Chemical Analysis. Francis Rouessac

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      1 We consider that volume V, leaving the column every second, is equal to the internal volume of the column over length ū. The section of the column is πID2/4 and thus V = ūπID2/4. Flow rate D expressed in ml/min (if ū and ID are in cm) is then D = 60V. If we choose to express ID in mm, we write D = 60ūπ(0.1)2 ID 2/4, or D = 0.47ūID 2.

      2 The thermodynamic study of a chemical equilibrium leads to the formula: or or If we assume that enthalpy and entropy variations are constant in the temperature range in question and if we switch to log, we can write log k = ‐ A + B/T if we assume the entropy variation < 0 (loss of disorder when the solute is fixed on the stationary phase) and the enthalpy variation < 0 (exothermic reaction).Coelution => k1 = k2, hence log k1 = log k2 and thus T1 = 421 K or t = 148°CFor T < T1 , k2 > k1, therefore 2 elutes after 1.If α = 2, this means that k2 / k1 = 2, or log k2 ‐ log k1 = log 2, and thus T = 343 K.T = 423 K thus k1 = 0.163 and k2 = 0.161; K1 = 40.73 and K2 = 40.16.

      3 The cancelation of the first derivative of H = f(u) enables us to find u corresponding to Hmin, thus u = 36.9 mm/s; Hmin = 0.295 mm.Nmax = L/Hmin = 40,674 plates.We are looking for the value of u that gives N= 0.95Nmax or N =38,641 theoretical plates, thus H= 0.310 mm; u = 50.2 mm/s or u = 27.3 mm/s.

      4 The order of elution follows that of the increasing boiling points. The alkenes, which are more polar than the corresponding alkanes, are less retained, which is expected from a nonpolar column.α = k2/k: α = 1.12 (at –35°C), α = 1.11 (at 25°C) and α = 1.09 (at 40°C).Since the column is the same, the retention time decreases following the reduction in the partition factor K = CS/CM with temperature.N = 138,493 theoretical plates.Hmin theo = 0.113 mm.

      5 L (m)a = √LtR (min)tR/LRR/a153.873.70.252.050.53305.487.50.252.910.53607.7515.30.254.150.53tR is proportional to length.R is proportional to .δ proportional to and N is proportional to L.δ = 0.0568 min, N60 = 148,166 plates.N30 = 74,083 plates and N15 = 37,041 plates.

      6 tM = L/u = 750 s = 12.5 min. For methane, k = 0, therefore K = 0 and cannot be calculated.The higher K is, the more solute is retained and the larger the absolute value of ∆ is. Therefore, peak 10 → 2,4 dimethyl pentane, 11 → benzene, 12 → 2‐methylhexane, 13 → 2,3 dimethyl pentane and 14 → 3‐methyl hexane.N = 5.54 (tR/δ)2 = 133,126 plates.d.β = ID/4df = 250. , K = k ∙ β = 853. These are very close values.

      7 Neff , k, α; GC: methane.The experiment described helps us write three equations for three unknowns, tM, a, and b: (1) log(271 – tM) = 6a + b; (2) log(311 ‐ tM) = 7a + b; (3) log (399 ‐ tM) = 8a + b, thus tM = 237.7 s.Ik = 749. The McReynolds constant for pyridine on this stationary phase is 54.

      8 The Kovats index of butanol is thus 645. Its McReynolds constant is: 645 – 590 = 55.

      9 tM = L/u = 3,000/50 = 60 s, i.e. 1 minute. Peak No.1 probably corresponds to the solvent peak.D = VM/tM = π ∙ L ∙ (ID)2/4 tM = 40.46 mm3∙s−1, i.e. 2.4 ml/min.β = (ID)/4df = 321/(4 x 0.25) = 321.HETP for peak No.2: H = L/N with N = 5.54tR 2 /δ 2 = 27,555 plates => H = 1.09 mm.k = (tR ‐ tM)/tM k2 = 0.93 and k3 = 0.99.K = k ∙ β, thus K2 = 0.93 × 321 = 298.5; K3 = 0.99 × 321 = 317.8.α2‐3 = k3 /k2 = 1.06.R2‐3 = 1.18(tR3 – tR2)/(δ2 + δ3) = 1.2. The resolution between two peaks for a good separation must be at least 1.5. The two peaks are therefore not well separated. To improve this separation, retention times must be increased by reducing either the carrier gas flow rate or the oven temperature.The higher the Kovats index is, the longer the retention time. Therefore, tR values are in the following order: octane < toluene < methyl pentanoate.These retention times are in the order of growing polarity of the analytes; therefore, this is a polar column.