Question

The optical rotation of cane sugar in 0.5 N lactic acid at \[25^\circ C\] at various time intervals are given below

Time (min)	0	1435	11360	\[\infty \]
Rotation (\[^\circ \])	34.50	31.10	13.98	-10.77

Show that the reaction is of first order

Answer 1

Hint: A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration. Hence only one component from the reactant side of the reaction is responsible for the rate of the entire reaction

Complete step by step answer:
The chemical reaction for the inversion of cane sugar or sucrose can be given as:
 \[
  {C_{12}}{H_{22}}{O_{11}} + {H_2}O\xrightarrow{{{H^ + }}}{C_6}{H_{12}}{O_6} + {C_6}{H_{12}}{O_6} \\
  \,( + 66.5^\circ )\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,( + 52.5^\circ )\,\,\,\,\,\,\,\,( - 92^\circ ) \\
 \]
The two products formed in this process are glucose (which is dextro-rotatory) and fructose (which is laevo-rotatory)
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
‘i‘ 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:
 \[(\;{r_0} - \;{r_t})\] \[\alpha \] amount of cane sugar that has hydrolysed
 \[(\;{r_0} - \;{r_t})\] \[\alpha \] (x)
We can also say that the value of angle of rotation 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
 \[(\;{r_0} - \;{r_t})\] \[\alpha \] (i)
From these equations, we can derive the following relation:
 \[\left( {i{\text{ }} - {\text{ }}x} \right)\]\[\alpha \] [\[(\;{r_0} - {r_\infty })\] - \[(\;{r_0} - \;{r_t})\]]
 \[\left( {i{\text{ }} - {\text{ }}x} \right)\] \[\alpha \] [\[(\;{r_0} - \;{r_t})\]]
This reaction is a pseudo first order reaction because of the relatively large quantity of water, which makes any changes in concentrations practically negligible. Representing this reaction in the form of a first order reaction we get:
K = \[\dfrac{{2.303}}{t}[\log (\dfrac{i}{{i - x}})]\] = \[\dfrac{{2.303}}{t}[\log (\dfrac{{{r_0} - {r_\infty }}}{{{r_t} - {r_\infty }}})]\]
This is the base equation that we will be using to validate the readings given to us. If the given readings satisfy this equation in all test cases, the given reaction can be considered as a first order reaction
 \[(\;{r_0} - {r_\infty })\] = (\[{a_0}\]) = \[34.5{\text{ }}-{\text{ }}\left( { - 10.77} \right){\text{ }} = {\text{ }}45.27\]
Substituting this value in the denominator of the logarithmic part of the reaction, we can calculate the value of K for the different readings as follows:

Time (min)	\[{r_t}\]	\[{r_t}\] - \[{r_\infty }\]	K (\[{\min ^{ - 1}}\])
1435	+31.10\[^\circ \]	+41.07\[^\circ \]	0.000056
11360	+13.98\[^\circ \]	+24.75\[^\circ \]	0.000056

Since the value of K remains constant for all readings, the given readings have managed to satisfy the equation of a first order reaction. hence, the reaction of hydrolysation of cane sugar can be identified as a first order reaction.

Note:
Sucrose is dextrorotatory, but the resulting mixture of glucose and fructose is slightly laevorotatory, because the laevorotatory fructose has a greater molar rotation than the dextrorotatory glucose. As the sucrose is used up and the glucose-fructose mixture is formed, the angle of rotation to the right (as the observer looks into the polarimeter tube) becomes less and less, and finally the light is rotated to the left.