By M. Albert Vannice
Describes how you can behavior kinetic experiments with heterogeneous catalysts, research and version the implications, and symbolize the catalysts certain research of mass move in liquid section reactions related to porous catalysts. vital to the fantastic chemical compounds and pharmaceutical industries so it has appeal to many researchers in either and academia (chemical engineering and chemistry departments
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Example text
From either a linear extrapolation of the high- P portions of the two isotherms in Figure 1 or using the difference between the two at 100 Torr CO pressure, the irreversible uptake is 580 umole CO g~~t. 8 Solution Step 2 defines the rate: =~m dNdt =~(dNdt m N 2oJ=k[N N2 r m 2 0*] Assuming all surface species are included, a site balance gives L = [*]+ lN 2 0 *J+ lo *J To remove the unknown [0 *], the SSA must be used: and Rearranging and squaring each side gives: and 59 The solution for this quadratic expression is: Substituting back and using the positive root gives: H= kKN20PN20/kl + 2L(1 + KN20PN20)+ ~2~_1 Po2/k1 ~(I+KN20PN2of] Because the rate is: rm = k[N 20 *] = kKN20PN20 [*], the final rate expression is: dP N20 {I + (a + C)PN20 + [aPN20 + (a2 + aC~N20 + bPo J1/2} =---'--------:-------:------------'-- r m (1 + CPN20 f -bPo2 parameters can be combined in various ways to give the values in Table 1.
K 2PBlS~ ] B·S , fA· S]2 C+S] , L= lc. S]= Pc [S] II K4 [A. S]+ [B. S]+ [C. S]+ [S] , K =~ [C. 2 PB +KBPB +Kcpcf K1 A2 + 2S T=@=:= 2A· S K2 2 [B + S T=@=:= B'S] K3 2 [A' S + B· S T=@=:= C'S + S] k4 2 [C' S A2+ 2 B ~4 ----- (reversible RDS) C+ S] 2C r = k 4[C. S][S] => rC. S] = K 3 [A . s I B . , dec (total) = O. Then, because steps 2 and 4 are very rapid: dt ~ and but e C2 H4 can be rewritten to give: so that which is notation consistent with reference 55. 54 k1 C2H6 + 2 S;;:=:- (b) ~l C2Hs .
U. ] lCH 4 jdt r][ ] (c) r 1. Literature solutionfor reaction as written = k 2 lCH 2ad H 2 & [C2H2ad][H2F K=--- [C 2H 6][S] [C2H 2ad] = K [C2H 6][S]I [H2]2 = K[C 2H 6 ] (L - [C2H 2ad]) [H 2 F [C2H2ad](1+K[C 2H6]1[H2]2)=LK [C2H 6 ] => [C H d]= [HJ2 @ SteadyState 2 2 a LK [C2H 6 ] r) [H2]2 (I+K[C 2H 6 ] / [H 2 d lCH 2ad J de 1 d lCH 2ad J r ][ ] =0 : - == k, lC2H2ad H 2 dt dt dt 2 42 II. , K C2 H6(g)+ S ~ C2H2-S + 2 H 2(g) (1) (2) k 1 C 2H2-S + H2(g) + S -=---t2 CH2-S k2 2 [CH 2-S + H 2 (g) ----=r CH4 (g)] (3) C2H6 + H2 ====} 2 CH4 Note that a CH 2 group almost certainly requires a site for itself as shown in step 2.