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Species A undergo a unimolecular reaction as follow:
A+A ↔A*+A(K1)
A*→P(K2)
for this reaction, the first order rate constant at high pressure is K∞. the first order rate constant becomes K∞/2 when pressure of [A]1/2.?
A+A ↔A*+A(K1)
A*→P(K2)
for this reaction, the first order rate constant at high pressure is K∞. the first order rate constant becomes K∞/2 when pressure of [A]1/2.?
Most Upvoted Answer
Species A undergo a unimolecular reaction as follow:A+A ↔A*+A(K1)A*→P(...
Unimolecular Reaction Overview
The reaction under consideration involves the transformation of species A to an excited state A* and subsequently to product P. The reaction can be summarized as follows:
- A + A ↔ A* + A (Equilibrium constant K1)
- A* → P (Rate constant K2)
Pressure Influence on Rate Constants
- At high pressures, the reaction exhibits a first-order rate constant denoted as K∞, indicating that the reaction is primarily dependent on the concentration of the excited state A*.
- As pressure decreases, particularly to a point where the concentration of A is halved, the first-order rate constant becomes K∞/2. This reflects the impact of pressure on the reaction kinetics.
Mechanism Behind Kinetics Changes
- Molecular Interactions: At high pressures, the molecules are closer together, favoring the formation of the excited state A*. The reaction is thus more efficient, yielding K∞.
- Reduced Concentration Effects: When the pressure is reduced to [A]1/2, the effective concentration of A decreases, leading to a reduced formation of A*. Consequently, the rate at which A* converts to P is also less efficient, resulting in K∞/2.
Conclusion
- The relationship between pressure and the rate constant in this unimolecular reaction exemplifies the sensitivity of molecular interactions to environmental conditions.
- Understanding these dynamics is crucial for optimizing reaction conditions in chemical processes.
The reaction under consideration involves the transformation of species A to an excited state A* and subsequently to product P. The reaction can be summarized as follows:
- A + A ↔ A* + A (Equilibrium constant K1)
- A* → P (Rate constant K2)
Pressure Influence on Rate Constants
- At high pressures, the reaction exhibits a first-order rate constant denoted as K∞, indicating that the reaction is primarily dependent on the concentration of the excited state A*.
- As pressure decreases, particularly to a point where the concentration of A is halved, the first-order rate constant becomes K∞/2. This reflects the impact of pressure on the reaction kinetics.
Mechanism Behind Kinetics Changes
- Molecular Interactions: At high pressures, the molecules are closer together, favoring the formation of the excited state A*. The reaction is thus more efficient, yielding K∞.
- Reduced Concentration Effects: When the pressure is reduced to [A]1/2, the effective concentration of A decreases, leading to a reduced formation of A*. Consequently, the rate at which A* converts to P is also less efficient, resulting in K∞/2.
Conclusion
- The relationship between pressure and the rate constant in this unimolecular reaction exemplifies the sensitivity of molecular interactions to environmental conditions.
- Understanding these dynamics is crucial for optimizing reaction conditions in chemical processes.
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Species A undergo a unimolecular reaction as follow:A+A ↔A*+A(K1)A*→P(K2)for this reaction, the first order rate constant at high pressure is K∞. the first order rate constant becomes K∞/2 when pressure of [A]1/2.?
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Species A undergo a unimolecular reaction as follow:A+A ↔A*+A(K1)A*→P(K2)for this reaction, the first order rate constant at high pressure is K∞. the first order rate constant becomes K∞/2 when pressure of [A]1/2.? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Species A undergo a unimolecular reaction as follow:A+A ↔A*+A(K1)A*→P(K2)for this reaction, the first order rate constant at high pressure is K∞. the first order rate constant becomes K∞/2 when pressure of [A]1/2.? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Species A undergo a unimolecular reaction as follow:A+A ↔A*+A(K1)A*→P(K2)for this reaction, the first order rate constant at high pressure is K∞. the first order rate constant becomes K∞/2 when pressure of [A]1/2.?.
Species A undergo a unimolecular reaction as follow:A+A ↔A*+A(K1)A*→P(K2)for this reaction, the first order rate constant at high pressure is K∞. the first order rate constant becomes K∞/2 when pressure of [A]1/2.? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Species A undergo a unimolecular reaction as follow:A+A ↔A*+A(K1)A*→P(K2)for this reaction, the first order rate constant at high pressure is K∞. the first order rate constant becomes K∞/2 when pressure of [A]1/2.? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Species A undergo a unimolecular reaction as follow:A+A ↔A*+A(K1)A*→P(K2)for this reaction, the first order rate constant at high pressure is K∞. the first order rate constant becomes K∞/2 when pressure of [A]1/2.?.
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Species A undergo a unimolecular reaction as follow:A+A ↔A*+A(K1)A*→P(K2)for this reaction, the first order rate constant at high pressure is K∞. the first order rate constant becomes K∞/2 when pressure of [A]1/2.?, a detailed solution for Species A undergo a unimolecular reaction as follow:A+A ↔A*+A(K1)A*→P(K2)for this reaction, the first order rate constant at high pressure is K∞. the first order rate constant becomes K∞/2 when pressure of [A]1/2.? has been provided alongside types of Species A undergo a unimolecular reaction as follow:A+A ↔A*+A(K1)A*→P(K2)for this reaction, the first order rate constant at high pressure is K∞. the first order rate constant becomes K∞/2 when pressure of [A]1/2.? theory, EduRev gives you an
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