5 replies to this topic

[chemtoli]

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:

 

CH₃COOH+CH₃CH₂OH <=> CH₃COOCH₂CH₃+H₂O

 

please see the attached document in page 7 it is highlighted with red color.

 

 

I would be very grateful if you could help me solve this problem.

 

 

 

Thank you very much.

Attached Files

•  CSTR Reactor.pdf   236.93KB   4 downloads

[breizh]

Hi,

Take a look at this document, you may find pointers.

Good luck

Breizh

[chemtoli]

Hi,

Take a look at this document, you may find pointers.

Good luck

Breizh

 

Dear breizh,

 

the rate of reaction in this case is -r_A=k₁(C_A)^0.5*(C_B)^0.5-k₂(C_R)*(C_S), and what is more important how can I find (Xe) when I don't know C_B0, C_R0 and C_S0 (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.

Attached Files

•  Reference 8.pdf   183.35KB   2 downloads

Edited by chemtoli, 15 April 2023 - 05:09 AM.

[breizh]

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 

[chemtoli]

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.

[breizh]

Hi,

Please read the document ,It is said that X=0.8*0.52 in the text .

Breizh