Posted by Robert on Friday, August 3, 2012 at 11:56am.
does it mean that because acetic acid has 0.03045 mol at equilibrium that x = 0.03045 mol? So then Ethanol would also have 0.03045 mol at eq and EtAc and H20 would both be their initial [conc] - x?
because it's a 1:1 ratio for all, then at rxn, we would have -x for both EtAc and H2O and + x on the right, so x = 0.03045.
Is this correct?
You have run your sentence together and I can't be sure about the concentrations. From the table, however, it appears initially you have EtOAc of 0.051 and H2O of 0.260 (so initially EtOH = 0 and acetic acid = 0).
Yes, if you have 0.03045 mols acetic acid at equilibrium you will have 0.03045 mols EtOH at equilibrium.
.......EtOAc + H2O ==> EtOH + HAc
I......0.051..0.260.....0......0
C.......-x.....-x.......x.......x
E.....0.051-x.0.260-x...x......x
So = 0.03045 which allows you to place numbers for all and calculate Keq.(at least I suppose that is what you are to do).
okay thanks Dr.Bob, that's exactly what I did too.
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