Problem 177 - A Complete Resource Book in Chemistry for JEE Main

Question

When sucrose is hydrolysed, it produces dextrorotatory "glucose" and laevorotatory "fructose" whereas the reactant itself is a dextrorotatory compound. The above a conversion process follows first order kinetics. Similarly, an optically active compound A is hydrolysed as follows. $$ \mathrm{A}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}^{+}}{\longrightarrow} 2 \mathrm{~B}+\mathrm{C} $$ The observed rotation of compound \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are \(60^{\circ}\), \(50^{\circ}\) and \(-80^{\circ}\) per mole respectively. The angles of rotation after 40 minutes and after the completion of reaction were \(26^{\circ}\) and \(10^{\circ}\) respectively. At \(27^{\circ} \mathrm{C}\) activation energy for conversion is \(27 \mathrm{~kJ} \mathrm{~mol}^{-1} .\) (Use: \(\log 1.25=0.0969, \log 14.97=\) 1.175) The value of \(t_{1 / 2}\) for the above process at \(127^{\circ} \mathrm{C}\) is (a) \(1.2\) hours (b) \(8.4\) minutes (c) \(1.5\) hours (d) \(5.68\) minutes

Step-by-Step Solution

Answer

Half-life at 127°C is 5.68 minutes.

Step 1: Identify the initial and final conditions

The initial rotation of compound A is given as \(60^{\circ}\) per mole, and the rotation after the completion of the reaction is \(10^{\circ}\). This change in angle is due to the reaction of A with water.

Step 2: Calculate the reaction progress at 40 minutes

The rotation after 40 minutes is given as \(26^{\circ}\). Using the formula for reaction progress, \( x = \theta_0 + \theta_t - 2\theta_\infty \) where \( \theta_0 = 60^{\circ}, \theta_t = 26^{\circ}, \theta_\infty = 10^{\circ} \). This helps us determine how much of A has reacted.

Step 3: Calculate reaction rate constant at 27°C

Using the first-order kinetics equation \( k = \frac{1}{t} \ln \left( \frac{a}{a-x} \right) \), substitute \( x \) and \( t = 40 \) minutes to solve for \( k \). We already know the value of \( \log 1.25 \).

Step 4: Calculate the activation energy effect on half-life

Using the Arrhenius equation \( k_2 = k_1 e^{- \frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) } \), find the new rate constant \( k_2 \) for the temperature change from 27°C to 127°C in Kelvin.

Step 5: Calculate the new half-life at the altered temperature

Substitute \( k_2 \) into the first-order half-life equation \( t_{1/2} = \frac{0.693}{k_2} \) to find \( t_{1/2} \) for the conversion process at 127°C.

Key Concepts

Optical Rotation

Optical rotation is a fascinating concept that involves the rotation of polarized light as it passes through a substance. Certain substances have the ability to rotate the plane of polarized light, making them optically active. In the exercise, compound A, B, and C have specific optical rotations: 60°, 50°, and -80° respectively. When studying reactions like the hydrolysis of sucrose, optical activity helps us track the progress of the reaction by measuring how much the angle of rotation changes over time.

Polarimeters are used to measure optical rotation. These devices work by passing polarized light through a sample and measuring the angle by which the light is rotated. By observing these changes, we can infer the concentration of optically active compounds before and after a reaction. This is crucial for understanding how much of the reactant has been transformed into products. This type of analysis forms the foundation of studying kinetic processes that are characterized by changes in optical properties.

Activation Energy

Activation energy is the energy barrier that must be overcome for a reaction to proceed. It is a fundamental concept in chemical kinetics. In the provided exercise, the activation energy for the conversion of compound A is 27 kJ/mol at 27°C. This energy determines the speed at which the reaction occurs, influencing the rate constant.

Activating a reaction requires that this barrier be met, allowing reactants to convert into products. Higher activation energy means that fewer molecules will have sufficient energy to react, leading to a slower reaction rate. Conversely, a lower activation energy increases the reaction rate because more molecules are capable of reacting.

The role of activation energy is significant in controlling the temperature dependence of reaction rates, as governed by the Arrhenius equation. Understanding activation energy helps chemists and students alike predict how changing conditions, such as temperature, affect the overall reaction speed, providing key insights into optimizing reactions for industrial and research applications.

Arrhenius Equation

The Arrhenius equation is a crucial part of understanding the relationship between temperature and reaction rates. Expressed as \( k = A e^{-\frac{E_a}{RT}} \), where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. This equation shows how even a small change in temperature can lead to a significant change in the reaction rate.

In relation to the exercise, by applying the Arrhenius equation, we can calculate how the rate constant \( k \) changes as the temperature increases from 27°C to 127°C. This allows us to determine the impact of temperature on the half-life \( t_{1/2} \) of the reaction, using the relationship between rate constants at different temperatures. This equation not only demonstrates how volatile reaction dynamics can be with slight thermal shifts but is also indispensable for designing and controlling chemical reactions.

Hydrolysis Reaction

Hydrolysis is a type of chemical reaction where water is used to break down a compound. In this exercise, an optically active compound A is hydrolyzed in the presence of water to produce compounds B and C. This process follows first-order kinetics, indicating that the rate of reaction depends linearly on the concentration of compound A.

Hydrolysis reactions are prevalent in both biological systems and industrial processes. For example, the hydrolysis of sucrose leads to glucose and fructose, a process essential for many metabolic pathways. The simplicity and efficiency of hydrolysis make it a favored reaction pathway for decomposing complex molecules.

Understanding hydrolysis through the study of its kinetics helps chemists determine the rate at which reactants transform into products, ascertain changes in optical activity, and predict how the reaction will behave under different conditions. Core to this understanding is tracking the concentration of reactants over time and seeing how external factors, such as temperature and catalysts, can affect the reaction speed and outcome.