1st order reversible reaction, according to page56 eq. 53b, we obtain
t??dt??0tXA0dXAk11?Lnk1?(k1?k2)XAk1?k2k1?(k1?k2)XA?1
?t?480sec?8min, XA?0.33, so we obtain eq(1)
480sec?8mink11Ln eq(1) k1?k2k1?(k1?k2)0.33Kc?Ck1CReM?XAe??, M?Ro?0, so we obtain eq(2) k2CAe1?XAeCAoKc?XAek1??k21?XAe2321?3?2,
?k1?2k2 eq(2)
Combining eq(1) and eq(2), we obtain
k2?0.02888min?1?4.8?10?4sec?1 k1?2k2?0.05776min?1?9.63?10?4sec?1
So the rate equation is ?rA??dCA?k1CA?k2(CAo?CA) dtc1(CA0?CA) ?4.8?10?4sec?1CA?9.63?10?4se?
3.10 Aqueous A reacts to form R (A→R) and in the first minute in a batch reactor its
concentration drops from CA0 = 2.03 mol/liter to CAf = 1.97 mol/liter. Find the rate equation from the reaction if the kinetics are second-order with respect to A.
Solution: It’s a irreversible second-order reaction system, according to page44 eq 12, we obtain
11??k1?1min, 1.972.03so k1?0.015L
mol?min2so the rate equation is ?rA?(0.015min?1)CA
3.15 At room temperature sucrose is hydrolyzed by the catalytic action of the enzyme
sucrase as follows:
10
sucrase Aucrose ???? products
Starting with a sucrose concentration CA0 = 1.0 millimol/liter and an enzyme concentration
CE0= 0.01 millimol/liter, the following kinetic data are obtained in a batch reactor (concentrations calculated from optical rotation measurements):
CA, millimol/liter 0.84 0.68 0.53 0.38 0.27 0.16 0.09 0.04 0.018 0.006 0.0025 t,hr
1
2
3
4
5
6
7
8
9
10
11
Determine whether these data can be reasonably fitted by a knietic equation of the Michaelis-Menten type, or -rA =
k3CACE0 where CM = Michaelis constant
CA?CM If the fit is reasonable, evaluate the constants k3 and CM. Solve by the integral method.
Solution: Solve the question by the integral method:
?rA??dCAk3CEoCAkC??4A, dtCA?CM1?k5CAk3CEo1, k5? CMCMk4?CAokCAt1 ?5??CAo?CAk4k4CAo?CALnCAoCA
CAo?CALn1.0897 1.2052 1.3508 1.5606 1.7936 2.1816 2.6461 3.3530 4.0910 5.1469
t,hr
CA,mmol/L
0.84 0.68 0.53 0.38 0.27 0.16 0.09 0.04 0.018 0.006
t
CAo?CA6.25 6.25 6.3830 6.4516 6.8493 7.1428 7.6923 8.3333 9.1650 10.0604
1 2 3 4 5 6 7 8 9 10
11
11 0.0025 6.0065 11.0276
CAoCAtSuppose y=, x=, thus we obtain such straight line graph
CAo?CACAo?CALn1210t/(Cao-Ca)86y = 0.9879x + 5.0497420012345Ln(Cao/Ca)/(Cao-Ca)67R2 = 0.998
Slope?kCM1??0.9879, intercept=5?5.0497 k4k3CEok4So CM?10.9879??0.1956(mmol/L), k55.0497k3?k4CM0.1956??19.80hr?1 CEo0.9879?0.01
3.18 Enzyme E catalyzes the transformation of reactant A to product R as follows:
??? R, -rA = A ?enzyme200CACE0mol
2?CAliter?min If we introduce enzyme (CE0 = 0.001 mol/liter) and reactant (CA0 = 10 mol/liter)
into a batch rector and let the reaction proceed, find the time needed for the concentration of reactant to drop to 0.025 mol/liter. Note that the concentration of enzyme remains unchanged during the reaction.. Solution:
?2?CA1dt10?????5 rAdCA200?0.001CACARearranging and integrating, we obtain:
12
t??dt???0t0.02510??C10(?5)dCA??10LnAo?5(CAo?CA)?CACA??0.02510?10Ln10?5(CAo?CA)?109.79min 0.025
3.20 M.Hellin and J.C. Jungers, Bull. soc. chim. France, 386(1957), present the data in Table P3.20 on thereaction of sulfuric acid with diethylsulfate in a aqueous solution at 22.9℃: H2SO4 + (C2H5)2SO4 → 2C2H5SO4H
Initial concentrations of H2SO4 and (C2H5)2SO4 are each 5.5 mol/liter. Find a rate equation for this reaction.
Table P3.20
t, min 0 41 48 55 75 96 127 146 162
C2H5SO4H, mol/liter 0 1.18 1.38 1.63 2.24 2.75 3.31 3.76 3.81
t, min 180 194 212 267 318 368 379 410 ∞
C2H5SO4H, mol/liter 4.11 4.31 4.45 4.86 5.15 5.32 5.35 5.42 (5.80)
Solution: It’s a constant-volume system, so we can use XA solving the problem: i) We postulate it is a 2nd order reversible reaction system A?B?2R The rate equation is: ?rA??dCA2?k1CACB?k2CR dtCAo?CBo?5.5mol/L, CA?CAo(1?XA), CB?CBo?CAoXA?CA, CR?2CAoXA
When t??, CRe?2CAoXAe?5.8mol/L So XAe?5.8?0.5273, 2?5.5/L CAe?CBe?CAo(1?XAe)?5.5?(1?0.5273)?2.6molAfter integrating, we obtain
13
LnXAe?(2XAe?1)XA1?2k1(?1)CAot eq (1)
XAe?XAXAeThe calculating result is presented in following Table.
t,
min 0 41 48 55 75 96 127 146 162 180 194 212 267 318 368 379 410
CR,mol/L
0 1.18 1.38 1.63 2.24 2.75 3.31 3.76 3.81 4.11 4.31 4.45 4.86 5.15 5.32 5.35 5.42 5.8
CA,mol/L
5.5 4.91 4.81 4.685 4.38 4.125 3.845 3.62 3.595 3.445 3.345 3.275 3.07 2.925 2.84 2.825 2.79 2.6
XA
0 0.1073 0.1254 0.1482 0.2036 0.25 0.3009 0.3418 0.3464 0.3736 0.3918 0.4045 0.4418 0.4682 0.4836 0.4864 0.4927 0.5273
LnXAe?(2XAe?1)XAX Ln(1?A)
XAe?XAXAe0 0.2163 0.2587 0.3145 0.4668 0.6165 0.8140 1.0089 1.0332 1.1937 1.3177 1.4150 1.7730 2.1390 2.4405 2.5047 2.6731 —
0 -0.2275 -0.2717 -0.3299 -0.4881 -0.6427 -0.8456 -1.0449 -1.0697 -1.2331 -1.3591 -1.4578 -1.8197 -2.1886 -2.4918 -2.5564 -2.7254 —
?
Draw LnXAe?(2XAe?1)XA~ t plot, we obtain a straight line:
XAe?XA 14