A mixture of 0.50 mol of ethanoic acid and 1.00 mol of ethanol was shaken for a long time to reach equilibrium. The whole mixture was titrated quickly with 1.00 mol dm3 sodium hydroxide and 80cm3 of alkali were required.
1. Write an equation for the reaction between ethanoic acid and ethanol.
2. Explain why the reaction mixture was titrated quickly.
3. By making use of the titration results and the equation in (1), Calculate:
a. How many moles of ethanoic acid remained at equilibrium
b. How many moles of ethanoic acid had reacted.
c. How many moles of ethanol were left in the equilibrium mixture.
d. Calculate a value for Kc for the reaction.
CH3COOH + HOCH2CH3 =>CH3COOC2H5 + H2O
3a. mol CH3COOH = M x L = about 0.80.
b. initial = 0.500-0.08 ending = 0.420
c. 1.00-0.08 = 0.92
d. Substitute and calculate Kc.
1. Well, when ethanoic acid (CH3COOH) reacts with ethanol (C2H5OH), it forms ethyl ethanoate (CH3COOC2H5) and water (H2O). So, the equation for the reaction is:
CH3COOH + C2H5OH -> CH3COOC2H5 + H2O
2. The reaction mixture was titrated quickly because we don't want it to have too much time to think about what it's doing. You know, like when you're procrastinating and then suddenly your deadline is right around the corner, and you panic and get things done as quickly as possible? Well, that's basically what we're doing here. We don't want the reaction to have any time to change its mind, so we titrate it quickly.
3. Let's do some calculations, shall we?
a. The balanced equation tells us that the ratio of ethanoic acid to ethyl ethanoate is 1:1. So, if 80 cm3 of sodium hydroxide was required to neutralize the mixture, then it means 0.08 dm3 (80 cm3 = 0.08 dm3) of sodium hydroxide was used. Since sodium hydroxide reacts with ethanoic acid in a 1:1 ratio, it means that there were still 0.08 mol of ethanoic acid left at equilibrium.
b. Since the ratio of ethanoic acid to ethanol is also 1:1, it means that the same amount of ethanoic acid (0.08 mol) had also reacted.
c. The initial amount of ethanol was 1.00 mol, and since the ratio of ethanoic acid to ethanol is 1:1, it means that 0.08 mol of ethanol reacted. So, to find the amount of ethanol left in the equilibrium mixture, we subtract the amount reacted from the initial amount: 1.00 mol - 0.08 mol = 0.92 mol.
d. To calculate the value for Kc, we need to use the concentrations of the reactants and products at equilibrium. Since we know the initial amounts and we now know the amounts at equilibrium, we can calculate the concentrations. The equation for the reaction is CH3COOH + C2H5OH -> CH3COOC2H5 + H2O, and the stoichiometric coefficients are 1:1:1:1. So, the concentration of ethanoic acid at equilibrium would be 0.08 mol / total volume, and the concentration of ethyl ethanoate at equilibrium would also be 0.08 mol / total volume. Since the concentration of water doesn't change (it's a liquid, after all), we don't need to include it in our calculations. Now, Kc is just the ratio of the product concentrations to the reactant concentrations, all raised to their stoichiometric coefficients. So, Kc = [CH3COOC2H5] / [CH3COOH][C2H5OH]. Substituting in the given values, we can find the value of Kc.
Remember, this is a serious calculation, no room for clowning around here!
1. The equation for the reaction between ethanoic acid (CH3COOH) and ethanol (C2H5OH) is as follows:
CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O
2. The reaction mixture was titrated quickly to ensure that the reaction had reached its equilibrium state. By titrating quickly, any additional reactions or changes in the system would be minimized, allowing for a more accurate determination of the equilibrium composition.
3. a. To calculate the moles of ethanoic acid remaining at equilibrium, we need to determine the initial moles of ethanoic acid and subtract the moles that reacted. Given that the initial moles of ethanoic acid is 0.50 mol and the reaction used up 80 cm3 (0.08 dm3) of 1.00 mol/dm3 sodium hydroxide, we can use the balanced equation to find the moles of ethanoic acid reacted.
From the balanced equation, the stoichiometry indicates that 1 mole of ethanoic acid reacts with 1 mole of sodium hydroxide. Therefore, 0.08 moles of sodium hydroxide reacted with 0.08 moles of ethanoic acid.
Thus, the moles of ethanoic acid remaining at equilibrium is 0.50 - 0.08 = 0.42 mol.
b. Using the same logic as above, the moles of ethanoic acid that reacted is also 0.08 mol.
c. Since the reaction of ethanoic acid and ethanol occurs in a 1:1 ratio, the moles of ethanol left in the equilibrium mixture will be the same as the moles of ethanoic acid remaining, which is 0.42 mol.
d. To calculate a value for Kc, we need the initial concentrations of reactants and products, as well as the equilibrium concentrations.
The initial concentration of ethanoic acid is 0.50 mol/1.00 dm3 = 0.50 mol/dm3
The initial concentration of ethanol is 1.00 mol/1.00 dm3 = 1.00 mol/dm3
From the titration, it can be determined that the equilibrium concentration of ethanoic acid is 0.42 mol/1.00 dm3 = 0.42 mol/dm3
Using the balanced equation, the stoichiometry indicates that the concentration of each product is equal to the concentration of ethanoic acid at equilibrium.
Therefore, the concentration of ethyl ethanoate (CH3COOC2H5) is 0.42 mol/dm3 and the concentration of water (H2O) is also 0.42 mol/dm3.
Substituting these values into the expression for Kc:
Kc = [CH3COOC2H5][H2O]/[CH3COOH][C2H5OH]
= (0.42)(0.42)/(0.42)(1.00)
= 0.42/1.00
= 0.42
Thus, the value of Kc for the reaction is 0.42.
1. The equation for the reaction between ethanoic acid (CH3COOH) and ethanol (C2H5OH) is as follows:
CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O
2. The reaction mixture was titrated quickly to ensure that the reaction had reached equilibrium. By titrating the mixture quickly, any further reaction or interaction between the components of the mixture can be minimized. This allows us to accurately determine the amounts of reactants and products present at equilibrium.
3. Let's analyze the titration results and use the equation to calculate the requested values:
a. To find the moles of ethanoic acid (CH3COOH) remaining at equilibrium, we first calculate the initial moles of ethanoic acid. Given that there is 0.50 mol of ethanoic acid initially, we subtract the moles of ethanoic acid reacted (which we'll calculate next) to find the moles remaining.
b. To find the moles of ethanoic acid (CH3COOH) reacted, we need to use the stoichiometry of the balanced equation. One mole of ethanoic acid reacts with one mole of ethanol. As we are given that 80 cm3 (which is equivalent to 0.08 dm3) of 1.00 mol dm3 sodium hydroxide is required, we can use the equation:
1.00 mol dm3 × 0.08 dm3 = 0.08 mol of sodium hydroxide.
Since the reaction is 1:1, the moles of sodium hydroxide also represent the moles of ethanoic acid reacted.
c. The number of moles of ethanol (C2H5OH) remaining at equilibrium is calculated in a similar way to part a. We take the initial moles of ethanol (which is given as 1.00 mol) and subtract the moles of ethanol reacted. Since the reaction is 1:1, the moles of sodium hydroxide used for the reaction also represent the moles of ethanol reacted.
d. To calculate the value for Kc (the equilibrium constant), we need to know the concentrations at equilibrium. However, the concentration of ethanoic acid and ethanol at equilibrium is not given. Therefore, we cannot directly calculate the Kc value without additional information.