Elementary Principles 143 - 5.49 (cont'd) c. A semilog plot...

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5- 31 5.49 (cont’d) c. A semilog plot of K p vs. 1 T is a straight line. Fitting the line to the exponential law yields ln K 7367 T 22.747 K 7.567 10 exp 7367 T a 7.567 10 atm b = 7367K pp 9 9 =− + = × F H G I K J 5.50 A + H 2 S Overall A balance kmol S 1 kmol A react h 1 kmol S form kmol A / h Overall H balance kmol S 1 kmol H react h 1 kmol S form kmol H / h 2 2 2 : . . : . . n n 1 2 500 == Extent of reaction equations n n A+H S ii 0i 2 : ±± ± =+ νξ A: n n H n n S: 5.00 = n n nn n = 5.00 n n .00 py P = n n P n- 5 . 0 0 n p y P = n n P 5 . 0 0 n 43 25 3 53 S tot 3 AA 4 tot 3 3 HH 5 tot 3 3 22 ± : ± ± . . ± ± ± ± ± . . ± ± ± ± . . =====> = U V | | W | | ⇒= = 3 35 0 0 45 3 0 0 10 0 0 0 10 0 ξ p y P = 5.00 n SS 3 = 0 0 10 0 ± . . K p 5.00 4n 10.0 3n n n k m o l H h p S AH 3 33 32 2 −− =⇒ = ± . ± . ± . . ± ./ 0100 1194 bg b g ± (. . ± ... / ± .( ) / n ) - 5.00 kmol A / h
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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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