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, (CH \(_{3} ) \mathrm{AuPH}_{3} .\) The following mechanism has been proposed for the decomposition of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} :\) $$ \quad Step \quad1.\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} \frac{\mathrm{k}_{1}}{\mathrm{k}_{-1}}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}+\mathrm{PH}_{3} $$ $$ Step\quad \quad2.\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au} \stackrel{k_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{Au} \quad$$ $$ Step\quad 3 :\left(\mathrm{CH}_{3}\right) \mathrm{Au}+\mathrm{PH}_{3} \stackrel{k_{3}}{\longrightarrow}\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3}$$ (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 ratedetermining 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 \(\left(\mathrm{CH}_{3}\right)_{3}\) AuPH \(_{3} ?\)

Step-by-Step Solution

Answer

The overall reaction is (CH3)3AuPH3 → C2H6 + (CH3)AuPH3. The intermediates are (CH3)3Au and (CH3)Au. The molecularity of each elementary step is 1, 1, and 2 for Steps 1, 2, and 3, respectively. The rate-determining step is Step 1, and the rate law is Rate = k'[(CH3)3AuPH3]/[PH3]. Adding PH3 to the solution of (CH3)3AuPH3 decreases the overall reaction rate.

Step 1: (a) Overall Reaction

To find the overall reaction, we need to add the elementary steps in the mechanism together and cancel out the intermediates appearing on both sides of the reaction. Step 1: (CH3)3AuPH3 <=> (CH3)3Au + PH3 Step 2: (CH3)3Au → C2H6 + (CH3)Au Step 3: (CH3)Au + PH3 → (CH3)AuPH3 After cancelling out intermediates, the overall reaction becomes: (CH3)3AuPH3 → C2H6 + (CH3)AuPH3

Step 2: (b) Intermediates

Intermediates are the species that are formed in one step but consumed in another step. From the given mechanism, we can identify two intermediates: 1. (CH3)3Au: It is formed in Step 1 and consumed in Step 2. 2. (CH3)Au: It is formed in Step 2 and consumed in Step 3.

Step 3: (c) Molecularity of Each Elementary Step

Molecularity is the number of molecules taking part in an elementary step of the reaction. There are three elementary steps: 1. Step 1: Molecularity of (CH3)3AuPH3 → (CH3)3Au + PH3 is 1, since it involves one reactant molecule. 2. Step 2: Molecularity of (CH3)3Au → C2H6 + (CH3)Au is 1, as it also has one reactant molecule. 3. Step 3: Molecularity of (CH3)Au + PH3 → (CH3)AuPH3 is 2, since it involves two reactant molecules.

Step 4: (d) Ratedetermining Step

The rate-determining step is the slowest step in the reaction mechanism. The information is not provided directly in the exercise, but we can determine the rate-determining step by examining the rate law for the overall reaction. In most cases, the rate-determining step is the one with the lowest rate constant. In order to answer this question, we need to calculate the rate law, which brings us to part (e).

Step 5: (e) Rate Law

The rate law can be determined from the rate-determining step (the slowest step), assuming steady-state approximation for the intermediates. Let's assume that Step 1 is the slowest step. In that case, we can write the equilibrium for Step 1 as: K_eq = [ (CH3)3Au][PH3] / [(CH3)3AuPH3] Next, let's write the rate law for the overall reaction: Rate = k2 [(CH3)3Au] Using the equilibrium expression from Step 1, we can substitute for the intermediate (CH3)3Au: [(CH3)3Au] = K_eq [(CH3)3AuPH3] / [PH3] Substitute in the rate law: Rate = k2 (K_eq [(CH3)3AuPH3] / [PH3]) The rate law for the overall reaction is now: Rate = k' [(CH3)3AuPH3] / [PH3], where k' = k2 * K_eq To determine the rate-determining step, we consider Step 1 in this case due to the relatively complex nature of the rate law. So, the rate-determining step is Step 1.

Step 6: (f) Effect on the Reaction Rate

The question asks about the effect on the reaction rate if PH3 is added to the solution of (CH3)3AuPH3. From the rate law we derived, we can see that the reaction rate is inversely proportional to the concentration of PH3: Rate = k' [(CH3)3AuPH3] / [PH3] Thus, adding more PH3 to the solution will decrease the overall reaction rate, as the denominator in the rate law expression increases.

Key Concepts

Hydrocarbon Solution

A hydrocarbon solution usually involves compounds composed of only carbon and hydrogen atoms. These solutions are commonly known for their ability to dissolve similar non-polar substances. In the context of reaction mechanisms, the hydrocarbon solution provides an inert medium where specific reactions can take place without interference. Such environments are crucial for studying the decomposition of compounds, such as gold compounds in this scenario. Hydrocarbon solutions allow reactions to proceed by stabilizing certain reaction intermediates and providing optimal conditions for reaction pathways to be studied and understood.

Gold Compound Decomposition

Gold compounds, due to their unique chemical properties, often undergo decomposition via distinct reaction mechanisms. Here, we explore the breakdown of the compound \( \left(\mathrm{CH}_3\right)_3 \mathrm{AuPH}_3 \), where it splits into smaller entities. These reactions usually proceed through a series of steps, including:

• the formation of intermediates
• the rearrangement of atoms
• the production of end-products like ethane and other gold complexes

During decomposition, intermediates such as \(\left(\mathrm{CH}_3\right)_3 \mathrm{Au}\) play a key role in the sequence of transformations, eventually leading to stable end-products. Understanding these pathways via gold compound decomposition helps chemists identify and manipulate reaction conditions to optimize outcomes for various industrial and research applications.

Rate-Determining Step

The rate-determining step in a reaction mechanism is the slowest elementary step, essentially bottlenecking the overall reaction speed. It dictates the reaction rate and is sometimes known as the 'rate-limiting' step. In the decomposition of \(\left(\mathrm{CH}_3\right)_3 \mathrm{AuPH}_3\), identifying the rate-determining step is critical as it informs us which stage of the reaction requires the most energy or involves the most complex transformation. This step can often be pinpointed by examining the rate law, as it typically involves the intermediate with the lowest kinetic rate constant. By targeting this step, chemists can devise strategies to increase the efficiency of the overall reaction under desired conditions.

Molecularity of Reactions

Molecularity refers to the number of reactant molecules participating in an elementary reaction step. It is a simplistic but useful concept used to describe steps within a reaction mechanism. For example, the decomposition reaction in this mechanism involves three steps with varying molecularities:

• Unimolecular: Steps involving a single reactant molecule, such as the decomposition of \(\left(\mathrm{CH}_3\right)_3 \mathrm{AuPH}_3\) in Step 1 and \(\left(\mathrm{CH}_3\right)_3 \mathrm{Au}\) in Step 2.
• Bimolecular: Steps involving two reactant molecules, like the reaction between \(\left(\mathrm{CH}_3\right) \mathrm{Au}\) and \(\mathrm{PH}_3\) in Step 3.

Understanding the molecularity assists in comprehending how collisions between molecules lead to product formation and helps researchers analyze reaction dynamics within more complex mechanisms.