Ethanoic acid or acetic acid (CH3COOH) is partially dimerized into dimers in the vapor phase. At a t — Analytical Chemistry Chemistry Question
Dimerization of Acetic Acid
Ethanoic acid or acetic acid (CH3COOH) is partially dimerized into dimers in the vapor phase. At a total pressure of 0.200 atm, ethanoic acid is 92.0% dimerized at 298 K. By increasing the temperature to 318 K, the degree of dimerization is lowered, with Kp = 37.3.
Calculate the enthalpy and the entropy changes for the reaction, assuming that ΔHo and ΔSo do not vary with temperature.
Model Answer
The reaction can be represented as 2 CH3COOH ⇌ (CH3COOH)2.
If we start with, say, 100.0 mol CH3COOH and 92.0% dimerizes, then 8.0 mol will be present at equilibrium. The 92.0 mol that react give rise to 46.0 mol dimers. The total number of moles present is therefore 54.0. Hence using the data at 298 K, the equilibrium constant, Kp, is given by
Kp = p(CH3COOH)2 / p^2(CH3COOH) = (x(CH3COOH)2 p_tot) / (x(CH3COOH) p_tot)^2 = ((46.0 / 54.0) × 0.200) / (((8.0 / 54.0) × 0.200)^2) = 194
From equation ΔGo = –RT ln Kp,
At 298 K, ΔGo = – 8.314 J K-1 mol-1 × 298 K × ln 194 = –13.0 kJ mol-1
At 318 K, ΔGo = – 8.314 J K-1 mol-1 × 318 K × ln (37.3) = –9.57 kJ mol-1
Applying the equation ΔGo = ΔHo – TΔSo :
The method for the calculation is to write the two equations
ΔGo (298 K)= –13.0 kJ mol-1 = ΔHo – 298 K × ΔSo
ΔGo (318 K)= –9.57 kJ mol-1 = ΔHo – 318 K × ΔSo
and solve the equations simultaneously.
Then ΔHo = – 64.1 kJ mol-1 and ΔSo = –170 J K-1 mol-1.
Applying the Le Chatelier’s principle, an increase of pressure should (select one correct answer)
⃝ favor the dimerization.
⃝ not favor the dimerization.
Model Answer
Applying the Le Chatelier’s principle, an increase of pressure should:
favor the dimerization.
Continued from question 1.2, the extent of dimerization (select one correct answer)
⃝ decreases with increasing the temperature.
⃝ increases with increasing the temperature.
Model Answer
The extent of dimerization:
decreases with increasing the temperature.