### Transcription for Educational Website:---### Writing the Rate Law for an Elementary ReactionFor the given chemical reaction:\[ \text{CH}_3\text{CH}_2\text{Br}(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{CH}_3\text{CH}_2\text{OH}(\text{aq}) + \text{Br}^-(\text{aq}) \]We are required to write the corresponding rate law using \( k_1 \) as the rate constant.### Given Reaction\[ \text{CH}_3\text{CH}_2\text{Br}(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{CH}_3\text{CH}_2\text{OH}(\text{aq}) + \text{Br}^-(\text{aq}) \]### InstructionUse \( k_1 \) to stand for the rate constant.In the provided space, write the rate law as:\[ \text{rate} = \]### Detailed Explanation- The box on the left is where you input the rate law expression.- The right section contains a set of icons presumably for additional functionalities like clearing the input (marked with an 'X'), refreshing (marked with a circular arrow), and possibly help or instructions (marked with a question mark).For an elementary reaction such as the one given, the rate law can be directly determined based on the reactants and their stoichiometry.---In the case of an elementary reaction, the rate law is formulated using the concentration of the reactants raised to their respective stoichiometric coefficients. For this reaction:\[ \text{CH}_3\text{CH}_2\text{Br}(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{CH}_3\text{CH}_2\text{OH}(\text{aq}) + \text{Br}^-(\text{aq}) \]the rate law can be written as:\[ \text{rate} = k_1 [\text{CH}_3\text{CH}_2\text{Br}][\text{OH}^-] \]where \( k_1 \) represents the rate constant for the reaction.#### Remember:

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Write the rate law for the following elementary reaction: CH3CH₂Br(aq) + OH (aq) → CH3CH₂OH(aq) + Br (aq) Use k₁ to stand for the rate constant. rate = 0/0 X Ś 00 ?

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

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Catalysis and Enzymatic Reactions

Catalysis is the kind of chemical reaction in which the rate (speed) of a reaction is enhanced by the catalyst which is not consumed during the process of reaction and afterward it is removed when the catalyst is not used to make up the impurity in the product. The enzymatic reaction is the reaction that is catalyzed via enzymes.

Lock And Key Model

The lock-and-key model is used to describe the catalytic enzyme activity, based on the interaction between enzyme and substrate. This model considers the lock as an enzyme and the key as a substrate to explain this model. The concept of how a unique distinct key only can have the access to open a particular lock resembles how the specific substrate can only fit into the particular active site of the enzyme. This is significant in understanding the intermolecular interaction between proteins and plays a vital role in drug interaction.

Question

### Transcription for Educational Website:

---

### Writing the Rate Law for an Elementary Reaction

For the given chemical reaction:

\[ \text{CH}_3\text{CH}_2\text{Br}(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{CH}_3\text{CH}_2\text{OH}(\text{aq}) + \text{Br}^-(\text{aq}) \]

We are required to write the corresponding rate law using \( k_1 \) as the rate constant.

### Given Reaction
\[ \text{CH}_3\text{CH}_2\text{Br}(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{CH}_3\text{CH}_2\text{OH}(\text{aq}) + \text{Br}^-(\text{aq}) \]

### Instruction
Use \( k_1 \) to stand for the rate constant.

In the provided space, write the rate law as:

\[ \text{rate} = \]

### Detailed Explanation
- The box on the left is where you input the rate law expression.
- The right section contains a set of icons presumably for additional functionalities like clearing the input (marked with an 'X'), refreshing (marked with a circular arrow), and possibly help or instructions (marked with a question mark).

For an elementary reaction such as the one given, the rate law can be directly determined based on the reactants and their stoichiometry.

---

In the case of an elementary reaction, the rate law is formulated using the concentration of the reactants raised to their respective stoichiometric coefficients. For this reaction:

\[ \text{CH}_3\text{CH}_2\text{Br}(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{CH}_3\text{CH}_2\text{OH}(\text{aq}) + \text{Br}^-(\text{aq}) \]

the rate law can be written as:

\[ \text{rate} = k_1 [\text{CH}_3\text{CH}_2\text{Br}][\text{OH}^-] \]

where \( k_1 \) represents the rate constant for the reaction.

#### Remember:

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