Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

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Unformatted text preview: 2 appears to be suspect, so discard it. The following results are q = k c1/n obtained using Eq. (15-35) for the Freundlich isotherm: and (15-36) for the Langmuir isotherm: Component Freundlich: k 74.0218 n 2.86424 q = qmKc/(1 + Kc) Langmuir: K 1.30434 qm 161.863 Toluene in water Water in 0.199624 1.42732 0.00214855 28.0673 toluene The plots of the fits are shown below and on the next page. For toluene in water, the Freundlich isotherm gives the best fit. For water in toluene, the Freundlich isotherm gives a very good fit. The linear isotherm gives a poor fit for both cases. Exercise 15.12 (continued) Analysis: (continued) Exercise 15.13 Subject: Ion exchange of sulfate ions with chloride ions, and regeneration of the resin. Given: 60 L of water with sulfate ion and chloride ion concentrations of 0.018 eq/L and 0.002 eq/L, respectively. Remove sulfate ion by chloride ion exchange on 1 L of a strong-base resin. Regeneration of the resin with 30 L of 10 wt% aq. NaCl. Resin ion-exchange capacity of 1.2 eq/L. From Table 15.6, molar selectivities are: 1.0 for Cl- and 0.15 for SO4=. Find: Derivation of Eq. (15-44). (a) Ion-exchange reaction. (b) Value of chemical equilibrium constant for KSO4=,Cl-. (c) Equilibrium concentrations of SO4= and Cl- ions in solution and in resin in eq/L for the ion-exchange step. (d) Concentration of Cl- in eq/L for the regenerating solution. (e) Equilibrium concentrations of SO4= and Cl- ions in solution and in resin in eq/L for the regeneration step. (f) Whether the steps are sufficiently selective. Analysis: Let A=SO = 4 and B=Cl− Eq. (15-44) is KA,B C = Q n −1 yA 1 − xA n xA 1 − yA n (1) This a case of unequal charges, so use Eq. (15-37) for the reaction: (a) SO 2(−) + 2Cl R (s) ↔ SO 4 R 2 (s) + 2Cl − (l ) 4l Thus, n = 2 and by the law of mass action: 2 qSO 4 R cCl − KA,B = 2 qClR cSO 2− (2) 4 The counterion valences are: zA = 2 and zB = 1 From Eq. (15-39), cA = CxA/2 and cB = CxB/1 From Eq. (15-40), qAR = QyA/2 and qBR = QyB/1 (3) (4) Substitution of Eqs. (2) and (3) into (1), gives: Q 2 yA C 2 x B 2 C yA 1 − xA C 2 KA,B = = = KA,B = 2 Q x A 1 − yA Q 2C Q 2 yB xA 2 since, n =2. This is Eq. (15-44). n −1 yA 1 − xA n xA 1 − yA n Exercise 15.13 (continued) Analysis: (continued) (b) From Eq. (15-45), using values of Ki from Table 15.6, KA,B = KA/KB = 0.15/1.0 = 0.15 (5) (c) For the initial ion-exchange step, Q = 1.2 eq/L and C = cA + cB = 0.018 + 0.002 = 0.02 eq/L Let: a = eq of SO4= exchanged. The equivalent fractions of the ions for 1.0 L of resin, 60 L of aq. solution, and n = 2 are: a 0.018 − 60 = 0.9 − 0.833a xSO= = (6) 4 0.02 a /1 a ySO= = = (7) 4 1.2 1.2 Combining Eqs (1), (5), (6), and (7), a 1 − 0.9 + 0.833a C 0.02 1.2 4 KA,B = = 015 = . n Q 1.2 a xSO= yCl 0.9 − 0.833a 1 − 4 12 . Solving Eq. (8) with a nonlinear solver, a = 0.755 eq. From Eqs. (6) and (7), xSO= = 0.9 − 0.833(0.755) = 0.271 n −1 ySO= xCl - n 2 −1 2 2 4 ySO= = 4 0.755 = 0.629 1.2 From Eq. (15-39), CxSO= 0.02(0.271) 4 cSO=...
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