Question

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

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

Predict the rate constant expression for the given first order reaction.

Here,
$r_0$ is the optical rotation at time = 0
$r_t$ is the optical rotation at time = t
$r_{\infty }$ is the optical rotation at time = $\infty$

• A. $\text{K}=\frac{2.303}{t}\left[log\right(\frac{r_0-r_{\infty }}{r_t-r_{\infty }}\left)\right]$
• B. $\text{K}=\frac{2.303}{t}\left[log\right(\frac{r_0-r_t}{r_t-r_{\infty }}\left)\right]$
• C. $\text{K}=\frac{2.303}{t}\left[log\right(\frac{r_0-r_{\infty }}{r_t-r_0}\left)\right]$
• D. $\text{K}=\frac{2.303}{t}\left[log\right(\frac{r_{\infty }-r_0}{r_t-r_0}\left)\right]$

Answer 1

The correct option is A $\text{K}=\frac{2.303}{t}\left[log\right(\frac{r_0-r_{\infty }}{r_t-r_{\infty }}\left)\right]$

The chemical reaction for the inversion of cane sugar or sucrose can be given as :
$\underset{(+{66.5}^o)}{C_{12}{H}_{22}{O}_{11}}+H_{2}O\stackrel{H^{+}}{⟶}\underset{(+{52.5}^o)}{C_6{H}_{12}{O}_6}+\underset{(-{92}^o)}{C_6{H}_{12}{O}_6}$

The two products formed in this process are glucose (which is dextro-rotatory) and fructose (which is laevorotatory)
In order to understand the kinetics of the given reaction, we must use the variations in the angle of rotation. Let us consider,
$r_0$ is the reading obtained at time = 0
$r_t$ is the reading obtained at time = t
$r_{\infty }$ is the reading obtained at time = $\infty$
'x' is the amount of cane sugar that has hydrolysed in the reaction
'a' is the initial concentration of the cane sugar used
We can say that the value of angle of rotation at a particular time instant is directly proportional the amount of cane sugar that has hydrolysed in the reaction. Representing this in a mathematical form:
$(r_0-r_t)\propto$ amount of cane sugar that has hydrolysed
$(r_0-r_t)\propto \text{(x)}$
We can also say that the value of angle of rotation at time = $\infty$ is directly proportional the initial concentration of the cane sugar used. Representing this in a mathematical form:
$(r_0-r_{\infty })\alpha$ initial concentration of the cane sugar used
$(r_0-r_{\infty })\alpha \text{(a)}$
From these equations, we can derive the following relation:
$(a-x)\propto (r_0-r_{\infty })-(r_0-r_t)$
$(a-x)\alpha (r_t-r_{\infty })$

Representing this reaction in the form of a first order reaction
we get:
$\text{K}=\frac{2.303}{t}\left[log\right(\frac{a}{a-x}\left)\right]=\frac{2.303}{t}\left[log\right(\frac{r_0-r_{\infty }}{r_t-r_{\infty }}\left)\right]$

Hence, option(a) is correct.