Problem 112 - Chemistry The Central Science In Si Units

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: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} \stackrel{\mathrm{k}_{\mathrm{L}}}{2}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}+\mathrm{PH}_{3}\) Step 2: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au} \stackrel{\mathrm{h}_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{Au} \quad\) (slow) Step 3: \(\left(\mathrm{CH}_{3}\right) \mathrm{Au}+\mathrm{PH}_{3} \stackrel{\mathrm{k}_{\mathrm{j}}}{\longrightarrow}\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3}\) (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 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 \((\mathrm{CH}_{3})_{3} \mathrm{AuPH}_{3} \rightarrow \mathrm{C}_{2} \mathrm{H}_{6} + (\mathrm{CH}_{3}) \mathrm{AuPH}_{3}\). Intermediates are \((\mathrm{CH}_{3})_{3} \mathrm{Au}\) and \((\mathrm{CH}_{3}) \mathrm{Au}\). Molecularity: Step 1 - unimolecular, Step 2 - unimolecular, Step 3 - bimolecular. Rate-determining step is Step 2. Rate law: Rate = \(k(\frac{k_L}{2})[(\mathrm{CH}_{3})_{3} \mathrm{AuPH}_{3}]\). Adding \(\mathrm{PH}_3\) has no effect on the rate.

Step 1: Determine the Overall Reaction

To find the overall reaction, we need to sum up all the steps and eliminate any species that appear both as reactants and products. From the given steps, adding them together gives:\[ (\mathrm{CH}_{3})_{3} \mathrm{AuPH}_{3} \rightarrow \mathrm{C}_{2} \mathrm{H}_{6} + (\mathrm{CH}_{3}) \mathrm{AuPH}_{3} \]Thus, the overall reaction is the conversion of \((\mathrm{CH}_{3})_{3} \mathrm{AuPH}_{3}\) into \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \((\mathrm{CH}_{3}) \mathrm{AuPH}_{3}\).

Step 2: Identify Intermediates

Intermediates are species that are produced in one step and consumed in another. By reviewing the mechanism steps:- \((\mathrm{CH}_{3})_{3} \mathrm{Au}\) is produced in Step 1 and consumed in Step 2.- \((\mathrm{CH}_{3}) \mathrm{Au}\) is produced in Step 2 and consumed in Step 3.Therefore, the intermediates are \((\mathrm{CH}_{3})_{3} \mathrm{Au}\) and \((\mathrm{CH}_{3}) \mathrm{Au}\).

Step 3: Determine Molecularity of Each Step

Molecularity refers to the number of molecules involved in an elementary reaction:- Step 1: Since it involves the reaction of one molecule of \((\mathrm{CH}_{3})_{3} \mathrm{AuPH}_{3}\), it's unimolecular.- Step 2: With a single molecule of \((\mathrm{CH}_{3})_{3} \mathrm{Au}\), it's also unimolecular.- Step 3: Involves \((\mathrm{CH}_{3}) \mathrm{Au}\) and \(\mathrm{PH}_{3}\), making it bimolecular.

Step 4: Identify Rate-Determining Step

The rate-determining step is the slowest step, which is provided in the problem statement as Step 2: \((\mathrm{CH}_{3})_{3} \mathrm{Au} \rightarrow \mathrm{C}_{2} \mathrm{H}_{6} + (\mathrm{CH}_{3}) \mathrm{Au}\).

Step 5: Write the Rate Law

The rate law is based on the rate-determining step (Step 2). Since Step 2 involves \((\mathrm{CH}_{3})_{3} \mathrm{Au}\), the rate law is:\[ \text{Rate} = k[(\mathrm{CH}_{3})_{3} \mathrm{Au}] \]Since \((\mathrm{CH}_{3})_{3} \mathrm{Au}\) is an intermediate, we replace it with its expression from the pre-equilibrium step (Step 1). Assuming Step 1 is at equilibrium:\[ k_L [(\mathrm{CH}_{3})_{3} \mathrm{AuPH}_{3}] = 2 [(\mathrm{CH}_{3})_{3} \mathrm{Au}] \\Rightarrow [(\mathrm{CH}_{3})_{3} \mathrm{Au}] = \frac{k_L}{2} [(\mathrm{CH}_{3})_{3} \mathrm{AuPH}_{3}] \]Thus, the rate law becomes:\[ \text{Rate} = k \frac{k_L}{2} [(\mathrm{CH}_{3})_{3} \mathrm{AuPH}_{3}] \]

Step 6: Analyze the Effect of Adding PH3

Adding \(\mathrm{PH}_3\) does not affect the rate of the rate-determining step, since \(\mathrm{PH}_3\) is involved in Step 3, not the slow step (Step 2). The reaction rate remains unchanged.

Key Concepts

Chemical Intermediates

Chemical intermediates are transient species that form temporarily during a reaction sequence. They are generated in one step and consumed in another. Understanding intermediates is crucial as they often dictate the possible pathways a reaction can take. In the given mechanism:

• \((\mathrm{CH}_3)_3 \mathrm{Au}\) is an intermediate because it is formed in the first step and used up in the second step.
• Similarly, \((\mathrm{CH}_3) \mathrm{Au}\) acts as another intermediate, appearing in the second step and being used in the third one.

These species do not appear in the final balanced equation but play a pivotal role in the mechanism.

Molecularity

Molecularity refers to the number of reactant molecules involved in an elementary step of a reaction mechanism. It describes how molecules come together in a single reaction step:

• **Unimolecular reactions** involve a single molecular entity. Steps 1 and 2 are both unimolecular in this mechanism, as they involve individual molecules decomposing into other units.
• **Bimolecular reactions** involve two different molecules or atoms. Step 3 involves two species, \((\mathrm{CH}_3) \mathrm{Au}\) and \(\mathrm{PH}_3\), thus making it bimolecular.

Why does molecularity matter? It determines the complexity of a reaction step, which can influence the reaction rate and the order of the reaction.

Rate-Determining Step

The rate-determining step is akin to the bottleneck in a series of reactions - it is the slowest step that limits the rate of the overall reaction. In this decomposition mechanism:

• Step 2 is identified as the rate-determining step, involving the slow transformation of \((\mathrm{CH}_3)_3 \mathrm{Au}\) into \(\mathrm{C}_2 \mathrm{H}_6\) and \((\mathrm{CH}_3) \mathrm{Au}\).

Understanding this step is crucial because it dictates the overall kinetics of the reaction. Factors affecting this step can dramatically change the rate at which the reaction proceeds.

Rate Laws

Rate laws express the relationship between the concentration of reactants and the rate of reaction. They are often derived from the rate-determining step. For our reaction:

• The rate law originates from Step 2, where the concentration of the intermediate \((\mathrm{CH}_3)_3 \mathrm{Au}\) is crucial.
• We find that \(\text{Rate} = k[(\mathrm{CH}_3)_3 \mathrm{Au}]\), which is modified by substituting the equilibrium expression from Step 1. This changes it to \(\text{Rate} = k \frac{k_L}{2} [(\mathrm{CH}_3)_3 \mathrm{AuPH}_3]\).

Rate laws are essential, allowing chemists to predict how changing conditions (like concentration) can speed up or slow down a chemical reaction.

Effect on Reaction Rate

The rate of reaction can be affected by various factors, including temperature, concentration of reactants, and the presence of catalysts. In the analyzed mechanism, adding \(\mathrm{PH}_3\) would not impact the rate since it doesn't participate in the rate-determining step.Here are some points to consider:

• Because \(\mathrm{PH}_3\) is involved in Step 3 (a fast step), its concentration has no direct impact on the slow step (Step 2).
• Typically, only reactants involved in the rate-limiting step will affect the reaction rate when their concentrations are changed.

Understanding how various factors affect the reaction rate is crucial for controlling industrial and laboratory reactions efficiently.