Problem 176 - A Complete Resource Book in Chemistry for JEE Main

Question

The following mechanism proposed for a reaction \(2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{D}+\mathrm{E}\) is as \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}\) (fast) \(\mathrm{A}+\mathrm{C} \rightarrow \mathrm{E}+\mathrm{D}\) (slow) The order of reaction is

Step-by-Step Solution

Answer

The overall reaction order is 3.

Step 1: Identify the Slow Step

The mechanism consists of two steps: \( \text{A} + \text{B} \rightarrow \text{C} \) (fast) and \( \text{A} + \text{C} \rightarrow \text{E} + \text{D} \) (slow). The slow step is the rate-determining step of the reaction.

Step 2: Write the Rate Law for the Slow Step

Since the second step \( \text{A} + \text{C} \rightarrow \text{E} + \text{D} \) is slow, the rate law will depend on the concentrations of \( \text{A} \) and \( \text{C} \). Thus, the rate law can be expressed as: \[ \text{Rate} = k[\text{A}][\text{C}] \]

Step 3: Express Intermediate Concentration in Terms of Reactants

\( \text{C} \) is an intermediate, formed and consumed during the reaction. From the fast equilibrium step \( \text{A} + \text{B} \rightleftharpoons \text{C} \), we assume it reaches equilibrium quickly. Thus, we can write: \[ [\text{C}] = k'[\text{A}][\text{B}] \] where \( k' \) is the equilibrium constant for the formation of \( \text{C} \).

Step 4: Substitute Back into the Rate Law

Substitute \( [\text{C}] = k'[\text{A}][\text{B}] \) into the rate law from Step 2: \[ \text{Rate} = k[\text{A}](k'[\text{A}][\text{B}]) = k k'[\text{A}]^2 [\text{B}] \]

Step 5: Determine Overall Reaction Order

The rate law \( \text{Rate} = k k'[\text{A}]^2 [\text{B}] \) indicates that the reaction order is the sum of the exponents: 2 with respect to \( \text{A} \) and 1 with respect to \( \text{B} \). Therefore, the overall order of the reaction is 2 + 1 = 3.

Key Concepts

Reaction Mechanism

A reaction mechanism is like a roadmap for a chemical reaction. It outlines the series of steps through which reactants transform into products. In our example, the mechanism consists of two main steps:

• First, a fast step \( \text{A} + \text{B} \to \text{C} \).
• Second, a slow step \( \text{A} + \text{C} \to \text{E} + \text{D} \).

The most important thing in a mechanism is identifying the slowest step, which acts like a bottleneck. We can think of this as the step that dictates how fast the entire reaction can proceed. It's often called the rate-determining step because it limits the speed of the reaction, similar to how the slowest person in a race sets the pace for everyone else. This step is critical, as it impacts the rate at which products are formed and ultimately shapes the entire reaction process.

Rate Law

The rate law expresses the relationship between the rate of a chemical reaction and the concentration of reactants. Specifically, for a multistep reaction, the rate law is determined from the rate-determining step. In the given mechanism, the rate-determining step is \(\text{A} + \text{C} \to \text{E} + \text{D}\). Thus, the rate law can be expressed as:\[\text{Rate} = k[\text{A}][\text{C}]\]Here, \(k\) is the rate constant, which is a specific value for a given reaction at a particular temperature. The rate law helps us understand how changes in concentrations affect the reaction rate. Knowing this allows chemists to control reactions, making them proceed faster or slower, depending on the desired outcome. It therefore plays a vital role in industries like pharmaceuticals and materials science, where precise reaction control is crucial.

Intermediate Concentration

In chemical reactions, intermediates are substances formed in one step and consumed in another. They are not present in the final product but are crucial for the reaction pathway. In this exercise, \(\text{C}\) is an intermediate.Since the first step \(\text{A} + \text{B} \to \text{C}\) is fast and reaches equilibrium quickly, we can relate the concentration of the intermediate to the reactants. This is achieved through an equation involving the equilibrium constant \(k'\):\[[\text{C}] = k'[\text{A}][\text{B}]\]Plugging this expression back into the rate law allows us to describe the rate in terms of only the initial reactants, \(\text{A}\) and \(\text{B}\), which can be directly measured or controlled. By understanding these concepts, students can predict how reactions will behave under different conditions, a powerful tool when designing experiments or industrial processes.

Reaction Order

The reaction order provides insight into the dependency of the reaction rate on concentration changes of each reactant. It is determined by adding up the exponents of the concentration terms in the rate law.From the modified rate law, \(\text{Rate} = kk'[\text{A}]^2[\text{B}]\), the exponents tell us the reaction order:

• The order with respect to \(\text{A}\) is 2, since its concentration is squared.
• The order with respect to \(\text{B}\) is 1.

Therefore, the overall order of the reaction is 3 (2 for \(\text{A}\) plus 1 for \(\text{B}\)). This tells us how the rate will change as concentrations of reactants change.Reaction order is crucial for understanding how sensitive a reaction is to changes in concentration. If you double \([\text{A}]\), the rate increases by a factor of four due to its square relationship. Understanding reaction order helps in predicting the behavior of reactions under different circumstances, which is essential in both laboratory and industrial settings.