[12%] N,N-Dimethylethanolamine ((CH3)2NCH2CH2OH, DMEA, M = 89.14) is a Brønsted base whose conjugate — Thermodynamics Chemistry Question
Problem Context
[12%] N,N-Dimethylethanolamine ((CH3)2NCH2CH2OH, DMEA, M = 89.14) is a Brønsted base whose conjugate acid, DMEAH+, has a pKa = 9.22 (Ka = 6.0 × 10^-10). Acetic acid (CH3COOH, M = 60.05) has a pKa = 4.75 (Ka = 1.8 × 10^-5).
Calculate ∆G°rxn at 298 K for the acid-base reaction of DMEA with CH3COOH.
Model Answer
Reaction is DMEA(aq) + CH3COOH(aq) → DMEAH+(aq) + CH3COO–(aq)
Keq = (Ka of CH3COOH)/(Ka of DMEAH+) = (1.8 × 10^-5)/(6.0 × 10^-10)
Keq = 3.0 × 10^4
∆G° = –RTln(Keq) = –(8.314 J mol^-1 K^-1)(298 K)ln(3.0 × 10^4)
∆G° = –25500 J mol^-1 = –25.5 kJ mol^-1
A solution consisting of 7.84 g CH3COOH and 107.17 g water is placed in a well-insulated (Dewar) flask. To this solution is added DMEA in small portions. After each portion of DMEA is added, the solution is stirred and the temperature measured with a digital thermometer. The data obtained are plotted at right. The solid line represents the best linear fit to the data with < 12 g added DMEA, the dashed line the average of the data with ≥ 12 g added DMMEA; both equations are given on the plot.
Calculate ∆H°rxn for the reaction of DMEA with CH3COOH. You may assume that all solutions have the same specific heat capacity as pure water.
Model Answer
At the intersection point, 1.141x + 19.03 = 31.84, so x = 11.23 g DMEA
At this point, qrxn = –qH2O = –mCp∆T
= –(7.84 g + 107.17 g + 11.23 g)(4.184 J mol^-1 K^-1)(31.84 °C – 19.03 °C)
qrxn = –6766 J
∆H°rxn = qrxn/(mol reacted)
At this point the number of moles of both reagents are equal; mol CH3COOH reacted = (7.84 g)/(60.05 g mol^-1) = 0.131 mol.
∆H°rxn = –6766 J/(0.131 mol) = –51600 J mol^-1 = –51.6 kJ mol^-1
Calculate ∆S°rxn for the reaction of DMEA with CH3COOH.
Model Answer
∆G° = ∆H° – T∆S°
–25500 J mol^-1 = –51600 J mol^-1 – (298 K)∆S°
∆S° = –87.6 J mol^-1 K^-1
The ∆S° calculated in part c. is negative. What features of the reaction of DMEA with CH3COOH cause it to have a negative entropy of reaction?
Model Answer
The reaction produces two charged particles from two neutral ones. Charged species in aqueous solution are strongly solvated by the water, which significantly constrains the orientations of the solvent molecules around the ions. This decreases the number of ways the solvent molecules can be arranged and so decreases the entropy.