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For a reaction, \(\mathrm{A} \rightarrow \mathrm{B}+\mathrm{C}\), it was found that at the end of \(10 \mathrm{~min}\) from the start the total optical, rotation of the system was \(50^{\circ} \mathrm{C}\) and when the reaction is complete it was \(100^{\circ}\). Assuming that only \(\mathrm{B}\) and \(\mathrm{C}\) are optically active and dextro rotator, the rate constant of this first order reaction would be (a) \(6.9 \mathrm{~min}^{-1}\) (b) \(0.069 \mathrm{~min}^{-1}\) (c) \(6.9 \times 10^{-2} \mathrm{~min}^{-1}\) (d) \(0.69 \mathrm{~min}^{-1}\)

Short Answer

Expert verified
Option (b) \( 0.069 \mathrm{~min}^{-1} \) is the correct answer.

Step by step solution

01

Understanding Initial Optical Rotation

The optical rotation at the start when the reaction has not proceeded is considered as \( 0^{\circ} \) since the total optical rotation is contributed only by products \( B \) and \( C \), which are optically active. Thus, initial optical rotation, \( R_0 = 0^{\circ} \).
02

Total Optical Rotation When Reaction Is Complete

When the reaction is complete, the total optical rotation \( R_\infty \) is \( 100^{\circ} \) since \( B \) and \( C \) are completely formed.
03

Optical Rotation After 10 Minutes

The total optical rotation after 10 minutes is given as \( 50^{\circ} \). Thus, at \( t = 10 \) minutes, \( R_t = 50^{\circ} \).
04

First Order Reaction Formula

For a first-order reaction, the equation for the optical rotation is given by:\[ R_t = R_\infty (1 - e^{-kt}) \].
05

Substitute Known Values

Substituting the values into the first-order equation: \[ 50 = 100 (1 - e^{-10k}) \].
06

Simplify the Equation

Divide both sides by 100: \[ 0.5 = 1 - e^{-10k} \]. This implies \[ e^{-10k} = 0.5 \].
07

Solve for k

Taking the natural logarithm of both sides gives: \[ -10k = \ln(0.5) \]. Therefore, \[ k = -\frac{\ln(0.5)}{10} \].
08

Calculate k

Calculate \( k \): \[ k = -\frac{-0.693}{10} = 0.0693 \].
09

Choose the Correct Option

The calculated rate constant \( k \) is approximately \( 0.069 \mathrm{~min}^{-1} \). Thus, option (b) \( 0.069 \mathrm{~min}^{-1} \) is the correct answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Optical Rotation
Optical rotation is a fascinating concept in chemistry that involves the rotation of polarized light as it passes through a substance. In the case of our reaction, the optical rotation is caused by the optically active products B and C. When light passes through these products, its plane of polarization is rotated. This is because such substances have asymmetric carbon centers that interact with the plane of light.
Understanding the optical rotation is different from the absorbance in other spectroscopic methods. Optical rotation specifically deals with the chiral nature of molecules. It's an important tool in determining the concentration of optically active substances in a solution.
  • Initial Optical Rotation: At the beginning of the reaction, the optical rotation is 0° because the reactant A does not contribute to any rotation.
  • Final Optical Rotation: Once the reaction is complete, the full optical rotation reaches 100°, representing the maximum concentration of B and C formed.
  • Role in Reaction Monitoring: By measuring optical rotation over time, you can infer reaction progress especially in reactions involving chiral substances.
Rate Constant Calculation
Calculating the rate constant is crucial to understanding the speed of a reaction. For first-order reactions, like our example, the concentration of the reactants affects the rate at which products are formed. This relationship is exponential, as shown by the integrated rate law equation used to solve this problem.
The rate constant, denoted as \( k \), helps us understand how fast the reaction proceeds. To find \( k \), we use the formula:
\[ R_t = R_\infty (1 - e^{-kt}) \]This formula relates the optical rotation at any time \( t \) to the maximum optical rotation when the reaction is complete.
  • Substituting Values: By substituting given values, such as \( R_t = 50 \) and \( R_\infty = 100 \), we rearrange the formula to solve for \( k \).
  • Using Logarithms: Natural logarithms are used to isolate and solve for \( k \), ensuring mathematical correctness.
  • Final Calculation: After performing calculations, we determine that \( k \approx 0.069 \, \mathrm{min}^{-1} \), indicating the moderate pace of the reaction.
Chemical Kinetics
Chemical kinetics is the study of reaction rates and provides valuable insights into the mechanisms of reactions. It helps us predict how changes in conditions affect the speed of reactions. For a first-order reaction like ours, the rate is determined by the concentration of a single reactant.
The study of chemical kinetics involves:
  • Understanding Reaction Order: First-order reactions are dependent on the concentration of one reactant. This perspective simplifies the complex web of chemical equilibria.
  • Time and Concentration: The rate equation for first-order reactions ties together time and concentration, allowing one to predict the concentration of reactants or products at any point in time.
  • Application of Formulas: Mastery of kinetic equations and concepts like half-life helps predict and control reactions in various industrial and laboratory settings.
Chemical kinetic studies, therefore, are essential for designing chemical processes and making informed decisions in both academic and industrial research.

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Most popular questions from this chapter

A catalyst is a substance which (a) supplies energy to the reaction (b) increases the equilibrium concentration of the product (c) changes the equilibrium constant of the reaction (d) shortens the time to each equilibrium

For a first-order reaction, (a) The degree of dissociation is equal to \(\left(1-\mathrm{e}^{-\mathrm{k}}\right)\) (b) The pre-exponential factor in the Arrhenius equation has the dimensions of time \(t^{-1}\). (c) The time taken for the completion of \(75 \%\) reaction is thrice the \(t_{1 / 2}\) of the reaction. (d) both (a) and (b)

A substance reacts according to first-order kinetics. The rate constant for the reaction is \(1 \times\) \(10^{-2} \mathrm{sec}^{1} .\) Its initial concentration is IM. Its initial rate is (a) \(2 \times 10^{2} \mathrm{Ms}^{-1}\) (b) \(1 \times 10^{2} \mathrm{Ms}^{-1}\) (c) \(1 \times 10^{-2} \mathrm{Ms}^{-1}\) (d) \(2 \times 10^{-2} \mathrm{Ms}^{-1}\)

For a \(1^{\text {* }}\) order reaction \(\mathrm{A} \longrightarrow \mathrm{P}\), the temperature (T) dependent rate constant \((\mathrm{K})\) was found to follow the equation \(\log \mathrm{k}=-(2000) \frac{1}{\mathrm{~T}}+6\) The pre- exponential factor \(\mathrm{A}\) and activation energy Ea are respectively? (a) \(1 \times 10^{6} \mathrm{~S}^{-1}\) and \(9.2 \mathrm{~kJ} / \mathrm{M}\) (b) \(1 \times 10^{6} \mathrm{~S}^{-1}\) and \(38.3 \mathrm{~kJ} / \mathrm{M}\) (c) \(1 \times 10^{6} \mathrm{~S}^{-1}\) and \(16.6 \mathrm{~kJ} / \mathrm{M}\) (d) \(6 \mathrm{~S}^{-1}\) and \(16.6 \mathrm{~kJ} / \mathrm{M}\)

The hypothetical reaction, \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}\) follows the following mechanism: \(\mathrm{A}_{2}=\mathrm{A}+\mathrm{A}\) (fast) \(\mathrm{A}+\mathrm{B}_{2} \longrightarrow \mathrm{AB}+\mathrm{B} \quad\) (slow) \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{AB} \quad\) (fast) The order of the overall reaction is: (a) \(\underline{1}\) (b) 2 (c) \(3 / 2\) (d) 0

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