In an esterification experiment, 2.0 mol of ethanol were mixed with 1.0 mol of ethanoic acid in a container of volume 30cm3 and the chemicals were allowed to come to equilibrium at 25°,
CH3CO2H + C2HOH <-> CH3COOC2H5 + H20
If the equilibrium mixture contained 0.845 mol of ethyl ethanoate, calculate the value of Kc at this temperature.
I have no idea where to start. I thought maybe I had to find the moles of H20 but then I wasn't sure what to do with the other values (like temp and volume).
Thanks for any help!
I can save some time by calling CH3 R and C2H5 R'.
RCOOH + HOR' ==> RCOOR' + H2O
Set up an ICE chart.
Initial:
RCOOH = 1.0 mol from the problem.
R'OH = 2.0 mol from the problem.
RCOOR' = 0 you know this and
H2O = 0 you know this because these are zero before the reaction begins..
change:
RCOOR' = x mol
H2O = +x mol
RCOOH = -x mol
R'OH = -x mol
equilibrium:
RCOOR' = 0.845 mol from the problem.
H2O = 0.845 mol (must be the same as RCOOR').
RCOOH = 1.0 - x
R'OH = 2.0 - x
If the equilibrium mixture is 0.845 mols H2O and 0.845 mols RCOOR', isn't that x? So now you know what the equilbrium mols are for RCOOH and R'OH because they are 1.0-x and 2.0-x and you know x.
So now you have the mols of each.
You need concentration in mols/L. These are moles and you know the volume (30 cc = 0.030 L) so figure the concn of each and plug in to the Kc expression. Go from there. Post your work if you get stuck.
To find the value of Kc, we need to start by writing the expression for the equilibrium constant in terms of the concentrations of the reactants and products. The equilibrium constant (Kc) expression for the given reaction is:
Kc = ([CH3COOC2H5] /[CH3CO2H] x [C2H5OH] /[H2O])
Given:
- Initial moles of ethanol (C2H5OH) = 2.0 mol
- Initial moles of ethanoic acid (CH3CO2H) = 1.0 mol
- Equilibrium moles of ethyl ethanoate (CH3COOC2H5) = 0.845 mol
- Volume of the container = 30 cm^3
To calculate the equilibrium concentrations, we need to divide the number of moles by the volume:
[CH3COOC2H5] = 0.845 mol / 30 cm^3
[CH3CO2H] = (1.0 mol - 0.845 mol) / 30 cm^3
[C2H5OH] = (2.0 mol - 0.845 mol) / 30 cm^3
Now, substitute these values into the equilibrium constant expression:
Kc = ([CH3COOC2H5] /[CH3CO2H] x [C2H5OH] /[H2O])
Kc = (0.845 mol / 30 cm^3) / ((1.0 mol - 0.845 mol) / 30 cm^3) x ((2.0 mol - 0.845 mol) / 30 cm^3)
Simplifying,
Kc = (0.845 / 30) / ((1 - 0.845) / 30) x ((2 - 0.845) / 30)
Kc = 0.0282 / (0.155 / 30) x (1.155 / 30)
Kc = 0.0282 / 0.00517
Finally,
Kc ≈ 5.459
So, the value of Kc at this temperature is approximately 5.459.
To find the value of Kc for this reaction, we need to set up an expression that represents the equilibrium constant in terms of concentrations of the reactants and products. Here's how you can approach this problem step by step:
Step 1: Determine the initial concentrations of the reactants and the product.
In this case, the initial concentration of ethanol (C2HOH) is given as 2.0 mol in a container of volume 30 cm3, meaning the initial concentration is 2.0 mol/30 cm3 = 0.0667 mol/cm3.
Similarly, the initial concentration of ethanoic acid (CH3CO2H) is given as 1.0 mol in a container of volume 30 cm3, giving us an initial concentration of 1.0 mol/30 cm3 = 0.0333 mol/cm3.
Since no ethyl ethanoate (CH3COOC2H5) is present initially, its initial concentration is zero.
Step 2: Determine the change in concentrations at equilibrium.
The balanced equation for the reaction shows that for every mole of ethanol and ethanoic acid that reacts, one mole of ethyl ethanoate and one mole of water are produced.
Therefore, the change in concentration of ethyl ethanoate (CH3COOC2H5) is 0.845 mol.
The change in concentration of water (H2O) is also 0.845 mol.
The change in concentration of ethanol (C2HOH) and ethanoic acid (CH3CO2H) is -0.845 mol because they are the reactants that are consumed.
Step 3: Calculate the equilibrium concentrations.
The equilibrium concentrations can be calculated by adding the initial concentrations to the change in concentrations.
For ethyl ethanoate (CH3COOC2H5): 0 + 0.845 mol = 0.845 mol
For water (H2O): 0 + 0.845 mol = 0.845 mol
For ethanol (C2HOH): 0.0667 mol - 0.845 mol = -0.7783 mol
For ethanoic acid (CH3CO2H): 0.0333 mol - 0.845 mol = -0.8117 mol
Note: Negative concentrations are used to show that these reactant concentrations have decreased at equilibrium.
Step 4: Calculate the Kc value using the equilibrium concentrations.
Now that we have the equilibrium concentrations, we can calculate the value of Kc using the formula:
Kc = ([CH3COOC2H5] * [H2O]) / ([C2HOH] * [CH3CO2H])
Substituting the values we obtained above:
Kc = (0.845 mol * 0.845 mol) / (0.7783 mol * 0.8117 mol)
Kc = 0.710 mol2 / 0.631 mol2
Kc ≈ 1.124 (since the values are slightly rounded)
Therefore, the value of Kc for this reaction at 25°C is approximately 1.124.