Question

Optical rotation sucrose in $1N$$HCl$ at various times was found as shown below:

Time (sec)	$0$	$7.18$	$18.0$	$27.05$	$\infty $
Rotation (deg)	$+24.09$	$+21.7$	$+17.7$	$+15.0$	$-10.74$

Show that inversion of sucrose is a first order reaction.

Answer 1

Hint: Chemical kinetics is basically deals with the different aspects of a chemical reaction. It deals with the rate of change of the reaction. It helps us to understand the rate of reaction and how it changes with certain conditions. It helps to analyse the mechanism of the reaction.

Complete step-by-step answer:In first order reaction, the rate constant value remains constant.
For inversion of sucrose to be the first order reaction, we should confirm the equation from above data
The equation is as follows:
$K=\dfrac{1}{t}\ln \dfrac{{{r}_{{}^\circ }}-{{r}_{\infty }}}{{{r}_{t}}-{{r}_{\infty }}}$
Where, ${{r}_{{}^\circ }}$ is denoted as rotation in time zero
${{r}_{t}}$ is denoted as rotation at time $t$
${{r}_{\infty }}$ is denoted as the rotation at completion of the reaction.
Now, we will substitute the above value in the equation ${{r}_{{}^\circ }}-{{r}_{\infty }}$ we get,
${{r}_{{}^\circ }}-{{r}_{\infty }}=+24.09-\left( -10.74 \right)$
${{r}_{{}^\circ }}-{{r}_{\infty }}=34.83$
Now we will calculate rate constant at different time ,
-At $t=7.18\min ,{{r}_{t}}=+21.7$
${{r}_{t}}-{{r}_{\infty }}=21.73-\left( -10.74 \right)$
${{r}_{t}}-{{r}_{\infty }}=32.44$
$K=\dfrac{1}{t}\ln \dfrac{{{r}_{{}^\circ }}-{{r}_{\infty }}}{{{r}_{t}}-{{r}_{\infty }}}$
Now we will substitute the above value in this equation we get,
$K=\dfrac{1}{7.18}\ln \dfrac{34.83}{32.44}$
$K=0.13\ln \dfrac{34.83}{32.44}$
On further solving we get,
$K=0.019{{\min }^{-1}}$
-At $t=18.0\min ,{{r}_{t}}=+17.7$
${{r}_{t}}-{{r}_{\infty }}=+17.7-\left( -10.74 \right)$
${{r}_{t}}-{{r}_{\infty }}=28.44$
$K=\dfrac{1}{t}\ln \dfrac{{{r}_{{}^\circ }}-{{r}_{\infty }}}{{{r}_{t}}-{{r}_{\infty }}}$
Now we will substitute the above value in this equation we get,
$K=\dfrac{1}{18}\ln \dfrac{34.83}{28.44}$
$K=0.05\ln \dfrac{34.83}{28.44}$
On further solving we get,
$K=0.011{{\min }^{-1}}$
-At $t=27.05\min ,{{r}_{t}}=+15.0$
${{r}_{t}}-{{r}_{\infty }}=+15.0-\left( -10.74 \right)$
${{r}_{t}}-{{r}_{\infty }}=25.74$
$K=\dfrac{1}{t}\ln \dfrac{{{r}_{{}^\circ }}-{{r}_{\infty }}}{{{r}_{t}}-{{r}_{\infty }}}$
Now we will substitute the above value in this equation we get,
$K=\dfrac{1}{27.05}\ln \dfrac{34.83}{25.74}$
$K=0.03\ln \dfrac{34.83}{28.44}$
On further solving we get,
$K=0.020{{\min }^{-1}}$
Here, we can see that the value of rate constant $\left( K \right)$ is fairly constant. Therefore, we can say that inversion of sucrose is a first order reaction.

Note:During a chemical reaction, when a reaction starts, it decreases the amount of the reactant and increases the amount product in the reaction. Rate of disappearance of reaction tells us about how an amount of reactant decreases with time.
Order of the reaction is defined as power of the concentration of the reactant in a chemical reaction. For example, the rate of first order reaction depends only on one concentration of the reactant.