Question

The inversion of a sugar follows a first-order rate equation which can be followed by noting the change in the rotation of the plane of polarization of light in the polarimeter. If \[{r_\alpha }\] , \[{r_1}\] and \[{r_0}\] are the rotations at \[t = \alpha \] , \[t = t\] and \[t = 0\] , then the first-order reaction can be written as:
a.) \[k = \dfrac{1}{t}\log \dfrac{{{r_1} - {r_\alpha }}}{{{r_0} - {r_\alpha }}}\]
b.) \[k = \dfrac{1}{t}\ln \dfrac{{{r_0} - {r_\alpha }}}{{{r_1} - {r_\alpha }}}\]
c.) \[k = \dfrac{1}{t}\ln \dfrac{{{r_\alpha } - {r_0}}}{{{r_\alpha } - {r_1}}}\]
d.) \[k = \dfrac{1}{t}\ln \dfrac{{{r_\alpha } - {r_1}}}{{{r_\alpha } - {r_0}}}\]

Answer 1

Hint: The expression of the rate constant for a first-rate order is \[k = \dfrac{1}{t}\ln \left[ {\dfrac{{{A_0}}}{A}} \right]\] . The concentration of sucrose is related to the optical rotation of the mixture, hence \[{A_0} = {r_0} - {r_\alpha }\] and \[A = {r_1} - {r_\alpha }\] .So the equation for the rate constant of a first-order reaction becomes \[k = \dfrac{1}{t}\ln \dfrac{{{r_0} - {r_\alpha }}}{{{r_1} - {r_\alpha }}}\] .

Complete step by step answer:
Let’s learn about the inversion of the sugar
Sucrose is dextrorotatory with a specific rotation of \[ + 66.5^\circ \] . On hydrolysis in the presence of \[HCl\] and enzymes invertase or sucrase, glucose with a specific rotation of \[ + 52.7^\circ \] and fructose with a specific rotation of \[ - 92.40^\circ \] is given as an equimolar mixture. Since D-(+) glucose’s basic rotation is lower than that of D-(-) fructose, the resulting solution becomes laevorotatory after hydrolysis. Since sucrose hydrolysis to an equimolar mixture of D-(+) glucose and D-(-) fructose is followed by a change in the optical rotation sign from dextrorotatory to levorotatory, sucrose hydrolysis is referred to as sugar inversion.
\[{C_{12}}{H_{22}}{O_{11}}\] (Sucrose) \[ + {H_2}O\xrightarrow{{{H^ + }}}{C_6}{H_{12}}{O_6}\] (D-glucose) \[ + {C_6}{H_{12}}{O_6}\] (D-fructose)
We know that for a first-order reaction, the expression of the rate constant is
\[k = \dfrac{1}{t}\ln \left[ {\dfrac{{{A_0}}}{A}} \right]\]
Here, \[k = \] The rate constant
\[t = \] Time
\[{A_0} = \] The initial concentration of Sucrose
\[A = \] The concentration of sucrose at time \[t\]
Now, as discussed earlier we know that in the inversion of sugar the sucrose mixture slowly changes from dextrorotatory to laevorotatory. So the concentration of sucrose is related to the optical rotation of the mixture.
So, \[{A_0} = {r_0} - {r_\alpha }\]
\[A = {r_1} - {r_\alpha }\]
So, the equation for the first-order reaction of inversion of sugar becomes
\[k = \dfrac{1}{t}\ln \dfrac{{{r_0} - {r_\alpha }}}{{{r_1} - {r_\alpha }}}\]
So, the correct answer is “Option B”.

Note: In the problem the term polarimeter was used. A polarimeter is a scientific instrument used to determine the angle of rotation caused by polarised light moving through an optically active material. As the angle of rotation is defined, the degree by which the light is rotated. The angle of rotation is defined as the angle observed.