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Chapter 14: Problem 95

In a basic aqueous solution, chloromethane undergoes a substitution reaction in which CI is replaced by \(\mathrm{OH}\) : $$ \mathrm{CH}_{3} \mathrm{Cl}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cl}^{-}(a q) $$ The equilibrium constant \(K_{c}\) is \(1 \times 10^{16} .\) Calculate the equilibrium concentrations of \(\mathrm{CH}_{3} \mathrm{Cl}, \mathrm{CH}_{3} \mathrm{OH}, \mathrm{OH}^{-}\), and \(\mathrm{Cl}^{-}\) in a solu- tion prepared by mixing equal volumes of \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{Cl}\) and \(0.2\) M \(\mathrm{NaOH}\). (Hint: In defining \(x\), assume that the reaction goes \(100 \%\) to completion, and then take account of a small amount of the reverse reaction.)

Short Answer

Expert verified
The equilibrium concentrations are \([\mathrm{CH}_3\mathrm{Cl}] = 2.5 \times 10^{-17} \ \mathrm{M}\), \([\mathrm{CH}_3\mathrm{OH}] = 0.05 \ \mathrm{M}\), \([\mathrm{OH}^-] = 0.05 \ \mathrm{M}\), \([\mathrm{Cl}^-] = 0.05 \ \mathrm{M}\).

Step by step solution

01

Set Up the Reaction Initial Conditions

Initially, the reaction is set up by mixing equal volumes of 0.1 M \( \mathrm{CH}_3 \mathrm{Cl} \) and 0.2 M \( \mathrm{NaOH} \). Since the volumes are equal, the concentrations are halved, resulting in initial concentrations of 0.05 M for \( \mathrm{CH}_3 \mathrm{Cl} \) and 0.1 M for \( \mathrm{OH}^- \).
02

Assume Full Reaction Completion

Assume the reaction goes to completion. \( \mathrm{CH}_3 \mathrm{Cl} \) converts to \( \mathrm{CH}_3 \mathrm{OH} \) and \( \mathrm{Cl}^- \). Hence, the \(\mathrm{CH}_3\mathrm{Cl}\) and \(\mathrm{OH}^-\) would both be reduced by their initial \(\mathrm{CH}_3\mathrm{Cl}\) concentration, which is 0.05 M.
03

Determine Final Concentrations at Completion

After full reaction to completion, the concentrations will be: \( [\mathrm{CH}_3\mathrm{Cl}] = 0 \), \( [\mathrm{CH}_3\mathrm{OH}] = 0.05 \mathrm{M} \), \( [\mathrm{OH}^-] = 0.05 \mathrm{M} \), and \( [\mathrm{Cl}^-] = 0.05 \mathrm{M} \).
04

Establish Equilibrium Conditions

Now consider a small reverse reaction where some \( x \) amount of \( \mathrm{CH}_3\mathrm{OH} \) converts back to \( \mathrm{CH}_3\mathrm{Cl} \). Equilibrium concentrations become \( [\mathrm{CH}_3\mathrm{Cl}] = x \), \( [\mathrm{CH}_3\mathrm{OH}] = 0.05 - x \), \( [\mathrm{OH}^-] = 0.05 - x \), and \( [\mathrm{Cl}^-] = 0.05 + x \).
05

Apply the Equilibrium Constant Expression

Using the equilibrium expression \( K_c = \frac{[\mathrm{CH}_3\mathrm{OH}][\mathrm{Cl}^-]}{[\mathrm{CH}_3\mathrm{Cl}][\mathrm{OH}^-]} = 1 \times 10^{16} \), substitute the equilibrium concentrations: \( \frac{(0.05-x)(0.05+x)}{x(0.05-x)} = 1 \times 10^{16} \).
06

Solve for \( x \)

Given the large \( K_c \), assume that \( x \ll 0.05 \), simplifying the equation to \( \frac{0.05^2}{0.05x} = 1 \times 10^{16} \). Solving gives \( x = 2.5 \times 10^{-17} \).
07

Calculate Equilibrium Concentrations

Substitute \( x = 2.5 \times 10^{-17} \) back into the equilibrium conditions to find \([\mathrm{CH}_3\mathrm{Cl}] = 2.5 \times 10^{-17} \ \mathrm{M}\), \([\mathrm{CH}_3\mathrm{OH}] = 0.05 - 2.5 \times 10^{-17} \approx 0.05 \ \mathrm{M}\), \([\mathrm{OH}^-] = 0.05 - 2.5 \times 10^{-17} \approx 0.05 \ \mathrm{M}\), and \([\mathrm{Cl}^-] = 0.05 + 2.5 \times 10^{-17} \approx 0.05 \ \mathrm{M}\).

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Key Concepts

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

Substitution Reaction
In the context of chemistry, a substitution reaction involves the replacement of one atom or group of atoms in a molecule with another. This type of reaction is common in organic chemistry.
In the given scenario, chloromethane (\(\text{CH}_3\text{Cl}\)) undergoes a substitution reaction with hydroxide ions (\(\text{OH}^-\)). In this process:
  • The chlorine atom in chloromethane is replaced by the hydroxide ion.
  • This results in the formation of methanol (\(\text{CH}_3\text{OH}\)) and a chloride ion (\(\text{Cl}^-\)).
This transformation can be represented by the equation:\[\text{CH}_3\text{Cl} + \text{OH}^- \leftrightarrow \text{CH}_3\text{OH} + \text{Cl}^-\]Such reactions are characterized by their mechanism, often relying on the presence of nucleophiles (like OH鈦� in this case) that "attack" the substrate (the organic compound).
Understanding substitution reactions is crucial for synthesizing new organic compounds and studying reaction pathways.
Equilibrium Concentrations
Equilibrium in chemical reactions is a state where the forward and reverse reactions occur at the same rate, resulting in constant concentrations of reactants and products. In a substitution reaction such as the one described, equilibrium determines how much of the chloromethane converts fully into methanol and chloride ions and how much might revert back.
The equilibrium constant (\(K_c\)) is a key factor in defining this balance. In our equation, this constant is extremely large (\(1 \times 10^{16}\)), indicating a strong favoring of products:
  • The larger the\(K_c\)value, the more the reaction lies to the right, favoring formation of methanol and chloride ions.
  • At equilibrium, the concentrations of reactants and products don鈥檛 change, but can be computed by taking the small reverse reaction into account.
As evident in the solution above,
by recognizing how the equilibrium constant represents the extent of reaction, and using it to solve for x, one determines the equilibrium concentrations of the substances involved.
Aqueous Solution
An aqueous solution is a solution where water is the solvent. The term "aqueous" comes from 'aqua' which means water. In this exercise, the substitution reaction occurs in such a solvent medium, which affects how reactants and ions interact.
Characteristics of an aqueous solution that impact reactions include:
  • Water's ability to dissolve a wide range of substances due to its polar nature, enabling the mixing of reactants like chloromethane and hydroxide ions.
  • The mobility of ions in water facilitates faster reaction rates, especially in substitution reactions where ions such as \(\text{Cl}^-\) and \(\text{OH}^-\) play a critical role.
In our case, this means that when equal volumes of chloromethane and sodium hydroxide are mixed, their initial concentrations are halved.
This sets up the stage for equilibrium as the reactants interact in the aqueous medium. Understanding these solutions is fundamental for studying reactions that occur in biological systems, as most are aqueous in nature.

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Most popular questions from this chapter

For the decomposition reaction \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\) \(\mathrm{Cl}_{2}(g), K_{\mathrm{p}}=381\) at \(600 \mathrm{~K}\) and \(K_{c}=46.9\) at \(700 \mathrm{~K}\) (a) Is the reaction endothermic or exothermic? Explain. Does your answer agree with what you would predict based on bond energies? (b) If \(1.25 \mathrm{~g}\) of \(\mathrm{PCl}_{5}\) is introduced into an evacuated \(0.500 \mathrm{~L}\) flask at \(700 \mathrm{~K}\) and the decomposition reaction is allowed to reach equilibrium, what percent of the \(\mathrm{PCl}_{5}\) will decompose and what will be the total pressure in the flask? (c) Write electron-dot structures for \(\mathrm{PCl}_{5}\) and \(\mathrm{PCl}_{3}\), and indicate whether these molecules have a dipole moment. Explain.

Gaseous indium dihydride is formed from the elements at elevated temperature: $$ \ln (g)+\mathrm{H}_{2}(g) \rightleftharpoons \ln \mathrm{H}_{2}(g) \quad K_{\mathrm{p}}=1.48 \text { at } 973 \mathrm{~K} $$ Partial pressures measured in a reaction vessel are: \(P_{\mathrm{n}}=0.0600 \mathrm{~atm}, P_{\mathrm{H}_{2}}=0.0350 \mathrm{~atm}, P_{\mathrm{nH}_{2}}=0.0760 \mathrm{~atm}\) (a) Calculate \(Q_{p}\), and determine the direction of reaction to attain equilibrium. (b) Determine the equilibrium partial pressures of all the gases.

An equilibrium mixture of \(\mathrm{O}_{2}, \mathrm{SO}_{2}\), and \(\mathrm{SO}_{3}\) contains equal concentrations of \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3} .\) Calculate the concentration of \(\mathrm{O}_{2}\) if \(K_{c}=2.7 \times 10^{2}\) for the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\) \(\rightleftharpoons 2 \mathrm{SO}_{3}(g) .\)

At \(100^{\circ} \mathrm{C}, K_{c}=4.72\) for the reaction \(2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g) .\) An empty \(10.0 \mathrm{~L}\) flask is filled with \(4.60 \mathrm{~g}\) of \(\mathrm{NO}_{2}\) at \(100^{\circ} \mathrm{C}\). What is the total pressure in the flask at equilibrium?

When air is heated at very high temperatures in an automobile engine, the air pollutant nitric oxide is produced by the reaction $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad \Delta H^{\circ}=+182.6 \mathrm{~kJ} $$ How does the equilibrium amount of NO vary with an increase in temperature?
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