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ICE table, esterification problem (just need work checked)

  1. Jul 26, 2007 #1
    1. The problem statement, all variables and given/known data
    The equilibrium constant for the esterification of benzoic acid with methanol is 3. Calculate the theoretical yield of benzoate expected using 0.082 mol of PhCO2H and 0.62 mol of CH3OH.

    I just need my work checked.

    2. Relevant equations
    See below.

    3. The attempt at a solution
    PhCO2Me MeOH PhCO2H
    I 0.082 mol 0.62 mol —
    C -x -x x
    E 0.082 - x 0.62 - x x

    Keq = ([PhCO2Me][H2O])/([PhCO2H][MeOH]) = 3
    3 = {(x)(x)}/{(0.082 - x)(0.62 - x)}
    3 = x²/(0.05084 - 0.702x + x²)
    0.15252 - 2.106x + 3x² = x²
    2x² - 2.106x + 0.15252 = 0
    x = {2.106 ± [4.4352 - (4)(2)(0.15252)]1/2}/4 = {2.106 ± [3.27504]1/2}/4 = {2.106 ± 1.7931}/4
    theoretical yield: x = 0.9747758 g or x = 0.078225 g


    Is my work correct? Are both values of x the theoretical yield, or should I reject one of them? I am a stumbled because there is no negative value to reject. Thanks.
     
  2. jcsd
  3. Jul 27, 2007 #2

    Gokul43201

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    Water has an activity of 1.
    The water will change the equations, but ask yourself if x can theoretically exceed 0.082 moles.

    Next time, use the homework forums.
     
  4. Jul 27, 2007 #3
    Oh, the original equation is what the teacher gave us. I basically just had to plug everything in, but it didn't seem correct. So my answer is 0.078225 mol, and I convert that to grams? So 0.078225 mol x 136 g/mol = 10.6386 g? And yes, this is in the homework forum section.
     
  5. Jul 27, 2007 #4

    Gokul43201

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    My bad - I need to get to bed.
     
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