Split_00061 - 8.86 (contd) b g b g H HCl, n = 5.75 = H s 25...

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8-61 ± . ± . . . Hn H n n mC dT sp HCl, C, kJ mol 1120 g 0.66 cal C J kJ 8 mols g C cal 10 J HCl 3 == ° =+ =− + −° ⋅° z 575 25 1 64 87 40 25 4184 25 40 bg b g Δ ± .. . . HT T T d T HCl, 20 C = -0.15 kJ / mol o 25 20 ej × + × × −−− z 0 02913 01341 10 0 9715 10 4 335 10 58 2 1 2 3 QH Δ 471 kJ L product c. H H T T = −− −= = + = 08 8 0 1 5 015 6487 1120 0 66 25 1 192 Δ ± . . ± . . g 8 mol cal gC C 4.184 J cal kJ 1000 J C o o o 8.87 Basis: Given solution feed rate 0.999 H O (mol air/min) 2 n a 200°C, 1.1 bars 150 mol/min solution 0.001 NaOH 25°C 0.95 H O (mol air/min) 2 n a (mol H O( )/min) 0.05 NaOH n 12 v saturated @ 50°C, 1 atm (mol/min) @ 50°C n 2 . . . . NaOH balance: 0001 150 0 05 30 22 ²² . b g =⇒ = nn mol min H O balance: mol H O min 2 2 0 999 150 0 95 30 147 11 . ² ² b g b g ⇒= Raoult’s law : yP n Pp n a n P a HO Table B.4 C mm Hg mol air min = + = = = = ² . ² ² 1 1 147 760 50 92 51 1061 1 ² . , V inlet air 1061 mol 22.4 L STP 473 K bars min 1 mol 273 K 1.1 bars Lm in 1013 37 900 References for enthalpy calculations:
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This note was uploaded on 01/07/2012 for the course CH E 210 at Pennsylvania State University, University Park.

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