Problem 165 - A Complete Resource Book in Chemistry for JEE Main

Question

The hypothetical reaction, \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}\) follows the following mechanism: \(\mathrm{A}_{2}=\mathrm{A}+\mathrm{A} \quad\) (fast) \(\mathrm{A}+\mathrm{B}_{2} \longrightarrow \mathrm{AB}+\mathrm{B} \quad\) (slow) \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{AB} \quad\) (fast) The order of the overall reaction is: (a) 1 (b) 2 (c) \(3 / 2\) (d) 0

Step-by-Step Solution

Answer

The order of the reaction is \(3/2\).

Step 1: Identify the Rate-Determining Step

Examine the given reaction mechanism and determine which step is the slowest. The second step, \( \mathrm{A} + \mathrm{B}_{2} \longrightarrow \mathrm{AB} + \mathrm{B} \), is the slowest and thus the rate-determining step.

Step 2: Write the Rate Law for the Slow Step

For the rate-determining step, the rate of the reaction is determined by the concentrations of the reactants involved in that step. Therefore, the rate law based on the step \( \mathrm{A} + \mathrm{B}_{2} \longrightarrow \mathrm{AB} + \mathrm{B} \) is \( \text{Rate} = k[\mathrm{A}][\mathrm{B}_2] \).

Step 3: Express Intermediate in Terms of Reactants

Since \( \mathrm{A} \) is an intermediate, we need to express its concentration in terms of the initial reactants. From the first step, \( \mathrm{A}_2 \rightarrow 2\mathrm{A} \), we assume it is fast to establish equilibrium, implying \( [\mathrm{A}] = K[\mathrm{A}_2]^{1/2} \), where \( K \) is the equilibrium constant.

Step 4: Substitute Intermediate Conc.

Substitute \( [\mathrm{A}] = K[\mathrm{A}_2]^{1/2} \) into the rate law from Step 2. This gives \( \text{Rate} = k' [\mathrm{A}_2]^{1/2} [\mathrm{B}_2] \), where \( k' = kK \).

Step 5: Determine Reaction Order

The overall order is the sum of the powers of the concentrations in the rate law. From \( \text{Rate} = k'[\mathrm{A}_2]^{1/2}[\mathrm{B}_2] \), the reaction is first order in \( [\mathrm{B}_2] \) and half-order in \( [\mathrm{A}_2] \). Thus, the overall order is \( \frac{1}{2} + 1 = \frac{3}{2} \).

Key Concepts

Rate-Determining Step

In the world of reaction mechanisms, the rate-determining step serves as the critical bottleneck. Envision it like the slow walker in a group that everyone else has to wait for. This particular step dramatically influences how quickly the entire reaction can occur.
For the reaction \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}\), three steps are outlined. Among them, the second step \( \mathrm{A} + \mathrm{B}_{2} \longrightarrow \mathrm{AB} + \mathrm{B} \) is noted to be the slowest. This step, therefore, is designated as the rate-determining step.

• Recognizing the rate-determining step is pivotal. It not only limits the reaction speed but also directly impacts the form of the rate law.
• All other fast steps are assumed to promptly reach equilibrium, setting up conditions for the slow step to proceed at the observed rate.

Now, equipped with this understanding, we can imagine the progression of particles in the reaction, critically analyzing the slow path that shapes the overall reaction speed.

Rate Law

The rate law is a mathematical expression that gives us insight into how the concentration of reactants affects the reaction rate. It is exclusively based on the rate-determining step since that is the tempo-setter for the sequence.
In the given example, after identifying the slow step \( (\mathrm{A} + \mathrm{B}_{2} \longrightarrow \mathrm{AB} + \mathrm{B}) \), the rate law is structured to reflect the concentrations of the reactants involved. Thus, it is expressed as: \[ \text{Rate} = k[\mathrm{A}][\mathrm{B}_2] \]

• Here, \( k \) is the rate constant, a measure of reaction speed.
• The concentrations \( [\mathrm{A}] \) and \( [\mathrm{B}_2] \) directly affect the rate when altered, emphasizing the pivotal role each reactant plays in speed determination.

Hence, the rate law provides a window into the dynamics of the reaction, quantifying how modifications in reactant concentrations can influence the overall rate.

Reaction Order

Understanding reaction order demands a breakdown of how each reactant influences the reaction rate. It is essentially the overall power to which reactant concentrations are raised in the rate law.
This links back to our rate law: \[ \text{Rate} = k'[\mathrm{A}_2]^{1/2}[\mathrm{B}_2] \]

• The term \( [\mathrm{A}_2]^{1/2} \) signifies a half-order dependence, indicating that doubling \( [\mathrm{A}_2] \) increases the rate by \( \sqrt{2} \).
• The term \( [\mathrm{B}_2] \) is straightforward first order, implying the rate doubles when \( [\mathrm{B}_2] \) is doubled.

The overall reaction order becomes the sum of these individual orders, computed as \[ \frac{1}{2} + 1 = \frac{3}{2} \].
This fractional order reveals complexities in the reaction dynamics, dictated by the presence of intermediates and non-integer dependencies.

Intermediate Concentration

Intermediates are like the unsung heroes in a chemical reaction. They're not part of the reaction's starting materials or end products, but they crucially bridge the gap between reactant molecules and products.
In the outlined mechanism, \( \mathrm{A} \) is an intermediate, transiently forming and subsequently reacting away.

• For the rate law involving \( [\mathrm{A}] \), we can't measure it directly. Instead, we express it using the equilibrium established in the fast steps preceding the rate-determining one.
• Using the fast equilibrium \( \mathrm{A}_2 \longrightarrow 2\mathrm{A} \), we derive \( [\mathrm{A}] = K[\mathrm{A}_2]^{1/2} \), coupling it with known reactant concentrations.

This approach allows us to incorporate the intermediate's role without needing to measure it directly, effectively simplifying complex reaction pathways in the pursuit of a viable rate equation.