/*! 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 116 Consider the gas-phase hydration... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the gas-phase hydration of hexafluoroacetone, \(\left(\mathrm{CF}_{3}\right)_{2} \mathrm{CO}=\) At \(76^{\circ} \mathrm{C}\), the forward and reverse rate constants are \(k_{\mathrm{f}}=0.13 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) and \(k_{\mathrm{r}}=6.2 \times 10^{-4} \mathrm{~s}^{-1}\). What is the value of the equilibrium constant \(K_{c} ?\)

Short Answer

Expert verified
The equilibrium constant \( K_c \) is approximately 209.68.

Step by step solution

01

Identify the Reaction

The reaction involves the hydration of hexafluoroacetone. We have forward rate constant \( k_f \) and reverse rate constant \( k_r \). The equilibrium constant \( K_c \) can be calculated using these rate constants.
02

Use the Formula for Equilibrium Constant

The equilibrium constant \( K_c \) in terms of rate constants for a reaction is given by the formula: \[ K_c = \frac{k_f}{k_r} \] where \( k_f \) and \( k_r \) are the forward and reverse rate constants, respectively.
03

Plug in the Rate Constants

Substitute \( k_f = 0.13 \, \text{M}^{-1} \text{s}^{-1} \) and \( k_r = 6.2 \times 10^{-4} \, \text{s}^{-1} \) into the equation:\[ K_c = \frac{0.13}{6.2 \times 10^{-4}} \]
04

Calculate the Equilibrium Constant

Perform the division: \[ K_c = \frac{0.13}{6.2 \times 10^{-4}} \approx 209.68 \]
05

Interpret the Result

The equilibrium constant \( K_c \) is approximately 209.68, indicating the equilibrium position favors the formation of products.

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Key Concepts

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

Rate Constants
Rate constants are crucial in understanding how fast a chemical reaction proceeds. These constants, denoted as \(k\), are specific to a given reaction at a certain temperature. For reversible reactions, we have two important rate constants: the forward rate constant \( k_f \), and the reverse rate constant \( k_r \). The forward rate constant tells us how quickly the reactants are converted into products. Conversely, the reverse rate constant indicates the speed at which products revert to reactants.

The units of the rate constants differ depending on the reaction order:
  • For a first-order reaction, the unit is \(s^{-1}\).
  • For a second-order reaction, like our hydration of hexafluoroacetone case, it typically uses \(M^{-1}s^{-1}\).
The magnitude of these constants provides insight into reaction dynamics, where larger values indicate faster reactions. By comparing these constants, we can foresee which direction the reaction is more likely to proceed.
Equilibrium Constant Kc
The equilibrium constant, \( K_c \), is a number that expresses the ratio of products to reactants at equilibrium for a reaction. It gives a snapshot of a reaction's status at equilibrium.

In chemical reactions, \( K_c \) can be derived using the rate constants from the forward and reverse reactions. This relationship is computed with the formula: \[ K_c = \frac{k_f}{k_r} \]In our example, with \( k_f = 0.13 \, M^{-1} s^{-1} \) and \( k_r = 6.2 \times 10^{-4} \, s^{-1} \), substituting these into the equation gives us:\[ K_c = \frac{0.13}{6.2 \times 10^{-4}} \approx 209.68 \]

With a \( K_c \) of approximately 209.68, the equilibrium strongly favors the products, as larger values of \( K_c \) suggest that at equilibrium, there's more product than reactant. This helps chemists understand the extent of a reaction and plan accordingly for yield optimization.
Gas-Phase Reactions
Gas-phase reactions are reactions where all reactants and products are in the gaseous state. These reactions are crucial in many industrial and atmospheric processes.

One important feature of gas-phase reactions is that they are often influenced by pressure and temperature. Because gases can compress or expand significantly more than liquids or solids, the concentration of gaseous reactants and products can change if the volume or pressure changes.
This is unlike reactions in liquid or solid states, where concentrations remain relatively stable under such conditions.

Additionally, the equation for the equilibrium constant \( K_c \) in gas-phase reactions is particularly useful because gas concentrations are directly proportional to pressure, according to the ideal gas law. In the problem with hexafluoroacetone hydration, understanding these dynamics helps predict how changes in conditions might shift the equilibrium, thereby assisting in controlling and optimizing the reaction conditions for desired outcomes.

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Most popular questions from this chapter

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.)

Ethyl acetate, a solvent used in many fingernail-polish removers, is made by the reaction of acetic acid with ethanol: \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{soln})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{soln}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}(\mathrm{soln})+\mathrm{H}_{2} \mathrm{O}(\) soln \() \quad \Delta H^{\circ}=-2.9 \mathrm{~kJ}\) \(\begin{array}{lll}\text { Acetic acid } & \text { Ethanol } & \text { Ethyl acetate }\end{array}\) Does the amount of ethyl acetate in an equilibrium mixture increase or decrease when the temperature is increased? How does \(K_{c}\) change when the temperature is decreased? Justify your answers using Le Châtelier's principle.

For each of the following equilibria, write the equilibrium constant expression for \(K_{c}\) : (a) \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)\) (b) \(3 \mathrm{~F}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{ClF}_{3}(g)\) (c) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\)

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.

At \(1000 \mathrm{~K}, K_{\mathrm{p}}=2.1 \times 10^{6}\) and \(\Delta H^{\circ}=-107.7 \mathrm{~kJ}\) for the reac- tion \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)\) (a) A \(0.974\) mol quantity of \(\mathrm{Br}_{2}\) is added to a \(1.00 \mathrm{~L}\) reaction vessel that contains \(1.22 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) gas at \(1000 \mathrm{~K}\). What are the partial pressures of \(\mathrm{H}_{2}, \mathrm{Br}_{2}\), and \(\mathrm{HBr}\) at equilibrium? (b) For the equilibrium in part (a), each of the following changes will increase the equilibrium partial pressure of \(\mathrm{HBr}\). Choose the change that will cause the greatest increase in the pressure of \(\mathrm{HBr}\), and explain your choice. (i) Adding \(0.10 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) (ii) Adding \(0.10 \mathrm{~mol}\) of \(\mathrm{Br}_{2}\) (iii) Decreasing the temperature to \(700 \mathrm{~K}\).

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