/*! 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 150 Acetic acid tends to form dimers... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Acetic acid tends to form dimers, \(\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)_{2}\), because of hydrogen bonding: The equilibrium constant \(K_{c}\) for this reaction is \(1.51 \times 10^{2}\) in benzene solution, but only \(3.7 \times 10^{-2}\) in water solution. (a) Calculate the ratio of dimers to monomers for \(0.100 \mathrm{M}\) acetic acid in benzene. (b) Calculate the ratio of dimers to monomers for \(0.100 \mathrm{M}\) acetic acid in water. (c) Why is \(K_{c}\) for the water solution so much smaller than \(K_{c}\) for the benzene solution?

Short Answer

Expert verified
(a) Ratio: 0.235, (b) Ratio: 18, (c) In water, strong H-bonding stabilizes monomers.

Step by step solution

01

Understanding Dimerization

The dimerization of acetic acid in solution can be described by the equilibrium reaction: \[2 \, \text{CH}_3\text{CO}_2\text{H} \rightleftharpoons (\text{CH}_3\text{CO}_2\text{H})_2\]The equilibrium constant \(K_c\) is defined as:\[K_c = \frac{[\text{Dimer}]}{[\text{Monomer}]^2}\]where \([\text{Dimer}]\) is the concentration of the dimer, and \([\text{Monomer}]\) is the concentration of acetic acid monomer.
02

Calculating Ratio in Benzene

In benzene, \(K_c = 1.51 \times 10^2\). Assume \([\text{Monomer}] = x\) and initial concentration is 0.100 M. Thus, \([\text{Dimer}] = 0.100 - 2x\) at equilibrium.At equilibrium, we have:\[1.51 \times 10^2 = \frac{0.100 - 2x}{(x)^2}\]Approximate \(x\) by assuming nearly complete dimerization, solve -> \(x = 0.081\) M.Therefore, \([\text{Dimer}] = 0.019\) M, so:\[\text{Ratio} = \frac{0.019}{0.081} \approx 0.235\]
03

Calculating Ratio in Water

In water, \(K_c = 3.7 \times 10^{-2}\). Now solve:\[3.7 \times 10^{-2} = \frac{0.100 - 2x}{(x)^2}\]Assume that \(x\) remains small to simplify (dominated by monomers), solve -> \(x = 0.005\) M.Therefore, \([\text{Dimer}] = 0.090\) M, so:\[\text{Ratio} = \frac{0.090}{0.005} = 18\]
04

Explaining Smaller Kc in Water

The value of \(K_c\) is smaller in water due to strong hydrogen bonding with the solvent, which stabilizes the monomer form. In benzene, a nonpolar solvent, the acetic acid tends to form dimers due to lack of competing hydrogen bonds with the solvent.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Acetic Acid Dimerization
Acetic acid is known for its tendency to dimerize, especially in nonpolar solvents like benzene. This dimerization occurs due to the formation of hydrogen bonds between two acetic acid molecules. The equilibrium reaction can be represented as:
  • \[2 \, \text{CH}_3\text{CO}_2\text{H} \rightleftharpoons (\text{CH}_3\text{CO}_2\text{H})_2\]
Dimerization is more favorable in an environment where the molecules can connect without interference from the solvent. The process can significantly reduce the number of free monomeric molecules. In benzene, such hydrogen bonding is facilitated due to the lack of competing interactions from the solvent. This results in a higher tendency for acetic acid molecules to pair up and form dimers.
Conversely, in polar solvents, the situation differs as the availability of hydrogen bonds with the solvent can compete with dimer formation.
Equilibrium Constant
The equilibrium constant, denoted as \(K_c\), is a crucial concept in understanding the position of a chemical equilibrium. For acetic acid dimerization, the equilibrium constant is defined as:
  • \[K_c = \frac{[\text{Dimer}]}{[\text{Monomer}]^2}\]
The value of \(K_c\) gives us a quantitative measure of the ratio of dimer to monomer concentrations at equilibrium. A larger \(K_c\) indicates that the equilibrium lies more towards the formation of dimers.
In benzene, the \(K_c\) is considerably high, \(1.51 \times 10^2\), suggesting a strong preferential formation of dimers.
On the other hand, in water, \(K_c = 3.7 \times 10^{-2}\), implies a significant dominance of monomers over dimers.
This reflects how different solvents affect the balance between dimer and monomer forms, showcasing the dependency of \(K_c\) on the chemical environment.
Hydrogen Bonding Effect in Solvents
Hydrogen bonding plays a pivotal role in determining the behavior of molecules in various solvents. In the case of acetic acid, hydrogen bonds within the dimer are crucial for stability, but interaction with the solvent also matters. In water, a polar solvent with extensive hydrogen bonding, acetic acid is more likely to remain as monomers.
Water forms strong hydrogen bonds with the acetic acid monomers, effectively competing with the dimerization process. This explains the much smaller \(K_c\) in water as compared to benzene. In contrast, benzene is nonpolar, lacking significant hydrogen bonding. This allows acetic acid molecules the freedom to form dimers more readily.
The role of hydrogen bonding emphasizes why the chemical nature of solvents dramatically influences the position of equilibrium in dimerization reactions. It is an intriguing interplay between solute molecules and the surrounding environment, illustrating the importance of solvent choice in chemical reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A sample of HI (9.30 \(\times 10^{-3} \mathrm{~mol}\) ) was placed in an empty \(2.00 \mathrm{~L}\) container at \(1000 \mathrm{~K}\). After equilibrium was reached, the concentration of \(\mathrm{I}_{2}\) was \(6.29 \times 10^{-4} \mathrm{M} .\) Calculate the value of \(K_{c}\) at 1000 \(\mathrm{K}\) for the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\).

Which direction will the equilibrium reaction of hemoglobin and oxygen shift as temperature increases in the muscles during intense exercise? Does the effect of temperature on equilibrium position help muscles acquire the oxygen they need? $$ \mathrm{Hb}+4 \mathrm{O}_{2} \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4} \quad \Delta \mathrm{H}=-200 \mathrm{~kJ} / \mathrm{mol} $$

A platinum catalyst is used in automobile catalytic converters to hasten the oxidation of carbon monoxide: $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \stackrel{\mathrm{Pt}}{\rightleftharpoons} 2 \mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=-566 \mathrm{~kJ} $$ Suppose that you have a reaction vessel containing an equilibrium mixture of \(\mathrm{CO}(\mathrm{g}), \mathrm{O}_{2}(\mathrm{~g})\), and \(\mathrm{CO}_{2}(\mathrm{~g})\). Under the following conditions, will the amount of CO increase, decrease, or remain the same? (a) A platinum catalyst is added. (b) The temperature is increased. (c) The pressure is increased by decreasing the volume. (d) The pressure is increased by adding argon gas. (e) The pressure is increased by adding \(\mathrm{O}_{2}\) gas.

Consider the endothermic reaction $$ \mathrm{Fe}^{3+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{FeCl}^{2+}(a q) $$ Use Le Châtelier's principle to predict how the equilibrium concentration of the complex ion \(\mathrm{FeCl}^{2+}\) will change when: (a) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) is added (b) \(\mathrm{Cl}^{-}\) is precipitated as \(\mathrm{AgCl}\) by addition of \(\mathrm{AgNO}_{3}\) (c) The temperature is increased (d) A catalyst is added

Calculate the equilibrium concentrations of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) at \(25^{\circ} \mathrm{C}\) in a vessel that contains an initial \(\mathrm{N}_{2} \mathrm{O}_{4}\) concentration of \(0.0500 \mathrm{M}\). The equilibrium constant \(K_{c}\) for the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) is \(4.64 \times 10^{-3}\) at \(25^{\circ} \mathrm{C} .\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.