Chap6Lect2 - 6.3 Algorithm for Solution to Complex Reactions 6.3.1 Mole Balances Because of parallel and consecutive reactions in complex reaction

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6.3 Algorithm for Solution to Complex Reactions 6.3.1 Mole Balances Because of parallel and consecutive reactions, in complex reaction systems N i or F i are more useful in solving rather than X. Table 6-1. Mole Balances for Multiple Reactions Reactor Gas Liquid Batch V r dt dN A A = A A r dt dC = CSTR A A Ao r F F V - - = A A Ao o r C C v V - - = PFR/PBR A A r dV dF = A A o r dt dC v = Semibatch (B is added) V r dt dN A A = V C v r dt dC A o A A - = Eq. (4-56) Bo B B F V r dt dN + = ( 29 V C C v r dt dC B Bo o B B - + = Eq.(4-58) 6.3.2 Net Rates of Reaction Let q reactions take place Reaction 1: A + B → A k 1 3C + D Reaction 2: A + 2C → A k 2 3E Reaction 3: 2B + 3E → B k 3 4F ………………………………………… Reaction q: A + 0.5B → qA k G The net rates of A and B are = = + + + + = q i iA qA A A A A r r r r r r 1 3 2 1 ... = = + + + + = q i iB qB B B B B r r r r r r 1 3 2 1 ... etc… Individual rates for species would be zero if they do not take part in a particular reaction: r 3A = r 2B = r 2D = 0, etc. Let A 1 = A, A 2 = B, etc. to arrive at Table 6-2
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Table 6-2 Reaction number Reaction stoichiometry i Σ v ij A j Σ v jp A p Thus: = = q i ij j r r 1 6.3.3 Rate Laws If reaction 2 follows an elementary reaction A + 2C 3E then 2 2 2 C A A A C C k r = - Or formation of A in reaction 2: 2 2 2 C A A A C C k r - = We need to determine the rate law for at least one species in a reaction. 6.3.4 Stoichiometry: Relative Rates of Reaction From Chapters 2 and 3 i iD i iC i iB i iA d r c r b r a r = = - = - Example: Reaction 2 : A + 2C → A k 2 3E 2 2 2 2 3 1 3 C A A A E C C k r r = - = The rate of formation of C 2 2 2 2 2 1 2 C A A A C C C k r r - = - - = Thus, for species in the same reaction ik ik ij ij v r v r = v ij is negative for reactants and positive for products.
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See Table 6.3 Table 6-3 Steps in analyzing multiple reactions 1. Number each reaction 2. Write mole balances for each species 3. Determine rate laws for each species in each reaction 4. Relate the rate of reaction of each species to the species for which the rate law is given for each reaction 5. Determine the net rate of generation for each species 6. Express rate laws as functions of C j for the case of no volume change 7. Express rate laws as functions of N j for batch reactors and F j for the flow reactors when there is volume change 8. Combine all the equations and solve coupled ODEs ( PFR, PBR, batch ) or algebraic equations ( CSTR )
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Example 6-4 Stoichiometry and Rate Laws for Multiple Reactions Write the net rates of generation for NO, O 2 , and N 2 for Reaction 1: 4NH 3 + 6NO 5N 2 + 6H 2 O 5 . 1 1 1 3 NO NH NO NO C C k r = - Reaction 2: 2NO N 2 + O 2 2 2 2 2 2 NO N N C k r = Reaction 3: N 2 + 2O 2 2NO 2 2 3 3 2 2 2 2 O N O O C C k r = - Solution: Rewrite the reactions in a normalized form NO + 2/3 NH
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This note was uploaded on 12/17/2011 for the course CHEMICAL E 304 taught by Professor Darwish during the Spring '11 term at American University of Sharjah.

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Chap6Lect2 - 6.3 Algorithm for Solution to Complex Reactions 6.3.1 Mole Balances Because of parallel and consecutive reactions in complex reaction

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