Question

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

Short answer

The reaction order is 2.

Step-by-step solution

Understand the Mechanism

The given reaction mechanism breaks down the formation of 2 AB via smaller steps. Note that the reaction specifies a slow step, essential for determining the rate law.

Determine the Rate-Determining Step

The rate-determining step is given as \( \mathrm{A} + \mathrm{B}_{2} \rightarrow \mathrm{AB} + \mathrm{B} \). This is the slow step and controls the reaction rate.

Write the Rate Law Based on the Slow Step

The rate law is determined by the rate-determining step. From the step \( \mathrm{A} + \mathrm{B}_{2} \rightarrow \mathrm{AB} + \mathrm{B} \), the rate law is \( \text{rate} = k [\mathrm{A}][\mathrm{B}_{2}] \).

Express Intermediate Concentrations

To express concentrations using initial reactants, note: \( \mathrm{A}_{2} \rightarrow 2 \mathrm{A} \), so \( [\mathrm{A}] = 2[\mathrm{A}_{2}] \). Substitute in the rate law to obtain \( \text{rate} = k (2[\mathrm{A}_{2}])[\mathrm{B}_{2}] \).

Simplify and Determine Reaction Order

Simplify the rate law: \( \text{rate} = 2k [\mathrm{A}_{2}][\mathrm{B}_{2}] \). The order with respect to \( \mathrm{A}_{2} \) is 1 and with respect to \( \mathrm{B}_{2} \) is 1, so the overall reaction order is 2.

Key concepts

Rate-Determining Step

In a multi-step reaction mechanism, not all steps occur at the same speed. The slowest step is called the rate-determining step because it controls the overall reaction rate. Picture a conveyor belt where several parts need to be assembled. If one station is slower than others, it holds back the overall production—similarly, in reactions, the slowest step limits how fast products can be generated.
In the provided exercise, the reaction \( \mathrm{A} + \mathrm{B}_{2} \rightarrow \mathrm{AB} + \mathrm{B} \) is identified as the slow step. This means that any modification in the reactants' concentration in this step significantly affects the reaction speed. Recognizing the rate-determining step allows chemists to focus on which intermediates to measure and control to influence the reaction rate effectively.

Rate Law

The rate law is a mathematical expression that relates the speed of a reaction to the concentration of the reactants. It usually takes the form of \( \text{rate} = k [ ext{Reactant 1}]^m [ ext{Reactant 2}]^n \), where \( k \) is the rate constant, and \( m \) and \( n \) are the orders of the reaction with respect to each reactant.
To derive the rate law from a mechanism, focus on the rate-determining step, because it dictates the overall rate. In our exercise, the rate law derived from \( \mathrm{A} + \mathrm{B}_{2} \rightarrow \mathrm{AB} + \mathrm{B} \) is \( \text{rate} = k [\mathrm{A}][\mathrm{B}_{2}] \). This indicates that the reaction rate increases as the concentrations of \( \mathrm{A} \) and \( \mathrm{B}_{2} \) increase. The exponents (1 in both cases here) reflect the stoichiometry and order deduced from the slow step.

Reaction Order

The reaction order gives insight into how reactant concentrations affect the reaction rate. It is the sum of the powers to which reactant concentrations are raised in the rate law. The overall reaction order is crucial for predictions and understanding reaction dynamics.
For our given reaction, \( \mathrm{A} + \mathrm{B}_{2} \rightarrow \mathrm{AB} + \mathrm{B} \), the exponents from the rate law \( \text{rate} = 2k [\mathrm{A}_{2}][\mathrm{B}_{2}] \) each contribute to the reaction order. Here, both \( \mathrm{A}_{2} \) and \( \mathrm{B}_{2} \) have an order of 1, summing to an overall reaction order of 2. This means if you double the concentration of either reactant, the rate of reaction doubles.

Intermediate Concentrations

Intermediates are species that appear in the reaction mechanism but not in the overall chemical equation. They are transient and do not accumulate, which complicates their use in rate laws directly. Thus, their concentrations must often be expressed in terms of initial reactants.
In the exercise, \( \mathrm{A} \) is an intermediate generated from \( \mathrm{A}_{2} \rightarrow 2 \mathrm{A} \). To express the concentration of \( \mathrm{A} \) in terms of \( \mathrm{A}_{2} \), note: \( [\mathrm{A}] = 2[\mathrm{A}_{2}] \). This substitution is crucial for expressing the rate law correctly. By using the initial concentrations, chemists bypass the direct measurement of intermediate concentrations, simplifying the application and analysis of rate laws effectively.