Question

Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
\(CH_3COOH (l) + C_2H_5OH (l) ⇋ CH_3COOC_2H_5 (l) + H_2O (l) \)

1. Write the concentration ratio (reaction quotient), Q_c , for this reaction (note: water is not in excess and is not a solvent in this reaction)
2. At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant. 
3. Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

Answer 1

Reaction:

CH₃COOH(l) + C₂H₅OH(l) ⇋ CH₃COOC₂H₅(l) + H₂O(l)

(All substances are liquids and water is neither a solvent nor in excess.)

---

(i) Reaction quotient, Q_c

For the given reaction,

Q_c = [CH₃COOC₂H₅][H₂O] / ([CH₃COOH][C₂H₅OH])

---

(ii) Calculation of equilibrium constant (K_c)

Given:

Initial moles:
Acetic acid = 1.00 mol
Ethanol = 0.18 mol
Ethyl acetate = 0 mol
Water = 0 mol

Moles of ethyl acetate at equilibrium = 0.171 mol

---

Step 1: Write equilibrium composition

CH₃COOH at equilibrium = 1.00 − 0.171 = 0.829 mol
C₂H₅OH at equilibrium = 0.18 − 0.171 = 0.009 mol
CH₃COOC₂H₅ at equilibrium = 0.171 mol
H₂O at equilibrium = 0.171 mol

---

Step 2: Write K_c expression

K_c = (0.171 × 0.171) / (0.829 × 0.009)

K_c = 0.0292 / 0.00746

K_c = 3.91

---

(iii) Has equilibrium been reached?

Given:

Initial moles:
Acetic acid = 1.0 mol
Ethanol = 0.5 mol

Moles of ethyl acetate formed = 0.214 mol

---

Step 1: Calculate reaction quotient Q_c

At this stage:

CH₃COOH = 1.0 − 0.214 = 0.786 mol
C₂H₅OH = 0.5 − 0.214 = 0.286 mol
CH₃COOC₂H₅ = 0.214 mol
H₂O = 0.214 mol

Q_c = (0.214 × 0.214) / (0.786 × 0.286)

Q_c = 0.0458 / 0.225

Q_c = 0.20

---

Step 2: Compare Q_c and K_c

Q_c = 0.20
K_c = 3.91

Since Q_c < K_c, the reaction has not reached equilibrium.

The reaction will proceed in the forward direction to form more ethyl acetate.

---

Final Answers:

(i) Q_c = [CH₃COOC₂H₅][H₂O] / ([CH₃COOH][C₂H₅OH])

(ii) K_c at 293 K = 3.91

(iii) Equilibrium has not been reached; reaction proceeds forward.