ICE table, esterification problem (just need work checked)

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The discussion revolves around calculating the theoretical yield of benzoate from the esterification of benzoic acid with methanol, given an equilibrium constant of 3. The user presents their ICE table and calculations but is uncertain about the validity of their results, particularly regarding two values of x obtained. Responses indicate that x cannot exceed the initial amount of benzoic acid, suggesting one of the values should be rejected. The user also confirms the conversion of moles to grams for the final yield. The conversation emphasizes the importance of correctly applying equilibrium concepts in chemical calculations.
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Homework Statement


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.

Homework Equations


See below.

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.
 
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FlipStyle1308 said:
Keq = ([PhCO2Me][H2O])/([PhCO2H][MeOH]) = 3
3 = {(x)(x)}/{(0.082 - x)(0.62 - x)}
Water has an activity of 1.
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 gIs 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.
The water will change the equations, but ask yourself if x can theoretically exceed 0.082 moles.

Next time, use the homework forums.
 
Gokul43201 said:
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.

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.
 
FlipStyle1308 said:
And yes, this is in the homework forum section.
My bad - I need to get to bed.
 

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