I can save some time by calling CH3 R and C2H5 R'.
RCOOH + HOR' ==> RCOOR' + H2O
Set up an ICE chart.
Initial:
RCOOH = 1.0 mol from the problem.
R'OH = 2.0 mol from the problem.
RCOOR' = 0 you know this and
H2O = 0 you know this because these are zero before the reaction begins..
change:
RCOOR' = x mol
H2O = +x mol
RCOOH = -x mol
R'OH = -x mol
equilibrium:
RCOOR' = 0.845 mol from the problem.
H2O = 0.845 mol (must be the same as RCOOR').
RCOOH = 1.0 - x
R'OH = 2.0 - x
If the equilibrium mixture is 0.845 mols H2O and 0.845 mols RCOOR', isn't that x? So now you know what the equilbrium mols are for RCOOH and R'OH because they are 1.0-x and 2.0-x and you know x.
So now you have the mols of each.
You need concentration in mols/L. These are moles and you know the volume (30 cc = 0.030 L) so figure the concn of each and plug in to the Kc expression. Go from there. Post your work if you get stuck.
In an esterification experiment, 2.0 mol of ethanol were mixed with 1.0 mol of ethanoic acid in a container of volume 30cm3 and the chemicals were allowed to come to equilibrium at 25°,
CH3CO2H + C2HOH <-> CH3COOC2H5 + H20
If the equilibrium mixture contained 0.845 mol of ethyl ethanoate, calculate the value of Kc at this temperature.
I have no idea where to start. I thought maybe I had to find the moles of H20 but then I wasn't sure what to do with the other values (like temp and volume).
Thanks for any help!
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