For the reaction $$2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ the experimental rate expression is rate \(=k[\mathrm{NO}]^{2} \times\left[\mathrm{H}_{2}\right] .\) The following mechanism is proposed: $$\begin{array}{cc}2 \mathrm{NO} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} & \text { (fast) } \\ \mathrm{N}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{N}_{2} \mathrm{O} & \text { (slow) } \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} & \text { (fast) } \end{array}$$ Is this mechanism consistent with the rate expression?

Understanding the rate-determining step (RDS) is crucial in the field of chemical kinetics. The RDS is the slowest step in a reaction mechanism, and it acts as a bottleneck, limiting the overall rate at which a reaction can proceed. In essence, the RDS governs the kinetics of the entire reaction, making its identification a key aspect in understanding and predicting the behavior of chemical processes.

In the context of our textbook exercise, we identify the second step of the proposed mechanism as the rate-determining step because it is labeled as 'slow'. This informs us that regardless of how quickly the other steps occur, the overall reaction cannot proceed faster than this slowest step. By analyzing the individual steps of a reaction, we gain insight into why the RDS impacts the rate so significantly and can adjust conditions to optimize the reaction's efficiency.

The rate expression, also known as the rate equation or rate law, is a mathematical description of how the reaction rate depends on the concentration of each reactant. It typically includes a rate constant (\( k \)), which is specific to the reaction at a given temperature, and the concentrations of the reactants raised to their respective orders, which indicate the stoichiometric coefficients in a balanced chemical equation.

Our exercise solution demonstrates how to derive a rate expression based on the rate-determining step of a reaction mechanism. We then use the equilibrium constant (\( K_{eq} \)) to express the concentration of an intermediate in terms of the concentrations of stable reactants. This refinement of the rate expression is key to matching it with the experimentally determined rate law. Understanding how to correctly write and interpret a rate expression enables students to connect the theoretical aspects of reaction mechanisms with observable kinetic data.

The equilibrium constant (\( K_{eq} \)) is a dimensionless number that expresses the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their respective stoichiometric coefficients in the balanced equation. It is a fixed value at a given temperature and provides insights into the extent of a reaction and the relative concentrations of reactants and products at equilibrium.

In relation to our exercise, we see that the equilibrium constant is used to derive the concentration of an intermediate (\( [\textrm{N}_2\textrm{O}_2] \)), which is essential for determining the rate expression of the rate-determining step. It's crucial to understand that while the equilibrium constant is involved in the kinetic analysis, it itself is not a measure of the rate of reaction but rather the position of equilibrium. This distinction helps clear up common misconceptions students might have about the role of equilibrium in dynamic chemical processes.