Consider the following elementary steps for a consecutive reaction: $$ \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{C} $$ (a) Write an expression for the rate of change of \(\mathrm{B}\). (b) Derive an expression for the concentration of \(\mathrm{B}\) under steady- state conditions; that is, when \(\mathrm{B}\) is decomposing to \(\mathrm{C}\) at the same rate as it is formed from \(A\).

Understanding the steady-state condition is crucial for grasping the dynamics of consecutive reactions in chemistry. The steady-state assumption applies when a reactive intermediate, such as substance B in the given reaction sequence, is formed and decomposed at equal rates, resulting in a constant concentration of this intermediate over time.

In the context of our exercise, steady-state conditions are achieved when the formation rate of B from A (first reaction) is exactly balanced by the decomposition rate of B to C (second reaction). Mathematically, this is denoted as the point at which the rate of change of B, or \( \frac{dB}{dt} \), is zero. At this point, despite ongoing reactions where B is both produced and consumed, the concentration of B remains unchanged because the input and output fluxes of B are equal.

The application of the steady-state condition simplifies complex reaction mechanisms by reducing the number of variables involved, making it easier to analyze and calculate the behavior of reactive intermediates.

The rate of change in chemistry refers to how fast a reactant is consumed or a product is formed over time. For a given chemical species, the rate of change is essential for predicting how the concentration of that species will evolve throughout the reaction.

In a consecutive reaction where A turns into B and then B into C, the rate at which B changes (\( \frac{dB}{dt} \) in our exercise) is a reflection of two competing processes: the formation of B from A, and its subsequent decomposition into C. By accounting for both the appearance and disappearance of B, we arrive at a comprehensive understanding of its kinetics, which is vital for controlling and optimizing chemical processes.

Furthermore, the accurate prediction of reaction rates is not only fundamental in academic exercises but also has practical applications in industries such as pharmaceuticals, where reaction speed can impact the quality and yield of product formation.

At the heart of our exercise lies the law of mass action, which governs the relationship between the reactants' concentrations and the reaction's rate. According to this law, the rate of a chemical reaction at a constant temperature is proportional to the product of the concentrations of the reactants, each raised to a power equal to their stoichiometric coefficients in the balanced chemical equation.

The rate at which B is formed from A (\(k_1[A]\)) and the rate at which B is transformed into C (\(k_2[B]\)) are both expressed in accordance with the law of mass action. When applying the law to a consecutive reaction, as shown in the exercise, it provides a quantitative framework for understanding the kinetics involved and developing the rate equations that describe the system's behavior.

Thus, the law of mass action is not only a cornerstone of chemical kinetics but also a powerful tool that aids in the design and analysis of chemical reactions. Through proper application of this law, chemists can predict reaction rates and optimize conditions for desired outcomes in a variety of chemical processes.