A mixture of 0.5 mole of ethanoic acid and 1 mole of ethanol was allowed to reach equilibrium at room temperature. The equilibrium mixture was quickly titrated and 40cm3 of 2moldm-3 sodium hydroxide was used in the titration.
a) Why is the reaction titrated quickly?
b) Using the titration results, calculate the number of moles of
i) ethanoic acid left in the equilibrium mixture
ii) ethanoic acid reacted
iii) ethanol reacted
iv)ethanol left in the equilibrium mixture
c) Calculate the equilibrium constant for the reaction between ethanoic acid and ethanol at room temperature.
This is an equilibrium reaction which means that it proceeds in the forward direction BUT ALSO in the reverse direction. If you change the concn of acetic acid by neutralizing part of it with the NaOH titrant you change the point of equilibrium. If you do it quickly you don't make much error since the change is slow.
b) Using the titration results, calculate the number of moles of
i) ethanoic acid left in the equilibrium mixture
ii) ethanoic acid reacted
iii) ethanol reacted
iv)ethanol left in the equilibrium mixture
c) Calculate the equilibrium constant for the reaction between ethanoic acid and ethanol at room temperature.
mols NaOH used = M x L = 2 x 0.04 = 0.08
mols HAc (CH3COOH) used = 0.08
mols HAc left at equilibrium = 0.5-0.08 = 0.43
Now set up and ICE chart.
....................CH3COOH + HOC2H5 --> CH3COOC2H5 + H2O
I....................0.5....................1.0..................0.......................0
C.................-0.08.................-0.08..............+0.08...............+0.08
E....................?..........................?....................?.......................?
Plug the E line into Kc expression and calculate Kc.
Post your work if you get stuck. Check my work.
a) The reaction is titrated quickly to capture the exact moment when the equivalence point is reached and determine the amount of acid and base reacted.
b) i) To calculate the number of moles of ethanoic acid left in the equilibrium mixture, we need to subtract the moles of sodium hydroxide used in the titration from the initial moles of ethanoic acid.
Let's calculate it step by step:
Moles of ethanoic acid initially = 0.5 mole (given)
Moles of sodium hydroxide used in titration = 40 cm3 x 2 mol dm-3 = 0.08 mole
Moles of ethanoic acid left = 0.5 mole - 0.08 mole = 0.42 mole
ii) To calculate the moles of ethanoic acid reacted, we need to subtract the final moles of ethanoic acid from the initial moles.
Moles of ethanoic acid reacted = 0.5 mole - 0.42 mole = 0.08 mole
iii) Since the reaction is 1:1 between ethanoic acid and ethanol, the moles of ethanol reacted will be the same as the moles of ethanoic acid reacted.
Moles of ethanol reacted = 0.08 mole
iv) To calculate the moles of ethanol left in the equilibrium mixture, we need to subtract the moles of ethanol reacted from the initial moles of ethanol.
Moles of ethanol initially = 1 mole (given)
Moles of ethanol left = 1 mole - 0.08 mole = 0.92 mole
c) To calculate the equilibrium constant (Kc), we can use the concentration of the products divided by the concentration of the reactants, raised to their respective stoichiometric coefficients.
The balanced equation for the reaction is:
CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O
Kc = [CH3COOC2H5] × [H2O] / [CH3COOH] × [C2H5OH]
However, we don't have the concentration values, only the moles. So we need to calculate the concentrations first.
Concentration of CH3COOH = moles / total volume
Concentration of CH3COOH = 0.42 mole / (0.04 L) = 10.5 moldm-3
Concentration of C2H5OH = moles / total volume
Concentration of C2H5OH = 0.92 mole / (0.04 L) = 23 moldm-3
Concentration of CH3COOC2H5 = moles / total volume
Concentration of CH3COOC2H5 = 0.08 mole / (0.04 L) = 2 moldm-3
Concentration of H2O = Not included in the calculation as it is the solvent.
Kc = (2 moldm-3) / (10.5 moldm-3 × 23 moldm-3)
Kc ≈ 0.0084 (approximately)
Remember, these calculations are based on the given data, and the values are approximations for illustrative purposes.
a) The reaction is titrated quickly to ensure that the equilibrium conditions at the moment of titration are maintained. If the reaction is allowed to proceed for too long, the equilibrium may shift and the titration results may not be accurate.
b) i) To calculate the number of moles of ethanoic acid left in the equilibrium mixture, we need to determine the amount of sodium hydroxide that reacted with it. From the balanced chemical equation, we know that the ratio of moles between ethanoic acid and sodium hydroxide is 1:1. Therefore, the number of moles of ethanoic acid left is equal to the number of moles of sodium hydroxide used in the titration.
Given that the concentration of sodium hydroxide is 2 mol/dm^3 and the volume used is 40 cm^3 (which we convert to dm^3 by dividing by 1000), we can calculate the number of moles of ethanoic acid left:
Moles of sodium hydroxide used = concentration * volume = 2 mol/dm^3 * (40 cm^3 / 1000 cm^3/dm^3) = 0.08 mol.
Therefore, the number of moles of ethanoic acid left in the equilibrium mixture is 0.08 mol.
ii) To calculate the number of moles of ethanoic acid reacted, we need to subtract the moles of ethanoic acid left from the initial moles of ethanoic acid. The initial mole ratio between ethanoic acid and ethanol is 1:2. Therefore, the initial number of moles of ethanoic acid is 0.5 mol * (1/3) = 0.167 mol (using the ratio 1/3 since the ratio between ethanoic acid and ethanol is 1:2).
Number of moles of ethanoic acid reacted = initial moles - moles left = 0.167 mol - 0.08 mol = 0.087 mol.
iii) The ratio between the number of moles of ethanol and ethanoic acid is 2:1. Therefore, the number of moles of ethanol reacted can be calculated as half of the number of moles of ethanoic acid reacted:
Number of moles of ethanol reacted = 0.087 mol / 2 = 0.044 mol.
iv) The number of moles of ethanol left in the equilibrium mixture can be calculated by subtracting the number of moles of ethanol reacted from the initial moles of ethanol. The initial moles of ethanol is 1 mol * (2/3) = 0.667 mol (using the ratio 2/3 since the ratio between ethanol and ethanoic acid is 2:1).
Number of moles of ethanol left = initial moles - moles reacted = 0.667 mol - 0.044 mol = 0.623 mol.
c) The equilibrium constant (Kc) for the reaction between ethanoic acid and ethanol can be calculated using the equation:
Kc = ([ethanoic acid]^x * [ethanol]^y) / ([acetic acid]^m * [ethanol]^n)
The balanced chemical equation for the reaction is:
CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O
From the balanced equation, we can see that the stoichiometric coefficients are 1 for ethanoic acid and ethanol, 1 for ethyl acetate, and 1 for water.
Therefore, the equilibrium constant expression can be written as:
Kc = ([CH3COOC2H5][H2O]) / ([CH3COOH][C2H5OH])
Since we are considering the reaction at room temperature, we assume that the concentration of water does not change significantly, so it can be treated as constant.
Let's assume that the concentration of ethyl acetate at equilibrium is [CH3COOC2H5] (mol/dm^3), the concentration of ethanoic acid at equilibrium is [CH3COOH] (mol/dm^3), and the concentration of ethanol at equilibrium is [C2H5OH] (mol/dm^3).
Given that the equilibrium mixture was reached using 0.5 mole of ethanoic acid and 1 mole of ethanol, we obtain the following relationship: [CH3COOH] = [C2H5OH] * 2.
Substituting these values into the equilibrium constant expression, we get:
Kc = ([CH3COOC2H5][H2O]) / ([C2H5OH] * 2 * [C2H5OH])
Kc = [CH3COOC2H5] / (2 * [C2H5OH])
To calculate the equilibrium constant, we need the concentrations of ethyl acetate and ethanol. Unfortunately, we do not have this information in the given scenario, so we cannot calculate the equilibrium constant.
a) The reaction is titrated quickly to ensure that the equilibrium is not disturbed. By quickly adding the sodium hydroxide, any further reaction or shifting of the equilibrium is minimized or prevented. This allows for accurate measurement of the reactants and products involved in the equilibrium mixture.
b) To calculate the number of moles of ethanoic acid and ethanol reacted or left in the equilibrium mixture, we can use the stoichiometry of the reaction and the given information from the titration.
i) To calculate the number of moles of ethanoic acid left in the equilibrium mixture, we need to determine the moles of sodium hydroxide reacted with it. Since the balanced equation of the reaction is not given, I will assume a simplified reaction:
Ethanoic acid + Sodium hydroxide → Ethanoate ion + Water
If 40 cm3 of 2 mol/dm3 sodium hydroxide was used, the number of moles of sodium hydroxide used in the titration can be calculated using the formula:
Number of moles = (concentration × volume) / 1000
Number of moles of sodium hydroxide = (2 mol/dm3 × 40 cm3) / 1000 = 0.08 mol
If we assume the reaction is 1:1 between ethanoic acid and sodium hydroxide, then the number of moles of ethanoic acid left in the equilibrium mixture would also be 0.08 mol.
ii) To calculate the number of moles of ethanoic acid reacted, we subtract the number of moles of ethanoic acid left from the initial moles present in the mixture. From the given information, we know that the initial mixture contained 0.5 mol of ethanoic acid. Thus,
Number of moles of ethanoic acid reacted = Initial moles - Moles left in equilibrium mixture
= 0.5 mol - 0.08 mol
= 0.42 mol
iii) Similar to calculating the moles of ethanoic acid reacted, we need to determine the moles of sodium hydroxide reacted with ethanol. Again, assuming a simplified reaction:
Ethanol + Sodium hydroxide → Ethoxide ion + Water
Using the same method as before, we find that the number of moles of sodium hydroxide used is 0.08 mol. Therefore, we can assume that 0.08 mol of ethanol has reacted.
iv) To calculate the moles of ethanol left in the equilibrium mixture, we subtract the moles of ethanol reacted from the initial moles present. From the given information, we know that the initial mixture contained 1 mol of ethanol. Thus,
Number of moles of ethanol left = Initial moles - Moles reacted
= 1 mol - 0.08 mol
= 0.92 mol
c) To calculate the equilibrium constant (Kc) for the reaction between ethanoic acid and ethanol, we would need the concentrations of the reactants and products at equilibrium. Unfortunately, the given information does not provide these concentrations. Therefore, it is not possible to calculate the equilibrium constant without additional data.