Hi,
How to find the rate of conversation at equilbrium (Xe) if the rate of conversation is 0.8 (X=0.8). For this reaction:
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Posted 14 April 2023 - 03:09 PM
Hi,
How to find the rate of conversation at equilbrium (Xe) if the rate of conversation is 0.8 (X=0.8). For this reaction:
Posted 15 April 2023 - 03:13 AM
Hi,
Take a look at this document, you may find pointers.
Good luck
Breizh
Posted 15 April 2023 - 05:05 AM Best Answer
Hi,
Take a look at this document, you may find pointers.
Good luck
Breizh
Dear breizh,
the rate of reaction in this case is -rA=k1(CA)0.5*(CB)0.5-k2(CR)*(CS), and what is more important how can I find (Xe) when I don't know CB0, CR0 and CS0 (If i don't know Xe I can't go further).
the data for k1 and k2 are taken from "Kinetics of the Esterification Reaction Between Ethanol and Acetic Acid" 1994. please see the attachment in page 2 (182). [reference 8]
But there is no any data about the rate of conversation at equilibrium (Xe).
So I'm very confused and I don't know what to do.
Anyway, I'm very grateful for the help you're giving me every time I asked for it.
Edited by chemtoli, 15 April 2023 - 05:09 AM.
Posted 15 April 2023 - 05:21 AM
Hi,
Ca=Ca0*(1-xa)
Cb=Ca0*(TetaB -xa)
Cc=Ca0*(xa)
Cd=Ca0*(TetaD+xa)
Note : at equilibrium ,-ra = 0 and xa=xe with Kequilibrium =k1/k2 =Cc*Cd/(Ca^0.5*Cb^0.5)
Breizh
Posted 15 April 2023 - 07:33 AM
Hi,
Ca=Ca0*(1-xa)
Cb=Ca0*(TetaB -xa)
Cc=Ca0*(xa)
Cd=Ca0*(TetaD+xa)
Note : at equilibrium ,-ra = 0 and xa=xe with Kequilibrium =k1/k2 =Cc*Cd/(Ca^0.5*Cb^0.5)
Breizh
But why they took value of Xe=0.52 so than X=0.416. because of Xe=0.52 it changes everything. I'm still confused, really don't know how to go further.
Edited by chemtoli, 15 April 2023 - 07:34 AM.
Posted 15 April 2023 - 07:38 AM
Hi,
Please read the document ,It is said that X=0.8*0.52 in the text .
Breizh
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