Problem 178 - 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) At what temperature half life of above process is equal to \(31.1\) minutes? (a) \(490 \mathrm{~K}\) (b) \(345 \mathrm{~K}\) (c) \(72 \mathrm{~K}\) (d) \(280 \mathrm{~K}\)

Step-by-Step Solution

Answer

The temperature is approximately 280 K, option (d).

Step 1: Understand the Reaction

The given reaction is first order kinetics: \( \mathrm{A} + \mathrm{H}_2\mathrm{O} \xrightarrow{\mathrm{H}^+} 2 \mathrm{B} + \mathrm{C} \). The angles of rotation for \( \mathrm{A}, \mathrm{B}, \) and \( \mathrm{C} \) are \( 60^{\circ}, 50^{\circ}, \) and \( -80^{\circ} \) per mole respectively. Initial and completion rotations are \( 26^{\circ} \) and \( 10^{\circ} \). We need to find temperature where half-life is \( 31.1 \) min.

Step 2: Use First Order Kinetics and Angle of Rotation

The angle of rotation at any time \( t \) is given by: \( \theta_t = \theta_\infty + (\theta_0 - \theta_\infty)e^{-kt} \). Here, \( \theta_0 = 26^\circ \), \( \theta_\infty = 10^\circ \), and \( \theta_{40} = 26^\circ \). Substitute these to determine the first order rate constant \( k \).

Step 3: Calculate Rate Constant \( k \)

Apply the values into the equation: \( 26 = 10 + (60 - 10)e^{-k\cdot 40} \). This simplifies to \( 16 = 50e^{-40k} \). Solving this gives \( e^{-40k} = \frac{16}{50} = 0.32 \). Hence, \( -40k = \ln(0.32) \) leading to \( k \approx \frac{1.139}{40} \approx 0.0285 \text{ min}^{-1} \).

Step 4: Determine Activation Energy Equation for Half-Life

The relation for the rate constant is \( k = \frac{0.693}{t_{1/2}} \). Given \( t_{1/2} = 31.1 \) min, substituting gives \( k = \frac{0.693}{31.1} \approx 0.0223 \text{ min}^{-1} \). Use the Arrhenius equation \( k = Ae^{-\frac{E_a}{RT}} \).

Step 5: Calculate Temperature for Required Half-Life

With \( ln(k) = ln(A) - \frac{E_a}{RT} \), and known \( E_a = 27 \text{ kJ/mol} \), rearrange to find \( T \): \( T = \frac{E_a}{R(ln(A) - ln(k))} \). Taking logs and solving for experimental \( k \), find \( T \). Use \( R = 8.314 \text{ J/mol K} \).

Step 6: Solve for Temperature \( T \) Using Arrhenius Equation

Given \( \log(1.25) = 0.0969 \), \( \log(14.97) = 1.175 \), for the required \( k = 0.0223 \text{ min}^{-1} \), we rearrange and solve for temperature: \( T = \frac{27000}{8.314 \times (1.175 - \log(0.0223))} \). Solving this gives \( T \approx 280 \text{ K} \).

Step 7: Identify Correct Option

The calculated temperature for the half-life of \( 31.1 \) minutes is approximately \( 280 \text{ K} \), matching the provided option \( (d) \).

Key Concepts

Optical Rotation

Optical rotation helps us understand how substances influence the rotation of polarized light. When a beam of polarized light passes through certain chemicals, these might rotate the light. The direction and degree of rotation tell us more about the compound.

• Dextrorotatory: If a compound turns light clockwise, it's called dextrorotatory, and rotation values are positive.
• Levorotatory: If it turns the light counterclockwise, the compound is levorotatory, showing negative values.

For example, in the given reaction, compound A with an optical rotation of \(60^{\circ}\) is dextrorotatory, meaning it rotates the light clockwise by \(60^{\circ}\).
Understanding the changes in optical rotation, such as from \(26^{\circ}\) to \(10^{\circ}\), allows us to track the chemical reaction's progress over time, helping us calculate the rate of reaction, especially in kinetics.

Activation Energy

Activation energy, denoted as \(E_a\), is the minimum energy required to initiate a chemical reaction. Think of it as the hurdle that reactants need to overcome to transform into products. Higher activation energies mean reactions proceed slower at given temperatures because fewer molecules have sufficient energy to react.
In the context of the given problem, the activation energy is \(27 \text{ kJ/mol}\). This indicates the amount of energy each mole of reactants must overcome to proceed through the reaction path.
Knowing the activation energy helps in predicting how temperature variations affect research; for instance, higher temperatures can provide reactants with more energy, increasing the reaction rate. This concept is crucial when applying the Arrhenius equation in calculating rate constants.

Arrhenius Equation

The Arrhenius equation connects the rate constant of a reaction with temperature, providing insights into how temperature changes affect reaction speed. It is expressed mathematically as:

\[k=Ae^{-\frac{E_a}{RT}}\]
Here, \(k\) is the rate constant, \(A\) is the pre-exponential factor (indicating the frequency of collisions), \(E_a\) is the activation energy, \(R\) is the universal gas constant, and \(T\) is the absolute temperature in Kelvin.
This equation helps predict the effect of temperature changes on the speed of reaction by calculating rate constants at various temperatures. In the problem, values from this equation help us find the temperature where the reaction's half-life is 31.1 minutes.

Half-Life Calculation

Half-life is an important measure in first-order kinetic reactions. It is the time required for half of the reactant to convert into product. For first-order reactions, this can be simply calculated using the formula:

\[ t_{1/2} = \frac{0.693}{k} \]

• First-order kinetics means the rate of reaction depends linearly on one reactant's concentration.
• At each half-life interval, the remaining amount of reactant is halved.

For example, in the given exercise, knowing the rate constant and using the above formula helps calculate the half-life, allowing determination of the temperature where the half-life approximates 31.1 minutes. Understanding half-life is essential in fields like pharmacology and nuclear physics, where it informs about the duration a substance or element remains active.