Problem 112 - Chemistry: The Central Science

Question

In a hydrocarbon solution, the gold compound \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\) decomposes into ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) and a different gold compound, \(\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3}\). The following mechanism has been proposed for the decomposition of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\) : Step 1: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftharpoons}}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}+\mathrm{PH}_{3}\) (fast) Step 2: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au} \stackrel{\mathrm{k}_{1}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{Au} \quad\) (slow) Step 3: \(\left(\mathrm{CH}_{3}\right) \mathrm{Au}+\mathrm{PH}_{3} \stackrel{k_{3}}{\longrightarrow}\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3}\) (fast) (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary steps? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism? (f) What would be the effect on the reaction rate of adding \(\mathrm{PH}_{3}\) to the solution of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\) ?

Step-by-Step Solution

Answer

The overall reaction is \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3}\), with intermediates \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\) and \(\left(\mathrm{CH}_{3}\right) \mathrm{Au\). Molecularity: Step 1 and 3 are bimolecular, while Step 2 is unimolecular. The rate-determining step is Step 2, and the rate law is \(Rate = k_{2}\frac{k_{1}}{k_{-1}}[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}][\mathrm{PH}_{3}]\). Adding \(\mathrm{PH}_{3}\) to the solution would increase the reaction rate.

Step 1: To find the overall reaction, we simply need to combine the individual steps given in the mechanism. In this case, the steps are: Step 1: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftharpoons}}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}+\mathrm{PH}_{3}\) Step 2: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au} \stackrel{k_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{Au}\) Step 3: \(\left(\mathrm{CH}_{3}\right) \mathrm{Au}+\mathrm{PH}_{3} \stackrel{k_{3}}{\longrightarrow}\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3}\) Combining all three steps, the overall reaction is: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3}\) #b) Intermediates in the mechanism#

An intermediate is a species that is produced and consumed within the reaction mechanism. In this case, the intermediates are \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\) and \(\left(\mathrm{CH}_{3}\right) \mathrm{Au}\). They appear as products in one step and reactants in another step. #c) Molecularity of each elementary step#

Step 2: Molecularity is the number of reacting species involved in an elementary step. In this mechanism, we have: Step 1: Bimolecular (two reactant molecules: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\) and \(\mathrm{PH}_{3}\)) Step 2: Unimolecular (one reactant molecule: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\)) Step 3: Bimolecular (two reactant molecules: \(\left(\mathrm{CH}_{3}\right) \mathrm{Au}\) and \(\mathrm{PH}_{3}\)) #d) Rate-determining step#

The rate-determining step is the slowest step in the reaction mechanism, as it limits the overall reaction rate. In this case, Step 2 is the slowest step, and therefore, it is the rate-determining step: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au} \stackrel{k_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{Au}\) #e) Rate law predicted by this mechanism#

Step 3: As Step 2 is the rate-determining step, the rate of the overall reaction will depend on the concentration of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\). Additionally, from Step 1, we can derive the equilibrium between reactants and products: \(k_{1}[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}] = k_{-1}[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}][\mathrm{PH}_{3}]\) Therefore, the concentration of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\) is given by: \([\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}] = \frac{k_{1}}{k_{-1}} [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}][\mathrm{PH}_{3}]\) The rate law for the overall reaction can be expressed as: Rate \(= k_{2}[\left(\mathrm{CH}_{3}\right)_3\mathrm{Au}] = k_{2}\frac{k_{1}}{k_{-1}}[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}][\mathrm{PH}_{3}]\) #f) Effect of adding \(\mathrm{PH}_{3}\) to the solution#

As seen in the rate law derived above, Rate \(= k_{2}\frac{k_{1}}{k_{-1}} [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}][\mathrm{PH}_{3}]\). The reaction rate is directly proportional to the concentration of \(\mathrm{PH}_{3}\). Therefore, adding \(\mathrm{PH}_{3}\) to the solution of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\) would increase the overall reaction rate.

Key Concepts

Chemical Kinetics

Chemical kinetics is the study of reaction rates and the factors influencing them. Here, it helps us understand how fast the gold compound decomposes. The decomposition involves multiple steps, each with its own speed. By investigating these steps, chemists learn how quickly a reaction proceeds and can manipulate conditions to either speed up or slow down the reaction. In this example, kinetics reveals why the breakdown of the gold compound occurs at a specific rate by considering the complex sequence of steps involved.

Molecularity

Molecularity refers to the number of molecules that participate in an elementary reaction step. It provides insight into the nature of a reaction and the mechanism it follows. For the given decomposition of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\), each step has its own molecularity classification:

• Step 1 is bimolecular, involving two reactants: the gold compound and \(\mathrm{PH}_{3}\).
• Step 2 is unimolecular, with only one reactant molecule, showcasing a change happening within a single entity.
• Step 3 returns to bimolecular, requiring two molecules to collide to form products.

This detailed look into reaction steps helps deduce how complex a reaction is and how molecules must align or interact to drive the process forward.

Reaction Intermediates

Intermediates are species formed during a reaction that do not appear in the overall balanced equation because they are consumed in subsequent steps. They provide essential clues to understanding complex reaction mechanisms. In this mechanism, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\) and \(\left(\mathrm{CH}_{3}\right) \mathrm{Au}\) are intermediates. They first appear as products before becoming reactants in subsequent steps, which exemplifies their transient role. Identifying intermediates is crucial for chemists to map out a reaction pathway and design targeted modifications to enhance reaction performance or selectivity.

Rate-Determining Step

The rate-determining step is the slowest step in a reaction mechanism, dictating the speed of the entire process. It's akin to a bottleneck in an assembly line. For \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\)'s decomposition, Step 2 serves as the rate-determining step. In this step, the transition from \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\) to product formation is slow, thus controlling the overall reaction rate. This insight is valuable for chemists aiming to enhance reaction efficiency. By targeting the rate-determining step, they can alter conditions to increase the reaction speed.

Rate Law

The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. For the mechanism we've explored, the rate law is influenced by the rate-determining step and can be expressed as:\[\text{Rate} = k_{2}\frac{k_{1}}{k_{-1}}[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}][\mathrm{PH}_{3}]\]This equation indicates the reaction rate depends on the concentrations of both \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\) and \(\mathrm{PH}_{3}\). Furthermore, it integrates constants \(k_{1}\), \(k_{-1}\), and \(k_{2}\), underscoring how equilibrium conditions of preceding steps influence the reaction.

Chemical Equilibrium

Chemical equilibrium involves the balance between forward and backward reactions, where no net change occurs in the concentration of reactants and products. In the mechanism discussed, Step 1 exemplifies this concept as it reaches a fast equilibrium, represented by the reversible reaction of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\) dissociating into \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\) and \(\mathrm{PH}_{3}\). Achieving equilibrium is vital for determining the concentrations of reactants available for subsequent steps, thereby affecting the rate and yield of the overall process. Understanding chemical equilibrium is foundational in designing and predicting the efficiency of reaction mechanisms.