Question

Time	t	$\infty$
Rotation of glucose and fructose	${{\text{r}}_{\text{t}}}$	${{\text{r}}_\infty }$

${\text{S}} \to {\text{G}} + {\text{F}}$
What is the value of k for the above reaction under given circumstances?
A.${\text{k}} = \dfrac{1}{{\text{t}}}\ln \dfrac{{{{\text{r}}_\infty }}}{{{{\text{r}}_\infty } - {{\text{r}}_{\text{t}}}}}$
B.${\text{k}} = \dfrac{1}{{\text{t}}}\ln \dfrac{{{{\text{r}}_{\text{t}}}}}{{{{\text{r}}_\infty } - {{\text{r}}_{\text{t}}}}}$
C.${\text{k}} = \dfrac{1}{{\text{t}}}\ln \dfrac{{{{\text{r}}_\infty }}}{{{{\text{r}}_{\text{t}}} - {{\text{r}}_\infty }}}$
D.${\text{k}} = \dfrac{1}{{\text{t}}}\ln \dfrac{{{{\text{r}}_{\text{t}}}}}{{{{\text{r}}_{\text{t}}} - {{\text{r}}_\infty }}}$

Answer 1

The above reaction occurs as a first order reaction. We have given the rotation of sucrose at time t and at time infinity. The initial concentration will depend on the rotation at infinity and the equilibrium concentration will depend on the rotation at time t.

Complete step by step answer:
The rotation of sucrose to form glucose and fructose follows the first order kinetics. Let us assume that initial amount of sucrose is ${{\text{a}}_{\text{o}}}$ and then we will write the amount at equilibrium.${\text{t}} = \infty$ Refers to the time when the reaction almost goes to completion that means all the reactant has been reacted. The amount of reactant left is zero at infinity:
${\text{ S }} \to {\text{ G }} + {\text{ F }}$
${\text{t}} = 0{\text{ }}\,{{\text{a}}_{\text{o}}}{\text{ }} \to {\text{ }}0{\text{ }} + {\text{ }}0$
Let x amount is reacted on equilibrium. So the amount at equilibrium will be:
${\text{t}} = 0{\text{ }}\,{{\text{a}}_{\text{o}}} - {\text{x }} \to {\text{ x }} + {\text{ x}}$
At infinity all the reactant is reacted so the equation will become:
${\text{t}} = \infty \,{\text{ 0 }} \to {\text{ }}\,{{\text{a}}_{\text{o}}}{\text{ }} + {\text{ }}\,{{\text{a}}_{\text{o}}}{\text{ }}$
Applying the rate formula for first order reaction we will get the value of rate constant:
${\text{k}} = \dfrac{1}{{\text{t}}}\ln \dfrac{{\,{{\text{a}}_{\text{o}}}{\text{ }}}}{{\,{{\text{a}}_{\text{o}}} - {\text{x}}}}$
The amount of sucrose reacted will depend upon the change in rotation of sucrose because more is the change in rotation of sucrose more will be the reacted amount of sucrose and hence the reacted amount x will be directly proportional to the change in rotation of sucrose between time zero to time t. In the same way the initial amount of sucrose present will depend upon the rotation at time 0 to time infinity. Hence we will get two equations as:
${\text{x}} \propto \left( {{{\text{r}}_0} - {{\text{r}}_{\text{t}}}} \right)$
${{\text{a}}_{\text{0}}} \propto \left( {{{\text{r}}_{\text{0}}} - {{\text{r}}_\infty }} \right)$
We will now substitute these two equations into the formula:
${\text{k}} = \dfrac{1}{{\text{t}}}\ln \dfrac{{\left( {{{\text{r}}_0} - {{\text{r}}_\infty }} \right)}}{{\left( {{{\text{r}}_{\text{0}}} - {{\text{r}}_\infty }} \right) - \left( {{{\text{r}}_0} - {{\text{r}}_{\text{t}}}} \right)}}$
Solving the above equation we will get:
${\text{k}} = \dfrac{1}{{\text{t}}}\ln \dfrac{{\left( {{{\text{r}}_0} - {{\text{r}}_\infty }} \right)}}{{{{\text{r}}_{\text{t}}} - {{\text{r}}_\infty }}}$
Since the $\,{{\text{r}}_{\text{o}}}$is not given to us so we can only consider that the initial concentration depends upon the rotation at infinite time. Hence keeping $\,{{\text{r}}_{\text{o}}}$ as zero we will get:
${\text{k}} = \dfrac{1}{{\text{t}}}\ln \dfrac{{{{\text{r}}_\infty }}}{{{{\text{r}}_\infty } - {{\text{r}}_{\text{t}}}}}$

Hence, the correct option is A.

Note:
First order reaction is that reaction in which the rate of reaction is directly proportional to the concentration of reactant. Sucrose is dextrorotatory, glucose is also dextrorotatory and fructose is laevorotatory.