Problem 78 - Chemistry: Principles and Reactions

Question

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?

Step-by-Step Solution

Answer

Answer: Yes, the proposed mechanism is consistent with the given rate expression, as both the rate expressions match: \(k[\mathrm{NO}]^2[\mathrm{H}_{2}]\).

Step 1: Identify the rate-determining step

In the proposed mechanism, we can see that the second step is labeled as slow. This means that this step is the rate-determining step: $$\mathrm{N}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{2}\mathrm{O}+\mathrm{N}_{2} \mathrm{O}$$

Step 2: Write the rate expression for the rate-determining step

Since the rate-determining step is the slowest step, the rate expression for the reaction is determined by this step. So the rate for this step would be: $$\text{rate} = k_{2}[\mathrm{N}_{2}\mathrm{O}_{2}][\mathrm{H}_{2}]$$ Where \(k_{2}\) is the rate constant for the second step.

Step 3: Find the equilibrium expression for the fast equilibrium step

Since the first step is a fast equilibrium step, we can write an equilibrium expression for it: $$K_{eq} = \frac{[\mathrm{N}_{2}\mathrm{O}_{2}]}{[\mathrm{NO}]^2}$$ Where \(K_{eq}\) is the equilibrium constant for the first step. Now, we need to solve for \([\mathrm{N}_{2}\mathrm{O}_{2}]\): $$[\mathrm{N}_{2}\mathrm{O}_{2}] = K_{eq}[\mathrm{NO}]^2$$

Step 4: Substitute the expression for \([\mathrm{N}_{2}\mathrm{O}_{2}]\) into the rate expression

Now, we substitute the expression for \([\mathrm{N}_{2}\mathrm{O}_{2}]\) from Step 3 into the rate expression from Step 2: $$\text{rate} = k_{2}[\mathrm{N}_{2}\mathrm{O}_{2}][\mathrm{H}_{2}] = k_{2}(K_{eq}[\mathrm{NO}]^2)[\mathrm{H}_{2}]$$

Step 5: Simplify the rate expression

Next, let's simplify the rate expression: $$\text{rate} = k_{2}K_{eq}[\mathrm{NO}]^2[\mathrm{H}_{2}]$$ Now, we can define a new rate constant, \(k = k_{2}K_{eq}\). So, we get the following rate expression for the proposed mechanism: $$\text{rate} = k[\mathrm{NO}]^2[\mathrm{H}_{2}]$$

Step 6: Compare the given rate expression with the overall rate expression for the proposed mechanism

We are given the rate expression as rate \(=k[\mathrm{NO}]^{2}\times\left[\mathrm{H}_{2}\right]\). The rate expression we derived from the proposed mechanism is also \(k[\mathrm{NO}]^2[\mathrm{H}_{2}]\). Since both expressions match, the proposed mechanism is consistent with the given rate expression.

Key Concepts

Rate-Determining Step

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.

Rate Expression

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.

Equilibrium Constant

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.