Elementary Principles 483 - 11.15 (cont'd) b. C A = 0.5C A0...

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11- 12 b. CC kt t kC AA A =⇒ −+ = = 05 1 11 0 00 12 0 . . ; but C n V P RT t RT kP A A 0 0 == = nn = 0 . nnA BA n A 2 2 .. mol react. mol mol react. bg b g CA n A 1 2 025 mol react. mol mol react. b g total moles = = 125 01 2 0 0 . nP nR T V P A A c. Plot t vs. 1 0 P on rectangular paper. Data fall on straight line (verifying 2 nd order decomposition) through tP 0 1060 1 1 0135 ,. di & 0 209 1 1 0 683 Slope: s atm K L a t m m o l K 143.2 s atm Lmo ls RT k k = =⋅ ⇒= ⋅⋅ 1060 209 1 0135 1 0 683 1432 1015 0 08206 0582 . . . b g d. t RT kP E RT RT k E RT 12 0 0 = F H G I K J F H G I K J =+ exp ln ln Plot tPR T (log scale) vs. 1 T (rect. scale) on semilog paper. R 0 1 0 08206 s atm, L atm / (mol K) T K b g , Data fall on straight line through T T 74 0 1 1 900 ., & T T 0 6383 1 1 1050 E R = = ln . . , 0 6383 74 0 1 1050 1 900 29 940 E 249 10 5 . J m o l ln ln . , 1 06383 29 940 1050 28 96 379 10 0 0 12 k k =− = = × L(mo ) e. Tk k E RT =
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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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