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Isopentyl alcohol reacts with pure acetic acid to form isopentyl acetate, the essence of banana oil: $$ \mathrm{C}_{5} \mathrm{H}_{11} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{5} \mathrm{H}_{11}+\mathrm{H}_{2} \mathrm{O} $$ A student adds a drying agent to remove \(\mathrm{H}_{2} \mathrm{O}\) and thus increase the yield of banana oil. Is this approach reasonable? Explain.

Short Answer

Expert verified
Yes, removing water will increase the yield by shifting equilibrium towards products.

Step by step solution

01

Understand the Reaction

The reaction is an esterification process where isopentyl alcohol reacts with acetic acid to form isopentyl acetate and water: \[ \mathrm{C}_{5} \mathrm{H}_{11} \mathrm{OH} + \mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{5} \mathrm{H}_{11} + \mathrm{H}_{2} \mathrm{O} \]
02

Identify the Reaction Type

This is a reversible chemical reaction, meaning it can proceed in both the forward and backward directions. The presence of water can affect the equilibrium position.
03

Effect of Removing Water

According to Le Chatelier's Principle, removing a product (in this case, water) from a reaction in equilibrium will shift the equilibrium towards the products to counteract the change.
04

Conclusion on Removing Water

By adding a drying agent to remove water, the equilibrium will shift to produce more isopentyl acetate to replace the removed water, thus increasing the yield of banana oil. Hence, the student's approach is reasonable.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Le Chatelier's Principle
When dealing with chemical reactions, understanding Le Chatelier's Principle is important. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. In simpler terms, a system at equilibrium will adjust itself to minimize the effect of any disturbance. This principle is crucial in predicting how changes to the system will affect the concentrations of the reactants and products. For example, in an esterification reaction, if we remove a product such as water, the equilibrium will shift to produce more water. This results in more banana oil being formed. Le Chatelier's Principle helps chemists control reaction yields by tweaking conditions like temperature, pressure, and the concentration of reactants or products.
Chemical Equilibrium
Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the backward reaction. In this state, the concentrations of reactants and products remain constant over time. It's a dynamic process; even though no net change is observed, molecules are still reacting. Take the esterification reaction as an example:
  • The forward reaction forms isopentyl acetate and water from isopentyl alcohol and acetic acid.
  • The backward reaction forms isopentyl alcohol and acetic acid from isopentyl acetate and water.
Understanding chemical equilibrium helps in realizing that not all reactants get converted to products. Instead, a balance is reached, which can be manipulated to favor the formation of more products or reactants. Factors like the concentration of reactants or products, temperature, and pressure influence this balance.
Removing Water in Reactions
In reactions where water is a byproduct, like esterification, removing water can be a critical tactic for shifting equilibrium. Water removal can be achieved using drying agents, chemicals that absorb water, or techniques like distillation. Drying agents, such as anhydrous magnesium sulfate or calcium chloride, are commonly used. Removing water shifts the equilibrium towards the products according to Le Chatelier's Principle:
  • Less water forces the reaction to produce more water, driving the forward reaction.
  • This creates more isopentyl acetate, increasing the yield of banana oil.
Removing water ensures that the reaction proceeds in the desired direction, maximizing product formation. This method is widely used in industrial and laboratory settings to enhance the efficiency of chemical processes.

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Most popular questions from this chapter

In a study of the thermal decomposition of lithium peroxide, $$ 2 \mathrm{Li}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2 \mathrm{Li}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g) $$ a chemist finds that, as long as some \(\mathrm{Li}_{2} \mathrm{O}_{2}\) is present at the end of the experiment, the amount of \(\mathrm{O}_{2}\) obtained in a given container at a given \(T\) is the same. Explain.

The water-gas shift reaction plays a central role in the chemical methods for obtaining cleaner fuels from coal: $$ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) $$ At a given temperature, \(K_{\mathrm{p}}=2.7 .\) If \(0.13 \mathrm{~mol}\) of \(\mathrm{CO}, 0.56 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}, 0.62 \mathrm{~mol}\) of \(\mathrm{CO}_{2},\) and \(0.43 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) are put in a \(2.0-\mathrm{L}\) flask, in which direction does the reaction proceed?

A key step in the extraction of iron from its ore is \(\mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) \quad K_{\mathrm{p}}=0.403\) at \(1000^{\circ} \mathrm{C}\) This step occurs in the \(700^{\circ} \mathrm{C}\) to \(1200^{\circ} \mathrm{C}\) zone within a blast furnace. What are the equilibrium partial pressures of \(\mathrm{CO}(g)\) and \(\mathrm{CO}_{2}(g)\) when \(1.00 \mathrm{~atm}\) of \(\mathrm{CO}(g)\) and excess \(\mathrm{FeO}(s)\) react in a sealed container at \(1000^{\circ} \mathrm{C} ?\)

Consider this equilibrium system: $$ \mathrm{CO}(g)+\mathrm{Fe}_{3} \mathrm{O}_{4}(s) \rightleftharpoons \mathrm{CO}_{2}(g)+3 \mathrm{FeO}(s) $$ How does the equilibrium position shift as a result of each of the following disturbances? (a) CO is added. (b) \(\mathrm{CO}_{2}\) is removed by adding solid \(\mathrm{NaOH}\). (c) Additional \(\mathrm{Fe}_{3} \mathrm{O}_{4}(s)\) is added to the system. (d) Dry ice is added at constant temperature.

Does \(Q\) for the formation of \(1 \mathrm{~mol}\) of NO from its elements differ from \(Q\) for the decomposition of \(1 \mathrm{~mol}\) of \(\mathrm{NO}\) to its elements? Explain and give the relationship between the two \(Q\) 's.

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