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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?

Short Answer

Expert verified
(a) At equilibrium, the products (C and D) predominate due to the very large equilibrium constant (K), which favors the forward reaction. (b) The forward reaction has a larger rate constant than the reverse reaction, as implied by the very large equilibrium constant (K).

Step by step solution

01

Recall the relationship between equilibrium constant and rate constants

The equilibrium constant (K) for a reaction is related to the rate constants of forward (kf) and reverse (kr) reactions as follows: \[K = \frac{k_f}{k_r}\] Since in this problem, we have been provided with the value of equilibrium constant (K) only, and it is very large, we need to use this information to deduce the concentrations of reactants and products at equilibrium and to compare the rate constants.
02

Determine the predominant species at equilibrium

If K is very large, that means the forward reaction is much more favored over the reverse reaction. In other words, the equilibrium lies heavily towards the products side, i.e., C and D. Therefore, the species that predominate at equilibrium are the products (C and D).
03

Compare rate constants of forward and reverse reactions

Since the equilibrium constant (K) is given by the ratio of the forward and reverse rate constants, a very large K indicates that the forward rate constant (kf) for the reaction must be much larger than the reverse rate constant (kr). Thus, the forward reaction has a larger rate constant than the reverse reaction. To summarize, (a) Products (C and D) predominate at equilibrium. (b) The forward reaction has a larger rate constant than the reverse reaction.

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Key Concepts

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

Understanding Chemical Equilibrium
Chemical equilibrium is a fundamental concept in the study of chemical reactions, referring to the state in which the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time. This balance is dynamic, meaning that reactions continue to occur, but without altering the overall concentrations of the substances involved.

In the equation \(\mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}\), when the system reaches equilibrium, the amount of reactants (A and B) and products (C and D) remain constant. The equilibrium constant (K) is a dimensionless value that quantifies the ratio of product concentrations to reactant concentrations at equilibrium, with each raised to the power of their stoichiometric coefficients. If K is very large, it implies that the concentration of products is much higher than that of reactants at equilibrium, suggesting that the forward reaction is favorable and the products will predominate.
Deciphering Rate Constants
Rate constants are crucial for understanding the kinetics of chemical reactions. They quantify the speed at which a reaction progresses and are denoted as \(k_f\) for the forward reaction and \(k_r\) for the reverse reaction in elementary processes. The magnitude of these constants determines how fast the reactants are converted to products and vice versa.

For the given equilibrium \(\mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}\), a large equilibrium constant indicates that \(k_f\) must be significantly greater than \(k_r\), as \(K = \frac{k_f}{k_r}\). The larger the value of K, the more the equilibrium is shifted towards the products, and the greater the rate at which A and B are being converted to C and D. This enables us to predict behavioral trends in reactions, such as which direction will be favored under certain conditions and how the reaction rates can be influenced by different variables.
Elementary Processes in Reactions
Elementary processes are the simple steps that comprise a complex reaction mechanism. They are single-step reactions with a specific rate law derived from the stoichiometry of the reaction. Elementary reactions can be unimolecular, involving a single molecule, bimolecular, involving two reactant species, or termolecular, involving three reacting species.

It's important to distinguish elementary processes from overall reactions since the rate laws for overall reactions are not always evident from their balanced chemical equations. Instead, rate laws for elementary steps can often be written directly from the reaction stoichiometry. In the context of the exercise, both the forward and reverse reactions are presumed to be elementary. This assumption simplifies the relationship between the equilibrium constant and the rate constants, as it suggests a direct connection without any intermediate steps influencing the rates of the forward and reverse reactions.

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Most popular questions from this chapter

For the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g), K_{c}=55.3\) at \(700 \mathrm{~K}\). In a 2.00-L flask containing an equilibrium mixture of the three gases, there are \(0.056 \mathrm{~g} \mathrm{H}_{2}\) and \(4.36 \mathrm{~g} \mathrm{} \mathrm{I}_{2}\). What is the mass of \(\mathrm{HI}\) in the flask?

Consider the following equilibrium: $$ 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g) \quad K_{c}=1.08 \times 10^{7} \text { at } 700{ }^{\circ} \mathrm{C} $$ (a) Calculate \(K_{p-}\) (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly \(\mathrm{H}_{2} \mathrm{~S}\) ? (c) Calculate the value of \(K_{c}\) if you rewrote the equation \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{~S}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)\).

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 \(298 \mathrm{~K}\) ): $$ \begin{array}{rll} 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) & \rightleftharpoons 2 \mathrm{NOBr}(g) & K_{c}=2.0 \\ 2 \mathrm{NO}(g) & \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K_{c}=2.1 \times 10^{30} \end{array} $$

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.15), \(K_{p}\) for the reaction $$ 2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) $$ is about \(1 \times 10^{-13}\). Assuming that the exhaust gas (total pressure \(1 \mathrm{~atm}\) ) contains \(0.2 \% \mathrm{CO}, 12 \% \mathrm{CO}_{2}\), and \(3 \% \mathrm{O}_{2}\) by volume, is the system at equilibrium with respect to the \(\mathrm{CO}_{2}\) reaction? Based on your conclusion, would the \(\mathrm{CO}\) concentration in the exhaust be decreased or increased by a catalyst that speeds up the \(\mathrm{CO}_{2}\) reaction? Recall that at a fixed pressure and temperature, volume \(\%=\mathrm{mol} \%\).

A mixture of \(1.374 \mathrm{~g}\) of \(\mathrm{H}_{2}\) and \(70.31 \mathrm{~g}\) of \(\mathrm{Br}_{2}\) is heated in a \(2.00\)-L vessel at \(700 \mathrm{~K}\). These substances react according to $$ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) $$ At equilibrium, the vessel is found to contain \(0.566 \mathrm{~g}\) of \(\mathrm{H}_{2}\). (a) Calculate the equilibrium concentrations of \(\mathrm{H}_{2}, \mathrm{Br}_{2}\), and \(\mathrm{HBr}\). (b) Calculate \(K_{\text {. }}\).

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