Problem 160 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction, \(2 \mathrm{NO}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{NOCl}\) The following mechanism has been proposed \(\mathrm{NO}+\rightleftharpoons \mathrm{Cl} \quad \mathrm{NOCl}_{2}\) (fast) \(\mathrm{NOCl}_{2}+\mathrm{NO} \longrightarrow 2 \mathrm{NOCl}\) (slow) (a) Rate \(=\mathrm{k}[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right]^{2}\) (b) Rate \(=\mathrm{k}[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right]\) (c) Rate \(=\mathrm{k}[\mathrm{NOCl}]^{2}\) (d) Rate \(=\mathrm{k}\left[\mathrm{NOCl}_{2}\right][\mathrm{NO}]\)

Step-by-Step Solution

Answer

(b) Rate \( = k [\mathrm{NO}]^2 [\mathrm{Cl}_2] \)

Step 1: Understanding the Reaction Mechanism

The reaction involves two steps. The first step is the equilibrium: \( \mathrm{NO} + \mathrm{Cl}_2 \rightleftharpoons \mathrm{NOCl}_2 \) (fast). The second step is the rate-determining step: \( \mathrm{NOCl}_2 + \mathrm{NO} \rightarrow 2 \mathrm{NOCl} \) (slow). The overall reaction is formed by combining these two steps.

Step 2: Formulation of the Rate Law

The rate law for the overall reaction depends on the slow step, often called the rate-determining step. This step is \( \mathrm{NOCl}_2 + \mathrm{NO} \rightarrow 2 \mathrm{NOCl} \). The rate law based on this step is: Rate = \( k [\mathrm{NOCl}_2][\mathrm{NO}] \).

Step 3: Expressing Intermediates in Terms of Reactants

Since \( \mathrm{NOCl}_2 \) is an intermediate, its concentration must be expressed in terms of the starting reactants \( \mathrm{NO} \) and \( \mathrm{Cl}_2 \). From the fast equilibrium step: \( \mathrm{NO} + \mathrm{Cl}_2 \rightleftharpoons \mathrm{NOCl}_2 \), we have \( [\mathrm{NOCl}_2] = K [\mathrm{NO}][\mathrm{Cl}_2] \) where \( K \) is the equilibrium constant for the first step.

Step 4: Substituting Back into the Rate Law

Substitute \( [\mathrm{NOCl}_2] = K [\mathrm{NO}][\mathrm{Cl}_2] \) into the rate law from Step 2. The new rate law becomes: Rate = \( k (K [\mathrm{NO}][\mathrm{Cl}_2])[\mathrm{NO}] \), simplifying to Rate = \( k' [\mathrm{NO}]^2 [\mathrm{Cl}_2] \), where \( k' = kK \).

Step 5: Match with Given Options

Review the given options: (a) Rate = \( k [\mathrm{NO}][\mathrm{Cl}_2]^2 \), (b) Rate = \( k [\mathrm{NO}]^2 [\mathrm{Cl}_2] \), (c) Rate = \( k [\mathrm{NOCl}]^2 \), (d) Rate = \( k [\mathrm{NOCl}_2][\mathrm{NO}] \). The correct expression derived is (b) Rate = \( k [\mathrm{NO}]^2 [\mathrm{Cl}_2] \).

Key Concepts

Understanding the Rate-Determining Step

In chemical reactions, the rate-determining step (RDS) is crucial. Think of it as the bottleneck of a reaction—a slower process limiting the overall speed of the reaction sequence. It is the hurdle that the reaction 'trips' over, making it decisive in determining the reaction rate.

In the NO and Cl2 reaction given, the slow step is identified as the rate-determining step. Here, the formation of 2 NOCl from NOCl2 and NO is painfully slow. This means it principally influences how fast the overall reaction will occur. Importantly, the rate law for the reaction can be derived from this rate-determining step. This is because at this step, whatever happens determines how quickly the reactants turn into products.

To sum it up, the rate-determining step defines the rate law which, for our reaction, was ultimately found to be Rate = k [NO]^2 [Cl2]. By understanding which step limits progress, we gain insight into controlling and predicting the behavior of chemical reactions.

Role of Intermediate Species

Intermediate species in a reaction mechanism are curious entities. They form during a reaction but do not appear in the final products. They simply 'disappear' as the reaction progresses. Nevertheless, they play a pivotal role, especially in multi-step reactions.

In the NO and Cl2 reaction, NOCl2 acts as an intermediate. Although not present in the initial reactants or final products, it forms temporarily. Its fleeting existence is crucial—allowing the reactants to finally transform into the desired products. But here's the challenge: intermediates are not stable enough to measure directly during reactions.

To manage this, scientists express intermediate concentrations in terms of reactants we can control. This is necessary to link the mechanism and rate law properly. For our reaction, the equilibrium of the fast step helped express [NOCl2] in terms of [NO] and [Cl2], thus facilitating the derivation of the rate law applicable to the reaction. Hence, while intermediates are transient, their impact on the reaction mechanism is far from trivial.

Understanding Equilibrium Constant

In a multi-step reaction mechanism, some steps reach an equilibrium rather quickly. The equilibrium constant (K) is a valuable tool that helps us quantify the ratio of reactants to products at this state of balance. It's like a map showing stable points at which the forward and backward reactions balance out.

For our reaction, the fast equilibrium step involves NO and Cl2 forming NOCl2. The equilibrium constant K relates these concentrations through the expression [NOCl2] = K [NO][Cl2]. Understanding K is quintessential for converting intermediate species into expressions manageable through reactants.

Equilibrium constants help bind the initial phase of reactions with rate laws, particularly when intermediates are involved. For instance, in deriving the rate law: by substituting the NOCl2 equilibrium expression into the rate-determining step’s rate law, we reached a more standard expression suitable for experimentation and observation (in our example, the rate law we derived was Rate = k' [NO]^2 [Cl2]). The beauty of the equilibrium constant lies in its ability to connect transient mechanistic details to tangible quantities we can measure and influence.