I've shortened the equation. HAc is acetic acid, RCH2OH is the alcohol and RCOOR is the ester.
..........HAc + RCH2OH --> RCOOR + H2O
I.........1.0....1.0........0.......0
C..........-x.....x.........x.......x
E.........1.0-x..1.0-x......x.......x
Substitute the E line into Kc and solve.
The equilibrium constant of the reaction CH3COOH (l)+C2H5OH (l)=CH3COOC2H5+H2O is 4.if one mole of each of acetic acid and ethyl alcohol are heated in presence of little concentrated H2SO4 at equilibrium the amount of ester present is?
5 answers
Ester present is 2 moles, because first of all,
4=x^2/(1-x)(1-x) solve it and it will turn out to be this
3x^2-8x+4
x=2
so the ester is 2 =
4=x^2/(1-x)(1-x) solve it and it will turn out to be this
3x^2-8x+4
x=2
so the ester is 2 =
It comes out 2 and 2/3 and we have to take 2/3
why not 2
Because (1-2) would be negative and concentration cannot be negative