By Richardson D.
Read or Download Solution of the identity problem for integral exponential functions PDF
Best nonfiction_6 books
Warpaint Series No. 48: Westland Lysander
Heritage of and squadron shades of the Westland Lysander airplane of WWII.
- Adaptive Navigation Control for Swarms of Autonomous Mobile Robots
- Terror Australis: Cthulhu Down Under, Australian Background and Adventures (Call of Cthulhu)
- Impedance of Accelerator Components
- Column shortening in tall structures : prediction and compensation
- Osmanli Hat Sanati. Sakip Sabanci Collection, Istanbul
- Metasomatism in oceanic and continental lithospheric mantle
Extra resources for Solution of the identity problem for integral exponential functions
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.