n one experiment, a mixture of 1.000 mol acetic acid and 0.5000 mol ethanol is brought to equilibrium. A sample containing exactly one-hundredth of the equilibrium mixture requires 28.80mL 0.1040M Ba(OH)2 for its titration.
Calculate the equilibrium constant, , for the ethanol-acetic acid reaction based on this experiment.
Ethanol + acetic acid ==> ethyl acetate + H2O
Ba(OH)2 + 2CH3COOH ==> Ba(CH3COO)2 + 2H2O
moles Ba(OH)2 = 0.02880*0.1040 = 0.002995.
moles CH3COOH remaining after the rxn (in the aliquot taken for analysis) = 2*0.002995 = 0.00599 moles.
Since that was exactly 1/100 of the equilibrium mixture, mol CH3COOH after the rxn = 0.599. I don't have enough room to write the equation unless we shorten how we write the chemicals. Therefore CH3COOH becomes HAc and ethanol becomes EtOH.
............HAc + EtOH ==> EtAc + H2O
initial..... 1.0...0.50.....0......0
change.......-x......-x......+x....+x
final........0.599....0.099..401...401
So if we start with 1.0 mol HAc and we have 0.599 at equilibrium, that means x, the amount reacted must be 0.401 and you can complete the ICE table as I've done above.
Then substitute equilibrium values into the Kc expression for the reaction and calculate the Keq.
To calculate the equilibrium constant for the ethanol-acetic acid reaction, we can use the concept of stoichiometry and the balanced equation of the reaction. Since the reaction involves the titration of acetic acid and ethanol with Ba(OH)2, the balanced equation for the reaction is:
CH3COOH + C2H5OH -> CH3COOC2H5 + H2O
We can see from this equation that the stoichiometric ratio of acetic acid to ethanol is 1:1. This means that one mole of acetic acid will react with one mole of ethanol to form one mole of ethyl acetate and one mole of water.
In our experiment, we have a mixture of 1.000 mol acetic acid and 0.5000 mol ethanol. Let's assume that x mol of acetic acid and x mol of ethanol reacted to form x mol of ethyl acetate and x mol of water.
From the balanced equation, we can see that the number of moles of acetic acid and ethanol that reacted is the same. Therefore, using the stoichiometric ratio, we can say:
1.000 mol acetic acid - x mol reacted
0.5000 mol ethanol - x mol reacted
Now, let's calculate the moles of Ba(OH)2 that reacted with the ethyl acetate formed. Since the sample containing one-hundredth of the equilibrium mixture required 28.80 mL of 0.1040 M Ba(OH)2 for its titration, we can calculate the moles of Ba(OH)2:
moles of Ba(OH)2 = volume of Ba(OH)2 (L) * concentration of Ba(OH)2 (M)
moles of Ba(OH)2 = (28.80 mL / 1000 mL/L) * 0.1040 M
Now, since the stoichiometric ratio between ethyl acetate and Ba(OH)2 in the balanced equation is 1:2, we need to multiply the moles of Ba(OH)2 by 2:
moles of ethyl acetate = 2 * moles of Ba(OH)2
Finally, we can calculate the equilibrium constant.
equilibrium constant = (moles of ethyl acetate) / (moles of acetic acid * moles of ethanol)
Substituting the values we obtained, we can calculate the equilibrium constant for the ethanol-acetic acid reaction based on this experiment.