Ok this is a three part questions but I think i correctly answered #1 but i get

Question

Ok this is a three part questions but I think i correctly answered #1 but i get lost after that..

1. An aqueous solution of ethyl acetate at room temperature has an initial concentration of 0.15 M. Calculate the equilibrium concentrations of ethyl acetate (CH3COOC2H5), ethanol (C2H5OH) and acetic acid if the equilibrium constant at this temperature is Kc = 3.00.
CH3COOC2H5(aq) + H2O ? CH3COOH(aq) + C2H5OH(aq)

Here I set up an ice table an ice table and got a quadratic of x^2+3.00x-.45=0
and i got x=.143.

Now here is what I get confused..

2) 2. A solution contains 0.15 M acetic acid and 0.15 ethanol. If the equilibrium constant defined by the reaction in Problem 1 is Kc = 3.00, calculate the equilibrium concentrations of ethyl acetate, acetic acid and ethanol.

Im guessing you just do products/reactants?
so
[0.15 M] x [0.15 M] / [CH3COOC2H5] = 3.00
and CH3COO2H5 = .0075 M.

I think im right up to this point then im not sure what to do next?

then there's

3. A solution contains 0.01 M ethyl acetate, 0.10 M acetic acid and 0.10 M ethanol. For the same equilibrium constant of 3.00, calculate the equilibrium concentrations of ethyl acetate, ethanol, and acetic acid in the solution.

Explanation / Answer

a. CH3COOC2H5 + H2O = CH3COOH + C2H5OH
0.15-x x x


Kc = x2/(0.15-x) =3

x2 +3x -0.45 = 0

x = -3+3.07 /2 = 0.286/2 = 0.143

equilibrium conc of acetic acid and ethanol is 0.143 M

of ethyl acetate is 6.83*10-3 M

b. CH3COOC2H5 + H2O = CH3COOH + C2H5OH
0.15-x 0.15+ x 0.15+x

Kc = (0.15+x)2/(0.15-x) = 3

x2 + 0.3x + .0225 = 0.45 - 3x

x2 +3.3x -0.4275 = 0

x = 0.124

equilibrium conc of acetic acid and ethanol is 0.124 + 0.15 = 0.274 M

of ethyl acetate is 0.0252 M

c.  CH3COOC2H5 + H2O = CH3COOH + C2H5OH
0.01-x 0.1+ x 0.1+x

Kc = (0.1+x)(0.1+x)/(0.01-x) = 3

x2 + 0.2x + 0.01 = 0.03 -3x

x2 + 3.2x -0.02 = 0

x = 6.237*10-3

equilibrium conc of acetic acid and ethanol is 6.237*10-3 + 0.1 = 0.106237 M

of ethyl acetate is 3.76*10-3 M

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