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

01

(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

02

(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.

03

(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.

04

(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).

05

(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.

06

(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.

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