Equilibrium Constant Calculation for the Reaction of Ethyl Alcohol and Acetic Acid

Given:
- Mass of ethyl alcohol (CH3CH2OH) = 92g
- Mass of acetic acid (CH3COOH) = 120g
- Mass of ester (ethyl acetate, CH3COOCH2CH3) formed at equilibrium = 117.34g
- Mass of water (H2O) formed at equilibrium = 24g

To determine the equilibrium constant (Kc) of the reaction:

1. Write the balanced chemical equation for the reaction:
CH3CH2OH + CH3COOH ⇌ CH3COOCH2CH3 + H2O

2. Calculate the initial moles of reactants:
- Moles of ethyl alcohol = 92g / 46.0684 g/mol = 1.997 mol
- Moles of acetic acid = 120g / 60.05 g/mol = 1.997 mol

3. Calculate the final moles of products:
- Moles of ester = 117.34g / 88.1051 g/mol = 1.332 mol
- Moles of water = 24g / 18.015 g/mol = 1.332 mol

4. Write the equilibrium expression for the reaction:
Kc = [CH3COOCH2CH3][H2O] / [CH3CH2OH][CH3COOH]

5. Substitute the equilibrium concentrations (in mol/L) into the equilibrium expression:
Kc = [(1.332 mol/L)(1.332 mol/L)] / [(1.997 mol/L)(1.997 mol/L)]
Kc = 0.750

6. Round the answer to the appropriate number of significant figures:
Kc = 0.75

7. Compare the calculated value of Kc to the answer choices given in the question:
- Choice 1: 4
- Choice 2: 8
- Choice 3: 10
- Choice 4: 15

Since the calculated value of Kc (0.75) does not match any of the answer choices exactly, we need to estimate which choice is closest. We can do this by comparing the order of magnitude of the calculated value to each answer choice:

- Choice 1: 4 = 10^0.602
- Choice 2: 8 = 10^0.903
- Choice 3: 10 = 10^1
- Choice 4: 15 = 10^1.176

The calculated value of Kc (0.75) is closest in magnitude to Choice 1 (4), which is approximately 10^0.602. Therefore, the answer to the question is Choice 1 (4).

Note: It is important to round the calculated value of Kc to the appropriate number of significant figures (in this case, two) before comparing it to the answer choices. Otherwise, we may choose the wrong answer due to rounding errors.