ICE table, esterification problem (just need work checked)

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Discussion Overview

The discussion revolves around a homework problem related to the equilibrium constant for the esterification of benzoic acid with methanol. Participants are checking the calculations for the theoretical yield of benzoate based on given initial amounts of reactants. The focus is on the mathematical reasoning and assumptions involved in the equilibrium expression and the resulting calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents their calculations for the equilibrium constant and expresses uncertainty about the validity of two potential values for x, questioning whether both should be considered as theoretical yields.
  • Another participant points out that the activity of water is 1 and suggests that the equations may be affected by this, prompting a consideration of whether x can exceed the initial amount of benzoic acid (0.082 moles).
  • A later reply acknowledges the original equation provided by the teacher and confirms the conversion of one of the values of x into grams, indicating a specific numerical result.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the correctness of the calculations and whether both values of x can be accepted as theoretical yields. There is no consensus on the resolution of these questions, and the discussion remains open-ended.

Contextual Notes

Participants note that the presence of water may influence the equilibrium calculations, and there is an implicit assumption regarding the maximum possible value of x based on the initial moles of benzoic acid.

Who May Find This Useful

Students working on equilibrium problems in chemistry, particularly those involving esterification reactions and the use of ICE tables.

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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.