At a certain temperature K = 4.0 for the esterification of ethanol and acetic ac
ID: 921450 • Letter: A
Question
At a certain temperature K = 4.0 for the esterification of ethanol and acetic acid:
C2H5OH (aq) + CH3COOH (aq) CH3COOC2H5 (aq) + H2O (l)
If initially 2.0 moles of C2H5OH and 1.0 mole of CH3COOH are in a 1.0 L aqueous solution, what are the equilibrium concentrations of the organic molecules.
Hint 1: Calculate the initial concentration of the reactants.
Hint 2: Set up your equilibrium table (Initial, Change, and Equilibrium concentrations) assuming that the concentration of each reactant decreases by X.
Hint 3: Ignore the solvent concentration.
Hint 4: Write an expression for the equilibrium constant.
Hint 5: Solve your quadratic equation. Hint 6: Always ignore solutions with negative values or values that are larger than the initial concentration.
Explanation / Answer
Answer – We are given, reaction –
C2H5OH (aq) + CH3COOH (aq) <----> CH3COOC2H5 (aq) + H2O (l)
Moles of C2H5OH (aq) = 2.0 , moles of CH3COOH (aq) = 1.0 moles
Volume = 1.0 L , K = 4.0
Now first we need to calculate the concentration of each
[C2H5OH (aq)] = 2.0 moles / 1.0 L = 2.0 M
[CH3COOH (aq)] = 1.0 moles / 1.0 L = 1.0 M
Now we need to put ICE chart
C2H5OH (aq) + CH3COOH (aq) <----> CH3COOC2H5 (aq) + H2O (l)
I 2.0 1.0 0
C -x -x +x
E 2.0-x 1.0-x +x
We know,
K = [CH3COOC2H5 (aq)] / [C2H5OH (aq)] [CH3COOH (aq)]
4.0 = x / (2.0-x) (1.0-x)
4.0 [(2.0-x) (1.0-x)] = x
4.0 ( x2 -3x+2) = x
4.0x2 -12x +8 = x
4.0x2 -13x +8 = 0
Using the quadratic equation
x = 0.825 M
so, at equilibrium .
[C2H5OH (aq)] = 2.0-x
= 2.0-0.825
= 1.175 M
[CH3COOH (aq)] = 1.0 – 0.825
= 0.175 M
[CH3COOC2H5 (aq)] = x = 0.825 M
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