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Chapter 15: Problem 12

Consider the reaction \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\). Assume that both the forward reaction and the reverse reaction are elementary processes and that the value of the equilibrium constant is very large. (a) Which species predominate at equilibrium, reactants or products? (b) Which reaction has the larger rate constant, the forward or the reverse? Explain.

Short Answer

Expert verified
(a) The products, C and D, predominate at equilibrium since the value of the equilibrium constant (K) is very large. (b) The forward reaction has the larger rate constant because a large K value means the forward rate constant (kf) is much larger than the reverse rate constant (kr).

Step by step solution

01

Relationship between equilibrium constant and concentrations

The equilibrium constant (K) for a given reaction can be expressed in terms of concentrations of reactants and products at equilibrium as follows: \[K = \frac{[\mathrm{C}]_{eq} [\mathrm{D}]_{eq}}{[\mathrm{A}]_{eq} [\mathrm{B}]_{eq}}\] Here, A and B are reactants, and C and D are products.
02

Determine the species that predominate at equilibrium

We are given that the value of the equilibrium constant is very large. If the value of K is large, this means that the ratio of products to reactants is high. In other words, the concentrations of C and D (products) are much higher than the concentrations of A and B (reactants) at equilibrium. Therefore, products predominate at equilibrium. Answer (a): The products, C and D, predominate at equilibrium.
03

Relationship between rate constants and equilibrium constant

We are given that both forward and reverse reactions are elementary processes. The rate constants of the forward (kf) and reverse (kr) reactions are related to the equilibrium constant as follows: \[K = \frac{k_f}{k_r}\]
04

Determine the reaction with the larger rate constant

We already know that the equilibrium constant (K) is very large. Since K is the ratio of the forward and reverse rate constants, a large value of K means that the forward rate constant (kf) is much larger than the reverse rate constant (kr). Hence, the forward reaction has the larger rate constant. Answer (b): The forward reaction has the larger rate constant.

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Most popular questions from this chapter

When the following reactions come to equilibrium, does the equilibrium mixture contain mostly reactants or mostly products? (a) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) ; K_{c}=1.5 \times 10^{-10}\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) ; K_{p}=2.5 \times 10^{9}\)

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)\). An equilibrium mix- ture in a 2.00-L vessel is found to contain \(0.0406 \mathrm{~mol}\) \(\mathrm{CH}_{3} \mathrm{OH}, 0.170 \mathrm{~mol} \mathrm{CO}\), and \(0.302 \mathrm{~mol} \mathrm{H}_{2}\) at \(500 \mathrm{~K}\). Cal- culate \(K_{c}\) at this temperature.

Consider the equilibrium $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \operatorname{NOBr}(g) $$ Calculate the equilibrium constant \(K_{p}\) for this reaction, given the following information (at \(298 \mathrm{~K}\) ): $$ \begin{aligned} 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) & \rightleftharpoons 2 \mathrm{NOBr}(g) & K_{c}=2.0 \\ 2 \mathrm{NO}(g) & \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K_{c}=2.1 \times 10^{30} \end{aligned} $$

As shown in Table 15.2, the equilibrium constant for the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is \(K_{p}=\) \(4.34 \times 10^{-3}\) at \(300^{\circ} \mathrm{C}\). Pure \(\mathrm{NH}_{3}\) is placed in a 1.00-L flask and allowed to reach equilibrium at this temperature. There are \(1.05 \mathrm{~g} \mathrm{NH}_{3}\) in the equilibrium mixture. (a) What are the masses of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) in the equilibrium mixture? (b) What was the initial mass of ammonia placed in the vessel? (c) What is the total pressure in the vessel?

The equilibrium \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) is established at \(500 \mathrm{~K}\). An equilibrium mixture of the three gases has partial pressures of \(0.095 \mathrm{~atm}, 0.171 \mathrm{~atm}\), and \(0.28\) atm for \(\mathrm{NO}, \mathrm{Cl}_{2}\), and \(\mathrm{NOCl}\), respectively. Calculate \(K_{p}\) for this reaction at \(500 \mathrm{~K}\).
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