When ethyl bromide is added to potassium tert -butoxide, the product is ethyl tert -butyl ether. CH 3 CH 2 –Br + (CH 3 ) 3 C–O – K + → (CH 3 ) 3 C–O–CH 2 CH 3 ethyl bromide potassium tert -butoxide ethyl ter t-butyl ether a. What happens to the reaction rate if the concentration of ethyl bromide is doubled?

Welcome back every once in another video, when methyl iodide reacts with sodium third betide, third beyl methyl ether is formed, we're given our elementary reaction. And the question is what will be the effect of doubling the concentration of methyl iodide on the reaction rate. First of all, we have to understand that since we're given an elementary reaction, this means that the psychometric coefficients will essentially be equal to the orders within our reaction rate. So we can say that they are equal to the orders. Now, let's remember that this is not applicable for an overall balanced reaction. Only when we verify that the given reaction is an elementary reaction. Then we can say that the stoichiometry efficient for each species will represent the orders. So in this case, we can say that the rate law of the given reaction will be rate equals the rate constant K multiplied by the concentration of methyl iodide. We're going to say CH three I and then multiplied by the concentration of sodium tur betide. So we can say CH three, three times con A now because our sty metric efficients are one for each of our reactants. That means our orders are also equal to one. So there's no necessity to add them because they are implicit. Now, essentially what we're performing here is doubling the concentration of methyl ionite. So using this rate expression, we can state that let's assume that the original rate are one is measured for specific concentrations of methyl iodide and sodium turbot oxide. And let's say that rate two is measured with a double concentration of methyl iodide. So we can say that we can essentially multiply the concentration of methyl iodide by two, we have that K in front. And what we essentially want to do is find the ratio between R two and R one. Essentially, if we substitute our expressions for R two, we're going to get K multiplied by two, multiplied by the concentration of methyl iodide and sodium turbot oxide. Now we're going to add our fraction sign and rewrite the expression for weight one, we may notice immediately that a lot of species cancel out. In particular, both of our reactants cancel out in terms of their concentrations, our weight constants cancel each other out and we obtain the ratio of two. What that means is that rate two is twice as large compared to rate one. So the rate doubles and we can state that our final answer is the rate of the reaction doubles. Now, what we can also say is that instead of performing all of this mass, we can say that rate is directly proportional to the first power of methyl iodide and the first power of sodium turbot oxide. So if we double any of our two concentrations, the rate will also double right, if we triple them, the rate will triple and so on because they are directly proportional. But we have also proved this mathematically. So our final answer is the rate doubles. Thank you for watching.

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6. Thermodynamics and Kinetics

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6. Thermodynamics and Kinetics

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Problem 37a

Textbook Question

When ethyl bromide is added to potassium tert-butoxide, the product is ethyl tert-butyl ether.
CH₃CH₂–Br + (CH₃)₃C–O^–K⁺ → (CH₃)₃C–O–CH₂CH₃ ethyl bromide potassium tert-butoxide ethyl tert-butyl ether
a. What happens to the reaction rate if the concentration of ethyl bromide is doubled?

Verified step by step guidance

Step 1: Recognize that this reaction is an example of a Williamson ether synthesis, which proceeds via an SN2 (bimolecular nucleophilic substitution) mechanism.

Step 2: Recall that the rate of an SN2 reaction depends on the concentrations of both the nucleophile (potassium tert-butoxide) and the electrophile (ethyl bromide). The rate law for an SN2 reaction is:

r = k \cdot [_{RBr} \cdot [_{Nu}

, where

RBr

is the alkyl halide (ethyl bromide) and

Nu

is the nucleophile (tert-butoxide).

Step 3: Analyze the effect of doubling the concentration of ethyl bromide. Since the rate law is directly proportional to the concentration of ethyl bromide, doubling its concentration will double the reaction rate, assuming the concentration of potassium tert-butoxide remains constant.

Step 4: Understand that this proportional relationship is a hallmark of the SN2 mechanism, where the reaction rate depends on the simultaneous interaction of the nucleophile and the electrophile in the rate-determining step.

Step 5: Conclude that if the concentration of ethyl bromide is doubled, the reaction rate will also double, as per the rate law for an SN2 reaction.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Reaction Rate

The reaction rate refers to the speed at which reactants are converted into products in a chemical reaction. It is influenced by factors such as concentration, temperature, and the presence of catalysts. In this context, understanding how the concentration of ethyl bromide affects the reaction rate is crucial for predicting the outcome of the reaction.

Concentration and Rate Law

The rate law expresses the relationship between the concentration of reactants and the rate of the reaction. For a reaction involving ethyl bromide, if the rate law indicates a first-order dependence on its concentration, doubling the concentration would double the reaction rate. This concept is essential for analyzing how changes in reactant concentrations impact the overall reaction kinetics.

Nucleophilic Substitution Mechanism

The reaction between ethyl bromide and potassium tert-butoxide involves a nucleophilic substitution mechanism, where the nucleophile (tert-butoxide) attacks the electrophilic carbon in ethyl bromide. Understanding this mechanism helps in predicting how the reaction proceeds and how the concentration of reactants influences the rate of product formation.

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Related Practice

Textbook Question

From the Arrhenius equation, predict how

a. increasing the experimental activation energy affects the rate constant for a reaction.

Textbook Question

From the Arrhenius equation, predict how

b. increasing the temperature affects the rate constant for a reaction.

Textbook Question

Consider the following reaction-energy diagram.

a. Label the reactants and the products. Label the activation energy for the first step and the second step.

b. Is the overall reaction endothermic or exothermic? What is the sign of ΔH°?

c. Which points in the curve correspond to intermediates? Which correspond to transition states?

d. Label the transition state of the rate-limiting step. Does its structure resemble the reactants, the products, or an intermediate?

Textbook Question

The rate of the reaction of methyl chloride with hydroxide ion is linearly dependent on both the concentration of methyl chloride and the concentration hydroxide ion. At 30 °C, the constant (k) for the reaction is 1.0 × 10^-5 M^-1 s^-1

b. If the concentration of methyl chloride is decreased to 0.010 M, what will be the effect on

1. the rate of the reaction?

2. the rate constant for the reaction?

Textbook Question

When ethyl bromide is added to potassium tert-butoxide, the product is ethyl tert-butyl ether.

CH₃CH₂–Br + (CH₃)₃C–O^–K⁺ → (CH₃)₃C–O–CH₂CH₃ ethyl bromide potassium tert-butoxide ethyl tert-butyl ether

b. What happens to the rate if the concentration of potassium tert-butoxide is tripled and the concentration of ethyl bromide is doubled?

c. What happens to the rate if the temperature is raised?

Textbook Question

Draw a reaction coordinate diagram for a one step reaction that has the following values of E_a and ∆H. (a) E_a = 9 kcal/mol ; ∆H° = + 4 kcal/mol

Multiple Choice

If the transition state has a high energy relative to the energies of the reactants and products, what does this say about the reaction?

Multiple Choice

Which equation would you use to calculate the rate of a reaction?

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When ethyl bromide is added to potassium tert-butoxide, the product is ethyl tert-butyl ether. CH3CH2–Br + (CH3)3C–O–K+ → (CH3)3C–O–CH2CH3 ethyl bromide potassium tert-butoxide ethyl tert-butyl ether a. What happens to the reaction rate if the concentration of ethyl bromide is doubled?

Key Concepts

Reaction Rate

Concentration and Rate Law

Nucleophilic Substitution Mechanism

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