Although Keqof Equation [1] in Problem 6.57 does not greatly favor formation of the product, it is sometimes possible to use Le Châtelier’s principle to increase the yield of ethyl acetate. Le Châtelier’s principle states that if an equilibrium is disturbed, a system will react to counteract this disturbance. How can Le Châtelier’s principle be used to drive the equilibrium to increase the yield of ethyl acetate? Another example of Le Châtelier’s principle is given in Section 9.8

Short Answer

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Answer

Increasing the concentration of acetic acid or ethanol leads to an increase in the yield of ethyl acetate.

Step by step solution

01

Step-by-Step SolutionStep 1: Le-Chatelier’s principle

Le Chatelier’s principle is used to describe the effect of changes in an equilibrium reaction.

It states that any change created on a system in equilibrium leads to the shift of reaction in a direction opposite to the change such that the induced change is neutralized.

02

The reaction of acetic acid with ethanol

The reaction of acetic acid with ethanol yields ethyl acetate.

Esterification reaction

03

Applying Le Chatelier’s principle to increase the yield

The increase in the concentration of the reactants leads to an increase in the concentration of the products as the reaction proceeds forwards to neutralize/counteract the increase in the concentration of the reactant.

Therefore, increasing the concentration of acetic acid or ethanol leads to an increase in the yield of ethyl acetate.

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Most popular questions from this chapter

Explain why HCCHis more acidic than CH3CH3, even though the C-H bond in HCCHhas a higher bond dissociation energy than the C-H bond in CH3CH3.

Consider the following two-step reaction:

a. How many bonds are broken and formed in Step [1]? Would you predict H°of Step [1] to be positive or negative?

b. How many bonds are broken and formed in Step [2]? Would you predict the H°of Step [2] to be positive or negative?

c. Which step is rate-determining?

d. Draw the structure for the transition state in both steps of the mechanism.

e. If H°overallis negative for this two-step reaction, draw an energy diagram illustrating all of the information in parts (a)–(d).

As we learned in Chapter 4, monosubstituted cyclohexanes exist as an equilibrium mixture of two conformations having either an axial or equatorial substituent. When R=CH2CH3, Keqfor this process is 23. When R=C(CH3)3,Keqfor this process is 4000.

a. When , which conformation is present in higher concentration?

b. Which R shows the higher percentage of equatorial conformation at equilibrium?

c. Which R shows the higher percentage of axial conformation at equilibrium?

d. For which R is more negative?

e. How is the size of R related to the amount of axial and equatorial conformations at equilibrium?

The use of curved arrows is a powerful tool that illustrates even complex reactions.

a. Add curved arrows to show how carbocation A is converted to carbocation B. Label each new bond formed. Similar reactions have been used in elegant synthesis of steroids.

b. Draw the product by following the curved arrows. This reaction is an example of a [3,3] sigma tropic rearrangement, as we will learn in chapter 27.

Question: Draw an energy diagram for a reaction in which the products are higher in energy than the starting materials and Eais large. Clearly label all of the following on the diagram: the axes, the starting materials, the products, the transition state, H°,and Ea.

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