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: \(\mathrm{A}_{2}=\mathrm{A}+\mathrm{A}\) (fast) \(\mathrm{A}+\mathrm{B}_{2} \longrightarrow \mathrm{AB}+\mathrm{B}\) (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 overall reaction order is \( \frac{3}{2} \); the answer is (c).

Step 1: Identify the Rate-Determining Step

The rate-determining step is the slowest step in the reaction mechanism. In this mechanism, it is the step \( \mathrm{A} + \mathrm{B}_{2} \rightarrow \mathrm{AB} + \mathrm{B} \). The rate of the overall reaction is dependent on this step.

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

For the rate-determining step \( \mathrm{A} + \mathrm{B}_{2} \rightarrow \mathrm{AB} + \mathrm{B} \), the rate law can be expressed as \( \text{Rate} = k[\mathrm{A}][\mathrm{B}_{2}] \). This is because the rate of reaction depends on the concentration of the reactants involved in this step.

Step 3: Express the Intermediate Concentration

\( \mathrm{A} \) is an intermediate formed in the fast first step \( \mathrm{A}_{2} \rightarrow \mathrm{A} + \mathrm{A} \) and consumed in the second step. We use the pre-equilibrium assumption: \( K = \frac{[\mathrm{A}]^2}{[\mathrm{A}_{2}]} \), where \( K \) is the equilibrium constant. Solving for \( [\mathrm{A}] \), we find \( [\mathrm{A}] = K [\mathrm{A}_{2}]^{1/2} \).

Step 4: Substitute Intermediate Concentration in Rate Law

Substitute \( [\mathrm{A}] \) into the rate law: \( \text{Rate} = k K^{1/2} [\mathrm{A}_{2}]^{1/2} [\mathrm{B}_{2}] \). This expression reflects the rate law accounting for the intermediate.

Step 5: Determine the Overall Reaction Order

The overall order of the reaction is the sum of the powers of the concentration terms in the rate law: \( \text{order} = \frac{1}{2} + 1 = \frac{3}{2} \). This is the total order considering both \( [\mathrm{A}_{2}]^{1/2} \) and \( [\mathrm{B}_{2}] \).

Key Concepts

Rate-Determining Step

In a complex reaction involving multiple steps, the rate-determining step is crucial in understanding the overall reaction rate. It is the slowest step of the reaction mechanism and typically acts as a bottleneck for the process. This step's pace dictates how quickly the entire reaction can proceed. In our given mechanism, the slowest step is \( \mathrm{A} + \mathrm{B}_{2} \rightarrow \mathrm{AB} + \mathrm{B} \). Since it is the slowest, it plays a major role in determining the rate of the overall reaction.
To identify the rate-determining step, look for the step characterized as slow in the reaction mechanism. Understanding this concept is vital, as it helps chemists manipulate and optimize conditions to accelerate reactions, particularly important in industrial and laboratory settings.

Reaction Order

The reaction order provides insight into how the concentration of reactants affects the rate of a chemical reaction. It is determined by summing the exponents of the concentration terms in the rate law expression. In the example provided, the rate law derived from the rate-determining step is \( \text{Rate} = k[\mathrm{A}][\mathrm{B}_{2}] \).
The reaction order for the individual components is:

• For \( \mathrm{A} \), the order is 1, as shown by the exponent of \([\mathrm{A}]\).
• For \( \mathrm{B}_{2} \), the order is 1, shown by the exponent of \([\mathrm{B}_{2}]\).

To find the overall reaction order, add these individual orders together:
\[ \text{Overall Order} = 1 + 1 = 2 \]However, after substituting and simplifying using intermediate concentrations, the order becomes \( \frac{3}{2} \). Reaction order can provide important information for predicting how changes in concentration can influence the reaction speed.

Intermediate Concentration

Intermediates are substances produced in one step of a reaction mechanism and consumed in a subsequent step. They do not appear in the overall chemical equation of the reaction. In the given mechanism, \( \mathrm{A} \) is formed quickly in the reaction \( \mathrm{A}_{2} \rightarrow \mathrm{A} + \mathrm{A} \) and consumed in the slow step.
To manage these intermediates in reaction equations, we use the pre-equilibrium assumption. It involves treating the formation and consumption of an intermediate as at equilibrium, allowing us to express its concentration in terms of the reactants or products' concentrations.
This yields:

• \[ K = \frac{[\mathrm{A}]^2}{[\mathrm{A}_{2}]} \]
• Solving for \([\mathrm{A}]\), we get \([\mathrm{A}] = K^{1/2} [\mathrm{A}_{2}]^{1/2}\).

This concentration expression is crucial since it directly enters our modified rate law, ensuring intermediate terms do not appear in the final rate law expression.

Rate Law

The rate law is a mathematical description that defines the relationship between the reaction rate and the concentrations of reactants. It is derived based on the rate-determining step, excluding the concentrations of intermediates. For our mechanism, the rate law initially is:

• \[ \text{Rate} = k[\mathrm{A}][\mathrm{B}_{2}] \]

However, \( \mathrm{A} \) is an intermediate, hence the need to substitute its value from the equilibrium expression. After substitution, the rate law becomes:

• \[ \text{Rate} = k K^{1/2} [\mathrm{A}_{2}]^{1/2} [\mathrm{B}_{2}] \]

Rate law expressions are pivotal because they help in predicting the change in reaction rates with varying concentrations. The final expression also helps determine the overall reaction order, indicating theoretical and practical insights into reaction management.