Textbook Question
The equilibrium constant Kc for the gas-phase thermal decomposition of cyclopropane to propene is 1.0 ⨉105 at 500 K:
(a) What is the value of Kp at 500 K?
Ch.15 - Chemical Equilibrium
Chapter 15, Problem 154
Acetic acid tends to form dimers, 1CH3CO2H22, because of hydrogen bonding: The equilibrium constant Kc for this reaction is 1.51 * 10^2 in benzene solution but only 3.7 * 10^-2 in water solution. Why is Kc for the water solution so much smaller than Kc for the benzene solution?
Verified step by step guidance1
Understand the concept of hydrogen bonding: Hydrogen bonds are a type of weak interaction that occurs between a hydrogen atom bonded to a highly electronegative atom (like oxygen or nitrogen) and another electronegative atom. In the case of acetic acid, hydrogen bonds can form between the hydrogen of one acetic acid molecule and the oxygen of another, leading to dimer formation.
Consider the role of the solvent: Benzene is a non-polar solvent, which does not interfere with hydrogen bonding between acetic acid molecules. This allows the dimers to form more readily, resulting in a higher equilibrium constant (Kc) for the dimerization reaction in benzene.
Analyze the effect of water as a solvent: Water is a polar solvent and can form hydrogen bonds with acetic acid molecules. This interaction competes with the hydrogen bonding between acetic acid molecules, reducing the tendency for dimer formation. Consequently, the equilibrium constant (Kc) for the dimerization reaction in water is lower.
Relate the equilibrium constant to the extent of reaction: The equilibrium constant (Kc) is a measure of the extent to which a reaction proceeds to form products. A higher Kc in benzene indicates that the reaction proceeds more towards dimer formation compared to in water, where the reaction is less favorable due to competing interactions.
Summarize the impact of solvent polarity: The difference in Kc values between benzene and water solutions highlights the impact of solvent polarity on chemical equilibria. Non-polar solvents like benzene favor dimer formation due to minimal interference, while polar solvents like water reduce dimer formation due to competing hydrogen bonding interactions.
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Hydrogen Bonding
Hydrogen bonding is a strong type of dipole-dipole interaction that occurs when a hydrogen atom covalently bonded to a highly electronegative atom, such as oxygen or nitrogen, experiences an attraction to another electronegative atom. In the case of acetic acid, hydrogen bonds facilitate the formation of dimers, where two acetic acid molecules associate. The strength and prevalence of these interactions can significantly influence the physical properties and behavior of substances in different solvents.
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Equilibrium Constant (Kc)
The equilibrium constant (Kc) quantifies the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. A higher Kc value indicates a greater tendency for products to form, while a lower Kc suggests that reactants are favored. In this context, the difference in Kc values for acetic acid in benzene versus water reflects how solvent properties affect the stability of the dimer formation.
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Solvent Effects on Dimerization
The nature of the solvent can greatly influence the interactions between solute molecules. In nonpolar solvents like benzene, acetic acid dimers are stabilized by hydrogen bonding without significant competition from solvent interactions. Conversely, in polar solvents like water, strong hydrogen bonding between water molecules can disrupt the dimer formation of acetic acid, leading to a lower Kc value. This illustrates how solvent polarity and hydrogen bonding capacity can affect chemical equilibria.
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Related Practice
Textbook Question
The equilibrium constant Kc for the gas-phase thermal decomposition of cyclopropane to propene is 1.0 * 105 at 500 K:
(c) Can you alter the ratio of the two concentrations at equilibrium by adding cyclopropane or by decreasing the volume of the container? Explain.
Textbook Question
Acetic acid tends to form dimers, (CH3CO2H2), because of hydrogen bonding: The equilibrium constant Kc for this reaction is 1.51⨉102 in benzene solution but only 3.7⨉10-2 in water solution. (a) Calculate the ratio of dimers to monomers for 0.100 M acetic acid in benzene.
Textbook Question
Acetic acid tends to form dimers, (CH3CO2H2), because of hydrogen bonding: The equilibrium constant Kc for this reaction is 1.51⨉102 in benzene solution but only 3.7⨉10-2 in water solution. (b) Calculate the ratio of dimers to monomers for 0.100 M acetic acid in water.