52. For a bimolecular elementary reaction, why does the reaction rate depend on the product of the reactant concentrations? Give a very simple example, using numbers, to illustrate your answer. 53. Why is there no such thing as a trimolecular elementary reaction?
52. For a bimolecular elementary reaction, why does the reaction rate depend on the product of the reactant concentrations? Give a very simple example, using numbers, to illustrate your answer. 53. Why is there no such thing as a trimolecular elementary reaction?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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![### Chemical Kinetics: Understanding Elementary Reactions
#### 52. For a bimolecular elementary reaction, why does the reaction rate depend on the product of the reactant concentrations? Give a very simple example, using numbers, to illustrate your answer.
**Explanation:**
In a bimolecular elementary reaction, two reactant molecules collide and react to form products. The rate of such reactions depends on how frequently the reactant molecules collide with each other, which in turn depends on their concentrations. Mathematically, the rate \( r \) of a bimolecular reaction involving reactants \( A \) and \( B \) can be expressed as:
\[ r = k[A][B] \]
Where:
- \( k \) is the rate constant.
- \([A]\) and \([B]\) are the concentrations of reactants A and B respectively.
**Example:**
Suppose \( [A] = 2 \) M and \( [B] = 3 \) M, and the rate constant \( k = 1 \, \text{L} \cdot \text{mol}^{-1} \cdot \text{s}^{-1} \).
Then, the reaction rate \( r \) can be calculated as:
\[ r = (1 \, \text{L} \cdot \text{mol}^{-1} \cdot \text{s}^{-1})(2 \, \text{M})(3 \, \text{M}) = 6 \, \text{M}^2 \cdot \text{L}^{-1} \cdot \text{s}^{-1} \]
This illustrates that the reaction rate is proportional to the product of the concentrations of \( A \) and \( B \).
#### 53. Why is there no such thing as a trimolecular elementary reaction?
**Explanation:**
In chemical kinetics, elementary reactions involve a small number of reactant molecules colliding simultaneously. While bimolecular reactions (involving two molecules) are quite common, trimolecular reactions (involving three molecules colliding simultaneously) are extremely rare. The reason is purely probabilistic: the likelihood of three molecules coming together at the exact same moment with the proper orientation and sufficient energy to react is very low. Such events are statistically unlikely in a typical chemical environment.
Instead, reactions that appear to involve three or more molecules usually proceed via a series of simpler bimolecular or unimolecular](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F03b7918c-d94c-4943-a460-522394a4ae27%2F1896563d-cd8b-4d9c-9e3a-c211b5219dc7%2F7igzsyb_processed.png&w=3840&q=75)
Transcribed Image Text:### Chemical Kinetics: Understanding Elementary Reactions
#### 52. For a bimolecular elementary reaction, why does the reaction rate depend on the product of the reactant concentrations? Give a very simple example, using numbers, to illustrate your answer.
**Explanation:**
In a bimolecular elementary reaction, two reactant molecules collide and react to form products. The rate of such reactions depends on how frequently the reactant molecules collide with each other, which in turn depends on their concentrations. Mathematically, the rate \( r \) of a bimolecular reaction involving reactants \( A \) and \( B \) can be expressed as:
\[ r = k[A][B] \]
Where:
- \( k \) is the rate constant.
- \([A]\) and \([B]\) are the concentrations of reactants A and B respectively.
**Example:**
Suppose \( [A] = 2 \) M and \( [B] = 3 \) M, and the rate constant \( k = 1 \, \text{L} \cdot \text{mol}^{-1} \cdot \text{s}^{-1} \).
Then, the reaction rate \( r \) can be calculated as:
\[ r = (1 \, \text{L} \cdot \text{mol}^{-1} \cdot \text{s}^{-1})(2 \, \text{M})(3 \, \text{M}) = 6 \, \text{M}^2 \cdot \text{L}^{-1} \cdot \text{s}^{-1} \]
This illustrates that the reaction rate is proportional to the product of the concentrations of \( A \) and \( B \).
#### 53. Why is there no such thing as a trimolecular elementary reaction?
**Explanation:**
In chemical kinetics, elementary reactions involve a small number of reactant molecules colliding simultaneously. While bimolecular reactions (involving two molecules) are quite common, trimolecular reactions (involving three molecules colliding simultaneously) are extremely rare. The reason is purely probabilistic: the likelihood of three molecules coming together at the exact same moment with the proper orientation and sufficient energy to react is very low. Such events are statistically unlikely in a typical chemical environment.
Instead, reactions that appear to involve three or more molecules usually proceed via a series of simpler bimolecular or unimolecular
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