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Chapter 15: Problem 69

(a) Is the dissociation of fluorine molecules into atomic fluorine, \(\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{~F}(g)\) an exothermic or endothermic process? (b) If the temperature is raised by \(100 \mathrm{~K}\), does the equilibrium constant for this reaction increase or decrease? (c) If the temperature is raised by \(100 \mathrm{~K},\) does the forward rate constant \(k_{f}\) increase by a larger or smaller amount than the reverse rate constant \(k_{r} ?\)

Short Answer

Expert verified
(a)The dissociation of fluorine molecules into atomic fluorine is an endothermic process. (b)If the temperature is raised by 100 K, the equilibrium constant for this reaction will increase. (c)If the temperature is raised by 100 K, the forward rate constant (kf) increases by a larger amount than the reverse rate constant (kr).

Step by step solution

01

Identify the nature of the reaction

Since breaking chemical bonds requires energy, dissociation of fluorine molecules into atomic fluorine requires an input of energy. Therefore, the given reaction is an endothermic process. Answer (a): The dissociation of fluorine molecules into atomic fluorine is an endothermic process.
02

Analyze the effect of a temperature change on the equilibrium constant

According to Le Ch芒telier's principle, if we increase the temperature of an endothermic reaction, the equilibrium will shift towards the endothermic side (the right side) in order to absorb the added energy. As the reaction shifts to the right, the concentration of products (atomic fluorine) will increase, leading to an increase in the equilibrium constant. Answer (b): If the temperature is raised by 100 K, the equilibrium constant for this reaction will increase.
03

Analyze the effect of a temperature change on the forward and reverse rate constants

According to the Arrhenius equation, the rate constants of both forward and reverse reactions will increase with an increase in temperature: \(k = Ae^{-\frac{Ea}{RT}}\) Where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. As the temperature increases, the exponent term becomes smaller, and k increases. However, the increase in the forward rate constant (kf) will be greater than that of the reverse rate constant (kr) since the forward reaction (an endothermic process) better absorbs the added thermal energy than the reverse reaction (an exothermic process). Answer (c): If the temperature is raised by 100 K, the forward rate constant (kf) increases by a larger amount than the reverse rate constant (kr).

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Key Concepts

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

Equilibrium Constant
The equilibrium constant (\( K \)) is a mathematical expression that quantifies the ratio of concentrations of products to reactants at equilibrium for a reversible reaction. \[K = \frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}}\]The value of \( K \) depends on temperature:
  • A high \( K \) indicates more products are present compared to reactants, meaning equilibrium favors the formation of products.
  • A low \( K \) signifies more reactants are present than products, implying the equilibrium leans towards the reactants.
In an endothermic reaction, like the dissociation of fluorine molecules into atomic fluorine, increasing the temperature shifts the equilibrium towards the products. This shift occurs because the system absorbs heat, favoring the endothermic side according to Le Ch芒telier's principle. This results in an increased value of the equilibrium constant.
Rate Constants
Rate constants characterize how fast a reaction proceeds. They are represented by \( k_f \) for the forward reaction and \( k_r \) for the reverse reaction. These constants are impacted by the following factors:
  • Temperature: Higher temperatures typically increase both \( k_f \) and \( k_r \) due to molecules moving faster and overcoming activation energy barriers more easily.
  • Activation Energy: According to the Arrhenius equation, reactions with a lower activation energy have larger rate constants because reactants convert to products more readily.
For the given endothermic reaction, when the temperature increases by 100 K, both \( k_f \) and \( k_r \) increase. However, \( k_f \) will increase to a greater extent than \( k_r \). This is because the endothermic forward reaction benefits more from the added heat compared to the exothermic reverse reaction.
Le Ch芒telier's Principle
Le Ch芒telier's principle describes how a system at equilibrium responds to external changes. It states that if an external condition, such as temperature, pressure, or concentration, is altered, the system will adjust to minimize that change and re-establish equilibrium.
In the context of an endothermic reaction:
  • Temperature Increase: For the reaction \( \mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{F}(g) \), raising the temperature causes the system to handle the additional heat by shifting towards the products, thus absorbing the added energy.
  • Equilibrium Shift: As the equilibrium shifts to favor the formation of atomic fluorine, the equilibrium constant \( K \) increases, reflecting the enhanced concentration of products.
This principle helps understand why certain reactions might favor products or reactants under different conditions and is a powerful tool for predicting the outcome of changes in reaction conditions.

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Most popular questions from this chapter

A flask is charged with \(152.0 \mathrm{kPa}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(101.3 \mathrm{kPa}\) \(\mathrm{NO}_{2}(g)\) at \(25^{\circ} \mathrm{C},\) and the following equilibrium is achieved: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ After equilibrium is reached, the partial pressure of \(\mathrm{NO}_{2}\) is \(51.9 \mathrm{kPa}\). (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?(\mathbf{b})\) Calculate the value of \(K_{p}\) for the reaction. (c) Calculate \(K_{c}\) for the reaction.

The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually \(\left.\mathrm{H}_{2} \mathrm{SO}_{4}\right)\). A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{solv})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{solv}) & \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\text { solv })+\mathrm{H}_{2} \mathrm{O}(\text { solv }) \end{aligned}$$ where "(solv)" indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\) is 6.68 . A pharmaceutical chemist makes up \(15.0 \mathrm{~L}\) of a solution that is initially \(0.275 \mathrm{M}\) in acetic acid and \(3.85 \mathrm{M}\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

Mercury(I) oxide decomposes into elemental mercury and elemental oxygen: \(2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\). (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibriumconstant expression in terms of molarities for the reaction, using (solv) to indicate solvation.

If \(K_{c}=0.013 \mathrm{~L} / \mathrm{mol}\) for \(2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)\) at \(1000 \mathrm{~K}\), what is the value of \(K_{p}\) for this reaction at this temperature?

The protein hemoglobin (Hb) transports \(\mathrm{O}_{2}\) in mammalian blood. Each Hb can bind \(4 \mathrm{O}_{2}\) molecules. The equilibrium constant for the \(\mathrm{O}_{2}\) binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygen-binding capacity, biochemists use a measure called the \(P 50\) value, defined as the partial pressure of oxygen at which \(50 \%\) of the protein is saturated. Fetal hemoglobin has a \(\mathrm{P} 50\) value of \(2.53 \mathrm{kPa},\) and adult hemoglobin has a P50 value of \(3.57 \mathrm{kPa}\). Use these data to estimate how much larger \(K_{c}\) is for the aqueous reaction \(4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \rightleftharpoons\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right]\) in a fetus, compared to \(K_{c}\) for the same reaction in an adult.
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