Question

The suggested mechanism for the reaction: \(\mathrm{CHCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CCl}_{4}(\mathrm{~g})\) \(+\mathrm{HCl}(\mathrm{g})\), is $$ \mathrm{Cl}_{2} \underset{K_{2}}{\stackrel{K_{1}}{\leftrightarrows}} 2 \mathrm{C} \mathrm{C} \text { (fast) } $$ \(\mathrm{CHCl}_{3}+\mathrm{Ci} \stackrel{K_{3}}{\longrightarrow} \mathrm{HCl}+\dot{\mathrm{C}} \mathrm{Cl}_{3}(\mathrm{slow})\) $$ \dot{\mathrm{C}} \mathrm{Cl}_{3}+\mathrm{C} \mathrm{\textrm{l }} \stackrel{K_{4}}{\longrightarrow} \mathrm{CCl}_{4} \text { (fast) } $$ The experimental rate law consistent with the mechanism is (a) rate \(=K_{3}\left[\mathrm{CHCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]\) (b) rate \(=K_{4}\left[\mathrm{CCl}_{3}\right][\mathrm{Cl}]\) (c) rate \(=K_{\text {eq }}\left[\mathrm{CHCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]\) (d) rate \(=K_{3} K_{\text {eq }}^{1 / 2}\left[\mathrm{CHCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]^{1 / 2}\)

Short answer

The experimental rate law consistent with the mechanism is (d) rate \(=K_3 K_{\text{eq}}^{1/2}[\mathrm{CHCl}_3][\mathrm{Cl}_2]^{1/2}\).

Step-by-step solution

Identify the Rate-Determining Step

Examine the given mechanism and identify the step that determines the overall reaction rate. This step is typically the slowest step in the reaction mechanism. In this case, the second step where \(\mathrm{CHCl}_3 + \mathrm{Cl}\bullet \rightarrow \mathrm{HCl} + \dot{\mathrm{C}}\mathrm{Cl}_3\) is labeled as slow and thus, it's the rate-determining step.

Write the Rate Law for the Rate-Determining Step

Since the rate-determining step is \(\mathrm{CHCl}_3 + \mathrm{Cl}\bullet \rightarrow \mathrm{HCl} + \dot{\mathrm{C}}\mathrm{Cl}_3\), the rate of the reaction is proportional to the concentrations of the reactants involved in this step. However, \mathrm{Cl}\bullet is an intermediate, so its concentration is not included directly in the rate law.

Express Intermediate Concentration in Terms of Reactant Concentrations

To express the concentration of the intermediate \mathrm{Cl}\bullet, we use the equilibrium expression for the first (fast) step of the reaction mechanism, which is in equilibrium. The equilibrium constant \(K_{\text{eq}}\) can be written as \(K_{\text{eq}} = [\mathrm{Cl}\bullet]^2 / [\mathrm{Cl}_2]\). Solving for \([\mathrm{Cl}\bullet]\) gives us \([\mathrm{Cl}\bullet] = \sqrt{K_{\text{eq}} [\mathrm{Cl}_2]}\).

Substitute the Intermediate Concentration into the Rate Law

Substitute \([\mathrm{Cl}\bullet] = \sqrt{K_{\text{eq}} [\mathrm{Cl}_2]}\) into the rate law, which yields the rate equation \(\text{rate} = K_3[\mathrm{CHCl}_3][\mathrm{Cl}\bullet]\) to become \(\text{rate} = K_3[\mathrm{CHCl}_3]\sqrt{K_{\text{eq}} [\mathrm{Cl}_2]}\).

Combine Constants

Combine the constants \(K_3\) and \(K_{\text{eq}}\) into a single rate constant to simplify the rate law. This simplifies to \(\text{rate} = K_3 K_{\text{eq}}^{1/2}[\mathrm{CHCl}_3][\mathrm{Cl}_2]^{1/2}\).

Compare to Given Rate Law Options and Select the Correct One

The simplified rate law we derived matches option (d): \(\text{rate} = K_3 K_{\text{eq}}^{1/2}[\mathrm{CHCl}_3][\mathrm{Cl}_2]^{1/2}\).

Key concepts

Reaction Rate Law

Understanding the rate at which chemical reactions occur is crucial for predicting how much time it will take for the reactants to convert into products. This is where the reaction rate law comes into play. The rate law is an expression that links the rate of a reaction to the concentration of its reactants, often symbolized by a relationship such as rate = k[A]ⁿ[B]ᵐ, where A and B represent the reactants, n and m are their respective orders, and k is the rate constant.

In the case of the mechanism we're addressing, the overall reaction rate is determined by the slowest step involving CHCl₃ and an intermediate Cl·. However, to craft a usable rate law, the rate is expressed exclusively in terms of reactant concentrations that are initially present. This implies that the concentration terms in the rate law need to be for CHCl₃ and Cl₂, as they are the original reactants. The solution shows that the correct rate law also involves the rate constant and the equilibrium constant, suggesting complex rate behavior influenced by an equilibrium step.

Rate-Determining Step

Within a reaction mechanism, some steps occur faster than others. The rate-determining step is the slowest step in a chemical reaction mechanism and effectively sets the pace for the overall reaction—it's the bottleneck, so to speak. You can think of it as the slowest runner in a relay race; no matter how fast the rest of the team is, the slowest runner determines the overall time. For the given reaction mechanism, the slow step is the combination of CHCl₃ and the chlorine radical Cl·.

Identifying the rate-determining step is essential, as it allows us to construct the rate law which dictates how the concentration and other factors will impact the overall reaction rate. This step-by-step approach shown in the exercise forms the backbone of understanding chemical kinetics and reaction rates.

Chemical Equilibrium

In the midst of dynamic reactions, some might reach a state called chemical equilibrium. This is a delicate balance where the forward and reverse reactions occur at the same rate, resulting in no net change in the concentration of reactants and products over time. It's important to recognize that chemical equilibrium does not mean the reactants and products are equal in concentration, but rather that their rates of formation are equal, leading to a constant ratio.

In our example, the fast initial step involving Cl₂ dissociating into two Cl· radicals is assumed to be in equilibrium, which is why we can use the equilibrium constant (K_eq) to relate the concentration of the intermediates to the reactants and simplify the rate law. This concept is a fundamental principle in understanding how reactions can be controlled and manipulated.

Reaction Intermediates

Some reactions occur in multiple steps and involve species that are formed in one step and consumed in another. These species are called reaction intermediates. An intermediate is not present in the overall balanced equation for the reaction, as it is not a reactant or a product but a transient species.

In our example, Cl· is an intermediate. It's generated in the first step and consumed in the second step. Dealing with intermediates can be tricky because they're not typically present in discernible concentrations in the mixture. Consequently, their concentrations are often determined by relating them to the concentrations of stable species using the state of equilibrium from other steps. By using the equilibrium constant from the fast initial step, we can substitute the intermediate concentration in the rate law to avoid unknown terms and accurately describe the rate using only the initial reactants' concentrations.