/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 When wine spoils, ethanol is oxi... [FREE SOLUTION] | 91Ó°ÊÓ

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When wine spoils, ethanol is oxidized to acetic acid as \(\mathrm{O}_{2}\) from the air reacts with the wine: $$ \underset{\text { Ethanol }}{\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(a q)}+\mathrm{O}_{2}(a q) \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ The value of \(K_{c}\) for this reaction at \(25^{\circ} \mathrm{C}\) is \(1.2 \times 10^{82}\). Will much ethanol remain when the reaction has reached equilibrium? Explain.

Short Answer

Expert verified
Very little ethanol will remain at equilibrium due to the large \(K_c\).

Step by step solution

01

Understand the Reaction Equation

First, understand that the given equation represents the oxidation of ethanol (C2H5OH) into acetic acid (CH3CO2H) and water. The reaction reaches equilibrium when the rates of the forward and reverse reactions are equal.
02

Analyze the Equilibrium Constant (Kc)

The equilibrium constant, denoted by \(K_{c}\), is \(1.2 \times 10^{82}\). This is an exceedingly large number, indicating that, at equilibrium, the concentration of products (acetic acid and water) is much greater compared to the reactants (ethanol and oxygen).
03

Predict Ethanol Concentration at Equilibrium

Given the exceedingly large \(K_c\), the system heavily favors the production of acetic acid and water. This implies that almost all of the ethanol will have reacted by the time equilibrium is reached, leaving only a negligible amount of ethanol remaining.

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Key Concepts

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

Ethanol Oxidation
Ethanol oxidation is a chemical process where ethanol (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\)) is converted into acetic acid (\(\mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{H}\)), typically in the presence of oxygen. This reaction is commonly associated with the spoilage of wine, as oxygen from the air interacts with ethanol in the wine.
During oxidation, ethanol (an alcohol) loses electrons and is transformed into acetic acid (a carboxylic acid), while oxygen is reduced. This type of chemical reaction is also known as an "oxidation-reduction" or "redox" reaction, where the transfer of electrons occurs.
Understanding oxidation is crucial in fields like chemistry and biochemistry, as it manifests in many forms, from metabolic processes in the body to the corrosion of metals. Ethanol's conversion to acetic acid can result in notable changes in taste and aroma in culinary contexts like wine and vinegar, being an essential reaction for both academic studies and industrial applications.
Equilibrium Constant (Kc)
The equilibrium constant, denoted as \(K_{c}\), is a fundamental concept in chemical equilibrium. It reflects the concentration of products divided by the concentration of reactants at equilibrium, each raised to the power of their coefficients in the balanced equation.
The formula for the equilibrium constant for a reaction \(aA + bB \rightleftharpoons cC + dD\) is:\[K_{c} = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]A large \(K_{c}\) value indicates that at equilibrium, the reaction strongly favors the products. Conversely, a small \(K_{c}\) value implies that the reactants are favored.
In the context of ethanol oxidation into acetic acid, a given \(K_{c}\) value of \(1.2 \times 10^{82}\) means nearly complete conversion of ethanol and oxygen into acetic acid and water. The magnitude of \(K_{c}\) is crucial for predicting the extent of the reaction, affecting practical applications like fermentation and chemical production.
Acetic Acid Production
Acetic acid is a significant chemical product derived from the oxidation of ethanol. Its production is not only essential in the food industry but also in manufacturing processes.
Acetic acid is widely known as vinegar when diluted, a staple in cooking and food preservation. The commercial production of acetic acid occurs predominantly through two methods:
  • Biological Fermentation: Utilizing acetic acid bacteria, ethanol in wine or other liquids can be transformed into acetic acid through a biological process. This is common in the traditional production of vinegars.
  • Chemical Synthesis: Acetic acid is industrially synthesized by directly oxidizing ethanol or by converting methanol through catalytic carbonylation processes.
This dual approach of using both biological and chemical methods highlights acetic acid's versatility and significance. From household kitchens to large-scale industrial settings, understanding its production allows for its effective utilization in various products, reinforcing the importance of chemical equilibrium in achieving desired results.

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Most popular questions from this chapter

When each of the following equilibria is disturbed by increasing the pressure as a result of decreasing the volume, does the number of moles of reaction products increase, decrease, or remain the same? (a) \(2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)\) (b) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\) (c) \(\mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SiCl}_{4}(g)\)

In the gas phase at \(400{ }^{\circ} \mathrm{C}\), isopropyl alcohol (rubbing alcohol) decomposes to acetone, an important industrial solvent: \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH}(g) \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}(g)+\mathrm{H}_{2}(g) \quad \Delta H^{\prime \prime}=+57.3 \mathrm{~kJ}\) Isopropyl akcahel Acetone Does the amount of acetone increase, decrease, or remain the same when an equilibrium mixture of reactants and products is subjected to the following changes? (a) The temperature is increased (b) The volume is increased (c) Argon is added (d) \(\mathrm{H}_{2}\) is added (e) A catalyst is added

At \(298 \mathrm{~K}, K_{c}\) is \(2.2 \times 10^{5}\) for the reaction \(\mathrm{F}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{O}_{2} \mathrm{~F}(g) .\) What is the value of \(K_{\mathrm{p}}\) at this temperature?

The following reaction, which has \(K_{c}=0.145\) at \(298 \mathrm{~K}\), takes place in carbon tetrachloride solution: $$ 2 \mathrm{BrCl}(\mathrm{soln}) \rightleftharpoons \mathrm{Br}_{2}(\mathrm{soln})+\mathrm{Cl}_{2}(\text { soln }) $$ A measurement of the concentrations shows \([\mathrm{BrCl}]=0.050 \mathrm{M}\), \(\left[\mathrm{Br}_{2}\right]=0.035 \mathrm{M}\), and \(\left[\mathrm{Cl}_{2}\right]=0.030 \mathrm{M}\) (a) Calculate \(\mathrm{Q}_{\circ}\) and determine the direction of reaction to attain equilibrium. (b) Determine the equilibrium concentrations of \(\mathrm{BrCl}, \mathrm{Br}_{\mathrm{t}}\), and \(\mathrm{Cl}_{2}\)

For each of the following equilibria, use Le Châtelier's principle to predict the direction of reaction when the volume is increased. (a) \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) (b) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g)\)

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