Problem 142 - A Complete Resource Book in Chemistry for JEE Main

Question

Reaction \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}\) is completed according to the following mechanism. \(\mathrm{A}_{2} \rightleftharpoons 2 \mathrm{~A}\) \(\mathrm{A}+\mathrm{B}_{2} \rightarrow \mathrm{AB}+\mathrm{B} \quad\) (slow step) \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{AB}\) The order of reaction is (a) 1 (b) \(3 / 2\) (c) \(1 / 2\) (d) 2

Step-by-Step Solution

Answer

The order of the reaction is \( \frac{3}{2} \) (Option b).

Step 1: Identify the Slow Step

The slow step in a reaction mechanism is the rate-determining step, which controls the overall reaction rate. For this mechanism, the slow step is given as \( \mathrm{A} + \mathrm{B}_{2} \rightarrow \mathrm{AB} + \mathrm{B} \).

Step 2: Write the Rate Expression for the Slow Step

The rate law depends only on the species that participate in the rate-determining step. From the slow step, the rate can be expressed as \( \text{Rate} = k[\mathrm{A}][\mathrm{B}_{2}] \), where \( k \) is the rate constant.

Step 3: Include Equilibrium Step Contributions

Since \( \mathrm{A}_{2} \rightleftharpoons 2 \mathrm{A} \) is at equilibrium, the concentration of \( \mathrm{A} \) is related to \( [\mathrm{A}_{2}] \) by the equilibrium condition. Generally, \( [\mathrm{A}] = K[\mathrm{A}_{2}]^{1/2} \), where \( K \) is the equilibrium constant.

Step 4: Substitute and Simplify Rate Law

Substitute \( [\mathrm{A}] = K[\mathrm{A}_{2}]^{1/2} \) into the rate expression from the slow step: \( \text{Rate} = k(K[\mathrm{A}_{2}]^{1/2})[\mathrm{B}_{2}] = kK[\mathrm{A}_{2}]^{1/2}[\mathrm{B}_{2}] \). Simplifying gives the form of the rate law \( \text{Rate} \propto [\mathrm{A}_{2}]^{1/2}[\mathrm{B}_{2}] \).

Step 5: Determine the Order of Reaction

The overall order of a reaction is the sum of the powers of the concentration terms in the rate law. In this case, it is the sum of the exponents in \( [\mathrm{A}_{2}]^{1/2}[\mathrm{B}_{2}]^{1} \), which is \( \frac{1}{2} + 1 = \frac{3}{2} \).

Key Concepts

Reaction Mechanism

In chemistry, understanding the reaction mechanism is crucial. It describes the step-by-step sequence of elementary reactions by which overall chemical change occurs. A mechanism provides insights into how molecules interact and transform.
For example, in the given reaction of \(\mathrm{A}_{2} + \mathrm{B}_{2} \rightarrow 2 \mathrm{AB}\), the mechanism consists of multiple steps. These include an initial equilibrium step \(\mathrm{A}_{2} \rightleftharpoons 2 \mathrm{~A}\) and subsequent reactions \(\mathrm{A} + \mathrm{B}_{2} \rightarrow \mathrm{AB} + \mathrm{B}\) and \(\mathrm{A} + \mathrm{B} \rightarrow \mathrm{AB}\).
This set of reactions shows how \(\mathrm{A}_2\) dissociates into \(\mathrm{A}\) atoms, which then engage in further reactions with \(\mathrm{B}_2\) and \(\mathrm{B}\) species to form \(\mathrm{AB}\). Each elementary step has its own rate and contributes to the overall reaction path, helping chemists understand the full reaction process.

Rate-determining Step

The rate-determining step is the slowest step in a reaction mechanism, much like the bottleneck in a production line. It determines the overall speed or rate of the chemical reaction because subsequent steps cannot proceed faster than this step.
For the reaction \(\mathrm{A}_{2} + \mathrm{B}_{2} \rightarrow 2 \mathrm{AB}\), the rate-determining step is identified as \(\mathrm{A} + \mathrm{B}_{2} \rightarrow \mathrm{AB} + \mathrm{B}\). This step is crucial because it most limits the formation rate of the product \(\mathrm{AB}\).
In analyzing reaction mechanisms, identifying the rate-determining step allows us to derive the rate law by focusing on the concentrations of the reactants involved, as their concentrations directly affect the overall rate of reaction.

Equilibrium in Chemical Reactions

Equilibrium plays a significant role in reaction mechanisms, especially when reversible reactions are involved. At equilibrium, the rates of forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products.
In our original exercise, the step \(\mathrm{A}_{2} \rightleftharpoons 2 \mathrm{~A}\) is at equilibrium. This affects how the concentration of \(\mathrm{A}\) is expressed. Specifically, the concentration of \(\mathrm{A}\) is linked to \(\mathrm{A}_{2}\) by the equilibrium condition, often represented as \([\mathrm{A}] = K[\mathrm{A}_{2}]^{1/2}\), where \(K\) is the equilibrium constant.
This relationship is crucial for deriving the rate law because it allows us to relate the concentrations of species present at equilibrium to those participating in the rate-determining step.

Rate Law Expression

The rate law expression is a mathematical representation of the reaction rate in terms of the concentration of reactants. It is determined experimentally and can be derived from the reaction mechanism, especially the rate-determining step.
For the given reaction, from the rate-determining step \(\mathrm{A} + \mathrm{B}_{2} \rightarrow \mathrm{AB} + \mathrm{B}\), the rate law is initially expressed as \(\text{Rate} = k[\mathrm{A}][\mathrm{B}_{2}]\), with \(k\) representing the rate constant. With the equilibrium consideration \([\mathrm{A}] = K[\mathrm{A}_{2}]^{1/2}\), it becomes \(\text{Rate} = kK[\mathrm{A}_{2}]^{1/2}[\mathrm{B}_{2}]\).
This formula highlights how the rate of reaction depends on the concentrations of \(\mathrm{A}_2\) and \(\mathrm{B}_2\), and helps us understand the overall reaction order, which in this case is \(\frac{3}{2}\). Understanding the rate law expression is essential for predicting how a change in reactant concentration will affect the reaction rate.