when 8.28g of ethanol were heated with 60g of ethanioc acid 49.74 of acid remain at equilibrium calcuma the value of k
mols C2H5OH = grams/molar mass = ?
mols CH3COOH = grams/molar mass = ?
..C2H5OH + CH3COOH ==> CH3COOC2H5 + H2O
I..0.18......1.0.........0...........0
C...-x.......-x..........x...........x
E.0.18-x....1.0-x........x...........x
49.74/60 = about 0.829
so 1-x = 0.829 which makes x = 0.171
Substitute the values into the Kc expression and solve for Kc.
Well, if I were to calculate the value of "k" for you, I'd probably end up with "k for clowning around"! Just kidding! But seriously, without any additional information or the specific chemical reaction involved, it's impossible for me to determine the value of "k" in this scenario.
To calculate the value of equilibrium constant, we need to know the balanced chemical equation for the reaction between ethanol and ethanoic acid. Assuming the reaction yields ethyl ethanoate and water, the balanced equation is:
CH3CH2OH + CH3COOH ↔ CH3COOCH2CH3 + H2O
We also need to know the initial and equilibrium concentrations of reactants and products. In this case, the initial concentration of ethanoic acid is 60g, and the equilibrium concentration is 49.74g.
Next, we need to calculate the initial and equilibrium concentrations of ethanol and ethyl ethanoate. Let's assume the initial concentration of ethanol is x g, and the equilibrium concentration is (8.28 - x) g. Similarly, the initial concentration of ethyl ethanoate is 0g, and the equilibrium concentration is x g.
To calculate the equilibrium constant (K), we use the following equation:
K = ([ethyl ethanoate] * [H2O]) / ([ethanoic acid] * [ethanol])
Substituting the given values:
K = (x * 0) / (49.74 * (8.28 - x))
Simplifying the equation:
K = 0
Therefore, the value of K is 0 for this reaction.
To calculate the value of K (equilibrium constant), we need to know the balanced chemical equation for the reaction between ethanol and ethanoic acid.
Let's assume the balanced equation is:
CH3CH2OH + CH3COOH ⇌ CH3COOCH2CH3 + H2O
The equation indicates that ethanol (CH3CH2OH) reacts with ethanoic acid (CH3COOH) to form ethyl acetate (CH3COOCH2CH3) and water (H2O).
Now, we are given that initially 8.28g of ethanol and 60g of ethanoic acid are present. We also know that at equilibrium, 49.74g of ethanoic acid remain.
To find the amount of ethanoic acid reacted, we can subtract the remaining amount from the initial amount:
Amount of ethanoic acid reacted = Initial amount - Remaining amount
= 60g - 49.74g
= 10.26g
Next, we need to calculate the number of moles of ethanoic acid reacted. To do this, we divide the mass of ethanoic acid reacted by its molar mass.
Molar mass of ethanoic acid (CH3COOH) = 60.05g/mol
Number of moles of ethanoic acid reacted = Mass / Molar mass
= 10.26g / 60.05g/mol
≈ 0.17 mol
Now, we can set up the expression of K, which is the ratio of products to reactants, with each concentration or partial pressure raised to the power of its stoichiometric coefficient.
K = [CH3COOCH2CH3] / ([CH3CH2OH] * [CH3COOH] * [H2O])
To find the equilibrium concentration of each compound (except water, as it is in liquid phase), we need to convert the moles of reactants and products to their concentrations. We do this by dividing the number of moles by the total volume.
Note: The volume is not given in the question, but assuming it remains constant, we can define it as 1L for simplicity.
[CH3CH2OH] = 8.28g / 46.07g/mol (molar mass of ethanol)
≈ 0.18 mol/L
[CH3COOH] = (60g - 10.26g) / 60.05g/mol
≈ 0.82 mol/L
[CH3COOCH2CH3] = 0 mol/L (as none is given remaining at equilibrium)
[H2O] = 1 mol/L (as it is in liquid phase and its concentration remains constant)
Now we can substitute these values into the expression for K:
K = (0 mol/L) / (0.18 mol/L * 0.82 mol/L * 1 mol/L)
= 0
Therefore, the value of K is 0 for this reaction.