Elementary Principles 127

Elementary Principles 127 - 1 5.32 NO + 2 O 2 NO 2 1 mol...

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5- 15 5.32 NO O NO +⇔ 1 2 22 1 mol 020 380 . m o l N O /m o l 0.80 mol air / mol 0.21 O 0.79 N P k P a 2 2 0 R S | T | U V | W | = n (mo l NO ) l O l N l P (kPa) 1 32 42 f ) ) ) a. Basis: 1.0 mol feed 90% 010 0 20 0 020 NO conversion: n mol NO NO reacted = 0.18 mol 1 = = .( .) . O balance: n mol NO mol O mol NO mol O 2 2 =− = 080 021 018 05 00780 .) . . N balance: n mol N 23 2 = = 080 0 79 0632 n mol NO mol NO 1 mol NO mol NO n n n n n mol 4 2 2f 1 2 3 4 == = + + + = 1 091 . .. y mol NO 0.91 mol mol NO mol y mol O mol y mol N mol y mol NO mol NO O 2 N 2 NO 2 2 = 0020 0022 0086 0695 0198 . . . PV = nRT PP n n kPa 0.91 mol mol f 0 f 0 f0 f 0 ⇒= = F H G I K J = 380 1 346 kPa b. n=n P P mol) 360 kPa mol f 0 (. 1 380 kPa 095 nn l ) nm o l O l N) n mol NO n mol NO n mol O n mol N n mol NO y , y y y ii 0i 1 f 12 2 3 2 4 2 NO O N NO 2 =+ E = = = ⇒ = = = =⇒ = = = = υξ ξ ξξ 021 080 079 080 10 5 0 9 5 0 1 0 010 0118 0105 0124 0 665 . ()
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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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