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Chapter 15: Problem 19

Which of the following statements are true and which are false? (a) The equilibrium constant can never be a negative number. (b) In reactions that we draw with a single-headed arrow, the equilibrium constant has a value that is very close to zero. (c) As the value of the equilibrium constant increases, the speed at which a reaction reaches equilibrium increases.

Short Answer

Expert verified
(a) True: The equilibrium constant can never be a negative number since it's calculated using the concentrations of products and reactants which are always positive values. (b) False: In reactions with single-headed arrows, the equilibrium constant has a value that is exceedingly large, not close to zero, as they are practically irreversible. (c) False: The equilibrium constant doesn't influence the speed at which a reaction reaches equilibrium. The rate of reaction is dictated by the rate constant, which depends on activation energy and temperature.

Step by step solution

01

Statement (a): The equilibrium constant can never be a negative number.

To determine the truth of this statement, recall that the equilibrium constant (K) for a reaction is calculated using the concentrations of products and reactants raised to the power of their stoichiometric coefficients in the balanced chemical equation. Since concentrations can never be negative, it's impossible for the equilibrium constant to be a negative number. Therefore, statement (a) is true.
02

Statement (b): In reactions that we draw with a single-headed arrow, the equilibrium constant has a value that is very close to zero.

Reactions drawn with a single-headed arrow are generally considered to be irreversible or practically irreversible, meaning they essentially proceed in only one direction, from reactants to products. This means that the concentration of reactants decreases to an extremely low value while the concentration of products increases until the reaction is almost complete. With little to no reactants remaining at equilibrium, the value of the equilibrium constant (K) will be exceedingly large, not close to zero. Thus, statement (b) is false.
03

Statement (c): As the value of the equilibrium constant increases, the speed at which a reaction reaches equilibrium increases.

The equilibrium constant (K) is related to the concentrations of products and reactants at equilibrium and does not provide information about the speed at which equilibrium is reached. The rate of the reaction is dictated by the rate constant, which depends on the activation energy and temperature. Therefore, there is no direct correlation between the equilibrium constant value and the speed at which a reaction reaches equilibrium. Statement (c) is false. In summary: - Statement (a) is true. - Statement (b) is false. - Statement (c) is false.

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Most popular questions from this chapter

For $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), K_{p}=3.0 \times 10^{2}\( at \)700 \mathrm{~K}$. In a 2.00-L vessel, the equilibrium mixture contains \(1.17 \mathrm{~g}\) of \(\mathrm{SO}_{3}\) and \(0.105 \mathrm{~g}\) of \(\mathrm{O}_{2}\). How many grams of \(\mathrm{SO}_{2}\) are in the vessel?

(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 \(k_{r} ?\)

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.

Consider the following equilibrium, for which \(K_{p}=7.62\) at $480^{\circ} \mathrm{C}:$ $$2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g)$$ (a) What is the value of \(K_{p}\) for the reaction $4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) ?$ (b) What is the value of \(K_{p}\) for the reaction $\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?$ (c) What is the value of \(K_{c}\) for the reaction in part (b)?

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?
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