Problem 211 - A Complete Resource Book in Chemistry for JEE Main

Question

The following mechanism has been proposed for the reaction of \(\mathrm{NO}\) with \(\mathrm{Br}_{2}\) to form NO Br \(\mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NOBr}_{2}(\mathrm{~g})\) \(\mathrm{NOBr}_{2}(\mathrm{~g})+\mathrm{NO}(\mathrm{g}) \longrightarrow 2 \mathrm{NOBr}(\mathrm{g})\) If the second step is the rate determining step, the order of the reaction with respect to \(\mathrm{NO}(\mathrm{g})\) is (a) 1 (b) 0 (c) 3 (d) 2

Step-by-Step Solution

Answer

The order of the reaction with respect to \(\mathrm{NO}\) is 2.

Step 1: Identify the Rate Determining Step

The second step, \(\mathrm{NOBr}_{2} + \mathrm{NO} \to 2\ \mathrm{NOBr}\), is given as the rate-determining step. This means it is the slowest step and limits the overall rate of the reaction.

Step 2: Determine the Rate-Law Expression for the Rate Determining Step

The rate law depends on the stoichiometry of the rate-determining step. For the step \(\mathrm{NOBr}_{2} + \mathrm{NO} \to \), the rate can be expressed as \(\text{Rate} = k[\mathrm{NOBr}_{2}][\mathrm{NO}]\). The reaction is first order with respect to both \(\mathrm{NOBr}_{2}\) and \(\mathrm{NO}\).

Step 3: Analyze the Formation of Intermediates

The intermediate \(\mathrm{NOBr}_{2}\) is formed in the first step \(\mathrm{NO} + \mathrm{Br}_{2} \rightleftharpoons \mathrm{NOBr}_{2}\). Since \(\mathrm{NOBr}_{2}\) does not appear in the overall reaction, its concentration needs to be expressed in terms of the reactants. From the equilibrium of the first step, \([\mathrm{NOBr}_{2}] = K_1 [\mathrm{NO}][\mathrm{Br}_{2}]\).

Step 4: Substitute the Intermediate Concentration

Substitute \([\mathrm{NOBr}_{2}]\) in the rate law of the rate-determining step with its expression in terms of \([\mathrm{NO}]\) and \([\mathrm{Br}_{2}]\):\[\text{Rate} = k(K_1 [\mathrm{NO}][\mathrm{Br}_{2}])[\mathrm{NO}] = kK_1 [\mathrm{NO}]^2[\mathrm{Br}_{2}]\]. This shows that the reaction is second order with respect to \(\mathrm{NO}\).

Step 5: Determine the Reaction Order with Respect to \(\text{NO}\)

The overall rate law, \(\text{Rate} = kK_1 [\mathrm{NO}]^2 [\mathrm{Br}_{2}]\), indicates that the reaction is second order with respect to \(\text{NO}\), because the exponent of \([\mathrm{NO}]\) is 2.

Key Concepts

Rate Determining Step

In chemical kinetics, the rate-determining step is a crucial concept. It refers to the slowest step in a reaction mechanism, which essentially governs the rate at which the overall reaction occurs. It's similar to a bottleneck in a traffic jam—no matter how fast the preceding cars move, the pace is controlled by the slowest car in front.
In our given reaction, the second step is identified as the rate-determining step, which involves \( \mathrm{NOBr}_{2} + \mathrm{NO} \rightarrow 2 \mathrm{NOBr} \). This step happens more slowly than the first. Here, the production of \( \mathrm{NOBr} \) is limited by the rate at which \( \mathrm{NOBr}_{2} \) reacts with \( \mathrm{NO} \). This directly influences the overall rate of the reaction.
The identification of this step is important because the rate law, describing how the reaction rate depends on the concentration of reactants, is derived from this slowest step. Thus, understanding which step is the slowest is key to predicting the reaction behavior.

Rate Law Expression

The rate law expression is vital in understanding how the concentrations of reactants affect the rate of a chemical reaction. Derived from the rate-determining step, it provides a mathematical expression to predict the reaction rate.

• For the rate-determining step \( \mathrm{NOBr}_{2} + \mathrm{NO} \rightarrow 2 \mathrm{NOBr} \), the rate law expression is given by \( \text{Rate} = k[\mathrm{NOBr}_{2}][\mathrm{NO}] \).
• This expression indicates that the reaction is first-order with respect to both \( \mathrm{NOBr}_{2} \) and \( \mathrm{NO} \) because each is raised to the power of one.
• Thus, any change in the concentration of \( \mathrm{NOBr}_{2} \) or \( \mathrm{NO} \) will directly affect the rate.

It's important to note that changes in concentration lead to proportional changes in the rate. For instance, doubling the concentration of \( \mathrm{NO} \) while keeping \( \mathrm{NOBr}_{2} \) constant will double the reaction rate. Deriving the rate law expression accurately is crucial for predicting how a reaction behaves under different conditions.

Chemical Intermediates

In reaction mechanisms, chemical intermediates are species that are formed in one step and consumed in another. They don't appear in the overall balanced equation for the reaction but play a critical role in the reaction process.
In this exercise, \( \mathrm{NOBr}_{2} \) is the chemical intermediate. It forms in the first step \( \mathrm{NO} + \mathrm{Br}_{2} \rightleftharpoons \mathrm{NOBr}_{2} \), and then reacts further in the rate-determining step.
Because intermediates like \( \mathrm{NOBr}_{2} \) are not part of the starting reactants or final products, their concentrations need to be expressed in terms of the stable molecules present—usually the original reactants or products. From the first step's equilibrium expression, we can express the concentration of \( \mathrm{NOBr}_{2} \) as \( [\mathrm{NOBr}_{2}] = K_1 [\mathrm{NO}][\mathrm{Br}_{2}] \). This allows us to substitute it back into the rate law for the rate-determining step to accurately depict how the concentration of intermediates affects the overall rate.
Understanding chemical intermediates provides insight into the mechanism's dynamics and helps in crafting precise rate laws.