Elementary - 4.47(cont'd d T(K 1223 1123 1023 923 823 723 623 673 698 688 1123 1123 1123 1123 x(CO 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.2 0.4

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Unformatted text preview: 4.47 (cont'd) d. T (K) 1223 1123 1023 923 823 723 623 673 698 688 1123 1123 1123 1123 x (CO) 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.2 0.4 0.3 0.5 x (H2O) 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.4 0.2 0.3 0.4 x (CO2) 0 0 0 0 0 0 0 0 0 0 0.1 0.1 0 0 Keq Keq (Goal Seek) Extent of Reaction 0.6610 0.6610 0.2242 0.8858 0.8856 0.2424 1.2569 1.2569 0.2643 1.9240 1.9242 0.2905 3.2662 3.2661 0.3219 6.4187 6.4188 0.3585 15.6692 15.6692 0.3992 9.7017 9.7011 0.3785 7.8331 7.8331 0.3684 8.5171 8.5177 0.3724 0.8858 0.8858 0.8858 0.8858 0.8863 0.8857 0.8856 0.8867 0.1101 0.1100 0.1454 0.2156 y (H2) 0.224 0.242 0.264 0.291 0.322 0.358 0.399 0.378 0.368 0.372 0.110 0.110 0.145 0.216 The lower the temperature, the higher the extent of reaction. An equimolar feed ratio of carbon monoxide and water also maximizes the extent of reaction. 4.48 a. A + 2B C ln K e = ln A0 + E T K ln K e1 / K e 2 E= 1 T1 - 1 T2 b b g g = lnd10.5 / 2.316 10 i = 11458 -4 1 373 - 1 573 ln A0 = ln K e1 - 11458 T1 = ln 10.5 - 11458 373 = -28.37 A0 = 4.79 10-13 Ke = 4.79 10 -13 exp 11458 T K atm-2 Ke (450K ) = 0.0548 atm-1 b. c b gh n A = n A0 - nB = nB 0 - 2 nC = nC 0 + nT = nT 0 - 2 U | | V | | W y A = n A0 - nT 0 - 2 y B = nB 0 - 2 nT 0 - 2 yC = nC 0 + nT 0 - 2 nT 0 = n A0 + nB 0 + nC 0 b b b b gb gb gb g g g g bg At equilibrium, n + e nT 0 - 2 e yC 1 = C0 2 2 y A yB P n A0 - e nB 0 - 2 e c. b b gb gb g g 2 2 1 = Ke T (substitute for K T from Part a.) e P2 bg Basis: 1 mol A (CO) n A0 = 1 nB 0 = 1 nC 0 = 0 nT 0 = 2 , P = 2 atm , T = 423K b1 - gb1 - 2 g e e e 2 - 2 e b g 2 2 1 . = K e 423 = 0.278 atm -2 2 - e + 01317 = 0 e 4 atm 2 b g 4- 38 ...
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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