In a hydrocarbon solution, the gold compound ( C H 3 ) 3 A u P H 3 decomposes in

Question

In a hydrocarbon solution, the gold compound (CH3)3AuPH3 decomposes into ethane (C2H6) and a different gold compound, (CH3)AuPH3. The following mechanism has been proposed for the decomposition of (CH3)3AuPH3:
Step 1: (CH3)3AuPH3?(CH3)3Au+PH3(fast)
Step 2: (CH3)3Au?C2H6+(CH3)Au(slow)
Step 3: (CH3)Au+PH3?(CH3)AuPH3(fast)

Overall reaction:

Explanation / Answer

if the second step is the rate determining step then mechanism can be written like this...

(CH3)3AuPH3 <--------> (CH3)3Au + PH3 ......(fast eqb)

(CH3)3Au ---------> C2H6 + (CH3)Au .....(slow)

(CH3)Au + PH3 <--------> (CH3)AuPH3 .....(fast eqb)

since rate law depends on slowest step ....hence rate wil be given as , r = k X [ (CH3)3Au ]

now (CH3)3Au is an intermediate as it is formed in first step and consumed in second step ...so its very difficult to measure the concentration of (CH3)3Au ....so we have to replace it by some species whose concentration can be measured ...

from first step .....

equailibrium constant Keq = [ (CH3)3Au] [ PH3] / [ (CH3)3AuPH3 ]

[ (CH3)3Au ] = Keq [ (CH3)3AuPH3 ] / [PH3]

putting in rate expression...

r = k X Keq [ (CH3)3AuPH3 ] / [PH3] = k X Keq X [ (CH3)3AuPH3 ] [PH3] ^-1 = k' [ (CH3)3AuPH3 ] [PH3]^-1

where k' = k X Keq

similarly if first step determines the rate thenn...


(CH3)3AuPH3 --------> (CH3)3Au + PH3 ......(slow)

(CH3)3Au <---------> C2H6 + (CH3)Au .....(fast eqb)

(CH3)Au + PH3 <--------> (CH3)AuPH3 .....(fast eqb)

so r = k X [ (CH3)3AuPH3 ]

here (CH3)3AuPH3 is not an intermediate as it is only consuming in first step so rate law is that only

feel free to ask any question

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