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Chapter 15: Problem 13

Suppose that the gas-phase reactions \(A \longrightarrow B\) and $B \longrightarrow A\( are both elementary processes with rate constants of \)2.5 \times 10^{-2} \mathrm{~min}^{-1}\( and \)2.5 \times 10^{-1} \mathrm{~min}^{-1}$, respectively. (a) What is the value of the equilibrium constant for the equilibrium $\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) ?(\mathbf{b})$ Which is greater at equilibrium, the partial pressure of A or the partial pressure of B?

Short Answer

Expert verified
(a) The equilibrium constant (K) for the equilibrium A(g) ⇌ B(g) is 0.1. (b) At equilibrium, the partial pressure of A is greater than the partial pressure of B.

Step by step solution

01

Write down the given information

We are given the rate constants for the reactions: \(k_{1} = 2.5 \times 10^{-2} \mathrm{~min}^{-1}\) for the reaction A → B \(k_{2} = 2.5 \times 10^{-1} \mathrm{~min}^{-1}\) for the reaction B → A
02

Write down the relationship between rate constants and the equilibrium constant

For an elementary reaction, the equilibrium constant (K) can be determined from the rate constants of the forward and reverse reactions: \[K = \frac{k_{1}}{k_{2}}\]
03

Calculate the equilibrium constant K

Using the given rate constants, we can now determine the value of K: \[K = \frac{2.5 \times 10^{-2} \mathrm{~min}^{-1}}{2.5 \times 10^{-1} \mathrm{~min}^{-1}}\] \[K = \frac{1}{10} = 0.1\] The equilibrium constant (K) = 0.1.
04

Determine the partial pressures of A and B at equilibrium

At equilibrium, the ratio of partial pressures of the reactants and products is equal to the equilibrium constant K: \(\frac{P_{B}}{P_{A}} = K\) Given K < 1, this indicates that the reaction is more favorable in the reverse direction. Therefore: \(P_{B} < P_{A}\)
05

Summarize the findings

(a) The equilibrium constant (K) for the equilibrium A(g) ⇌ B(g) is 0.1. (b) At equilibrium, the partial pressure of A is greater than the partial pressure of B.

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Most popular questions from this chapter

A sample of nitrosyl bromide (NOBr) decomposes according to the equation $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g).$$ An equilibrium mixture in a 5.00-L vessel at \(100^{\circ} \mathrm{C}\) contains \(3.22 \mathrm{~g}\) of NOBr, \(3.08 \mathrm{~g}\) of \(\mathrm{NO},\) and $4.19 \mathrm{~g}\( of \)\mathrm{Br}_{2}\(. (a) Calculate \)K_{c}$. (b) What is the total pressure exerted by the mixture of gases? (c) What was the mass of the original sample of NOBr?

At \(1285^{\circ} \mathrm{C}\), the equilibrium constant for the reaction \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\) is $K_{c}=1.04 \times 10^{-3} .\( A \)1.00-\mathrm{L}$ vessel containing an equilibrium mixture of the gases has \(1.50 \mathrm{~g}\) \(\mathrm{Br}_{2}(g)\) in it. What is the mass of \(\mathrm{Br}(g)\) in the vessel?

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?

At \(120^{\circ} \mathrm{C}, K_{c}=0.090\) for the reaction $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ In an equilibrium mixture of the three gases, the concentrations of \(\mathrm{SO}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{SO}_{2}\) are $0.100 \mathrm{M}\( and \)0.075 \mathrm{M}$, respectively. What is the partial pressure of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

True or false: When the temperature of an exothermic reaction increases, the rate constant of the forward reaction decreases, which leads to a decrease in the equilibrium constant, \(K_{c}\)
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