Problem 111 - Chemistry The Molecular Science

Question

Which of these mechanisms (more than one may be chosen) is compatible with the rate law? Rate \(=k\left[\mathrm{Cl}_{2}\right]^{3 / 2}[\mathrm{CO}]\) $$ \text { (a) } \frac{1}{2} \mathrm{Cl}_{2} \rightleftharpoons \mathrm{Cl} $$ \(\mathrm{Cl}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{Cl}_{3}\) fast \(\mathrm{Cl}_{3}+\mathrm{CO} \longrightarrow \mathrm{COCl}_{2}+\mathrm{Cl} \quad\) slow $$ \mathrm{Cl} \rightleftharpoons \frac{1}{2} \mathrm{Cl}_{2} $$ (b) \(\mathrm{Cl}_{2}+\mathrm{CO} \longrightarrow \mathrm{CCl}_{2}+\mathrm{O}\) slow \(\mathrm{O}+\mathrm{Cl}_{2} \longrightarrow \mathrm{Cl}_{2} \mathrm{O}\) fast $$ \mathrm{Cl}_{2} \mathrm{O}+\mathrm{CCl}_{2} \longrightarrow \mathrm{COCl}_{2}+\mathrm{Cl}_{2} $$ fast s (c) \(\frac{1}{2} \mathrm{Cl}_{2} \rightleftharpoons \mathrm{Cl}\) $$ \mathrm{Cl}+\mathrm{CO} \rightleftharpoons \mathrm{COCl} $$ fast $$ \begin{array}{l} \mathrm{COCl}+\mathrm{Cl}_{2} \longrightarrow \mathrm{COCl}_{2}+\mathrm{Cl} \\\ \mathrm{Cl} \rightleftharpoons \frac{1}{2} \mathrm{Cl}_{2} \end{array} $$ slow (d) \(\mathrm{Cl}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{COCl}+\mathrm{Cl}\) fast \(\mathrm{COCl}+\mathrm{Cl}_{2} \longrightarrow \mathrm{COCl}_{2}+\mathrm{Cl}\) slow \(\mathrm{Cl}+\mathrm{Cl} \longrightarrow \mathrm{Cl}_{2}\) fast

Step-by-Step Solution

Answer

Mechanisms (a) and (d) are compatible with the rate law.

Step 1: Understand the Rate Law

The given rate law is \( k[\mathrm{Cl}_{2}]^{3/2}[\mathrm{CO}] \). It means the reaction rate is dependent on the concentration of \([\mathrm{Cl}_{2}]\) raised to the power of \(3/2\) and \([\mathrm{CO}]\) to the power of 1.

Step 2: Analyze Mechanism (a)

Mechanism (a) involves the intermediary \(\mathrm{Cl}_{3}\). The rate-determining step is slow and involves \(\mathrm{Cl}_{3}\) and \(\mathrm{CO}\). However, this step does not produce a rate law with a \([\mathrm{Cl}_{2}]^{3/2}\) term, as it does not match intermediates or products expected by the rate law.

Step 3: Analyze Mechanism (b)

In mechanism (b), the slow step is the formation of \(\mathrm{CCl}_{2}\) and \(\mathrm{O}\) from \(\mathrm{Cl}_{2}\) and \(\mathrm{CO}\). This slow step suggests a rate law \(k[\mathrm{Cl}_{2}][\mathrm{CO}]\), which does not match the given rate law's \([\mathrm{Cl}_{2}]^{3/2}\) term.

Step 4: Analyze Mechanism (c)

Mechanism (c) has a rate-determining step forming \(\mathrm{COCl}\) from \(\mathrm{Cl}\) and \(\mathrm{CO}\) quickly, followed by a slow step. This doesn't match the given rate law \(k[\mathrm{Cl}_{2}]^{3/2}[\mathrm{CO}]\) since there isn't a mechanism for achieving the \([\mathrm{Cl}_{2}]^{3/2}\) exponent.

Step 5: Analyze Mechanism (d)

In mechanism (d), the fast equilibrium step involving \(\mathrm{COCl}\) from \(\mathrm{Cl}_{2}\) and \(\mathrm{CO}\) can produce a rate law consistent with \(k[\mathrm{Cl}_{2}]^{1/2}[\mathrm{CO}]\) after considering intermediates and fast reactions. This partial dependence on intermediates and \([\mathrm{Cl}_{2}]\) aligns with the rate law requirement of \([\mathrm{Cl}_{2}]^{3/2}[\mathrm{CO}]\).

Key Concepts

Rate Law

Understanding rate laws is crucial in chemical kinetics as they provide insights into how the concentrations of substances affect the speed of a reaction. A rate law is expressed using a mathematical equation that relates the rate of a reaction to the concentration of reactants. In this case, the rate law is \( k[\mathrm{Cl}_{2}]^{3/2}[\mathrm{CO}] \). This indicates that the reaction rate is directly proportional to the concentration of \([\mathrm{Cl}_{2}]\) raised to the power of \(3/2\) and the concentration of \([\mathrm{CO}]\) raised to the power of 1. The constants and exponents in a rate law are determined experimentally and provide crucial information about the reaction pathway and mechanism.

Rate laws can vary greatly depending on the mechanism of the reaction. Therefore, it is important to analyze the mechanism carefully to see which step determines the rate and matches with the given rate law. This often involves considering the slowest step in the reaction mechanism or the step involving intermediates.

Chemical Kinetics

Chemical kinetics is a field of chemistry that studies the rates of chemical processes. It delves into understanding how different conditions affect the speed of reactions. The goal is to identify the rate-determining step, which is often the slowest step in a series of reactions and controls the overall rate of the process.

In multi-step reactions, not all steps occur at the same rate. Some are fast, while others are slow. The slowest step acts as a bottleneck, determining the speed of the entire reaction. When analyzing mechanisms, attention should be paid to this rate-determining step as it often dictates the structure of the rate law. By understanding chemical kinetics, chemists can manipulate conditions to optimize reaction speeds, thereby improving efficiency in industrial and laboratory settings.

Intermediates

Intermediates are transient molecules that are formed in one step of a reaction mechanism and consumed in another. They play a crucial role in understanding reaction mechanisms as they often appear in the elementary steps of a reaction but not in the overall balanced equation.

Analyzing intermediates helps chemists verify the validity of proposed reaction mechanisms. For example, in mechanism (a), the intermediate \(\mathrm{Cl}_{3}\) does not contribute to forming a rate law consistent with the one provided in the problem. However, in mechanism (d), the presence of intermediates along with fast equilibrium steps involving them allows the creation of a rate law that matches \( k[\mathrm{Cl}_{2}]^{3/2}[\mathrm{CO}] \).

Understanding how intermediates interact within a reaction mechanism is key to linking the experimental rate law with the theoretical pathway. This connection provides critical insights into how reactions proceed and can assist in predicting catalysts or conditions to enhance reaction speed.