Question

Consider the following reaction, which takes place in a single elementary step: $$2 \mathrm{~A}+\mathrm{B} \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftarrows}} \mathrm{A}_{2} \mathrm{~B}$$ If the equilibrium constant \(K_{\mathrm{c}}\) is 12.6 at a certain temperature and if \(k_{1}=5.1 \times 10^{-2} \mathrm{~s}^{-1},\) calculate the value of \(k_{-1}\).

Short answer

\(k_{-1} \approx 4.05 \times 10^{-3} \text{ s}^{-1}\)

Step-by-step solution

Understand the reaction and equilibrium constant

The reaction given is an equilibrium reaction where the rate constant for the forward reaction is \(k_1\), and for the reverse reaction is \(k_{-1}\). The equilibrium constant at this given temperature is \(K_c = 12.6\). This constant expresses the ratio of the rate of the forward reaction to the rate of the reverse reaction.

Apply the expression for equilibrium constant

The equilibrium constant \(K_c\) is defined as the ratio of the rate constant of the forward reaction \(k_1\) to the rate constant of the reverse reaction \(k_{-1}\). The equation can be written as: \[ K_c = \frac{k_1}{k_{-1}} \] Substitute the given values into this equation.

Substitute and solve for \(k_{-1}\)

Substitute \(K_c = 12.6\) and \(k_1 = 5.1 \times 10^{-2} \text{ s}^{-1}\) into the expression: \[ 12.6 = \frac{5.1 \times 10^{-2}}{k_{-1}} \]To solve for \(k_{-1}\), rearrange and calculate: \[ k_{-1} = \frac{5.1 \times 10^{-2}}{12.6} \approx 4.05 \times 10^{-3} \text{ s}^{-1} \]

Key concepts

Understanding Elementary Reactions

An elementary reaction is a fundamental chemical reaction that occurs in a single step. In these processes, reactants directly convert into products, with no intermediate steps involved. This makes elementary reactions unique, as they provide insight into the molecular level of reaction mechanisms. The stoichiometry of an elementary reaction is depicted in its balanced chemical equation, which directly shows the reactants turning into products.
In the given exercise's reaction, \(2 \mathrm{A} + \mathrm{B} \rightarrow \mathrm{A}_2 \mathrm{B}\), the reaction is not decomposed into multiple steps. Instead, it signifies one straightforward molecular rearrangement. The simplicity of elementary reactions allows us to directly relate their stoichiometric coefficients to the reaction order, making the analysis more straightforward, especially when discussing rates and equilibria.

The Role of Rate Constants in Chemical Reactions

The rate constant, often denoted as \(k\), is a fundamental component in the study of chemical kinetics. It quantifies the speed of a chemical reaction under a given set of conditions. The value of the rate constant is influenced by factors such as temperature and the nature of the reactants.
There are separate rate constants for the forward and reverse reactions in reversible processes. For a forward reaction, indicated as \(k_1\), it measures how rapidly reactants are converted to products. Conversely, the reverse reaction rate constant, \(k_{-1}\), measures the speed at which products revert to reactants.
The magnitude of these constants directly influences the position of equilibrium in a reaction. For example, larger \(k_1\) compared to \(k_{-1}\) suggests a predominance of products at equilibrium. Understanding these constants helps us predict reaction behavior over time and under varying conditions.

Exploring Forward and Reverse Reactions

In chemical equilibrium, the concepts of forward and reverse reactions play a critical role. A forward reaction is the initial process where reactants turn into products. This is represented by \(k_1\) in the equation, indicating the rate at which \(2 \mathrm{A} + \mathrm{B}\) forms \(\mathrm{A}_2 \mathrm{B}\).
The reverse reaction, conversely, depicts the products decomposing back into reactants, described by the rate constant \(k_{-1}\). In the given exercise, both reactions are in balance when the system reaches equilibrium.
At equilibrium, both the forward and reverse reactions continue to occur, but at rates that make the concentrations of reactants and products constant over time. This dynamic balance is pivotal in understanding how reaction systems respond to changes, like concentration shifts or temperature variations.

Equilibrium Calculation and its Importance

Equilibrium in a chemical reaction is reached when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain stable. The equilibrium constant, \(K_c\), quantifies this balance. It is expressed as the ratio of the rate constant for the forward reaction, \(k_1\), to that of the reverse reaction, \(k_{-1}\).
In our exercise, by knowing \(K_c\) and \(k_1\), we calculated \(k_{-1}\) with the formula \( K_c = \frac{k_1}{k_{-1}}\). This highlights why equilibrium calculations are critical: they allow us to understand and predict the proportions of substances present in equilibrium states.
Moreover, such calculations aid in designing reactions for desired outcomes, optimizing production processes in industrial chemistry, and devising strategies to control reaction conditions effectively.