane sugar) Glucose Fructose

(dextro-rotatory) (dextro-rotatory) (laevo-rotatory)

$(+{66.5}^{\circ })$ $(+{52.5}^{\circ })$ $(-{92}^{\circ })$

{Laevo- rotatory}

The kinetics of this reaction is studied by noting the angle of rotation by polarimeter.

Suppose,

Reading of polarimeter at zero time= $r_o$

Reading of polarimeter at time $t$ = $r_t$

Reading of polarimeter at infinite time (when the reaction gets completed)= $r_{\infty }$

(Refer to Image)

Now we can observe that

Angle of rotation at any instant of time

$=(r_o-r_t)$ $\propto$ amount of cane sugar hydrolysed $\left(x\right)$

$\Rightarrow$ $x\propto (r_o-r_t)$ $-\left(i\right)$

Angle of rotation at infinite time=$(r_o-r_{\infty })$

$\propto$ initial concentration of cane sugar $\left(a\right)$

$\Rightarrow$ $a\propto (r_o-r_{\infty })$ $-\left(ii\right)$

From equations $\left(i\right)$ & $\left(ii\right)$ :-

$(a-x)\propto (r_o-r_{\infty })-(r_o-r_t)$

$\Rightarrow$ $(a-x)\propto (r_t-r_{\infty })$ $-\left(iii\right)$

This reaction is pseudo first order reaction because water $\left(H_{2}O\right)$ is present in large excess so the change in its' concentration is negligible.

So, for first order reaction we have:-

$K=\frac{2.303}{t}$ $\log \frac{a}{a-x}$ $-\left(iv\right)$

where, $K$= rate constant of the reaction

$t$= time

$a$= initial concentration of the reactant

$x$= amount of reactant reacted in time $t$

Substituting the value from $\left(ii\right)$ and $\left(iii\right)$ to $\left(iv\right)$ :-

$K=\frac{2.303}{t}$ $\log \frac{r_o-r_{\infty }}{r_t-r_{\infty }}$ $-\left(v\right)$

If the given data confirms this equation (v) then the reaction will be of first order.

Now, $a_o=r_o-r_{\infty }={34.50}^{\circ }-(-{10.77}^{\circ })$

$={34.50}^{\circ }+{10.77}^{\circ }={45.27}^{\circ }$

The value of $K$ at different times is calculated by the equation $\left(V\right)$ as follows:

Time(min) $r_t$ $r_t-r_{\infty }$ $K$

1. 1435 $+{31.10}^{\circ }$ $+41.07$ $\frac{2.303}{1435}$ $\log \frac{45.27}{41.07}$

$=0.000056$ $mi{n}^{-1}$

2. 11360 $+{13.98}^{\circ }$ $+24.75$ $\frac{2.303}{11360}$ $\log \frac{45.27}{24.75}$

$=0.000056$ $mi{n}^{-1}$

The constant values of $K$ depicts that the given reaction is of first order.