IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry - Equilibrium-please reply

rahulraj1

K_c for the equilibrium of

CH₃CO₂H(l) + C₂H₅OH(l)---------------CH₃CO₂C₂H₅(l) +H₂O(l) is 4.

How much ester is formed if 2 mol of the acid and 1 mol of the alcohol are taken initially to bring the reaction to equilibrium?

abrambenny

Consider the no. of moles of ester formd. as x.
Then Eqbm. conc. of the components are.

CH3CO2H- 1-X; C2H50H- 2-X ; CH3CO2C2H5 - X; H20-x

Then Kc= (X*X) / (1-x)(2-x)=4
----which gives the eqtn. 3x^2-12x+8=0
solving x= 19/6 and x=5/6.
As x can't be greater than one, x is 5/6. (i've done some approx. with the roots of the quadratic eqtn)

rahulraj1

But the answer is .25 please explain once again

rahulraj1

I think no body there to give my answer

destinationIIT09

delta n = o,so volume not needed 2 solve

intial conc in mol/L = 2 1 0 0

at equilibrium =2-x 1-x x x

Kc = ( x * x) / (2-x)(1-x)

Kc is given as 4

on solving for x u will get the required answer

rate if useful

cheeeeeeersssssssss!!!!!!!

abrambenny

Another guy has asked this same question and gotten the same ans.(0.85) Pls check up on ur question .Eithr there's more to the questn or the answr's wrng.

I hope this is not rude :)

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