Question

In a hydrocarbon solution, the gold compound (CH3)3AuPH3 decomposes into ethane (C2H6) and a different gold compound, (CH3)AuPH3. The following mechanism is proposedfor the decomposition of (CH3)3AuPH3.

Step 1:(CH3)3AuPH <-----> (CH3)3Au + PH3 (fast)
Step 2:(CH3)3Au -----> C2H6 + (CH3)Au (slow)
Step 3:(CH3)Au + PH3 ------>(CH3)AuPH3 (fast)

A)What is the overall reaction?
B)What are the intermediates in the mechanism?
C)What is the rate law predicted by this mechanism?
Use a for :(CH3)3AuPH , b for (CH3)3Au, c for PH3, d for C2H6 , e for (CH3)Au and K for the rate constant.

Answer 1

The following mechanism is written as follows:

$$ \text { step } 1:\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH} \underset{k_{-1}}{\stackrel{k_{1}}{\longrightarrow}}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}+\mathrm{PH}_{3} \text { (fast) } $$

step \(2:\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au} \stackrel{k_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{Au}(\) 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) Add up the above three equations to get the overall reaction as follows:

step \(1:\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH} \underset{k_{-1}}{\stackrel{k_{1}}{\longrightarrow}}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}+\mathrm{PH}_{3}\) (fast)

step \(2:\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au} \stackrel{k_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{Au}(\) 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)

Overall: \({\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH} \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3}}\)

B) Intermediates are the species that are produced during reaction and consumed during reaction. Here are in the above mechanism, \(\left(\mathbf{C H}_{3}\right)_{3} \mathbf{A} \mathbf{u}\) and \(\mathbf{P} \mathbf{H}_{3}\) are the intermediates.

C) The rate law is written from the slow step as follows:

Rate \(=k_{2}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\right]\)

Apply steady state approximation on \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\) as follows:

$$ \begin{array}{l} k_{1}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\right]-k_{-1}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\right]\left[\mathrm{PH}_{3}\right]-k_{2}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\right]=0 \\ {\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\right]=\frac{k_{1}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\right]}{k_{-1}\left[\mathrm{PH}_{3}\right]+k_{2}}} \end{array} $$

Apply steady-state approximation on \(\mathrm{PH}_{3}\) as follows:

$$ k_{1}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\right]-k_{-1}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\right]\left[\mathrm{PH}_{3}\right]-k_{3}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\right]\left[\mathrm{PH}_{3}\right]=0 \quad \ldots \ldots(2) $$

Subtract equation (1) from equation (2) as follows:

$$ \left[\mathrm{PH}_{3}\right]=\frac{k_{2}}{k_{3}} $$

Substitute this in \(\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\right]\) as follows:

$$ \begin{aligned} \left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\right] &=\frac{k_{1}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\right]}{k_{-1}\left[\mathrm{PH}_{3}\right]+k_{2}} \\ &=\frac{k_{1}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\right]}{k_{-1}\left(\frac{k_{2}}{k_{3}}\right)+k_{2}} \end{aligned} $$

The rate is written as follows:

$$ \begin{aligned} \text { Rate } &=k_{2}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}\right] \\ &=\frac{k_{1}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\right]}{k_{-1}\left(\frac{k_{2}}{k_{3}}\right)+k_{2}} \end{aligned} $$