Question

At T(K), 3 moles of ethyl alcohol is mixed with 3 moles of acetic acid in a 1L vessel. At equilibrium, 2 moles of ester are formed. Calculate the equilibrium constant for the reaction.

• A. 4
• B. 2/9
• C. 2
• D. 4/9

Hint:

The equilibrium constant (K) helps predict the extent of reaction under specified conditions and is crucial for understanding reaction dynamics.

Answer 1

The Correct Option is A

Step 1: Establish initial conditions and changes at equilibrium. Initial conditions set at 3 moles each for reactants with no ester or water. At equilibrium, formation of 2 moles of ester and water reduces reactants by 2 moles each. 
Step 2: Calculate equilibrium concentrations and derive K. Given the reaction in a 1L vessel, equilibrium concentrations are directly equal to the moles at equilibrium: \[ [C_2H_5OH] = 1 { M}, [CH_3COOH] = 1 { M}, [CH_3COOC_2H_5] = 2 { M}, [H_2O] = 2 { M} \] \[ K = \frac{[CH_3COOC_2H_5][H_2O]}{[C_2H_5OH][CH_3COOH]} = \frac{2 \times 2}{1 \times 1} = 4 \]

Answer 2

To solve the problem, we need to calculate the equilibrium constant (K) for the reaction given the initial moles of reactants and the number of moles of ester formed at equilibrium.

1. Understanding the Reaction and ICE Table:
The given reaction is the esterification of ethyl alcohol with acetic acid: \[ \text{Ethyl Alcohol (C}_2\text{H}_5\text{OH)} + \text{Acetic Acid (CH}_3\text{COOH)} \rightleftharpoons \text{Ester (CH}_3\text{COOC}_2\text{H}_5) + \text{Water (H}_2\text{O)} \] At equilibrium, we are given that 2 moles of ester are formed in a 1 L vessel. We are also given the initial moles of ethyl alcohol and acetic acid as 3 moles each. We can use an ICE table (Initial, Change, Equilibrium) to determine the equilibrium concentrations:
Initial:
Ethyl Alcohol: 3 moles
Acetic Acid: 3 moles
Ester: 0 moles
Water: 0 moles

Change:
The change in moles of ethyl alcohol and acetic acid is -2 moles (since 2 moles of ester are formed).
The change in moles of ester and water is +2 moles.

Equilibrium:
Ethyl Alcohol: 1 mole (3 - 2)
Acetic Acid: 1 mole (3 - 2)
Ester: 2 moles (0 + 2)
Water: 2 moles (0 + 2)

2. Calculating the Equilibrium Constant:
The equilibrium constant for the reaction is given by: \[ K = \frac{[\text{Ester}] [\text{Water}]}{[\text{Ethyl Alcohol}] [\text{Acetic Acid}]} \] Substituting the equilibrium concentrations: \[ K = \frac{(2)(2)}{(1)(1)} = \frac{4}{1} = 4 \]

3. Identifying the Correct Answer:
The equilibrium constant \( K \) is 4, which corresponds to Option 1.

Final Answer:
The correct answer is Option (A): 4.