Chemical Eq Part #3 Post-Module Assessment

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gabbym
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Joined: Wed Apr 11, 2018 3:00 am

Chemical Eq Part #3 Post-Module Assessment

Postby gabbym » Wed Jan 23, 2019 9:21 am

15. 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.
Can someone please work this question out for mw and show me how you did it? I'm not sure I'm setting it up right.

ChathuriGunasekera1D
Posts: 78
Joined: Fri Sep 28, 2018 12:17 am

Re: Chemical Eq Part #3 Post-Module Assessment

Postby ChathuriGunasekera1D » Wed Jan 23, 2019 10:20 am

So [C2H5OH] is 2 M and [CH3COOH] is 1 M and K is 4 so the ICE table looks like

C2H5OH (aq) + CH3COOH (aq) ⇌ CH3COOC2H5 (aq) + H2O (l)

2 M 1 M 0 M
-x -x +x
2-x 1-x x

The equation is (x)/((2-x)(1-x)) = 4 and because K isn't that small, you have to do the whole quadratic and all that.

x/(2-3x+x^2) = 4
x = 4x^2 - 12x +8
0 = 4x^2 -13x +8
After you do quadratic you get two answers but one is too large and I think negative too so it can't work and the other is 0.825 M
[C2H5OH] = 2-0.825 = 1.175 M
[CH3COOH] = 1-0.825 = 0.175 M
[CH3COOC2H5] = 0.825 M

You can check you answer by plugging it into the K expression and if you do you get about 4


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