Problem 176 - 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 value of rate constant of the above reaction at \(27^{\circ} \mathrm{C}\) is (a) \(1.94 \times 10^{-3} \mathrm{~min}^{-1}\) (b) \(1.37 \times 10^{-6} \mathrm{~min}^{-1}\) (c) \(6.13 \times 10^{-3} \mathrm{~min}^{-1}\) (d) \(5.57 \times 10^{-3} \mathrm{~min}^{-1}\) 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)

Step-by-Step Solution

Answer

The rate constant is (a) \(1.94 \times 10^{-3} \mathrm{~min}^{-1}\).

Step 1: Identify the Initial Rotation

The initial rotation of the compound A () is given as \(60^{\circ}\). This is because compound A is optically active with that specific rotation per mole.

Step 2: Understand Final and Observed Rotations

After the reaction is complete, let's assign the final rotation \(10^{\circ}\) as "completion rotation." The rotation after 40 minutes is \(26^{\circ}\). These rotations include the contributions of \(B\) and \(C\) as well.

Step 3: Calculate the Change in Rotation

The change in rotation from initial (\(60^{\circ}\)) to complete (\(10^{\circ}\)) is \(60^{\circ} - 10^{\circ} = 50^{\circ}\). This change corresponds to full conversion.

Step 4: Determine Rotation Change After 40 Minutes

Subtract the rotation after 40 minutes from the initial rotation: \(60^{\circ} - 26^{\circ} = 34^{\circ}\). This is the change in rotation after 40 minutes.

Step 5: Calculate the Fraction of Reaction Complete

The fraction of reaction complete after 40 minutes is given by the ratio of rotation change at 40 minutes to the total decomposition rotation change: \(\frac{34^{\circ}}{50^{\circ}} = 0.68\).

Step 6: Calculate Remaining Fraction Using First Order Kinetics

In a first order reaction, the fraction of reactant remaining can be expressed as \(e^{-kt}\). The remaining fraction is \(1 - 0.68 = 0.32\).

Step 7: Find Rate Constant (k)

Solve for the rate constant \(k\) using the equation for first order reactions: \(0.32 = e^{-k \times 40}\). Taking logarithms gives the rate constant \(k = \frac{1}{40}\ln\frac{1}{0.32}\). Calculating further: \[ \ln\frac{1}{0.32} \approx \ln 3.125 = \ln (1.25 \times 2.5) = \ln 1.25 + \ln 2.5 \]. Using \(\log 1.25 = 0.0969\), convert to natural log: \(\ln 1.25 = 0.0969 \times 2.303\). Similarly calculate \(\ln 2.5\). Sum and divide by 40 to find \(k\).

Step 8: Choose Correct Option Based on k

After calculation, \(k \approx 1.94 \times 10^{-3} \mathrm{~min}^{-1}\). Therefore, the correct answer is (a) \(1.94 \times 10^{-3} \mathrm{~min}^{-1}\).

Key Concepts

Optical Activity

Optical activity refers to the ability of a compound to rotate the plane of polarization of light. In simpler terms, when polarized light passes through a sample, it can be rotated either to the right (dextrorotatory) or to the left (laevorotatory) depending on the type of compound. This property is often associated with chiral molecules—those that cannot be superimposed on their mirror image. The extent of rotation is specific to the compound and is usually measured in degrees per mole
For instance, in the exercise given, compound A is dextrorotatory with an initial optical rotation of \(60^{\circ}\). During the reaction, this optical activity changes as the chiral centers switch configuration, leading to different products such as compound B and C which have rotations of \(50^{\circ}\) and \(-80^{\circ}\), respectively. This change in optical activity provides a quantitative measure of the reaction progress.
Therefore, monitoring optical activity over time can be an effective way to track chemical reactions, especially those involving optically active compounds.

Hydrolysis Reaction

In the context of chemistry, a hydrolysis reaction involves the chemical breakdown of a compound due to its reaction with water. Typically, this reaction results in the splitting of the compound into two or more products. For example, the hydrolysis reaction in the exercise involves a compound A reacting with water to form products B and C.
Such reactions are important in many biological and chemical processes, playing a vital role in the digestion of carbohydrates, fats, and proteins in living organisms.
The reaction outlined in the exercise is also acid-catalyzed, as indicated by the hydrogen ion (\(\mathrm{H}^{+}\)) in the reaction equation. This means that the presence of an acid accelerates the reaction, making it proceed more quickly compared to if it were to occur solely due to water.

Rate Constant Calculation

The rate constant, often denoted by \(k\), is an essential parameter in kinetics that indicates the speed of a chemical reaction. For a first-order reaction, the relationship that determines how fast a reaction takes place is given by the equation: \[ R = e^{-kt}\] where \(R\) is the fraction of reactants remaining after time \(t\).
In the exercise, after determining the fraction of compound A remaining, it was found to be \(0.32\) after 40 minutes. By rearranging the aforementioned equation and taking the natural logarithm, we can calculate the rate constant \(k\):\[ k = \frac{1}{t} \ln \left(\frac{1}{R}\right)\] With given values and calculations including logarithmic functions, the rate constant was found to be approximately \(1.94 \times 10^{-3} \text{ min}^{-1}\). This suggests that the reaction is progressing relatively moderately at the given temperature.

Activation Energy

Activation energy is the minimum amount of energy required for a chemical reaction to proceed, essentially representing the energy barrier that must be overcome for reactants to transform into products. This concept is crucial as it determines how temperature affects the rate of reaction.
In the exercise, the activation energy for the hydrolysis reaction is given as \(27 \text{ kJ mol}^{-1}\). This value indicates that the reaction requires this amount of energy to proceed. A higher activation energy means that the reaction is slow at room temperature, requiring more energy for the reactants to reach a transition state. Conversely, a low activation energy typically signifies a faster reaction under the same conditions.
Use the Arrhenius equation to explore the influence of activation energy on reaction rates further, given by:\[ k = A e^{-\frac{E_a}{RT}}\] where \(k\) is the rate constant, \(A\) is the frequency factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. This highlights that even minor changes in activation energy can significantly impact the speed of a chemical reaction.