ICE table, esterification problem (just need work checked)

In summary, the equilibrium constant for the esterification of benzoic acid with methanol is 3. PhCO2Me MeOH PhCO2H takes 0.082 mol of PhCO2H and 0.62 mol of CH3OH to reach equilibrium. The theoretical yield of benzoate is expected using 0.082 mol of PhCO2H and 0.62 mol of CH3OH.
  • #1
FlipStyle1308
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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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  • #2
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.
 
  • #3
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.
 
  • #4
FlipStyle1308 said:
And yes, this is in the homework forum section.
My bad - I need to get to bed.
 

1. What is an ICE table?

An ICE table is a tool used in chemistry to keep track of the initial, change, and equilibrium concentrations of species in a reaction. It helps to determine the equilibrium constant (K) for the reaction.

2. How do I set up an ICE table for an esterification problem?

To set up an ICE table for an esterification problem, first write out the balanced chemical equation for the reaction. Then, list all the species involved in the reaction and their initial concentrations. Next, determine the changes in concentration for each species based on the stoichiometry of the reaction. Finally, use the equilibrium constant expression to calculate the equilibrium concentrations.

3. What is the purpose of an ICE table in an esterification problem?

The purpose of an ICE table in an esterification problem is to determine the equilibrium concentrations of reactants and products in the reaction. This information can then be used to calculate the equilibrium constant for the reaction.

4. How do I solve for the equilibrium constant using an ICE table?

To solve for the equilibrium constant (K) using an ICE table, divide the equilibrium concentrations of the products by the equilibrium concentrations of the reactants, each raised to their respective stoichiometric coefficients. This will give you the equilibrium constant expression, which can then be solved for K.

5. Can an ICE table be used for any type of chemical reaction?

Yes, an ICE table can be used for any type of chemical reaction as long as the reaction is in equilibrium and the equilibrium concentrations of the species are known or can be calculated. However, some more complex reactions may require additional considerations and modifications to the ICE table setup.

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